[ { "problem_id": 1, "question": "In the given solid shapes, which can be obtained by rotating a plane figure around a certain straight line for a full revolution?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0006_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0006_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0006_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0006_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Since a solid obtained by rotating a plane figure around a certain straight line must have a curved surface,\n\nTherefore, option B fits the requirement;\n\nHence, the answer is: B.\n\n[Key Insight] This question examines the concepts of points, lines, surfaces, and solids. Understanding that \"a moving point forms a line, a moving line forms a surface, and a moving surface forms a solid\" is crucial to solving the problem." }, { "problem_id": 2, "question": "Which of the following plane figures, when folded, cannot form a cube?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0008_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0008_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0008_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0008_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: According to the common forms of unfolded cube diagrams that cannot form a cube, namely \"a line should not exceed four, and shapes resembling '田' or concave should be discarded,\"\n\nit can be concluded that option C cannot form a cube.\n\nTherefore, the answer is: C.\n\n[Key Point] This question examines the unfolded diagrams of a cube. Memorizing the 11 common forms of unfolded diagrams and the common forms that cannot form a cube, such as \"a line should not exceed four, and shapes resembling '田' or concave should be discarded,\" is crucial for solving the problem." }, { "problem_id": 3, "question": "As shown in the figure, if a plane cuts the left cone, which of the following cross-sections is impossible to obtain?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0009_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0009_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0009_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0009_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0009_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: If a plane is used to cut a cone, the cross-section obtained when the plane passes through the apex of the cone is a triangle; if it does not pass through the apex and is cut vertically, the resulting figure is the one shown in option B.\n\nIf the plane does not pass through the apex and is parallel to the base, the cross-section is a circle. If the plane is not parallel to the base, the cross-section is an ellipse, but it cannot be a square.\n\nTherefore, the correct answer is: D.\n\n[Key Insight] This question primarily examines the concept of cutting a geometric solid. The shape of the cross-section depends not only on the geometric solid being cut but also on the angle and direction of the cut. For such problems, it is best to combine hands-on practice with mental analysis, allowing one to learn and apply analytical and inductive thinking methods through direct experience." }, { "problem_id": 4, "question": "Construct a 3D solid using small cubes, such that its views from two directions are as shown in the figure. It requires a minimum of ( ) small cubes and a maximum of ( ) small cubes.\n\n\n\nView from the front\n\n\n\nView from above\nA. 9,14\nB. 9,16\nC. 8,16\nD. 10,14", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0015_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0015_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The minimum number of such geometric units required is distributed as shown in the figure:\n\n\n\nViewed from above\n\nThe maximum number of such geometric units required is distributed as shown in the figure:\n\n\n\nViewed from above\n\nIn summary, a minimum of 9 units is required; a maximum of 14 units is required;\n\nTherefore, the correct choice is: A.\n\n【Key Point】This question primarily examines the view of a geometric figure from different perspectives. The key to solving the problem lies in assigning numbers to the corresponding positions of the figure as seen from above." }, { "problem_id": 5, "question": "As shown in the figure, there is a cube-shaped cardboard box on the ground. Each face of the box is painted with dye. Without lifting the box off the ground, it is cut open so that the dye on each face can be printed on the ground. The figures that can be formed on the ground are ( ).\n\n\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. 0\nB. 1\nC. 2\nD. 3", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0032_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0032_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0032_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0032_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By observing Figures (1), (2), and (3), it can be concluded that only Figure (1) is the unfolded diagram of a cube-shaped carton, meaning there is only one figure that can be formed on the ground.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question tests the understanding of the unfolded diagrams of a cube. Mastering the characteristics of a cube's unfolded diagram is crucial for solving the problem." }, { "problem_id": 6, "question": "The lateral surface expanded viewof a geometric solid is shown in the figure. Determine the shape of the base of the solid.\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0037_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0037_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0037_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0037_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0037_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, this geometric figure is a triangular prism.\n\nTherefore, the shape of the base of this geometric figure is a triangle,\n\nHence, the correct choice is: D.\n\n[Key Insight] This question examines the lateral development of geometric figures. Familiarity with the lateral developments of common geometric figures is crucial for solving the problem." }, { "problem_id": 7, "question": "Among the following diagrams, which one cannot be folded into a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0038_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0038_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0038_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0038_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: C, after folding, one face overlaps and one face is missing, so it cannot be folded into a cube;\n\nA, B, and D can all be folded into a cube.\n\nTherefore, the answer is: C.\n\n[Key Point] This question tests the knowledge of folding a net into a cube. The key to solving the problem is to understand the characteristics of a cube or to memorize the 11 types of nets for a cube. Any net that contains a \"田\" (field) or \"凹\" (concave) pattern is not a valid net for the surface of a cube." }, { "problem_id": 8, "question": "Among the following geometric solids, which can be obtained by rotating a plane figure around a straight line for a full revolution?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0044_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0044_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0044_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0044_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Option B can be obtained by rotating a rectangle around its longer side for one full rotation,\n\nTherefore, choose B.\n\n[Key Point] This question primarily tests the preliminary understanding of solid geometry. The key to solving the problem is the ability to proficiently identify which three-dimensional shape can be obtained through rotation." }, { "problem_id": 9, "question": "As shown in the figure, the unfolding pattern of the cube is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0056_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0056_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0056_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0056_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0056_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "A: The unfolded diagram is correct;\n\nB: The checkmark face and the equal sign face are opposite each other, which does not match the problem's requirement;\n\nC: The direction of the checkmark is incorrect, hence it is not correct;\n\nD: The orientation of the three symbols does not match, hence it is not correct;\n\nTherefore, the correct answer is A.\n\n[Key Point] This question mainly tests the understanding of the unfolded diagram of a cube. Accurately judging the direction of the symbols is the key to solving the problem." }, { "problem_id": 10, "question": "During a project study, the teacher asked students to make an open-top rectangular box using a rectangular sheet of paper measuring 12 cm by 8 cm. Three students each cut off two corners (shaded areas in the figures) of the rectangular sheet in the following ways, and then folded the sheet along the dotted lines to form an open-top rectangular box.\n\nStudent A: As shown in Figure 1, the base of the box, quadrilateral $\\mathrm{ABCD}$, is a square.\n\nStudent B: As shown in Figure 2, the base of the box, quadrilateral $\\mathrm{ABCD}$, is a square.\n\nStudent C: As shown in Figure 3, the base of the box, quadrilateral $\\mathrm{ABCD}$, is a rectangle, with $\\mathrm{AB}=2 \\mathrm{AD}$.\n\nArrange the volumes of the boxes made by these three students in descending order. The correct order is\n\n\nFigure 1\n\n\nFigure 2\n\n\nFigure 3\nA. A $>$ B $>$ C\nB. A $>$ C $>$ B\nC. C $>$ A $>$ B\nD. C $>$ B $>$", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0068_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0068_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0068_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "A: The length of the cuboid is $5 \\mathrm{~cm}$, the width is $3 \\mathrm{~cm}$, and the height is $3 \\mathrm{~cm}$. The volume is $5 \\times 3 \\times 3=45 \\mathrm{~cm}^{3}$.\n\nB: The length of the cuboid is $10 \\mathrm{~cm}$, the width is $2 \\mathrm{~cm}$, and the height is $2 \\mathrm{~cm}$. The volume is $10 \\times 2 \\times 2=40 \\mathrm{~cm}^{3}$.\n\nC: The length of the cuboid is $6 \\mathrm{~cm}$, the width is $4 \\mathrm{~cm}$, and the height is $2 \\mathrm{~cm}$. The volume is $6 \\times 4 \\times 2=48 \\mathrm{~cm}^{3}$.\n\nTherefore, C $>$ A $>$ B.\n\nHence, choose C.\n\n【Key Point】This question mainly examines the volume of a cuboid. Mastering the volume formula of a cuboid is the key to solving the problem." }, { "problem_id": 11, "question": "As shown in the figure, this is a model that can be folded into a cuboid. Which of the following figures is the cuboid formed by folding it?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0084_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0084_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0084_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0084_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0084_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Analysis:** From the unfolded diagram, the colors of the opposite faces of the cuboid can be deduced, allowing us to infer the correct option.\n\n**Detailed Explanation:** Based on the unfolded diagram, it can be deduced that the colors of the opposite faces of the cuboid should be the same, and both the front and back faces have shading. Therefore, options A, B, and C are incorrect.\n\n**Answer:** D\n\n**Key Point:** This question tests the knowledge of unfolded diagrams. The key to solving it lies in analyzing the colors of the opposite faces of the cube." }, { "problem_id": 12, "question": "In the following figures, which pair of angles, $\\angle 1$ and $\\angle 2$, are complementary? ( )\n\n\n\nA\n\n\n\nB\n\n\n\nC\n\n\n\nD\nA. A\nB. B\nC. C\nD. D", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0091_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0091_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0091_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0091_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "In option A, $\\angle 1$ and $\\angle 2$ are corresponding angles, which does not meet the requirement of the question; in option B, $\\angle 1$ and $\\angle 2$ are vertical angles, which also does not meet the requirement; in option C, $\\angle 1$ and $\\angle 2$ are supplementary angles, which satisfies the requirement; in option D, $\\angle 1$ and $\\angle 2$ are not supplementary, which does not meet the requirement. Therefore, the correct choice is C.\n\n【Key Point】This question mainly tests the knowledge of supplementary angles. The key to solving the problem is to be familiar with the concept of supplementary angles." }, { "problem_id": 13, "question": "Fold a square piece of paper as shown in Figure (1) and Figure (2), then cut along the dotted line in Figure (3). Finally, unfold the paper from Figure (4). The resulting pattern is ( ).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0100_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0100_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0100_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0100_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0100_5.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0100_6.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0100_7.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0100_8.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Follow the sequence of operations as strictly shown in the diagram, and the resulting unfolded figure is as depicted in option B.\n\nTherefore, choose B.\n\n[Key Insight] This question tests the paper-cutting problem, which primarily assesses students' hands-on skills and spatial imagination. For such problems, if students personally perform the operations, the answer will become intuitively clear." }, { "problem_id": 14, "question": "Given a right-angled triangle paper, first fold it along the side containing the right angle $\\angle A B C$ (as shown in Figure 1), so that point $A$ lands on the hypotenuse $B C$ at point $A^{\\prime}$, and the crease intersects side $A C$ at point $D$. Then fold along the other acute angle $\\angle D C B$ (as shown in Figure 2), so that $C D$ also lands on the hypotenuse, and the crease intersects $A^{\\prime} D$ at point $P$. It is measured that $D P = \\frac{3}{2} A^{\\prime} P = 3$. Find the distance from point $P$ to $C D$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 4\nB. 3\nC. 2\nD. $\\frac{3}{2}$", "input_image": [ "batch10-2024_06_14_1fc8b75729663205496ag_0029_1.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0029_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Through point \\( P \\), draw \\( PM \\perp CD \\) intersecting at \\( M \\), as shown in the figure:\n\n\n\nSince one of the acute angles \\( \\angle ABC \\) is folded such that point \\( A \\) falls on the hypotenuse \\( BC \\) at \\( A' \\),\n\nTherefore, \\( \\angle BA'D = \\angle BAD = 90^\\circ \\),\n\nHence, \\( PA' \\perp BC \\),\n\nSince the other acute angle \\( \\angle DCB \\) is folded such that \\( CD \\) also falls on the hypotenuse \\( BC \\),\n\nTherefore, \\( \\angle DCP = \\angle BCP \\), meaning \\( CP \\) is the angle bisector of \\( \\angle BCD \\),\n\nBecause \\( PA' \\perp BC \\) and \\( PM \\perp CD \\),\n\nTherefore, \\( PM = A'P \\),\n\nSince \\( \\frac{3}{2} A'P = 3 \\),\n\nTherefore, \\( A'P = 2 \\),\n\nHence, \\( PM = 2 \\), which means the distance from point \\( P \\) to \\( CD \\) is 2,\n\nTherefore, the answer is: C.\n\n【Key Insight】This problem examines the folding properties in right-angled triangles. The key to solving it lies in understanding the properties of folding." }, { "problem_id": 15, "question": "In the following figures, which one is not an axisymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_1fc8b75729663205496ag_0065_1.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0065_2.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0065_3.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0065_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "From the analysis, it can be determined that among the given figures, the one that does not belong to the axisymmetric figures is Figure D.\nTherefore, the answer is D.\n\n【Key Point】Understand the concept of axisymmetric figures. The key to identifying an axisymmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide when folded along the axis." }, { "problem_id": 16, "question": "The following compass and straightedge constructions can determine that $A D$ is the angle bisector of $\\triangle A B C$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_1fc8b75729663205496ag_0085_1.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0085_2.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0085_3.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0085_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Option $\\mathrm{A}$ involves using a compass and straightedge to construct the perpendicular bisector of side $BC$, thus making $AD$ the median of $\\triangle ABC$. Therefore, option $\\mathrm{A}$ does not meet the requirement of the question.\n\nOption $\\mathrm{B}$ involves using a compass and straightedge to construct the perpendicular bisector of side $AB$, which implies $AD = BD$. Therefore, option $\\mathrm{B}$ does not meet the requirement of the question.\n\nOption $\\mathrm{C}$ involves constructing the ray $AD$ as the angle bisector of $\\angle BAC$, so option $\\mathrm{C}$ meets the requirement of the question.\n\nIn option $\\mathrm{D}$, it cannot be confirmed that $AD$ is the angle bisector of $\\triangle ABC$, so option $\\mathrm{D}$ does not meet the requirement of the question.\n\nTherefore, the correct choice is:\nC.\n\n【Key Point】This question primarily tests the ability to construct an angle bisector using a compass and straightedge. Mastering the steps of compass and straightedge constructions is crucial for solving such problems." }, { "problem_id": 17, "question": "In a math class, there was a drawing problem like this: As shown in Figure 1, points $P$ and $Q$ are on the same side of the straight line $l$. Please determine a point $R$ on the straight line $l$ such that the perimeter of $\\triangle P Q R$ is minimized. Xiaoming's approach is shown in Figure 2:\n\n(1) Draw the point $Q$'s reflection about the straight line $l$, denoted as $Q^{\\prime}$;\n\n(2) Connect $P Q^{\\prime}$ and intersect it with the straight line $l$ at point $R$;\n\n(3) Connect $R Q$ and $P Q$. Then point $R$ is the point that minimizes the perimeter of $\\triangle P Q R$.\n\nXiaoming's approach utilizes a theorem we have learned. Which of the following four theorems was NOT used by Xiaoming?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. If two figures are symmetric about a certain straight line, then the symmetry axis is the perpendicular bisector of the line connecting the corresponding points.\n\nB. The altitude of an isosceles triangle drawn from the base is also the median of the vertex angle.\n\nC. Points on the perpendicular bisector of a line segment are equidistant from the endpoints of the line segment.\n\nD. Among all paths connecting two points, a straight line segment has the shortest length.", "input_image": [ "batch10-2024_06_14_2434588caaee122ec85bg_0048_1.jpg", "batch10-2024_06_14_2434588caaee122ec85bg_0048_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the construction, it can be seen that the line \\( l \\) is the perpendicular bisector of the segment \\( QQ' \\),\n\n\\[\n\\therefore RQ = RQ',\n\\]\n\n\\[\n\\therefore \\text{The perimeter of } \\triangle PRQ = PQ + RQ + RP = PQ + PR + RQ' = PQ + PQ',\n\\]\n\nAccording to the principle that the shortest distance between two points is a straight line: at this point, the perimeter of \\( \\triangle PRQ \\) is minimized.\n\nTherefore, options A, C, and D do not meet the requirement,\n\nHence, the correct choice is: B.\n\n【Key Insight】This problem examines the concept of the shortest path using symmetry, the principle that the shortest distance between two points is a straight line, and the properties of the perpendicular bisector of a segment. The key to solving the problem lies in understanding the question and flexibly applying the learned knowledge to find the solution." }, { "problem_id": 18, "question": "The figure below is divided into two congruent figures, which of the following is correct ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0097_1.jpg", "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0097_2.jpg", "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0097_3.jpg", "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0097_4.jpg", "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0097_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The figure is divided into two congruent figures, as shown in the diagram:\n\n\n\nTherefore, choose B.\n\n[Key Point] This question primarily tests the recognition of congruent figures. The key to solving it lies in understanding the properties of congruence." }, { "problem_id": 19, "question": "The number of prisms among the following plane figures is ( )\n\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\n\n\n(5)\n\n\n(6)\n\n\n(7)\nA. 2\nB. 3\nC. 4\nD. 5", "input_image": [ "batch10-2024_06_14_329d65dd392595b77024g_0029_1.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0029_2.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0029_3.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0029_4.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0029_5.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0029_6.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0029_7.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "According to the definition, among the given geometric shapes,\n\n(1) Cube, (2) Rectangular prism (quadrilateral prism), (6) Quadrilateral prism, and (7) Triangular prism are all types of prisms; the others, namely (3) Sphere, (4) Cylinder, and (5) Cone, are not prisms.\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the definition of a prism. It is crucial to be familiar with the concepts of prisms, cones, and spheres." }, { "problem_id": 20, "question": "The number of correct angle representations in the following figures is ( )\n\n\n$\\angle A B C$\n\n\n$\\angle A O B$ is a straight angle\n\n\nRay $A B$ is a full angle\n\n\n$\\angle C A B$\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch10-2024_06_14_3a73c762f5eb2b761854g_0009_1.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0009_2.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0009_3.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0009_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: In Figure 1, the vertex of the angle is at $\\mathrm{A}$, and it should be denoted as $\\angle CAB$;\n\nFigure 2 is correctly represented;\n\nIn Figure 3, a ray and a full angle are two different concepts; a ray cannot represent a full angle;\n\nFigure 4 is correctly represented.\n\nTherefore, the number of correct representations is 2.\n\nHence, the correct choice is: B.\n\n[Key Insight] This question primarily examines the methods of representing angles, straight angles, rays, and full angles. Understanding and mastering these concepts is crucial for solving the problem." }, { "problem_id": 21, "question": "Line $l$ represents a river, and $P, Q$ are two villages on the same side of $l$. During the pandemic, to facilitate management, a temporary observation point needs to be constructed at point $M$ on the line $l$. The shortest distance scheme from $M$ to both $P$ and $Q$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0081_1.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0081_2.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0081_3.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0081_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: According to the principle that the shortest distance between two points is a straight line and the characteristics of axial symmetry, to determine the shortest path from point $M$ to locations $P$ and $Q$, first find the symmetric point of $P$ with respect to line $l$. Then, connect point $Q$ to the symmetric point of $P$. The intersection of this connecting line with line $l$ is point $M$, which means option D is correct.\n\nTherefore, the answer is: D.\n\n[Key Insight] This question tests the understanding of the characteristics of axially symmetric figures. The key to solving this problem lies in the ability to proficiently use these characteristics to construct the required figure." }, { "problem_id": 22, "question": "As shown in the figure, take a square piece of paper, fold it twice to form an isosceles right triangle. Xiaoming cuts out a small square from this isosceles right triangle, then unfolds the folded paper and lays it flat. The resulting pattern is ( )\n\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_4ce737a200cf8fbade32g_0017_1.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0017_2.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0017_3.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0017_4.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0017_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: Unfolding once yields:\n\n\nFurther unfolding results in: The final figure obtained\n\nTherefore, the answer is: B.\n\n【Key Point】This question examines the folding and unfolding of shapes; the key to solving the problem lies in correctly unfolding the folded shape." }, { "problem_id": 23, "question": "As shown in the figure, inside $\\angle A O B$, from the vertex $O$ in Figure (1), 1 ray is drawn, resulting in 3 angles in the figure. From the vertex $O$ in Figure (2), 2 rays are drawn, resulting in 6 angles in the figure. Following this pattern, if 29 rays are drawn from the vertex $O$, how many angles will there be in the figure?\n\n\n\n( 1)\n\n\n\n( 2)\nA. 465\nB. 450\nC. 425\nD. 300", "input_image": [ "batch10-2024_06_14_55584d21b09f596e3cf6g_0053_1.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0053_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: Inside $\\angle AOB$, drawing 1 ray from the vertex $O$ in Figure (1) results in a total of $1+2=3$ angles; drawing 2 rays from the vertex $O$ in Figure (2) results in a total of $1+2+3=6$ angles;\n\n.....\n\nIf $n$ rays are drawn from the vertex of the angle, the total number of angles in the figure is $1+2+3+\\cdots+(n+1)=\\frac{1}{2}(n+2)(n+1)$;\n\n$\\therefore$ Drawing 29 rays from the vertex of the angle results in a total of $=\\frac{1}{2}(29+2)(29+1)=465$ angles;\n\nTherefore, the correct choice is: A\n\n【Key Insight】This problem examines the concept of angles, with the key being to deduce from the given conditions that drawing $n$ rays from the vertex of an angle results in a total of $\\frac{1}{2}(n+2)(n+1)$ angles in the figure." }, { "problem_id": 24, "question": "As shown in the figure, given that $B C$ is the diameter of the base of the cylinder and $A B$ is the height of the cylinder, a shortest path of metal wire is wound around the cylinder's lateral surface, passing through points $A$ and $C$. If the cylinder's lateral surface is cut along $A B$, the resulting lateral surface expanded view is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_55584d21b09f596e3cf6g_0094_1.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0094_2.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0094_3.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0094_4.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0094_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since the lateral surface of the cylinder unfolds into a rectangle, $AC$ should unfold into two line segments with a common point $C$.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question primarily tests the understanding of the lateral unfolding of a cylinder and the student's spatial thinking ability." }, { "problem_id": 25, "question": "Given rectangle $\\mathrm{ABCD}$ with one side length of 20 and the other side length of $\\mathrm{a}(\\mathrm{a}<20)$, a square is cut out, leaving a rectangle, which is called the first operation. In the remaining rectangular paper, another square is cut out, leaving another rectangle, which is called the second operation. If after the third operation, the remaining rectangle is a square, then the value of $\\mathrm{a}$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 5\nB. $5, 8$\nC. 5, 8, 15\nD. $5, 8, 12, 15$", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0029_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0029_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "**Question Analysis:** Based on the given information, there are 4 possible scenarios that meet the conditions. A diagram can be drawn to illustrate this.\n\n**Solution:** The schematic diagram of the cutting line is as follows:\n\n\n$a=12$\n\n\n\nTherefore, the correct answer is D.\n\n**Review:** This question tests the properties of rectangles and squares, the application of pattern recognition, and primarily assesses students' transformation and comprehension abilities. Note: Classification and discussion are necessary." }, { "problem_id": 26, "question": "In life, there are the following two phenomena. Which of the following is the correct explanation for these two phenomena? \n\n\n\nPhenomenon 1: Ink lines on a wooden board\n\n\n\nPhenomenon 2: The curved river becomes straight\n\nA. Both are explained by the shortest line segment between two points\n\nB. Both are explained by the fact that there is only one straight line passing through two points\n\nC. Phenomenon 1 is explained by the shortest line segment between two points, and Phenomenon 2 is explained by the fact that there is only one straight line passing through two points\n\nD. Phenomenon 1 is explained by the fact that there is only one straight line passing through two points, and Phenomenon 2 is explained by the shortest line segment between two points\n\n", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0043_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0043_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: Phenomenon 1: The act of snapping a chalk line on a wooden board can be explained by the principle that \"two points determine a straight line.\"\n\nPhenomenon 2: Straightening a meandering river channel to shorten the sailing distance can be explained by the principle that \"the shortest distance between two points is a straight line.\"\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question tests the understanding of the principles that two points determine a straight line and that the shortest distance between two points is a straight line. Proficiency in applying these concepts is crucial for solving the problem." }, { "problem_id": 27, "question": "Among the four figures below, which one can represent the same angle using all three methods of $\\angle 1, \\angle A O B, \\angle O$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0070_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0070_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0070_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0070_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: The figure in option A is the one that can represent the same angle using the three notations $\\angle 1$, $\\angle AOB$, and $\\angle O$. The figures in options B, C, and D cannot represent the same angle using these three notations. Therefore, the correct choice is: A.\n\n【Key Point】This question primarily tests the concept of an angle. An angle is formed by two rays with a common endpoint, where this common endpoint is the vertex of the angle, and the two rays are the sides of the angle. An angle can be represented by three uppercase letters, with the vertex letter in the middle. Only when there is a single angle at the vertex can the angle be denoted by a single letter at the vertex; otherwise, it would be unclear which angle the letter refers to." }, { "problem_id": 28, "question": "The following geometric figures correspond to the corresponding language descriptions:\n\n\nFigure 1\n\n\nFigure 2\n\n\nFigure 3\n\n\nFigure 4\nA. As shown in Figure 1, extend segment $B A$ to point $C$\nB. As shown in Figure 2, ray $B C$ passes through point $A$\nC. As shown in Figure 3, lines $a$ and $b$ intersect at point $A$\nD. As shown in Figure 4, ray $C D$ and segment $A B$ have no intersection points", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0073_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0073_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0073_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0073_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. As shown in Figure 1, extending line segment $BA$ to point $C$ implies that there should be no line to the left of point $C$. Therefore, the geometric figure in this option does not match the corresponding verbal description.\n\nB. As shown in Figure 2, the ray $BC$ does not pass through point $A$. Therefore, the geometric figure in this option does not match the corresponding verbal description.\n\nC. As shown in Figure 3, line $a$ and line $b$ intersect at point $A$. Therefore, the geometric figure in this option matches the corresponding verbal description.\n\nD. As shown in Figure 4, the ray $CD$ can be infinitely extended, thus it will intersect with line segment $AB$. Therefore, the geometric figure in this option does not match the corresponding verbal description.\n\nHence, the correct choice is: C.\n\n【Key Insight】This question tests the properties of line segments, rays, and lines. A thorough understanding and application of these properties are crucial for solving this problem." }, { "problem_id": 29, "question": "As shown in Figure (1), a square-gridded right-angled isosceles triangle is given. The triangle is folded along the arrow direction into Figure (2), then again folded into Figure (3). Next, the triangle above the line $l$ is cut along $l$. Finally, the remaining part is unfolded. The resulting figure is ( ).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0088_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0088_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0088_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0088_4.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0088_5.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0088_6.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0088_7.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, after cutting away the necessary parts, the remaining figure is\n\n\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question tests the folding of triangles, which is a fundamental concept. Mastering the relevant knowledge is crucial for solving the problem." }, { "problem_id": 30, "question": "As shown in the figure, a pair of triangle rulers is placed in different positions. The arrangement in which $\\angle \\alpha$ is not equal to $\\angle \\beta$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_89c51f0358ecdde09f0fg_0009_1.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0009_2.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0009_3.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0009_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: \n\nA. According to the sum and difference relationship of angles, we have $\\angle \\alpha = \\angle \\beta = 45^{\\circ}$, so this option does not meet the requirement of the question;\n\nB. According to the property that complementary angles of the same angle are equal, we have $\\angle \\alpha = \\angle \\beta$, so this option does not meet the requirement of the question;\n\nC. According to the property that supplementary angles of equal angles are equal, we have $\\angle \\alpha = \\angle \\beta$, so this option does not meet the requirement of the question;\n\nD. From the figure, it can be seen that $\\angle \\alpha$ is not necessarily equal to $\\angle \\beta$, so this option meets the requirement of the question.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question mainly examines the calculation of angles and the properties of complementary and supplementary angles, where complementary angles of equal angles are equal, and supplementary angles of equal angles are equal." }, { "problem_id": 31, "question": "Given: As shown in Figure 1, points $A, O, B$ are on the straight line $MN$ in sequence. Now let the ray $OA$ rotate around point $O$ in the clockwise direction at a speed of $2^{\\circ}$ per second; at the same time, the ray $OB$ rotates around point $O$ in the counterclockwise direction at a speed of $4^{\\circ}$ per second. As shown in Figure 2, let the rotation time be $t$ seconds $(0 \\leq t \\leq 90)$. Which of the following statements is correct ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. During the entire motion, there is no situation where $\\angle A O B=90^{\\circ}$\n\nB. When $\\angle A O B=60^{\\circ}$, the rotation time $t$ of the two rays must be 20 seconds\n\nC. When When the $t$ value is 36 seconds, the ray $OB$ just bisects $\\angle M O A$\n\nD. When $\\angle A O B=60^{\\circ}$, the rotation time $t$ of the two rays must be 40 seconds\n\n", "input_image": [ "batch10-2024_06_14_89c51f0358ecdde09f0fg_0074_1.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0074_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we know that $\\angle MOA = 2t^\\circ$ and $\\angle NOA = 180^\\circ - 2t^\\circ$. When $0 \\leq t \\leq 45$, $\\angle NOB = 4t^\\circ$; when $45 < t \\leq 90$, $\\angle NOB = 360^\\circ - 4t^\\circ$.\n\nLet $\\angle AOB = \\angle NOA - \\angle NOB = 90^\\circ$, which means $180^\\circ - 2t^\\circ - 4t^\\circ = 90^\\circ$. Solving this gives $t = 15$ seconds.\n\nTherefore, there exists a scenario where $\\angle AOB = 90^\\circ$.\n\nHence, option A is incorrect and does not meet the requirement.\n\nLet $\\angle AOB = \\angle NOA - \\angle NOB = 60^\\circ$, which means $180^\\circ - 2t^\\circ - 4t^\\circ = 60^\\circ$. Solving this gives $t = 20$ seconds.\n\nLet $\\angle AOB = \\angle NOB - \\angle NOA = 60^\\circ$, which means $4t^\\circ - (180^\\circ - 2t^\\circ) = 60^\\circ$. Solving this gives $t = 40$ seconds.\n\nTherefore, when $\\angle AOB = 60^\\circ$, the rotation time $t$ of the two rays is not necessarily 20 seconds.\n\nHence, options B and D are incorrect and do not meet the requirement.\n\nWhen $t = 36$, $\\angle MOA = 72^\\circ$, $\\angle NOA = 108^\\circ$, and $\\angle NOB = 144^\\circ$.\n\nTherefore, $\\angle AOB = \\angle NOB - \\angle NOA = 144^\\circ - 108^\\circ = 36^\\circ$.\n\nSince $\\angle AOB = \\frac{1}{2} \\angle MOA$,\n\nthe ray $OB$ exactly bisects $\\angle MOA$.\n\nHence, option C is correct and meets the requirement.\n\nThe correct choice is C.\n\n[Key Insight] This problem mainly tests the knowledge of angle operations and angle bisectors. The key to solving the problem lies in correctly representing each angle." }, { "problem_id": 32, "question": "As shown in Figure 1, a pair of right-angled triangles are placed with one side overlapping. In Figure 2, the right-angled triangle $A C D$ with a $45^\\circ$ angle is rotated clockwise around point $A$ by $30^\\circ$ to form $\\triangle A C'^{\\prime} D'^{\\prime}$. If $B C = 2$, what is the area of $\\triangle B C C'^{\\prime}$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $2 \\sqrt{3}-3$\nB. $3-\\sqrt{3}$\nC. $4 \\sqrt{3}-6$\nD. $6-2 \\sqrt{3}$", "input_image": [ "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0010_1.jpg", "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0010_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular from point $C$ to $AB$, meeting at point $H$,\n\n\n\nFigure 2\n\nGiven that $BC = 2$, $\\angle CAB = 30^\\circ$, and $\\angle ACB = 90^\\circ$,\n\nWe find that $AB = 2BC = 4$, and $AC = \\sqrt{3} \\times BC = 2\\sqrt{3}$,\n\nSince the area of triangle $ABC$ can be expressed as $\\frac{1}{2} \\times AC \\times BC = \\frac{1}{2} \\times AB \\times CH$,\n\nIt follows that $CH = \\frac{AC \\times BC}{AB} = \\sqrt{3}$,\n\nWhen the right triangular plate $ACD$ is rotated clockwise by $30^\\circ$ around point $A$, we obtain triangle $AC'D'$,\n\nThus, $AC' = AC = 2\\sqrt{3}$,\n\nTherefore, $BC' = AB - AC' = 4 - 2\\sqrt{3}$,\n\nThe area of triangle $BCC'$ is $\\frac{1}{2} \\times (4 - 2\\sqrt{3}) \\times \\sqrt{3} = 2\\sqrt{3} - 3$,\n\nHence, the correct choice is: $A$.\n\n【Key Insight】This problem tests the properties of rotation, the characteristics of right-angled triangles, and the formula for the area of a triangle. The key to solving this problem lies in determining the length of $CH$." }, { "problem_id": 33, "question": "In Figure (1), $\\triangle \\mathrm{ABC}$ and $\\triangle \\mathrm{ADE}$ are both isosceles right triangles, with $\\angle \\mathrm{ACB}$ and $\\angle \\mathrm{D}$ both being right angles. Point $\\mathrm{C}$ is on $\\mathrm{AE}$. After $\\triangle \\mathrm{ABC}$ rotates counterclockwise around point $\\mathrm{A}$, it coincides with $\\triangle \\mathrm{ADE}$. Then, the Figure (1) is used as the \"base figure\" and rotated counterclockwise around point $\\mathrm{A}$ to obtain Figure (2). The angles of these two rotations are ( )\n\n\n\nFigure 1\n\n\nA. $45^\\circ, 90^\\circ$\nB. $90^\\circ, 45^\\circ$\nC. $60^\\circ, 30^\\circ$\nD. $30^\\circ, 60^\\circ$", "input_image": [ "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0079_1.jpg", "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0079_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "According to Figure 1,\n\n$\\because \\triangle \\mathrm{ABC}$ and $\\triangle \\mathrm{ADE}$ are isosceles right triangles,\n\n$\\therefore \\angle \\mathrm{CAB}=45^{\\circ}$,\n\nwhich means $\\triangle \\mathrm{ABC}$ can be rotated $45^{\\circ}$ counterclockwise around point $\\mathrm{A}$ to reach $\\triangle \\mathrm{ADE}$;\n\nAs shown in the figure,\n\n\n\n$\\because \\triangle \\mathrm{ABC}$ and $\\triangle \\mathrm{ADE}$ are isosceles right triangles,\n\n$\\therefore \\angle \\mathrm{DAE}=\\angle \\mathrm{CAB}=45^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{FAB}=\\angle \\mathrm{DAE}+\\angle \\mathrm{CAB}=90^{\\circ}$,\n\nwhich means Figure 1 can be rotated $90^{\\circ}$ counterclockwise to obtain Figure 2.\n\nTherefore, the correct choice is: A.\n\n【Key Insight】This question tests the properties of rotation and the properties of isosceles right triangles. The key to solving the problem is understanding the properties of rotation and correctly identifying the center of rotation and the angle of rotation." }, { "problem_id": 34, "question": "The angles on the clock faces in Figure (1) and Figure (2) are ( ) degrees, respectively.\n\n\n\n$1: 00$\n\n(1)\n\n\n\n$2: 15$\n\n(2)\nA. $20 ; 15$\nB. $30 ; 25$\nC. $30 ; 22.5$\nD. $22.5 ; 15$", "input_image": [ "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0007_1.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0007_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: The degree of the angle on the clock face in Figure (1) is: $\\frac{360^{\\circ}}{12}=30^{\\circ}$;\n\nSince the degree between the hour hand at $2:00$ and the hour hand at $2:15$ is $30^{\\circ} \\times \\frac{15}{60}=7.5^{\\circ}$,\n\nTherefore, the degree between the hour hand and the minute hand at $2:15$ is $30^{\\circ}-7.5^{\\circ}=22.5^{\\circ}$,\n\nThus, the degree of the angle on the clock face in Figure (2) is: $22.5^{\\circ}$.\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the application of clock face angles, and mastering the degree of each major division on the clock face is key to solving the problem." }, { "problem_id": 35, "question": "Tangram is an ancient Chinese traditional puzzle toy. Using the tangram pieces as shown in Figure 1, arrange them to form the parallelogram and rectangle as shown in Figure 2. Which of the following statements is correct ( )\n\n\n\nFigure 1\n\n\nFigure 2\n\nA. Can form a parallelogram, but cannot form a rectangle.\n\nB. Can form a rectangle, but cannot form a parallelogram.\n\nC. Can form both a rectangle and a parallelogram.\n\nD. Cannot form either a rectangle or a parallelogram.", "input_image": [ "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0024_1.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0024_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure below,\n\n\nFrom the figure, it can be seen that the tangram shown in Figure 1 can be assembled into both a rectangle and a parallelogram. Therefore, the correct choice is C.\n\n[Key Point] This question tests the application of the tangram. The key to solving this problem lies in the flexible use of the knowledge learned." }, { "problem_id": 36, "question": "Given: A triangle paper $\\triangle A B C$, fold the paper in the following two ways:\n\n(1) As shown in Figure 1, fold $\\triangle A B D$ along the bisector $A D$ of $\\angle B A C$, resulting in $\\triangle A E D$. Let the perimeter of $\\triangle C D E$ be $m$.\n\n(2) As shown in Figure 2, fold $\\triangle B F G$ along the perpendicular bisector of $A B$, resulting in $\\triangle A F G$. Let the perimeter of $\\triangle A G C$ be $n$. The length of segment $A B$ can be expressed as an algebraic expression containing $m$ and $n$ as ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $n-m$\nB. $\\frac{m+n}{2}$\nC. $m$\nD. $\\frac{n}{2}$", "input_image": [ "batch10-2024_06_14_becff04b6620d764874eg_0031_1.jpg", "batch10-2024_06_14_becff04b6620d764874eg_0031_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since folding $\\triangle ABD$ along the angle bisector $AD$ of $\\angle BAC$ results in $\\triangle AED$,\n\n$\\therefore \\triangle ABD \\cong \\triangle AED$,\n\n$\\therefore AB = AE, BD = DE$,\n\nSince the perimeter of $\\triangle CDE$ is $m$,\n\n$\\therefore CD + DE + CE = CD + BD + CE = BC + CE = m$,\n\nThat is, $BC = m - CE$,\n\nSince folding $\\triangle BFG$ along the perpendicular bisector of $AB$ results in $\\triangle AFG$,\n\n$\\therefore \\triangle BFG \\cong \\triangle AFG$,\n\n$\\therefore AG = BG$,\n\nSince the perimeter of $\\triangle AGC$ is $n$,\n\n$\\therefore AG + CG + AC = BG + CG + AC = BC + AC = n$,\n\nThat is, $BC = n - AC$,\n\n$\\therefore m - CE = n - AC$,\n\n$\\therefore AC - CE = n - m$,\n\nSince $AC - CE = AE = AB$\n$\\therefore AB = n - m$.\n\nTherefore, the answer is: A\n\n【Key Insight】This problem primarily examines the folding properties of shapes and the properties of congruent triangles. Mastering the folding properties of shapes and the properties of congruent triangles is crucial for solving the problem." }, { "problem_id": 37, "question": "In $\\triangle A B C$ as shown, $\\angle B A C = 90^\\circ$ and $\\angle B = 60^\\circ$. Using a straightedge without markings and a compass, find a point $D$ on side $B C$ such that $\\triangle A B D$ is an equilateral triangle. Which of the following methods is incorrect? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_becff04b6620d764874eg_0037_1.jpg", "batch10-2024_06_14_becff04b6620d764874eg_0037_2.jpg", "batch10-2024_06_14_becff04b6620d764874eg_0037_3.jpg", "batch10-2024_06_14_becff04b6620d764874eg_0037_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution:\n\nA. According to the construction, point $D$ is the intersection of the perpendicular bisector of $A C$ with $B C$, so $D A = D C$. Therefore, $\\angle D A C = \\angle C = 30^{\\circ}$, and $\\angle B A D = 60^{\\circ}$. This makes $\\triangle A B D$ an equilateral triangle, so option A does not meet the requirement;\n\nB. According to the construction, $B A = B D$, and since $\\angle B = 60^{\\circ}$, $\\triangle A B D$ is an equilateral triangle. Thus, option B does not meet the requirement;\n\nC. According to the construction, point $D$ is the intersection of the perpendicular bisector of $A B$ with $B C$, so $D A = D B$. Given that $\\angle B = 60^{\\circ}$, $\\triangle A B D$ is an equilateral triangle. Therefore, option C does not meet the requirement;\n\nD. According to the construction, $A D$ bisects $\\angle B A C$, so $\\angle B A D = 45^{\\circ}$. This means $\\triangle A B D$ is not an equilateral triangle, making option D the correct choice.\n\nTherefore, the correct answer is: D.\n\n[Key Insight] This question tests basic geometric constructions and the identification of equilateral triangles. Mastering basic constructions is crucial for solving such problems." }, { "problem_id": 38, "question": "As shown in Figure 1, a small cube with an edge length of 1 is placed on a horizontal table. Figures 2 and 3 are geometric solids formed by stacking such small cubes. Following this pattern, what is the exposed surface area of the solid in the 5th stacked figure? ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. 117\nB. 118\nC. 119\nD. 120", "input_image": [ "batch10-2024_06_14_d6e7887b4e796c8b833cg_0085_1.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0085_2.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0085_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, the number of faces exposed outside the tabletop is: $1+3+5+\\ldots+(2n-1)=\\frac{n(2n-1+1)}{2}=n^{2}$. Therefore, from the front, back, left, and right views, the number of faces exposed outside the tabletop is $4n^{2}$.\n\nFrom the top view, the number of faces exposed outside the tabletop is: $2(2n-1)-1=4n-3$.\n\nThus, in the $n$th stacked figure, the total number of faces exposed outside the tabletop is: $4n^{2}+4n-3=(2n+1)^{2}-4$.\n\nThe surface area exposed outside the tabletop is $(2n+1)^{2}-4$.\n\n$\\therefore$ In the 5th stacked figure, the surface area of the geometric body exposed outside the tabletop is $(5 \\times 2+1)^{2}-4=117$.\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This problem tests the pattern of graphical changes. The three-dimensional figure is complex, and it is crucial to determine the number of cubes and the number of faces exposed outside the tabletop in a systematic manner to avoid repetition and omission, which is also the key to solving the problem." }, { "problem_id": 39, "question": "Which of the following statements is false ( ）\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\n\nA. As shown in Figure (1), when construction workers build walls, they often insert wooden stakes at the two corners and then pull a straight reference line. This practice can be explained geometrically as: two points determine a line.\n\nB. As shown in Figure (2), when two rulers, a and b, are placed together with their ends aligned, if ruler a is calibrated as straight, then ruler b is not straight. The basis for judgment is: through two points, there is exactly one straight line.\n\nC. As shown in Figure (3), to measure the angle $\\angle A O B$ between two walls, but a person cannot enter the walls, one can first extend $B O$ to get $\\angle A O C$, then measure the angle $\\angle A O C$, and finally calculate $\\angle A O B$. The principle used here is: the complement of equal angles is equal.\n\nD. As shown in Figure (4), there were originally three routes from Xiao Ming's home $\\mathrm{A}$ to school $B$: route (1) $A-D-B$; route (2) $A-E-B$; route (3) $A-C-B$. A straight path was later opened, route (4) $A-B$. Among these four routes, route (4) has the shortest distance, which is based on the principle that: the shortest distance between two points is a straight line.", "input_image": [ "batch10-2024_06_14_d6e7887b4e796c8b833cg_0089_1.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0089_2.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0089_3.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0089_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: \n\nA. The knowledge \"two points determine a straight line\" is applied, hence it is correct and does not match the question's requirement;\n\nB. The knowledge \"through two points, there is one and only one straight line\" is applied, hence it is correct and does not match the question's requirement;\n\nC. The knowledge of adjacent supplementary angles is applied, rather than the equality of the complements of equal angles, hence it is incorrect and matches the question's requirement;\n\nD. The knowledge \"the line segment between two points is the shortest\" is applied, hence it is correct and does not match the question's requirement;\n\nTherefore, the answer is: C.\n\n[Key Point] This question tests knowledge such as two points determining a straight line, through two points there is one and only one straight line, complementary angles, adjacent supplementary angles, and the shortest line segment between two points. The key to solving the problem lies in understanding the actual scenario and interpreting it with mathematical knowledge." }, { "problem_id": 40, "question": "As shown in the figure, there is a straight river $\\mathrm{L}$. A shepherd starts from point $\\mathrm{P}$, takes his horses to drink at point $\\mathrm{M}$ on the riverbank, and then goes to point $\\mathrm{Q}$. Among the following four options, which one will minimize the shepherd's total path length?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_fd61d246744e8785fd79g_0045_1.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0045_2.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0045_3.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0045_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Find the symmetric point $\\mathrm{P}^{\\prime}$ of point $\\mathrm{P}$ with respect to line 1, then connect $\\mathrm{QP}^{\\prime}$ and let it intersect line 1 at point $\\mathrm{M}$.\n\nAs shown in the figure,\n\n\n\nAccording to the principle that the shortest distance between two points is a straight line, it can be concluded that option B results in the shortest path for the herder.\n\nTherefore, choose D.\n\n【Key Point】This question examines the mathematical problem of the shortest path. The solution to such problems is based on the principle that \"the shortest distance between two points is a straight line.\"" }, { "problem_id": 41, "question": "The Tangram, known to Westerners as the \"Magic Square of the East.\" The two figures below are formed by the same set of Tangrams. If the side length of the square formed by the Tangram (as shown in Figure 1) is 8, then the perimeter of the \"clothing shape\" (as shown in Figure 2) is ( )\n\n\n\n$⿺_{*}$\n\n\n\n$8 / 2$\nA. 10\nB. 12\nC. $10+4 \\sqrt{2}$\nD. $12+20 \\sqrt{2}$", "input_image": [ "batch11-2024_06_14_03bf5911dff0ff531efeg_0056_1.jpg", "batch11-2024_06_14_03bf5911dff0ff531efeg_0056_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nSince all the triangles within the tangram are isosceles right triangles, all acute angles are equal to $45^{\\circ}$. Given that the side length of the square is 8, according to the Pythagorean theorem, we have: $O A=O D=O C=E F=4 \\sqrt{2}$, $A E=B E=B F=C F=H I=4$, $G A=G E=E I=O G=O I=O H=H C=2 \\sqrt{2}$.\n\nAs illustrated; when the tangram is assembled into the \"clothing shape\",\n\n\n\nThe perimeter of the \"clothing shape\" (as shown in Figure 2) is:\n\n$$\n\\begin{aligned}\n& A B+B C+C D+D E+E F+F G+G H+H I+I J+J K+K A \\\\\n& =4 \\sqrt{2}+4 \\sqrt{2}+4+2 \\sqrt{2}+4+2 \\sqrt{2}+2 \\sqrt{2}+4+2 \\sqrt{2}+4+4 \\sqrt{2} \\\\\n& =12+20 \\sqrt{2}\n\\end{aligned}\n$$\n\nTherefore, the correct choice is: D.\n\n【Key Insight】This problem primarily tests the Pythagorean theorem and the properties of isosceles right triangles. Determining the side lengths of each shape within the tangram is crucial for solving the problem." }, { "problem_id": 42, "question": "The \"Pythagorean diagram\" has a long history and has piqued the interest of many people. In 1955, Greece issued a stamp with the Pythagorean diagram as its background (as shown in Figure 1). Euclid studied this diagram in depth in his work \"Elements.\" As shown in Figure 2, in triangle $A B C$, $\\angle A C B = 90^\\circ$. Square regions are constructed outwardly on each side of the triangle, and lines $E B$, $C M$, $D G$, and $C M$ are drawn, with $D G$ and $C M$ intersecting $A B$ and $B E$ at points $P$ and $Q$, respectively. If $\\angle A B E = 30^\\circ$, then the value of $\\frac{D G}{Q M}$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{\\sqrt{3}}{2}$\nB. $\\frac{\\sqrt{5}}{3}$\nC. $\\frac{4}{5}$\nD. $\\sqrt{3}-1$", "input_image": [ "batch11-2024_06_14_03bf5911dff0ff531efeg_0069_1.jpg", "batch11-2024_06_14_03bf5911dff0ff531efeg_0069_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: \n\nIn triangles $\\triangle EAB$ and $\\triangle CAM$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\nAE = AC \\\\\n\\angle EAB = \\angle CAM, \\\\\nAB = AM\n\\end{array}\n\\right.\n\\]\n\nTherefore, $\\triangle EAB \\cong \\triangle CAM$ (by SAS congruence),\n\nHence, $\\angle EBA = \\angle CMA = 30^\\circ$,\n\nThus, $\\angle BPQ = \\angle APM = 60^\\circ$,\n\nAnd $\\angle BQP = 90^\\circ$,\n\nSo, $PQ = \\frac{1}{2} PB$,\n\nAssuming $AP = 1$, then $AM = \\sqrt{3}$, $PM = 2$, $PB = \\sqrt{3} - 1$, $PQ = \\frac{\\sqrt{3} - 1}{2}$,\n\nTherefore, $QM = QP + PM = \\frac{\\sqrt{3} - 1}{2} + 2 = \\frac{\\sqrt{3} + 3}{2}$;\n\nIn right triangles $\\triangle ACB$ and $\\triangle DCG$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\nCG = BC \\\\\nAC = CD\n\\end{array}\n\\right.\n\\]\n\nThus, $\\triangle ACB \\cong \\triangle DCG$ (by HL congruence),\n\nHence, $DG = AB = \\sqrt{3}$\n\nTherefore, $\\frac{DG}{GM} = \\frac{\\sqrt{3}}{\\frac{\\sqrt{3} + 3}{2}} = \\sqrt{3} - 1$\n\nThe correct choice is D.\n\n[Highlight] This problem primarily tests knowledge of the Pythagorean theorem, the criteria for triangle congruence, and properties of congruent triangles." }, { "problem_id": 43, "question": "As shown in Figure 1, a ball rolls down an incline on the left and continues to roll along a flat surface until it comes to a stop. During this process, the ball's motion speed $v$ (in units of $\\mathrm{m} / \\mathrm{s}$) is plotted against the motion time $t$ (in units of $\\mathrm{s}$) as shown in Figure 2. The approximate function graph of the ball's motion distance $y$ (in units of $\\mathrm{m}$) against the motion time $t$ (in units of $\\mathrm{s}$) is ( )\n\n\n\nFigure 1\n\nA.\n\n\n\n\n\nFigure 2\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch11-2024_06_14_96b2f24d25c83e6985b9g_0006_1.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0006_2.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0006_3.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0006_4.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0006_5.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0006_6.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, when the ball rolls down the slope, let \\( V_{1} = k_{1} t \\) (where \\( k_{1} > 0 \\)).\n\nThen, \\( y = V_{1} t = k_{1} t^{2} \\).\n\nTherefore, the distance traveled \\( y \\) is a quadratic function of \\( t \\), with the graph opening upwards, and the trend of the graph changes from gradual to steep.\n\nWhen the ball rolls on the horizontal surface, let \\( V_{2} = k_{2} t \\) (where \\( k_{2} < 0 \\)).\n\nThen, \\( y = V_{2} t = k_{2} t^{2} \\).\n\nTherefore, the distance traveled \\( y \\) is a quadratic function of \\( t \\), with the graph opening downwards, and the trend of the graph changes from steep to gradual.\n\nHence, the correct choice is C.\n\n[Key Insight] This problem mainly examines the function graph of a moving point. The key to solving it is to understand the problem, formulate the function expression, and flexibly apply the knowledge learned to solve the problem." }, { "problem_id": 44, "question": "A cubical wooden block, fixed on the ground (as shown in Figure (1)), has edges of $4 \\mathrm{~cm}$. A corner is removed along the diagonals of three adjacent faces (as shown by the dashed lines in Figure (1)), resulting in the geometric wooden block shown in Figure (2). The shortest distance the ant can travel along the surface of the block from point $A$ to point $B$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $2(\\sqrt{2}+\\sqrt{6})$\nB. $4 \\sqrt{2}+4$\nC. $4 \\sqrt{2}+2$\nD. $2 \\sqrt{6}+4$", "input_image": [ "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0051_1.jpg", "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0051_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, the diagonal on the top surface of the cube is $CD$. By unfolding figure (2) and connecting $AB$ to intersect $CD$ at point $E$, the length of segment $AB$ represents the shortest distance the ant must crawl.\n\n\n\nFrom the problem statement, we know that $ACD$ is an equilateral triangle, and $\\triangle CBD$ is an isosceles right triangle.\n\nSince $AC = AD$, $BC = BD$, and $AB = AB$,\n\n$\\triangle ACB \\cong \\triangle ADB$ (by SSS congruence),\n\nTherefore, $\\angle CBE = \\angle DBE$,\n\nHence, $AB \\perp CD$.\n\nGiven that the edge length of the cube is $4 \\mathrm{~cm}$,\n\n$BC = BD = 4 \\mathrm{~cm}$, and $AC = AD = CD = \\sqrt{4^{2} + 4^{2}} = 4 \\sqrt{2}$.\n\nIn right triangle $\\triangle CEB$, $BE = CE = \\frac{1}{2} CD = 2 \\sqrt{2}$.\n\nIn right triangle $\\triangle CEA$, $AE = \\sqrt{AC^{2} - CE^{2}} = 2 \\sqrt{6}$.\n\nThus, $AB = AE + CE = 2 \\sqrt{2} + 2 \\sqrt{6} = 2(\\sqrt{2} + \\sqrt{6})$.\n\nTherefore, the correct answer is A.\n\n【Key Insight】This problem tests the application of the Pythagorean theorem. The key to solving it lies in unfolding the three-dimensional figure and determining the shortest path based on the principle that the shortest distance between two points is a straight line." }, { "problem_id": 45, "question": "As shown in Figure (1), a supermarket has installed a doorbell A on the wall, which is $4.5 \\mathrm{~m}$ above the ground. The doorbell automatically speaks \"Welcome\" when a person is within $4 \\mathrm{~m}$ of the entrance. As shown in Figure (2), a student with a height of $1.5 \\mathrm{~m}$ just arrived at point D and the doorbell rang. What is the distance from the top of the student's head C to the doorbell A?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $7 \\mathrm{~m}$\nB. $6 \\mathrm{~m}$\nC. $5 \\mathrm{~m}$\nD. $4 \\mathrm{~m}$", "input_image": [ "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0073_1.jpg", "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0073_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement:\n\n$BE = CD = 1.5 \\mathrm{~m}$,\n\n$AE = AB - BE = 4.5 - 1.5 = 3 \\mathrm{~m}, \\quad CE = 4 \\mathrm{~m}$,\n\nUsing the Pythagorean theorem, we find:\n\n$$\nAC = \\sqrt{AE^{2} + CE^{2}} = \\sqrt{3^{2} + 4^{2}} = 5 \\quad(\\mathrm{~m})\n$$\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This problem tests the application of the Pythagorean theorem. Being adept at observing the information provided in the problem is crucial for solving it and excelling in mathematics." }, { "problem_id": 46, "question": "A rectangular container is placed as shown in Figure (1), with a length and width of 2 and a height of 8. It is filled with water up to a height of 5. If the container is tilted (without spilling the water), the front view of the tilted rectangular container is shown in Figure (2). What is the length of segment $C D$ in the figure?\n\n\n\nFigure 1\n\n\nA. $2 \\sqrt{5}$\nB. $3 \\sqrt{5}$\nC. $3 \\sqrt{10}$\nD. $2 \\sqrt{10}$", "input_image": [ "batch11-2024_06_14_d5d1bb66c8fb9cc6a3d9g_0015_1.jpg", "batch11-2024_06_14_d5d1bb66c8fb9cc6a3d9g_0015_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nLet $DE = x$, then $AD = 8 - x$.\n\nAccording to the problem, we have:\n\\[\n\\frac{1}{2}(8 - x + 8) \\times 2 \\times 2 = 2 \\times 2 \\times 5\n\\]\nSolving this equation gives:\n\\[\nx = 6\n\\]\nTherefore, $DE = 6$.\n\nSince $\\angle E = 90^\\circ$, by the Pythagorean theorem:\n\\[\nCD = \\sqrt{DE^2 + CE^2} = \\sqrt{6^2 + 2^2} = 2\\sqrt{10}\n\\]\nThus, the length of $CD$ is $2\\sqrt{10}$.\n\nThe correct answer is: D.\n\n【Key Insight】This problem examines the application of the Pythagorean theorem, the volume of a rectangular prism, and the calculation method for the area of a trapezoid. Mastering the Pythagorean theorem and deriving the equation from the volume of water in the rectangular container is crucial to solving the problem." }, { "problem_id": 47, "question": "As shown in Figure 1, squares are constructed outwardly on the sides of a right-angled triangle with areas $S_{1}, S_{2}, S_{3}$; as shown in Figure 2, semicircles are constructed outwardly with the lengths of the sides of the right-angled triangle as their diameters, with areas $S_{4}, S_{5}, S_{6}$. Given that $S_{1}=1, S_{2}=3, S_{5}=2, S_{6}=4$, what is the value of $S_{3}+S_{4}$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 10\nB. 9\nC. 8\nD. 7", "input_image": [ "batch12-2024_06_15_02413583c06955ad7983g_0032_1.jpg", "batch12-2024_06_15_02413583c06955ad7983g_0032_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n$\\because S_{1}=a^{2}, S_{2}=b^{2}, S_{3}=c^{2}, \\quad a^{2}+b^{2}=c^{2}$,\n\n$\\therefore S_{1}+S_{2}=S_{3}$,\n\nSimilarly, we can obtain, $S_{5}+S_{6}=S_{4}$,\n\n$\\because S_{1}=1, S_{2}=3, S_{5}=2, S_{6}=4$\n\n$\\therefore S_{3}+S_{4}=(1+3)+(2+4)=10$,\n\nTherefore, the answer is: $A$.\n\n\n\n【Key Insight】This question tests the application of the Pythagorean theorem. The key to solving the problem lies in correctly understanding the relationships between the different parts of the figure and the calculation formula of the Pythagorean theorem." }, { "problem_id": 48, "question": "In the ancient Chinese mathematical text \"Zhou Bi Suan Jing,\" a \"chord diagram\" is described as being formed by four right-angled triangles surrounding a hollow square. In a math class, the teacher leads the students to use four right-angled triangles with legs $a$ and $b$ as shown in Figure 1 to form two \"chord diagrams\" (as shown in Figure 2 and Figure 3). Given that the area of the large square $A B C D$ is 48 and the area of the small square $E F G H$ is 12, what is the area of one right-angled triangle?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. 3\nB. 4.5\nC. 6\nD. 9", "input_image": [ "batch12-2024_06_15_02413583c06955ad7983g_0060_1.jpg", "batch12-2024_06_15_02413583c06955ad7983g_0060_2.jpg", "batch12-2024_06_15_02413583c06955ad7983g_0060_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: The area of the large square \\( A B C D \\) is 48, and the area of the small square \\( E F G H \\) is 12.\n\n\\[\n\\therefore (a + b)^2 = 48, \\quad (a - b)^2 = 12,\n\\]\n\n\\[\n\\therefore a^2 + b^2 + 2ab = 48, \\quad a^2 + b^2 - 2ab = 12.\n\\]\n\nSubtracting the two equations gives: \\( 4ab = 36 \\),\n\n\\[\n\\therefore \\frac{1}{2}ab = \\frac{9}{2}, \\text{ which means the area of one right triangle is } 4.5.\n\\]\n\nTherefore, the correct answer is B.\n\n【Key Insight】This problem tests the application of the Pythagorean theorem and the perfect square formula. The key to solving it lies in deriving the equations based on the given geometric figure." }, { "problem_id": 49, "question": "As shown in the figure, Figure 1 is a diagram of the famous ancient Chinese \"Zhao Shuang Chord Diagram,\" which is composed of four congruent right-angled triangles. If $A C = 6$ and $B C = 5$, extend each of the 6-unit legs of the four right-angled triangles outward by a factor of two to form the \"Mathematical Windmill\" as shown in Figure 2. Then the outer circumference of this windmill is ( )\n\n\n\nFigure 1\n\n\nA. 24\nB. 52\nC. 61\nD. 76", "input_image": [ "batch12-2024_06_15_02413583c06955ad7983g_0068_1.jpg", "batch12-2024_06_15_02413583c06955ad7983g_0068_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, $\\angle \\mathrm{ACB}$ is a right angle, and $\\mathrm{AD}=6$,\n\n$\\therefore C D=6+6=12$,\n\nBy the Pythagorean theorem, $\\mathrm{BD}^{2}=\\mathrm{BC}^{2}+\\mathrm{CD}^{2}$,\n\n$\\therefore \\mathrm{BD}^{2}=12^{2}+5^{2}=169$,\n\nTherefore, $\\mathrm{BD}=13$,\n\nSo the perimeter of the \"mathematical windmill\" is: $(13+6) \\times 4=76$.\n\nHence, the answer is: D.\n\n\n\n【Insight】This problem applies the Pythagorean theorem in a practical scenario. Mastering the Pythagorean theorem is crucial for solving this problem. In a right-angled triangle, if the two legs are $\\mathrm{a}$ and $\\mathrm{b}$, and the hypotenuse is $\\mathrm{c}$, then $\\mathrm{a}^{2}+\\mathrm{b}^{2}=\\mathrm{c}^{2}$." }, { "problem_id": 50, "question": "There is a square with an area of 1. After one \"growth\", two smaller squares are born on its left and right shoulders (as shown in Figure 1), forming a right-angled triangle with the original square. After another \"growth\", four more squares are born (as shown in Figure 2). If this pattern continues, the square will become \"bushy\". After 2017 growths, the sum of the areas of all the squares in the resulting figure is $(\\quad)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 2015\nB. 2016\nC. 2017\nD. 2018", "input_image": [ "batch12-2024_06_15_22894a68bf357c1c0757g_0012_1.jpg", "batch12-2024_06_15_22894a68bf357c1c0757g_0012_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the three sides of the right triangle be $a, b, c$. According to the Pythagorean theorem, we have: $a^{2}+b^{2}=c^{2}$, which means the area of square $A$ plus the area of square $B$ equals the area of square $C$, which is $1$.\n\nExtending this, after \"growing\" 2017 times, the total area of all the squares formed in the resulting figure is $2018 \\times 1=2018$. Therefore, the correct answer is D.\n\n\n\n[Insight] This question tests the properties of squares and the Pythagorean theorem. The key to solving this problem lies in understanding the relationship between the area of the new squares obtained each time and the area of the original square, based on the Pythagorean theorem." }, { "problem_id": 51, "question": "The Pythagorean theorem is one of the greatest scientific discoveries in human history, and it was recorded in ancient Chinese mathematical text \"Zhou Bi Suan Jing\" long ago. As shown in Figure 1, squares are constructed outwardly on each side of a right-angled triangle, and then the two smaller squares are rotated around the vertices on the hypotenuse to form Figure 2. The shaded area in Figure 2 is equal to ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. the area of the right-angled triangle\nB. the area of the largest square\nC. the sum of the areas of the largest square and the right-angled triangle\nD. the area of the overlapping part of the two smaller squares", "input_image": [ "batch12-2024_06_15_22894a68bf357c1c0757g_0055_1.jpg", "batch12-2024_06_15_22894a68bf357c1c0757g_0055_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let the shorter leg of the right triangle be \\( a \\), the longer leg be \\( b \\), and the hypotenuse be \\( c \\).\n\n\n\nBy the Pythagorean theorem, we have \\( c^{2} = a^{2} + b^{2} \\).\n\nThe area of the shaded region \\( = c^{2} - b^{2} - a(c - b) = a^{2} - a(c - b) = a(a + b - c) \\).\n\nThe area of the overlapping part of the two smaller squares \\( = a(a + b - c) \\).\n\n\\(\\therefore\\) The area of the shaded region equals the area of the overlapping part of the two smaller squares.\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This problem primarily tests knowledge of the Pythagorean theorem. The key to solving it lies in analyzing the problem using the mathematical concept of combining numbers and shapes." }, { "problem_id": 52, "question": "Five small sticks are given. Arrange them to form two right-angled triangles. Which of the following arrangements is correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n15", "input_image": [ "batch12-2024_06_15_2e56bd186f393949ff46g_0004_1.jpg", "batch12-2024_06_15_2e56bd186f393949ff46g_0004_2.jpg", "batch12-2024_06_15_2e56bd186f393949ff46g_0004_3.jpg", "batch12-2024_06_15_2e56bd186f393949ff46g_0004_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Incorrect placement, because as seen from the figure: $\\sqrt{24^{2}+15^{2}} \\neq 20$, which does not satisfy the converse of the Pythagorean theorem, thus it does not meet the requirement;\n\nB. Incorrect placement, because as seen from the figure: $\\sqrt{20^{2}+15^{2}} \\neq 24$, which does not satisfy the converse of the Pythagorean theorem, thus it does not meet the requirement;\n\nC. Incorrect placement, because as seen from the figure: $\\sqrt{20^{2}+7^{2}} \\neq 25$, which does not satisfy the converse of the Pythagorean theorem, thus it does not meet the requirement;\n\nD. Correct placement, because as seen from the figure: $\\sqrt{20^{2}+15^{2}}=25$ and $\\sqrt{24^{2}+7^{2}}=25$, which satisfy the converse of the Pythagorean theorem, thus it meets the requirement;\n\nTherefore, the correct choice is: D.\n\n[Highlight] This question tests the converse of the Pythagorean theorem. The key to solving it is to understand the converse of the Pythagorean theorem and analyze it in conjunction with the figure." }, { "problem_id": 53, "question": "As shown in the figure, a water pumping station is to be built on the riverbank $l$ to supply water to Villages $A$ and $B$. It is known that the distances from Villages $A$ and $B$ to the riverbank are $1 \\mathrm{~km}$ and $3 \\mathrm{~km}$, respectively, and that $C D$ is $3 \\mathrm{~km}$ apart. What is the shortest length of the pipeline in $(\\quad) \\mathrm{km}$?\n\n\n\nA. 5\nB. 4\nC. 3\nD. 6", "input_image": [ "batch12-2024_06_15_2e56bd186f393949ff46g_0016_1.jpg", "batch12-2024_06_15_2e56bd186f393949ff46g_0016_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Construct the symmetric point $E$ of $A$ with respect to the river, connect $A$ to $E$, and connect $B$ to $E$. Then, $BE$ represents the shortest distance required. Draw $AG \\perp BF$ at $G$ from $A$, and draw $EF \\perp BF$ at $F$ from $E$.\n\nGiven that $AC = 1 \\text{ km}$ and $BD = 3 \\text{ km}$,\n\nwe have $BG = 3 - 1 = 2 \\text{ km}$, $GD = AC = 1 \\text{ km}$, $CE = DF = AC = 1 \\text{ km}$, $BF = 3 + 1 = 4 \\text{ km}$, and $EF = CD = 3 \\text{ km}$.\n\nIn the right triangle $\\triangle BEF$, $BE = \\sqrt{EF^2 + BF^2} = \\sqrt{4^2 + 3^2} = 5 \\text{ km}$.\n\nTherefore, the shortest length for laying the water pipe is $5 \\text{ km}$,\n\nand the correct choice is A.\n\n\n\n【Key Insight】This problem primarily examines the shortest path problem using symmetry and the Pythagorean theorem. Correctly drawing the auxiliary lines is crucial to solving the problem." }, { "problem_id": 54, "question": "As shown in Figure 1, squares are constructed outside each side of a right-angled triangle. Then, the two smaller squares are arranged inside the larger square as shown in Figure 2. Let the area of quadrilateral $A B C D$ be $S_{1}$, the area of quadrilateral $D C E G$ be $S_{2}$, the area of quadrilateral $H G F P$ be $S_{3}$, and the area of triangle $G E F$ be $S_{4}$. If the area of the shaded region in the figure is known, then it is definitely possible to find ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $S_{1}$\nB. $S_{2}$\nC. $S_{3}$\nD. $S_{4}$", "input_image": [ "batch12-2024_06_15_2e56bd186f393949ff46g_0076_1.jpg", "batch12-2024_06_15_2e56bd186f393949ff46g_0076_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, let the area of the smallest square be $a$, the area of square $HBEP$ be $b$, and the area of the largest square be $c$.\n\nAs shown in Figure 2, since $S_{3} + S_{\\text{shaded area}} = \\frac{1}{2}(c - a)$ and $S_{3} + S_{4} = \\frac{1}{2}b$,\n\nand since $c = a + b$,\n\nit follows that $b = c - a$.\n\nTherefore, $S_{3} + S_{\\text{shaded area}} = S_{3} + S_{4}$,\n\nwhich implies $S_{\\text{shaded area}} = S_{4}$.\n\nThus, knowing the area of the shaded part in the figure allows us to determine $S_{4}$.\n\nHence, the correct choice is: D.\n\n[Key Insight] This problem examines the Pythagorean theorem and the operations of algebraic expressions. Mastering the Pythagorean theorem is crucial for solving this problem." }, { "problem_id": 55, "question": "A paper model of an isosceles triangle $A B C$ is cut along the perpendicular segments $A D$ and $D E$. The resulting triangles (1), (2), and (3) are rearranged as shown in Figure 2, with $E C$ collinear with $B D$. If $B D = 6$, what is the length of $A B$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{22}{3}$\nB. $\\frac{15}{2}$\nC. $\\sqrt{50}$\nD. 7", "input_image": [ "batch12-2024_06_15_395683fc667a063e6e86g_0081_1.jpg", "batch12-2024_06_15_395683fc667a063e6e86g_0081_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let $\\angle B$ be $\\angle 1$, $\\angle C$ be $\\angle 2$, and $\\angle CDE$ be $\\angle 3$. In Figure 2, the complementary angle of $\\angle 1$ is $\\angle 4$. Since $\\triangle ABC$ is an isosceles triangle and $BD=6$,\n\n$\\therefore \\angle 1 = \\angle 2$, and $CD=6$.\n\n$\\because \\angle 2 + \\angle 3 = \\angle 1 + \\angle 4 = 90^{\\circ}$,\n\n$\\therefore \\angle 3 = \\angle 4$.\n\nCombining the two figures, we get $AB = AD + \\frac{1}{2} CD$.\nLet $AB$ be $x$.\n\nAccording to the Pythagorean theorem, $AD = \\sqrt{x^{2} - 6^{2}} = \\sqrt{x^{2} - 36}$.\n\n$\\therefore x = \\sqrt{x^{2} - 36} + \\frac{1}{2} \\times 6$.\n\nSolving the equation gives: $x = \\frac{15}{2}$.\n\n$\\therefore AB = \\frac{15}{2}$.\n\nTherefore, the correct answer is: B.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Insight】This problem examines the properties of isosceles triangles and the Pythagorean theorem. Identifying the graphical relationships and equal quantities is key to solving the problem." }, { "problem_id": 56, "question": "As shown in the figure, in square $\\square \\mathrm{ABCD}$, the diagonals $\\mathrm{AC}$ and $\\mathrm{BD}$ intersect at point $\\mathrm{E}, \\angle \\mathrm{AEB}=45^{\\circ}$, and $\\mathrm{BD}=2$. The triangle $\\triangle \\mathrm{ABC}$ is folded along the line containing $\\mathrm{AC}$ for $180^{\\circ}$ to its original plane. If the point $\\mathrm{B}$'s landing point is denoted as $\\mathrm{B}^{\\prime}$, then the length of $\\mathrm{DB}^{\\prime}$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 1\nB. $\\sqrt{2}$\nC. $\\frac{3}{2}$\nD. $\\sqrt{3}$", "input_image": [ "batch12-2024_06_15_52a97c4d589071dddfc6g_0052_1.jpg", "batch12-2024_06_15_52a97c4d589071dddfc6g_0052_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since quadrilateral $\\mathrm{ABCD}$ is a parallelogram, and $\\mathrm{BD}=2$,\n\n$\\therefore \\mathrm{BE}=\\frac{1}{2} \\mathrm{BD}=1$.\n\nAs shown in Figure 2, connect $\\mathrm{BB}^{\\prime}$.\n\nAccording to the properties of folding, $\\angle \\mathrm{AEB}=\\angle \\mathrm{AEB}^{\\prime}=45^{\\circ}$, and $\\mathrm{BE}=\\mathrm{B}^{\\prime} \\mathrm{E}$.\n\n$\\therefore \\angle \\mathrm{BEB}^{\\prime}=90^{\\circ}$,\n\n$\\therefore \\triangle \\mathrm{BB}^{\\prime} \\mathrm{E}$ is an isosceles right triangle, hence $\\mathrm{BB}^{\\prime}=\\sqrt{2} \\mathrm{BE}=\\sqrt{2}$,\n\nMoreover, since $\\mathrm{BE}=\\mathrm{DE}$, and $\\mathrm{B}^{\\prime} \\mathrm{E} \\perp \\mathrm{BD}$,\n\n$\\therefore \\mathrm{DB}^{\\prime}=\\mathrm{BB}^{\\prime}=\\sqrt{2}$.\n\n\n\nFigure 2\n\nTherefore, choose B.\n\n【Highlight】This problem examines the properties of parallelograms and isosceles right triangles. The difficulty is moderate, and attention should be paid to the method of drawing auxiliary lines and the application of the combination of numbers and shapes." }, { "problem_id": 57, "question": "As shown in Figure (1), there is a right-angled triangle paper, with $\\angle C = 90^\\circ, AB = 13 \\text{ cm}, BC = 5 \\text{ cm}$. It is folded so that point C falls on the hypotenuse at point $C^\\prime$, and the crease is BD (as shown in Figure (2)). The length of DC is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $\\frac{10}{3} \\text{ cm}$\nB. $\\frac{8}{3} \\text{ cm}$\nC. $\\frac{5}{2} \\text{ cm}$\nD. $\\sqrt{5} \\text{ cm}$", "input_image": [ "batch12-2024_06_15_52a97c4d589071dddfc6g_0071_1.jpg", "batch12-2024_06_15_52a97c4d589071dddfc6g_0071_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Given that \\(\\angle C = 90^\\circ\\), \\(AB = 13 \\text{ cm}\\), and \\(BC = 5 \\text{ cm}\\), we can find \\(AC\\) using the Pythagorean theorem:\n\n\\[\nAC = \\sqrt{AB^2 - BC^2} = \\sqrt{13^2 - 5^2} = \\sqrt{169 - 25} = \\sqrt{144} = 12 \\text{ cm}\n\\]\n\nAdditionally, since \\(BC' = BC = 5 \\text{ cm}\\) and \\(DC' = DC\\), we have:\n\n\\[\nAC' = AB - BC' = 13 - 5 = 8 \\text{ cm}\n\\]\n\nLet \\(DC = x\\), then \\(AD = AC - DC = 12 - x\\).\n\nSince \\(DC' = x\\), in the right triangle \\(\\triangle AC'D\\), applying the Pythagorean theorem gives:\n\n\\[\nAC'^2 + DC'^2 = AD^2\n\\]\n\nSubstituting the known values:\n\n\\[\n8^2 + x^2 = (12 - x)^2\n\\]\n\nExpanding and simplifying:\n\n\\[\n64 + x^2 = 144 - 24x + x^2\n\\]\n\n\\[\n64 = 144 - 24x\n\\]\n\n\\[\n24x = 80\n\\]\n\n\\[\nx = \\frac{80}{24} = \\frac{10}{3}\n\\]\n\nTherefore, \\(DC = \\frac{10}{3} \\text{ cm}\\), and the correct choice is A.\n\n**Key Insight:** Mastery of flip transformations and the Pythagorean theorem is essential. The key to solving this problem lies in recognizing that the lengths of corresponding segments and angles remain equal after flipping, and using the Pythagorean theorem to set up the equation." }, { "problem_id": 58, "question": "As shown in Figure 1, if a point $P$ inside $\\triangle A B C$ satisfies $\\angle P A C = \\angle P B A = \\angle P C B$, then point $P$ is the Brocard point of $\\triangle A B C$. The Brocard point of a triangle was first discovered by French mathematician and mathematics educator Closse in 1816, but his discovery was not noticed by people at the time. In 1875, the Brocard point was rediscovered by a mathematics enthusiast, French military officer Brocard, and named after him. Problem: Given an isosceles right triangle $D E F$ as shown in Figure 2, where $\\angle E D F = 90^\\circ$, if point $Q$ is the Brocard point of $\\triangle D E F$, and $D Q = 2$, then $E Q + F Q = $ ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 6\nB. 4\nC. $4+\\sqrt{2}$\nD. $4+2 \\sqrt{2}$", "input_image": [ "batch12-2024_06_15_5e8a141fcd4a4c59eae2g_0055_1.jpg", "batch12-2024_06_15_5e8a141fcd4a4c59eae2g_0055_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Rotate $\\triangle QFD$ clockwise by $90^\\circ$ around point $D$ to obtain $\\triangle MED$, as shown in the figure:\n\n\n\n$\\therefore \\triangle MED \\cong \\triangle QFD$, and $\\angle MDQ = 90^\\circ$,\n\n$\\therefore ME = QF$, $MD = QD = 2$, $\\angle MED = \\angle QFD$, $\\angle MDE = \\angle QDF$,\n\n$\\therefore \\angle DQM = 45^\\circ$, $MQ = \\sqrt{2^2 + 2^2} = 2\\sqrt{2}$,\n\n$\\because$ point $Q$ is the \"Brocard point\" of $\\triangle DEF$,\n\n$\\therefore \\angle QFE = \\angle QDF = \\angle QED$,\n\n$\\therefore \\angle MDE = \\angle QED$,\n\n$\\therefore QE \\parallel DM$,\n\n$\\therefore \\angle DQE = 90^\\circ$,\n\n$\\therefore \\angle MQE = \\angle DQE - \\angle DQM = 90^\\circ - 45^\\circ = 45^\\circ$,\n\n$\\because \\angle QFD + \\angle QFE = 45^\\circ$, and $\\angle MED = \\angle QFD$,\n\n$\\therefore \\angle MED + \\angle QED = \\angle QEM = 45^\\circ$,\n\n$\\therefore \\triangle MQE$ is an isosceles right triangle,\n$\\therefore ME = MQ = 2\\sqrt{2}$,\n\nBy the Pythagorean theorem: $QE = \\sqrt{(2\\sqrt{2})^2 + (2\\sqrt{2})^2} = 4$,\n\nAlso, since $ME = FQ = 2\\sqrt{2}$,\n\n$\\therefore EQ + FQ = 4 + 2\\sqrt{2}$.\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This problem examines the properties of congruent triangles, the determination and properties of isosceles right triangles, the determination and properties of parallel lines, and the Pythagorean theorem. The key to solving the problem lies in constructing an isosceles right triangle through rotation." }, { "problem_id": 59, "question": "As shown in the figure, there are four triangles, each with one side of length 6 and one side of length 8. If the third side lengths are $6,8,10$, and 12, respectively, which triangle has the largest area?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch12-2024_06_15_64fa0b45fb92075e65bbg_0065_1.jpg", "batch12-2024_06_15_64fa0b45fb92075e65bbg_0065_2.jpg", "batch12-2024_06_15_64fa0b45fb92075e65bbg_0065_3.jpg", "batch12-2024_06_15_64fa0b45fb92075e65bbg_0065_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "As shown in the figure, draw the altitude on the side of each triangle with a length of 8. According to the principle that the perpendicular segment is the shortest, in options A, B, and D, the altitude on the side with a length of 8 is less than 6.\n\nIn option C, since \\(6^{2} + 8^{2} = 10^{2}\\), this triangle is a right-angled triangle, so the altitude on the side with a length of 8 is 6. Therefore, among these four options, when the base is 8, the altitude in option C is the largest, making the area of option C the largest. Hence, the correct choice is C.\n\n\n\nA\n\n\n\nB\n\n\n\nC\n\n\n\nD" }, { "problem_id": 60, "question": "Given a triangle with side lengths 4, 9, and 12, the correct way to construct the altitude from the longest side is:\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch12-2024_06_15_64fa0b45fb92075e65bbg_0087_1.jpg", "batch12-2024_06_15_64fa0b45fb92075e65bbg_0087_2.jpg", "batch12-2024_06_15_64fa0b45fb92075e65bbg_0087_3.jpg", "batch12-2024_06_15_64fa0b45fb92075e65bbg_0087_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Since \\(4^{2} + 9^{2} = 97 < 12^{2}\\),\n\nTherefore, the triangle is an obtuse-angled triangle,\n\nThus, the altitude on the longest side is drawn from the vertex opposite the longest side, perpendicular to the opposite side, with the foot of the perpendicular lying on the longest side.\nHence, the correct choice is: C.\n\n[Key Insight] This question examines the method of drawing altitudes in a triangle. When the triangle is acute-angled, all three altitudes lie inside the triangle; when the triangle is right-angled, two of the altitudes are the legs of the triangle, and one lies inside; when the triangle is obtuse-angled, two of the altitudes lie outside the triangle, and one lies inside." }, { "problem_id": 61, "question": "The Tangram is an outstanding creation of our ancestors. The following four figures were formed by thoughtful Xiaohong using the Tangram shown. Arrange the four figures in descending order of their perimeters:\n\n\n\n\nA\n\n\nB\n\n\nC\n\n\nD\nA. B $>$ C $>$ A $>$ D\nB. B $>$ A $>$ C $>$ D\nC. C $>$ B $>$ A $>$ D\nD. C $>$ B $>$ D $>$ A", "input_image": [ "batch12-2024_06_15_6caee0709bd1c52a47c4g_0063_1.jpg", "batch12-2024_06_15_6caee0709bd1c52a47c4g_0063_2.jpg", "batch12-2024_06_15_6caee0709bd1c52a47c4g_0063_3.jpg", "batch12-2024_06_15_6caee0709bd1c52a47c4g_0063_4.jpg", "batch12-2024_06_15_6caee0709bd1c52a47c4g_0063_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the smallest side length in the tangram be 1. According to the Pythagorean theorem,\n\nthe lengths of the other sides can be determined as $2, \\sqrt{2}, 2 \\sqrt{2}$.\n\nCalculate the sum of the lengths of the overlapping line segments in each figure; the larger the sum, the smaller the perimeter.\n\nIn figure A, the sum of the overlapping line segments is: $7+2 \\sqrt{2}$;\n\nIn figure B, the sum is: $5+2 \\sqrt{2}$;\n\nIn figure C, the sum is: $5+3 \\sqrt{2}$;\n\nIn figure D, the sum is: $6+3 \\sqrt{2}$;\n\nSince $6+3 \\sqrt{2} > 7+2 \\sqrt{2} > 5+3 \\sqrt{2} > 5+2 \\sqrt{2}$,\n\nTherefore, the order is: B > C > A > D.\n\nHence, the correct choice is: $A$.\n\n【Key Insight】This problem tests the Pythagorean theorem and the perimeter of irregular shapes. The key to solving it is understanding that with a fixed total perimeter, the larger the sum of overlapping line segments, the smaller the perimeter." }, { "problem_id": 62, "question": "As shown in the figure, a rectangular iron block hanging on a spring scale is fully immersed in water. The spring scale is lifted uniformly upward until the iron block floats in the air (ignoring air resistance). The graph of the reading $F(\\text{N})$ of the spring scale versus time $t(\\text{s})$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch12-2024_06_15_7367a6691403d7f10b90g_0001_1.jpg", "batch12-2024_06_15_7367a6691403d7f10b90g_0001_2.jpg", "batch12-2024_06_15_7367a6691403d7f10b90g_0001_3.jpg", "batch12-2024_06_15_7367a6691403d7f10b90g_0001_4.jpg", "batch12-2024_06_15_7367a6691403d7f10b90g_0001_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: When the top surface of the iron block is still submerged in water, the reading on the spring scale remains unchanged;\n\nWhen the top surface of the iron block emerges from the water while the bottom surface remains submerged, as the iron block floats upward, the reading on the spring scale gradually increases;\n\nWhen the bottom surface of the iron block emerges from the water, the reading on the spring scale remains unchanged.\n\nTherefore, the correct choice is: A.\n\n[Key Point] This question examines the function's graph. The key to solving the problem lies in identifying the function's graph based on the decreasing buoyant force as more of the iron block emerges from the water." }, { "problem_id": 63, "question": "In a grid of unit squares where each vertex of a small square is called a lattice point, a move called a \"jump\" is defined as moving from one lattice point to another that is $\\sqrt{5}$ units away. For example, in a $4 \\times 4$ square grid (as shown in Figure 1), point A can reach points B, C, D, E, etc. with a single jump. Given a $10 \\times 10$ square grid (as shown in Figure 2), the minimum number of jumps required to move from a vertex M to its opposite vertex N, N, is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 7\nB. 8\nC. 9\nD. 10", "input_image": [ "batch12-2024_06_15_7cf6608e234119db2455g_0052_1.jpg", "batch12-2024_06_15_7cf6608e234119db2455g_0052_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in Figure 1, connect $AD$ and $DF$, then $AF=3\\sqrt{2}$. Therefore, the two transformations are equivalent to moving 3 units to the right and 3 units upwards.\n\nSince $MN=10\\sqrt{2}$, then $10\\sqrt{2} \\div 3\\sqrt{2}=\\frac{10}{3}$ (which is not an integer). Therefore, after performing the transformation in the direction of $A-D-F$ four times consecutively, it is equivalent to moving $4 \\div 2 \\times 3=6$ units to the right and $4 \\div 2 \\times 3=6$ units upwards. At this point, $M$ is located at point $G$ on the square grid as shown in Figure 2. Then, by performing the transformation four more times as shown in the figure, it can reach point $N$. Therefore, the minimum number of knight transformations required to move from vertex $M$ of the square to its opposite vertex $N$ is $4+4=8$ times.\n\nHence, the answer is B.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Insight】This question mainly examines the types of geometric transformations and the application of the Pythagorean theorem. When solving, note that under translation transformations, corresponding line segments are parallel and equal in length, and the line segment connecting two corresponding points is parallel (collinear) and equal in length to the given directed line segment. The key to solving the problem lies in identifying the pattern of the transformations." }, { "problem_id": 64, "question": "Figure 1 is the emblem of the 7th International Congress on Mathematical Education (ICME.7), where the main pattern evolves from a sequence of right-angled triangles depicted in Figure 2. In this sequence, $O A_{1}=A_{1} A_{2}=A_{2} A_{3}=\\cdots=A_{n-1} A_{n}=1$. If the product $O A_{5} \\cdot O A_{n}$ is an integer and $1 \\leq n \\leq 50$, then how many possible values of $n$ are there that satisfy this condition?\n\n\n\nICME. 7\n\nFigure 1\n\n\n\nFigure 2\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch12-2024_06_15_7d60d88ae8f737fcbaf3g_0013_1.jpg", "batch12-2024_06_15_7d60d88ae8f737fcbaf3g_0013_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "According to the problem, we can derive:\n\n$\\therefore O A_{2}=\\sqrt{O A_{1}^{2}+A_{1} A_{2}^{2}}=\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$,\n\n$O A_{3}=\\sqrt{O A_{2}{ }^{2}+A_{2} A_{3}{ }^{2}}=\\sqrt{(\\sqrt{2})^{2}+1^{2}}=\\sqrt{3}$,\n\nThen $O A_{4}=\\sqrt{4}$,\n\n$O A_{5}=\\sqrt{5}$\n\n$O A_{n}=\\sqrt{n}$,\n\n$\\because O A_{5} \\cdot O A_{n}$ is an integer, and $1 \\leq n \\leq 50$,\n\n$\\therefore \\sqrt{5} \\times \\sqrt{n}=\\sqrt{5 n}$,\n\n$\\therefore n=5$ or 20 or 45,\n\nThere are 3 values of $n$ that meet the condition.\n\nTherefore, the answer is: C.\n\n【Key Point】This problem examines the Pythagorean theorem, the pattern of graphical changes, and the simplification of square roots, among other related concepts. The key to solving the problem lies in identifying the pattern of numerical changes and applying the knowledge of the Pythagorean theorem to find the solution." }, { "problem_id": 65, "question": "As shown in Figure 1, point $P$ is a moving point on the side of the rectangle $A B C D$. Point $P$ starts from $A$ and moves along the four sides of the rectangle, eventually returning to $A$. Let the distance traveled by point $P$ be $x$, and the area of triangle $A B P$ be $y$. The graph in Figure 2 represents the relationship between $y$ and $x$. Determine the length of the diagonal $B D$ of the rectangle $A B C D$ ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\sqrt{34}$\nB. $\\sqrt{41}$\nC. 8\nD. 10", "input_image": [ "batch12-2024_06_15_7d60d88ae8f737fcbaf3g_0057_1.jpg", "batch12-2024_06_15_7d60d88ae8f737fcbaf3g_0057_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: When point $P$ moves along side $AB$, it does not form a triangle, and the area of $ABP$ is 0. From the function graph, we can see that $AB=5$.\n\nWhen point $P$ moves along side $CD$, the area of $ABP$ reaches its maximum of 10.\n\nAt this point, $S_{\\triangle ABP}=\\frac{1}{2} AB \\times BC=\\frac{1}{2} \\times 5 \\times BC=10$,\n\nSolving for $BC$, we get $BC=4$.\n\nTherefore, the diagonal $BD=\\sqrt{AB^{2}+BC^{2}}=\\sqrt{5^{2}+4^{2}}=\\sqrt{41}$.\n\nHence, the correct answer is: B.\n\n【Highlight】This question tests the Pythagorean theorem and the ability to interpret graphs. Understanding the graph and mastering the Pythagorean theorem are key to solving the problem." }, { "problem_id": 66, "question": "Pythagorean theorem is one of the greatest scientific discoveries in human history, and it was recorded in ancient Chinese mathematical book \"Zhou Bi Suan Jing\" much earlier. As shown in Figure 1, squares are constructed outside each side of a right-angled triangle, and the two smaller square papers are placed inside the largest square as shown in Figure 2. If the area of the shaded region in Figure 2 is 2, and $A B + A C = 8$, then the length of $B C$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $4 \\sqrt{2}$\nB. 6\nC. $\\frac{25}{4}$\nD. $\\frac{13}{2}$", "input_image": [ "batch12-2024_06_15_8372d55620e8759965f8g_0010_1.jpg", "batch12-2024_06_15_8372d55620e8759965f8g_0010_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in Figure 2:\n\n\n\nLet $\\mathrm{AC}=\\mathrm{a}, \\mathrm{AB}=\\mathrm{b}, \\mathrm{BC}=\\mathrm{c}$, then $\\mathrm{a}+\\mathrm{b}=8, \\mathrm{c}^{2}=\\mathrm{a}^{2}+\\mathrm{b}^{2}, \\mathrm{HG}=\\mathrm{c}-\\mathrm{b}, \\mathrm{DG}=\\mathrm{c}-\\mathrm{a}$,\n\nThe area of the shaded region $\\mathrm{S}=\\mathrm{HG} \\cdot \\mathrm{DG}=（ \\mathrm{c}-\\mathrm{b} ）(\\mathrm{c}-\\mathrm{a})=2$,\n\n$\\because(\\mathrm{a}+\\mathrm{b})^{2}=\\mathrm{a}^{2}+\\mathrm{b}^{2}+2 \\mathrm{ab}=64$,\n\n$\\therefore \\mathrm{ab}=32-\\frac{c^{2}}{2}$,\n\n$\\therefore \\mathrm{S}=\\mathrm{c}^{2}-\\mathrm{c}(\\mathrm{a}+\\mathrm{b})+\\mathrm{ab}=\\mathrm{c}^{2}-8 \\mathrm{c}+32-\\frac{c^{2}}{2}=2$,\n\nSolving gives $c_{1}=6, c_{2}=10$ (discarded).\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem tests the Pythagorean theorem, which states that if the lengths of the two legs of a right triangle are $\\mathrm{a}$ and $\\mathrm{b}$, and the hypotenuse is $\\mathrm{c}$, then $\\mathrm{a}^{2}+\\mathrm{b}^{2}=\\mathrm{c}^{2}$." }, { "problem_id": 67, "question": "As shown in the figure, a rectangle with a length of $2 \\mathrm{~cm}$ and a width of $1 \\mathrm{~cm}$ is translated horizontally from left to right along a straight line according to the four cases below (the dotted lines in the figure are all horizontal). Which of the following cases requires the shortest translation distance?\n\nA.\n\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch12-2024_06_15_8372d55620e8759965f8g_0069_1.jpg", "batch12-2024_06_15_8372d55620e8759965f8g_0069_2.jpg", "batch12-2024_06_15_8372d55620e8759965f8g_0069_3.jpg", "batch12-2024_06_15_8372d55620e8759965f8g_0069_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution:\n\nA. The distance of translation $= 2 + 1 = 3$,\n\nB. The distance of translation $= 2 + 1 = 3$,\n\nC. The distance of translation $= \\sqrt{2^{2} + 2^{2}} = 2\\sqrt{2}$,\n\nD. The distance of translation $= \\sqrt{1^{2} + 2^{2}} = \\sqrt{5}$,\n\nSince $\\sqrt{5} < 2\\sqrt{2} < 3$,\n\nTherefore, the option with the shortest required translation distance is $\\mathrm{D}$.\n\nHence, choose D.\n\n[Key Insight] This question examines the Pythagorean theorem, the comparison of real numbers, and the properties of translation: moving a figure along a straight line results in a new figure that is identical in shape and size to the original. Each point in the new figure corresponds to a point in the original figure that has been moved, and these corresponding points are connected by parallel and equal-length segments. The key to solving this problem lies in utilizing the properties of isosceles right triangles and the Pythagorean theorem to calculate the translation distances for each figure." }, { "problem_id": 68, "question": "Pythagoras' theorem was the first theorem to connect numbers with shapes, and its proof marked the beginning of synthetic geometry. Among the four diagrams below, which one does NOT prove the Pythagorean theorem?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0039_1.jpg", "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0039_2.jpg", "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0039_3.jpg", "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0039_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem statement:\n\nA. The sum of the areas of the four small figures equals the area of the large square, $2ab + a^{2} + b^{2} = (a + b)^{2}$. This can be used to prove the perfect square formula based on the figure, but it cannot prove the Pythagorean theorem. Therefore, this option is correct.\n\nB. The sum of the areas of two right-angled triangles with legs $a$ and $b$ and an isosceles right-angled triangle with leg $c$ equals the area of a trapezoid with upper base $a$, lower base $b$, and height $(a + b)$. \n\nThus, $\\frac{1}{2}ab + \\frac{1}{2}ab + \\frac{1}{2}c^{2} = \\frac{1}{2}(a + b)^{2}$,\n\nSimplifying, we get: $c^{2} = a^{2} + b^{2}$,\n\nThis proves the Pythagorean theorem, so this option is incorrect.\n\nC. The sum of the areas of four congruent right-angled triangles with legs $a$ and $b$ and a small square with side length $(b - a)$ equals the area of a square with side length $c$, $4 \\times \\frac{1}{2}ab + (b - a)^{2} = c^{2}$,\n\nSimplifying, we get: $c^{2} = a^{2} + b^{2}$,\n\nThis proves the Pythagorean theorem, so this option is incorrect.\n\nD. The sum of the areas of two congruent right-angled triangles with legs $a$ and $b$ and a square with side length $c$ equals the sum of the areas of two trapezoids: one with upper base $a$, lower base $(a + b)$, and height $a$, and the other with upper base $b$, lower base $(a + b)$, and height $b$,\n\nThus, $2 \\times \\frac{1}{2}ab + c^{2} = \\frac{1}{2}a(2a + b) + \\frac{1}{2}b(2b + a)$,\n\nSimplifying, we get: $c^{2} = a^{2} + b^{2}$,\n\nThis proves the Pythagorean theorem, so this option is incorrect.\n\nTherefore, the correct choice is: A.\n\n【Key Point】This problem examines the derivation of the Pythagorean theorem and the perfect square formula using area. Mastering the derivation of the Pythagorean theorem and the perfect square formula using area is crucial." }, { "problem_id": 69, "question": "In the diagram, the side length of the square is 12. Which of the following is correct ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0067_1.jpg", "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0067_2.jpg", "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0067_3.jpg", "batch12-2024_06_15_8bdff7ea2d5b14de969fg_0067_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Given that the side length of the square is 12, the area is calculated to be 144.\n\n$\\mathrm{A}$: The area of the square in the figure is $400-225=175$, so $\\mathrm{A}$ is incorrect and does not meet the requirement;\n\nB: The area of the square in the figure is $400+225=625$, so B is incorrect and does not meet the requirement;\n\nC: The area of the square in the figure is $256-112=144$, so C is correct and meets the requirement;\n\nD: The area of the square in the figure is $400-120=280$, so D is incorrect and does not meet the requirement;\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the Pythagorean theorem. Understanding that the area of the square is the square of one side of the right triangle is the key to solving the problem." }, { "problem_id": 70, "question": "As shown in the figure, there are two different rectangular sheets of paper, A and B. After cutting along the dotted lines, each is to be assembled into a square with the same area as the original. Then ( )\n\n\nSheet A\n\n\nSheet B\nA. Both A and B can.\nB. Neither A nor B can.\nC. A cannot, but B can.\nD. A can, but B cannot.", "input_image": [ "batch12-2024_06_15_90b4b38c4dedc2c47116g_0004_1.jpg", "batch12-2024_06_15_90b4b38c4dedc2c47116g_0004_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Question Analysis: Cut and assemble as shown in the figure below:\n\n\n\nA\n\n\n\nB\nTherefore, choose A\n\nKey Points: Cutting and assembling, area invariance, square roots" }, { "problem_id": 71, "question": "As shown in Figure 1, a ladder AB of length $5 \\mathrm{~m}$ leans against a wall. The bottom end B of the ladder is $3 \\mathrm{~m}$ away from the wall. If the top end A of the ladder slides down by $1 \\mathrm{~m}$ (as shown in Figure 2), the distance BD that the bottom end of the ladder slides horizontally is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $1 \\mathrm{~m}$\nB. Greater than $1 \\mathrm{~m}$\nC. Between $0 \\mathrm{~m}$ and $0.5 \\mathrm{~m}$\nD. Between\n$0.5 \\mathrm{~m}$ and $1 \\mathrm{~m}$", "input_image": [ "batch12-2024_06_15_918e80c1491a267ee8fag_0010_1.jpg", "batch12-2024_06_15_918e80c1491a267ee8fag_0010_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In Figure (1), $AB=5 \\mathrm{~m}$, $BC=3 \\mathrm{~m}$, and by the Pythagorean theorem, $AC=4 \\mathrm{~m}$. Since the ladder has slipped down by $1 \\mathrm{~m}$,\n\n$\\therefore AE=1 \\mathrm{~m}$, hence $EC=3 \\mathrm{~m}$,\n\n\n\n(1)\n\n\n\n(2)\n\nIn Figure (2), $EC=3 \\mathrm{~m}$, $ED=5 \\mathrm{~m}$, and by the Pythagorean theorem, $CD=4 \\mathrm{~m}$. Therefore, the ladder has slipped outward by $1 \\mathrm{~m}$. Thus, the correct choice is A.\n\nKey Point: This question tests the application of the Pythagorean theorem, requiring a thorough understanding and proficiency." }, { "problem_id": 72, "question": "As shown in Figure (a), in right triangle $A B C$, the sides $a, b, c$ satisfy the relationship $a^{2}+b^{2}=c^{2}$. Use this relationship to investigate the following problem: As shown in Figure (b), $O A B$ is an isosceles right triangle with a leg length of 1, where $\\angle O A B=90^{\\circ}$. Extend $O A$ to $B_{1}$ such that $A B_{1}=O A$. Construct an isosceles right triangle $O A_{1} B_{1}$ outside of $O A B$ with $O B_{1}$ as the base, and then extend $O A_{1}$ to $B_{2}$ such that $A_{1} B_{2}=O A_{1}$. Construct an isosceles right triangle $O A_{2} B_{2}$ outside of $O A_{1} B_{1}$ with $O B_{2}$ as the base, and so on, following this pattern to construct isosceles right triangles $O A_{n} B_{n}\\left(n \\geq 1, n\\right.$ is a positive integer). Then, the length of $A_{2} B_{2}$ and the area of $O A_{2021} B_{2021}$ are respectively ( )\n\n\n\n(a)\n\n\n\n(b)\nA. $2,2^{2020}$\nB. $4,2^{2021}$\nC. $2 \\sqrt{2}, 2^{2020}$\nD. $2,2^{2019}$", "input_image": [ "batch12-2024_06_15_b02da2a1aee0515fe6e4g_0015_1.jpg", "batch12-2024_06_15_b02da2a1aee0515fe6e4g_0015_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the problem statement, we have: $OA = AB = AB_1 = 1$, and $OB_1 = 2$.\n\nSince $OA_1B_1$ is an isosceles right triangle, and given that \"for a right triangle $ABC$ with sides $a$, $b$, and $c$, the relation $a^2 + b^2 = c^2$ holds\",\n\nTherefore, according to the problem statement: $OA_1 = A_1B_1 = \\sqrt{2}$,\n\nHence, $OB_2 = 2OA_1 = 2\\sqrt{2}$,\n\nThus, $OA_2 = A_2B_2 = (\\sqrt{2})^2 = 2$,\n\nAnd so on,\n\nTherefore, we can summarize that $OA_n = (\\sqrt{2})^n$,\n\nGiven that the area $S_{\\triangle OAB} = \\frac{1}{2} \\times 1 \\times 1 = \\frac{1}{2}$, $S_{\\triangle OA_1B_1} = \\frac{1}{2} \\times \\sqrt{2} \\times \\sqrt{2} = 1$, $S_{\\triangle OA_2B_2} = \\frac{1}{2} \\times 2 \\times 2 = 2$,\n\nWe can induce the general rule: $S_{OA_nB_n} = \\frac{1}{2} \\times (\\sqrt{2})^n \\times (\\sqrt{2})^n = 2^{n-1}$,\n\nTherefore, $S_{OA_{2021}B_{2021}} = 2^{2020}$,\n\nHence, the correct choice is: A.\n\n[Insight] This problem tests the properties of isosceles right triangles and the exploration of patterns in graphical changes. Understanding the problem statement and flexibly applying the properties of isosceles right triangles to induce general rules is key to solving the problem." }, { "problem_id": 73, "question": "As shown in the figure, in right-angled triangle paper $A B C$, $A B = 6$, $A C = 8$, and $D$ is the midpoint of the hypotenuse $B C$. In the first fold, the paper is folded so that point $A$ coincides with point $D$, and the crease intersects $A D$ at point $P_{1}$. Set the midpoint of $P_{1} D$ as $D_{1}$. In the second fold, the paper is folded again so that point $A$ coincides with point $D_{1}$, and the crease intersects $A D$ at point $P_{2}$. Set the midpoint of $P_{2} D_{1}$ as $D_{2}$. In the third fold, the paper is folded so that point $A$ coincides with point $D_{2}$, and the crease intersects $A D$ at point $P_{3}$. Then the length of $A P_{3}$ is ( )\n\n\n\nFirst Fold\n\n\n\nSecond Fold\nA. $\\frac{3^{4}}{2^{6} \\times 5}$\nB. $\\frac{3^{3} \\times 5}{2^{6}}$\nC. $\\frac{3^{3}}{2^{5} \\times 5}$\nD. $\\frac{3^{2} \\times 5}{2^{3}}$", "input_image": [ "batch12-2024_06_15_b02da2a1aee0515fe6e4g_0020_1.jpg", "batch12-2024_06_15_b02da2a1aee0515fe6e4g_0020_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since \\(\\angle \\mathrm{BAC} = 90^\\circ\\), \\(\\mathrm{AB} = 6\\), and \\(\\mathrm{AC} = 8\\),\n\\[\n\\therefore \\mathrm{BC} = \\sqrt{\\mathrm{AB}^2 + \\mathrm{AC}^2} = 10.\n\\]\n\nSince \\(\\mathrm{D}\\) is the midpoint of the hypotenuse \\(\\mathrm{BC}\\),\n\\[\n\\therefore \\mathrm{AD} = \\frac{1}{2} \\mathrm{BC} = 5.\n\\]\n\nFrom the folding, we know:\n\\[\n\\mathrm{AD}_1 = \\frac{3}{4} \\mathrm{AD}, \\quad \\mathrm{AP}_1 = \\frac{1}{2} \\mathrm{AD},\n\\]\n\\[\n\\therefore \\mathrm{AP}_1 = \\frac{2}{3} \\mathrm{AD}_1.\n\\]\n\n\\[\n\\mathrm{AD}_2 = \\frac{3}{4} \\mathrm{AD}_1 = \\frac{9}{16} \\mathrm{AD}, \\quad \\mathrm{AP}_2 = \\frac{1}{2} \\mathrm{AD}_1 = \\frac{3}{8} \\mathrm{AD},\n\\]\n\\[\n\\therefore \\mathrm{AP}_2 = \\frac{2}{3} \\mathrm{AD}_2.\n\\]\n\nIt follows that:\n\\[\n\\mathrm{AP}_3 = \\frac{2}{3} \\mathrm{AD}_3.\n\\]\n\n\\[\n\\mathrm{AD}_1 = \\frac{3}{4} \\mathrm{AD} = \\frac{3 \\times 5}{4},\n\\]\n\\[\n\\mathrm{AD}_2 = \\frac{3}{4} \\mathrm{AD}_1 = \\frac{9}{16} \\mathrm{AD} = \\frac{3^2 \\times 5}{2^4},\n\\]\n\\[\n\\therefore \\mathrm{AD}_3 = \\frac{3}{4} \\mathrm{AD}_2 = \\frac{3}{4} \\times \\frac{3^2 \\times 5}{2^4} = \\frac{3^3 \\times 5}{2^6},\n\\]\n\\[\n\\therefore \\mathrm{AP}_3 = \\frac{2}{3} \\mathrm{AD}_3 = \\frac{3^2 \\times 5}{2^5}.\n\\]\n\nTherefore, the correct answer is D.\n\n**Key Insight:** This problem primarily examines the properties of folding transformations, the Pythagorean theorem, and the properties of right triangles. The key is to flexibly apply the properties of folding transformations and correctly identify the implicit quantitative relationships in the problem. It also requires a high level of computational and problem-solving skills." }, { "problem_id": 74, "question": "As shown in Figure 1, there are two squares with side lengths $a$ and $b$. By cutting and pasting as shown in Figure 2, a square with side length $c$ can be obtained, and its area equals the sum of the areas of the two squares before cutting and pasting. This method can be used to derive or verify the Pythagorean theorem. Now, please point out which of the following understandings of the cutting and pasting process in Figure 2 is incorrect ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Cut (5) and paste (6)\nB. Cut (3) and paste (1)\nC. Cut (1) and paste (4)\nD. Cut (3) and paste (2)", "input_image": [ "batch12-2024_06_15_ed1911a30eded62729c0g_0053_1.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0053_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement:\n\nTo form a square, one should cut (5) and add (6), cut (1) and add (4), cut (3) and add (2).\n\nTherefore, the correct choice is B.\n\n[Key Point] This question primarily tests the design of shapes. The key to solving the problem lies in correctly understanding that the sum of the areas of the small squares equals the area of the assembled square." }, { "problem_id": 75, "question": "On a $4 \\times 4$ grid with unit side lengths (as shown in Figure A), a Cartesian coordinate system is established. In the first quadrant, draw the graphs of the inverse proportion functions $y = \\frac{16}{x}, y = \\frac{6}{x}, y = \\frac{4}{x}$, which respectively pass through one, two, and three grid points. On a $10 \\times 10$ grid with unit side lengths (as shown in Figure B), establish a Cartesian coordinate system in the first quadrant and draw the graphs of inverse proportion functions that pass through three or four grid points. How many graphs can be drawn at most?\n\n\n\nA\n\n\n\nB\nA. 12\nB. 13\nC. 25\nD. 50", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0011_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0011_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, to ensure that the $k$ in $y=\\frac{k}{x}$ is a composite number between $1$ and $100$, and that the two numbers obtained from factoring these composite numbers do not exceed $10$ when used as coordinates, through experimental methods, we find that for $k \\leq 40$, there are only $27$ such composite numbers. After factoring these $27$ numbers, the values of $k$ that meet the criteria are: $4, 6, 8, 9, 10, 12, 16, 18, 20, 24, 30, 36$." }, { "problem_id": 76, "question": "Among the following figures, the shaded areas have equal areas are ( )\n\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\nA. (1)(2)\nB. (2)(3)\nC. (3)(4)\nD. (1)(4)", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0035_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0035_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0035_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0035_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: \n\n(1) The intersection points of the line \\( y = -x + 2 \\) with the coordinate axes are \\( (2, 0) \\) and \\( (0, 2) \\). Therefore, the area \\( S_{\\text{冽}} = \\frac{1}{2} \\times 2 \\times 2 = 2 \\).\n\n(2) The intersection point of the line \\( y = 3x \\) with \\( x = 1 \\) is \\( (1, 3) \\). Thus, the area \\( S_{\\text{明的}} = \\frac{1}{2} \\times 1 \\times 3 = \\frac{3}{2} \\).\n\n(3) The function \\( y = |x| - 1 \\) intersects the coordinate axes at \\( (-1, 0) \\), \\( (1, 0) \\), and \\( (0, -1) \\). The shaded region forms an isosceles right triangle with an area \\( S = \\frac{1}{2} \\times 2 \\times 1 = 1 \\).\n\n(4) This function is a reciprocal function, and the area of the shaded region is \\( S = \\frac{1}{2} \\times |2| = 1 \\). Therefore, the correct answer is **C**.\n\n**Key Insight**: First, determine the intersection points of each graph's function with the coordinate axes based on their respective equations. Then, calculate the area of each shaded region to compare their sizes. This problem primarily tests the understanding of the graphs and properties of linear and reciprocal functions, as well as the application of the triangle area formula. Mastery of the graphs and properties of these three types of functions is essential to solve the problem." }, { "problem_id": 77, "question": "As shown in the figure, the base area of container A is $30 \\mathrm{~cm}^{2}$ and its height is $8 \\mathrm{~cm}$. The base area of container B is $\\mathrm{xcm}^{2}$. If container A is filled with water and then all the water in container A is poured into container B without spilling, the approximate graph of the relationship between the water level height $y$ (in $\\mathrm{cm}$) in container B and the base area $x$ (in $\\mathrm{cm}^{2}$) of container B is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0070_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0070_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0070_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0070_4.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0070_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem statement, we have\n\n$y=\\frac{30 \\times 8}{x}=\\frac{240}{x}$.\n\nWhen $x=40$, $y=6$,\nTherefore, the correct choice is C.\n\n【Key Insight】This question examines the graph of an inverse proportional function. The key to solving this problem lies in formulating the function's expression based on the given conditions." }, { "problem_id": 78, "question": "The area of triangle $ABC$ is 2, the length of side $BC$ is $x$, and the height from side $BC$ is $y$. The graphical representation of the relationship between $y$ and $x$ would be approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0089_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0089_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0089_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0089_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "According to the problem, we have:\n\n$$\n\\frac{1}{2} x y=2\n$$\n\n$\\therefore y=\\frac{4}{x}$\n\n$\\because x>0, y>0$\n\n$\\therefore$ The relationship between $y$ and $x$ can be roughly represented by the graph:\n\n\n\nTherefore, the correct answer is: A.\n\n【Key Point】This question examines the graph of an inverse proportional function. Understanding the properties of inverse proportional function graphs is crucial for solving the problem." }, { "problem_id": 79, "question": "Xiaoming learned the principle of lever balance in physics and discovered that the resistance times the resistance arm equals the force times the effort arm. Given that the resistance and resistance arm of a certain lever are $2400 \\mathrm{~N}$ and $1 \\mathrm{~m}$ respectively, the graph of the force $F$ (in units of $\\mathrm{N}$) with respect to the effort arm $l$ (in units of $\\mathrm{m}$) would approximately be\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0026_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0026_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0026_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0026_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the resistance multiplied by the resistance arm equals the effort multiplied by the effort arm, and given that the resistance and resistance arm are $2400 N$ and $1 m$ respectively,\n\nTherefore, the functional relationship of the effort $F$ (unit: $N$) with respect to the effort arm $l$ (unit: $m$) is: $2400 \\times 1 = F l$, which simplifies to $F = \\frac{2400}{l}$. This is an inverse proportion function, and option $A$ matches this description,\n\nHence, the answer is: $A$.\n\n[Insight] This question primarily tests the application of inverse proportion functions. Correctly understanding the problem and deriving the relationship is key to solving it." }, { "problem_id": 80, "question": "Given the relationship between voltage $U$, current $I$, and resistance $R$: $U = I R$. When one of these quantities is a constant, the graph of the relationship between the other two variables cannot be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0037_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0037_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0037_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0037_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since \\( U = IR \\),\n\nwhen \\( I \\) is a constant, \\( U \\) is directly proportional to \\( R \\), so option A is possible;\n\nsince \\( U = IR \\),\n\nthen \\( I = \\frac{U}{R} \\),\n\nwhen \\( U \\) is a constant, \\( I \\) is inversely proportional to \\( R \\), so option B is possible;\n\nsince \\( U = IR \\),\n\nthen \\( R = \\frac{U}{I} \\),\n\nwhen \\( U \\) is a constant, \\( R \\) is inversely proportional to \\( I \\), so option C is possible;\n\nsince \\( U = IR \\),\n\nthen \\( I = \\frac{U}{R} \\),\n\nwhen \\( R \\) is a constant, \\( I \\) is directly proportional to \\( U \\), so option D is not possible;\n\nTherefore, the correct answer is D.\n\n[Key Insight] This question mainly tests the determination of direct and inverse proportionality functions. Accurate analysis and judgment are key to solving the problem." }, { "problem_id": 81, "question": "As shown in the figure, the side $B C$ of triangle $A B C$ is $y$, the height from $A$ to side $B C$ is $x$, and the area of triangle $A B C$ is 3. The rough graph of the function $y$ in terms of $x$ is\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0050_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0050_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0050_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0050_4.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0050_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the area of triangle \\( ABC \\) is 3, we have \\( 3 = \\frac{1}{2} x \\cdot y \\),\n\nThus, \\( y = \\frac{6}{x} \\),\n\nTherefore, \\( y \\) and \\( x \\) are inversely proportional functions,\n\nHence, the graph of the function is a hyperbola,\n\nSince \\( x > 0 \\) and \\( y > 0 \\),\n\nThe graph of this inverse proportional function lies in the first quadrant.\n\nTherefore, the correct choice is: A\n\n[Key Insight] This question tests the understanding of the graph of an inverse proportional function. The key to solving the problem is to derive the functional relationship based on the given information, and to note the implicit condition that the side lengths are greater than 0." }, { "problem_id": 82, "question": "A new operation is defined as follows: $a \\oplus b = \\left\\{\\begin{array}{l}\\frac{a}{b} \\quad (b > 0) \\\\ -\\frac{a}{b} \\quad (b < 0)\\end{array}\\right.$. For example, $3 \\oplus 4 = \\frac{3}{4}$ and $3 \\oplus (-4) = -\\frac{3}{4}$. The graph of the function $y = 5 \\oplus x (x \\neq 0)$ is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0077_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0077_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0077_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0077_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem, we have:\n\n\\[ y = 5 \\oplus x = \\begin{cases} \n\\frac{5}{x} & (x > 0) \\\\\n-\\frac{5}{x} & (x < 0)\n\\end{cases} \\]\n\nWhen \\( x > 0 \\), the inverse proportional function \\( y = \\frac{5}{x} \\) lies in the first quadrant.\n\nWhen \\( x < 0 \\), the inverse proportional function \\( y = -\\frac{5}{x} \\) lies in the second quadrant.\n\nSince the graph of an inverse proportional function is a hyperbola, option B is correct.\n\nTherefore, the answer is: **B**.\n\n**Key Insight**: This question primarily tests the properties of inverse proportional functions, emphasizing that the graph of such a function is a hyperbola." }, { "problem_id": 83, "question": "For a certain task, it is known that each person completes the same amount of work per day, and that one person takes 10 days to finish. If $m$ people work together and finish in $n$ days, six pairs of numbers $(m, n)$ are selected and plotted on a coordinate system. Which of the following is correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_24af9853dd19771e4af2g_0033_1.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0033_2.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0033_3.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0033_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem statement, $\\frac{1}{10} m n=1$.\n\nTherefore, $m n=10$,\n\nHence, $n=\\frac{10}{m}$, where $m, n>0$ and are integers.\n\nThus, the correct choice is C.\n\n【Insight】This problem examines the practical application of inverse proportionality. The key to solving it lies in establishing a functional model based on the given conditions." }, { "problem_id": 84, "question": "As shown in the figure, given points $A$ and $B$ on the graph of the inverse proportion function $y = \\frac{k}{x} (k > 0, x > 0)$. Line segment $BC$ is parallel to the $y$-axis and intersects the $x$-axis at point $C$. Point $P$ starts from the origin $O$ and moves along the path $O \\rightarrow A \\rightarrow B \\rightarrow C$ at a constant speed until it reaches point $C$. Line segment $PQ$ is drawn perpendicular to the $x$-axis at point $Q$. Let the area of triangle $OPQ$ be $S$, and the time taken by point $P$ to move be $t$. The graph of $S$ as a function of $t$ would approximately look like ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_24af9853dd19771e4af2g_0047_1.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0047_2.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0047_3.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0047_4.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0047_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When point \\( P \\) moves along the segment \\( OA \\), let the equation of line \\( OA \\) be \\( y = a x \\). The coordinates of point \\( P \\) satisfy \\( y = a x \\), thus \\( S = \\frac{1}{2} a x^{2} \\) (where \\( a \\) is a positive constant and \\( x > 0 \\)). The graph is a part of a parabola.\n\nWhen point \\( P \\) moves along \\( AB \\), the area \\( S \\) of triangle \\( OPQ \\) remains constant at \\( S = \\frac{1}{2} k \\) (where \\( k > 0 \\)).\n\nWhen point \\( P \\) moves along \\( BC \\), let the total distance of the path \\( O \\rightarrow A \\rightarrow B \\rightarrow C \\) be \\( l \\), and the speed of point \\( P \\) be \\( b \\). Then, \\( S = \\frac{1}{2} OC \\times CP = \\frac{1}{2} OC \\times (l - b t) \\). Since \\( l \\), \\( OC \\), and \\( b \\) are constants, \\( S \\) is a linear function of \\( t \\).\n\nIn summary, the graph of \\( S \\) as a function of \\( t \\) is approximately as shown in option B.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This problem examines the comprehensive function and the function graph of moving point problems. The key to solving it is to determine the type of function based on the movement of the point, thereby determining its graph." }, { "problem_id": 85, "question": "If the graph of the linear function $y = ax + b$ and the reciprocal function $y = -\\frac{c}{x}$ have two intersection points in the second quadrant, and one of the intersection points has a x-coordinate of -1, then the graph of the quadratic function $y = ax^2 + bx + c$ could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_24af9853dd19771e4af2g_0099_1.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0099_2.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0099_3.jpg", "batch13-2024_06_15_24af9853dd19771e4af2g_0099_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the graph of the linear function \\( y = ax + b \\) and the graph of the inverse proportional function \\( y = -\\frac{c}{x} \\) have two intersection points in the second quadrant,\n\nthe graph of the inverse proportional function is located in the second and fourth quadrants, and the graph of the linear function passes through the first, second, and third quadrants.\n\nTherefore, \\( a > 0 \\), \\( b > 0 \\), and \\( -c < 0 \\),\n\nwhich implies \\( c > 0 \\).\n\nThus, the graph of the quadratic function opens upwards and intersects the positive half of the y-axis.\n\nSince the x-coordinate of one of the intersection points is -1,\n\nwhen \\( x = -1 \\), \\( -a + b = -c \\), that is, \\( a - b + c = 0 \\).\n\nTherefore, one of the intersection points of the quadratic function \\( y = ax^2 + bx + c \\) with the x-axis is \\( (-1, 0) \\).\n\nHence, the correct option is A.\n\nAnswer: A\n\n[Insight] This question mainly examines the properties of the graphs of linear and inverse proportional functions, as well as the properties of quadratic functions. The key to solving the problem lies in determining the direction in which the quadratic function opens and the positions of its intersections with the coordinate axes based on the given conditions." }, { "problem_id": 86, "question": "As shown in the figure, in the same Cartesian coordinate system, the graph of the linear function $y=ax+b(a \\neq 0, b \\neq 0)$ and the graph of the reciprocal function $y=\\frac{ab}{x}(a \\neq 0, b \\neq 0)$ can roughly be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_2fc1de4d32eb05e42d26g_0080_1.jpg", "batch13-2024_06_15_2fc1de4d32eb05e42d26g_0080_2.jpg", "batch13-2024_06_15_2fc1de4d32eb05e42d26g_0080_3.jpg", "batch13-2024_06_15_2fc1de4d32eb05e42d26g_0080_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: \n\nA. From the linear function graph, it is known that $a$ and $b$ have opposite signs, while from the inverse proportion function graph, $a$ and $b$ have the same sign. Therefore, this option is incorrect and does not match the intended meaning.\n\nB. From the linear function graph, it is known that $a$ and $b$ have the same sign, while from the inverse proportion function graph, $a$ and $b$ have opposite signs. Therefore, this option is incorrect and does not match the intended meaning.\n\nC. From the linear function graph, it is known that $a$ and $b$ have opposite signs, and from the inverse proportion function graph, $a$ and $b$ also have opposite signs. Therefore, this option is correct and matches the intended meaning.\n\nD. From the linear function graph, it is known that $a$ and $b$ have opposite signs, while from the inverse proportion function graph, $a$ and $b$ have the same sign. Therefore, this option is incorrect and does not match the intended meaning.\n\nHence, the correct choice is: C.\n\n[Key Insight] This question examines the relationship between the coefficients and the graphs of linear and inverse proportion functions. The key to solving it lies in determining the signs of $a$ and $b$ and understanding the relationship between the coefficients and the function graphs." }, { "problem_id": 87, "question": "In Figure (1), in quadrilateral $A B C D$, $A B \\parallel C D$, and $\\angle A = 90^\\circ$. Point $P$ starts from point $A$ and moves towards point $B$ at a speed of $1 \\text{~cm/s}$; simultaneously, point $Q$ starts from point $C$ and moves towards point $D$ at a speed of $2 \\text{~cm/s}$. The motion of the points stops when one of them reaches its destination. Let the time of motion be $x(\\text{~s})$, and the length of segment $P Q$ be $y(\\text{~cm})$. The relationship between $y$ and $x$ is shown in Figure (2), with the lowest point being $(2,3)$. Given the following statements: (1) $A B = 4 \\text{~cm}$, (2) $C D = 6 \\text{~cm}$, (3) $B C = 3\\sqrt{5} \\text{~cm}$, (4) When $x = \\frac{4}{3}$, $P Q \\parallel B C$. How many of these statements are correct?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch13-2024_06_15_3ef2fb8ea53e7839469cg_0003_1.jpg", "batch13-2024_06_15_3ef2fb8ea53e7839469cg_0003_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: From the graph passing through the point $(0,3 \\sqrt{5})$, we know that when $x=0$, $y=PQ=3 \\sqrt{5}$.\n\n$\\therefore AC=3 \\sqrt{5}$.\n\nFrom the lowest point on the graph $(2,3)$, we know that when $x=2$, $y=PQ=3$.\n\nAt this moment, $PQ \\perp CD$.\n\n$\\because AB / / CD, \\angle A=90^{\\circ}$,\n\n$\\therefore$ at this moment, quadrilateral $ADQP$ is a rectangle,\n\n$\\therefore AD=3$.\n\n$\\therefore$ according to the Pythagorean theorem, $CD=\\sqrt{AC^{2}-AD^{2}}=\\sqrt{(3 \\sqrt{5})^{2}-3^{2}}=6$, hence (2) is correct.\n\n$\\therefore$ point $Q$ moves for a maximum of $3 \\mathrm{~s}$.\n\nFrom the last point $(3,3 \\sqrt{2})$, we know that after moving for $3 \\mathrm{~s}$, $PQ=3 \\sqrt{2}$.\n\nAt this moment, $Q$ coincides with $D$, $AP=\\sqrt{PQ^{2}-AD^{2}}=\\sqrt{(3 \\sqrt{2})^{2}-3^{2}}=3$.\n\n$\\therefore$ the length of $AB$ cannot be determined.\n\n$\\therefore$ (1)(3)(4) cannot be judged as correct or incorrect.\n\nTherefore, the answer is: A.\n\n【Key Insight】This problem examines the function graph of a moving point, primarily utilizing the Pythagorean theorem, with a key focus on understanding the coordinates of three points on the graph." }, { "problem_id": 88, "question": "During an after-school class, Wang Lin used four wooden sticks, each $4 \\mathrm{~cm}$ in length, to create the square shown in Figure 1. He then fixed the side $B C$ and pushed it flat to form the figure in Figure 2, measuring $\\angle B=60^{\\circ}$. Which of the following conclusions is incorrect during this transformation?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. The length of $A B$ remains unchanged at $4 \\mathrm{~cm}$.\nB. The length of $A C$ decreases, reducing by $4(\\sqrt{2}-1) \\mathrm{cm}$.\nC. The length of $B D$ increases, adding $4(\\sqrt{3}-\\sqrt{2}) \\mathrm{cm}$.\nD. The area of $A B C D$ decreases, reducing by $8(\\sqrt{3}-1) \\mathrm{cm}^{2}$.", "input_image": [ "batch13-2024_06_15_3ef2fb8ea53e7839469cg_0015_1.jpg", "batch13-2024_06_15_3ef2fb8ea53e7839469cg_0015_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Connect $AC$ and $BD$,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n$\\because$ Quadrilateral $ABCD$ is a square,\n\n$\\therefore \\angle B=90^{\\circ}, AB=CB, \\quad AB=BC=4 \\mathrm{~cm}, \\quad AB^{2}+CB^{2}=AC^{2}$,\n\n$\\therefore AC^{2}=2 \\times 4^{2}$, the area of square $ABCD$ $=4^{2}=16\\left(\\mathrm{~cm}^{2}\\right)$,\n\n$\\therefore AC=BD=4 \\sqrt{2}(\\mathrm{~cm})$,\n\nIn rhombus $ABCD$, connect $AC$ and $BD$, and draw $AH \\perp BC$ at point $H$,\n\n$AB=CB=4, BO=DO, AO=CO$,\n\n$\\because \\angle B=60^{\\circ}$,\n\n$\\therefore ABC$ is an equilateral triangle,\n\n$\\therefore AC=AB=BC=4 \\mathrm{~cm}, AO=2, \\quad BH=CH=2$,\n\n$\\therefore BD=2 BO=2 \\sqrt{AB^{2}-AO^{2}}=4 \\sqrt{3}, AH=\\sqrt{AB^{2}-BH^{2}}=2 \\sqrt{ }$\n\nThe area of rhombus $ABCD$ $=4 \\times 2 \\sqrt{3}=8 \\sqrt{3}\\left(\\mathrm{~cm}^{2}\\right)$,\nTherefore, option A does not meet the requirement;\n\n$\\because 4 \\sqrt{2}-4=4(\\sqrt{2}-1)(\\mathrm{cm})$,\n\nTherefore, option B does not meet the requirement;\n\n$\\because 4 \\sqrt{3}-4 \\sqrt{2}=4(\\sqrt{3}-\\sqrt{2})(\\mathrm{cm})$,\n\nTherefore, option C does not meet the requirement;\n\n$\\because 16-8 \\sqrt{3}=8(2-\\sqrt{3})\\left(\\mathrm{cm}^{2}\\right)$\n\nTherefore, option D meets the requirement;\n\nHence, the answer is: D.\n\n【Insight】This question examines the properties of squares and rhombuses, as well as the Pythagorean theorem. Mastering the properties of squares and rhombuses is key to solving the problem." }, { "problem_id": 89, "question": "In rectangle $A B C D$ shown in Figure (1), $B C = x$, $C D = y$, and $y$ is inversely proportional to $x$ as shown in Figure (2). The isosceles right triangle $A E F$ has hypotenuse $E F$ passing through point $C$, and $M$ is the midpoint of $E F$. Which of the following conclusions is correct?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\nA. When $x = 3$, $E C < E M$\n\nB. When $y = 9$, $E C > E M$\n\nC. As $x$ increases, the value of $E C \\cdot C F$ increases\n\nD. As $x$ varies, the area of quadrilateral $B C D A$ remains constant", "input_image": [ "batch13-2024_06_15_408a01c96134b7549006g_0004_1.jpg", "batch13-2024_06_15_408a01c96134b7549006g_0004_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a rectangle,\n\n\\(\\therefore AB = CD\\).\n\nSince \\(\\triangle AEF\\) is an isosceles right triangle,\n\n\\(\\therefore \\angle E = \\angle F = 45^\\circ\\),\n\n\\(\\therefore \\triangle BEC\\) and \\(\\triangle CDF\\) are both isosceles right triangles.\n\nSince \\( BC = x \\), \\( CD = y \\),\n\n\\(\\therefore AE = x + y\\),\n\n\\(\\therefore EC = \\sqrt{2}x\\), \\( CF = \\sqrt{2}y \\), \\( EF = \\sqrt{2}(x + y) \\).\n\nSince \\( y \\) and \\( x \\) satisfy an inverse proportional function relationship, and the point \\( (3, 3) \\) lies on the graph of this function,\n\n\\(\\therefore xy = 9\\).\n\nA. When \\( x = 3 \\), \\( y = \\frac{9}{3} = 3 \\), \\( EC = 3\\sqrt{2} \\), \\( EF = 6\\sqrt{2} \\).\n\nAlso, since \\( M \\) is the midpoint of \\( EF \\),\n\n\\(\\therefore EM = 3\\sqrt{2} = EC\\), so option A does not meet the condition;\n\nB. When \\( y = 9 \\), \\( x = 1 \\),\n\n\\(\\therefore EC = \\sqrt{2} \\), \\( EM = \\frac{1}{2}EF = 5\\sqrt{2} \\),\n\n\\(\\therefore EC < EM \\), so option B does not meet the condition;\n\nC. Since \\( EC = \\sqrt{2}x \\), \\( CF = \\sqrt{2}y \\),\n\n\\(\\therefore EC \\cdot CF = 2xy = 2 \\times 9 = 18 \\), so option C does not meet the condition;\n\nD. Since the area of quadrilateral \\( BCDA \\) is \\( xy = 9 \\),\n\n\\(\\therefore\\) when \\( x \\) changes, the area of quadrilateral \\( BCDA \\) remains unchanged, so option D meets the condition.\n\nTherefore, the correct answer is: D.\n\n[Key Insight] This problem examines the properties of rectangles, the characteristics of points on the graph of an inverse proportional function, isosceles right triangles, and the area of rectangles. The key to solving the problem lies in using the method of elimination to analyze the correctness of each option one by one." }, { "problem_id": 90, "question": "If the expression for the inverse proportion function in the diagram is $y = \\frac{4}{x}$ for all, then the shaded area is 2 for ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\nA. Figure 1\nB. Figure 2\nC. Figure 3\nD. Figure 4", "input_image": [ "batch13-2024_06_15_4193f00ebf98327944fdg_0036_1.jpg", "batch13-2024_06_15_4193f00ebf98327944fdg_0036_2.jpg", "batch13-2024_06_15_4193f00ebf98327944fdg_0036_3.jpg", "batch13-2024_06_15_4193f00ebf98327944fdg_0036_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "In option A, the area of the shaded region \\(= xy = 4 \\neq 2\\), so option A does not meet the requirement;\n\nIn option B, the area of the shaded region is \\(\\frac{1}{2} xy = \\frac{1}{2} \\times 4 = 2\\), so option B meets the requirement;\n\nIn option C, the area of the shaded region is \\(2 \\times \\frac{1}{2} xy = 2 \\times \\frac{1}{2} \\times 4 = 4\\), so option C does not meet the requirement;\n\nIn option D, the area of the shaded region is \\(4 \\times \\frac{1}{2} xy = 4 \\times \\frac{1}{2} \\times 4 = 8\\), so option D does not meet the requirement;\n\nTherefore, the correct choice is: B.\n\n【Insight】This question examines the geometric meaning of \\(k\\) in the inverse proportionality function \\(y = \\frac{k}{x}\\), that is, the area of the rectangle formed by drawing perpendiculars from any point on the hyperbola to the \\(x\\)-axis and \\(y\\)-axis is \\(|k|\\). This is a frequently tested concept. It also reflects the idea of combining numbers with shapes. When solving such problems, it is essential to correctly understand the geometric meaning of \\(k\\). The question also tests the symmetry of the inverse proportionality function and the area of a triangle." }, { "problem_id": 91, "question": "The graph of the quadratic function $y = ax^2 + bx + c$ is shown in the figure. Its axis of symmetry is the straight line $x = \\frac{1}{2}$, and point A has coordinates $(1, 0)$. AB is perpendicular to the x-axis, and CB is connected. Which of the following statements is definitely true ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\n\nA. As shown in Figure (1), quadrilateral $ABCO$ is a rectangle.\n\nB. In the same Cartesian coordinate system, the graphs of the quadratic function $y = ax^2 + bx$, the linear function $y = ax + b$, and the reciprocal function $y = \\frac{ab}{x}$ are approximately as shown in Figure (2).\n\nC. In the same Cartesian coordinate system, the graphs of the quadratic function $y = -x(ax + b) + c$ and the reciprocal function $y = \\frac{b}{x}$ are approximately as shown in Figure (3).\n\nD. In the same Cartesian coordinate system, the graphs of the linear function $y = bx - ac$ and the reciprocal function $y = \\frac{2a + c}{x}$ are approximately as shown in Figure (4).\n\n##", "input_image": [ "batch13-2024_06_15_4193f00ebf98327944fdg_0069_1.jpg", "batch13-2024_06_15_4193f00ebf98327944fdg_0069_2.jpg", "batch13-2024_06_15_4193f00ebf98327944fdg_0069_3.jpg", "batch13-2024_06_15_4193f00ebf98327944fdg_0069_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "From Figure (1), it is known that \\( a > 0 \\), \\( c < 0 \\), and \\( -\\frac{b}{2a} > 0 \\), so \\( b < 0 \\).\n\nA. Since the axis of symmetry of the quadratic function \\( y = ax^2 + bx + c \\) is the line \\( x = \\frac{1}{2} \\), the points \\( O(0,0) \\) and \\( A(1,0) \\) are symmetric about the axis of symmetry. Since \\( AB \\) is perpendicular to the \\( x \\)-axis, points \\( B \\) and \\( C \\) are also symmetric about the axis of symmetry. Therefore, quadrilateral \\( ABCO \\) is a rectangle, making option A correct.\n\nB. Since \\( a > 0 \\) and \\( b < 0 \\), the graph of the linear function \\( y = ax + b \\) passes through the first, third, and fourth quadrants, making option B incorrect.\n\nC. Since \\( c < 0 \\), the graph of the quadratic function \\( y = -x(ax + b) + c = -ax^2 - bx + c \\) does not pass through the origin, making option C incorrect.\n\nD. Since \\( b < 0 \\) and \\( -ac > 0 \\), the graph of the linear function \\( y = bx - ac \\) passes through the first, second, and fourth quadrants, making option D incorrect.\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This question tests the understanding of the graphs and properties of quadratic, inverse proportional, and linear functions. The key to solving the problem lies in mastering the properties of these functions and applying them flexibly." }, { "problem_id": 92, "question": "As shown in Figure 1, in triangle $O A B$, $\\angle A O B = 45^\\circ$, the coordinates of point $B$ are $(3 \\sqrt{2}, 0)$, and point $A$ is on the graph of the inverse proportion function $y = \\frac{2}{x}$. Let the area of triangle $O A B$ be $S_1$. In Figure 2, in triangle $A B C$, $A B = A C$, and $B C$ is on the $x$-axis. Also, $O B : B C = 1 : 2$, and point $A$ is on the graph of the inverse proportion function $y = \\frac{3}{x}$. Let the area of triangle $A B C$ be $S_2$. Then the value of $S_1 + S_2$ is $($ $)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{9}{2}$\nB. 5\nC. $\\frac{5}{2}$\nD. $\\frac{15}{4}$", "input_image": [ "batch13-2024_06_15_41e6ed55a219b8916817g_0004_1.jpg", "batch13-2024_06_15_41e6ed55a219b8916817g_0004_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Since in Figure 1, within triangle $OAB$, $\\angle AOB = 45^\\circ$,\n\nTherefore, setting point $A(c, c)$ where $c > 0$ and substituting into $y = \\frac{2}{x}$ yields: $c^2 = 2$,\n\nSolving gives: $c = \\sqrt{2}$,\n\nThus, $S_1 = \\frac{1}{2} \\times OB \\times y_A = \\frac{1}{2} \\times 3\\sqrt{2} \\times \\sqrt{2} = 3$;\n\nSince in Figure 2, within triangle $ABC$, $AB = AC$, $BC$ lies on the x-axis, and $OB : BC = 1 : 2$, with point $A$ on the graph of the inverse proportional function $y = \\frac{3}{x}$,\n\nTherefore, point $A$ lies on the perpendicular bisector of segment $BC$,\n\nThus, setting point $B(a, 0)$, then point $C(3a, 0)$, and setting point $A(2a, b)$,\n\nTherefore, $b = \\frac{3}{2a}$,\n\nThus, $S_2 = \\frac{1}{2} \\times BC \\times y_A = \\frac{1}{2} \\times 2a \\times \\frac{3}{2a} = \\frac{3}{2}$,\n\nTherefore, $S_1 + S_2 = 3 + \\frac{3}{2} = \\frac{9}{2}$,\n\nHence, the answer is: A.\n\n[Key Insight] This problem primarily examines the application of inverse proportional functions, and mastering the method of finding coordinates using the inverse proportional function's equation is crucial for solving it." }, { "problem_id": 93, "question": "As shown in Figure 1, a desk lamp with adjustable brightness can change its brightness by regulating the total resistance to control the change in current. As shown in Figure 2, the current $I(\\text{A})$ of the lamp is inversely proportional to the resistance $R(\\Omega)$, and the graph passes through point $P(880, 0.25)$. According to the graph, which of the following statements is correct ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. When $R<0.25$, $I<880$\nB. The function relationship between $I$ and $R$ is $I=\\frac{200}{R}(R>0)$\nC. When $R>1000$, $I>0.22$\nD. When $880 0 \\)).\n\nSince the graph passes through the point \\( P(880, 0.25) \\),\n\n\\[\n\\frac{U}{880} = 0.25,\n\\]\n\n\\[\n\\therefore U = 220.\n\\]\n\nThus, the functional relationship between \\( I \\) and \\( R \\) is \\( I = \\frac{220}{R} \\) (where \\( R > 0 \\)), so option **B** does not fit the context.\n\nWhen \\( R = 0.25 \\), \\( I = 880 \\); when \\( R = 1000 \\), \\( I = 0.22 \\).\n\nSince the inverse proportionality function \\( I = \\frac{U}{R} \\) (where \\( R > 0 \\)) implies that \\( I \\) decreases as \\( R \\) increases,\nwhen \\( R < 0.25 \\), \\( I > 880 \\); when \\( R > 1000 \\), \\( I < 0.22 \\). Therefore, options **A** and **C** do not fit the context.\n\nWhen \\( R = 880 \\), \\( I = 0.25 \\); when \\( R = 1000 \\), \\( I = 0.22 \\).\n\nThus, when \\( 880 < R < 1000 \\), the range of \\( I \\) is \\( 0.22 < I < 0.25 \\), so **D** fits the context.\n\nThe correct choice is: **D**.\n\n[Key Insight] This problem primarily examines the application of inverse proportionality functions. Determining the inverse proportionality function's expression using the method of undetermined coefficients is key to solving the problem." }, { "problem_id": 94, "question": "A new operation is defined as: $a ※ b = \\left\\{\\begin{array}{l}a - 1 (a \\leq b) \\\\ -\\frac{a}{b} (a > b \\text{ and } b \\neq 0)\\end{array}\\right.$, then the graph of the function $y = 3 ※ x$ is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_58474443df6e195863d8g_0094_1.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0094_2.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0094_3.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0094_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "From the problem statement, we have \\( y = 3 \\oplus x = \\left\\{\\begin{array}{l}2 \\quad (x \\geq 3) \\\\ -\\frac{3}{x} \\quad (x < 3 \\text{ and } x \\neq 0)\\end{array}\\right. \\)\n\nWhen \\( x \\geq 3 \\), \\( y = 2 \\);\n\nWhen \\( x < 3 \\) and \\( x \\neq 0 \\), \\( y = -\\frac{3}{x} \\).\n\nThe graph is as shown:\n\n\n\nTherefore, the correct choice is: B\n\n【Key Insight】This problem primarily examines the properties of the graphs of inverse proportional functions and linear functions. Mastery of their properties is essential to solve such problems flexibly." }, { "problem_id": 95, "question": "The approximate graph of the quadratic function $y = -kx^2 - k^2$ and the reciprocal function $y = \\frac{k}{x} (k \\neq 0)$ in the same Cartesian coordinate system could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_5d227d10b18c15d49d57g_0008_1.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0008_2.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0008_3.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0008_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: When \\( k > 0 \\), the graph of the quadratic function \\( y = -k x^{2} - k^{2} \\) opens downward and intersects the negative half of the y-axis. At this time, the graph of the inverse proportional function \\( y = \\frac{k}{x} \\) (where \\( k \\neq 0 \\)) is located in the first and third quadrants. Therefore, option A is correct, and option D is incorrect.\n\nWhen \\( k < 0 \\), the graph of the quadratic function \\( y = -k x^{2} - k^{2} \\) opens upward and intersects the negative half of the y-axis. At this time, the graph of the inverse proportional function \\( y = \\frac{k}{x} \\) (where \\( k \\neq 0 \\)) is located in the second and fourth quadrants. Therefore, options B and C are incorrect.\n\nHence, the correct choice is A.\n\n[Key Insight] This question examines the properties of the graphs of quadratic and inverse proportional functions. Categorizing the cases based on the value of \\( k \\) is crucial for solving this problem." }, { "problem_id": 96, "question": "As shown in the figure, in rectangle $A B C D$, $A B = 3$ and $B C = 4$. Point $P$ moves along side $B C$, and line segment $D P$ is drawn. Through point $A$, line $A E$ is drawn perpendicular to $D P$, with point $E$ as the foot of the perpendicular. Let $D P = x$ and $A E = y$. Which of the following graphs approximately represents the relationship between $y$ and $x$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_6232eddd3fe50c545eddg_0069_1.jpg", "batch13-2024_06_15_6232eddd3fe50c545eddg_0069_2.jpg", "batch13-2024_06_15_6232eddd3fe50c545eddg_0069_3.jpg", "batch13-2024_06_15_6232eddd3fe50c545eddg_0069_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the area of triangle \\( \\triangle APD \\) can be expressed as:\n\\[\nS \\triangle APD = \\frac{1}{2} \\times PD \\times AE = \\frac{1}{2} \\times AD \\times AB,\n\\]\nwe have:\n\\[\nxy = 3 \\times 4.\n\\]\nThus:\n\\[\nxy = 12, \\quad y = \\frac{12}{x},\n\\]\nwhich is an inverse proportionality function. Therefore, the correct choice should be between options \\( C \\) and \\( D \\). Since the minimum value of \\( x \\) should not be less than \\( CD \\) and the maximum value should not exceed \\( BD \\), we have:\n\\[\n3 \\leq x \\leq 5.\n\\]\nHence, the correct choice is: **C**.\n\n**Key Insight**: The problem requires using the different expressions for the area of \\( \\triangle APD \\) to determine the functional relationship between \\( y \\) and \\( x \\)." }, { "problem_id": 97, "question": "In the Cartesian coordinate system $\\mathrm{xoy}$, point $\\mathrm{P}$ is a moving point on the graph of the inverse proportion function $\\mathrm{y}=\\frac{1}{x}(\\mathrm{x}>0)$. Point $\\mathrm{A}$ is on the $\\mathrm{x}$-axis, and $\\mathrm{PO}=\\mathrm{PA}$. Line segment $\\mathrm{AB}$ is the altitude of $\\triangle \\mathrm{PAO}$ on side $\\mathrm{OP}$. Given $\\mathrm{OA}=\\mathrm{m}$ and $\\mathrm{AB}=\\mathrm{n}$, which of the following graphs approximately represents the relationship between $\\mathrm{n}$ and $\\mathrm{m}$?\n\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_6232eddd3fe50c545eddg_0072_1.jpg", "batch13-2024_06_15_6232eddd3fe50c545eddg_0072_2.jpg", "batch13-2024_06_15_6232eddd3fe50c545eddg_0072_3.jpg", "batch13-2024_06_15_6232eddd3fe50c545eddg_0072_4.jpg", "batch13-2024_06_15_6232eddd3fe50c545eddg_0072_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Question Analysis: According to the problem statement, $\\mathrm{m}$ and $\\mathrm{h}$ have an inverse proportional relationship, which means only option $\\mathrm{D}$ satisfies the condition.\n\nKey Point: Function Graphs" }, { "problem_id": 98, "question": "By cutting along a straight line on a quadrilateral piece of paper, the two resulting figures have equal interior angle sums. Which of the following options is correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_72def6568d0ee26f353eg_0031_1.jpg", "batch13-2024_06_15_72def6568d0ee26f353eg_0031_2.jpg", "batch13-2024_06_15_72def6568d0ee26f353eg_0031_3.jpg", "batch13-2024_06_15_72def6568d0ee26f353eg_0031_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nOption $\\mathrm{A}$, after cutting, the two shapes are a triangle and a quadrilateral, and the sum of their internal angles is not equal. Therefore, option $\\mathrm{A}$ is incorrect and does not meet the requirement;\n\nOption $\\mathrm{B}$, after cutting, the two shapes are a triangle and a quadrilateral, and the sum of their internal angles is not equal. Therefore, option $\\mathrm{B}$ is incorrect and does not meet the requirement;\n\nOption $\\mathrm{C}$, after cutting, the two shapes are a triangle and a quadrilateral, and the sum of their internal angles is not equal. Therefore, option $\\mathrm{C}$ is incorrect and does not meet the requirement;\n\nOption $\\mathrm{D}$, after cutting, both shapes are quadrilaterals, and the sum of their internal angles is equal. Therefore, option D is correct and meets the requirement; hence, the answer is: D.\n\n[Key Insight] This question mainly tests the theorem of the sum of internal angles of polygons. Understanding and mastering the theorem and calculation method of the sum of internal angles of polygons is the key to solving the problem." }, { "problem_id": 99, "question": "If the graphs of the functions $y = \\frac{k}{x}$ and $y = ax^2 + bx + c$ are as shown in the figure, then the approximate graph of the function $y = kx - b$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_84512915d0d2c440e574g_0024_1.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0024_2.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0024_3.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0024_4.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0024_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Based on the graph of the inverse proportional function located in the second and fourth quadrants, we know that $\\mathrm{k}<0$.\n\nFrom the graph of the quadratic function, it is confirmed that $\\mathrm{a}>0$ and $\\mathrm{b}<0$.\n\nTherefore, the approximate graph of the function $\\mathrm{y}=\\mathrm{kx}+\\mathrm{b}$ passes through the second, third, and fourth quadrants.\n\nHence, the correct choice is: B.\n\n[Key Insight] This question tests the understanding of the graphs of inverse proportional functions, quadratic functions, and linear functions. The key to solving it lies in mastering the graphs of these functions." }, { "problem_id": 100, "question": "Given that the point $(1,1)$ lies on the graph of the inverse proportion function $y = \\frac{k}{x}$ (where $k$ is a constant, $k \\neq 0$), which of the following is a rough sketch of the graph of this inverse proportion function?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_84512915d0d2c440e574g_0083_1.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0083_2.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0083_3.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0083_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since this function is an inverse proportionality function,\n\nTherefore, the graph of this function is a hyperbola,\n\nHence, options A and B are incorrect;\n\nSince the point $(1,1)$ lies on the graph of the inverse proportionality function $y=\\frac{k}{x}$ (where $k$ is a constant, $k \\neq 0$),\n\nTherefore, $k=1 \\times 1=1$,\n\nThus, the graph of this inverse proportionality function lies in the first and third quadrants,\n\nTherefore, option C is correct.\n\nThe correct choice is C.\n\n[Insight] This question examines the image and properties of inverse proportionality functions. Mastering the method of undetermined coefficients to find the inverse proportionality function and understanding the properties of inverse proportionality functions are key to solving the problem." }, { "problem_id": 101, "question": "To standardize market order and ensure public welfare projects, regulatory authorities continuously monitor the price of a certain commodity. The price of the commodity, denoted as $y_{1}$ (yuan per piece), changes with time $t$ (days) as shown in the figure. Let $y_{2}$ (yuan per piece) represent the average price of the commodity from the first day to the $t$th day. The graph of $y_{2}$ against $t$ would approximately look like ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0046_1.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0046_2.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0046_3.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0046_4.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0046_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Statistics", "image_relavance": "1", "analysis": "Solution: \n\nGiven the problem conditions, we have:\n\n- For \\(1 \\leq t \\leq 6\\), \\(y_{1} = 2t + 3\\),\n- For \\(6 < t \\leq 25\\), \\(y_{1} = 15\\),\n- For \\(25 < t \\leq 30\\), \\(y_{1} = -2t + 65\\).\n\nTherefore:\n\n- For \\(1 \\leq t \\leq 6\\), \\(y_{2} = \\frac{(5 + 2t + 3)t}{2} \\div t = t + 4\\),\n- For \\(6 < t \\leq 25\\), \\(y_{2} = \\left[\\frac{(5 + 15) \\times 6}{2} + 15(t - 6)\\right] \\div t = 15 - \\frac{30}{t}\\),\n- For \\(25 < t \\leq 30\\), \\(y_{2} = \\left[\\frac{(5 + 15) \\times 6}{2} + 15 \\times (25 - 6) + \\frac{[13 + (-2t + 65)] \\times (t - 25)}{2}\\right] \\div t = -t - \\frac{630}{t} + 64\\).\n\nThus, when \\(t = 30\\), \\(y_{2} = 13\\), and the only option that fits is A.\n\nTherefore, the correct choice is A.\n\n[Key Insight] This problem primarily tests the understanding of function graphs and their analytical expressions. Mastering the method of undetermined coefficients and the significance of coordinates on a function graph is crucial for solving such problems." }, { "problem_id": 102, "question": "In recent times, balloons in various shapes are very popular among children. As shown in Figure 1, this is a \"Bing Dwen Dwen\" shaped balloon, which is filled with a certain mass of gas. When the temperature remains constant, the pressure of the gas inside the balloon, $p(\\text{kPa})$, is inversely proportional to the volume of the balloon, $V(\\text{m}^3)$. The graph of this relationship is shown in Figure 2. If the pressure inside the balloon exceeds $200 \\text{kPa}$, the balloon will explode. For safety reasons, the range of the balloon's volume, $V$, should be *\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $V > 0.48 \\text{ m}^3$\nB. $V < 0.48 \\text{ m}^3$\nC. $V \\geq 0.48 \\text{ m}^3$\nD. $V \\leq 0.48 \\text{ m}^3$", "input_image": [ "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0075_1.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0075_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Let the functional relationship between \\( P \\) and \\( V \\) be \\( P = \\frac{k}{V} \\).\n\nGiven that \\( \\frac{k}{2} = 48 \\),\n\nwe solve for \\( k \\) to get \\( k = 96 \\).\n\nTherefore, the functional relationship is \\( P = \\frac{96}{V} \\).\n\nWhen \\( P > 200 \\) kPa, the balloon will explode.\n\nThus, \\( P \\leq 200 \\), which means \\( \\frac{96}{V} \\leq 200 \\).\n\nSolving this inequality gives \\( V \\geq 0.48 \\) m³.\n\nHence, the correct choice is C.\n\n【Key Insight】This problem examines the practical application of inverse proportionality functions, emphasizing the establishment of the functional relationship and its use to solve the given problem." }, { "problem_id": 103, "question": "A school uses a drug vapor disinfection method to disinfect classrooms. The measured data of $y$ and $x$ at different times are shown in the table:\n\n| Time $x$ (minutes) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 16 | 20 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Drug Concentration $y$ (milligrams) | 0 | 1.5 | 3 | 4.5 | 6 | 4.8 | 4 | 3 | 2.4 |\n\nWhich of the following graphs could represent the functional relationship between $y$ and $x$? (A)\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0091_1.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0091_2.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0091_3.jpg", "batch13-2024_06_15_8ff8ea6aee1f816a0b95g_0091_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Statistics", "image_relavance": "0", "analysis": "Solution: From the data in the table, we can derive the following:\n\nFor the interval \\(0 \\leq x < 8\\), the data shows proportional growth, indicating a direct proportionality relationship. We can express this relationship with the equation: \\(y = kx\\).\n\nSubstituting the point \\((2, 1.5)\\) into the equation gives: \\(1.5 = 2k\\),\n\nSolving for \\(k\\) yields: \\(k = \\frac{3}{4}\\),\n\nThus, the function equation is: \\(y = \\frac{3}{4}x\\) for \\(0 \\leq x < 8\\).\n\nFor the interval \\(8 \\leq x\\), the data shows inverse proportionality, indicating an inverse proportionality relationship. We can express this relationship with the equation: \\(y = \\frac{a}{x}\\).\n\nSubstituting the point \\((12, 4)\\) into the equation gives: \\(a = 48\\),\n\nThus, the function equation is: \\(y = \\frac{48}{x}\\) for \\(x \\geq 8\\).\n\nTherefore, the correct graph of the function is D.\n\nThe correct choice is: D.\n\n[Insight] This problem primarily tests the application of direct and inverse proportionality functions. Correctly deriving the function equations is key to solving the problem." }, { "problem_id": 104, "question": "Given a parabola $y = a x^2 + b x + 3$ with a symmetry axis of $y$-axis, it intersects the $x$-axis at two points with $x$-coordinates $x_1$ and $x_2$. If the point $(x_1, x_2)$ lies on the graph of the inverse proportional function $y = \\frac{-3}{x}$, the number of lattice points (points with integer coordinates) enclosed by the parabola and the $x$-axis (excluding the boundary) is $k$. Then, the graph of the inverse proportional function $y = \\frac{k}{x} (x > 0)$ is:\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_c2b339e9e67c87712768g_0081_1.jpg", "batch13-2024_06_15_c2b339e9e67c87712768g_0081_2.jpg", "batch13-2024_06_15_c2b339e9e67c87712768g_0081_3.jpg", "batch13-2024_06_15_c2b339e9e67c87712768g_0081_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the parabola \\( y = ax^2 + bx + 3 \\) has its axis of symmetry along the \\( y \\)-axis and intersects the \\( x \\)-axis at two points with coordinates \\( x_1 \\) and \\( x_2 \\),\n\n\\[\n\\therefore b = 0, \\quad x_1 = -x_2.\n\\]\n\nGiven that the point \\( (x_1, x_2) \\) lies on the graph of the inverse proportional function \\( y = \\frac{-3}{x} \\),\n\n\\[\n\\therefore x_2 = \\frac{-3}{x_1},\n\\]\n\nwhich implies,\n\n\\[\n- x_1 = \\frac{-3}{x_1}.\n\\]\n\nSolving this, we find,\n\n\\[\nx_1 = \\pm \\sqrt{3}.\n\\]\n\nAssuming \\( x_1 < x_2 \\), then \\( x_1 = -\\sqrt{3} \\) and \\( x_2 = \\sqrt{3} \\).\n\n\\[\n\\therefore \\text{The points of intersection of the parabola with the } x\\text{-axis are } (-\\sqrt{3}, 0) \\text{ and } (\\sqrt{3}, 0).\n\\]\n\nSubstituting into the parabola equation,\n\n\\[\n0 = a \\times (\\sqrt{3})^2 + 3,\n\\]\n\nwe find,\n\n\\[\na = -1.\n\\]\n\nThus, the equation of the parabola is,\n\n\\[\ny = -x^2 + 3.\n\\]\n\nThe integer points within the closed region bounded by the parabola and the \\( x \\)-axis (excluding the boundary) are:\n\n\\[\n(-1, 1), \\quad (0, 1), \\quad (0, 2), \\quad (1, 1).\n\\]\n\n\\[\n\\therefore \\text{There are 4 integer points within the region.}\n\\]\n\n\\[\n\\therefore k = 4.\n\\]\n\n\\[\n\\therefore \\text{The graph of the inverse proportional function } y = \\frac{k}{x} \\text{ for } x > 0 \\text{ is } y = \\frac{4}{x} \\text{ for } x > 0.\n\\]\n\nTherefore, the correct choice is D.\n\n**Key Insight:** This problem examines the coordinate characteristics of points on an inverse proportional function's graph, the graph of the inverse proportional function itself, the properties of quadratic functions, and the intersection points of a parabola with the \\( x \\)-axis. The key to solving it lies in understanding the given conditions and determining the value of \\( k \\) accordingly, using the properties of inverse proportional functions to find the solution." }, { "problem_id": 105, "question": "Xiao Ming uses some identical $A B C$ pieces of paper. It is known that six $A B C$ pieces of paper can be spliced ​​as shown in Figure 1 to obtain a regular hexagonal outline. If $n$ $A B C$ pieces of paper are spliced ​​as shown in Figure 2, then the outer contour pattern can be obtained ()\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Regular dodecagon\nB. Regular decagon\nC. Regular nonagon\nD. Regular octagon", "input_image": [ "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0027_1.jpg", "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0027_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Each interior angle of a regular hexagon is: $\\frac{1}{6} \\times(6-2) \\times 180^{\\circ}=120^{\\circ}$,\n\nSince $\\angle ABC=80^{\\circ}$,\n\nTherefore, $\\angle ACB=120^{\\circ}-80^{\\circ}=40^{\\circ}$,\n\nThus, $\\angle CAB=180^{\\circ}-\\angle ABC-\\angle ACB=60^{\\circ}$,\n\nAccording to the problem, the new pattern is a regular polygon,\n\nTherefore, one interior angle of the new polygon is $\\angle ABC+\\angle CAB=140^{\\circ}$,\n\nLet the number of sides of the new polygon be $n$,\n\n$(n-2) \\times 180^{\\circ}=140^{\\circ} n$,\n\n## Solving gives $\\boldsymbol{n}=9$.\n\nHence, the answer is: C.\n\n【Key Insight】This problem tests the knowledge that the sum of the interior angles of a triangle is $180^{\\circ}$, the formula for the interior angle of a regular polygon, and the formula for the sum of the interior angles of a polygon. Understanding the problem and calculating one interior angle of the regular polygon is the key to solving the problem." }, { "problem_id": 106, "question": "As shown in Figure 1, make the reverse extension line $P A$ of the bisector of $\\angle B P C$. Now, we need to make regular polygons with $\\angle A P B, \\angle A P C, and \\angle B P C$ as the inner angles, and the side lengths are all 1. Fill the three regular polygons with different patterns to form a pattern. For example, if $\\angle B P C$ is used as the inner angle, a square with a side length of 1 can be made. At this time, $\\angle B P C=90^{\\circ}$, and $\\frac{90}{2}=45^{\\circ}$ is $\\frac{1}{8}$ of $360^{\\circ}$ (the sum of the outer angles of the polygon). In this way, two regular octagons with side lengths of 1 can be made. After filling the pattern, a pattern that meets the requirements can be obtained. As shown in Figure 2, the perimeter of the outer contour of the pattern in Figure 2 is 14. Among all the patterns that meet the requirements, choose the one with the largest outer contour circumference as the monogram. The outer contour circumference of the monogram is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 14\nB. 16\nC. 19\nD. 21", "input_image": [ "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0040_1.jpg", "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0040_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let $\\angle BPC = 2x$,\n\nTherefore, the number of sides of the regular polygon with $\\angle BPC$ as an internal angle is: $\\frac{360}{180-2x} = \\frac{180}{90-x}$,\n\nThe number of sides of the regular polygon with $\\angle APB$ as an internal angle is: $\\frac{360}{x}$,\n\nThus, the perimeter of the outer contour of the pattern is:\n\n$\\frac{180}{90-x} - 2 + \\frac{360}{x} - 2 + \\frac{360}{x} - 2 = \\frac{180}{90-x} + \\frac{720}{x} - 6$,\n\nAccording to the problem, the value of $2x$ can only be $60^\\circ, 90^\\circ, 120^\\circ, 144^\\circ$,\n\nThe smaller $x$ is, the larger the perimeter,\n\nTherefore, when $x = 30$, the perimeter is maximized, and the pattern is designated as the emblem,\n\nThen the perimeter of the outer contour of the emblem is: $\\frac{180}{90-30} + \\frac{720}{30} - 6 = 21$, so option D is correct.\n\nHence, the answer is: D.\n\n[Key Insight] This problem mainly tests the comprehension of the relationship between the number of sides of a regular polygon and its internal and external angles. It is crucial to understand that in a regular polygon, all internal angles are equal, all external angles are equal, and the sum of the external angles is $360^\\circ$. The problem is solved using a combination of numerical and geometric approaches." }, { "problem_id": 107, "question": "As shown in Figure 1, this is a hexagonal ruler used in woodworking table saws, providing common angle measurements. In the hexagonal ruler diagram in Figure 2, the value of $x$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $135^{\\circ}$\nB. $120^{\\circ}$\nC. $112.5^{\\circ}$\nD. $112^{\\circ}$", "input_image": [ "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0048_1.jpg", "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0048_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: According to the problem, we have:\n\\[135^{\\circ} + x + (x + 9)^{\\circ} + 126^{\\circ} + 120^{\\circ} + (2x - 120)^{\\circ} = (6 - 2) \\times 180^{\\circ}\\]\nSolving the equation yields:\n\\[x = 112.5^{\\circ}\\]\n\nTherefore, the correct answer is: C.\n\n[Key Insight] This problem primarily tests the theorem of the sum of interior angles of a polygon and the method of solving a linear equation in one variable. Mastering the theorem of the sum of interior angles of a polygon is crucial for solving this problem." }, { "problem_id": 108, "question": "As shown in the figure, in quadrilateral $A B C D$, $\\angle B=90^{\\circ}, A C=6, A B \\parallel C D$, and $A C$ bisects $\\angle D A B$. Let $A B=x$ and $A D=y$. The graph that can roughly represent the functional relationship between $y$ and $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0092_1.jpg", "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0092_2.jpg", "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0092_3.jpg", "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0092_4.jpg", "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0092_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since \\( AB \\parallel CD \\), it follows that \\( \\angle ACD = \\angle BAC \\).\n\nSince \\( AC \\) bisects \\( \\angle DAB \\), we have \\( \\angle BAC = \\angle CAD \\).\n\nTherefore, \\( \\angle ACD = \\angle CAD \\), which implies that \\( CD = AD = y \\), making \\( \\triangle ACD \\) an isosceles triangle.\n\nDraw a perpendicular from point \\( D \\) to \\( AC \\), meeting \\( AC \\) at point \\( E \\).\n\n\n\nThen, \\( DE \\) is the perpendicular bisector of \\( AC \\), so \\( AE = CE = \\frac{1}{2} AC = 3 \\), and \\( \\angle AED = 90^\\circ \\).\n\nSince \\( \\angle BAC = \\angle CAD \\) and \\( \\angle B = \\angle AED = 90^\\circ \\), it follows that \\( \\triangle ABC \\sim \\triangle AED \\).\n\nThus, \\( \\frac{AC}{AD} = \\frac{AB}{AE} \\),\n\nwhich gives \\( \\frac{6}{y} = \\frac{x}{3} \\),\n\nso \\( y = \\frac{18}{x} \\).\n\nSince in \\( \\triangle ABC \\), \\( AB < AC \\),\n\nit follows that \\( x < 6 \\).\n\nTherefore, the correct choice is D.\n\n【Key Insight】This problem tests the understanding of the definition of an angle bisector, the properties of isosceles triangles, the properties of similar triangles, and the graph of inverse proportional functions. The key to solving this problem is proving that \\( \\triangle ABC \\sim \\triangle AED \\)." }, { "problem_id": 109, "question": "A carpenter wants to enclose a garden with a 40-meter-long wooden board. He is considering one of the following garden designs. Which of the following designs cannot be enclosed with a 40-meter-long wooden board?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_00a109ed8135f00b577eg_0023_1.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0023_2.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0023_3.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0023_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: The perimeters of the shapes in options A, C, and D are all:\n\n$$\n\\begin{aligned}\n& (12 + 8) \\times 2 \\\\\n= & 20 \\times 2 \\\\\n= & 40 \\text{ meters}\n\\end{aligned}\n$$\n\nIn option B, the length of the parallelogram is 12 meters, and its height is 8 meters. Clearly, the perimeter of this shape is greater than 40 meters. Therefore, option B is correct.\n\nThus, the answer is: B.\n\n[Key Insight] By using the method of translation, it's easy to see that the perimeters of the shapes in options A, C, and D are the same. This makes solving the problem straightforward." }, { "problem_id": 110, "question": "As shown in the figure, a portion of a square with side length 2 is transformed as in Figure (1) to Figure (4) and assembled into Figure (5). The area of Figure (5) is ( )\n\n\n(1)\n\n\n(2)\n\n(3)\n\n\n(4)\n\n\n(5)\nA. 18\nB. 16\nC. 12\nD. 8", "input_image": [ "batch14-2024_06_15_00a109ed8135f00b577eg_0080_1.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0080_2.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0080_3.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0080_4.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0080_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: The area of a single square is \\(2 \\times 2 = 4\\). When transforming a square from (1) to (4), the area remains unchanged. Therefore, figure (5), which is composed of 4 figures of type (4), has an area of \\(4 \\times 4 = 16\\). Hence, option B is correct.\n\nAnswer: B.\n\n[Key Insight] This problem primarily examines the transformation of shapes through splicing and translation. The key to solving it lies in understanding that translation does not alter the shape or size of a figure, meaning the area remains unchanged." }, { "problem_id": 111, "question": "Among the following patterns, which one is not a central symmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0001_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0001_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0001_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0001_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Question Analysis: A centrally symmetric figure is one that coincides with itself upon a 180-degree rotation. Figure B coincides with itself upon a 120-degree rotation, hence it is not a centrally symmetric figure. Therefore, the correct choice is B.\n\nExam Focus: Identification of Centrally Symmetric Figures" }, { "problem_id": 112, "question": "Among the following mathematical curves, which ones are centrosymmetric figures? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_252e90a086ab6e027c70g_0042_1.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0042_2.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0042_3.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0042_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. It is not a centrally symmetric figure, so this option does not meet the requirement;\n\nB. It is not a centrally symmetric figure, so this option does not meet the requirement;\n\nC. It is not a centrally symmetric figure, so this option does not meet the requirement;\n\nD. It is a centrally symmetric figure, so this option meets the requirement;\n\nTherefore, the correct choice is: D\n\n[Key Point] This question mainly tests the definition of a centrally symmetric figure. It is crucial to understand that in a plane, if a figure can be rotated 180 degrees around a certain point and the rotated figure coincides with the original figure, then it is called a centrally symmetric figure." }, { "problem_id": 113, "question": "The following patterns are made up of regular polygons, of which the one that is both an axisymmetric and a centrosymmetric figure is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_252e90a086ab6e027c70g_0056_1.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0056_2.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0056_3.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0056_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \nA. The figure is axially symmetric but not centrally symmetric, so this option does not meet the requirement; \n\nB. The figure is both axially symmetric and centrally symmetric, so this option meets the requirement; \n\nC. The figure is axially symmetric but not centrally symmetric, so this option does not meet the requirement; \n\nD. The figure is axially symmetric but not centrally symmetric, so this option does not meet the requirement; \n\nTherefore, the correct choice is: B. \n\n**Key Insight:** \nThis question primarily tests the concepts of centrally symmetric and axially symmetric figures. A figure is called centrally symmetric if it can be rotated 180 degrees around a point and the rotated figure coincides with the original figure. A figure is called axially symmetric if it can be folded along a straight line, and the two parts on either side of the line coincide. Mastering these concepts is crucial for solving the problem." }, { "problem_id": 114, "question": "The following are logos of four types of vehicles. Among them, which one is both center-symmetric and axis-symmetric?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_252e90a086ab6e027c70g_0059_1.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0059_2.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0059_3.jpg", "batch14-2024_06_15_252e90a086ab6e027c70g_0059_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: \n\nA. It is an axisymmetric figure, not a centrally symmetric figure, hence this option is incorrect;\n\nB. It is an axisymmetric figure, not a centrally symmetric figure, hence this option is incorrect;\n\nC. It is an axisymmetric figure and also a centrally symmetric figure, hence this option is correct;\n\nD. It is an axisymmetric figure, not a centrally symmetric figure, hence this option is incorrect.\n\nTherefore, the correct choice is: C.\n\n[Highlight] This question primarily tests the concepts of centrally symmetric figures and axisymmetric figures. The key to identifying an axisymmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide upon folding. For a centrally symmetric figure, the focus is on finding the center of symmetry, where the two parts of the figure coincide upon a 180-degree rotation." }, { "problem_id": 115, "question": "As shown in the figure, pattern (6) is formed by combining two of the five basic shapes (1)(2)(3)(4)(5). These two basic shapes are ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\n\n\n\n(5)\n\n\n\n(6)\nA. (1)(5)\nB. (2)(5)\nC. (3)(5)\nD. (2)(4)", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0046_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0046_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0046_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0046_4.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0046_5.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0046_6.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "**Question Analysis:** \nPattern (6) can be transformed as shown in the figure below. By observing the figure, it can be deduced that the basic shapes composing pattern (6) are (2) and (5). Therefore, the correct answer is **B**.\n\n\n\n(6)\n\n**Key Point:** Translation of shapes." }, { "problem_id": 116, "question": "Among the four patterns below, which one cannot be obtained by translating the Pattern 1 to get Pattern 2?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0073_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0073_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0073_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0073_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: By observing each option, it can be determined that the pattern in option D cannot be obtained by translating figure 1 to figure 2.\n\nTherefore, choose D.\n\n[Key Insight] This question tests the understanding of graphic translation. The key to solving the problem lies in recognizing that translation only changes the position of the figure, not its shape or size." }, { "problem_id": 117, "question": "Stack figure (1) on top of figure (2), and the possible resulting figures are ( )\n\n\n\n(1)\n\n\n\n(2)\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0092_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0092_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0092_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0092_4.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0092_5.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0092_6.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: By superimposing figure (1) onto figure (2), the possible resulting figure is:\n\n\n\nTherefore, choose $B$.\n\n[Key Insight] This question primarily tests the understanding of translation transformations. Correctly utilizing the properties of graphical transformations is crucial for solving the problem." }, { "problem_id": 118, "question": "Among the following patterns, which one is both an axisymmetric figure and a centrosymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_3987e06d627f73b8dfaeg_0062_1.jpg", "batch14-2024_06_15_3987e06d627f73b8dfaeg_0062_2.jpg", "batch14-2024_06_15_3987e06d627f73b8dfaeg_0062_3.jpg", "batch14-2024_06_15_3987e06d627f73b8dfaeg_0062_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "A. It is both an axisymmetric figure and a centrally symmetric figure, so this option is correct;\n\nB. It is an axisymmetric figure, but not a centrally symmetric figure, so this option is incorrect;\n\nC. It is not an axisymmetric figure, but it is a centrally symmetric figure, so this option is incorrect;\n\nD. It is an axisymmetric figure, but not a centrally symmetric figure, so this option is incorrect.\n\nTherefore, the correct choice is A.\n\n[Highlight] This question tests the concepts of centrally symmetric figures and axisymmetric figures. The key to identifying an axisymmetric figure is to find the axis of symmetry, where the two parts of the figure can coincide when folded. For a centrally symmetric figure, the key is to find the center of symmetry, where the two parts of the figure coincide after a 180-degree rotation." }, { "problem_id": 119, "question": "As shown in the figure, there are two shaded shapes, $M$ and $N$, in a $6 \\times 6$ grid. In (1), the shape $M$ is translated to the position shown in (2). Which of the following descriptions of the translation of shape $M$ is correct? \n\n\n\n(1)\n\n\n\n(2)\n\nA. Translate right by 2 units, down by 3 units.\nB. Translate right by 1 unit, down by 3 units.\n\nC. Translate right by 1 unit, down by 4 units.\n\nD. Translate right by 2 units, down by 4 units.\n\n##", "input_image": [ "batch14-2024_06_15_42999df1f8dd14bc18ffg_0002_1.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0002_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Based on the positional changes of corresponding points before and after the translation of figure $M$, it is determined that a translation of 1 unit to the right and 3 units downward is required.\n\nTherefore, the correct choice is: B.\n\n[Highlight] This question primarily examines the translation of figures, with the key to solving it lying in identifying the corresponding points in the figure before and after the translation." }, { "problem_id": 120, "question": "The following traffic sign patterns can be viewed as obtained by translating the shape are ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_42999df1f8dd14bc18ffg_0008_1.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0008_2.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0008_3.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0008_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \nA. It can be obtained by translating a \"basic pattern,\" so this option aligns with the question's requirement; \nB. It is not obtained by translating a \"basic pattern,\" so this option does not align with the question's requirement; \nC. It can be obtained by rotating a \"basic pattern,\" so this option does not align with the question's requirement; \nD. It can be obtained by rotating a \"basic pattern,\" so this option does not align with the question's requirement. \n\nTherefore, the correct answer is: A. \n\n**[Key Insight]** This question primarily examines the understanding of translation and rotation of shapes. Accurate analysis and judgment are crucial to solving the problem." }, { "problem_id": 121, "question": "Among the figures in the options below, which one has the longest perimeter?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_42999df1f8dd14bc18ffg_0011_1.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0011_2.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0011_3.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0011_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. From the figure, it can be deduced that its perimeter is greater than \\(12 \\mathrm{~cm}\\).\n\nB. From the figure, it can be deduced that its perimeter is \\(12 \\mathrm{~cm}\\).\n\nC. From the figure, it can be deduced that its perimeter is \\(12 \\mathrm{~cm}\\).\n\nD. From the figure, it can be deduced that its perimeter is \\(12 \\mathrm{~cm}\\).\n\nTherefore, the longest perimeter is that of A.\n\nHence, the correct choice is: A.\n\n【Key Insight】This question primarily tests the understanding of the translation of shapes, and the correct application of the properties of translation is crucial to solving the problem." }, { "problem_id": 122, "question": "The following patterns are obtained by translating the pattern shown in the figure ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_42999df1f8dd14bc18ffg_0050_1.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0050_2.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0050_3.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0050_4.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0050_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: \n\n$A$ - The pattern shown in the figure is formed by rotation, so this option is incorrect;\n\n$B$ - The pattern shown in the figure is formed by translation, so this option is correct;\n\n$C$ - The pattern shown in the figure is formed by rotation, so this option is incorrect;\n\n$D$ - The pattern shown in the figure is formed by reflection, so this option is incorrect.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question primarily tests the properties of translation. It is crucial to understand that the shape and size of the figure remain unchanged before and after translation; only the position is altered." }, { "problem_id": 123, "question": "Examine the following subway symbols, which of them are centrosymmetric patterns?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_44c1bd239a547bfdce70g_0071_1.jpg", "batch14-2024_06_15_44c1bd239a547bfdce70g_0071_2.jpg", "batch14-2024_06_15_44c1bd239a547bfdce70g_0071_3.jpg", "batch14-2024_06_15_44c1bd239a547bfdce70g_0071_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\n$A$ is a centrally symmetric figure, therefore this option meets the requirement of the question;\n\n$B$ is not a centrally symmetric figure, therefore this option does not meet the requirement of the question;\n\n$C$ is not a centrally symmetric figure, therefore this option does not meet the requirement of the question;\n\n$D$ is not a centrally symmetric figure, therefore this option does not meet the requirement of the question.\n\nHence, the correct choice is: $A$.\n\n[Key Insight] This question tests the understanding of centrally symmetric figures. Accurately grasping the characteristics of centrally symmetric figures is crucial for solving the problem." }, { "problem_id": 124, "question": "Among the following figures, which one is a central symmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_44c1bd239a547bfdce70g_0097_1.jpg", "batch14-2024_06_15_44c1bd239a547bfdce70g_0097_2.jpg", "batch14-2024_06_15_44c1bd239a547bfdce70g_0097_3.jpg", "batch14-2024_06_15_44c1bd239a547bfdce70g_0097_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Not a centrally symmetric figure;\n\nB. Not a centrally symmetric figure;\n\nC. Is a centrally symmetric figure;\n\nD. Not a centrally symmetric figure;\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the concept of a centrally symmetric figure. If a figure is rotated $180^{\\circ}$ around a certain point and the rotated figure coincides with the original figure, then it is called a centrally symmetric figure." }, { "problem_id": 125, "question": "Among the following figures, which one is a central symmetric figure? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_4cae88dad4057d404d1dg_0003_1.jpg", "batch14-2024_06_15_4cae88dad4057d404d1dg_0003_2.jpg", "batch14-2024_06_15_4cae88dad4057d404d1dg_0003_3.jpg", "batch14-2024_06_15_4cae88dad4057d404d1dg_0003_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. It is not a centrally symmetric figure, so option A does not meet the requirement;\n\nB. It is not a centrally symmetric figure, so option B does not meet the requirement;\n\nC. It is not a centrally symmetric figure, so option C does not meet the requirement;\n\nD. It is a centrally symmetric figure, so option D meets the requirement;\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question tests the definition of a centrally symmetric figure. Mastering the definition is crucial for solving the problem." }, { "problem_id": 126, "question": "During the Spring Festival, pasting couplets and sending blessings are our time-honored traditions. In China's traditional culture, the \"Fortune, Blessing, Longevity, Happiness\" picture (as shown in the figure) is composed of four patterns. Among these four patterns, which one is a central symmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_4cae88dad4057d404d1dg_0011_1.jpg", "batch14-2024_06_15_4cae88dad4057d404d1dg_0011_2.jpg", "batch14-2024_06_15_4cae88dad4057d404d1dg_0011_3.jpg", "batch14-2024_06_15_4cae88dad4057d404d1dg_0011_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the definition of a centrally symmetric figure, it can be deduced that: Figure B is a centrally symmetric figure, while figures A, C, and D are not.\n\nTherefore, the answer is: B.\n\n[Key Point] This question primarily tests the understanding of the definition of a centrally symmetric figure. Fully comprehending the definition of a centrally symmetric figure is crucial for solving the problem." }, { "problem_id": 127, "question": "Among the following figures, which one is both an axisymmetric and a centrosymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_4d71f51b84e8e9829619g_0008_1.jpg", "batch14-2024_06_15_4d71f51b84e8e9829619g_0008_2.jpg", "batch14-2024_06_15_4d71f51b84e8e9829619g_0008_3.jpg", "batch14-2024_06_15_4d71f51b84e8e9829619g_0008_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \nA. It is an axisymmetric figure but not a centrally symmetric figure, so this option does not meet the requirement; \nB. It is not an axisymmetric figure but is a centrally symmetric figure, so this option does not meet the requirement; \nC. It is an axisymmetric figure but not a centrally symmetric figure, so this option does not meet the requirement; \nD. It is both an axisymmetric figure and a centrally symmetric figure, so this option meets the requirement. \n\nTherefore, the correct answer is: D. \n\n**Key Point:** \nThis question tests the concepts of axisymmetric and centrally symmetric figures. \n- **Definition of an axisymmetric figure:** A figure that can be folded along a straight line so that the two parts on either side of the line completely coincide. \n- **Definition of a centrally symmetric figure:** A figure that, when rotated 180 degrees around a point, coincides with its original position. \nUnderstanding these two concepts is crucial." }, { "problem_id": 128, "question": "Given that the point $\\mathrm{M}(1-2 \\mathrm{~m}, \\mathrm{~m}-1)$ is symmetrical about the $\\mathrm{x}$ axis in the second quadrant, the correct range of values ​​for $\\mathrm{m}$ on the number axis is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0025_1.jpg", "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0025_2.jpg", "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0025_3.jpg", "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0025_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Since the symmetric point of point $\\mathrm{M}(1-2m, m-1)$ with respect to the $x$-axis lies in the second quadrant,\n\n$\\therefore$ point $\\mathrm{M}$ is in the third quadrant,\n\n$\\therefore\\left\\{\\begin{array}{l}1-2m<0 \\\\ m-1<0\\end{array}\\right.$,\n\n$\\therefore 0.5\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\nA. $(-m, n)$\nB. $(-m,-n)$\nC. $(m,-n)$\nD. $(m, n)$", "input_image": [ "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0064_1.jpg", "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0064_2.jpg", "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0064_3.jpg", "batch14-2024_06_15_50897e63ebf3f45f7b3eg_0064_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: After the first reflection of point $A$ across the $y$-axis, it lies in the first quadrant.\n\nAfter the second reflection of point $A$ across the $x$-axis, it lies in the fourth quadrant.\n\nAfter the third reflection of point $A$ across the $y$-axis, it lies in the third quadrant.\n\nAfter the fourth reflection of point $A$ across the $x$-axis, it lies in the second quadrant,\n\nwhich means point $A$ returns to its original position.\n\nTherefore, every four reflections form a cycle that repeats in sequence.\n\nSince $2020 \\div 4 = 505$,\n\nthe position of point $A$ after the 2020th transformation is the same as after the first transformation, located in the first quadrant with coordinates $(m, n)$.\n\nHence, the correct answer is: D.\n\n【Key Insight】This problem examines the properties of symmetry and the transformation rules of point coordinates. Understanding the information in the problem and observing that every four reflections form a repeating cycle is the key to solving the problem, and it is also the challenging aspect of this question." }, { "problem_id": 130, "question": "Among the following figures, how many are centro-symmetric?\n\n\n\nSquare\n\n\n\nHexagon\n\n\n\nTriangle\n\n\n\nCircle\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0037_1.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0037_2.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0037_3.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0037_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: A square, a regular hexagon, and a circle can all have a point such that when the shape is rotated by $180^{\\circ}$ around this point, it coincides with the original shape. Therefore, they are all centrally symmetric figures.\n\nThus, there are 3 centrally symmetric figures.\n\nHence, the correct choice is: C.\n\n[Key Insight] This question tests the concept of centrally symmetric figures, which are defined by the existence of a center of symmetry such that a rotation of 180 degrees around this center results in the figure coinciding with its original position." }, { "problem_id": 131, "question": "As shown in the figure, the ray of light enters from point $P$. After reflecting off the mirror $E F$, the point it passes through is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Point $A$\nB. Point $B$\nC. Point $C$\nD. Point $D$", "input_image": [ "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0047_1.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0047_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, the ray passing through point $P$ and point $B$ intersects at a point $O$,\n\n\n\nFigure 1-1\n\n\n\nFigure 1-2\n\nTherefore, the answer is: B.\n\n[Key Insight] This question tests the properties of axial symmetry transformation. The key to solving the problem lies in understanding the question and flexibly applying the knowledge learned to solve it." }, { "problem_id": 132, "question": "Observe the data marked in the figure on the right, and determine which of the following statements is correct ( )\n\n\n\nA\n\n\n\nB\n\nC\n A. A, B, and C are all axisymmetric figures.\nB. Only A is an axisymmetric figure.\nC. Only C is not an axisymmetric figure.\nD. Only A and C are axisymmetric figures.", "input_image": [ "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0061_1.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0061_2.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0061_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Triangle A is an isosceles triangle and is an axisymmetric figure;\n\nAs shown in the figure,\n\n\n\n$\\because \\angle A + \\angle B = 144^{\\circ} + 36^{\\circ} = 180^{\\circ}$,\n\n$\\therefore AD \\parallel BC$,\n$\\because \\angle A + \\angle D = 144^{\\circ} + 36^{\\circ} = 180^{\\circ}$,\n\n$\\therefore AB \\quad CD$,\n\n$\\therefore$ Figure B is a parallelogram, but not an axisymmetric figure;\n\nAs shown in the figure,\n\n\n\n$\\angle D = 360^{\\circ} - \\angle A - \\angle B - \\angle C = 360^{\\circ} - 88^{\\circ} - 90^{\\circ} - 90^{\\circ} = 92^{\\circ}$,\n\n$\\therefore$ Figure C is not an axisymmetric figure;\n\nTherefore, the correct choice is: B\n\n[Key Insight] This question primarily examines the properties of isosceles triangles, the determination and properties of parallelograms, the interior angle sum theorem of quadrilaterals, and axisymmetric figures. Mastering the relevant knowledge points is crucial for solving the problem." }, { "problem_id": 133, "question": "As shown in the figure, fold the square paper $A B C D$ along the crease $B E$ so that point $A$ falls on segment $B C$, which is labeled (1). Unfold and then fold the square paper again along the crease $C F$ so that point $D$ falls on segment $B C$, which is labeled (2). Unfold again and continue to fold the square paper so that segment $A D$ falls on line segment $B C$, which is labeled (3). Unfold once more, and label this as (4). If $A B = 4$ and $B C = 7$, what is the length of segment $G H$ in Figure (4)?\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. $\\frac{5}{2}$\nB. $\\frac{7}{2}$\nC. 3\nD. 4", "input_image": [ "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0069_1.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0069_2.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0069_3.jpg", "batch14-2024_06_15_553a3d3b134f0d1ad9a7g_0069_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (4), connect $EH$ and extend $EH$ to intersect $BC$ at point $M$.\n\n\n\n(4)\n\nFrom the problem statement, it is easy to deduce: $AB = AE = 4$, $CD = DF = 4$, and $GH$ is the midline of $\\triangle EBM$.\n\nSince $AD = BC = 7$,\n\nTherefore, $AF = DE = 3$, and $EF = 1$.\n\nSince $EH = HM$, $\\angle EFH = \\angle MCH$, and $\\angle EHF = \\angle CHM$,\n\nThus, $\\triangle EFH \\cong \\triangle MCH$ (by AAS),\n\nTherefore, $EF = CM = 1$.\n\nHence, $BM = BC - CM = 6$.\n\nSince $GH$ is the midline of $\\triangle EBM$,\n\nTherefore, $GH = \\frac{1}{2} BM = 3$.\n\nThus, the correct answer is: C.\n\n[Key Insight] This problem examines the properties of folding transformations, the properties of parallelograms, the midline theorem of triangles, and the properties of congruent triangles. The key to solving the problem lies in understanding the problem statement and knowing how to add commonly used auxiliary lines." }, { "problem_id": 134, "question": "By joining 4 congruent regular octagons so that two adjacent octagons share a common side, a circle is formed with a square in the middle (as shown in Figure 1). If $n$ congruent regular hexagons are joined in the same manner, as shown in Figure 2, and a regular polygon is formed in the middle after forming a circle, what is the value of $n$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 5\nB. 6\nC. 8\nD. 10", "input_image": [ "batch14-2024_06_15_637191b7c30eeed99c12g_0074_1.jpg", "batch14-2024_06_15_637191b7c30eeed99c12g_0074_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The measure of each interior angle of a regular hexagon is: $\\frac{180^{\\circ} \\times(6-2)}{6}=120^{\\circ}$.\n\nThen, the measure of the interior angle of the polygon formed by the regular hexagon is: $360^{\\circ} - 2 \\times 120^{\\circ} = 120^{\\circ}$. According to the problem, we have: $180(n-2) = 120n$,\n\nSolving for $n$, we get: $\\mathrm{n}=6$.\n\nTherefore, the correct choice is: B.\n\n【Key Point】This question tests the theorem of the sum of interior angles of a polygon. Correctly understanding the theorem and determining the measure of the interior angle of the formed polygon is crucial." }, { "problem_id": 135, "question": "As shown in the figure, among the four tile patterns, which one cannot tile the ground?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_637191b7c30eeed99c12g_0079_1.jpg", "batch14-2024_06_15_637191b7c30eeed99c12g_0079_2.jpg", "batch14-2024_06_15_637191b7c30eeed99c12g_0079_3.jpg", "batch14-2024_06_15_637191b7c30eeed99c12g_0079_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "$\\because$ The shapes that can tile the plane are those whose internal angles can sum up to $360^{\\circ}$.\n\n$\\because$ An equilateral triangle has an internal angle of $60^{\\circ}$, a square has an internal angle of $90^{\\circ}$, a regular pentagon has an internal angle of $108^{\\circ}$, and a regular hexagon has an internal angle of $120^{\\circ}$. Only the regular pentagon cannot sum up to $360^{\\circ}$.\n\nTherefore, the answer is: $C$.\n\n【Key Point】This question tests the concept of tiling (tessellation). Mastering the conditions for tiling is crucial for solving such problems." }, { "problem_id": 136, "question": "Which of the following figures can be obtained by a translation transformation from a basic figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_64cba4825319390cfdd5g_0070_1.jpg", "batch14-2024_06_15_64cba4825319390cfdd5g_0070_2.jpg", "batch14-2024_06_15_64cba4825319390cfdd5g_0070_3.jpg", "batch14-2024_06_15_64cba4825319390cfdd5g_0070_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Translation, rotation, and symmetry are key topics in the high school entrance examination. Options A, C, and D all require rotation. Therefore, choose B." }, { "problem_id": 137, "question": "As shown in Figure 1, equilateral triangles $\\triangle \\mathrm{ABD}$ and $\\triangle \\mathrm{CBD}$ both have side lengths of 2. Triangle $\\triangle \\mathrm{ABD}$ is translated to the right along the $\\mathrm{AC}$ direction by $k$ units to form triangle $\\triangle A^{\\prime} B^{\\prime} D^{\\prime}$ in Figure 2. The following statements are given: (1) The perimeter of the shaded region is 4; (2) When $k=\\frac{\\sqrt{3}}{2}$, the shaded region in the figure is a regular hexagon; (3) When $k=\\frac{\\sqrt{3}}{2}$, the area of the shaded region in the figure is $\\frac{5}{8} \\sqrt{3}$. Which of the following combinations of statements are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. (1)\nB. (1)(2)\nC. (1)(3)\nD. (1)(2)(3)", "input_image": [ "batch14-2024_06_15_64cba4825319390cfdd5g_0077_1.jpg", "batch14-2024_06_15_64cba4825319390cfdd5g_0077_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since both equilateral triangles $\\triangle \\mathrm{ABD}$ and $\\triangle \\mathrm{CBD}$ have side lengths of 2, and $\\triangle \\mathrm{ABD}$ is translated to the right along $\\mathrm{AC}$ to the position of $\\triangle \\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{D}^{\\prime}$,\n\nTherefore, $\\mathrm{A}^{\\prime} \\mathrm{M}=\\mathrm{A}^{\\prime} \\mathrm{N}=\\mathrm{MN}$, $\\mathrm{MO}=\\mathrm{DM}=\\mathrm{DO}$, $\\mathrm{OD}^{\\prime}=\\mathrm{D}^{\\prime} \\mathrm{E}=\\mathrm{OE}$, $\\mathrm{EG}=\\mathrm{EC}=\\mathrm{GC}$, $\\mathrm{B}^{\\prime} \\mathrm{G}=\\mathrm{RG}=\\mathrm{RB}^{\\prime}$,\n\nThus, $\\mathrm{OM}+\\mathrm{MN}+\\mathrm{NR}+\\mathrm{GR}+\\mathrm{EG}+\\mathrm{OE}=\\mathrm{A}^{\\prime} \\mathrm{D}^{\\prime}+\\mathrm{CD}=2+2=4$,\n\nHence, statement (1) is correct;\n\nSince $\\mathrm{k}=\\frac{\\sqrt{3}}{2}$\n\nTherefore, $\\mathrm{A}^{\\prime} \\mathrm{F}=\\frac{\\sqrt{3}}{2}$\n\nThus, $\\mathrm{A}^{\\prime} \\mathrm{M}=\\mathrm{A}^{\\prime} \\mathrm{F} \\div \\cos 30^{\\circ}=1$, and $\\mathrm{MN}=1$\n\nTherefore, $M O=\\frac{1}{2}(2-1)=\\frac{1}{2}$.\n\nThus, $\\mathrm{MO} \\neq \\mathrm{MN}$,\n\nTherefore, the shaded part is not a regular hexagon,\n\nHence, statement (2) is incorrect;\n\nThe area of the shaded part $=$ area of $\\triangle A^{\\prime} B^{\\prime} D^{\\prime}$ $-$ area of $\\triangle A^{\\prime} M N$ $-$ area of $\\triangle O D^{\\prime} E$ $-$ area of $\\triangle R G B^{\\prime}$\n\n$=\\frac{\\sqrt{3}}{4} \\times\\left(2^{2}-1^{2}-2 \\times\\left(\\frac{1}{2}\\right)^{2}\\right]$\n\n$=\\frac{5 \\sqrt{3}}{8}$.\n\nHence, statement (3) is correct,\n\nTherefore, the correct choice is C.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n[Highlight] This problem mainly examines the properties of equilateral triangles and the application of translation properties. The key to solving the problem lies in determining the side lengths and areas of the smaller equilateral triangles based on the translation distance." }, { "problem_id": 138, "question": "As shown in Figure 1, in square $A B C D$, point $E$ is the midpoint of a side. Point $P$ starts from point $A$ and moves along the edges of the square at a constant speed in the direction $A \\rightarrow B \\rightarrow E$ until it reaches point $E$. Let the distance traveled by point $P$ be $x$, and let $y=P D-P B$. The graph in Figure 2 represents the relationship between $y$ and $x$ as point $P$ moves. Based on the data in the graph, the coordinates of point $Q$ can be determined as ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $(3,3 \\sqrt{2})$\nB. $\\left(3, \\frac{3}{2} \\sqrt{5}-\\frac{3}{2}\\right)$\nC. $\\left(\\frac{9}{2}, \\frac{3}{2} \\sqrt{5}-\\frac{3}{2}\\right)$\nD. $\\left(\\frac{9}{2}, \\frac{3}{2} \\sqrt{5}\\right)$", "input_image": [ "batch14-2024_06_15_66746bbb4e0e1d91de17g_0035_1.jpg", "batch14-2024_06_15_66746bbb4e0e1d91de17g_0035_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to Figure 2,\n\nWhen point \\( P \\) reaches point \\( A \\), \\( y = PD - PB = 0 \\).\n\nWhen point \\( P \\) reaches point \\( B \\), \\( y = PD - PB = 3\\sqrt{2} \\), and \\( PB = 0 \\), which means \\( PD = BD = 3\\sqrt{2} \\). Therefore, \\( AB = AD = 3 \\).\n\nWhen point \\( P \\) reaches point \\( E \\), the distance traveled by point \\( P \\) is \\( x = \\frac{9}{2} \\), and \\( PB = BE = EC = \\frac{3}{2} \\). By the Pythagorean theorem,\n\n\\( PD = DE = \\sqrt{3^{2} + \\left(\\frac{3}{2}\\right)^{2}} = \\frac{3\\sqrt{5}}{2} \\), so \\( y = PD - PB = \\frac{3\\sqrt{5}}{2} - \\frac{3}{2} \\).\n\nThus, the coordinates of point \\( Q \\) are \\( \\left(\\frac{9}{2}, \\frac{3}{2}\\sqrt{5} - \\frac{3}{2}\\right) \\).\n\nTherefore, the correct choice is: \\( C \\).\n\n[Key Insight] This problem primarily examines the moving point issue within a square. Identifying the key points in the figure and their corresponding critical values is essential for solving the problem." }, { "problem_id": 139, "question": "As shown in Figure 1, a \"windmill\" pattern is formed by four congruent right-angled triangles, where $\\angle A O B = 90^{\\circ}$. By extending the hypotenuse of each right-angled triangle, it exactly meets at the midpoint of the hypotenuse of another right-angled triangle, as shown in Figure 2. If $I J = 2$, then the area of the \"windmill\" is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $2 \\sqrt{2}+1$\nB. $2 \\sqrt{2}$\nC. $4+\\sqrt{2}$\nD. $4 \\sqrt{2}$", "input_image": [ "batch14-2024_06_15_66746bbb4e0e1d91de17g_0088_1.jpg", "batch14-2024_06_15_66746bbb4e0e1d91de17g_0088_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Connect $BH$.\n\n\n\nGiven that quadrilateral $IJKL$ is a square.\n\nSince $IJ = 2$,\n\nThe area of square $IJKL = 4$,\n\nTherefore, the area of quadrilateral $IBOH = \\frac{1}{4} \\times 4 = 1$,\n\nSince $HI$ is the perpendicular bisector of $AB$,\n\nThus, $HA = HB$,\n\nSince $OH = OB$ and $\\angle BOH = 90^\\circ$,\n\nTherefore, $HA = BH = \\sqrt{2} \\times OH$,\n\nThus, the ratio of the areas $S_{\\triangle ABH} : S_{\\triangle BOH} = \\sqrt{2}$,\n\nSince $S_{\\triangle AIH} = S_{\\triangle IBH}$,\n\nTherefore, the ratio $S_{\\triangle IBH} : S_{\\triangle BOH} = \\sqrt{2} : 2$,\n\nThus, $S_{\\triangle AHI} = S_{\\triangle IBH} = \\frac{\\sqrt{2}}{\\sqrt{2} + 2} \\times S_{\\text{quadrilateral } IBOH} = \\frac{\\sqrt{2}}{\\sqrt{2} + 2} \\times 1 = \\sqrt{2} - 1$,\n\nTherefore, $S_{\\triangle AOB} = S_{\\triangle AIH} + S_{\\text{quadrilateral } IBOH} = \\sqrt{2} - 1 + 1 = \\sqrt{2}$,\n\nThus, the area of the \"windmill\" $= 4 \\times S_{\\triangle AOB} = 4\\sqrt{2}$.\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This problem examines the properties of congruent triangles and the properties of isosceles right triangles. The key to solving this problem lies in understanding and applying the properties of congruent triangles." }, { "problem_id": 140, "question": "In the Cartesian coordinate system $x O y$, which of the following pairs of triangles, $A B C$ and $A^{\\prime} B^{\\prime} C^{\\prime}$, are centrally symmetric with respect to the origin $O$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_67b9e2592a4ca695d018g_0011_1.jpg", "batch14-2024_06_15_67b9e2592a4ca695d018g_0011_2.jpg", "batch14-2024_06_15_67b9e2592a4ca695d018g_0011_3.jpg", "batch14-2024_06_15_67b9e2592a4ca695d018g_0011_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Since $ABC$ and $A'B'C'$ are symmetric about the $y$-axis, option A does not meet the requirement;\n\nB. Since $ABC$ and $A'B'C'$ are symmetric about the $x$-axis, option B does not meet the requirement;\n\nC. Since $ABC$ and $A'B'C'$ are symmetric about the point $\\left(-\\frac{1}{2}, 0\\right)$, option C does not meet the requirement;\n\nD. Since $ABC$ and $A'B'C'$ are symmetric about the origin $O$, option D meets the requirement;\n\nTherefore, the correct choice is: D.\n\n[Highlight] This question examines the concept of central symmetry: rotating a figure $180^{\\circ}$ around a certain point, if it can coincide with another figure, then the two figures are said to be symmetric about that point or centrally symmetric, with the point called the center of symmetry. The corresponding points in the two figures are called symmetric points about the center. Properties of central symmetry: two figures symmetric about a center can completely coincide; for two figures symmetric about a center, the lines connecting corresponding points all pass through the center of symmetry and are bisected by it." }, { "problem_id": 141, "question": "As shown in the figure, line $\\mathrm{m} / \\mathrm{n}$, point $\\mathrm{A}$ is on line $\\mathrm{m}$, and $\\mathrm{BC}$ is on line $\\mathrm{n}$, forming triangle $\\mathrm{ABC}$. Triangle $\\mathrm{ABC}$ is translated to the right by half the length of $\\mathrm{BC}$ to obtain $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ (as shown in Figure (1)). Then, $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is translated to the right by half the length of $\\mathrm{BC}$ to obtain $\\triangle A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime}$ (as shown in Figure (2)). This translation process is continued to obtain Figure (3), ..., By observing, we know that there are 4 triangles in Figure (1), 8 triangles in Figure (2), and the number of triangles in the 2020th figure is ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. 4040\nB. 6060\nC. 6061\nD. 8080", "input_image": [ "batch14-2024_06_15_761fd3a7e6741c0d3db1g_0079_1.jpg", "batch14-2024_06_15_761fd3a7e6741c0d3db1g_0079_2.jpg", "batch14-2024_06_15_761fd3a7e6741c0d3db1g_0079_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: By observing the diagram, we can see that in the first figure, there are 2 large triangles and 2 small triangles. In the second figure, there are 4 large triangles and 4 small triangles. In the third figure, there are 6 large triangles and 6 small triangles, and so on.\n\nFollowing this pattern, the nth figure will have 2n large triangles and 2n small triangles.\n\nTherefore, the number of triangles in the 2019th figure is: \\(2 \\times 2020 + 2 \\times 2020 = 8080\\).\n\nHence, the correct answer is: D.\n\n[Insight] This question tests pattern recognition, the properties of parallel lines, and translation transformations. The key to solving it lies in understanding how to explore patterns, a common type of question in middle school exams." }, { "problem_id": 142, "question": "As shown in the figure, in rectangle $A B C D$, $A B = 1$, $A D = 2$, and $M$ is the midpoint of $A D$. Point $P$ is on the edges of the rectangle, starting from point $A$ and moving along $A \\rightarrow B \\rightarrow C \\rightarrow D$ until it reaches point $D$ when the movement terminates. Let the area of triangle $A P M$ be $y$, and the distance traveled by point $P$ be $x$. Which of the following graphs correctly represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_82623ec71bb7bc51c88bg_0017_1.jpg", "batch14-2024_06_15_82623ec71bb7bc51c88bg_0017_2.jpg", "batch14-2024_06_15_82623ec71bb7bc51c88bg_0017_3.jpg", "batch14-2024_06_15_82623ec71bb7bc51c88bg_0017_4.jpg", "batch14-2024_06_15_82623ec71bb7bc51c88bg_0017_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Question Analysis: According to the area formula of a triangle, we discuss by cases: When point $P$ moves along $AB$, the area of the triangle increases; when $P$ moves along $BC$, the area of the triangle remains constant at $\\frac{1}{2} \\times 1 \\times 1=\\frac{1}{2}$; and when $P$ moves along $CD$, the area of the triangle decreases.\n\nTherefore, the correct choice is: A." }, { "problem_id": 143, "question": "As shown in the figure, a pair of angle rulers is placed in different positions. Among the following arrangements, the ones where $\\angle \\alpha$ is definitely equal to $\\angle \\beta$ are ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)\nB. (1)(3)\nC. (2)(4)\nD. (3)(4)", "input_image": [ "batch14-2024_06_15_84c339bc1ff9aab9aa3ag_0020_1.jpg", "batch14-2024_06_15_84c339bc1ff9aab9aa3ag_0020_2.jpg", "batch14-2024_06_15_84c339bc1ff9aab9aa3ag_0020_3.jpg", "batch14-2024_06_15_84c339bc1ff9aab9aa3ag_0020_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In Figure (1), according to the property that the complements of the same angle are equal, we have $\\angle \\alpha = \\angle \\beta$;\n\nIn Figure (2), $\\angle \\alpha = 135^{\\circ}$ and $\\angle \\beta = 120^{\\circ}$, therefore $\\angle \\alpha \\neq \\angle \\beta$;\n\nIn Figure (3), according to the property that the supplements of equal angles are equal, we have $\\angle \\alpha = \\angle \\beta$;\n\nIn Figure (4), $\\angle \\alpha + \\angle \\beta = 180^{\\circ} - 90^{\\circ} = 90^{\\circ}$, which means they are complementary.\n\nThus, $\\angle \\alpha$ and $\\angle \\beta$ must be equal in Figure (1) and Figure (3).\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question tests the understanding of complementary and supplementary angles. It is a basic problem, and mastering the concepts and properties is crucial for solving it." }, { "problem_id": 144, "question": "A school has a rectangular garden with a length of $a$ and a width of $b$. There are two identical rectangular paths, as shown in Figure 1, running across the garden. Now, grass is being planted in the shaded areas as shown in Figure 2, and a fence is to be built around the grass. Using your knowledge, calculate the total length of the fence required (in terms of $a$ and $b$).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $2 b$\nB. $4 b$\nC. $2(a+b)$\nD. $4(a-b)$", "input_image": [ "batch14-2024_06_15_a2af22e8ba137bc6684fg_0016_1.jpg", "batch14-2024_06_15_a2af22e8ba137bc6684fg_0016_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Let the length of the small rectangle be \\( m \\) and the width be \\( n \\). From Figure (2), it is easy to see that \\( m + n = a \\). By using the translation method to align the lengths of the two shaded parts together, it is not difficult to find that the sum of the lengths of the shaded parts above and below is \\( 2a \\). The width of the lower-left shaded part is \\( b - n \\), and the width of the upper-right shaded part is \\( b - m \\). Therefore,\n\nThe perimeter of the shaded region is:\n\n\\[\n\\begin{aligned}\n& 2a + 2(b - n) + 2(b - m) \\\\\n& = 2a + 2b - 2n + 2b - 2m \\\\\n& = 2a + 4b - 2(m + n)\n\\end{aligned}\n\\]\n\n\n\nSubstituting \\( m + n = a \\) into the equation, we get:\n\nOriginal expression \\( = 4b \\).\n\nTherefore, the correct answer is B.\n\n【Key Point】This problem tests the ability to calculate the perimeter using algebraic expressions. Mastering the method of using translation to find the perimeter and identifying the equal relationships in the figure is the key to solving this problem." }, { "problem_id": 145, "question": "During a mathematics club activity, Student Xiaohong designed and created the three different shapes shown in the figure using wire. Please observe the shapes A, B, and C, and determine the relationship between the lengths of wire used to create them:\n\n\n\nA\n\n\n\nB\n\n\n\nC\nA. The wire used to make Shape A is the longest.\nB. The wire used to make Shape B is the longest.\nC. The wire used to make Shape C is the longest.\nD. The same amount of wire was used to make all three shapes.", "input_image": [ "batch14-2024_06_15_a2af22e8ba137bc6684fg_0023_1.jpg", "batch14-2024_06_15_a2af22e8ba137bc6684fg_0023_2.jpg", "batch14-2024_06_15_a2af22e8ba137bc6684fg_0023_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, we can deduce that the length of the wire used by A is: $2a + 2b$,\n\nThe length of the wire used by B is: $2a + 2b$,\n\nThe length of the wire used by C is: $2a + 2b$,\n\nTherefore, the lengths of wire used in all three schemes are the same.\n\nHence, the correct choice is: D.\n\n[Key Insight] This question primarily examines the phenomenon of translation in real life, and determining the length of the wire in each figure is crucial to solving the problem." }, { "problem_id": 146, "question": "If point $P(2x+2, x-2)$ is translated 4 units to the right and 2 units down, and the resulting point $P$ is in the fourth quadrant of the Cartesian coordinate system, then the range of $x$ on the number line can be represented as ( )\n\nA. \nB. \nC. \nD. ", "input_image": [ "batch14-2024_06_15_a2af22e8ba137bc6684fg_0060_1.jpg", "batch14-2024_06_15_a2af22e8ba137bc6684fg_0060_2.jpg", "batch14-2024_06_15_a2af22e8ba137bc6684fg_0060_3.jpg", "batch14-2024_06_15_a2af22e8ba137bc6684fg_0060_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: The point \\( P(2x + 2, x - 2) \\) is translated 4 units to the right and then 2 units downward. After the translation, the coordinates of point \\( P \\) become \\( P(2x + 6, x - 4) \\). After the translation, point \\( P \\) lies in the fourth quadrant.\n\n\\[\n\\begin{cases}\n2x + 6 > 0 \\\\\nx - 4 < 0\n\\end{cases}\n\\]\n\nSolving the inequalities gives: \\( -3 < x < 4 \\), which is represented on the number line as shown in the figure:\n\n\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem tests the understanding of how points change during translation, representing the solution set of an inequality group on a number line, solving a system of linear inequalities, and the coordinates of points. The key breakthrough in this problem is setting up the inequality group." }, { "problem_id": 147, "question": "For the problem: As shown in Figure 1, in a plane, there is a right triangle with legs of lengths 2 and 5 inside a square. The triangle can be freely moved within the square and along its boundary by \"translating\" (i.e., translating or rotating). Determine the smallest integer value $n$ for the side length of the square. Students A and B each constructed a square with the smallest possible side length, first calculating the side length $x$, then taking the smallest integer value $n$.\n\nFigure 1\n\n\nFigure 2\n\n\nFigure 3\n\n A: As shown in Figure 2, when $x$ is the length of the hypotenuse of the right triangle, it can be translated. The result is $n=6$.\n\n B: As shown in Figure 3, when $x$ is $\\frac{\\sqrt{2}}{2}$ times the sum of the lengths of the two legs of the right triangle, it can be translated. The result is $n=5$. Which of the following statements is correct?\nA. Student A's approach is correct, but his value of $n$ is incorrect.\nB. Student A's approach and his value of $n$ are both correct.\nC. Student B's approach is correct, but his value of $n$ is incorrect.\nD. The approaches of both Student A and Student B are incorrect.", "input_image": [ "batch14-2024_06_15_a76a212ead7e23c003beg_0007_1.jpg", "batch14-2024_06_15_a76a212ead7e23c003beg_0007_2.jpg", "batch14-2024_06_15_a76a212ead7e23c003beg_0007_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the right-angled triangle has side lengths of 2 and 5,\n\nTherefore, the length of the hypotenuse of the right-angled triangle is: $\\sqrt{2^{2}+5^{2}}=\\sqrt{29}$.\n\nSince the right-angled triangle can freely transform from a horizontal to a vertical position within and on the boundary of the square through translation or rotation,\n\nTherefore, the side length of the square must be no less than $\\sqrt{29}$.\n\nSince $5<\\sqrt{29}<6$,\n\nTherefore, the smallest integer $n$ for the side length of the square is 6.\n\nThus, Person A's reasoning is correct; the hypotenuse is the longest side, and as long as the hypotenuse can pass through, $n=6$.\n\nPerson B's reasoning and calculations are both incorrect, as the illustrated situation does not represent the longest side.\n\nHence, the correct choice is: B.\n\n[Key Insight] This problem examines the properties of translation and rotation, the Pythagorean theorem, and the estimation of irrational numbers. Proficiency in applying the Pythagorean theorem and estimating irrational numbers is crucial for solving the problem." }, { "problem_id": 148, "question": "Go, originating in China, was historically referred to as \"wan\" and has a history of over 4,000 years. Which of the following patterns formed by black and white chess pieces is both an axisymmetric and a centrosymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_a76a212ead7e23c003beg_0055_1.jpg", "batch14-2024_06_15_a76a212ead7e23c003beg_0055_2.jpg", "batch14-2024_06_15_a76a212ead7e23c003beg_0055_3.jpg", "batch14-2024_06_15_a76a212ead7e23c003beg_0055_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. The item is neither an axisymmetric figure nor a centrally symmetric figure;\n\nB. The item is an axisymmetric figure but not a centrally symmetric figure;\n\nC. The item is not an axisymmetric figure but is a centrally symmetric figure;\n\nD. The item is both an axisymmetric figure and a centrally symmetric figure.\n\nTherefore, the correct choice is D.\n\n[Key Point] This question tests the concepts of axisymmetric and centrally symmetric figures. Familiarity with these concepts is crucial for solving the problem." }, { "problem_id": 149, "question": "Yongzhuo Temple Twin Towers, also known as the Lingshao Twin Towers, are the tallest buildings among the existing ancient structures in Taiyuan, Shanxi Province. Each of the thirteen stories is an octagonal pavilion-style hollow brick tower, as shown in Figure 1. As depicted in Figure 2, the octagon represents the floor plan of one of the stories. The measure of each interior angle of this octagon is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $80^{\\circ}$\nB. $100^{\\circ}$\nC. $120^{\\circ}$\nD. $135^{\\circ}$", "input_image": [ "batch14-2024_06_15_c3477cdfc979a549e610g_0001_1.jpg", "batch14-2024_06_15_c3477cdfc979a549e610g_0001_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: Since the sum of the exterior angles of a polygon is $360^{\\circ}$, and an octagon is a regular polygon,\n\nTherefore, each exterior angle of a regular octagon is $360^{\\circ} \\div 8=45^{\\circ}$,\n\nHence, the measure of each interior angle of a regular octagon is $180^{\\circ}-45^{\\circ}=135^{\\circ}$.\n\nThus, the correct choice is: D.\n\n[Highlight] This question tests the knowledge of the interior and exterior angles of polygons. In a regular polygon, each interior angle is equal, and each exterior angle is equal. The key to solving the problem lies in understanding the sum of interior angles, the sum of exterior angles, and the properties of regular polygons." }, { "problem_id": 150, "question": "The following are rhombuses drawn by 4 students. According to the marked data, the one that is not necessarily a rhombus is ()\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n", "input_image": [ "batch14-2024_06_15_c3477cdfc979a549e610g_0049_1.jpg", "batch14-2024_06_15_c3477cdfc979a549e610g_0049_2.jpg", "batch14-2024_06_15_c3477cdfc979a549e610g_0049_3.jpg", "batch14-2024_06_15_c3477cdfc979a549e610g_0049_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution:\n\nOption A: Since all four sides of the quadrilateral are 2, and a quadrilateral with all sides equal is a rhombus, therefore option A is correct.\n\nOption B: Because three sides of the quadrilateral are 2, the adjacent sides are equal. Since \\(50^\\circ + 130^\\circ = 180^\\circ\\), one pair of opposite sides of the quadrilateral are parallel. However, since the other pair of opposite sides are both 2, it cannot be proven that the quadrilateral is a parallelogram. Therefore, a quadrilateral with one pair of adjacent sides equal is not necessarily a rhombus, so option B is incorrect.\n\nOption C: Because the adjacent sides of the quadrilateral are both 2, they are equal. Since the sum of the interior angles of a quadrilateral is \\(360^\\circ\\), the remaining angle is \\(50^\\circ\\), making the two pairs of opposite angles equal. Therefore, the quadrilateral is a parallelogram. According to the property that a parallelogram with one pair of adjacent sides equal is a rhombus, option C is correct.\n\nOption D: Because \\(60^\\circ + 120^\\circ = 180^\\circ\\), one pair of opposite sides of the quadrilateral are parallel. Since this pair of opposite sides are both 2, the quadrilateral is a parallelogram. Because the adjacent sides are both 2, according to the property that a parallelogram with one pair of adjacent sides equal is a rhombus, option D is correct.\n\n[Key Insight] This question tests the determination of a rhombus. A thorough understanding of the four determining theorems of a rhombus is crucial for solving the problem." }, { "problem_id": 151, "question": "As shown in the figure, Xiaocong used the set of Tangram in Figure 1 to form the \"bird\" as shown in Figure 2. Given that the side length of the square $A B C D$ is 4, the distance between points $E$ and $F$ in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\sqrt{26}$\nB. $2 \\sqrt{13}$\nC. $\\sqrt{10}$\nD. 4", "input_image": [ "batch14-2024_06_15_c3477cdfc979a549e610g_0058_1.jpg", "batch14-2024_06_15_c3477cdfc979a549e610g_0058_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $EG \\perp FH$ extending $FH$ to point $G$,\n\n\n\nFigure 2\n\nGiven that the side length of square $ABCD$ is 4,\n\n$\\therefore AC=4 \\sqrt{2}$,\n\n$\\therefore$ The side length of the square where point $E$ is located is $\\frac{1}{4} \\times 4 \\sqrt{2}=\\sqrt{2}$,\n\n$\\therefore EG=HG=\\frac{\\sqrt{2}}{\\sqrt{2}}=1$,\n\nFrom the properties of the tangram and the square, we know: $FG=GH+FH=1+4=5$,\n\nIn right triangle $\\triangle FEG$, by the Pythagorean theorem, $EF=\\sqrt{1^{2}+5^{2}}=\\sqrt{26}$, thus option A is correct.\n\nTherefore, the correct choice is: A.\n\n[Insight] This problem mainly examines the properties of squares, the characteristics of the tangram, and the Pythagorean theorem. The key to solving the problem lies in using the characteristics of the tangram to determine that $EG=1$ and $FG=1+4=5$." }, { "problem_id": 152, "question": "As shown in the figure, on a grid paper, triangle $A O B$ is rotated clockwise around point $O$ by $90^{\\circ}$ to form triangle $A^{\\prime} O B^{\\prime}$. Which of the following four figures is correct?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_c4cc2d5f01acdabd3ae3g_0012_1.jpg", "batch14-2024_06_15_c4cc2d5f01acdabd3ae3g_0012_2.jpg", "batch14-2024_06_15_c4cc2d5f01acdabd3ae3g_0012_3.jpg", "batch14-2024_06_15_c4cc2d5f01acdabd3ae3g_0012_4.jpg", "batch14-2024_06_15_c4cc2d5f01acdabd3ae3g_0012_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Rotate triangle $AOB$ $90^\\circ$ clockwise around point $O$ to obtain triangle $A^{\\prime}OB^{\\prime}$.\n\nThe correct figure is:\n\n\n\nNone of the other figures meet the requirements of the problem.\n\nTherefore, the answer is: A.\n\n[Key Insight] This question tests the understanding of rotational transformations. The key to solving it lies in determining the direction and degree of rotation." }, { "problem_id": 153, "question": "Among the following figures, which one is both an axisymmetric figure and a centrosymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_c672595237947d6cada1g_0044_1.jpg", "batch14-2024_06_15_c672595237947d6cada1g_0044_2.jpg", "batch14-2024_06_15_c672595237947d6cada1g_0044_3.jpg", "batch14-2024_06_15_c672595237947d6cada1g_0044_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "A. It is neither an axisymmetric figure nor a centrally symmetric figure, so this option does not meet the requirement;\n\nB. It is an axisymmetric figure but not a centrally symmetric figure, so this option meets the requirement;\n\nC. It is not an axisymmetric figure but is a centrally symmetric figure, so this option does not meet the requirement;\n\nD. It is both an axisymmetric figure and a centrally symmetric figure, so this option does not meet the requirement; Therefore, the correct choice is: D.\n\n[Highlight] This question tests the concepts of centrally symmetric figures and axisymmetric figures: The key to identifying an axisymmetric figure is to find the axis of symmetry, where the two parts of the figure can coincide when folded along the axis; A centrally symmetric figure requires finding the center of symmetry, where the figure coincides with the original after a 180-degree rotation." }, { "problem_id": 154, "question": "As shown in the figure, (1) is a type of window grilles found in ancient Chinese architecture, and the ice crack pattern symbolizes the appearance of cracks in the ice and its beginning to melt. The shape follows no specific rule, representing a natural and harmonious beauty. In Figure (2), a graph composed of five line segments is extracted from the ice crack window grilles pattern in Figure (1). Then, $\\angle 1 + \\angle 2 + \\angle 3 + \\angle 4 + \\angle 5 = $ ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $72^{\\circ}$\nB. $108^{\\circ}$\nC. $360^{\\circ}$\nD. $540^{\\circ}$", "input_image": [ "batch14-2024_06_15_ce590429da467201533fg_0006_1.jpg", "batch14-2024_06_15_ce590429da467201533fg_0006_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since the sum of the exterior angles of a polygon equals $360^{\\circ}$, we know that\n\n$\\angle 1 + \\angle 2 + \\angle 3 + \\angle 4 + \\angle 5 = 360^{\\circ}$.\n\nTherefore, the correct choice is: C.\n【Key Point】This question tests the understanding of the interior and exterior angles of a polygon. The key to solving the problem is to grasp that the sum of the exterior angles of a polygon is equal to $360^{\\circ}$." }, { "problem_id": 155, "question": "As shown in Figure 1, the Yingxian Wooden Pagoda is located in Yingxian County, Shuozhou City, Shanxi Province, and it is the oldest and tallest pure wooden structured pavilion-type building in China. It has been measured that the pagoda is built on a platform about four meters high, with the bottom layer of the platform designed in the shape of a regular polygon. As shown in Figure 2, a part of the regular polygon at the bottom layer of the platform is depicted, with its exterior angles being $45^\\circ$. The regular polygon is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Regular pentagon\nB. Regular hexagon\nC. Regular heptagon\nD. Regular octagon", "input_image": [ "batch14-2024_06_15_ce590429da467201533fg_0010_1.jpg", "batch14-2024_06_15_ce590429da467201533fg_0010_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the number of sides of the regular polygon be $n$. According to the theorem of the sum of exterior angles of a regular polygon, we have $45n = 360$.\n\nSolving for $n$ gives: $n = 8$.\n\nTherefore, the regular polygon is a regular octagon.\n\nHence, the correct choice is: D.\n\n[Key Insight] This question tests the understanding of the sum of exterior angles of a regular polygon. Mastering the theorem of the sum of exterior angles of a regular polygon is crucial for solving the problem." }, { "problem_id": 156, "question": "Using a wide and long enough strip of paper, tie a knot as shown in Figure 1, and then gently pull it tight and flatten it to obtain the regular pentagon $A B C D E$ as shown in Figure 2, where $\\angle A F B = ($ )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $54^{\\circ}$\nB. $63^{\\circ}$\nC. $72^{\\circ}$\nD. $81^{\\circ}$", "input_image": [ "batch14-2024_06_15_ce590429da467201533fg_0014_1.jpg", "batch14-2024_06_15_ce590429da467201533fg_0014_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since the sum of the interior angles of a regular pentagon $ABCDE$ is $180^{\\circ}(5-2)=540^{\\circ}$, and $AB = AE$, $AE = ED$,\n\nTherefore, $\\angle BAE = \\angle AED = \\frac{540^{\\circ}}{5} = 108^{\\circ}$.\nSince $AB = AE$, $AE = ED$,\n\n$\\angle AEB = \\angle ABE = \\frac{1}{2}\\left(180^{\\circ} - \\angle EAB\\right) = 36^{\\circ}$, and $\\angle EAD = \\angle ADE = \\frac{1}{2}\\left(180^{\\circ} - \\angle AED\\right) = 36^{\\circ}$.\n\nThus, $\\angle EAF = \\angle AEF = 36^{\\circ}$,\n\nTherefore, $\\angle AFB = \\angle EAF + \\angle AEF = 72^{\\circ}$.\n\nHence, the correct choice is: C.\n\n[Key Insight] This problem primarily examines the properties of a regular pentagon and an isosceles triangle, where an exterior angle of a triangle equals the sum of the two non-adjacent interior angles. The key to solving this problem lies in the proficient application of the properties of a regular pentagon and an isosceles triangle." }, { "problem_id": 157, "question": "The interior angles of the following polygons add up to $720^\\circ$ are ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_ce590429da467201533fg_0017_1.jpg", "batch14-2024_06_15_ce590429da467201533fg_0017_2.jpg", "batch14-2024_06_15_ce590429da467201533fg_0017_3.jpg", "batch14-2024_06_15_ce590429da467201533fg_0017_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Let the sum of the interior angles of an $n$-sided polygon be $720^{\\circ}$,\n\nthen $(n-2) \\times 180^{\\circ}=720^{\\circ}$\n\nSolving for $n$: $n=6$,\n\n$\\therefore$ The sum of the interior angles of a hexagon is $720^{\\circ}$,\n\nTherefore, the correct choice is: C\n\n[Key Insight] This problem examines the calculation of the sum of interior angles of a polygon. The sum of the interior angles of an $n$-sided polygon is $(n-2) \\times 180^{\\circ}$." }, { "problem_id": 158, "question": "As shown in the figure, Xiao Fan cut off a portion (dashed lines) from several hexagonal paper pieces to form a new polygon. If the sum of the interior angles of the new polygon is twice the sum of its exterior angles, then the corresponding figure is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_ce590429da467201533fg_0019_1.jpg", "batch14-2024_06_15_ce590429da467201533fg_0019_2.jpg", "batch14-2024_06_15_ce590429da467201533fg_0019_3.jpg", "batch14-2024_06_15_ce590429da467201533fg_0019_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Let the number of sides of the new polygon be \\( n \\) (where \\( n \\geq 3 \\)).\n\nAccording to the problem, we have:\n\\[ 180^{\\circ}(n - 2) = 2 \\times 360^{\\circ} \\]\n\nSolving for \\( n \\):\n\\[ n - 2 = \\frac{2 \\times 360^{\\circ}}{180^{\\circ}} \\]\n\\[ n - 2 = 4 \\]\n\\[ n = 6 \\]\n\nTherefore, the number of sides of the new polygon is 6.\n\nThe correct answer is: **B**.\n\n**Key Insight**: This problem tests the understanding of the interior and exterior angles of polygons. Mastering the formulas for the sum of interior and exterior angles is crucial for solving such problems." }, { "problem_id": 159, "question": "As shown in the figure, while exploring the number of triangles formed by drawing diagonals from a vertex of a polygon, the following diagrams were created:\n\n\n\nFirst Diagram\n\n\n\nSecond Diagram\n\n\n\nThird Diagram\n\nBased on the diagrams, the number of triangles formed by drawing diagonals from a vertex of an $n$-sided polygon is ( )\nA. $(n-3)$\nB. $(n-2)$\nC. $(n-1)$\nD. $(n+1)$", "input_image": [ "batch14-2024_06_15_ce590429da467201533fg_0063_1.jpg", "batch14-2024_06_15_ce590429da467201533fg_0063_2.jpg", "batch14-2024_06_15_ce590429da467201533fg_0063_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By observing the figures, we can deduce the following:\n\nIn the first figure, drawing 1 diagonal from one vertex of a quadrilateral divides it into 2 triangles. In the second figure, drawing 2 diagonals from one vertex of a pentagon divides it into 3 triangles. In the third figure, drawing 3 diagonals from one vertex of a hexagon divides it into 4 triangles.\n\n...\n\nIn the $(n-3)$th figure, drawing $(n-3)$ diagonals from one vertex of an $n$-sided polygon divides it into $(n-2)$ triangles.\n\nTherefore, the correct answer is: B.\n\n[Key Insight] This question tests the ability to identify patterns in geometric transformations. Carefully observing the figures and identifying the pattern of change is crucial to solving the problem." }, { "problem_id": 160, "question": "As shown in the figure, the polygon is divided into triangles. In Figure (1), 2 triangles can be formed, in Figure (2), 3 triangles can be formed, and in Figure (3), 4 triangles can be formed. Based on this, you can infer that an $n$-sided polygon can be divided into ( ) triangles.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. $(n-2)$ triangles\nB. $(n-1)$ triangles\nC. $n$ triangles\nD. infinitely many triangles", "input_image": [ "batch14-2024_06_15_ce590429da467201533fg_0065_1.jpg", "batch14-2024_06_15_ce590429da467201533fg_0065_2.jpg", "batch14-2024_06_15_ce590429da467201533fg_0065_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Analyzing the pattern based on the figures:\n\n(1) A triangle is divided into two triangles;\n\n(2) A quadrilateral is divided into three triangles;\n\n(3) Following this pattern, an n-sided polygon is divided into (n-1) triangles.\n\nThat is, an n-sided polygon can be divided into (n-1) triangles.\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question examines the problem of polygons. The key to solving it lies in analyzing specific numerical values to identify the pattern." }, { "problem_id": 161, "question": "Little student Lei Lei wants to make an open-top box with an equilateral triangular base and equal height using an equilateral triangular cardboard (as shown in Figure (a)). Inside triangle $A B C$, he first drew another equilateral triangle $D E F$, and then planned to cut off three corners (such as quadrilateral $A M D N$). However, after pondering for a while, he still doesn't know how to proceed. Using the knowledge you've learned, determine that if Lei Lei wants to exactly cut off three corners, the measure of $\\angle M D N$ should be ( )\n\n\n\n(a)\n\n\n\n(b)\nA. $100^{\\circ}$\nB. $110^{\\circ}$\nC. $120^{\\circ}$\nD. $130^{\\circ}$", "input_image": [ "batch14-2024_06_15_d000ab92c0a58289133dg_0058_1.jpg", "batch14-2024_06_15_d000ab92c0a58289133dg_0058_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Since $\\triangle ABC$ is an equilateral triangle,\n\n$\\therefore \\angle A=60^{\\circ}$.\n\nWe need to make a lidless box with an equilateral triangle base and equal height.\n\n$\\therefore \\angle AMD=\\angle DNA=90^{\\circ}$.\n\nFrom the sum of the interior angles of a quadrilateral, we get: $\\angle MDN=360^{\\circ}-\\angle AMD-\\angle DNA-\\angle A$\n\n$=360^{\\circ}-90^{\\circ}-90^{\\circ}-60^{\\circ}$\n\n$=120^{\\circ}$.\n\nTherefore, the answer is: C.\n\n【Key Insight】This question tests the sum of the interior angles of a quadrilateral and the definition of an equilateral triangle. According to the problem, deducing that $\\angle AMD=\\angle DNA=90^{\\circ}$ is the key to solving the problem." }, { "problem_id": 162, "question": "In Figure 1, a hexagonal ring is given, where $S = \\angle A_{1} + \\angle A_{2} + \\ldots + \\angle A_{6} = 360^\\circ$. In Figure 2, a square ring is given, where $S = \\angle A_{1} + \\angle A_{2} + \\ldots + \\angle A_{8} = 720^\\circ$. In Figure 3, a pentagonal ring is given, where $S = \\angle A_{1} + \\angle A_{2} + \\ldots + \\angle A_{10} = 1080^\\circ \\ldots$ \n\nSmart students, please directly write the value of $S$, the sum of the angles in a decagonal ring, as $(\\quad)$ degrees.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. 1440\nB. 1800\nC. 2880\nD. 3600", "input_image": [ "batch14-2024_06_15_d000ab92c0a58289133dg_0072_1.jpg", "batch14-2024_06_15_d000ab92c0a58289133dg_0072_2.jpg", "batch14-2024_06_15_d000ab92c0a58289133dg_0072_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: According to the problem, for a two-ring triangle, the sum of interior angles \\( S = 360 \\) degrees;\n\nFor a two-ring quadrilateral, \\( S = 720 = 360 \\times 2 = 360 \\times (4 - 2) \\) degrees;\n\nFor a two-ring pentagon, \\( S = 1080 = 360 \\times 3 = 360 \\times (5 - 2) \\) degrees;\n\n\\(\\cdots\\)\n\n\\(\\therefore\\) For a two-ring decagon, \\( S = 360 \\times (10 - 2) = 2880 \\) degrees.\n\nTherefore, the correct answer is: \\( C \\).\n\n【Key Insight】This problem examines the sum of interior angles of polygons. The pattern can be directly identified based on the given values of \\( S \\), and then applied to determine the sum of interior angles for a two-ring decagon." }, { "problem_id": 163, "question": "The window in Figure A is a common type found in daily life. If the window is opened from a fully closed position to the position in Figure A, without considering the aluminum frame (as shown in Figure B), the relationship between the overlapping area of the two glass panels, $S_{1}$, and the total ventilation area of the window, $S_{2}$, is correctly described as ( )\n\n\n\nA\n\n\nA. $S_{1}>S_{2}$\nB. $S_{1}=S_{2}$\nC. $S_{1}\n\nA\n\n\n\nB\n\n【Key Insight】This question tests the properties of translation, and correctly identifying the figure is crucial to solving the problem." }, { "problem_id": 164, "question": "Read the following material:\n\nTheorem: The median of a triangle is parallel to the third side and is half the length of the third side. Given: In triangle $A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$, respectively.\n\n\n\nProve: $D E \\parallel B C$, and $D E = \\frac{1}{2} B C$.\n\nProof: Extend $D E$ to point $F$ such that $E F = D E$, and connect $C F, \\ldots$\n\nPerson A and Person B have the following ideas for the subsequent proof:\n\nPerson A: As shown in Figure 1, first prove that $A D E \\cong C F E$, then deduce that quadrilateral $D B C F$ is a parallelogram.\n\nPerson B: As shown in Figure 2, connect $D C$ and $A F$. Prove that quadrilaterals $A D C F$ and $D B C F$ are parallelograms, respectively.\n\nWhich of the following judgments is correct ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Person A's idea is correct, and Person B's idea is incorrect.\nB. Person A's idea is incorrect, and Person B's idea is correct.\nC. Both Person A and Person B's ideas are correct.\nD. Both Person A and Person B's ideas are incorrect.", "input_image": [ "batch14-2024_06_15_e140a62ba5fbff1efa45g_0017_1.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0017_2.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0017_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to Person A's approach, the proof is as follows:\n\nExtend $DE$ to point $F$ such that $EF = DE$, and connect $CF$, as shown in Figure 1.\n\n\n\nFigure 1\n\nSince $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively,\n\n$\\therefore AD = BD$, $AE = CE$.\n\nIn triangles $ADE$ and $CFE$,\n\n\\[\n\\left\\{\n\\begin{aligned}\nDE &= EF, \\\\\n\\angle AED &= \\angle CEF, \\\\\nAE &= CE\n\\end{aligned}\n\\right.\n\\]\n\n$\\therefore \\triangle ADE \\cong \\triangle CFE$ (by SAS congruence),\n\n$\\therefore AD = CF$, $\\angle A = \\angle ACF$,\n\n$\\therefore CF = BD$, and $CF \\parallel BD$,\n\n$\\therefore$ quadrilateral $DBCF$ is a parallelogram,\n\n$\\therefore DF \\parallel BC$, and $DF = BC$,\n\nAlso, since $DE = \\frac{1}{2} DF$,\n\n$\\therefore DE \\parallel BC$, and $DE = \\frac{1}{2} BC$.\n\nAccording to Person B's approach, the proof is as follows:\n\nAs shown in Figure 2, extend $DE$ to point $F$ such that $EF = DE$, and connect $CF$, $DC$, and $AF$.\n\n\n\nFigure 2\n\nSince $D$ and $E$ are the midpoints of sides $AB$ and $AC$, respectively,\n\n$\\therefore AD = BD$, $AE = CE$.\n\nSince $AE = CE$ and $DE = EF$,\n\n$\\therefore$ quadrilateral $ADCF$ is a parallelogram, $CF \\parallel DA$, and $CF = DA$,\n\n$\\therefore CF \\parallel BD$, and $CF = BD$,\n\n$\\therefore$ quadrilateral $DBCF$ is a parallelogram,\n\n$\\therefore DF \\parallel BC$, and $DF = BC$,\n\nAlso, since $DE = \\frac{1}{2} DF$,\n\n$\\therefore DE \\parallel BC$, and $DE = \\frac{1}{2} BC$.\n\nIn conclusion, both Person A and Person B's approaches are correct.\n\nTherefore, the correct answer is: C\n\n[Key Insight] This problem tests the knowledge of the properties and criteria of congruent triangles and parallelograms. Mastering the properties and criteria of congruent triangles and parallelograms is crucial for solving such problems." }, { "problem_id": 165, "question": "After cutting a quadrilateral paper along a straight line, which of the following pairs of resulting figures have equal interior angle sums?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_e140a62ba5fbff1efa45g_0019_1.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0019_2.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0019_3.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0019_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nFor option $\\mathrm{A}$, after cutting, the two shapes formed are a triangle and a quadrilateral, and their internal angle sums are not equal. Therefore, option $\\mathrm{A}$ is incorrect and does not meet the requirement.\n\nFor option $\\mathrm{B}$, after cutting, the two shapes formed are a triangle and a quadrilateral, and their internal angle sums are not equal. Therefore, option $\\mathrm{B}$ is incorrect and does not meet the requirement.\n\nFor option $\\mathrm{C}$, after cutting, the two shapes formed are a triangle and a quadrilateral, and their internal angle sums are not equal. Therefore, option $\\mathrm{C}$ is incorrect and does not meet the requirement.\n\nFor option $\\mathrm{D}$, after cutting, both shapes formed are quadrilaterals, and their internal angle sums are equal. Therefore, option $\\mathrm{D}$ is correct and meets the requirement.\n\nHence, the correct choice is: D.\n\n[Key Insight] This question primarily tests the theorem of the sum of internal angles of polygons. Understanding and mastering the theorem and calculation method for the sum of internal angles of polygons is crucial for solving the problem." }, { "problem_id": 166, "question": "As shown in the figure, there are two squares, $A$ and $B$. If these two squares are stacked together, we get Figure (1), where the shaded area is 1. If $A$ and $B$ are placed side by side to form a new square, we get Figure (2), where the shaded area is 24. What is the area of the newly constructed square?\n\n\n\n\nFigure (1)\n\n\nFigure (2)\nA. 49\nB. 65\nC. 78\nD. 97", "input_image": [ "batch14-2024_06_15_e140a62ba5fbff1efa45g_0059_1.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0059_2.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0059_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the side length of square $A$ be $a$, and the side length of square $B$ be $b$. From the diagram, we can see that the side length of the small square in (1) is $b - a$, and its area is 1.\n\nThus, $(b - a)^2 = 1$.\n\nSince $b > a$, we have $b - a = 1$.\n\nFrom the diagram, we can see that the side length of the newly constructed square in (2) is $a + b$.\n\nTherefore, its area is $(a + b)^2$.\n\nSo, $(a + b)^2 - a^2 - b^2 = 24$.\n\nThis simplifies to $ab = 12$.\n\nWe have the system of equations:\n\\[\n\\begin{cases}\nb - a = 1 \\\\\nab = 12\n\\end{cases}\n\\]\nSolving this system, we get $a = 3$ or $a = -4$ (which is discarded).\n\nWhen $a = 3$, $b = 4$.\n\nTherefore, the area of the newly constructed square is $(a + b)^2 = (3 + 4)^2 = 49$.\n\nHence, the correct answer is: A.\n\n【Key Insight】This problem tests the application of the perfect square formula. Mastering the perfect square formula and its transformations is key to solving the problem." }, { "problem_id": 167, "question": "In the figure, there are shapes formed by connecting 1, 2, and $n$ (where $n$ is a positive integer) squares. In Figure 1, $x = 70^\\circ$; in Figure 2, $y = 28^\\circ$. Based on these calculations, express the sum $a + b + c + \\ldots + d$ in Figure 3 as a function of $n$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $45^\\circ n$\nB. $90^\\circ n$\nC. $135^\\circ n$\nD. $180^\\circ n$", "input_image": [ "batch14-2024_06_15_e2d05115e255d3be2d31g_0068_1.jpg", "batch14-2024_06_15_e2d05115e255d3be2d31g_0068_2.jpg", "batch14-2024_06_15_e2d05115e255d3be2d31g_0068_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect the diagonals of each small square,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nIn Figure 1, $61^{\\circ}+119^{\\circ}+20^{\\circ}+x+45^{\\circ} \\times 2=360^{\\circ}$,\n\nwhich simplifies to $20^{\\circ}+x=90^{\\circ}$,\n\nIn Figure 2, $61^{\\circ}+119^{\\circ}+31^{\\circ}+121^{\\circ}+y+45^{\\circ} \\times 4=360^{\\circ}+180^{\\circ}$,\n\nwhich simplifies to $31^{\\circ}+121^{\\circ}+y=180^{\\circ}=2 \\times 90^{\\circ}$,\n\n$a+b+c+\\ldots+d=n \\times 90^{\\circ}=90^{\\circ} n$,\nTherefore, the answer is B.\n\n【Key Insight】This problem examines the formula for the sum of interior angles of a polygon and the property that the diagonals of a square bisect a pair of opposite angles. The key to solving the problem lies in constructing the polygon by drawing auxiliary lines." }, { "problem_id": 168, "question": "As shown in Figure 1, two right-angled triangles $A B C$ and $D E F$ are placed as depicted in Figure 2. Given that $\\angle F = 30^\\circ$, in the position shown in Figure 3, $D E F$ rotates counterclockwise around point $D$ until $D F$ coincides with $B C$. During the rotation, when $E F$ is parallel to the sides of $A B C$, the angle of rotation is (1) $30^\\circ$; (2) $45^\\circ$; (3) $75^\\circ$; (4) $135^\\circ$; (5) $165^\\circ$. Which of these are correct? ( )\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $(1)(3)(4)$\nB. $(1)(3)(5)$\nC. $(1)(4)(5)$\nD. $(2)(3)(5)$", "input_image": [ "batch14-2024_06_15_eb2de245c7c39ee503bdg_0044_1.jpg", "batch14-2024_06_15_eb2de245c7c39ee503bdg_0044_2.jpg", "batch14-2024_06_15_eb2de245c7c39ee503bdg_0044_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: When \\( EF \\parallel BC \\),\n\n\n\n\\(\\therefore \\angle FDC = \\angle F = 30^\\circ\\), which means \\( DEF \\) is rotated \\( 30^\\circ \\) counterclockwise around point \\( D \\);\n\nWhen \\( EF \\parallel AB \\),\n\n\n\n\\(\\therefore \\angle E = \\angle AGD\\),\n\n\\(\\because \\angle F = 30^\\circ, \\angle EDF = 90^\\circ\\),\n\n\\(\\therefore \\angle E = 60^\\circ = \\angle AGD\\),\n\nAlso, \\(\\angle B = 45^\\circ\\),\n\n\\(\\therefore \\angle BDG = \\angle AGD - \\angle B = 15^\\circ\\),\n\n\\(\\therefore \\angle CDF = 180^\\circ - \\angle BDG - \\angle EDF = 75^\\circ\\), which means \\( DEF \\) is rotated \\( 75^\\circ \\) counterclockwise around point \\( D \\);\n\nWhen \\( EF \\parallel AC \\),\n\n\n\nExtend \\( CB \\) to intersect \\( EF \\) at point \\( G \\),\n\n\\(\\therefore \\angle EGC = \\angle C = 45^\\circ\\),\n\n\\(\\therefore \\angle FDG = \\angle EGD - \\angle F = 15^\\circ\\),\n\n\\(\\therefore \\angle CDF = 180^\\circ - \\angle FDG = 165^\\circ\\), which means \\( DEF \\) is rotated \\( 165^\\circ \\) counterclockwise around point \\( D \\);\n\nIn summary, when \\( DEF \\) is rotated \\( 30^\\circ \\) or \\( 75^\\circ \\) or \\( 165^\\circ \\) counterclockwise around point \\( D \\), \\( EF \\) is parallel to one of the sides of \\( ABC \\). Therefore, the correct answer is: B.\n\n【Key Insight】This question examines the properties of rotation and parallel lines. Mastering the properties of rotation is crucial for solving this problem." }, { "problem_id": 169, "question": "As shown in the figure, in the rectangle $A B C D$, a moving point $P$ starts from point $B$ and moves along $B C, C D, D A$ until it reaches point $A$. Let the distance traveled by point $P$ be $x$, and the area of triangle $A B P$ be $y$. The graph of $y$ against $x$ is shown in Figure 2. If $b - 2a = 5$, then the perimeter of rectangle $A B C D$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 20\nB. 18\nC. 16\nD. 24", "input_image": [ "batch14-2024_06_15_f1c27c1996e9a93f4957g_0022_1.jpg", "batch14-2024_06_15_f1c27c1996e9a93f4957g_0022_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the point $(a, 10)$ in Figure 2, we know that $BC = a$ and $\\frac{1}{2} AB \\times BC = 10$.\n\n$\\therefore AB = \\frac{20}{a}$,\n$\\therefore BC + CD + DA = 2a + \\frac{20}{a} = b$,\n\n$\\therefore b - 2a = \\frac{20}{a}$,\n\n$\\because b - 2a = 5$,\n\n$\\therefore \\frac{20}{a} = 5$,\n\n$\\therefore a = 4$,\n\n$\\therefore AB = 5, \\quad BC = 4$,\n\n$\\therefore$ The perimeter of rectangle $ABCD$ is $2 \\times (5 + 4) = 18$.\n\nTherefore, the correct answer is: B.\n\n[Key Insight] This problem mainly examines the function graph of moving point problems. The key to solving this problem is to determine the lengths of the relevant line segments from the function graph and then calculate the perimeter of the rectangle." }, { "problem_id": 170, "question": "In rectangle $A B C D$ as shown in Figure 1, point $E$ is the midpoint of $B C$. Point $P$ moves along $B C$ from point $B$ to point $C$. Let the distance between points $B$ and $P$ be $x$, and let $y$ be the difference between $P A$ and $P E$. The graph in Figure 2 shows the relationship between $y$ and $x$ as point $P$ moves. What is the length of $B C$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 6\nB. 7\nC. 8\nD. 9", "input_image": [ "batch14-2024_06_15_f1c27c1996e9a93f4957g_0047_1.jpg", "batch14-2024_06_15_f1c27c1996e9a93f4957g_0047_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the function graph, we know that when $x=0$, i.e., when point $P$ is at point $B$, $BA - BE = 1$.\n\nUsing the principle that the shortest distance between two points is a straight line, we obtain $PA - PE \\leq AE$.\n\nTherefore, the maximum value of $y$ is $AE$,\n\nHence, $AE = 5$.\n\nIn the right triangle $\\triangle ABE$, by the Pythagorean theorem, we have: $BA^{2} + BE^{2} = AE^{2} = 25$,\n\nLet the length of $BE$ be $t$,\n\nThen, $AB = t + 1$,\n\nThus, $(t + 1)^{2} + t^{2} = 25$,\n\nWhich simplifies to $t^{2} + t - 12 = 0$,\n\nFactoring gives $(t - 3)(t + 4) = 0$,\n\nSolving yields $t = 3$ or $t = -4$,\n\nSince $t > 0$,\nTherefore, $t = 3$.\n\nHence, $BE = 3$,\n\nSince point $E$ is the midpoint of $BC$,\n\nTherefore, $BC = 6$\n\nThus, the correct choice is: A.\n\n【Key Insight】This problem examines the function graph of a moving point. Effectively extracting information from the graph is crucial for solving the problem." }, { "problem_id": 171, "question": "As shown in the figure, in right triangle $\\triangle A B C$, $\\angle C$ is a right angle, $D E$ is the median, and point $P$ starts from point $D$ and moves along the direction $D \\rightarrow C \\rightarrow B$ at a speed of $1.5 \\mathrm{~cm} / \\mathrm{s}$ until it reaches point $B$. Figure 2 shows the graph of the area $y\\left(\\mathrm{~cm}^{2}\\right)$ of $\\triangle D E P$ against time $x(\\mathrm{~s})$ as point $P$ moves. What is the value of $a$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 3\nB. $\\frac{3}{2}$\nC. $\\frac{4}{3}$\nD. 4.5", "input_image": [ "batch14-2024_06_15_f1c27c1996e9a93f4957g_0048_1.jpg", "batch14-2024_06_15_f1c27c1996e9a93f4957g_0048_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the graph, it is observed that when point \\( P \\) moves on \\( BC \\), the area of \\( \\triangle DEP \\) remains unchanged.\n\n\\[\n\\therefore BC = (a + 4 - a) \\times 1.5 = 6 \\text{ cm}.\n\\]\n\nSince \\( DE \\) is the midline,\n\n\\[\nDE = \\frac{1}{2} BC = 3 \\text{ cm}.\n\\]\n\nWhen point \\( P \\) is on segment \\( DC \\), the area of \\( \\triangle DEP \\) is\n\n\\[\nS_{\\triangle DEP} = \\frac{1}{2} DE \\cdot PD = \\frac{1}{2} \\times 3 \\times 1.5x.\n\\]\n\nFrom the graph, when point \\( P \\) coincides with point \\( C \\), i.e., when \\( x = a \\), the area of \\( \\triangle DEP \\) is 3.\n\n\\[\n\\therefore \\frac{1}{2} \\times 3 \\times 1.5a = 3.\n\\]\n\nSolving for \\( a \\):\n\n\\[\na = \\frac{4}{3}.\n\\]\n\nTherefore, the correct answer is: C.\n\n**Key Insight:** This problem examines the function graph of a moving point and involves the mid-segment theorem of a triangle. The key is to determine the length of \\( BC \\) based on Figure 2." }, { "problem_id": 172, "question": "A rectangle with a length of 20 and a width of 16 is folded first from left to right, and then from bottom to top, resulting in the rectangle shown in Figure (1). By cutting along the line connecting the midpoints of two adjacent sides (dashed line), and then unfolding, the figure shown in Figure (2) is obtained. The area of Figure (2) is ( )\n\n\n\n(1)\n\n\n\n(2)\nA. $80 \\mathrm{~cm}^{2}$\nB. $40 \\mathrm{~cm}^{2}$\nC. $20 \\mathrm{~cm}^{2}$\nD. $10 \\mathrm{~cm}^{2}$", "input_image": [ "batch14-2024_06_15_f1c27c1996e9a93f4957g_0068_1.jpg", "batch14-2024_06_15_f1c27c1996e9a93f4957g_0068_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: According to the problem, the area of the rhombus is calculated as $S_{\\text{rhombus}} = \\frac{1}{2} \\times 10 \\times 8 = 40 \\left(\\text{cm}^2\\right)$.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question primarily tests the properties of a rhombus and the problem of paper cutting. Determining the length of the diagonals of the rhombus is the key to solving the problem." }, { "problem_id": 173, "question": "In the square $A B C D$ as shown in Figure (1), point $M$ is the midpoint of $A B$, and point $N$ is a moving point on the diagonal $B D$. Let $D N = x$ and $A N + M N = y$. Given that the graph of the function $y$ as a function of $x$ is shown in Figure (2), where point $E(a, 2 \\sqrt{5})$ is the lowest point on the graph, what is the value of the side length of the square?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 2\nB. $2 \\sqrt{2}$\nC. 4\nD. $2 \\sqrt{5}$", "input_image": [ "batch14-2024_06_15_f1c27c1996e9a93f4957g_0097_1.jpg", "batch14-2024_06_15_f1c27c1996e9a93f4957g_0097_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $AC$ intersecting $BD$ at point $O$, connect $NC$, and connect $MC$ intersecting $BD$ at point $N^{\\prime}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n$\\because$ Quadrilateral $ABCD$ is a square,\n\n$\\therefore O$ is the midpoint of $BD$,\n\n$\\because$ Point $M$ is the midpoint of $AB$,\n\n$\\therefore N^{\\prime}$ is the centroid of triangle $ABC$,\n\n$\\therefore N^{\\prime}O = \\frac{1}{3}BO$,\n$\\therefore N^{\\prime}D = \\frac{2}{3}BD$,\n\n$\\because$ Points $A$ and $C$ are symmetric about $BD$,\n\n$\\therefore NA = NC$,\n\n$\\therefore AN + MN = NC + MN$,\n\n$\\because$ When points $M$, $N$, and $C$ are collinear, the value of $y$ is minimized,\n\n$\\therefore$ The minimum value of $y$ is the length of $MC$,\n\n$\\therefore MC = 2\\sqrt{5}$,\n\nLet the side length of the square be $m$, then $BM = \\frac{1}{2}m$,\n\nIn right triangle $BCM$, by the Pythagorean theorem: $MC^{2} = BC^{2} + MB^{2}$,\n\n$\\therefore 20 = m^{2} + \\left(\\frac{1}{2}m\\right)^{2}$,\n\n$\\therefore m = 4$ (negative value discarded),\n\nThus, the side length of the square is 4.\n\nTherefore, the answer is: C.\n\n【Key Insight】This problem examines the moving point image problem, involving the properties of squares, the properties of centroids, and the use of the Pythagorean theorem to determine the length of a segment, which is key to solving the problem." }, { "problem_id": 174, "question": "As shown in the figure above, $A B C$ and $D E F$ are congruent isosceles right triangles, where $\\angle A B C = \\angle D E F = 90^\\circ$ and $A B = 4 \\text{ cm}$. The sides $B C$ and $E F$ are on the straight line $l$. Initially, point $C$ coincides with point $E$. Let $A B C$ slide along the straight line $l$ to the right until point $B$ coincides with point $F$. Let the area of the overlapping part (the shaded region in the figure) between $A B C$ and $D E F$ be $y \\text{ cm}^2$, and the length of $C E$ be $x \\text{ cm}$. The graph of the relationship between $y$ and $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch15-2024_06_15_1ded49d55890835ffae2g_0003_1.jpg", "batch15-2024_06_15_1ded49d55890835ffae2g_0003_2.jpg", "batch15-2024_06_15_1ded49d55890835ffae2g_0003_3.jpg", "batch15-2024_06_15_1ded49d55890835ffae2g_0003_4.jpg", "batch15-2024_06_15_1ded49d55890835ffae2g_0003_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since triangles $ABC$ and $DEF$ are congruent isosceles right triangles,\n\nthe overlapping area of $ABC$ and $DEF$ is also an isosceles right triangle.\n\nWhen $ABC$ is translated along the line $l$ from point $E$ to point $F$, i.e., $0 \\leq x \\leq 4$,\n\nthe area $y$ of the overlapping part of $ABC$ and $DEF$ is $y=\\frac{1}{2} x^{2}$.\n\nWhen $4 \\leq x \\leq 8$, the area $y$ of the overlapping part of $ABC$ and $DEF$ is $y=\\frac{1}{2}(x-8)^{2}$.\n\nThus, the function graph between $y$ and $x$ is approximately option C.\n\nTherefore, the correct choice is C.\n\n[Highlight] This problem examines the function graph of a moving point, utilizing the properties of isosceles right triangles and the method of calculating area, as well as the graph of a quadratic function. The key is to determine the functional relationship between $y$ and $x$ based on the given conditions, paying attention to the range of $x$." }, { "problem_id": 175, "question": "As shown in the figure, a square $A B C D$ with side length $4 \\mathrm{~cm}$ is cut along its diagonal $A C$. Then, triangle $A B C$ is translated along direction $A D$ to form triangle $A^{\\prime} B^{\\prime} C^{\\prime}$. If the area of the overlapping part between the two triangles is $4 \\mathrm{~cm}^{2}$, what is the distance $A A^{\\prime}$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $3 \\mathrm{~cm}$\nB. $2.5 \\mathrm{~cm}$\nC. $1.5 \\mathrm{~cm}$\nD. $2 \\mathrm{~cm}$", "input_image": [ "batch15-2024_06_15_24e5afd6a917894d08c0g_0037_1.jpg", "batch15-2024_06_15_24e5afd6a917894d08c0g_0037_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nLet $A^{\\prime} B^{\\prime}$ intersect $A C$ at point $E$.\n\nFrom the given conditions, we know that $\\angle A=45^{\\circ}$,\n\nTherefore, $A A^{\\prime}=A E$.\n\nLet $A A^{\\prime}=x \\mathrm{~cm}$, then $A^{\\prime} E=x$, and $A^{\\prime} D=(4-x) \\mathrm{cm}$.\n\nSince the area of the overlapping part of the two triangles is $4 \\mathrm{~cm}^{2}$,\n\nWe have $x(4-x)=4$,\n\nSolving for $x$, we get $x=2$,\n\nThus, the distance of translation is $2 \\mathrm{~cm}$,\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This problem mainly examines the properties of squares and the nature of translation. The key to solving the problem lies in using $A A^{\\prime}$ to express the area of the overlapping part." }, { "problem_id": 176, "question": "As shown in the figure, circle $\\odot \\mathrm{A}$ is translated to circle $\\odot \\mathrm{O}$ (as shown in Figure 2). If the coordinates of a point $\\mathrm{P}$ on circle $\\odot \\mathrm{A}$ in Figure (1) are $(\\mathrm{m}, \\mathrm{n})$, what are the coordinates of the corresponding point $\\mathrm{P}^{\\prime}$ in Figure (2) after the translation?\n\n\n\n(1)\n\n\n\nA. $(\\mathrm{m}+2, \\mathrm{n}+1)$\n\nB. $(\\mathrm{m}-2, \\mathrm{n}-1)$\n\nC. $(\\mathrm{m}-2, \\mathrm{n}+1)$\n\nD. $(m+2, n-1)$", "input_image": [ "batch15-2024_06_15_63cc2bb499d687c2b4bdg_0099_1.jpg", "batch15-2024_06_15_63cc2bb499d687c2b4bdg_0099_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "The center of the circle moves from $\\mathrm{A}(-2,1)$ to $\\mathrm{O}(0,0)$, shifting 2 units to the right and 1 unit downward. Therefore, the coordinates of the corresponding point $\\mathrm{P}^{\\prime}$ of $\\mathrm{P}(m, n)$ are $(m+2, n-1)$." }, { "problem_id": 177, "question": "Among the four figures below, which one is both an axisymmetric and a centrosymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch15-2024_06_15_74b29a087c3d1dad0202g_0029_1.jpg", "batch15-2024_06_15_74b29a087c3d1dad0202g_0029_2.jpg", "batch15-2024_06_15_74b29a087c3d1dad0202g_0029_3.jpg", "batch15-2024_06_15_74b29a087c3d1dad0202g_0029_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. It is a centrally symmetric figure but not an axially symmetric figure, therefore this option does not meet the requirement;\n\nB. It is both a centrally symmetric figure and an axially symmetric figure, therefore this option meets the requirement;\n\nC. It is a centrally symmetric figure but not an axially symmetric figure, therefore this option does not meet the requirement;\n\nD. It is a centrally symmetric figure but not an axially symmetric figure, therefore this option does not meet the requirement.\n\nHence, the correct choice is: B.\n\n[Key Insight] This question tests the recognition of centrally symmetric and axially symmetric figures. Understanding the concepts of centrally symmetric and axially symmetric figures is crucial for solving the problem." }, { "problem_id": 178, "question": "Among the four patterns A, B, C, and D below, which one cannot be obtained by translating the pattern?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch15-2024_06_15_74b29a087c3d1dad0202g_0043_1.jpg", "batch15-2024_06_15_74b29a087c3d1dad0202g_0043_2.jpg", "batch15-2024_06_15_74b29a087c3d1dad0202g_0043_3.jpg", "batch15-2024_06_15_74b29a087c3d1dad0202g_0043_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the properties of translation, among the four patterns, the one that cannot be obtained by translating the pattern is B, hence the answer is: B.\n\n[Key Insight] This question tests the properties of translation. It is crucial to be familiar with these properties to solve the problem. The shape and size of the figure remain unchanged before and after translation, but its position changes." }, { "problem_id": 179, "question": "As shown in Figure 1, it is the Parthenon Temple from ancient Greece. Draw the rectangle indicated by the dashed lines in Figure 1 as the rectangle $A B C D$ in Figure 2. Surprisingly, when a square $A B E F$ is constructed with the width $A B$ of the rectangle $A B C D$ as a side, it is found that the rectangle $C D F E$ is similar to the rectangle $A B C D$. Then, $\\frac{B E}{E C}$ equals ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{\\sqrt{5}-1}{2}$\nB. $\\frac{3}{2}$\nC. $\\frac{\\sqrt{3}+1}{2}$\nD. $\\frac{\\sqrt{5}+1}{2}$", "input_image": [ "batch16-2024_06_15_0e1373c2ed89fa5d0610g_0008_1.jpg", "batch16-2024_06_15_0e1373c2ed89fa5d0610g_0008_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since rectangle \\( C D F E \\) is similar to rectangle \\( A B C D \\),\n\n\\[\n\\frac{E F}{B C} = \\frac{E C}{A B}.\n\\]\n\nSince quadrilateral \\( A B E F \\) is a square,\n\n\\[\nA B = E F = B E.\n\\]\n\nGiven that \\( B C = B E + C E \\),\n\n\\[\n\\frac{B E}{B E + C E} = \\frac{E C}{B E}.\n\\]\n\nThus,\n\n\\[\nB E^{2} = E C \\cdot B E + C E^{2},\n\\]\n\nwhich leads to\n\n\\[\n\\frac{B E^{2}}{C E^{2}} = \\frac{B E}{C E} + 1,\n\\]\n\nor equivalently,\n\n\\[\n\\frac{B E^{2}}{C E^{2}} - \\frac{B E}{C E} - 1 = 0.\n\\]\n\nLet \\( \\frac{B E}{E C} = x \\), then:\n\n\\[\nx^{2} - x - 1 = 0.\n\\]\n\nSolving this quadratic equation yields:\n\n\\[\nx = \\frac{1 + \\sqrt{5}}{2} \\quad (\\text{the negative solution is discarded}).\n\\]\n\nTherefore,\n\n\\[\n\\frac{B E}{E C} = \\frac{1 + \\sqrt{5}}{2}.\n\\]\n\nHence, the correct choice is D.\n\n**Key Insight:** This problem tests the properties of similar polygons and squares, as well as solving quadratic equations. Mastery of the proportional relationships in similar polygons is crucial for solving such problems." }, { "problem_id": 180, "question": "As shown in the figure, point $\\mathrm{P}$ is the midpoint of the semicircle with center $\\mathrm{O}$ and diameter $\\mathrm{AB}$, where $\\mathrm{AB}=2$. A right-angled isosceles triangle with a $45^\\circ$ angle at its vertex is placed with that vertex coinciding with point $\\mathrm{P}$. As the triangle rotates around point $\\mathrm{P}$, its hypotenuse and one leg intersect the diameter $\\mathrm{AB}$ at points $\\mathrm{C}$ and $\\mathrm{D}$, respectively. Let the length of segment $\\mathrm{AD}$ be $\\mathrm{x}$ and the length of segment $\\mathrm{BC}$ be $\\mathrm{y}$. Which of the following graphs approximately represents the relationship between $\\mathrm{y}$ and $\\mathrm{x}$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch16-2024_06_15_ba83abc9d3af5badfa3bg_0034_1.jpg", "batch16-2024_06_15_ba83abc9d3af5badfa3bg_0034_2.jpg", "batch16-2024_06_15_ba83abc9d3af5badfa3bg_0034_3.jpg", "batch16-2024_06_15_ba83abc9d3af5badfa3bg_0034_4.jpg", "batch16-2024_06_15_ba83abc9d3af5badfa3bg_0034_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Question Analysis:** As shown in the figure, connect PO, AP, BP.\n\nSince point P is the midpoint of the semicircle with O as the center and AB as the diameter, and AB = 2,\n\nTherefore, triangles APO and BPO are isosceles right triangles.\n\nThus, AP = BP = √2.\n\nLet CD = a,\n\nSince AD = x, and BC = y,\n\nTherefore,\n\nOC = OA - AC = OA - (AB - BC) = 1 - (2 - y) = y - 1, \nOD = OB - BD = OB - (AB - AD) = 1 - (2 - x) = x - 1.\n\nThus, \nPC = √(OC² + OP²) = √((y - 1)² + 1²), \nPD = √(OD² + OP²) = √((x - 1)² + 1²).\n\nSince ∠PAD = ∠CPD = 45°, and ∠PDA = ∠CDP,\n\nTherefore, triangle PAD is similar to triangle CPD.\n\nThus, \nPA / CP = AD / PD, \ni.e., √2 / √((y - 1)² + 1²) = x / √((x - 1)² + 1²) \n⇒ √((x - 1)² + 1²) / √((y - 1)² + 1²) = x / √2.\n\nSince ∠PBC = ∠DPC = 45°, and ∠PCB = ∠DCP,\n\nTherefore, triangle PBC is similar to triangle DPC.\n\nThus, \nPB / DP = BC / PC, \ni.e., √2 / √((x - 1)² + 1²) = y / √((y - 1)² + 1²) \n⇒ √((x - 1)² + 1²) / √((y - 1)² + 1²) = √2 / y.\n\nTherefore, \nx / √2 = √2 / y \n⇒ y = 2 / x (1 ≤ x ≤ 2).\n\nThus, the graph that represents the functional relationship between y and x is approximately C.\n\n**Answer:** C.\n\n**Key Points:** \n1. Surface rotation problems." }, { "problem_id": 181, "question": "As shown in Figure 1, given sector $A O B$, point $P$ starts from point $O$ and moves along the path $O-A-B-O$ at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$. Let the time of point $P$'s movement be $x \\mathrm{~s}$, and the distance from $O$ to $P$ be $y \\mathrm{~cm}$. The graph of $y$ as a function of $x$ is shown in Figure 2. Determine the area of sector $A O B$ ( $\\quad$ ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $3 \\pi \\mathrm{cm}^{2}$\nB. $\\pi \\mathrm{cm}^{2}$\nC. $2 \\pi \\mathrm{cm}^{2}$\nD. $1.5 \\pi \\mathrm{cm}^{2}$", "input_image": [ "batch17-2024_06_14_0c602413ec66cf6e570cg_0005_1.jpg", "batch17-2024_06_14_0c602413ec66cf6e570cg_0005_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From the graph, it can be observed that the time taken for point $P$ to move from point $B$ to point $O$ is $\\pi+6-(\\pi+3)=3$,\n\n$\\therefore O B=3 \\mathrm{~cm}$, which means the radius of the sector is $3 \\mathrm{~cm}$,\n\nFrom the graph, it can be seen that the time taken for point $P$ to move from point $O$ to point $B$ is $\\pi+3$,\n\n$\\therefore$ the length of $A B$ is $\\pi \\mathrm{cm}$, which means the arc length is $\\pi \\mathrm{cm}$,\n\nLet the central angle of the sector be $n$, according to the arc length formula: $\\frac{n \\times 3 \\pi}{180}=\\pi$,\n\nSolving gives $n=60^{\\circ}$,\n\nFrom the sector area formula, the area of sector $A O B$ is $\\frac{60 \\times 3^{2} \\pi}{360}=1.5 \\pi \\mathrm{cm}^{2}$.\n\nTherefore, choose D.\n\n【Insight】This problem belongs to the category of moving point function graph problems, mainly testing knowledge points such as the arc length of a sector and the sector area formula. Determining the radius and arc length of the sector from the graph is key to solving this problem." }, { "problem_id": 182, "question": "Given that the radii of $\\odot O_{1}$ and $\\odot O_{2}$ are 2 and 5, respectively, if the circles are externally tangent, the correct representation of the range of the distance between the centers, $\\mathrm{O}_{1} \\mathrm{O}_{2}$, on the number line is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch17-2024_06_14_0c602413ec66cf6e570cg_0016_1.jpg", "batch17-2024_06_14_0c602413ec66cf6e570cg_0016_2.jpg", "batch17-2024_06_14_0c602413ec66cf6e570cg_0016_3.jpg", "batch17-2024_06_14_0c602413ec66cf6e570cg_0016_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Since the radii of circles $\\odot O_{I}$ and $\\odot O_{2}$ are 2 and 5 respectively, and the positional relationship between the two circles is that they are separate from each other,\n\nTherefore, the range of the distance between the centers $O_{I} O_{2}$ is $d > 2 + 5$, which means $d > 7$.\n\nHence, the correct choice is C.\n\n[Key Insight] This question tests the knowledge of the positional relationship between circles and the representation of inequality solutions on a number line. It is crucial to determine the quantitative relationship between the radii and the distance between the centers based on the positional relationship of the two circles." }, { "problem_id": 183, "question": "Among the angles in the following figures, which ones are central angles of a circle?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0011_1.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0011_2.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0011_3.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0011_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Option A is the central angle;\n\nOption B is incorrect because its vertex is not at the center of the circle;\n\nOption C is incorrect because its vertex is not at the center of the circle;\n\nOption D is incorrect because its vertex is not at the center of the circle.\n\nTherefore, the correct choice is: A\n\n[Key Insight] This question primarily tests the concept of a central angle. Understanding the concept of a central angle is crucial for solving the problem." }, { "problem_id": 184, "question": "Among the following four patterns, which pattern's shaded area is different from the others ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0018_1.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0018_2.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0018_3.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0018_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Let the side length of the square be \\(2a\\),\n\n\\(\\therefore\\) The area of the shaded part in option A is: \\(2a \\times 2a - \\pi a^{2} = 4a^{2} - \\pi a^{2}\\);\n\nLet the radius of the sector be \\(x\\),\n\n\\(\\therefore\\) The area of the shaded part in option B is: \\(2a \\times 2a - \\frac{1}{4} \\times \\pi x^{2} \\times 2 = 4a^{2} - \\frac{\\pi x^{2}}{2}\\);\n\n\\(\\therefore\\) The area of the shaded part in option C is: \\(2a \\times 2a - \\frac{1}{4} \\times \\pi a^{2} \\times 4 = 4a^{2} - \\pi a^{2}\\);\n\n\\(\\therefore\\) The area of the shaded part in option D is: \\(2a \\times 2a - \\frac{1}{2} \\pi a^{2} \\times 2 = 4a^{2} - \\pi a^{2}\\);\n\nTherefore, choose option B.\n\n【Insight】This question examines the area of a square, the area of a circle, and the area of a sector. Correctly dividing the figure is key to solving the problem." }, { "problem_id": 185, "question": "As shown in Figure 1 and 2, the radii of the two circles are equal, with $O_{1}$ and $O_{2}$ being the centers of the circles. The shaded area in Figure 1 is $S_{1}$, and the shaded area in Figure 2 is $S_{2}$. What is the relationship between $S_{1}$ and $S_{2}$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $S_{1}S_{2}$\nD. Cannot be determined", "input_image": [ "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0021_1.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0021_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the radii of both circles be $r$. Then, the side length of the square in Figure 1 is $2r$.\n\nSince the area of the square in Figure 2 is the sum of the areas of four right-angled triangles: $4 \\times \\frac{1}{2} r \\times r = 2 r^{2}$,\n\nTherefore, the side length of the square in Figure 2 is $\\sqrt{2} r$.\n\nThen, $S_{1} = 2r \\times 2r - \\pi r^{2} = 4r^{2} - \\pi r^{2}$, and $S_{2} = \\pi r^{2} - \\sqrt{2} r \\times \\sqrt{2} r = \\pi r^{2} - 2r^{2}$.\n\n$S_{1} - S_{2} = \\left(4r^{2} - \\pi r^{2}\\right) - \\left(\\pi r^{2} - 2r^{2}\\right) = 6r^{2} - 2\\pi r^{2} = (6 - 2\\pi) r^{2}$.\n\nSince $6 - 2\\pi < 0$,\n\nTherefore, $S_{1} - S_{2} < 0$,\n\nHence, $S_{1} < S_{2}$,\n\nThus, the correct choice is: A.\n\n[Key Insight] This problem mainly tests the method of comparing the sizes of real numbers. The key to solving it is to understand how to calculate the areas of squares and circles." }, { "problem_id": 186, "question": "Given that the radii of the two semicircles in Figure 1 and Figure 2 are equal, with $O_{1}$ and $O_{2}$ being the centers of the circles, the shaded area in Figure 1 is $S_{1}$, and the shaded area in Figure 2 is $S_{2}$. The relationship between $S_{1}$ and $S_{2}$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $S_{1}S_{2}$\nD. Cannot be determined", "input_image": [ "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0029_1.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0029_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the radius of both semicircles be \\( r \\). Then, in Figure 1, the length of the rectangle is \\( 2r \\), and the width is \\( r \\). The area \\( S_{1} \\) is calculated as:\n\n\\[\nS_{1} = 2r \\times r - \\frac{1}{2} \\pi r^{2} = 2r^{2} - \\frac{1}{2} \\pi r^{2} = \\left(2 - \\frac{1}{2} \\pi\\right) r^{2} \\approx 0.43 r^{2}\n\\]\n\nIn Figure 2, the base of the triangle is \\( 2r \\), and the height is \\( r \\). The area \\( S_{2} \\) is calculated as:\n\n\\[\nS_{2} = \\frac{1}{2} \\pi r^{2} - \\frac{1}{2} \\times 2r \\times r = \\frac{1}{2} \\pi r^{2} - r^{2} = \\left(\\frac{1}{2} \\pi - 1\\right) r^{2} \\approx 0.57 r^{2}\n\\]\n\nTherefore, \\( S_{1} < S_{2} \\).\n\nHence, the correct choice is A.\n\n**Key Insight:** This problem tests the ability to calculate the area of shaded regions, utilizing properties of circles, semicircles, rectangles, and triangles. The key to solving it lies in understanding how to find the areas of semicircles, rectangles, and triangles, and applying these methods to determine the area of the shaded regions." }, { "problem_id": 187, "question": "There are four figures below, with equal-sized squares. Which of the following figures have equal areas of the shaded regions?\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)(3)\nB. (2)(3)(4)\nC. (1)(3)(4)\nD. (1)(2)(4)", "input_image": [ "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0038_1.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0038_2.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0038_3.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0038_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Given the side length of the square is 2, the area of the shaded parts in Figures (1), (2), and (3) is calculated as \\(2^{2} - \\pi \\times 1^{2} = 4 - \\pi\\);\n\nThe area of the shaded part in Figure (4) is \\(\\frac{1}{4} \\times \\pi \\times 1^{2} = \\frac{1}{4} \\pi\\);\n\nTherefore, the areas of the shaded parts in Figures (1), (2), and (3) are equal.\n\nHence, the correct choice is A.\n\n[Key Insight] This question tests the method of finding the area of irregular shapes within a circle. The key to solving the problem lies in converting these shapes into familiar regular shapes." }, { "problem_id": 188, "question": "As shown in the figure, the three views of a geometric solid are given. Determine the lateral surface area of the solid ( ).\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View\nA. $\\sqrt{10} \\pi$\nB. $2 \\sqrt{10} \\pi$\nC. $3 \\pi$\nD. $6 \\pi$", "input_image": [ "batch19-2024_05_24_045e9cdbec7f0275c3eeg_0043_1.jpg", "batch19-2024_05_24_045e9cdbec7f0275c3eeg_0043_2.jpg", "batch19-2024_05_24_045e9cdbec7f0275c3eeg_0043_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, we can see that the geometric figure is a cone. The diameter of the base of the cone is 2, and its height is 3. By the Pythagorean theorem, the slant height of the cone is $\\sqrt{\\left(2 \\times \\frac{1}{2}\\right)^{2}+3^{2}}=\\sqrt{10}$.\n\nTherefore, the lateral surface area of the cone is: $S=\\frac{1}{2} \\pi \\times 2 \\times \\sqrt{10}=\\sqrt{10} \\pi$.\n\nHence, the correct choice is: A.\n\n【Key Insight】This problem mainly tests the understanding of the three views of geometric figures and the lateral surface area of a cone, as well as the Pythagorean theorem. Mastering the calculation formula for the area of a sector is crucial for solving the problem." }, { "problem_id": 189, "question": "As shown in the figure, the front view of the object is $(\\quad)$\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_0a1ad522308a2b92a15dg_0073_1.jpg", "batch19-2024_05_24_0a1ad522308a2b92a15dg_0073_2.jpg", "batch19-2024_05_24_0a1ad522308a2b92a15dg_0073_3.jpg", "batch19-2024_05_24_0a1ad522308a2b92a15dg_0073_4.jpg", "batch19-2024_05_24_0a1ad522308a2b92a15dg_0073_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "From the front view, it can be seen that there is a diagonal line inside the quadrilateral, and the diagonal line starts from the angle at the bottom right side, hence option B is chosen.\n\n【Key Point】This question tests the understanding of three-view drawings, and the key to solving it lies in mastering the method of determining the front view." }, { "problem_id": 190, "question": "We know that the centroid $G$ of triangle $A B C$ is the intersection point of the medians $A D, B E, C F$. In Figure 1, $\\frac{G D}{A D} = \\frac{G E}{B E} = \\frac{G F}{C F} = \\frac{1}{3}$. In Figure 2, right triangle $A B C$ has $\\angle C = 90^\\circ, A C = 4, B C = 8$. When right triangle $A B C$ is rotated around its centroid $G$, the corresponding points of $A, B, C$ are $A_1, B_1, C_1$. The value closest to the maximum length of $C A_1$ is ( )\n\n(Reference data: $\\sqrt{2} \\approx 1.41, \\sqrt{3} \\approx 1.73, \\sqrt{5} \\approx 2.24$ )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 5.5\nB. 6.5\nC. 7.5\nD. 8.5", "input_image": [ "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0017_1.jpg", "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0017_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the problem, we know that point \\( D \\) is the midpoint of \\( BC \\),\n\n\\[\n\\therefore CD = \\frac{1}{2} BC = 4,\n\\]\n\n\\[\n\\therefore AD = \\sqrt{AC^{2} + CD^{2}} = 4\\sqrt{2}, \\quad AB = \\sqrt{AC^{2} + BC^{2}} = 4\\sqrt{5},\n\\]\n\n\\[\n\\because \\text{point } F \\text{ is the midpoint of } AB,\n\\]\n\n\\[\n\\therefore CF = \\frac{1}{2} AB = 2\\sqrt{5},\n\\]\n\nFrom the problem statement, we have:\n\n\\[\nAG = \\frac{2}{3} AD = \\frac{8}{3}\\sqrt{2}, \\quad CG = \\frac{2}{3} CF = \\frac{4}{3}\\sqrt{5},\n\\]\n\n\\[\n\\because \\text{point } A \\text{ moves on a circle with center at point } G \\text{ and radius } AG,\n\\]\n\n\\[\n\\therefore \\text{the maximum value of } CA_{1} \\text{ is } CG + AG,\n\\]\n\n\\[\n\\text{which is } \\frac{8}{3}\\sqrt{2} + \\frac{4}{3}\\sqrt{5} \\approx \\frac{8}{3} \\times 1.41 + \\frac{4}{3} \\times 2.24 \\approx 6.5;\n\\]\n\nTherefore, the correct answer is \\( B \\).\n\n【Key Insight】This problem mainly examines the application of the centroid of a triangle. The key to solving it lies in combining the Pythagorean theorem and the properties of right triangles." }, { "problem_id": 191, "question": "As shown in Figure 1, the rectangular box contains a modern tangram puzzle, which consists of two identical semicircles (1) and (7), an isosceles right triangle (2), an irregular angular figure (3), a right trapezoid (4), an irregular circular figure (5), and a circle (6). It is known that $A J=B K$. To celebrate the Beijing Winter Olympics, Xiao Ming put the tangram puzzle into a skater pattern, as shown in Figure 2. If $A B$ is parallel to the ground $M N, A J=2$, then the height of the pattern is $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 8\nB. $9-\\frac{\\sqrt{2}}{2}$\nC. $7+\\sqrt{2}$\nD. $10-\\sqrt{2}$\n\n", "input_image": [ "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0062_1.jpg", "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0062_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the data in Figure 1, it is known that the heights of (3), (5), and (6) are all 2. \\( MN \\) is the tangent to the semicircle (7), and the figure shows a right trapezoid (4) and semicircle (7). Construct \\( RP \\perp AB \\), \\( OT \\perp MN \\), and \\( RQ \\perp OT \\).\n\n\n\nIn Figure 1, since \\( \\angle KBA = 45^\\circ \\), it follows that \\( \\angle GBC = 45^\\circ = \\angle HGB \\), meaning the acute angle of the right trapezoid (4) is \\( 45^\\circ \\). Therefore, \\( \\angle PRO = \\angle ORQ = 45^\\circ \\).\n\nSince the distance from \\( B \\) to \\( HG \\) in Figure 1 is 2, it follows that \\( RP = 2 \\) in the figure. Given that the diameter of semicircle (7) is 2, \\( OR = 1 \\), and \\( OQ = \\frac{\\sqrt{2}}{2} OR = \\frac{\\sqrt{2}}{2} \\). Thus, \\( QT = 1 - \\frac{\\sqrt{2}}{2} \\).\n\nTherefore, the height is \\( 6 + PR + QT = 9 - \\frac{\\sqrt{2}}{2} \\).\n\nThe correct answer is: B.\n\n【Key Insight】This problem tests the properties of isosceles right triangles and rectangles. Understanding the relationships between the line segments is crucial for solving the problem." }, { "problem_id": 192, "question": "Famous Shanxi craftwork, Pingyao Pushuang lacquerware, has an antique and elegant appearance, with a shiny and glittering finish, and golden and magnificent paintings, named after the technique of pushing the surface to a shine with the palm. Figure 1 shows a pattern of Pingyao Pushuang lacquerware, and Figure 2 is a magnified diagram of a part of it. Quadrilateral $A B C D$ is a square with a side length of 2. Arcs are drawn with the four vertices of the square as centers and half the length of the diagonal as radii. The four arcs intersect at point $O$. The area of the shaded region in the figure is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $2 \\pi-4$\nB. $\\pi-2$\nC. $2 \\pi$\nD. $\\frac{1}{4} \\pi$", "input_image": [ "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0064_1.jpg", "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0064_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a square with side length 2,\n\nTherefore, the length of the diagonal of the square is \\( 2\\sqrt{2} \\).\n\nThus, the radius is \\( \\sqrt{2} \\).\n\nSince the area of the shaded region equals the area of the circle minus the area of the square,\n\nTherefore, the area of the shaded region is \\( \\pi(\\sqrt{2})^{2} - 2^{2} = 2\\pi - 4 \\).\n\nHence, the correct choice is: A.\n\n【Key Insight】This problem examines the properties of squares and circles, as well as the area of composite shapes. The key to solving it lies in transforming the irregular shape into a relationship between the areas of regular shapes." }, { "problem_id": 193, "question": "As shown in Figure 1, a schematic diagram of an overpass (road widths are ignored), $A$ is the entrance, $F$ and $G$ are the exits. The straight roads are $A B, C G, E F$, and $A B=C G=E F$. The curved roads are segments of a circle with center $O$, where the central angles corresponding to $B C$, $C D$, and $D E$ are all $90^\\circ$. Cars A and B start simultaneously from entrance $A$ and travel at a speed of $10 \\mathrm{~m} / \\mathrm{s}$, exiting through different exits. The relationship between the distance $y(\\mathrm{~m})$ of the cars to point $O$ and time $x$ (s) is shown in Figure 2. Based on the information provided, which of the following statements is correct $(\\quad)$?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. Car A exits from $F$ and Car B exits from $G$.\n\nB. When Car A exits the overpass, Car B is on segment $D E$.\n\nC. Cars A and B are on the overpass simultaneously for $10 \\mathrm{~s}$.\n\nD. The total length of the overpass in the diagram is $140 \\mathrm{~m}$.", "input_image": [ "batch19-2024_05_24_2722bd9afb6a19be889cg_0015_1.jpg", "batch19-2024_05_24_2722bd9afb6a19be889cg_0015_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "A. Analyzing Figure 2, it can be seen that Car A exits the interchange first, followed by Car B. Therefore, Car A exits from exit $G$, and Car B exits from exit $F$. The original statement is incorrect and does not match the intended meaning.\n\nB. According to the graph of Car A in Figure 2, Car A takes a total of $5+3=8 \\mathrm{~s}$ to travel from $A$ to $G$ on the interchange, with $2 \\mathrm{~s}$ spent from $B$ to $C$. Since both cars have the same speed, Car B takes $3+2 \\times 2=7 \\mathrm{~s}$ to travel from $A$ to $D$ and $3+3 \\times 2=9 \\mathrm{~s}$ from $A$ to $E$. Therefore, at the $8 \\mathrm{~s}$ mark, Car B is on segment $DE$. The original statement is correct and matches the intended meaning.\n\nC. Based on the analysis in option B, the time both cars are simultaneously on the interchange is $8 \\mathrm{~s}$. The original statement is incorrect and does not match the intended meaning.\n\nD. According to the problem, the total length of the interchange is: $10 \\times 3 \\times 3+10 \\times 2 \\times 3=150 \\mathrm{~m}$. The original statement is incorrect and does not match the intended meaning. Therefore, the correct choice is B.\n\n【Key Point】This question tests the understanding of function graphs in relation to actual travel problems, including related knowledge points such as central angles. Understanding the actual meaning corresponding to the function graph is the key to solving the problem." }, { "problem_id": 194, "question": "As shown in the figure, $O$ is the circumcenter of triangle $A B C$. Find a point $M$ on the arc $B C$ that divides the arc into two equal parts. Here are two different methods provided by Jiajia and Qiqi:\n\nJiajia: As shown in Figure 1, draw the bisector of $\\angle B A C$, denoted as $A F$, which intersects the arc $B C$ at point $M$. Then point $M$ is the desired point.\n\nQiqi: As shown in Figure 2, draw the perpendicular bisector of segment $B C$, denoted as $P Q$, which intersects the arc $B C$ at point $M$. Then point $M$ is the desired point.\n\nWhich of the following statements about the above construction methods is true?\nA. Jiajia's method is correct.\nB. Qiqi's method is correct.\nC. Both Jiajia's and Qiqi's methods are incorrect.\nD. Both Jiajia's and Qiqi's methods are correct.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch19-2024_05_24_2722bd9afb6a19be889cg_0092_1.jpg", "batch19-2024_05_24_2722bd9afb6a19be889cg_0092_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: Since $AF$ bisects $\\angle BAC$,\n\n$\\therefore \\angle BAM = \\angle CAM$\n$\\therefore BM = CM$,\n\n$\\therefore$ Point $M$ is the desired point,\n\nThus, Jiajia's method is correct;\n\nSince $PQ$ is the perpendicular bisector of chord $BC$,\n\n$\\therefore PQ$ must pass through the center $O$,\n\n$\\therefore OM \\perp BC$\n\n$\\therefore OM$ bisects $BC$,\n\n$\\therefore BM = CM$,\n\nThen point $M$ is the desired point,\n\nThus, Qiqi's method is correct,\n\nTherefore, both Jiajia and Qiqi's methods are correct.\n\nHence, the answer is: D.\n\n【Key Point】This question tests the understanding of the Perpendicular Chord Bisector Theorem and the Inscribed Angle Theorem. Understanding the problem and the basis of the construction is key to solving it." }, { "problem_id": 195, "question": "As shown in Figure 1, a moving point $P$ starts from point $A$ of a regular hexagon and moves along the path $A \\rightarrow B \\rightarrow C \\rightarrow D \\rightarrow E$ at a constant speed of $1 \\mathrm{~cm} / \\mathrm{s}$. Figure 2 shows the relationship between the area $y\\left(\\mathrm{~cm}^{2}\\right)$ of triangle $A P E$ and the time $x(\\mathrm{~s})$ as point $P$ moves, where $y$ increases with $x$. The value of $m$ in Figure 2 is $(\\quad)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{8 \\sqrt{3}}{3} \\mathrm{~cm}^{2}$\nB. $3 \\sqrt{3} \\mathrm{~cm}^{2}$\nC. $\\frac{9 \\sqrt{3}}{4} \\mathrm{~cm}^{2}$\nD. $4 \\sqrt{3} \\mathrm{~cm}^{2}$", "input_image": [ "batch19-2024_05_24_29ba9504abd9939bfb56g_0026_1.jpg", "batch19-2024_05_24_29ba9504abd9939bfb56g_0026_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Connect $AC$ and $EC$, draw $BG \\perp AC$ at point $G$, and draw $AH \\perp EP$ at point $H$.\n\n\n\nWhen point $P$ moves to point $C$, the area of $APE$ reaches its maximum.\n\nFrom Figure 2, we know that $AB + BC = 2\\sqrt{3}$.\n\nSince hexagon $ABCDEF$ is a regular hexagon,\n\n$\\therefore AB = BC = CD = DE = EF = FA$, and $\\angle ABC = \\angle F = \\angle D = 120^{\\circ}$.\n\n$\\therefore AB = \\sqrt{3}$.\n\n$\\therefore AG = AB \\cdot \\sin 60^{\\circ} = \\frac{\\sqrt{3}}{2} \\times \\sqrt{3} = \\frac{3}{2}$.\n\n$\\therefore AC = 2AG = 3$.\n\nSince $AB = BC = CD = DE = EF = FA$, and $\\angle ABC = \\angle F = \\angle D = 120^{\\circ}$,\n\n$\\therefore ABC \\cong EDC \\cong AFE$.\n\n$\\therefore AC = EC = AE$.\n\n$\\therefore AEC$ is an equilateral triangle.\n\n$\\therefore AH = \\frac{\\sqrt{3}}{2} AC$.\n\n$\\therefore S_{AEC} = \\frac{1}{2} EC \\cdot AH = \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} AC \\times AC = \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} \\times 3^{2} = \\frac{9\\sqrt{3}}{4}$.\n\n$\\therefore m = \\frac{9\\sqrt{3}}{4}$.\n\nTherefore, the answer is: C.\n\n【Key Insight】This problem examines the properties of a regular hexagon and an equilateral triangle. The key to solving the problem is identifying the position where the area is maximized." }, { "problem_id": 196, "question": "In acute triangle $A B C$, $\\angle B = 45^\\circ$. On side $A B$, a point $D$ is to be constructed such that $\\triangle B C D$ is an isosceles right triangle. Which of the construction trace shown in the figure does not meet the requirement? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_2a3fabb2e3002647b180g_0014_1.jpg", "batch19-2024_05_24_2a3fabb2e3002647b180g_0014_2.jpg", "batch19-2024_05_24_2a3fabb2e3002647b180g_0014_3.jpg", "batch19-2024_05_24_2a3fabb2e3002647b180g_0014_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "$\\mathrm{A}$. From the construction, point $D$ is the intersection of the perpendicular bisector of $B C$ with $A B$, so $D B = D C$. Therefore, $\\angle D C B = \\angle B = 45^{\\circ}$, making $\\triangle B C D$ an isosceles right triangle. Thus, option $\\mathrm{A}$ does not meet the requirement;\n\nB. From the construction, point $D$ is the intersection of the angle bisector of $\\angle A C B$ with $A B$, so $\\angle D C B = \\frac{1}{2} \\angle A C B < 45^{\\circ}$. Therefore, $\\triangle B C D$ is not an isosceles right triangle, making option $\\mathrm{B}$ meet the requirement;\n\nC. From the construction, point $D$ is the intersection of the circle with diameter $B C$ and $A B$, so $\\angle B D C = 90^{\\circ}$. Therefore, $\\triangle B C D$ is an isosceles right triangle, making option $\\mathrm{C}$ not meet the requirement;\n\nD. From the construction, $\\angle D C B = \\angle B = 45^{\\circ}$, so $\\triangle B C D$ is an isosceles right triangle, making option $\\mathrm{D}$ not meet the requirement;\n\nTherefore, the correct choice is B.\n\n【Key Insight】This question tests the ability to perform complex constructions: the key to solving such problems is to be familiar with the properties of basic geometric shapes and to break down complex constructions into basic steps by leveraging the fundamental properties of geometric shapes. It also examines the properties of isosceles right triangles, the properties of perpendicular bisectors, the definition of angle bisectors, and the corollary of the inscribed angle theorem." }, { "problem_id": 197, "question": "In the following figures, which one has $\\angle A = \\frac{1}{2} \\angle B$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_2a3fabb2e3002647b180g_0053_1.jpg", "batch19-2024_05_24_2a3fabb2e3002647b180g_0053_2.jpg", "batch19-2024_05_24_2a3fabb2e3002647b180g_0053_3.jpg", "batch19-2024_05_24_2a3fabb2e3002647b180g_0053_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "A. In this figure, according to the principle that angles subtended by the same or equal arcs are equal, we know that $\\angle A = \\angle B$. Therefore, this option does not meet the requirement of the question;\n\nB. In this figure, $\\angle B + \\frac{1}{2} \\angle A = 180^{\\circ}$. Therefore, this option does not meet the requirement of the question;\n\nC. In this figure, according to the property that opposite angles of a cyclic quadrilateral are supplementary, we have $\\angle A + \\angle B = 180^{\\circ}$. Therefore, this option does not meet the requirement of the question;\n\nD. In this figure, $\\angle A = \\frac{1}{2} \\angle B$. Therefore, this option meets the requirement of the question;\n\nHence, the correct choice is: D.\n\n【Key Point】This question tests the understanding of the theorem of angles in a circle and the properties of cyclic quadrilaterals. Proficiency in applying the principles that \"angles subtended by the same or equal arcs are equal\" and \"opposite angles of a cyclic quadrilateral are supplementary\" is crucial for solving this problem." }, { "problem_id": 198, "question": "As shown in the figure, there is a chord $C D$ of fixed length on the semicircle $O$. If $C D < O A$ and $C E \\perp C D$ meets $A B$ at point $E$, and $D F \\perp C D$ meets $A B$ at point $F$, when $C D$ moves along the arc $A B$ from point $A$ to point $B$ (points $C$ and $A$ do not coincide, and points $D$ and $B$ do not coincide), if the area of quadrilateral $C D E F$ is $y$ and the time of movement is $x$, then the graph of $y$ against $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_30e46a32398ce13ab826g_0097_1.jpg", "batch19-2024_05_24_30e46a32398ce13ab826g_0097_2.jpg", "batch19-2024_05_24_30e46a32398ce13ab826g_0097_3.jpg", "batch19-2024_05_24_30e46a32398ce13ab826g_0097_4.jpg", "batch19-2024_05_24_30e46a32398ce13ab826g_0097_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "As shown in the figure, let lines $CE$ and $DF$ intersect circle $O$ at points $G$ and $H$. Connect $CH$,\n\n\n\n$\\because CD \\perp DH$\n\n$\\therefore CH$ is the diameter, passing through the center $O$ of the circle.\n\n$\\because CD$ is of fixed length, and the circle is fixed, $CD^{2}+DH^{2}=CH^{2}$,\n\n$\\therefore DH$ is of fixed length.\n\n$\\therefore$ The area of quadrilateral $CDHG$ is constant.\n\nMoreover, $\\because EF$ passes through the center $O$ of rectangle $CDHG$,\n\n$\\therefore$ The area of quadrilateral $CDFE$ is half the area of quadrilateral $CDHG$, which is also constant.\n\nTherefore, the answer is: A.\n 【Key Insight】This problem examines the function graph of moving point problems. The key to solving it is to draw auxiliary lines and discover that the area of quadrilateral CDEF remains unchanged." }, { "problem_id": 199, "question": "In triangle $\\mathrm{ABC}$, $\\angle \\mathrm{B}=90^{\\circ}$ and $\\mathrm{AC}=10$. A circle inscribed in triangle $\\mathrm{ABC}$ is drawn, touching $\\mathrm{AB}$, $\\mathrm{BC}$, and $A C$ at points $D$, $E$, and $F$, respectively. Let $A D=x$ and the area of triangle $\\mathrm{ABC}$ be $S$. The graph of $S$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_3a96f8de3116dd4987dbg_0096_1.jpg", "batch19-2024_05_24_3a96f8de3116dd4987dbg_0096_2.jpg", "batch19-2024_05_24_3a96f8de3116dd4987dbg_0096_3.jpg", "batch19-2024_05_24_3a96f8de3116dd4987dbg_0096_4.jpg", "batch19-2024_05_24_3a96f8de3116dd4987dbg_0096_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Connect $\\mathrm{OD}$ and $\\mathrm{OE}$ as shown in the figure, with the radius of $O$ being $\\mathrm{r}$.\n\n\n\nSince the incircle $\\mathrm{O}$ of $\\triangle \\mathrm{ABC}$ is tangent to $\\mathrm{AB}$, $\\mathrm{BC}$, and $\\mathrm{AC}$ at points $\\mathrm{D}$, $\\mathrm{E}$, and $\\mathrm{F}$ respectively,\n\n$\\therefore O D \\perp A B$, $O E \\perp B C$, $\\mathrm{AF}=\\mathrm{AD}=\\mathrm{x}$, and $\\mathrm{CE}=\\mathrm{CF}=10-\\mathrm{x}$.\n\nIt is easy to see that quadrilateral DBEO is a square,\n\n$\\therefore D B=B E=O D=r$,\n\n$\\because S_{\\triangle A B C}=\\frac{1}{2} r(A B+B C+A C)=\\frac{1}{2} r(x+r+r+10-x+10)=r^{2}+10 r$,\n\n$\\because A B^{2}+B C^{2}=A C^{2}$\n\n$\\therefore(x+r)^{2}+(10-x+r)^{2}=10^{2}$,\n\n$\\therefore r^{2}+10 r=-x^{2}+10 x$\n\n$\\therefore S=-x^{2}+10 x=-(x-5)^{2}+25(0<\\mathrm{x}<10)$.\n\nTherefore, the correct answer is A.\n\n【Key Insight】This problem mainly examines the properties of tangents, the incircle of a triangle and its center, and function graphs. Accurate analysis and judgment are key to solving the problem." }, { "problem_id": 200, "question": "As shown in the figure, point $P$ is a point on circle $O$. Using a straightedge and compass, a line is to be drawn through point $P$ such that it is tangent to circle $O$ at point $P$. Here are the methods of A and B:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n A: As shown in Figure 1, connect $O P$. Draw an arc with center $P$ and radius $O P$ intersecting circle $O$ at point $A$. Connect and extend $O A$. Then, on line segment $O A$, cut off $A B$ equal to $O P$. Line segment $P B$ is the required line.\n\n B: As shown in Figure 2, draw a diameter $P A$. Take a point $B$ on circle $O$ (different from points $P$ and $A$), and connect $A B$ and $B P$. Draw an angle $\\angle B P C$ equal to $\\angle A$ at point $P$. Then, line segment $P C$ is the required line.\n\nWhich of the following statements about the methods of A and B is correct?\nA. Both A and B's methods are correct.\nB. Both A and B's methods are incorrect.\nC. A's method is correct, and B's method is incorrect.\nD. A's method is incorrect, and B's method is correct.\n\n##", "input_image": [ "batch19-2024_05_24_3e7887329f9cb8035ce0g_0069_1.jpg", "batch19-2024_05_24_3e7887329f9cb8035ce0g_0069_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: A is correct.\n\nReason: As shown in Figure 1, connect $PA$.\n\n$\\because AP = PO = AO$,\n\n$\\therefore \\triangle AOP$ is an equilateral triangle,\n\n$\\therefore \\angle OPA = \\angle OAP = 60^\\circ$,\n\n$\\because AB = OP = AP$,\n\n$\\therefore \\angle APB = \\angle ABP$,\n\n$\\because \\angle OAP = \\angle APB + \\angle ABP$,\n\n$\\therefore \\angle APB = \\angle ABP = 30^\\circ$,\n\n$\\therefore \\angle OPB = 90^\\circ$,\n\n$\\therefore OP \\perp PB$,\n\n$\\therefore PB$ is the tangent to circle $O$;\n\nB is correct.\n\nReason: $\\because AP$ is the diameter,\n\n$\\therefore \\angle ABP = 90^\\circ$,\n\n$\\therefore \\angle APB + \\angle PAB = 90^\\circ$,\n\n$\\because \\angle BPC = \\angle BAP$,\n\n$\\therefore \\angle APB + \\angle BPC = 90^\\circ$,\n\n$\\therefore \\angle OPB = 90^\\circ$,\n\n$\\therefore OP \\perp PB$,\n\n$\\therefore PB$ is the tangent to circle $O$,\n\nTherefore, the answer is: A.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Point】This question examines the knowledge of complex geometric constructions and the determination of tangents. The key to solving the problem lies in understanding the given information and flexibly applying the learned knowledge, which is a common type of question in middle school exams." }, { "problem_id": 201, "question": "Among the following figures, the regular polygon is inscribed in a circle with equal radii. Which regular polygon has the smallest perimeter?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_47cb1a69323446b3f097g_0017_1.jpg", "batch19-2024_05_24_47cb1a69323446b3f097g_0017_2.jpg", "batch19-2024_05_24_47cb1a69323446b3f097g_0017_3.jpg", "batch19-2024_05_24_47cb1a69323446b3f097g_0017_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: The more sides an inscribed regular polygon of a circle has, the closer its perimeter approximates the circumference of the circle, and the longer the perimeter of the regular polygon becomes. Therefore, the correct choice is: A.\n\n[Key Insight] This question primarily examines the relationship between regular polygons and circles. The key to solving it lies in understanding that \"the more sides an inscribed regular polygon of a circle has, the closer its perimeter approximates the circumference of the circle, and the longer the perimeter of the regular polygon becomes.\"" }, { "problem_id": 202, "question": "Chinese cuisine emphasizes color, aroma, taste, and beauty, and an elegant arrangement can add to the charm of the food. In the arrangement shown in Figure (1), its shape is a part of a sector. Figure (2) is a geometric representation of the arrangement (the shaded area represents the arrangement). By measuring, we find that $A C=B D=10 \\text{ cm}$, the distance between points $C$ and $D$ is $2 \\text{ cm}$, and the central angle is $60^\\circ$. Then the area of the arrangement in Figure (2) is $(\\quad)$\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $\\frac{70}{6} \\pi \\text{ cm}^{2}$\nB. $\\frac{40}{3} \\pi \\text{ cm}^{2}$\nC. $\\frac{70}{3} \\pi \\text{ cm}^{2}$\nD. $40 \\pi \\text{ cm}^{2}$", "input_image": [ "batch19-2024_05_24_47cb1a69323446b3f097g_0094_1.jpg", "batch19-2024_05_24_47cb1a69323446b3f097g_0094_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $CD$.\n\n\n\nFigure (2)\n\nSince $OC = OD$ and $\\angle O = 60^\\circ$,\n\n$\\triangle OCD$ is an equilateral triangle,\nTherefore, $OC = OD = CD = 2 \\text{ cm}$,\n\nThus, the shaded area $S_{\\text{shaded}} = S_{\\text{sector } AAB} - S_{\\text{sector } OCD}$\n\n$= \\frac{60 \\pi \\times 12^{2}}{360} - \\frac{60 \\pi \\times 2^{2}}{360} = \\frac{70}{3} \\pi \\text{ cm}^{2}$,\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem tests the calculation of the area of a sector and the properties of an equilateral triangle. The key to solving this problem is to understand the given information and use a combination of geometric shapes to find the solution." }, { "problem_id": 203, "question": "As shown in the figure, given $C A = C B$, point $D$ is the midpoint of the arc with chord $A B$, $A B = 4$, and points $E$ and $F$ are moving points on segments $C D$ and $A B$, respectively. If $A F = x$ and $A E^2 - F E^2 = y$, then the graph that represents the functional relationship between $y$ and $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_49482659984335c99957g_0064_1.jpg", "batch19-2024_05_24_49482659984335c99957g_0064_2.jpg", "batch19-2024_05_24_49482659984335c99957g_0064_3.jpg", "batch19-2024_05_24_49482659984335c99957g_0064_4.jpg", "batch19-2024_05_24_49482659984335c99957g_0064_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Extend $DC$ to intersect $AB$ at point $H$,\n\n\n\nSince point $D$ is the midpoint of the arc with chord $AB$,\n\n$\\therefore DH \\perp AB$, and $AH = BH = \\frac{1}{2} AB = 2$,\n\n$\\therefore FH = AH - AF = 2 - x$,\n\n$\\therefore$ In right triangles $EAH$ and $EFH$, $EH^{2} = AE^{2} - AH^{2} = AE^{2} - 2^{2}$,\n\n$EH^{2} = EF^{2} - FH^{2} = EF^{2} - (2 - x)^{2}$,\n\n$\\therefore AE^{2} - 2^{2} = EF^{2} - (2 - x)^{2}$,\n\n$\\therefore AE^{2} - EF^{2} = 2^{2} - (2 - x)^{2}$, which is $y = 2^{2} - (2 - x)^{2}$,\n\nSimplifying, we get $y = -x^{2} + 4x$,\n\n$\\therefore$ It can be seen that the function of $y$ with respect to $x$ is a quadratic function, whose graph is a parabola opening downward and passing through the origin.\n\nTherefore, the correct choice is: A.\n\n【Key Insight】This problem mainly examines the function image of a moving point, involving knowledge points such as the corollary of the perpendicular bisector theorem, the Pythagorean theorem, and the image of quadratic functions. The key to solving the problem is to skillfully use the Pythagorean theorem to combine geometric figures with functions." }, { "problem_id": 204, "question": "As shown in Figure 1, Xiao Ke performed the following consecutive operations on a circular paper:\n\n(1) Fold the circular paper along the left-right and top-bottom directions, obtaining perpendicular creases $A B$ and $C D$ with their intersection at point $M$, as shown in Figure 2.\n\n(2) Fold the circular paper along $E F$ so that points $B$ and $M$ coincide. The crease $E F$ intersects $A B$ at $N$. Connect $A E$, $A F$, $B E$, and $B F$, as shown in Figure 3.\nXiao Ke derived the following conclusions: (1) $C D \\parallel E F$; (2) $\\angle E A F = \\frac{1}{2} \\angle E B F$; (3) $\\triangle A E F$ is an equilateral triangle; (4) $E N \\times F N = A M^2 - B N^2$. How many of these conclusions are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch19-2024_05_24_5d4aa0794529d4a200c4g_0009_1.jpg", "batch19-2024_05_24_5d4aa0794529d4a200c4g_0009_2.jpg", "batch19-2024_05_24_5d4aa0794529d4a200c4g_0009_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nFrom the folding, we have $\\angle AMD = \\angle ANF = 90^\\circ$.\n\n$\\therefore CD \\parallel EF$, so statement (1) is correct.\n\nFrom the folding, we have $\\angle EMF = \\angle EBF$.\n\nSince $\\angle EAF = \\frac{1}{2} \\angle EMF$,\n\n$\\therefore \\angle EAF = \\frac{1}{2} \\angle EBF$, so statement (2) is correct.\n\nSince $ME = MB = 2MN$,\n\n$\\therefore \\angle MEN = 30^\\circ$,\n\n$\\therefore \\angle EMN = 60^\\circ$.\n\nSince $AM = ME$,\n\n$\\therefore \\angle EAN = \\angle AEM = \\frac{1}{2} \\angle EMN = 30^\\circ$,\n\n$\\therefore \\angle AEF = \\angle AEM + \\angle MEN = 60^\\circ$.\n\nSimilarly, $\\angle AFE = 60^\\circ$,\n\n$\\therefore \\triangle AEF$ is an equilateral triangle; so statement (3) is correct.\n\nIn right triangle $\\triangle EMN$, $EN^2 = EM^2 - MN^2$.\n\nSince $EM = AM$, $MN = BN$, and $EN = NF$,\n\n$\\therefore EN^2 = AM^2 - BN^2$,\n\n$\\therefore EN \\times FN = AM^2 - BN^2$. So statement (4) is correct.\n\nTherefore, the correct answer is: D.\n\n**Key Insight:** This problem tests the properties of folding, the equality of radii in a circle, the inscribed angle theorem, the determination of an equilateral triangle, and the Pythagorean theorem. Mastering these concepts and applying them comprehensively is crucial for solving the problem." }, { "problem_id": 205, "question": "As shown in Figure (1), it is an ancient wheel exhibited in the museum. To measure the radius of the wheel, as shown in Figure (2), points $A$ and $B$ are taken on the wheel, with the center of the circle containing $A B$ being $O$. A perpendicular to chord $A B$ is drawn through $O$, with $D$ being the foot of the perpendicular. Then $D$ is the midpoint of $A B$. After measurement: $A B = 90 \\text{ cm}, C D = 15 \\text{ cm}$, what is the length of $O A$?\n\n\n\n(1)\n\n\n\n(2)\nA. $60 \\text{ cm}$\nB. $65 \\text{ cm}$\nC. $70 \\text{ cm}$\nD. $75 \\text{ cm}$", "input_image": [ "batch19-2024_05_24_6bcbb8f3b990d3c7c6d0g_0085_1.jpg", "batch19-2024_05_24_6bcbb8f3b990d3c7c6d0g_0085_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( OC \\perp AB \\) and \\( AB = 90 \\, \\text{cm} \\),\n\n\\[ AD = \\frac{1}{2} AB = 45 \\, \\text{cm}. \\]\n\nLet the radius of the wheel be \\( r \\). According to the problem, \\( OD = (r - 15) \\, \\text{cm} \\).\n\nIn the right triangle \\( \\triangle OAD \\), by the Pythagorean theorem:\n\n\\[ r^{2} = 45^{2} + (r - 15)^{2}, \\]\n\nSolving this equation yields:\n\n\\[ r = 75. \\]\n\nTherefore, the length of \\( OA \\) is \\( 75 \\, \\text{cm} \\).\n\nThe correct choice is: D\n\n[Key Insight] This problem tests the Perpendicular Chord Bisector Theorem: a diameter that is perpendicular to a chord bisects the chord and its corresponding arcs. It also involves the application of the Pythagorean theorem." }, { "problem_id": 206, "question": "In a tiling pattern, the tiles need to be arranged such that the areas of the gray and white colors are roughly equal. Which of the following options best meets this requirement?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_6e79566ad49bd46d2380g_0093_1.jpg", "batch19-2024_05_24_6e79566ad49bd46d2380g_0093_2.jpg", "batch19-2024_05_24_6e79566ad49bd46d2380g_0093_3.jpg", "batch19-2024_05_24_6e79566ad49bd46d2380g_0093_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Let the side length of the square be \\(2a\\), then:\n\nA. The area of the gray region = area of the square - area of the circle = \\((2a)^2 - \\pi a^2 = (4 - \\pi)a^2\\), the area of the white region = area of the circle = \\(\\pi a^2\\), the difference between the two is significant.\n\nB. The area of the gray region = area of the square - area of the circle = \\((2a)^2 - \\pi a^2 = (4 - \\pi)a^2\\), the area of the white region = area of the circle = \\(\\pi a^2\\), the difference between the two is significant.\n\nC. The area of the gray region = area of the square - area of the circle = \\((2a)^2 - \\pi a^2 = (4 - \\pi)a^2\\), the area of the white region = area of the circle = \\(\\pi a^2\\), the difference between the two is significant.\n\nD. The area of the gray region = area of the semicircle - area of the square = \\(\\frac{1}{2}\\pi(2a)^2 - (2a)^2 = (2\\pi - 4)a^2\\), the area of the white region = area of the square - area of the gray region = \\((2a)^2 - (2\\pi - 4)a^2 = (8 - 2\\pi)a^2\\), the two areas are relatively close.\n\nTherefore, the correct answer is D.\n\n[Key Insight] This question tests the formulas for the area of a square and a circle. Carefully observing the diagram and understanding the relationships between the areas of the gray region, white region, square, and circle is key to solving this problem." }, { "problem_id": 207, "question": "As shown in the figure, $A B$ is a diameter of $\\odot O$, chord $C D$ intersects $A B$ at $E$, and $\\angle A E C = 45^\\circ$. Given that $A B = 2$, let $A E = x$ and $C E^2 + D E^2 = y$. Which of the following graphs correctly represents the function relationship between $y$ and $x$? $(\\quad)$\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_77dd29084539ddc516d5g_0017_1.jpg", "batch19-2024_05_24_77dd29084539ddc516d5g_0017_2.jpg", "batch19-2024_05_24_77dd29084539ddc516d5g_0017_3.jpg", "batch19-2024_05_24_77dd29084539ddc516d5g_0017_4.jpg", "batch19-2024_05_24_77dd29084539ddc516d5g_0017_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Draw $OF \\perp CD$ at point $F$, and connect $OD$.\n\nSince $OF \\perp CD$,\n\nTherefore, $CF = DF$.\n\nGiven that $\\angle OEF = \\angle AEC = 45^\\circ$,\n\nIt follows that $OF = EF$.\n\nLet $DF = CF = a$, and $EF = OF = b$.\n\nThus, $CE = a - b$, and $DE = a + b$.\n\nIn the right triangle $OFD$, $OF^2 + DF^2 = OD^2$,\n\nTherefore, $a^2 + b^2 = R^2 = 1$.\n\nHence, $y = CE^2 + DE^2 = (a - b)^2 + (a + b)^2 = 2(a^2 + b^2) = 2$.\n\nThis means that the value of $y$ is a constant and does not change with the value of $x$.\n\nTherefore, the correct choice is A.\n\n\n\n【Key Insight】This problem examines the function image of a moving point. Accurately drawing the auxiliary line and obtaining $CE^2 + DE^2 = 2R^2$ is the key to solving the problem." }, { "problem_id": 208, "question": "The water wheel is an ancient Chinese invention for irrigation, showcasing the wisdom of ancient Chinese laborers. In Figure 1, point $M$ represents a water bucket of the water wheel. As shown in Figure 2, when the water wheel operates, the path of the water bucket forms a circle with the center $O$ as the origin and a radius of $5 \\text{ m}$, with the center above the water surface. If the chord $A B$ of the circle intercepted by the water surface has a length of $8 \\text{ m}$, then the maximum depth of the water bucket below the water surface while the water wheel is working is $(\\quad)$\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\nA. 1 meter\nB. 2 meters\nC. 3 meters\nD. 4 meters", "input_image": [ "batch19-2024_05_24_77dd29084539ddc516d5g_0028_1.jpg", "batch19-2024_05_24_77dd29084539ddc516d5g_0028_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw the radius $O D \\perp A B$ at point $E$, as shown in the figure.\n\n$\\therefore A E = B E = \\frac{1}{2} A B = \\frac{1}{2} \\times 8 = 4$.\n\nIn the right triangle $A E O$, $O E = \\sqrt{O A^{2} - A E^{2}} = \\sqrt{5^{2} - 4^{2}} = 3$.\n\n$\\therefore E D = O D - O E = 5 - 3 = 2 \\mathrm{~m}$.\n\n$\\therefore$ When the water wheel is operating, the maximum depth of the water bucket below the water surface is $2 \\mathrm{~m}$.\n\nTherefore, the correct answer is: B.\n\n\n\n【Key Insight】This problem examines the Perpendicular Chord Bisector Theorem, which states that the diameter perpendicular to a chord bisects the chord and the arcs it subtends. Proficiency in applying the Perpendicular Chord Bisector Theorem is crucial for solving this problem." }, { "problem_id": 209, "question": "In Figure 1, the floor plan of a school auditorium is a rectangle $A B C D$, with a width of $A B = 20$ meters and a length of $B C = 24$ meters. To monitor the interior of the auditorium, a camera $M$ needs to be installed on the back wall $C D$ to observe the front wall $A B$. Additionally, the viewing angle from point $M$ to $A B$, denoted as $\\angle A M B$, should be $45^\\circ$ for optimal observation. Students A and B propose different methods to find point $M$ and calculate the length $M C$. Which of the following statements is correct $(\\quad)$?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n A: In Figure 2, a point $O$ is taken in rectangle $A B C D$ such that $O A = O B = O M$, and $M$ is the desired point. In this case, $C M = 10$ meters;\n\n B: In Figure 3, a point $O$ is taken in rectangle $A B C D$ such that $O A = O B$ and $\\angle A O B = 90^\\circ$. An arc is drawn with $O$ as the center and $O A$ as the radius, intersecting $C D$ at points $M_{1}$ and $M_{2}$. Both $M_{1}$ and $M_{2}$ satisfy the conditions, and in this case, $M C = 8$ or $12$ meters.\nA. Student A's approach is incorrect, but the value of $M C$ is correct.\nB. Student B's approach is correct, and both values of $M C$ are correct and complete.\nC. The values of $M C$ obtained by Student A and B together are complete.\nD. Student A's approach is correct, but the value of $M C$ is incorrect.", "input_image": [ "batch19-2024_05_24_7bbaf69124e1595f4e32g_0076_1.jpg", "batch19-2024_05_24_7bbaf69124e1595f4e32g_0076_2.jpg", "batch19-2024_05_24_7bbaf69124e1595f4e32g_0076_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Using $AB$ as one side, construct an isosceles right triangle $AOB$ inside rectangle $ABCD$, with $\\angle AOB = 90^\\circ$. Draw $HG \\perp AB$ through point $O$, intersecting $CD$ at point $G$.\n\n\n\n$\\therefore$ Quadrilaterals $HBCG$ and $AHGD$ are rectangles.\n\n$\\because AB = 20$, $BC = 24$,\n\n$\\therefore HG = BC = 24$, and $HG \\perp CD$,\n\n$\\therefore BH = AH = OH = 10$,\n\n$\\therefore OA = \\sqrt{AH^2 + OH^2} = \\sqrt{10^2 + 10^2} = 10\\sqrt{2}$,\n\n$\\therefore OG = HG - OH = 24 - 10 = 14$.\n\n$\\because 10\\sqrt{2} > 14$,\n\n$\\therefore$ The circle centered at $O$ with radius $OA$ intersects $CD$,\n\n$\\therefore$ There exists a point $M$ on $O$ such that $\\angle AMB = 45^\\circ$. There are two such points $M$, namely $M_1$ and $M_2$,\n\n$\\therefore GM_1 = GM_2$.\n\nDraw $M_1F \\perp AB$ at $F$, and $EO \\perp M_1F$ at $E$, then connect $OF$.\n\n\n\n$\\therefore$ Quadrilaterals $HFEO$, $EOGM_1$, and $FBCM_1$ are rectangles,\n\n$\\therefore EF = OH = 10$, $OM_1 = OA_1 = 10\\sqrt{2}$,\n\n$\\therefore EM_1 = OG = 14$,\n$\\therefore OE = \\sqrt{OM_1^2 - EM_1^2} = \\sqrt{(10\\sqrt{2})^2 - 14^2} = 2$,\n\n$\\therefore GM_2 = GM_1 = FH = OE = 2$.\n\n$\\therefore CM_1 = CG - GM_1 = BH - GM_1 = 10 - 2 = 8$ (meters),\n\n$CM_2 = CG - GM_2 = BH + GM_2 = 10 + 2 = 12$ (meters),\n\n$\\therefore$ The length of $MC$ is either 8 meters or 12 meters,\n\n$\\therefore$ Person B's reasoning is correct, and the values of $MC$ are both correct and complete.\n\nTherefore, choose: B.\n\n【Key Insight】This problem is a comprehensive circle problem, mainly testing the determination and properties of rectangles, properties of isosceles right triangles, the Pythagorean theorem, and the chord theorem. The key to solving the problem is transforming the quadrilateral problem into a circle problem." }, { "problem_id": 210, "question": "Given that the radius of circle $O$ is $O A = 2$, if $O B = \\sqrt{5}$, which of the following correct figures could be obtained?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_8204df6bb88539a906dfg_0013_1.jpg", "batch19-2024_05_24_8204df6bb88539a906dfg_0013_2.jpg", "batch19-2024_05_24_8204df6bb88539a906dfg_0013_3.jpg", "batch19-2024_05_24_8204df6bb88539a906dfg_0013_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Since the radius $OA$ of circle $O$ is 2, and $OB = \\sqrt{5}$,\n\n$\\therefore OA < OB$,\n\n$\\therefore$ point $B$ is outside the circle,\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question examines the positional relationship between a point and a circle. The key to solving the problem lies in determining the relationship between the distance from the point to the center and the radius of the circle based on the given data. The difficulty level is not high." }, { "problem_id": 211, "question": "In a maritime activity, the area where $O$ is located is the activity area, and the area enclosed by the chord $AB$ and the major arc $AB$ is the area that the sonar needs to detect. Now a sonar device is installed at $A$, and its detection area is shown in the shadow in the figure. If a sonar device of the same model is installed at $B$, it can just complete the detection of all areas, as shown in the shadow in Figure 2.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nAs shown in Figure 3, the placement of the sonar device is now changed to points $D, E, and F$ on the circle $O$, and three plans are designed:\n\n(1) Place two sonar devices of this model at point $D$\n\n(2) Place one sonar device of this model at point $D$ and point $E$ respectively\n\n(3) Place one sonar device of this model at point $F$ Place two sonar devices of this model\n\nIf all areas can be detected, the correct solution is ( )\nA. (1)(3)\nB. (1)(2)(3)\nC. (2)(3)\nD. (1)(2)", "input_image": [ "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0036_1.jpg", "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0036_2.jpg", "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0036_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, when a sonar is installed at points $A$ and $B$ respectively, it can precisely cover all areas for detection. If two sonar devices of the same model are installed at the same point, oriented as shown in Figure 1 for $A$ and $B$, they can also precisely cover all areas for detection. Therefore, option (1) is feasible. However, option (3) is oriented opposite to (1), so it cannot fully cover all areas for detection, making option (3) unsuitable. If a sonar device of this model is placed at points $D$ and $E$ respectively, the situation is the same as with $D$ and $E$, and it can also precisely cover all areas for detection, so option (2) is also feasible.\n\nIn summary, options (1) and (2) meet the requirements, while option (3) does not.\n\nTherefore, the correct choice is D.\n\n【Key Point】This question is a material-based problem, primarily testing the student's comprehension ability." }, { "problem_id": 212, "question": "The teacher wrote this exercise on the blackboard:\n\nAs shown in Figure 1, quadrilateral $A B C D$ is an inscribed quadrilateral in circle $\\odot O$. The segments $A C$ and $B D$ are drawn. Segment $B C$ is a diameter of $\\odot O$, and $A B = A C$. Please explain the relationship between segments $A D$, $B D$, and $C D$.\n\n\n\nFigure 1\n\nThe following is a fragment of Wang Lin's solution to the problem. Which of the following options is incorrect? ( )\n\nAs shown in Figure 2, a perpendicular line $A M$ is drawn from point $A$ to $A D$ and intersects $B D$ at point $M$. $\\star$ indicates that $\\angle A B M$ and $\\angle A C D$ are both inscribed angles subtended by the same arc on $A D$. Therefore, $\\triangle A B M \\cong \\triangle A C D$ by @, which implies $A M = A D$ and $B M = C D$. It follows that $\\triangle M A D$ is a right isosceles triangle, and we can deduce $\\sqrt{2} A D + C D = B D$.\n\n\n\nFigure 2\n\nA. $\\star$ indicates that $\\angle A B M$ and $\\angle A C D$ are both inscribed angles subtended by the same arc on $A D$\n\nB. The direct reference to @ indicates $A A S$ congruence\n\nC. $\\triangle M A D$ is a right isosceles triangle\n\nD. The auxiliary line method in the diagram could also be to take $B M = C D$ on segment $B D$\n", "input_image": [ "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0041_1.jpg", "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0041_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Detailed Explanation: The complete proof is as follows:\n\nSince both $\\angle ABM$ and $\\angle ACD$ are angles subtended by the chord $AD$ in the circle, option A is correct;\n\nTherefore, $\\angle ABM = \\angle ACD$.\n\nSince $BC$ is the diameter of circle $O$, and $AM \\perp AD$,\n\nTherefore, $\\angle BAC = \\angle MAD = 90^{\\circ}$,\n\nHence, $\\angle BAM = \\angle DAC = 90^{\\circ} - \\angle MAC$.\n\nIn triangles $ABM$ and $ACD$,\n\n$\\left\\{\\begin{array}{c}\\angle ABM = \\angle ACD \\\\ AB = AC \\\\ \\angle BAM = \\angle DAC\\end{array}\\right.$,\n\nTherefore, $\\triangle ABM \\cong \\triangle ACD$ (ASA), so option B is incorrect;\n\nThus, $AM = AD$, $BM = CD$, and $\\triangle MAD$ is an isosceles right triangle, so option C is correct.\n\nTherefore, $MD = \\sqrt{2} AD$,\n\nHence, $BD = BM + DM = CD + \\sqrt{2} AD$,\n\nThat is, $\\sqrt{2} AD + CD = BD$.\n\nTaking $BM = CD$ on $BD$, we can prove $\\triangle ABM \\cong \\triangle ACD$ based on \"SAS\", with everything else remaining the same. Therefore, option D is correct.\n\nThus, the correct choice is B.\n\n[Key Insight] This question examines the theorem of angles in a circle, the properties and criteria of congruent triangles, and the properties of right triangles. The key to solving this problem lies in the ability to integrate these concepts for reasoning." }, { "problem_id": 213, "question": "Complete the following drawings using only a straightedge without markings (dashed lines indicate drawing steps):\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) As shown in Figure 1, $A D \\parallel B C, A D = B C$, and $E$ is the midpoint of $A D$. Find the third point $F$ that divides $B D$ into three equal parts;\n\n(2) As shown in Figure 2, $A D \\parallel B C, A D = 2 B C$, and $E$ is the midpoint of $A D$. Draw the median $D F$ of $\\triangle A D C$;\n\n(3) As shown in Figure 3, $A B$ is the diameter of a semicircle, and point $C$ is inside the semicircle. Draw the altitude $C F$ of $\\triangle A B C$.\n\nThe number of correct drawings is ( $\\quad$)\n\nA. 0\nB. 1\nC. 2\nD. 3", "input_image": [ "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0059_1.jpg", "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0059_2.jpg", "batch19-2024_05_24_9f1ab8ef12cdd6028748g_0059_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "(1) Since \\( AD = BC \\) and \\( E \\) is the midpoint of \\( AD \\),\n\n\\[ DE = \\frac{1}{2} AD = \\frac{1}{2} BC, \\]\n\nSince \\( AD \\parallel BC \\),\n\n\\[ \\frac{DE}{BC} = \\frac{DF}{BF}, \\]\n\n\\[ \\frac{DF}{BF} = \\frac{\\frac{1}{2} BC}{BC} = \\frac{1}{2}, \\]\n\n\\[ \\frac{DF}{BD} = \\frac{DF}{BF + DF} = \\frac{1}{3}, \\]\n\nThus, (1) is correct.\n\n(2) Since \\( E \\) is the midpoint of \\( AD \\),\n\n\\[ AE = ED, \\] which means \\( AD = 2ED \\),\n\nSince \\( AD = 2BC \\),\n\n\\[ ED = BC = AE, \\]\n\nSince \\( AD \\parallel BC \\),\n\n\\[ \\frac{AE}{BC} = \\frac{AF}{FC}, \\]\n\n\\[ \\frac{AF}{FC} = 1, \\]\n\nThus, point \\( F \\) is the midpoint of \\( AC \\),\n\nTherefore, \\( DF \\) is the median of \\( \\triangle ADC \\),\n\nThus, (2) is correct.\n\n(3) According to the construction marks, the construction method is as follows: Extend \\( BC \\) and \\( AC \\) to intersect the circle at points \\( D \\) and \\( E \\), connect \\( AD \\) and \\( BE \\), and extend \\( AD \\) and \\( BE \\) to intersect at point \\( P \\), then connect \\( PC \\) to intersect \\( AB \\) at \\( F \\).\n\nSince \\( AB \\) is the diameter of the circle,\n\n\\[ \\angle ADB = \\angle AEB = 90^\\circ, \\]\n\n\\[ BD \\perp AP, \\quad AE \\perp BP, \\]\n\nThus, \\( BD \\) and \\( AE \\) are the altitudes of \\( \\triangle APB \\),\n\nTherefore, point \\( C \\) is the orthocenter of \\( \\triangle APB \\),\n\nThus, \\( PF \\) is also an altitude of \\( \\triangle APB \\),\n\n\\[ PF \\perp AB, \\]\n\nTherefore, \\( CF \\) is the altitude of \\( \\triangle ABC \\), thus (3) is correct.\n\nThe correct answer is: D.\n\n【Insight】This question examines complex constructions, the properties of parallelograms, the properties of parallel lines, and the properties of the orthocenter of a triangle. Understanding the construction method based on the construction marks is key to solving this problem." }, { "problem_id": 214, "question": "In the following figures, $\\angle 1$ is definitely greater than $\\angle 2$ in $(\\quad)$\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_ab879b02decac11d39c7g_0014_1.jpg", "batch19-2024_05_24_ab879b02decac11d39c7g_0014_2.jpg", "batch19-2024_05_24_ab879b02decac11d39c7g_0014_3.jpg", "batch19-2024_05_24_ab879b02decac11d39c7g_0014_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "A. According to the property that vertically opposite angles are equal, $\\angle 1 = \\angle 2$, so this option is incorrect;\n\nB. When two lines are parallel, $\\angle 1 = \\angle 2$; when two lines are not parallel, the relationship between $\\angle 1$ and $\\angle 2$ is uncertain, so this option is incorrect;\n\nC. According to the property that an exterior angle is equal to the sum of the two non-adjacent interior angles, $\\angle 1 > \\angle 2$, so this option is correct;\n\nD. According to the property of angles in a circle, $\\angle 1 = \\angle 2$, so this option is incorrect.\n\nTherefore, the correct answer is C.\n\n[Key Insight] This question examines the comparison of the sizes of two angles, involving properties such as vertically opposite angles, exterior angles of a triangle, and the inscribed angle theorem. Mastering these related properties is key to solving the problem." }, { "problem_id": 215, "question": "Given point $P$ on circle $O$ in Figure 1 and Figure 2, a straight line is to be constructed using a straightedge and compass that is tangent to $O$ at point $P$. Here are the methods of A and B:\n\n A: As shown in Figure 1, draw segment $O P$, and then draw an arc with center $P$ and radius $O P$ that intersects $O$ at point $A$. Draw and extend segment $O A$. Next, on ray $O A$, mark off segment $A B$ such that $A B = O P$. Construct line $P B$, which is the required line.\n\n B: As shown in Figure 2, draw diameter $P A$. Take a point $B$ on circle $O$ (distinct from points $P$ and $A$), and draw segments $A B$ and $B P$. Draw $\\angle B P C$ at point $P$ such that $\\angle B P C = \\angle A$. Line $P C$ is then the required line.\n\nRegarding the methods of A and B, which of the following statements is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Both A and B's methods are correct.\nB. Both A and B's methods are incorrect.\nC. A's method is correct, and B's method is incorrect.\nD. A's method is incorrect, and B's method is correct.", "input_image": [ "batch19-2024_05_24_ac1176a5e0a575ef65c4g_0030_1.jpg", "batch19-2024_05_24_ac1176a5e0a575ef65c4g_0030_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, connect $PA$.\n\n$\\because AP = PO = AO$,\n\n$\\therefore \\triangle AOP$ is an equilateral triangle,\n\n$\\therefore \\angle OPA = \\angle OAP = 60^\\circ$,\n\n$\\because AB = OP = AP$,\n\n$\\therefore \\angle APB = \\angle ABP$,\n\n$\\because \\angle OAP = \\angle APB + \\angle ABP$,\n\n$\\therefore \\angle APB = \\angle ABP = 30^\\circ$,\n\n$\\therefore \\angle OPB = 90^\\circ$, which means $OP \\perp PB$,\n\n$\\therefore PB$ is the tangent to $O$, so statement A is correct;\n\n\n\nFigure 1\n\nAs shown in Figure 2, $\\because AP$ is the diameter,\n\n$\\therefore \\angle ABP = 90^\\circ$,\n\n$\\therefore \\angle APB + \\angle PAB = 90^\\circ$,\n\n$\\because \\angle BPC = \\angle BAP$,\n\n$\\therefore \\angle APB + \\angle BPC = 90^\\circ$,\n\n$\\therefore \\angle OPB = 90^\\circ$, which means $OP \\perp PB$,\n\n$\\therefore PB$ is the tangent to $O$, so statement B is correct;\n\nTherefore, the correct choice is: A.\n\n【Key Insight】This question mainly examines the determination of tangents, the properties and determination of equilateral triangles, the properties of exterior angles of triangles, the fact that the angle subtended by the diameter is a right angle, and the complementary nature of the acute angles in a right triangle. The key to solving the problem lies in understanding the question and flexibly applying the learned knowledge to solve the problem, which is a common type of question in middle school exams." }, { "problem_id": 216, "question": "Reading Comprehension: As shown in Figure 1, in a plane, choose a fixed point $O$, draw a directed ray $O x$, and then select a unit length. The position of any point $M$ in the plane can be determined by the angle $\\angle M O x$ in degrees, denoted as $\\theta$, and the length $O M$, denoted as $m$. The ordered pair $(\\theta, m)$ is called the \"polar coordinates\" of point $M$. This system of coordinates is called a \"polar coordinate system\".\n\nApplication: In the polar coordinate system shown in Figure 2, if the side length of the regular hexagon is 2, with one side $O A$ on the ray $O x$, what should the polar coordinates of the hexagon's vertex $C$ be recorded as $(\\quad)$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\left(60^{\\circ}, 4\\right)$\nB. $\\left(45^{\\circ}, 4\\right)$\nC. $\\left(60^{\\circ}, 2 \\sqrt{2}\\right)$\nD. $\\left(50^{\\circ}, 2 \\sqrt{2}\\right)$", "input_image": [ "batch19-2024_05_24_ac1176a5e0a575ef65c4g_0062_1.jpg", "batch19-2024_05_24_ac1176a5e0a575ef65c4g_0062_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let the center of the regular hexagon be $D$, and connect $A D$.\n\nSince $\\angle A D O = 360^{\\circ} \\div 6 = 60^{\\circ}$ and $O D = A D$,\n\n$\\triangle A O D$ is an equilateral triangle,\n\nTherefore, $O D = O A = 2$, and $\\angle A O D = 60^{\\circ}$,\n\nThus, $O C = 2 O D = 2 \\times 2 = 4$,\n\nHence, the polar coordinates of vertex $C$ of the regular hexagon should be recorded as $\\left(60^{\\circ}, 4\\right)$.\n\nTherefore, the correct choice is A.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Insight】This problem examines the properties of regular polygons and circles, as well as the determination of coordinates. It primarily utilizes the properties of a regular hexagon. Understanding the problem's information and the definition of \"polar coordinates\" is crucial for solving the problem." }, { "problem_id": 217, "question": "In a school's science and technology club, Xiaoming follows these steps to make a practical device: (1) Xiaoming takes a circular thin iron ring provided by the teacher, finds the center $O$ using the knowledge learned in the chapter on circles, and then chooses an arbitrary diameter, which he marks as $\\mathrm{AB}$ (as shown in Figure 1). He measures $\\mathrm{AB}$ to be 4 decimeters; (2) He folds the ring so that point $\\mathrm{B}$ falls on the center $O$, creating intersection points $\\mathrm{C}$ and $\\mathrm{D}$ between the folded and unfolded parts of the ring (as shown in Figure 2); (3) He connects points $\\mathrm{C}$ and $\\mathrm{D}$ with a thin rubber rod (as shown in Figure 3); (4) He calculates the length of the rubber rod $\\mathrm{CD}$.\n\n\n\nFigure 1.\n\n\n\nFigure 2.\n\n\nFigure 3.\nXiaoming calculates the length of the rubber rod CD as $(\\quad)$\nA. $2 \\sqrt{2}$ decimeters\nB. $2 \\sqrt{3}$ decimeters\nC. $3 \\sqrt{2}$ decimeters\nD. $3 \\sqrt{3}$ decimeters\n\n##", "input_image": [ "batch19-2024_05_24_aedceadbca49a09d5943g_0047_1.jpg", "batch19-2024_05_24_aedceadbca49a09d5943g_0047_2.jpg", "batch19-2024_05_24_aedceadbca49a09d5943g_0047_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Connect $\\mathrm{OC}$ and draw $\\mathrm{OE} \\perp \\mathrm{CD}$, as shown in Figure 3.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nGiven that $\\mathrm{AB}=4$ decimeters,\n\n$\\therefore \\mathrm{OC}=2$ decimeters.\n\nSince the ring is folded such that point $\\mathrm{B}$ coincides with the center $\\mathrm{O}$,\n\n$\\therefore O E=\\frac{1}{2} O C=1$ decimeter.\n\nIn the right triangle $\\triangle \\mathrm{OCE}$, $\\mathrm{CE}=\\sqrt{O C^{2}-O E^{2}}=\\sqrt{3}$ decimeters,\n\n$\\therefore C D=2 \\sqrt{3}$ decimeters.\n\nTherefore, the correct answer is: B.\n\n【Key Insight】This problem comprehensively applies the Pythagorean theorem and the perpendicular chord theorem. Note the construction of a right triangle formed by the radius, half-chord, and the distance from the center to the chord for relevant calculations." }, { "problem_id": 218, "question": "In the Cartesian coordinate system $x O y$, a circle $\\odot P$ is drawn with center $(3,0)$. $\\odot P$ intersects the $x$-axis at points $A$ and $B$, and it intersects the $y$-axis at point $C(0,2)$. Point $Q$ is any point on $\\odot P$ other than $A$ and $B$. The line segments $QA$ and $QB$ are drawn, and perpendiculars $PE$ and $PF$ are drawn from point $P$ to $QA$ and $QB$, respectively. Let the $x$-coordinate of point $Q$ be $x$, and let $PE^2 + PF^2 = y$. As point $Q$ moves counterclockwise along $\\odot P$ from point $A$ to point $B$, which of the following graphs represents the part of the graph of the function $y$ as a function of $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_aedceadbca49a09d5943g_0060_1.jpg", "batch19-2024_05_24_aedceadbca49a09d5943g_0060_2.jpg", "batch19-2024_05_24_aedceadbca49a09d5943g_0060_3.jpg", "batch19-2024_05_24_aedceadbca49a09d5943g_0060_4.jpg", "batch19-2024_05_24_aedceadbca49a09d5943g_0060_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, connect $\\mathrm{PC}$ and $\\mathrm{EF}$, then\n\n\n\n$\\because$ point $\\mathrm{P}$ is at $(3,0)$, and point $\\mathrm{C}$ is at $(0,2)$,\n\n$\\therefore P C=\\sqrt{2^{2}+3^{2}}=\\sqrt{13}$,\n\n$\\therefore$ the radius $r=P C=\\sqrt{13}$,\n$\\therefore A B=2 \\sqrt{13} ;$\n\n$\\because P E \\perp Q A$ at $E$, and $P E \\perp Q B$ at $F$,\n\n$\\therefore$ point $\\mathrm{E}$ is the midpoint of $\\mathrm{AQ}$, and point $\\mathrm{F}$ is the midpoint of $\\mathrm{BQ}$,\n\n$\\therefore \\mathrm{EF}$ is the midline of $\\triangle \\mathrm{QAB}$,\n\n$\\therefore E F=\\frac{1}{2} A B=\\frac{1}{2} \\times 2 \\sqrt{13}=\\sqrt{13}$ is a constant;\n\n$\\because \\mathrm{AB}$ is the diameter, then $\\angle \\mathrm{AQB}=90^{\\circ}$,\n\n$\\therefore$ quadrilateral $\\mathrm{PFQE}$ is a rectangle,\n\n$\\therefore E F^{2}=P E^{2}+P F^{2}=y=13$ ，is a constant;\n\n$\\therefore$ as point $Q$ moves clockwise from point $\\mathrm{A}$ to point $B$ on $\\odot P$, the value of $\\mathrm{y}$ remains unchanged;\n\nTherefore, choose: A.\n\n 【Key Insight】This problem examines the properties of circles, the perpendicular chord theorem, the determination and properties of rectangles, the Pythagorean theorem, and the midline theorem of triangles. Correctly drawing auxiliary lines and solving based on the learned properties, accurately finding\n\n$E F^{2}=P E^{2}+P F^{2}=y=13$ is the key to solving the problem." }, { "problem_id": 219, "question": "Given that the radius $r$ of $\\odot O$ is 8, and the distance from point $O$ to the line $l$ is 4, which of the following diagrams correctly depicts the position relationship between the circle and the line?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0044_1.jpg", "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0044_2.jpg", "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0044_3.jpg", "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0044_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Since \\( r = 8 > 4 \\),\n\nTherefore, the line intersects the circle and does not pass through the center of the circle.\n\nHence, the correct choice is D.\n\n**Key Point:** This question tests knowledge about the positional relationship between a line and a circle. The key to solving this problem is to be familiar with the positional relationships and their names. When determining the positional relationship between a line and a circle, it is usually necessary to find the distance \\( d \\) from the center of the circle to the line and the radius \\( r \\) of the circle. Then, the relationship between \\( d \\) and \\( r \\) is used to make the judgment. The positional relationships between a line and a circle are as follows: (1) When \\( d > r \\), the line is separate from the circle; (2) When \\( d = r \\), the line is tangent to the circle; (3) When \\( d < r \\), the line intersects the circle." }, { "problem_id": 220, "question": "Given that the radius of circle $O$ is $1$ and the length of $O A$ is $1.2$, which of the following figures could be correct ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0073_1.jpg", "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0073_2.jpg", "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0073_3.jpg", "batch19-2024_05_24_b0b9f062b2d178a3a8f6g_0073_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Since the radius \\( r \\) of circle \\( O \\) is 1, and \\( OA = 1.2 \\),\n\n\\(\\therefore r < OA\\),\n\n\\(\\therefore\\) point \\( A \\) lies outside the circle,\n\nTherefore, the correct choice is: D.\n\n【Key Insight】This question examines the positional relationship between a point and a circle. The key to solving it lies in understanding that the position of a point relative to a circle is determined by the distance from the point to the center of the circle and the radius of the circle." }, { "problem_id": 221, "question": "As shown in the figure, if $\\angle 1 = \\angle 2$, then $AB = CD$ is ( ).\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_b3eaff0bd129c384e98bg_0086_1.jpg", "batch19-2024_05_24_b3eaff0bd129c384e98bg_0086_2.jpg", "batch19-2024_05_24_b3eaff0bd129c384e98bg_0086_3.jpg", "batch19-2024_05_24_b3eaff0bd129c384e98bg_0086_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: According to the principle that in the same or equal circles, equal arcs subtend equal central angles, only option C fits the question;\n\n\n\nSince $\\angle 1 = \\angle 2$,\n\nTherefore, $AB = CD$.\n\nHence, the correct choice is: C.\n\n[Key Insight] This question tests the relationship between central angles and arcs. Understanding that in the same or equal circles, equal central angles subtend equal arcs is crucial for solving the problem." }, { "problem_id": 222, "question": "A park plans to build a fountain with a shape as shown in Figure (1). Later, someone suggests changing it to the shape in Figure (2), keeping the diameter of the outer circle unchanged, and maintaining the width and height of the fountain edge. Which of the following is true about the amount of material needed to build the fountain edge ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. Figure (1) requires more material.\nB. Figure (2) requires more material.\nC. Figures (1) and (2) require the same amount of material.\nD. It cannot be determined.", "input_image": [ "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0046_1.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0046_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Let the diameter of the large circle be \\( D \\), and the diameters of the three small circles in Figure (2) be \\( d_1, d_2, d_3 \\), respectively.\n\n\\[\n\\therefore d_1 + d_2 + d_3 = D\n\\]\n\nAccording to the circumference formula of a circle, in Figure (1), the required length is \\( 2\\pi D \\).\n\nIn Figure (2), the required length is:\n\\[\n\\pi D + \\pi d_1 + \\pi d_2 + \\pi d_3 = \\pi D + \\pi(d_1 + d_2 + d_3) = 2\\pi D\n\\]\n\nTherefore, the correct choice is: C.\n\n**Key Point:** In the second figure, when calculating the circumferences of the three small circles, factor out \\( \\pi \\), and the sum of all diameters equals the diameter of the large circle." }, { "problem_id": 223, "question": "As shown in the figure, point $C$ is a moving point on the diameter $A B$ of circle $\\odot O$. A line passing through point $C$ intersects circle $\\odot O$ at points $D$ and $E$, with $\\angle A C D = 45^\\circ$. Line $D F$ is perpendicular to $A B$ at point $F$, and line $E G$ is perpendicular to $A B$ at point $G$. As point $C$ moves along $A B$, let $A F = x$ and $D E = y$. Which of the following graphs approximately represents the relationship between $y$ and $x$ as a function?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0082_1.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0082_2.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0082_3.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0082_4.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0082_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "The movement of point $\\mathrm{C}$ from point $\\mathrm{A}$ to point $\\mathrm{B}$ is illustrated in the diagram below,\n\n\nDuring the movement of point $\\mathrm{C}$ from point $\\mathrm{A}$ to point $\\mathrm{B}$, the value of $\\mathrm{x}$ gradually increases, and the length of $\\mathrm{DE}$, denoted as $\\mathrm{y}$, initially increases and then decreases as $\\mathrm{x}$ changes. When $\\mathrm{C}$ coincides with $\\mathrm{O}$, $\\mathrm{y}$ reaches its maximum value.\n\nWhen $\\mathrm{x}=0$, as shown in the diagram below,\n\n\n$\\because \\angle \\mathrm{AGE}=90^{\\circ}, \\mathrm{OD}=\\mathrm{OE}$\n\n$\\therefore \\triangle \\mathrm{DEO}$ is an isosceles right triangle\n\n$\\therefore \\mathrm{DE}^{2}=\\mathrm{OA}^{2}+\\mathrm{OE}^{2}$\n$\\mathrm{DE}^{2}=\\left(\\frac{1}{2} \\mathrm{AB}\\right)^{2}+\\left(\\frac{1}{2} \\mathrm{AB}\\right)^{2}$\n\n$\\mathrm{DE}^{2}=\\frac{1}{2} \\mathrm{AB}^{2}$\n\n$\\therefore \\mathrm{DE}=\\frac{\\sqrt{2}}{2} \\mathrm{AB}$\n\n$\\therefore$ when $\\mathrm{x}=0$, $\\mathrm{y}=\\frac{\\sqrt{2}}{2} \\mathrm{AB}$,\n\nWhen $\\mathrm{DE}$ passes through point $\\mathrm{O}$, as shown in the diagram below,\n\n\nAt this point: $\\mathrm{DE}=\\mathrm{AB}$\n\n$\\because \\angle \\mathrm{AOD}=45^{\\circ}, \\angle \\mathrm{OFD}=90^{\\circ}$,\n\n$\\therefore \\triangle \\mathrm{DFO}$ is an isosceles right triangle\n\n$\\mathrm{DF}^{2}+\\mathrm{OF}^{2}=\\mathrm{OD}^{2}$\n\n$2 \\mathrm{OF}^{2}=\\left(\\frac{1}{2} \\mathrm{AB}\\right)^{2}, \\mathrm{OF}=\\frac{\\sqrt{2}}{4} \\mathrm{AB}$\n\n$\\therefore \\mathrm{AF}=\\frac{1}{2} \\mathrm{AB}-\\frac{\\sqrt{2}}{4} \\mathrm{AB}$\n\n$\\therefore$ when $x=\\frac{1}{2} A B-\\frac{\\sqrt{2}}{4} A B$, $y=D E=A B$ reaches its maximum value,\n\nWhen $x=A B$, as shown in the diagram below,\n\n\n$\\because \\triangle \\mathrm{EOB}$ is a right triangle\n\n$\\therefore \\mathrm{DE}^{2}=\\mathrm{OB}^{2}+\\mathrm{OE}^{2}$\n\n$\\mathrm{DE}^{2}=\\left(\\frac{1}{2} \\mathrm{AB}\\right)^{2}+\\left(\\frac{1}{2} \\mathrm{AB}\\right)^{2}$\n\n$\\mathrm{DE}^{2}=\\frac{1}{2} \\mathrm{AB}^{2}$\n\n$\\mathrm{DE}^{2}=\\frac{\\sqrt{2}}{2} \\mathrm{AB}$\n\n$x=A B, y=\\frac{\\sqrt{2}}{2} \\quad A B$\n\nTherefore, as $\\mathrm{x}$ increases, $\\mathrm{y}$ first increases and then decreases, resembling a parabolic curve,\nHence, the correct choice is: A.\n\n【Key Insight】This problem examines the function image of a moving point. The key to solving this problem lies in analyzing the trend of how $\\mathrm{y}$ changes with $\\mathrm{x}$." }, { "problem_id": 224, "question": "A circular paper was given to Xiaofang, who performed the following consecutive operations:\n\n(1) Fold the circular paper horizontally, creating a crease $\\mathrm{AB}$, as shown in Figure (2).\n\n(2) Fold the circular paper vertically, so that points A and B overlap, creating a crease $\\mathrm{CD}$ that intersects $\\mathrm{AB}$ at point M, as shown in Figure (3).\n\n(3) Fold the circular paper along $\\mathrm{EF}$, so that points B and M overlap, with crease $\\mathrm{EF}$ intersecting $\\mathrm{AB}$ at point N, as shown in Figure (4).\n\n(4) Connect points $A E, A F, B E, B F$, as shown in Figure (5).\n\nAfter these operations, Xiaofang concluded:\n\n(1) $\\mathrm{CD} \\parallel \\mathrm{EF}$; (2) Quadrilateral $\\mathrm{MEBF}$ is a rhombus; (3) Triangle $\\mathrm{AEF}$ is equilateral; (4) The area ratio of quadrilateral $\\mathrm{AEFB}$ to rhombus $\\mathrm{EMF}$ is $3 \\sqrt{3}: \\pi$. How many of these conclusions are correct? $($ )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\n\n\n\nFigure (5)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch19-2024_05_24_b5a33cb1b8c467c0e128g_0036_1.jpg", "batch19-2024_05_24_b5a33cb1b8c467c0e128g_0036_2.jpg", "batch19-2024_05_24_b5a33cb1b8c467c0e128g_0036_3.jpg", "batch19-2024_05_24_b5a33cb1b8c467c0e128g_0036_4.jpg", "batch19-2024_05_24_b5a33cb1b8c467c0e128g_0036_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since the paper is folded vertically, points A and B coincide,\n\n$\\therefore \\angle \\mathrm{BMD}=90^{\\circ}$,\n\nSince the paper is folded along $\\mathrm{EF}$, points $\\mathrm{B}$ and $\\mathrm{M}$ coincide,\n\n$\\therefore \\angle \\mathrm{BNF}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{BMD}=\\angle \\mathrm{BNF}=90^{\\circ}$,\n\n$\\therefore \\mathrm{CD} / / \\mathrm{EF}$, thus (1) is correct;\n\nAccording to the perpendicular bisector theorem, $\\mathrm{BM}$ is the perpendicular bisector of $\\mathrm{EF}$,\n\nAlso, since the paper is folded along $\\mathrm{EF}$, points $\\mathrm{B}$ and $\\mathrm{M}$ coincide,\n\n$\\therefore \\mathrm{BN}=\\mathrm{MN}, \\quad \\therefore \\mathrm{BM}$ and $\\mathrm{EF}$ bisect each other perpendicularly,\n\n$\\therefore$ quadrilateral MEBF is a rhombus, thus (2) is correct;\n\n$\\because \\mathrm{ME}=\\mathrm{MB}=2 \\mathrm{MN}$,\n$\\therefore \\angle \\mathrm{MEN}=30^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{EMN}=90^{\\circ}-30^{\\circ}=60^{\\circ}$\n\nAlso, since $\\mathrm{AM}=\\mathrm{ME}$ (both are radii),\n\n$\\therefore \\angle \\mathrm{AEM}=\\angle \\mathrm{EAM}$,\n\n$\\therefore \\angle \\mathrm{AEM}=\\frac{1}{2} \\angle \\mathrm{EMN}=\\frac{1}{2} \\times 60^{\\circ}=30^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{AEF}=\\angle \\mathrm{AEM}+\\angle \\mathrm{MEN}=30^{\\circ}+30^{\\circ}=60^{\\circ}$,\n\nSimilarly, $\\angle \\mathrm{AFE}=60^{\\circ}, \\quad \\therefore \\angle \\mathrm{EAF}=60^{\\circ}$,\n\n$\\therefore \\triangle \\mathrm{AEF}$ is an equilateral triangle, thus (3) is correct;\n\nLet the radius of the circle be $\\mathrm{r}$, then $\\mathrm{EN}=\\frac{\\sqrt{3}}{2} r, \\quad \\therefore \\mathrm{EF}=2 \\mathrm{EN}=\\sqrt{3} r$,\n\n$\\therefore \\mathrm{S}_{\\text {quadrilateral } \\mathrm{AEBF}}: \\mathrm{S}_{\\text {sector } \\mathrm{BEMF}}=\\left(\\frac{1}{2} \\times \\sqrt{3} r \\times 2 r\\right):\\left(\\frac{120}{360} \\pi r^{2}\\right)=3 \\sqrt{3}: \\pi$,\n\nThus (4) is correct,\n\nIn summary, the correct conclusions are (1)(2)(3)(4), totaling 4.\n\nTherefore, the answer is: D.\n\n【Key Insight】This problem is a comprehensive circle problem, mainly examining the properties of folding transformations, the determination of parallel lines, the property that a quadrilateral with diagonals bisecting each other perpendicularly is a rhombus, and the determination and properties of equilateral triangles. It is crucial to note the corresponding relationships of the figures before and after folding." }, { "problem_id": 225, "question": "In the following figures, which one has $\\angle B A C$ as a central angle of a circle?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_b72b576918335d3d02e7g_0076_1.jpg", "batch19-2024_05_24_b72b576918335d3d02e7g_0076_2.jpg", "batch19-2024_05_24_b72b576918335d3d02e7g_0076_3.jpg", "batch19-2024_05_24_b72b576918335d3d02e7g_0076_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of a central angle, the angle $\\angle BAC$ that qualifies as a central angle is: $\\mathrm{B}$. The angles that do not qualify as central angles are: $\\mathrm{A}$, C, D.\n\nTherefore, the correct choice is B.\n\n【Key Point】This question tests the understanding of the definition of a central angle. The question is relatively simple, and the key to solving it lies in comprehending the definition of a central angle." }, { "problem_id": 226, "question": "Given that the radii of $\\odot \\mathrm{O}_{1}$ and $\\odot \\mathrm{O}_{2}$ are 1 and 4, respectively, if the circles are in an intersecting position, which of the following correctly represents the range of the distance between the centers, $\\mathrm{O}_{1} \\mathrm{O}_{2}$, on a number line?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_c146e2aaeb1beea8235eg_0003_1.jpg", "batch19-2024_05_24_c146e2aaeb1beea8235eg_0003_2.jpg", "batch19-2024_05_24_c146e2aaeb1beea8235eg_0003_3.jpg", "batch19-2024_05_24_c146e2aaeb1beea8235eg_0003_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Let the distance between the centers of the two circles be $\\mathrm{d}$. Since the condition for the two circles to intersect is $\\left|r_{1}-r_{2}\\right|\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_ce4c68b4e78efc164f53g_0017_1.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0017_2.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0017_3.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0017_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: \nA: The radius of circle $O$ is $r=6$, and the distance from point $O$ to line $l$ is 3. Therefore, A is correct. \n\nB: The distance from point $O$ to line $l$ is 6, so B is incorrect. \n\nC: The distance from point $O$ to line $l$ is greater than 6, so C is incorrect. \n\nD: The distance from point $O$ to line $l$ is 0, so D is incorrect. \n\nThus, the correct answer is: A. \n\n**[Key Insight]** This question examines the positional relationship between a line and a circle. The key to solving it lies in comparing the distance from the center of the circle to the line with the radius of the circle." }, { "problem_id": 228, "question": "As shown in the figure, point $A$ is on the circle $\\odot O$ with radius 2. A line $l$ is drawn through a point $P$ on segment $O A$, intersecting the tangent to $\\odot O$ at point $A$ at point $B$, such that $\\angle A P B = 60^\\circ$. Let $O P = x$, then the area $y$ of triangle $P A B$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_ce4c68b4e78efc164f53g_0070_1.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0070_2.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0070_3.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0070_4.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0070_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Since line segment \\( AB \\) is tangent to circle \\( \\odot O \\),\n\nit follows that \\( \\angle BAP = 90^\\circ \\).\n\nGiven that \\( OP = x \\), \\( AP = 2 - x \\), and \\( \\angle BPA = 60^\\circ \\),\n\nwe can deduce that \\( AB = \\sqrt{3}(2 - x) \\).\n\nTherefore, the area \\( y \\) of triangle \\( \\triangle APB \\) is \\( y = \\frac{\\sqrt{3}}{2}(2 - x)^2 \\), for \\( 0 \\leq x \\leq 2 \\).\n\nThus, the graph of the area \\( y \\) of triangle \\( \\triangle PAB \\) as a function of \\( x \\) is a parabola passing through the point \\( (2, 0) \\) over the interval \\( 0 \\leq x \\leq 2 \\).\n\nHence, the correct choice is D." }, { "problem_id": 229, "question": "A school held a school anniversary party. The main stage was a circular stage with $O as the center. A and B$ are two fixed positions on the edge of the stage. The area enclosed by the line segment $AB$ and the major arc $AB$ is the performance area. If a lighting device of a certain model is installed at $A$, the illuminated area is shown as the shaded area in Figure 1. If another lighting device of the same model is installed at $B$, it can just illuminate the entire performance area, as shown as the shaded area in Figure 2\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nIf the lighting device is placed at point $M, N$ or $P$ as shown in Figure 3, the solution to completely illuminate the performance area may be $(\\quad)$\n\n(1)Place 2 lighting devices of the same model at $M$\n\n(2)Place 1 lighting device of the same model at $B$\n\n(3)Place 2 lighting devices of the same model at $M$\n\n(4)Place 2 lighting devices of the same model at $M$\n\n(5)Place 2 lighting devices of the same model at $B$\n\n(6)Place 2 lighting devices of the same model at $M$\n\n(7)Place 2 lighting devices of the same model at $B$\n\n(8)Place 2 lighting devices of the same model at $M$\n\n(9)Place 2 lighting devices of the same model at $B$\n\n Place one lighting device of the same model at each of $M and N$\n\n(3) Place two lighting devices of the same model at $P$\nA. (1)\nB. (1)(2)\nC. (2)(3)\nD. (1)(2)", "input_image": [ "batch19-2024_05_24_d3a2c3783a8ca4a9acccg_0041_1.jpg", "batch19-2024_05_24_d3a2c3783a8ca4a9acccg_0041_2.jpg", "batch19-2024_05_24_d3a2c3783a8ca4a9acccg_0041_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Place 2 units of this model lighting device at point $M$, as shown in the figure below.\n\n\n\n$\\because$ Installing one unit of a certain model lighting device at each of points $A$ and $B$ can exactly illuminate the entire performance area,\n\n$\\therefore \\angle C A B + \\angle C B A =$ the central angle corresponding to the major arc $A B$.\n\nTo illuminate the entire performance area, the two lighting devices must cover an angle of $\\angle E M F$, and $\\angle E M F = \\angle A M B$.\n\n$\\therefore \\angle A M B$ is the central angle corresponding to the major arc $A B$.\n\n$\\therefore \\angle A M B = \\angle C A B + \\angle C B A$, which means scheme (1) is valid;\n\nPlace 1 unit of this model lighting device at each of points $M$ and $N$, and connect $A M$, $B M$, $A N$, $B N$, $C M$, $C N$ respectively, as shown in the figure below.\n\n\n\n$\\because \\angle A N C = \\angle A B C$, $\\angle B M C = \\angle B A C$,\n\n$\\therefore$ the scheme is valid;\n\nPlace 2 units of this model lighting device at point $P$, as shown in the figure below, where $E F$ is tangent to $O$ at point $P$.\n\n\n\nTo illuminate the entire performance area, the two lighting devices must cover a total angle of $\\angle E P F = 180^{\\circ}$.\n\nAccording to the problem statement, $\\angle C A B + \\angle C B A < 180^{\\circ}$, which means the total angle covered by the two lighting devices is $< 180^{\\circ}$.\n\n$\\therefore$ scheme (3) is not valid;\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem examines knowledge of circles and the sum of angles in a triangle; the key to solving it lies in a thorough understanding of the properties of central angles, thereby enabling the solution to be found." }, { "problem_id": 230, "question": "As shown in Figure (1), point $C$ is a moving point on the semicircle with center $O$ and diameter $A B$ (which can coincide with points $A$ or $B$). A perpendicular line $C D$ is drawn from point $C$ to $A B$ at point $D$. The segment $C A$ is joined, and the length of $C A$ is denoted as $x$, while the length of $C D$ is denoted as $y$. Figure (2) represents the graph of the function relationship between $y$ and $x$ as point $C$ moves, where the highest point $M$ has coordinates ( ).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $(2,2)$\nB. $(2 \\sqrt{2}, 2)$\nC. $(2,3)$\nD. $(2 \\sqrt{2}, 3)$", "input_image": [ "batch19-2024_05_24_dd341a611d5d139e3378g_0010_1.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0010_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the graph, we have $AB = 4$,\n\n$\\therefore$ the radius of circle $O$ is 2.\n\nWhen point $D$ coincides with point $O$, $CD$ is at its maximum,\n\nAt this point, $CD$ is equal to the radius of circle $O$,\n\n$\\therefore y = 2$,\n\nWhen $y = 2$, $x = \\sqrt{AD^{2} + CD^{2}} = \\sqrt{2^{2} + 2^{2}} = 2\\sqrt{2}$,\n\n$\\therefore$ the coordinates of point $M$ are $(2\\sqrt{2}, 2)$,\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem mainly examines the function graph of moving point problems. The key to solving it is to determine the radius of the circle based on the graph and to identify the position of point $C$ when $CD$ is at its maximum." }, { "problem_id": 231, "question": "In the following figures, $\\angle A O B$ is the central angle in $(\\quad)$\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_dd341a611d5d139e3378g_0011_1.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0011_2.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0011_3.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0011_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of a central angle:\n\nA. The vertex is not at the center of the circle, so option A does not meet the requirement;\n\nB. The vertex is on the circle, $\\angle AOB$ is an inscribed angle, so option B does not meet the requirement;\n\nC. The vertex of $\\angle AOB$ is at the center of the circle, and both sides intersect the circle, so option C meets the requirement;\n\nD. The vertex is on the circle, $\\angle AOB$ is an inscribed angle, so option D does not meet the requirement.\n\nTherefore, the correct choice is: C.\n\n【Key Point】This question tests the definitions of central and inscribed angles. The key to solving this problem is to understand the definitions of central and inscribed angles." }, { "problem_id": 232, "question": "As shown in the figure, $M$ is a fixed point on line segment $O$. A right-angled triangle with a $30^\\circ$ angle is placed so that its right-angle vertex coincides with point $M$, and its sides intersect $O$ at points $A$ and $B$. The triangle is rotated clockwise around point $M$ until one of the intersection points coincides with point $M$ again. Let $AB = y$ and the angle of rotation be $\\alpha$. Which of the following graphs can represent the relationship between $y$ and $\\alpha$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_dd341a611d5d139e3378g_0015_1.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0015_2.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0015_3.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0015_4.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0015_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "According to the problem, $\\angle BMA$ is a circumferential angle, and chord $AB$ is the chord subtended by $\\angle BMA$.\n\nWhen the triangle is rotated clockwise around point $M$, the size of $\\angle BMA$ remains unchanged, so the length of chord $AB$ does not change, meaning $y$ does not vary with $\\alpha$.\n\nTherefore, the correct choice is A.\n\n【Key Insight】This question primarily tests the properties of circumferential angles. The key to solving it lies in understanding the definition of a circumferential angle." }, { "problem_id": 233, "question": "As shown in the figure, given that $A B$ is the diameter of the semicircle $O$, $A B=6$, point $P$ is a point on the semicircle $O$ (not coinciding with points $A$ or $B$), $P C$ is perpendicular to $A B$ at point $C$, $O D$ is perpendicular to $B P$, and $O E$ is perpendicular to $A P$, with the feet of the perpendiculars being points $D$ and $E$, respectively. If $A C=x$ and $O E^2+O D^2=y$, then the approximate graph of $y$ versus $x$ is $(\\quad)$\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_e1df54a2873d1b01739ag_0025_1.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0025_2.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0025_3.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0025_4.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0025_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Connect $DE$ and $OP$ as shown in the diagram:\n\n\n\nSince $AB$ is the diameter of the semicircle,\n\n$\\therefore \\angle APB = 90^{\\circ}$,\n\nSince $OE \\perp AP$ and $OD \\perp BP$,\n\n$\\therefore$ quadrilateral $ODPE$ is a rectangle,\n\n$\\therefore DE = OP = 3$,\n\nIn right triangle $ABC$,\n\n$OE^{2} + OD^{2} = DE^{2}$,\n\n$\\therefore y = OE^{2} + OD^{2}$\n\n$= OP^{2}$\n\n$= 3^{2}$\n\n$= 9$ (for $0 < x < 6$),\n\nTherefore, the answer is: A.\n\n【Key Insight】This problem primarily examines the inscribed angle theorem and the function graph of a moving point. The key to solving it is to understand that the angle subtended by the diameter is a right angle." }, { "problem_id": 234, "question": "Draw a perpendicular line $A D$ to line $M N$ through point $A$ using a ruler and compass, as shown in the figure. Which of the following methods are correct ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)(3)\nB. (2)(3)(4)\nC. (1)(2)(4)\nD. (1)(2)(3)(4)", "input_image": [ "batch19-2024_05_24_f70a56b6ae318beb8943g_0055_1.jpg", "batch19-2024_05_24_f70a56b6ae318beb8943g_0055_2.jpg", "batch19-2024_05_24_f70a56b6ae318beb8943g_0055_3.jpg", "batch19-2024_05_24_f70a56b6ae318beb8943g_0055_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: In Figure (1), according to the Inscribed Angle Theorem, $\\angle ADN = 90^{\\circ}$, which satisfies the given condition;\n\nIn Figure (2), from the construction, it is known that $AD \\perp MN$, which satisfies the given condition;\n\nIn Figure (3), from the construction, it is known that $MN$ is the perpendicular bisector of segment $AD$, which satisfies the given condition;\n\nIn Figure (4), based on the property that the three lines of an isosceles triangle coincide, it is known that $AD \\perp MN$, which satisfies the given condition.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question tests the knowledge of complex constructions and perpendicular lines. The key to solving the problem lies in mastering the five basic constructions, and it is a common type of question in middle school exams." }, { "problem_id": 235, "question": "From the following positions of a right-angled triangle ruler and an arc, which indicates that the arc is a semicircle?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_1508ae9282a0e8806141g_0015_1.jpg", "batch20-2024_05_23_1508ae9282a0e8806141g_0015_2.jpg", "batch20-2024_05_23_1508ae9282a0e8806141g_0015_3.jpg", "batch20-2024_05_23_1508ae9282a0e8806141g_0015_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Since the angle subtended by the diameter on the circumference is a right angle,\n\nTherefore, from the positional relationship between the right-angled triangle and the arc, it can be determined that the arc forming a semicircle is B.\n\nHence, the correct choice is: B\n[Key Insight] This question tests the corollary of the Inscribed Angle Theorem. Understanding that the angle subtended by the diameter is a right angle is crucial for solving the problem." }, { "problem_id": 236, "question": "As shown in Figure 1, a moving point $P$ starts from point $A$ of a regular hexagon and moves along $A \\rightarrow B \\rightarrow C \\rightarrow D \\rightarrow E$ at a constant speed of $1 \\mathrm{~cm} / \\mathrm{s}$ until it reaches point $E$. Figure 2 shows the graph of the area $y\\left(\\mathrm{~cm}^{2}\\right)$ of triangle $A P E$ as a function of time $x(\\mathrm{~s})$ while point $P$ is moving. The value of $m$ in Figure 2 is $(\\quad)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{8 \\sqrt{3}}{3} \\mathrm{~cm}^{2}$\nB. $3 \\sqrt{3} \\mathrm{~cm}^{2}$\nC. $\\frac{9 \\sqrt{3}}{4} \\mathrm{~cm}^{2}$\nD. $4 \\sqrt{3} \\mathrm{~cm}^{2}$", "input_image": [ "batch20-2024_05_23_2453e1228cee9a0540beg_0007_1.jpg", "batch20-2024_05_23_2453e1228cee9a0540beg_0007_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Connect $AC$ and $EC$, draw $BG \\perp AC$ at point $G$, and draw $AH \\perp EP$ at point $H$.\n\n\n\nWhen point $P$ moves to point $C$, the area of $APE$ is maximized.\n\nFrom Figure 2, we know that $AB + BC = 2\\sqrt{3}$.\n\nSince the hexagon $ABCDEF$ is a regular hexagon,\n\n$$\n\\therefore AB = BC = CD = DE = EF = FA, \\quad \\angle ABC = \\angle F = \\angle D = 120^\\circ,\n$$\n\n$\\therefore AB = \\sqrt{3}$,\n\n$\\therefore AG = AB \\cdot \\sin 60^\\circ = \\frac{\\sqrt{3}}{2} \\times \\sqrt{3} = \\frac{3}{2}$,\n\n$\\therefore AC = 2AG = 3$,\n\nSince $AB = BC = CD = DE = EF = FA$, and $\\angle ABC = \\angle F = \\angle D = 120^\\circ$,\n\n$\\therefore ABC \\cong EDC \\cong AFE$,\n\n$\\therefore AC = EC = AE$,\n\n$\\therefore AEC$ is an equilateral triangle,\n\n$\\therefore AH = \\frac{\\sqrt{3}}{2} AC$,\n\n$\\therefore S_{AEC} = \\frac{1}{2} EC \\cdot AH = \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} AC \\times AC = \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} \\times 3^2 = \\frac{9\\sqrt{3}}{4}$,\n\n$\\therefore m = \\frac{9\\sqrt{3}}{4}$.\n\nTherefore, the answer is: C.\n\n【Key Insight】This problem examines the properties of a regular hexagon and an equilateral triangle. The key to solving the problem is identifying the position where the area is maximized." }, { "problem_id": 237, "question": "The front view of a \"non-toppling cup\" (Figure 1) is shown in Figure 2. Segments $P A$ and $P B$ are tangent to the circle $O$ containing $\\triangle A M B$ at points $A$ and $B$, respectively. If the radius of this circle is $3 \\mathrm{~cm}$ and $\\angle P = 60^\\circ$, then the length of $\\triangle A M B$ is ( )\n\n\n\nFront View\n\n\n\nFigure 2\nA. $6 \\pi \\mathrm{cm}$\nB. $4 \\pi \\mathrm{cm}$\nC. $3 \\pi \\mathrm{cm}$\nD. $2 \\pi \\mathrm{cm}$", "input_image": [ "batch20-2024_05_23_2453e1228cee9a0540beg_0031_1.jpg", "batch20-2024_05_23_2453e1228cee9a0540beg_0031_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $O A, O B$\n\n\n\nFigure 2\n\n$\\because P A, P B$ are tangent to the circle at points $A, B$ respectively.\n\n$\\therefore \\angle P A O=\\angle P B O=90^{\\circ}$,\n\n$\\because \\angle P=60^{\\circ}$,\n\n$\\therefore \\angle A O B=360^{\\circ}-90^{\\circ}-90^{\\circ}-60^{\\circ}=120^{\\circ}$,\n\n$\\because$ the radius of the circle is $3 \\mathrm{~cm}$,\n\n$\\therefore A M B=\\frac{360-120}{180} \\pi \\times 3=4 \\pi \\mathrm{cm}$,\n\nTherefore, the answer is: B.\n\n[Key Insight] This problem tests the properties of tangents and the calculation of arc length. Remembering the arc length formula is crucial for solving the problem." }, { "problem_id": 238, "question": "As shown in the figure, the three views and related data of a geometric solid are given. Determine the surface area of the solid ( ).\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View\nA. $16 \\pi$\nB. $20 \\pi$\nC. $12 \\pi$\nD. $15 \\pi$", "input_image": [ "batch20-2024_05_23_36bcbfd945dbcb1ba197g_0033_1.jpg", "batch20-2024_05_23_36bcbfd945dbcb1ba197g_0033_2.jpg", "batch20-2024_05_23_36bcbfd945dbcb1ba197g_0033_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: The surface area of the geometric figure $=\\frac{1}{2} \\times 4 \\pi \\times 8 \\pi \\times 2^{2}=20 \\pi$.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question examines the calculation of the surface area of a cone: the lateral surface of a cone unfolds into a sector of a circle, where the arc length of the sector equals the circumference of the cone's base, and the radius of the sector equals the slant height of the cone. It also tests the recognition of three-dimensional views." }, { "problem_id": 239, "question": "Among the following figures, which one is called a sector? $(\\quad)$\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0007_1.jpg", "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0007_2.jpg", "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0007_3.jpg", "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0007_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of a sector (a figure enclosed by two radii of a central angle and the arc subtended by that angle), the correct option is a sector, while the other options are not.\n\nTherefore, the answer is: B.\n\n[Key Point] This question tests the understanding of a sector, and mastering the definition of a sector is crucial to answering this question correctly." }, { "problem_id": 240, "question": "As shown in the figure, $O B C A$ is a square. In Figure 1, a semicircle is drawn with $A B$ as the diameter, and the shaded area is denoted as $S_{1}$. In Figure 2, an arc is drawn with $O$ as the center and $O A$ as the radius, and the shaded area is denoted as $S_{2}$. Then ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $S_{1}S_{2}$\nD. It cannot be determined", "input_image": [ "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0018_1.jpg", "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0018_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Let the side length of the square be \\( a \\), then \\( AB = \\sqrt{2}a \\).\n\nAccording to the problem, \n\\[\nS_{1} = \\frac{1}{2} \\cdot \\pi \\cdot \\left(\\frac{1}{2} \\cdot \\sqrt{2}a\\right)^{2} - \\frac{1}{2} \\cdot a \\cdot a = \\frac{1}{4} \\pi a^{2} - \\frac{1}{2}a^{2},\n\\]\n\\[\nS_{2} = \\frac{1}{4} \\cdot \\pi \\cdot a^{2} - \\frac{1}{2} \\cdot a \\cdot a = \\frac{1}{4} \\pi a^{2} - \\frac{1}{2}a^{2}.\n\\]\nThus, \\( S_{1} = S_{2} \\).\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem tests the calculation of the area of a sector. The key to solving it lies in mastering the formula for the area of a sector and the method for calculating the area of composite shapes." }, { "problem_id": 241, "question": "As shown in the figure, on a horizontal ground, there is a gray sector $O A B$ with an area of $30 \\pi \\text{cm}^2$. The length of $O A$ is $6 \\text{~cm}$, and $O A$ is perpendicular to the ground. If the sector, as shown in Figure ( A), is rolled without slipping to the right until point $A$ touches the ground again, as shown in Figure ( B), how far has point $O$ moved in $\\text{cm}$?\n\n\n\nFigure ( A)\n\n\n\nFigure ( B)\nA. $11 \\pi$\nB. $12 \\pi$\nC. $10 \\pi+2 \\sqrt{3}$\nD. $11 \\pi+\\sqrt{3}$", "input_image": [ "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0071_1.jpg", "batch20-2024_05_23_3f2b16bce4ad53bd5fa5g_0071_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "It can be divided into two processes: \n\nThe first process involves the sector moving to the right, where the distance moved by point $O$ equals the arc length of $\\frac{2 S}{r}=\\frac{2 \\times 30 \\pi}{6}=10 \\pi$.\n\nThe second process involves the major angle $\\angle B O A=\\frac{30 \\pi \\times 360^{\\circ}}{\\pi \\times 6^{2}}=300^{\\circ}$,\n\nthus the acute angle $\\angle B O A=60^{\\circ}, \\frac{30 \\pi \\times 6}{180}=\\pi$.\n\nTherefore, point $O$ has moved a total of $11 \\pi \\mathrm{cm}$.\n\nHence, the correct choice is A.\n\n【Key Insight】This question tests the calculation of arc length. Understanding the movement process of point $O$ and mastering the arc length formula are crucial for solving the problem." }, { "problem_id": 242, "question": "Among the figures we have studied, the distance between any pair of parallel tangents to a circle is always equal, making the circle an \"equal-width curve.\" In addition to circles, there are other geometric figures that are also \"equal-width curves,\" such as the Reuleaux triangle (as shown in Figure 1), which is formed by drawing a circular arc of radius equal to the side length of an equilateral triangle, centered at each vertex, connecting the arcs at the other two vertices. Figure 2 shows a cross-sectional view of an equal-width Reuleaux triangle and a circular log.\n\nThere are the following four conclusions:\n\n(1) The Reuleaux triangle is a centrally symmetric figure;\n\n(2) In Figure 1, if the side length of the equilateral triangle is 2, then the perimeter of the Reuleaux triangle is $2 \\pi$;\n\n(3) In Figure 2, the perimeter of the Reuleaux triangle is equal to the circumference of the circle;\n\n(4) Using a log with a cross-section in the shape of a Reuleaux triangle to transport items will not cause any up-and-down shaking.\n\nWhich of the above conclusions are correct? The sequence numbers of all correct conclusions are ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. (1)(2)\nB. (2)(3)\nC. (3)(4)\nD. (2)(3)(4)", "input_image": [ "batch20-2024_05_23_41326cc1be232a221581g_0083_1.jpg", "batch20-2024_05_23_41326cc1be232a221581g_0083_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: (1) The Reuleaux triangle is not a centrally symmetric figure, so statement ① is incorrect and does not meet the requirements of the question.\n\n(2) In Figure 1, the side length of the equilateral triangle is 2, so the perimeter of the Reuleaux triangle is $\\frac{60 \\times \\pi \\times 2}{180} \\times 3=2 \\pi$. Therefore, statement (2) is correct and meets the requirements of the question.\n\n(3) In Figure 2, let the side length of the equilateral triangle in the Reuleaux triangle be $a$. Then, the diameter of the circle is $a$, so the perimeter of the Reuleaux triangle is $\\frac{60 \\times \\pi \\times a}{180} \\times 3=a \\pi$, and the circumference of the circle is also $a \\pi$. Therefore, in Figure 2, the perimeter of the Reuleaux triangle is equal to the circumference of the circle, so statement (3) is correct and meets the requirements of the question.\n\n(4) The Reuleaux triangle sandwiched between parallel lines will always maintain a constant distance between the lines no matter how it rolls. For example, in Figure 1, the distance from point $\\mathrm{A}$ to any point on $BC$ is the same. Therefore, using a rolling log with a Reuleaux triangle cross-section to move objects will not cause any vertical jitter, so statement (4) is correct and meets the requirements of the question.\n\nThus, the numbers of all correct statements are: (2)(3)(4).\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question introduces a new definition and primarily examines the distance between parallel lines, the properties of equilateral triangles and circles, central symmetry, and the arc length formula. Correctly understanding the new definition and being proficient with related concepts and properties are crucial for answering this question." }, { "problem_id": 243, "question": "When the following four regular polygons are simultaneously rotated around their centers by an equal angle, the one that will first overlap with its original shape is $(\\quad)$\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_52428f73168146a0a8c3g_0034_1.jpg", "batch20-2024_05_23_52428f73168146a0a8c3g_0034_2.jpg", "batch20-2024_05_23_52428f73168146a0a8c3g_0034_3.jpg", "batch20-2024_05_23_52428f73168146a0a8c3g_0034_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "The measure of the central angle of a square \\( = \\frac{360^{\\circ}}{4} = 90^{\\circ} \\); \nThe measure of the central angle of a regular pentagon \\( = \\frac{360^{\\circ}}{5} = 72^{\\circ} \\); \nThe measure of the central angle of a regular hexagon \\( = \\frac{360^{\\circ}}{6} = 60^{\\circ} \\); \nThe measure of the central angle of a regular octagon \\( = \\frac{360^{\\circ}}{8} = 45^{\\circ} \\). \n\nSince \\( 45^{\\circ} < 60^{\\circ} < 72^{\\circ} < 90^{\\circ} \\), \n\nthe first shape to coincide with the original figure is the regular octagon. \n\nTherefore, the correct answer is: D. \n\n【Key Insight】This question primarily tests the calculation of the central angle of regular polygons. Mastering the method for calculating the central angle of regular polygons is the key to solving the problem." }, { "problem_id": 244, "question": "Among the following life and production phenomena, those that cannot be explained by the basic fact 'two points determine a straight line' are ()\n\n\n\nA single ink line is drawn on a flat plate\n\n\n\nConstruction workers build walls\n\n\n\nTeacups are placed in the meeting place\n\n\n\nA curved river is straightened\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch20-2024_05_23_688a28ab97635c6136b7g_0008_1.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0008_2.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0008_3.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0008_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "The phenomena of life and production depicted in the first, second, and third images can be explained by the fundamental fact that \"two points determine a straight line.\" The knowledge utilized in the fourth image is that \"the shortest distance between two points is a straight line segment.\"\n\nTherefore, the correct choice is: A.\n\n[Key Point] This question primarily examines the properties of straight lines and line segments. Correctly understanding these properties is crucial for solving the problem." }, { "problem_id": 245, "question": "Among the four figures below, which segments (or lines, rays) can intersect?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_688a28ab97635c6136b7g_0021_1.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0021_2.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0021_3.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0021_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "**Question Analysis:** Based on the definitions of lines, rays, and line segments, analyze each option and use the elimination method to solve.\n\n**Solution:**\n\nA. The two lines $\\mathrm{AB}$ and $\\mathrm{CD}$ can intersect, so this option is correct.\n\nB. The ray $CD$ cannot intersect with the line $AB$, so this option is incorrect.\n\nC. The ray $CD$ cannot intersect with the line segment $AB$, so this option is incorrect.\n\nD. The two line segments $\\mathrm{AB}$ and $\\mathrm{CD}$ do not have an intersection point, so this option is incorrect.\n\nTherefore, the correct answer is **A**.\n\n**Key Concepts:** Lines, rays, and line segments." }, { "problem_id": 246, "question": "Given points $A, B, C, D$, draw the figures according to the given statements: (1) straight line $A B$; (2) ray $C B$; (3) segment $C D$, and extend segment $C D$. Which of the following four diagrams is correct $(\\quad)$?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_688a28ab97635c6136b7g_0029_1.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0029_2.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0029_3.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0029_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "In options A and C, the ray is BC, not CB, so options A and C are incorrect. In option B, the extension is not of segment CD but of segment DC, so option B is also incorrect.\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question tests the definitions of lines, rays, and segments. A thorough understanding of these definitions is crucial for solving the problem." }, { "problem_id": 247, "question": "Place a pair of right-angled triangles as shown in the figure. Which of the following pairs have $\\angle \\alpha = \\angle \\beta$?\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)\nB. (2)(3)\nC. (1)(4)\nD. (2)(4)", "input_image": [ "batch20-2024_05_23_688a28ab97635c6136b7g_0091_1.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0091_2.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0091_3.jpg", "batch20-2024_05_23_688a28ab97635c6136b7g_0091_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "In figure $A$, according to the equality of complementary angles of the same angle, we have $\\angle \\alpha = \\angle \\beta$;\n\nIn figure $B$, $\\angle \\alpha > \\angle \\beta$;\n\nIn figure $C$, $\\angle \\alpha < \\angle \\beta$;\n\nIn figure $D$, $\\angle \\alpha = \\angle \\beta = 45^{\\circ}$.\n\nTherefore, the figures where $\\angle \\alpha = \\angle \\beta$ are (1) and (4).\n\nHence, the correct choice is: C.\n\n【Key Point】This question tests the understanding of complementary and supplementary angles. Mastering the concepts of complementary and supplementary angles and correctly comparing the sizes of angles are key to solving this problem." }, { "problem_id": 248, "question": "Question: \"As shown in Figure 1, in a plane, there is a rectangular paper with a length of 12 and a width of 6 inside a square. It can freely be transferred (either by translation or rotation) from a horizontal position to a vertical position within the square. Find the smallest integer n for the side length of the square.\"\n\nStudents A, B, C each attempted to find the smallest square with side length $x$ and then took the smallest integer $n$.\n\n A: As shown in Figure 2, the idea is that when $x$ is the diagonal length of the rectangle, it can be transferred; the result is $n=13$.\n\n B: As shown in Figure 3, the idea is that when $x$ is the diameter of the rectangle's circumscribed circle, it can be transferred; the result is $n=14$.\n\n C: As shown in Figure 4, the idea is that when $x$ is the sum of the length and width of the rectangle, it can be transferred; the result is $n=18$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nWhich of the following evaluations of A, B, C is correct?\nA. Student A's idea is wrong, but the $n$ value is correct.\nB. Student B's idea is correct, and the $n$ value is correct.\nC. Student C's idea is correct, and the $n$ value is correct.\nD. The ideas of students A and B are wrong, but C's idea is correct.", "input_image": [ "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0003_1.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0003_2.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0003_3.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0003_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of the diagonal of the rectangle be $m$, which is $\\sqrt{6^{2}+12^{2}}=\\sqrt{180}$,\n$\\therefore m=\\sqrt{6^{2}+12^{2}}=\\sqrt{180}$\n\n$\\because 13^{2}=169 \\quad, 14^{2}=196$\n\n$\\therefore 13\n\nFigure 1\n\n\n\nFigure 2\nA. $2 \\mathrm{~mm}$\nB. $2 \\sqrt{2} \\mathrm{~mm}$\nC. $2 \\sqrt{3} \\mathrm{~mm}$\nD. $4 \\mathrm{~mm}$", "input_image": [ "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0013_1.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0013_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Connect $CF$ and $AD$ intersecting at point $O$,\n\nSince $ABCDEF$ is a regular hexagon,\n\n$\\therefore \\angle COD = \\frac{360^{\\circ}}{6} = 60^{\\circ}$, $CO = DO$, $AO = DO = \\frac{1}{2} AD = 4 \\mathrm{~mm}$,\n\n$\\therefore \\triangle COD$ is an equilateral triangle,\n\n$\\therefore CD = CO = DO = 4 \\mathrm{~mm}$,\n\nThat is, the side length of the regular hexagon $ABCDEF$ is $4 \\mathrm{~mm}$,\n\nTherefore, the answer is: D.\n\n\n【Key Insight】This question examines the properties of regular polygons and circles. The key to solving the problem lies in correctly understanding the relationship between the central angle, radius, and side length of a regular hexagon." }, { "problem_id": 250, "question": "A regular polygon with an interior angle of $108^{\\circ}$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0029_1.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0029_2.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0029_3.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0029_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Since each interior angle of the regular polygon is $108^{\\circ}$, each exterior angle is: $180^{\\circ} - 108^{\\circ} = 72^{\\circ}$.\n\nTherefore, the number of sides of the regular polygon is: $\\frac{360^{\\circ}}{72^{\\circ}} = 5$.\n\nHence, the correct choice is B.\n\n[Key Insight] This problem tests the understanding that the interior and exterior angles of a regular polygon are supplementary, and the key to solving for the number of sides of a regular polygon lies in utilizing the fact that the sum of the exterior angles is $360^{\\circ}$." }, { "problem_id": 251, "question": "As shown in Figure 1, a regular hexagon (denoted as \"Polygon $P_{1}$\") with a side length of 6 units is given. Divide each side into three equal parts, and cut out 6 small triangles (the shaded triangles in Figure 1) by connecting the two points of each division adjacent to a vertex. These 6 small triangles are assembled to form a regular hexagon with the outline shown in Figure 2 (denoted as \"Polygon $P_{2}$\"). Draw the incircle $\\odot O$ of Polygon $P_{2}$, as shown in Figure 3, leading to the following conclusions:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) The remaining polygon in Figure 1 (the white part) is a regular dodecagon;\n\n(2) Let the blank part in Figure 2 be denoted as \"Polygon $P_{3}$\", then the ratio of the perimeters of Polygon $P_{1}$, $P_{2}$, and $P_{3}$ is $3: 2: \\sqrt{3}$;\n\n(3) The maximum distance from any point on the side of the regular hexagon in Figure 3 to any point on $\\odot O$ is $4+\\sqrt{3}$.\n\nThe correct conclusions are $(\\quad)$\nA. (2)(3)\nB. (1)(3)\nC. (2)\nD. (1)\n\n##", "input_image": [ "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0033_1.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0033_2.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0033_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Label the letters as shown in the figure. Draw a perpendicular from point $B$ to $AC$ at point $N$.\n\n\n\nFigure 1\n\nGiven that $BE = 6$, points $C$ and $D$ are the trisection points of $BE$, and points $A$ and $F$ are the trisection points of $BG$.\n\nThus, $BC = CD = DE = AB = 2$.\n\nSince each internal angle of a regular hexagon is $\\left(180 - \\frac{360}{6}\\right)^\\circ = 120^\\circ$.\n\nIn triangle $ABC$, $AB = BC = 2$, and $\\angle B = 120^\\circ$.\n\nTherefore, $\\angle BAN = 30^\\circ$.\n\nIn right triangle $ABN$,\n\n$BN = \\frac{1}{2} AB = 1$.\n\nThus, $AN = \\sqrt{AB^2 - BN^2} = \\sqrt{2^2 - 1^2} = \\sqrt{3}$.\n\nSince $BN \\perp AC$,\n\n$AN = NC = \\frac{1}{2} AC$,\n\nTherefore, $AC = 2AN = 2\\sqrt{3}$.\n\nSince $CD = 2 \\neq AC$,\n\nStatement (1) is incorrect.\n\nFigure $P_1$ has a side length of 6, so the perimeter of $P_1$ is $6 \\times 6 = 36$.\n\nAs shown in the figure, according to the given conditions, $AB = BC = CD = 2$.\n\nThen, $BD = 4$. According to the conditions, $P_2$ is a regular hexagon.\n\n\n\nFigure 2\n\nThus, the perimeter of $P_2$ is $4 \\times 6 = 24$.\n\nThe blank part in Figure 2 is denoted as \"Figure $P_3$\". From (1), we have $AC = 2\\sqrt{3}$.\n\n$P_3$ is a regular hexagon,\n\nSo, the perimeter of $P_3$ is $2\\sqrt{3} \\times 6 = 12\\sqrt{3}$.\n\nTherefore, the ratio of the perimeters of $P_1$, $P_2$, and $P_3$ is $36:24:12\\sqrt{3} = 3:2:\\sqrt{3}$;\n\nHence, statement (2) is correct.\n\nAs shown in the figure, draw a perpendicular from point $O$ to $AB$ at point $D$, intersecting the incircle at point $E$. Then, $AE$ is the required length.\n\n\n\nFigure 3\n\nAccording to the properties of a regular hexagon, triangle $AOB$ is equilateral.\n\nThus, $OD = \\frac{\\sqrt{3}}{2} AO = OE$.\n\nSince $AB = 4$,\n\n$AO = AB = 4$, and $OD = \\frac{\\sqrt{3}}{2} AO = 2\\sqrt{3}$.\n\nTherefore, $AE = AO + EO = AO + OD = AB + OD = 4 + 2\\sqrt{3}$.\n\nHence, statement (3) is correct.\n\nThe correct choice is A.\n\n【Key Insight】This problem examines the properties of a regular hexagon and its incircle, the Pythagorean theorem, and the properties of a 30-degree right triangle. Understanding the problem and calculating the lengths of the segments is crucial to solving it." }, { "problem_id": 252, "question": "Four quadrilaterals as shown in Figure 1 and four rhombi as shown in Figure 2 are combined to form a regular octagon (as shown in Figure 3). The ratio of the area of the shaded part in Figure 3 to the area of the blank part is $(\\quad)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $\\frac{1}{2}$\nB. $\\frac{\\sqrt{2}}{2}$\nC. $\\frac{\\sqrt{3}}{2}$\nD. $\\sqrt{2}$", "input_image": [ "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0041_1.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0041_2.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0041_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Through vertex \\( B \\) of the rhombus in Figure 2, draw \\( BE \\perp AD \\) intersecting at \\( E \\). Let the center point of the regular octagon in Figure 3 be point \\( O \\), with one side as \\( MN \\). Connect \\( OM \\) and \\( ON \\), and through point \\( M \\), draw \\( MP \\perp ON \\) intersecting at \\( P \\).\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nLet the side length of the regular octagon be \\( a \\). Then, \\( AB = AD = MN = a \\).\n\nFrom the properties of a regular octagon, we have:\n\\[\n\\angle ABC = \\frac{(8-2) \\times 180^\\circ}{8} = 135^\\circ,\n\\]\n\\[\n\\angle MON = \\frac{360^\\circ}{8} = 45^\\circ.\n\\]\n\nSince \\( AD \\parallel BC \\),\n\\[\n\\angle BAE = 45^\\circ,\n\\]\n\\[\nBE = \\frac{\\sqrt{2}}{2} AB = \\frac{\\sqrt{2}}{2} a,\n\\]\n\\[\nS_{\\text{rhombus}} = AD \\times BE = \\frac{\\sqrt{2}}{2} a^2.\n\\]\n\nThe area of the blank part is:\n\\[\n4 \\times \\frac{\\sqrt{2}}{2} a^2 = 2\\sqrt{2} a^2.\n\\]\n\nSince \\( \\angle MON = 45^\\circ \\),\n\\[\nOP = PM.\n\\]\n\nLet \\( OP = PM = x \\), then \\( OM = ON = \\sqrt{2} x \\),\n\\[\nPN = (\\sqrt{2} - 1) x.\n\\]\n\nSince \\( PM^2 + PN^2 = MN^2 \\),\n\\[\nx^2 + (\\sqrt{2} - 1)^2 x^2 = a^2,\n\\]\n\\[\nx^2 = \\frac{2 + \\sqrt{2}}{4} a^2,\n\\]\n\\[\nS_{\\text{OMN}} = \\frac{1}{2} \\times ON \\times PM = \\frac{\\sqrt{2}}{2} x^2 = \\frac{1 + \\sqrt{2}}{4} a^2.\n\\]\n\nThus, the area of the regular octagon is:\n\\[\n8 \\times \\frac{1 + \\sqrt{2}}{4} a^2 = 2(1 + \\sqrt{2}) a^2.\n\\]\n\nTherefore, the area of the shaded part is:\n\\[\n2(1 + \\sqrt{2}) a^2 - 2\\sqrt{2} a^2 = 2a^2.\n\\]\n\nThe ratio of the shaded area to the blank area is:\n\\[\n\\frac{2a^2}{2\\sqrt{2} a^2} = \\frac{\\sqrt{2}}{2}.\n\\]\n\nHence, the correct answer is: B.\n\n【Key Insight】This problem primarily examines the properties of regular polygons and rhombuses. The key is to correctly construct right triangles and express the areas of various parts using the side length of the regular polygon." }, { "problem_id": 253, "question": "As shown in the figure, the plane figure $A B D$ is composed of an isosceles right triangle $A O D$ with legs of length 1 and a sector $B O D$. Point $P$ is on segment $A B$, and $P Q \\perp A B$, intersecting $A D$ or $D B$ at point $Q$. Let $A P = x (0 < x < 2)$, and the area of the shaded plane figure $A P Q$ (or $A P Q D$) is $y$. The approximate graph of the function $y$ with respect to $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch20-2024_05_23_88703de1aac697eff4e1g_0040_1.jpg", "batch20-2024_05_23_88703de1aac697eff4e1g_0040_2.jpg", "batch20-2024_05_23_88703de1aac697eff4e1g_0040_3.jpg", "batch20-2024_05_23_88703de1aac697eff4e1g_0040_4.jpg", "batch20-2024_05_23_88703de1aac697eff4e1g_0040_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When point \\( Q \\) is on \\( AD \\), that is, when point \\( P \\) is on \\( AO \\), we have \\( 0 < x \\cdot 1 \\).\n\nAt this time, the shaded area is an isosceles right triangle,\n\n\\[\n\\therefore y = \\frac{1}{2} \\cdot x \\cdot x = \\frac{1}{2} x^{2},\n\\]\n\nThis function is a quadratic function, opening upwards, eliminating options \\( B \\) and \\( C \\);\n\nWhen point \\( Q \\) is on arc \\( BD \\), the figure is completed as shown in the diagram,\n\n\n\nThe area of the shaded region equals the area of the isosceles right triangle \\( \\triangle AOD \\) plus the area of sector \\( BOD \\), minus the area of the plane figure \\( PBQ \\), which is minus half the area of the segment \\( QBF \\).\n\nLet \\( \\angle QOB = \\theta \\), then \\( \\angle QOF = 2\\theta \\),\n\n\\[\n\\therefore S_{\\triangle AOD} = \\frac{1}{2} \\times 1 \\times 1 = \\frac{1}{2}, \\quad S_{\\text{segment } QBF} = \\frac{\\theta \\pi r^{2}}{180} - S_{\\triangle QOF},\n\\]\n\nWhen \\( \\theta = 45^\\circ \\), \\( AP = x = 1 + \\frac{\\sqrt{2}}{2} \\approx 1.7 \\), \\( S_{\\text{segment } QBF} = \\frac{\\pi}{4} - \\frac{1}{2} \\times \\sqrt{2} \\times \\frac{\\sqrt{2}}{2} = \\frac{\\pi}{4} - \\frac{1}{2} \\),\n\n\\[\ny = \\frac{1}{2} + \\frac{\\pi}{4} - \\frac{1}{2}\\left(\\frac{\\pi}{4} - \\frac{1}{2}\\right) = \\frac{3}{4} + \\frac{\\pi}{8} \\approx 1.15,\n\\]\n\nWhen \\( \\theta = 30^\\circ \\), \\( AP = x = 1.86 \\), \\( S_{\\text{segment } QBF} = \\frac{\\pi}{6} - \\frac{1}{2} \\times \\frac{1}{2} \\times \\sqrt{3} = \\frac{\\pi}{6} - \\frac{\\sqrt{3}}{4} \\),\n\n\\[\ny = \\frac{1}{2} + \\frac{\\pi}{4} - \\frac{1}{2}\\left(\\frac{\\pi}{6} - \\frac{\\sqrt{3}}{4}\\right) = \\frac{1}{2} + \\frac{\\sqrt{3}}{8} + \\frac{\\pi}{6} \\approx 1.45,\n\\]\n\nBy locating these two specific values in options \\( A \\) and \\( D \\), it is found that option \\( D \\) matches the description.\n\nTherefore, the answer is: D.\n\n[Highlight] This question mainly examines the graph and properties of quadratic functions and the area of shapes. Using specific values to solve problems is a common method in multiple-choice questions. Constructing the figure to express the area of the shaded region is the key to solving this problem." }, { "problem_id": 254, "question": "A construction team for a subway in a city began tunnel excavation work. As shown in Figure 1, curved concrete segments are essential components for forming a circular tunnel. As shown in Figure 2, there is a curved concrete segment placed on a horizontal ground, with its bottom secured by two identical rectangular wooden blocks. To estimate the size of the tunnel excavation face, Group A and Group B measured related data, as shown in the table below. The group that can estimate the tunnel's outer diameter $(O B)$ using the data is ( )\n\n| Group | Measured Content |\n| :---: | :---: |\n| A | Lengths of $A B, A D, B C$ |\n| B | Lengths of $H G, G N_{\\text {long }}$ |\n\n\n\nFigure 1\n\n\n\nA. Neither group's data is sufficient\nB. Group A\nC. Group B\nD. Both groups", "input_image": [ "batch20-2024_05_23_88703de1aac697eff4e1g_0056_1.jpg", "batch20-2024_05_23_88703de1aac697eff4e1g_0056_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Both Group A and Group B's methods are valid.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nReason for Group B's method: As shown in Figure 2, based on the measurement data, we know that $HG = KL$ and $GN = HM$. Using the Perpendicular Chord Theorem, we can determine $HK$. In the right triangle $OHK$, by applying the Pythagorean Theorem, we can find $OH$, and subsequently determine $OL$, thus solving the problem.\n\nReason for Group A's method: As shown in Figure 1, since $AB$ is known, we can set the radius of the outer circle as $R$, and then express the radius of the inner circle $OA$. By setting up a system of equations using the arc length formula, we can solve for $R$.\n\nTherefore, both Group A and Group B's methods are valid.\n\nHence, the correct choice is: $D$.\n\n[Key Insight] This problem tests the understanding of the Perpendicular Chord Theorem, the relationship between the sides and angles of a right triangle, and the calculation of arc lengths. Mastering the Perpendicular Chord Theorem, the Pythagorean Theorem, and the method for calculating arc lengths is crucial for solving the problem." }, { "problem_id": 255, "question": "The front of an electric toy consists of a small disc with radius $10 \\mathrm{~cm}$ and a large disc with radius $20 \\mathrm{~cm}$ connected as shown in the figure on the right. The small disc rolls without slipping along the circumference of the large disc for one full revolution and returns to its original position. Among the three circles shown in the dotted line positions, which one correctly depicts the positions of the hair, eyes, and mouth when the toy is in this position (try it out yourself!)?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_8c3329fc8d8b0c814fd8g_0092_1.jpg", "batch20-2024_05_23_8c3329fc8d8b0c814fd8g_0092_2.jpg", "batch20-2024_05_23_8c3329fc8d8b0c814fd8g_0092_3.jpg", "batch20-2024_05_23_8c3329fc8d8b0c814fd8g_0092_4.jpg", "batch20-2024_05_23_8c3329fc8d8b0c814fd8g_0092_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "When the small disk rolls around the large disk for one full rotation, the small disk rotates on its own axis $\\frac{2 \\pi R}{2 \\pi r}+1=\\frac{2 \\times 20 \\pi}{2 \\times 10 \\pi}+1=3$ times. The three long pointers inside the large disk just divide the angle of the large disk equally,\n\nTherefore, the images pointed to by the three long pointers are exactly the same, thus eliminating options $C$ and $D$.\n\nThe images pointed to by two adjacent long pointers are definitely not the same as the image in between them, thus eliminating option $A$.\n\nOnly option $B$ fits.\n\nTherefore, the answer is: $B$.\n\n[Insight] This question mainly examines the calculation of the circumference and arc length of a circle. Calculating that the small disk rotates 3 times on its own axis when rolling around the large disk for one full rotation and understanding that the long pointers divide the large disk into 3 equal parts are key to solving this problem." }, { "problem_id": 256, "question": "As shown in Figure 1, a closed cylindrical bucket is filled with half a bucket of water (the thickness of the bucket is ignored). The diameter of the circular base of the cylinder is equal to its slant height. When the bucket is placed horizontally as shown in Figure 2, let the surface areas of the geometric solids formed by the water in Figures 1 and 2 be $S_{1}$ and $S_{2}$, respectively. Then, the relationship between $S_{1}$ and $S_{2}$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $S_{1} \\leq S_{2}$\nB. $S_{1}S_{2}$\nD. $S_{1} \\geq S_{2}$", "input_image": [ "batch20-2024_05_23_8c3329fc8d8b0c814fd8g_0093_1.jpg", "batch20-2024_05_23_8c3329fc8d8b0c814fd8g_0093_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Let the radius of the base of the cylinder be \\( r \\). The surface area of the water in Figure 1 is:\n\\[ S_{I} = 2 \\pi r^{2} + 2 \\pi r \\cdot r = 4 \\pi r^{2}. \\]\n\nFor Figure 2:\n\n- The length of the upper rectangle is \\( 2r \\), and the width is \\( 2r \\). Thus, its area is \\( 4r^{2} \\).\n- The length of the rectangle formed by unfolding the curved surface is \\( \\pi r \\), and the width is \\( 2r \\). Thus, its area is \\( 2 \\pi r^{2} \\).\n- The combined area of the top and bottom surfaces is \\( \\pi \\times r^{2} \\).\n\nTherefore, the surface area of the water in Figure 2 is:\n\\[ S_{2} = (4 + 3 \\pi) r^{2}. \\]\n\nClearly, \\( S_{1} < S_{2} \\).\n\nHence, the correct choice is: \\( B \\).\n\n**Insight:** This problem primarily examines calculations related to cylinders. The key to solving it lies in mastering the concept of transforming curved surfaces into flat ones." }, { "problem_id": 257, "question": "As shown in the figure, in right triangle $O A B$, $\\angle A O B = 90^\\circ, O A = 4, O B = 3$. The radius of circle $O$ is 2. Point $P$ is a moving point on segment $A B$. A tangent line $P Q$ to circle $O$ is drawn through point $P$, with $Q$ as the point of tangency. If $A P = x$ and $P Q^2 = y$, then the graph of the function $y$ versus $x$ is approximately ( ).\n\n\n\nA. A\nB. B\nC. C\nD. D", "input_image": [ "batch20-2024_05_23_911de7006c8db00d3cf9g_0036_1.jpg", "batch20-2024_05_23_911de7006c8db00d3cf9g_0036_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, draw $PC \\perp OA$, with the foot of the perpendicular at $C$,\n\n\n\nSince $PC \\parallel BO$,\n\n$\\therefore \\triangle ABO \\sim \\triangle APC$,\n\n$\\therefore \\frac{AP}{AB} = \\frac{PC}{OB} = \\frac{AC}{OA}$\n\nGiven $AP = x$, $OA = 4$, $OB = 3$,\n\n$\\therefore PC = \\frac{3}{5}x$, $AC = \\frac{4}{5}x$\n\n$\\therefore OC = 4 - \\frac{4}{5}x$\n\n$\\therefore OP^{2} = \\left(4 - \\frac{4}{5}x\\right)^{2} + \\left(\\frac{3}{5}x\\right)^{2} = x^{2} - \\frac{32}{5}x + 16$\n\n$\\therefore y = OP^{2} - OQ^{2} = x^{2} - \\frac{32}{5}x + 16$\n\nWhen $x = 0$, $y = 12$,\n\nWhen $x = 5$, $y = 5$.\n\nTherefore, the correct choice is A.\n\n【Key Insight】This problem primarily examines the graph of a function and the formulation of a function expression. Understanding the functional relationship described in the problem is crucial for making the correct judgment." }, { "problem_id": 258, "question": "As shown in the figure, $A B$ and $C D$ are two perpendicular diameters of circle $O$. Point $P$ starts from point $O$ and moves along the route $O \\rightarrow C \\rightarrow B \\rightarrow O$ at a constant speed. Let $\\angle A P D = y$ (unit: degrees). The graph that represents the relationship between $y$ and the time (unit: seconds) that point $P$ has been moving is $(\\quad)$.\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_911de7006c8db00d3cf9g_0047_1.jpg", "batch20-2024_05_23_911de7006c8db00d3cf9g_0047_2.jpg", "batch20-2024_05_23_911de7006c8db00d3cf9g_0047_3.jpg", "batch20-2024_05_23_911de7006c8db00d3cf9g_0047_4.jpg", "batch20-2024_05_23_911de7006c8db00d3cf9g_0047_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) When point $\\mathrm{P}$ moves along $\\mathrm{O} \\rightarrow \\mathrm{C}$,\n\nWhen point $\\mathrm{P}$ is at the position of point $\\mathrm{O}$, $y=90^{\\circ}$,\n\nWhen point $\\mathrm{P}$ is at the position of point $\\mathrm{C}$,\n\nSince $\\mathrm{OA}=\\mathrm{OC}$,\n\nTherefore, $y=45^{\\circ}$,\n\nThus, $y$ gradually decreases from $90^{\\circ}$ to $45^{\\circ}$;\n\n(2) When point $P$ moves along $C \\rightarrow B$,\n\nAccording to the inscribed angle theorem, we have\n\n$\\mathrm{y} \\equiv 90^{\\circ} \\div 2=45^{\\circ} ;$\n\n(3) When point $P$ moves along $B \\rightarrow O$,\n\nWhen point $\\mathrm{P}$ is at the position of point $\\mathrm{B}$, $y=45^{\\circ}$, when point $\\mathrm{P}$ is at the position of point $\\mathrm{O}$, $y=90^{\\circ}$, thus $y$ gradually increases from $45^{\\circ}$ to $90^{\\circ}$.\n\nTherefore, the answer is: B.\n\n[Key Insight] This question mainly examines the function graph of moving point problems and the inscribed angle theorem. The key to solving the problem is to obtain information by reading the graph and to solve practical problems in life. When using graphs to solve problems, it is essential to understand the meaning of the graph, i.e., to learn how to interpret the graph." }, { "problem_id": 259, "question": "As shown in Figure (1), points $A$ and $B$ are fixed points on $\\odot O$. A moving point $P$ on the circle starts from a fixed point $B$ and moves along the circle in the counterclockwise direction at a uniform speed until it reaches point $A$. The time of movement is $x(s)$, and the length of segment $A P$ is $y(\\mathrm{~cm})$. The graph in Figure (2) shows the relationship between $y$ and $x$. What is the value of $m$ in the graph?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $\\frac{9}{2} s$\nB. $4 \\sqrt{2} s$\nC. $5 s$\nD. $\\frac{7}{3} \\mathrm{~s}$", "input_image": [ "batch20-2024_05_23_95ac08c4de9b76a9854fg_0010_1.jpg", "batch20-2024_05_23_95ac08c4de9b76a9854fg_0010_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From Figure (2), when \\( x = 1 \\), \\( y = AP = 6 \\), which means that at this moment, points \\( A \\), \\( O \\), and \\( P \\) are collinear. Therefore, the radius of the circle is \\( \\frac{1}{2} AP = 3 \\).\n\nWhen \\( x = 0 \\), \\( OB^2 + OA^2 = AP^2 \\), \n\nso \\( \\triangle OAB \\) is a right-angled triangle with \\( OA \\perp OB \\).\n\nAs point \\( P \\) moves from point \\( B \\) to the position where \\( A \\), \\( O \\), and \\( P \\) are collinear,\n\nas shown in the figure,\n\n\n\nFigure (1)\n\nAt this moment, \\( x = 1 \\), and the angle traversed is \\( 90^\\circ \\). Therefore, the length of the arc traversed is \\( \\frac{1}{4} \\) of the circumference, which is \\( \\frac{3p}{2} \\).\n\nThus, the speed of point \\( P \\) is \\( \\frac{3p}{2} \\) cm/s.\n\nWhen \\( t = m \\), \\( AP = OA = OP = 3 \\), meaning \\( \\triangle OAP \\) is an equilateral triangle.\n\nTherefore, \\( \\angle AOP = 60^\\circ \\).\n\nThe central angle corresponding to the arc traversed by \\( P \\) is \\( 360^\\circ - 90^\\circ - 60^\\circ = 210^\\circ \\).\n\nAt this moment, the length of the arc traversed by point \\( P \\) is: \\( \\frac{210 \\times p \\times 3}{180} = \\frac{7p}{2} \\).\n\nTherefore, \\( m = \\frac{7p}{2} \\div \\frac{3p}{2} = \\frac{7}{3} \\).\n\nHence, the correct answer is: D.\n\n[Key Insight] This problem examines the motion of a point on a graph, the fundamental properties of a circle, the application of the Pythagorean theorem, and the calculation of arc lengths. The key to solving such problems lies in understanding the correspondence between the graph and the geometric figure at different time intervals, and then proceeding to solve accordingly." }, { "problem_id": 260, "question": "As shown in Figure (1), it is a diagram of the emblem of the 2022 Hangzhou Asian Games. If $A O=5, B O=2, \\angle A O D=120^{\\circ}$, then the area of the shaded region is $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $14 \\pi$\nB. $7 \\pi$\nC. $\\frac{25}{3} \\pi$\nD. $2 \\pi$", "input_image": [ "batch20-2024_05_23_95ac08c4de9b76a9854fg_0057_1.jpg", "batch20-2024_05_23_95ac08c4de9b76a9854fg_0057_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: The area of the shaded region, \\( S_{\\text{shaded}} \\), is equal to the area of sector \\( AOD \\) minus the area of sector \\( BOC \\). This can be calculated as:\n\\[\nS_{\\text{shaded}} = \\frac{1}{3} \\times \\pi \\times 5^{2} - \\frac{1}{3} \\times \\pi \\times 2^{2} = 7\\pi\n\\]\nTherefore, the correct answer is: **B**.\n\n**Key Insight**: This problem tests the calculation of the area of a shaded region related to sectors. The key to solving it lies in correctly identifying that the shaded area is the difference between the areas of two sectors and accurately applying the sector area formula." }, { "problem_id": 261, "question": "One morning, Xiaohua walked around the garden at a constant speed in a clockwise direction. Given that Xiaohua started from point $A$ in the garden, and Xiaohua is regarded as a moving point $B$, with the garden's center being $O$. Let the angle $(\\angle A O B)$ formed by Xiaohua, point $O$, and point $A$ be $y^\\circ$ (where $y \\leq 180$), and the graph of $y$ as a function of time $x$ is as shown in the figure. Which of the following shapes could the garden possibly be $(\\quad)$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_95ac08c4de9b76a9854fg_0084_1.jpg", "batch20-2024_05_23_95ac08c4de9b76a9854fg_0084_2.jpg", "batch20-2024_05_23_95ac08c4de9b76a9854fg_0084_3.jpg", "batch20-2024_05_23_95ac08c4de9b76a9854fg_0084_4.jpg", "batch20-2024_05_23_95ac08c4de9b76a9854fg_0084_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the angle ($\\angle AOB$) formed by Xiao Wan, point $O$, and point $A$ has a measure of $y^{\\circ}$, by observing the graph, it is deduced that the relationship between $y$ and time $x$ is a linear function.\n\nAccording to the arc length formula: $l=\\frac{n \\pi r}{180}$,\n\nwe can derive: $n=\\frac{180}{\\pi r} l$.\n\nAssuming Xiao Wan's walking speed is $v$,\nthe expression for $y$ in terms of $x$ during the first half of the journey is $y=\\frac{180}{\\pi r} l=\\frac{180 v}{\\pi r} x$,\n\nand during the second half of the journey, the expression for $y$ in terms of $x$ is $y=180-\\frac{180 v}{\\pi r} x$.\n\nTherefore, the shape of the garden is circular,\n\nHence, the correct choice is: A.\n\n[Insight] This problem examines the function graph of a moving point. The key to solving it is to understand the given conditions and apply the concept of combining numbers with shapes to find the solution." }, { "problem_id": 262, "question": "As shown in the figure, the complement of $\\angle 1$ could be ( ) in the figure.\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_c8f7fa0db5b83322a3c4g_0019_1.jpg", "batch20-2024_05_23_c8f7fa0db5b83322a3c4g_0019_2.jpg", "batch20-2024_05_23_c8f7fa0db5b83322a3c4g_0019_3.jpg", "batch20-2024_05_23_c8f7fa0db5b83322a3c4g_0019_4.jpg", "batch20-2024_05_23_c8f7fa0db5b83322a3c4g_0019_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Two angles are complementary if their sum is $90^{\\circ}$. Among the given options, only option $\\mathrm{C}$ satisfies this condition.\n\nTherefore, the correct choice is C.\n\n[Key Insight] This question tests the definition of complementary angles. The key to solving it lies in understanding the definition of complementary angles." }, { "problem_id": 263, "question": "In triangle $A B C$, $\\angle A C B = 90^\\circ$, and $A C + B C = 8$. Semicircles are drawn with $A B$, $A C$, and $B C$ as diameters. If the area of the shaded region is denoted as $y$, and $A C$ is denoted as $x$, which of the following graphs correctly represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0022_1.jpg", "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0022_2.jpg", "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0022_3.jpg", "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0022_4.jpg", "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0022_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since \\( AC + BC = 8 \\) and \\( AC = x \\),\n\n\\[\n\\therefore BC = 8 - x.\n\\]\n\nMoreover, in triangle \\( \\triangle ABC \\), \\( \\angle ACB = 90^\\circ \\),\n\n\\[\n\\therefore AB = \\sqrt{x^{2} + (8 - x)^{2}}.\n\\]\n\nThus, the area \\( S \\) is given by:\n\n\\[\nS = \\frac{1}{2} \\pi \\left(\\frac{x}{2}\\right)^{2} + \\frac{1}{2} \\pi \\left(\\frac{8 - x}{2}\\right)^{2} - \\frac{1}{2} \\pi \\left(\\frac{\\sqrt{x^{2} + (8 - x)^{2}}}{2}\\right)^{2} + \\frac{1}{2} x(8 - x) = -\\frac{1}{2} x^{2} + 4x,\n\\]\n\nwhich simplifies to:\n\n\\[\ny = -\\frac{1}{2} x^{2} + 4x \\quad (0 < x < 8).\n\\]\n\nThe graph of this function is a downward-opening parabola, with the independent variable \\( x \\) ranging from 0 to 8.\n\nTherefore, the correct answer is A.\n\n**Key Insight:** This problem tests the understanding of function graphs related to moving points, the Pythagorean theorem, and the area formula of a circle. The key to solving it lies in correctly expressing the area of the shaded region." }, { "problem_id": 264, "question": "The Fibonacci spiral, also known as the \"golden spiral,\" can be constructed by drawing successive semicircles with radii $1, 1, 2, 3, 5, \\ldots$ and central angles of $90^\\circ$ (as shown in the figure). In the first step, the semicircle has a radius of $1 \\mathrm{~cm}$. Following the method shown in the figure, the arc length of the semicircle drawn in the eighth step is ( )\n\n\n\nStep 1\n\n\n\nStep 3\n\n\n\nStep 4\nA. $4 \\pi$\nB. $\\frac{21}{2} \\pi$\nC. $17 \\pi$\nD. $\\frac{55}{2} \\pi$", "input_image": [ "batch20-2024_05_23_e0fb99ec31043881873eg_0009_1.jpg", "batch20-2024_05_23_e0fb99ec31043881873eg_0009_2.jpg", "batch20-2024_05_23_e0fb99ec31043881873eg_0009_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Arithmetic", "image_relavance": "0", "analysis": "Solution: The Fibonacci sequence is $1, 1, 2, 3, 5, \\ldots$\n\nThe radius at the 6th step is $3 + 5 = 8 \\text{ cm}$;\n\nThe radius at the 7th step is $5 + 8 = 13 \\text{ cm}$;\n\nThe radius at the 8th step is $8 + 13 = 21 \\text{ cm}$;\n\nAccording to the problem, the radius of the sector drawn at the 8th step is $21 \\text{ cm}$,\n\n$\\therefore$ The arc length of the sector drawn at the 8th step $= \\frac{90 \\pi \\times 21}{180} = \\frac{21}{2} \\pi \\text{ cm}$,\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This problem tests the calculation of arc length and the pattern of number changes. The key to solving the problem is to identify the pattern of radius changes based on the given information." }, { "problem_id": 265, "question": "As shown in Figure (1), given sector $A O B$, point $P$ starts from point $O$ and moves along $O \\rightarrow A \\rightarrow B \\rightarrow O$ at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$. Let the time of point $P$'s movement be $x s$. The length of $O P$ is $y \\mathrm{~cm}$, and the graph of $y$ as a function of $x$ is shown in Figure (2). Then, the area of sector $A O B$ is $(\\quad)$\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $3 \\pi \\mathrm{cm}^{2}$\nB. $2.5 \\pi \\mathrm{cm}^{2}$\nC. $2 \\pi \\mathrm{cm}^{2}$\nD. $\\pi \\mathrm{cm}^{2}$", "input_image": [ "batch20-2024_05_23_e0fb99ec31043881873eg_0076_1.jpg", "batch20-2024_05_23_e0fb99ec31043881873eg_0076_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From the graph, it can be determined that the radius of the sector is 3, and the arc length is \\(2\\pi\\). Let the central angle of the sector be \\(n\\). According to the arc length formula, we have:\n\n\\[\n\\frac{2n \\times 3\\pi}{360} = 2\\pi\n\\]\n\nSolving for \\(n\\), we get \\(n = 120^\\circ\\).\n\nUsing the area formula for a sector, the area of sector \\(AOB\\) is:\n\n\\[\n\\frac{120 \\times 3^2 \\pi}{360} = 3\\pi \\text{ cm}^2\n\\]\n\nTherefore, the correct answer is A.\n\n**Key Insight**: This problem falls under the category of moving point function graph problems, primarily testing knowledge of the sector's arc length and area formulas. The key to solving this problem lies in determining the sector's radius and arc length from the graph." }, { "problem_id": 266, "question": "Among the figures we have studied, the distance between any pair of parallel tangents to a circle is always equal, making the circle an \"equal-width curve.\" In addition to circles, there are other geometric figures that are also \"equal-width curves,\" such as the Reuleaux triangle (as shown in Figure 1), which is formed by drawing a circular arc of the same radius as the side length of an equilateral triangle, centered at each vertex, between the other two vertices, creating a curvilinear triangle. Figure 2 shows a cross-sectional view of an equal-width Reuleaux triangle and a circular log.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nThe following four conclusions are given:\n\n(1) The Reuleaux triangle is a centrally symmetric figure.\n\n(2) In Figure 1, the distance from point A to any point on line segment BC is equal.\n\n(3) In Figure 2, the perimeter of the Reuleaux triangle is equal to the circumference of the circle.\n\n(4) When using a log with a cross-sectional shape of a Reuleaux triangle to transport items, there will be an up-and-down shaking. \n\nWhich of the above conclusions are correct? The sequence numbers of all correct conclusions are ( )\nA. (1)(2)\nB. (2)(3)\nC. (2)(4)\nD. (3)(4)", "input_image": [ "batch20-2024_05_23_edee79dd21e93b933810g_0093_1.jpg", "batch20-2024_05_23_edee79dd21e93b933810g_0093_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "(1) The Reuleaux triangle is not a centrally symmetric figure, so statement (1) is incorrect;\n\n(2) In Figure 1, the distance from point A to any point on BC is equal, so statement (2) is correct;\n\n(3) In Figure 2, let the radius of the circle be r.\nTherefore, the perimeter of the Reuleaux triangle = 3 × (120πr)/180 = 2πr.\nThe circumference of the circle is 2πr.\nThus, the perimeter of the Reuleaux triangle is equal to the circumference of the circle, so statement (3) is correct;\n\n(4) Using a rolling log with a Reuleaux triangle cross-section to move objects will not cause up-and-down jitter, so statement (4) is incorrect. Therefore, the correct choice is B.\n\n[Key Insight] This question mainly examines the concepts of centrally symmetric figures and the arc length formula. Mastering these concepts is key to solving the problem." }, { "problem_id": 267, "question": "The area of a sector with central angle $60^\\circ$ is $\\mathrm{S}$, and the radius is $\\mathrm{r}$. Which of the following graphs can roughly describe the function relationship between $\\mathrm{S}$ and $\\mathrm{r}$? $(\\quad)$\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_f566d85bc9fcb0b0c2b8g_0059_1.jpg", "batch20-2024_05_23_f566d85bc9fcb0b0c2b8g_0059_2.jpg", "batch20-2024_05_23_f566d85bc9fcb0b0c2b8g_0059_3.jpg", "batch20-2024_05_23_f566d85bc9fcb0b0c2b8g_0059_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the area of a sector with a central angle of $60^{\\circ}$ is $\\mathrm{S}$ and the radius is $\\mathrm{r}$,\n\nTherefore, $S=\\frac{60 \\pi r^{2}}{360}=\\frac{\\pi r^{2}}{6}$.\n\nThus, $\\mathrm{S}$ is a quadratic function of $\\mathrm{r}$, and $\\mathrm{r}>0$,\n\nHence, options $\\mathrm{C}$ and $\\mathrm{D}$ are incorrect;\n\nWhen $r=1$, $S=\\frac{\\pi}{6}<1$;\n\nWhen $\\mathrm{r}=2$, $\\mathrm{S}=\\frac{4 \\pi}{6} \\approx 2.09$,\n\nTherefore, the correct choice is A.\n\n[Key Insight] This question tests the understanding of the graph and properties of quadratic functions and the calculation of the area of a sector. The key to solving the problem lies in deriving the functional relationship between $\\mathrm{S}$ and $\\mathrm{r}$." }, { "problem_id": 268, "question": "The following is a proof of the median theorem in a triangle in a slide, which requires supplement the language and symbols on the horizontal line.\n\nGiven: In $\\triangle A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$, respectively. Prove that $D E / / B C$ and $D E = \\frac{1}{2} B C$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nProof: In Figure 2, extend $D E$ to point $F$ such that ※, and connect $C F$.\n\n$\\because A E = C E, \\angle A E D = \\angle C E F$,\n\n$\\therefore \\triangle A D E \\cong \\triangle C F E$ ( ○ )\n\n$\\therefore A D = C F, \\angle A = *$,\n\n$\\therefore A D / / C F$, which implies $B D / / C F$,\n\nAlso $\\because B D = A D = C F$,\n\n$\\therefore$ Quadrilateral $D B C F$ is\n\n$\\therefore D F / / B C$, which implies $D E / / B C$ and $D F = B C = 2 D E$,\n\n$\\therefore D E / / B C$, and $D E = \\frac{1}{2} B C$.\nThe correct options are ( )\nA. ※ represents $E F = D E$\nB. © represents ASA\nC. * represents $\\angle E F C$\nD. $\\odot$ represents rhombus", "input_image": [ "batch22-2024_06_14_19d3318758f805445608g_0043_1.jpg", "batch22-2024_06_14_19d3318758f805445608g_0043_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "**Proof:** Extend $DE$ to point $F$ such that $EF = DE$, and connect $CF$.\n\nSince $AE = CE$ and $\\angle AED = \\angle CEF$, \n$\\triangle ADE \\cong \\triangle CFE$ (by the SAS congruence criterion). \n\nTherefore, $AD = CF$ and $\\angle A = \\angle ECF$. \nThis implies that $AD \\parallel CF$, which means $BD \\parallel CF$. \n\nFurthermore, since $BD = AD = CF$, \nquadrilateral $DBCF$ is a parallelogram. \n\nThus, $DF \\parallel BC$, which means $DE \\parallel BC$, and $DF = BC = 2DE$. \nHence, $DE \\parallel BC$, and $DE = \\frac{1}{2}BC$. \n\nThe correct answer is: **A**. \n\n**Key Insight:** This problem examines the properties of parallelograms, the congruence of triangles, and the midsegment theorem of triangles. Mastering the properties of parallelograms is crucial for solving this problem." }, { "problem_id": 269, "question": "As shown in the figure, these are the auxiliary lines drawn by individuals A and B separately. Which of the auxiliary line constructions can be used to prove the theorem of the median of a triangle?\n\n| Method of A | Method of B |\n| :--- | :--- |\n| As shown, extend $D E$ to $F$, such that $E F = D E$, and connect | As shown, draw $G E \\parallel A B$ through point $E$, and draw |\n| $D C, A F, F C$. | $A F \\parallel B C$ through point $A$. $G E$ intersects $A F$ at point $F$. |\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Both the auxiliary line constructions of A and B are applicable.\nB. Neither the auxiliary line construction of A nor B is applicable.\nC. The auxiliary line construction of A is applicable, but not that of B.\nD. The auxiliary line construction of B is applicable, but not that of A.", "input_image": [ "batch22-2024_06_14_19d3318758f805445608g_0062_1.jpg", "batch22-2024_06_14_19d3318758f805445608g_0062_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: \n\n**Part A:** In triangles $\\triangle ADE$ and $\\triangle CFE$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\nAE = CE \\\\\n\\angle AED = \\angle CEF, \\\\\nDE = EF\n\\end{array}\n\\right.\n\\]\n\nTherefore, $\\triangle ADE \\cong \\triangle CFE$ (by SAS congruence).\n\nHence, $AD = CF$ and $\\angle ADE = \\angle CFE$,\n\nwhich implies $AB \\parallel CF$.\n\nSince $AD = BD$,\n\nit follows that $BD = CF$.\n\nThus, quadrilateral $BCFD$ is a parallelogram,\n\nand so $DF = BC$ and $DF \\parallel BC$.\n\nGiven that $DE = \\frac{1}{2} DF$,\n\nwe have $DE = \\frac{1}{2} BC$ and $DE \\parallel BC$.\n\nTherefore, the auxiliary line construction in Part A successfully proves the Midsegment Theorem of a triangle.\n\n**Part B:** Since $AF \\parallel BC$,\n\nwe have $\\angle AFE = \\angle CGE$.\n\nIn triangles $\\triangle AEF$ and $\\triangle CEG$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle AFE = \\angle CGE \\\\\n\\angle AEF = \\angle CEG, \\\\\nAE = CE\n\\end{array}\n\\right.\n\\]\n\nTherefore, $\\triangle AEF \\cong \\triangle CEG$ (by AAS congruence).\n\nHence, $AF = CG$ and $EF = EG$.\n\nSince $AF \\parallel BC$ and $AB \\parallel GF$,\n\nquadrilateral $ABGF$ is a parallelogram.\n\nThus, $AB = GF$ and $AF = BG$.\n\nGiven that $AF = \\frac{1}{2} BC$,\n\nand since $AD = \\frac{1}{2} AB$ and $EF = \\frac{1}{2} GF$,\n\nwe have $AD = EF$.\n\nMoreover, since $AD \\parallel EF$,\n\nquadrilateral $ADEF$ is a parallelogram.\n\nTherefore, $AF \\parallel DE$ and $AF = DE$,\n\nwhich implies $DE \\parallel BC$ and $DE = \\frac{1}{2} BC$.\n\nThus, the auxiliary line construction in Part B also successfully proves the Midsegment Theorem of a triangle.\n\n**Conclusion:** The correct answer is **A**.\n\n**Key Insight:** This problem examines the proof of the Midsegment Theorem of a triangle, utilizing properties of congruent triangles and parallelograms. Mastery of these geometric principles is essential for solving such problems." }, { "problem_id": 270, "question": "Figure 1 is a right-angled isosceles triangle with the length of the legs being $2 \\mathrm{~cm}$. After cutting along the line connecting the midpoints of a leg and the hypotenuse (the dashed line in the figure), the resulting pieces are rearranged to form the quadrilateral as shown in Figure 2. The perimeter of this quadrilateral is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $6 \\mathrm{~cm}$\nB. $4 \\mathrm{~cm}$\nC. $(4+2 \\sqrt{2}) \\mathrm{cm}$\nD. $(4+\\sqrt{2}) \\mathrm{cm}$", "input_image": [ "batch22-2024_06_14_300e12ec623424d720acg_0031_1.jpg", "batch22-2024_06_14_300e12ec623424d720acg_0031_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nFrom the problem statement, we can deduce the following: In Figure 1, $BC = AC = 2 \\text{ cm}$.\n\nSince $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively,\n$\\therefore DE$ is the midline of $\\triangle ABC$,\n\n$\\therefore AE = DE = \\frac{1}{2} BC = 1 \\text{ cm}$.\n\nSince $\\triangle ABC$ is an isosceles right triangle with the length of the legs being $2 \\text{ cm}$,\n\n$\\therefore$ by the Pythagorean theorem: $AB = \\sqrt{BC^2 + AC^2} = \\sqrt{2^2 + 2^2} = 2\\sqrt{2} \\text{ cm}$.\n\nSince $D$ is the midpoint of $AB$,\n\n$\\therefore AD = BD = \\frac{1}{2} AB = \\sqrt{2} \\text{ cm}$.\n\nIn Figure 2, $AC = 1 \\text{ cm}$,\n\n$\\therefore$ the perimeter of the quadrilateral is: $AB + BD + DE + AE = AC + BC + BD + DE + AE = 1 + 2 + \\sqrt{2} + 1 + \\sqrt{2} = (4 + 2\\sqrt{2}) \\text{ cm}$.\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem tests the understanding of graphic dissection, the properties and determination of isosceles right triangles, the Pythagorean theorem, the midline theorem of triangles, and the application of the midline theorem of triangles is crucial." }, { "problem_id": 271, "question": "Given: Connecting the midpoints of the sides of a rectangle consecutively results in a diamond, as shown in Figure (1); then connecting the midpoints of the sides of the diamond consecutively results in a new rectangle, as shown in Figure (2); then connecting the midpoints of the sides of the new rectangle consecutively results in a new diamond, as shown in Figure (3); and so on. If this operation is repeated indefinitely, the number of right-angled triangles in the 2012th figure is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA. 8048 triangles\nB. 4024 triangles\nC. 2012 triangles\nD. 1066 triangles", "input_image": [ "batch22-2024_06_14_40dfaec763ee71d319b8g_0094_1.jpg", "batch22-2024_06_14_40dfaec763ee71d319b8g_0094_2.jpg", "batch22-2024_06_14_40dfaec763ee71d319b8g_0094_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The first figure has 4 right-angled triangles,\n\nThe second figure has 4 right-angled triangles,\n\nThe third figure has 8 right-angled triangles,\n\nThe fourth figure has 8 right-angled triangles,\n\nAnd so on. When \\( n \\) is odd, the number of triangles is \\( 2(n+1) \\), and when \\( n \\) is even, the number of triangles is \\( 2n \\).\n\nTherefore, the number of right-angled triangles in the 2012th figure is \\( 2 \\times 2012 = 4024 \\).\n\nHence, the correct choice is B." }, { "problem_id": 272, "question": "As shown in Figure (1), in the square $A B C D$, point $E$ is the midpoint of $A B$, and point $P$ is a moving point on the diagonal $A C$. Let $P C = x$ and $P E + P B = y$. The graph in Figure (2) represents the function $y$ as a function of $x$. The lowest point $Q$ on the graph has coordinates $(4 \\sqrt{2}, 3 \\sqrt{5})$. What is the length of the side of the square $A B C D$?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 6\nB. $3 \\sqrt{5}$\nC. $4 \\sqrt{2}$\nD. 4", "input_image": [ "batch22-2024_06_14_432983757ec833e0afc7g_0064_1.jpg", "batch22-2024_06_14_432983757ec833e0afc7g_0064_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, point $D$ is the symmetric point of point $B$ with respect to the line $AC$. Connecting $DE$ intersects $AC$ at point $P$, at which point $y$ reaches its minimum value.\n\n\n\nAccording to the symmetry of points, $PB = PD$, thus $y = PE + PB = PD + PE = DE$ is minimized.\n\nTherefore, $ED = 3\\sqrt{5}$.\n\nLet the side length of the square be $x$, then $AE = \\frac{1}{2}x$.\n\nIn the right triangle $\\triangle ADE$, by the Pythagorean theorem: $DE^{2} = AD^{2} + AE^{2}$,\n\nThat is, $x^{2} + \\left(\\frac{1}{2}x\\right)^{2} = (3\\sqrt{5})^{2}$, solving gives: $x = 6$ (the negative value has been discarded).\n\nHence, the answer is: $A$.\n【Key Insight】This problem mainly examines the properties of squares, the equation of a linear function, the Pythagorean theorem, and the solution of a quadratic equation. Accurate calculation is the key to solving the problem." }, { "problem_id": 273, "question": "As shown in Figure 1, in rectangle $A B C D$ (with $A B < A D$), point $E$ is the midpoint of side $A D$. A moving point $F$ travels along the broken line $B \\rightarrow C \\rightarrow D$, starting from point $B$ and reaching point $D$. Let the distance traveled by $F$ be $x$, and the area of triangle $A E F$ be $y$. The graph of the relationship between $y$ and $x$ is shown in Figure 2. When $x = 5$, the value of $y$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 2\nB. 3\nC. 4\nD. 5", "input_image": [ "batch22-2024_06_14_432983757ec833e0afc7g_0071_1.jpg", "batch22-2024_06_14_432983757ec833e0afc7g_0071_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement, we know that\n\nwhen point $\\mathrm{F}$ is on segment $\\mathrm{BC}$, $\\mathrm{y}=3$ remains unchanged.\n\nSince point $\\mathrm{E}$ is the midpoint of $\\mathrm{AD}$, then\n\n$\\frac{1}{2} AE \\bullet AB = \\frac{1}{2} \\times \\frac{1}{2} AD \\bullet AB = 3$,\n$\\therefore AD \\bullet AB = 12$.\n\nWhen point $\\mathrm{F}$ moves from $\\mathrm{C}$ to $\\mathrm{D}$, $\\mathrm{y}$ gradually decreases, thus we have\n\n$AD + AB = 7$,\n\nand since $(AB < AD)$,\n\n$\\therefore AD = 4, \\quad AB = 3$.\n\nWhen $x=5$, point $\\mathrm{F}$ is on $\\mathrm{CD}$, so $CF = 5 - 4 = 1$,\n\n$\\therefore DF = 3 - 1 = 2$,\n\n$\\therefore y = \\frac{1}{2} AE \\bullet DF = \\frac{1}{2} \\times \\frac{1}{2} \\times 4 \\times 2 = 2$.\n\nTherefore, the correct answer is: A.\n\n[Insight] This problem examines the function graph of moving points and the properties of rectangles. The key to solving it is to understand the problem statement and to use a categorical discussion approach to think through the problem, which is a common type of question in middle school exams." }, { "problem_id": 274, "question": "As shown in Figure (1), point $E$ starts from vertex $A$ of the rhombus $A B C D$ and moves along $A \\rightarrow C \\rightarrow D$ at a constant speed of $1 \\mathrm{~cm} / \\mathrm{s}$ until it reaches point $D$. Figure (2) is a graph showing the relationship between the area $y\\left(\\mathrm{~cm}^{2}\\right)$ of $\\triangle A B E$ and the time $x(s)$ as point $E$ moves. Determine the length of the sides of the rhombus.\n\n\n\n(1)\n\n\n\n(2)\nA. 3\nB. 4\nC. $\\frac{25}{6}$\nD. 5", "input_image": [ "batch22-2024_06_14_432983757ec833e0afc7g_0084_1.jpg", "batch22-2024_06_14_432983757ec833e0afc7g_0084_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "(1)\n\nAs shown in the figure, draw a perpendicular from point $\\mathrm{C}$ to $\\mathrm{AB}$, intersecting $\\mathrm{AB}$ at $\\mathrm{E}$.\n\nGiven: The area of $\\triangle \\mathrm{ABC}$ is $2a$, $\\mathrm{AC}=5$, and $\\mathrm{DC}=a$.\n\nSince quadrilateral $\\mathrm{ABCD}$ is a rhombus,\n\n$\\therefore \\mathrm{AB}=\\mathrm{DC}=\\mathrm{BC}=a$,\n\n$\\therefore$ the area of $\\triangle \\mathrm{ABC} = \\frac{1}{2} \\cdot AB \\cdot CE = 2a$,\n\n$\\therefore \\mathrm{CE}=4$.\n\n$\\therefore$ In right triangle $\\triangle \\mathrm{AEC}$, $\\mathrm{AE}=\\sqrt{AC^{2}-CE^{2}}=3$,\n\n$\\therefore \\mathrm{BE}=a-3$,\n\n$\\therefore$ In right triangle $\\triangle \\mathrm{BEC}$, $BC^{2}=CE^{2}+BE^{2}$,\n\ni.e., $a^{2}=4^{2}+(a-3)^{2}$,\n\nSolving gives: $a=\\frac{25}{6}$.\n\n$\\therefore$ The side length of the rhombus is $\\frac{25}{6}$.\n\nTherefore, the correct answer is: C\n\n【Key Point】This problem mainly examines the comprehensive application of rhombus and triangle dynamic point problems. Mastering the relevant properties is the key to solving the problem." }, { "problem_id": 275, "question": "A square paper with side length $\\mathrm{a}$ is placed inside the rectangle $\\mathrm{ABCD}$ as shown in Figure 1. A rectangle with length $\\mathrm{b}$ ($\\mathrm{b} < \\mathrm{a}$) and width $\\frac{\\mathrm{b}}{2}$ is placed inside the rectangle in two ways as shown in Figures 2 and 3. The uncovered parts of the rectangle are indicated by shading. Let the area of the shaded part in Figure 2 be $\\mathrm{S}_{1}$ and the area of the shaded part in Figure 3 be $\\mathrm{S}_{2}$. If $\\mathrm{S}_{2}-\\mathrm{S}_{1}=2 \\mathrm{~b}$, then the value of $\\mathrm{AD}-\\mathrm{AB}$ is $($ $)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. 1\nB. 2\nC. 4\nD. Indeterminate", "input_image": [ "batch22-2024_06_14_44926fcff4cc631e81bag_0052_1.jpg", "batch22-2024_06_14_44926fcff4cc631e81bag_0052_2.jpg", "batch22-2024_06_14_44926fcff4cc631e81bag_0052_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Because \\( \\mathrm{S}_{1} = (CD - a) \\cdot a + \\left(AB - \\frac{1}{2}b\\right)(BC - a) \\)\n\n\\( = (AB - a) \\cdot a + \\left(AB - \\frac{1}{2}b\\right)(AD - a) \\),\n\n\\( \\mathrm{S}_{2} = AB(AD - a) + \\left(a - \\frac{1}{2}b\\right)(AB - a) \\),\n\nTherefore, \\( \\mathrm{S}_{1} - \\mathrm{S}_{2} = (AB - a) \\cdot a + \\left(AB - \\frac{1}{2}b\\right)(AD - a) - AB(AD - a) - \\left(a - \\frac{1}{2}b\\right)(AB - a) \\)\n\n\\( = (AD - a)\\left(AB - \\frac{1}{2}b - AB\\right) + (AB - a)\\left(a - a + \\frac{1}{2}b\\right) \\)\n\n\\( = \\frac{b}{2}(AB - AD) \\),\n\nBecause \\( \\mathrm{S}_{2} - \\mathrm{S}_{1} = 2b \\),\n\nTherefore, \\( \\frac{b}{2}(AB - AD) = -2b \\),\n\nBecause \\( b \\neq 0 \\),\n\nTherefore, \\( AD - AB = 4 \\).\n\nHence, the answer is: C.\n\n【Highlight】This question examines the properties of squares, the formulation of algebraic expressions, and the mixed operations of polynomials. The holistic approach is quite common in polynomial operations, and adopting it at the right time can simplify problems and lead to quick solutions. It is important to note that algebraic expressions treated as a whole should usually be enclosed in parentheses." }, { "problem_id": 276, "question": "As shown in the figure, in quadrilateral $A B C D$, the diagonals $A C$ and $B D$ intersect at point $O$, with $A C \\perp B C$. Given that $B C = 4$ and $\\angle A B C = 60^\\circ$, if $E F$ passes through point $O$ and intersects sides $A B$ and $C D$ at points $E$ and $F$, respectively, with $B E = x$ and $O E^2 = y$, the graph of $y$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch22-2024_06_14_47fa32e48ddb8124f149g_0056_1.jpg", "batch22-2024_06_14_47fa32e48ddb8124f149g_0056_2.jpg", "batch22-2024_06_14_47fa32e48ddb8124f149g_0056_3.jpg", "batch22-2024_06_14_47fa32e48ddb8124f149g_0056_4.jpg", "batch22-2024_06_14_47fa32e48ddb8124f149g_0056_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, draw a perpendicular from point $O$ to $AB$, intersecting $AB$ at point $M$.\n\n\n\nSince $AC \\perp BC$ and $\\angle ABC = 60^\\circ$,\n\n$\\therefore \\angle BAC = 30^\\circ$.\n\nGiven that $BC = 4$,\n\n$\\therefore AB = 8$, $AC = 4\\sqrt{3}$.\n\nSince quadrilateral $ABCD$ is a parallelogram,\n\n$\\therefore AO = \\frac{1}{2} AC = 2\\sqrt{3}$,\n\n$\\therefore OM = \\frac{1}{2} AO = \\sqrt{3}$,\n\n$\\therefore AM = \\sqrt{AO^2 - OM^2} = 3$.\n\nLet $BE = x$, $OE^2 = y$, then $EM = AB - AM - BE = 8 - 3 - x = 5 - x$.\n\nSince $OE^2 = OM^2 + EM^2$,\n\n$\\therefore y = (x - 5)^2 + 3$.\n\nWhen $0 \\leq x < 3$, $3 < y \\leq 28$.\n\nWhen $3 \\leq x \\leq 8$, $3 \\leq y \\leq 12$.\n\nThe graph is a part of a quadratic function.\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem mainly tests the properties of parallelograms, the Pythagorean theorem, the properties of right triangles with a $30^\\circ$ angle, and the knowledge of quadratic function graphs. The key to solving the problem lies in determining the function's analytical expression and the range of its values." }, { "problem_id": 277, "question": "As shown in the figure, in square $A B C D$, the diagonals $A C$ and $B D$ intersect at point $E$, $\\angle A D E = 15^\\circ$, and $B D = 2\\sqrt{2}$. The triangle $A B C$ is folded along the line containing $A C$ for $180^\\circ$ into its original plane. If the point where $B$ lands is labeled $B^\\prime$, it exactly happens that $B E \\perp B^\\prime E$. If point $F$ is on segment $B C$, what is the shortest distance from $B^\\prime$ to $F$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 1\nB. $\\sqrt{2}$\nC. $\\sqrt{3}$\nD. $\\sqrt{5}$", "input_image": [ "batch22-2024_06_14_6243a99ff96f9ef4f503g_0063_1.jpg", "batch22-2024_06_14_6243a99ff96f9ef4f503g_0063_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nFrom the properties of folding, we can derive:\n- $\\angle BCB' = 2\\angle ACB$,\n- $\\angle AEB = \\angle AEB'$,\n- $BC = B'C$.\n\nSince $BE \\perp B'E$,\n- $\\angle BEB' = 90^\\circ$,\n- Therefore, $\\angle AEB' = 45^\\circ$.\n\nGiven that $\\angle ADE = 15^\\circ$,\n- $\\angle DAE = 30^\\circ$.\n\nSince quadrilateral $ABCD$ is a parallelogram,\n- $AD \\parallel BC$,\n- Therefore, $\\angle ACB = \\angle DAC = 30^\\circ$,\n- Hence, $\\angle BCB' = 2\\angle ACB = 60^\\circ$.\n\nConnecting $BB'$ and drawing $B'F \\perp BC$ (as shown in the figure),\n- Triangle $BB'C$ is equilateral,\n- Therefore, $\\angle B'BC = 60^\\circ$.\n\nSince quadrilateral $ABCD$ is a parallelogram,\n- $DE = BE = \\frac{1}{2} BD = \\frac{1}{2} \\times 2\\sqrt{2} = \\sqrt{2} = B'E$.\n\nIn right triangle $BB'E$,\n- $BE = B'E = \\sqrt{2}$,\n- Therefore, $BB' = \\sqrt{BE^2 + B'E^2} = 2$.\n\nFrom the property that the perpendicular segment is the shortest,\n- When $B'F \\perp BC$, $B'F$ is the shortest.\n\nIn right triangle $BB'F$,\n- $\\angle B'BF = 60^\\circ$, $B'B = 2$,\n- Therefore, $BF = \\frac{1}{2} BB' = 1$,\n- Hence, $B'F = \\sqrt{B'B^2 - BF^2} = \\sqrt{3}$.\n\n**Answer:** C.\n\n**Key Points:**\nThis problem primarily examines the properties of parallelograms, folding properties, the Pythagorean theorem, properties of right triangles with a $30^\\circ$ angle, and the concept that the perpendicular segment is the shortest. Mastery of these theorems is crucial for solving the problem." }, { "problem_id": 278, "question": "As shown in the figure, in the quadrilateral $A B C D$, point $P$ moves along the direction $A \\rightarrow B \\rightarrow C$ from point $A$ to point $C$. Let the distance traveled by point $P$ be $x$, and the length of segment $A P$ be $y$. Figure 2 is a graph showing the relationship between $y$ and $x$ as point $P$ moves. Then the length of $B C$ is $(\\quad)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 4\nB. 4.8\nC. 5\nD. 10", "input_image": [ "batch22-2024_06_14_6243a99ff96f9ef4f503g_0080_1.jpg", "batch22-2024_06_14_6243a99ff96f9ef4f503g_0080_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure below,\n\n\n\nAccording to Figure 2,\n\nWhen point $P$ reaches point $B$, $AP = AB = 3$,\n\nWhen $AP \\perp BC$, $AB + BP = 4.8$,\n\n$\\therefore BP = BE = 1.8$,\n\n$\\therefore AE = \\sqrt{AB^{2} - BE^{2}} = \\sqrt{3^{2} - 1.8^{2}} = \\frac{12}{5}$,\n\nWhen point $P$ reaches point $C$, $AP = AC = 4$,\n\n$\\therefore EC = \\sqrt{AC^{2} - AE^{2}} = \\sqrt{4^{2} - \\frac{12^{2}}{5}} = \\frac{16}{5}$,\n\n$\\therefore BC = BE + EC = 1.8 + \\frac{16}{5} = 5$.\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This question mainly examines the properties of parallelograms and the Pythagorean theorem. Mastering the properties of parallelograms and solving the problem based on the movement pattern of point $P$ and the relationship diagram is key." }, { "problem_id": 279, "question": "As shown in the figure, in parallelogram $\\mathrm{ABCD}$, the quadrilateral $\\mathrm{EFGH}$ may not be a parallelogram under the following conditions ( )\n\nA.\n\n\n\nB.\n\n\n\n$E, F, G, H$ are the midpoints of the sides\n\nC.\n\n\n\n$AF, BH, CH, DF$ are the angle bisectors\n\nD.\n\n\n\n$EG, FH$ are two line segments passing through the intersection point of the diagonals\n\n##", "input_image": [ "batch22-2024_06_14_73271a234763f51d2516g_0041_1.jpg", "batch22-2024_06_14_73271a234763f51d2516g_0041_2.jpg", "batch22-2024_06_14_73271a234763f51d2516g_0041_3.jpg", "batch22-2024_06_14_73271a234763f51d2516g_0041_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Option A: Since the given conditions only involve the relationships between angles and there are no equal relationships between the sides of the triangles, it is impossible to prove that the triangles are congruent or that the sides are parallel. Therefore, it cannot be proven that quadrilateral $\\mathrm{EFGH}$ is a parallelogram.\n\nOption B: By connecting $\\mathrm{AC}$ and applying the Midsegment Theorem of triangles, it is easy to prove that $\\mathrm{EF} = \\mathrm{GH}$ and $\\mathrm{EF} \\parallel \\mathrm{GH}$. Thus, quadrilateral $\\mathrm{EFGH}$ is a parallelogram.\n\nOption C: Using the fact that $\\mathrm{AD} \\parallel \\mathrm{BC}$ and that $\\mathrm{AE}$ and $\\mathrm{BE}$ are angle bisectors, it is easy to prove that $\\angle \\mathrm{ABE} = 90^{\\circ}$. Consequently, $\\angle \\mathrm{HEF} = 90^{\\circ}$, and similarly, $\\angle \\mathrm{EFH} = \\angle \\mathrm{FGH} = \\angle \\mathrm{EHG} = 90^{\\circ}$. Therefore, quadrilateral $\\mathrm{EFGH}$ is a rectangle, and hence, it is a parallelogram.\n\nOption D: Let the intersection point of the diagonals of parallelogram $\\mathrm{ABCD}$ be $\\mathrm{O}$. Since $\\mathrm{ABCD}$ is a parallelogram, $\\angle \\mathrm{EAO} = \\angle \\mathrm{GCO}$, $\\angle \\mathrm{AOE} = \\angle \\mathrm{COG}$, and $\\mathrm{OA} = \\mathrm{OC}$. Using the ASA (Angle-Side-Angle) criterion, it can be proven that $\\triangle \\mathrm{AOE} \\cong \\triangle \\mathrm{COG}$, so $\\mathrm{OE} = \\mathrm{OG}$. Similarly, $\\mathrm{OH} = \\mathrm{OF}$. Thus, quadrilateral $\\mathrm{EFGH}$ is a parallelogram. Therefore, the correct answer is A.\n\nKey Point: This question tests the determination and properties of congruent triangles and the determination and properties of parallelograms. The key to solving the problem lies in mastering various methods to prove that a quadrilateral is a parallelogram." }, { "problem_id": 280, "question": "Fold a square piece of paper along the dashed lines as shown in steps 1 and 2, and then cut off a corner along the dashed line in step 3. The unfolded and flattened shape is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch22-2024_06_14_77881459dbacbd0f1d19g_0004_1.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0004_2.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0004_3.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0004_4.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0004_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the problem statement, it is known that the shape cut out is a quadrilateral with diagonals that are perpendicular bisectors of each other, which is a rhombus. Additionally, the vertices of the rhombus lie on the fold lines.\n\nTherefore, the correct choice is A.\n\n[Highlight] This question tests the ability to fold shapes and hands-on operation. For such problems, when visualization is difficult, hands-on operation can be an effective method to solve the problem." }, { "problem_id": 281, "question": "Fold a square piece of paper as shown in the figure, and then cut out a diamond-shaped hole on it. After unfolding the paper, ( )\n\nA.\n\n\nB.\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch22-2024_06_14_77881459dbacbd0f1d19g_0093_1.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0093_2.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0093_3.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0093_4.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0093_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Question Analysis: By integrating spatial thinking, analyze the folding process and the position where the rhombus is cut, paying attention to the symmetry of the figure. It is easy to deduce the unfolded shape.\n\nQuestion Explanation: When a square piece of paper is folded twice along its diagonal to form a right-angled triangle, cutting a rhombus at a position perpendicular to the hypotenuse leaves the right-angled vertex intact, meaning the center of the original square remains undamaged, and the rhombus is symmetrical about the diagonal. Therefore, the correct choice is C.\n\nExam Focus: Paper-cutting problems." }, { "problem_id": 282, "question": "As shown in the figure, a rhombus with side length 10 is divided into four congruent right-angled triangles along its diagonals. One of the diagonals of the rhombus has a length of 16. These four right-angled triangles are rearranged to form a square as shown in Figure 2. Determine the area of the shaded region in Figure 2.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 2\nB. 4\nC. 9\nD. 16", "input_image": [ "batch22-2024_06_14_860e4b332fd6ae9f3c23g_0003_1.jpg", "batch22-2024_06_14_860e4b332fd6ae9f3c23g_0003_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nGiven that quadrilateral \\( ABCD \\) is a rhombus, with \\( AC = 16 \\) and \\( AD = 10 \\),\n\nTherefore, \\( OA = OC = 8 \\), \\( OB = OD \\), and \\( AC \\perp BD \\),\n\nThus, \\( OB = OD = \\sqrt{AD^{2} - OA^{2}} = 6 \\),\n\nHence, \\( BD = 2 \\times OD = 12 \\),\n\nTherefore, the area of the rhombus \\( = \\frac{1}{2} \\times 12 \\times 16 = 96 \\),\n\nThe area of the square in Figure 2 \\( = 10^{2} = 100 \\),\n\nThus, the area of the shaded region \\( = 100 - 96 = 4 \\).\n\nTherefore, the correct answer is: B.\n\n【Key Insight】This question tests the understanding of graphic cutting and splicing, the properties of a rhombus, and the properties of a square. The key to solving this problem lies in mastering the properties of a rhombus." }, { "problem_id": 283, "question": "In each of the following figures, a $4 \\mathrm{~cm} \\times 4 \\mathrm{~cm}$ square overlaps with a $8 \\mathrm{~cm} \\times 2 \\mathrm{~cm}$ rectangle. Which figure has the largest overlapping area?\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch22-2024_06_14_960c2f9512da7775d41eg_0012_1.jpg", "batch22-2024_06_14_960c2f9512da7775d41eg_0012_2.jpg", "batch22-2024_06_14_960c2f9512da7775d41eg_0012_3.jpg", "batch22-2024_06_14_960c2f9512da7775d41eg_0012_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution:\n\nA. The area of the shaded region is calculated as \\( S_{\\text{shaded}} = 2 \\times 4 = 8 \\, \\text{cm}^2 \\).\n\nB. As shown in the figure, according to the Pythagorean theorem, \\( 2x^2 = 4 \\), so \\( x = \\sqrt{2} \\). The area of the shaded region is:\n\\[\nS_{\\text{shaded}} = 4 \\times 4 - 2 \\times \\frac{1}{2} \\times (4 - \\sqrt{2}) \\times (4 - \\sqrt{2}) = 8\\sqrt{2} - 2 \\, \\text{cm}^2.\n\\]\n\nC. For Figure C, rotating it \\( 90^\\circ \\) counterclockwise and viewing it from the back, it can be compared with Figure D. Since the slope of Figure C is less than that of Figure D, the base of Figure C is smaller than that of Figure D. Both figures have the same height but different bases, so the area of the shaded region in Figure C is less than that in Figure D.\n\nD. As shown in the figure, let the base of the shaded parallelogram be \\( x \\). Therefore, the shorter leg of the right triangle is \\( \\sqrt{x^2 - 16} \\). Since the area of the square equals the area of the shaded region plus the areas of the two blank triangles:\n\\[\n\\frac{1}{2} \\times 4 \\times \\sqrt{x^2 - 16} \\times 2 + 2x = 16.\n\\]\nSolving for \\( x \\), we get \\( x = \\frac{8(\\sqrt{7} - 1)}{3} \\). The area of the shaded region is:\n\\[\nS_{\\text{shaded}} = 2 \\times \\frac{8(\\sqrt{7} - 1)}{3} = \\frac{16(\\sqrt{7} - 1)}{3}.\n\\]\nGiven that \\( \\sqrt{2} \\approx 1.414 \\) and \\( \\sqrt{7} \\approx 2.646 \\), we have:\n\\[\n8\\sqrt{2} - 2 \\approx 9.312, \\quad \\frac{16(\\sqrt{7} - 1)}{3} \\approx 8.775.\n\\]\nThus, \\( 8\\sqrt{2} - 2 > \\frac{16(\\sqrt{7} - 1)}{3} \\), meaning the area of the shaded region in Figure B is greater than that in Figure D.\n\nAdditionally, since the shaded quadrilaterals in Figures A, C, and D have the same height but different bases, and Figure D has the greatest slope, the base of the shaded region in Figure D is the largest.\n\nTherefore, the correct answer is B.\n\n**Key Insight:** This problem tests the calculation of areas of rectangles and triangles. The key to solving it lies in identifying that the shaded quadrilaterals in Figures A, B, and D have the same height but different bases, and that the area increases with the slope." }, { "problem_id": 284, "question": "As shown in Figure 1, this is a figure created by the ancient Chinese mathematician Zhao Shang to prove the Pythagorean Theorem. It is later called \"Zhao Shang's String Diagram.\" Figure 2 is derived from the string diagram, consisting of eight congruent right-angled triangles. Let the areas of the squares $A B C D$, $E F G H$, and $M N K T$ be denoted as $S_{1}$, $S_{2}$, and $S_{3}$, respectively. If\n\n$S_{1}+S_{2}+S_{3}=24$, then the value of $S_{2}$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 10\nB. 9\nC. 8\nD. 7", "input_image": [ "batch22-2024_06_14_960c2f9512da7775d41eg_0014_1.jpg", "batch22-2024_06_14_960c2f9512da7775d41eg_0014_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since the eight right-angled triangles are congruent, the quadrilaterals \\(ABCD\\), \\(EFGH\\), and \\(MNKT\\) are squares.\n\nTherefore, \\(CG = NG\\), \\(BG = NF\\), and \\(FG = GH\\).\n\nGiven:\n\\[\nS_{1} = (CG + BG)^{2} = CG^{2} + BG^{2} + 2 \\cdot CG \\cdot BG = GH^{2} + 2 \\cdot CG \\cdot BG,\n\\]\n\\[\nS_{2} = GH^{2},\n\\]\n\\[\nS_{3} = (NF - NG)^{2} = NG^{2} + NF^{2} - 2 \\cdot NG \\cdot NF.\n\\]\n\nThus:\n\\[\nS_{1} + S_{2} + S_{3} = 24 = GH^{2} + 2 \\cdot CG \\cdot BG + GH^{2} + NG^{2} + NF^{2} - 2 \\cdot NG \\cdot NF = 3GH^{2}.\n\\]\n\nTherefore:\n\\[\nGH^{2} = 8, \\quad \\text{which means} \\quad S_{2} = 8.\n\\]\n\nHence, the correct choice is: C.\n\n**Key Insight:** This problem tests the application of the Pythagorean theorem, involving the properties of squares and congruent triangles. Mastering these concepts is crucial for solving the problem." }, { "problem_id": 285, "question": "When three square sheets of paper with side length $m$ are arranged as shown in Figure 1 to form a large square with side length $n$, the overlapping area of the three small squares is two squares with sides of length 1. When they are arranged as shown in Figure 2 to form a rectangle with adjacent sides of length $3m$ and $n$, the area of the rectangle is 35. What is the perimeter of the rectangle in Figure 2?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 24\nB. 26\nC. 28\nD. 30", "input_image": [ "batch22-2024_06_14_96aff5d95dec44dfd892g_0001_1.jpg", "batch22-2024_06_14_96aff5d95dec44dfd892g_0001_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "According to the problem, from Figure 1, we have \\(3m = n + 2\\), and from Figure 2, we have \\(3mn = 35\\).\n\nTherefore, \\(n(n + 2) = 35\\),\n\nwhich simplifies to \\(n^{2} + 2n + 1 = 36\\).\n\nSolving this, we find \\(n = 5\\) or \\(n = -7\\) (discarded).\n\nThus, when \\(n = 5\\), \\(3m = 7\\).\n\nThen, the perimeter of the rectangle in Figure 2 is \\(2(3m + n) = 2 \\times (7 + 5) = 24\\).\n\nTherefore, the correct answer is A.\n\n【Key Insight】This problem tests the ability to solve equations using factorization. Identifying the correct relationships and setting up the equations are crucial to solving the problem." }, { "problem_id": 286, "question": "As shown in the figure, two people, A and B, play a game of folding a rectangle to make a rhombus. A folds along $B E$ so that point $A$ falls on $B D$, and then folds along $D F$ so that point $C$ falls on $B D$. A claims that the quadrilateral $B E D F$ is a rhombus. B folds along $M N$ so that $A B$ coincides with $D C$, and then folds out $B M$ and $D N$. B claims that the quadrilateral $B M D N$ is a rhombus. Which of the following statements is correct ( )\n\n\n\n A\n\n\n\n B\nA. A's claim is definitely true, and B's claim may be true.\nB. A's claim may be true, and B's claim is definitely false.\nC. A's claim is definitely true, and B's claim is definitely false.\nD. A's claim may be true, and B's claim may also be true.", "input_image": [ "batch22-2024_06_14_adeb8a4d57e8df5fe295g_0041_1.jpg", "batch22-2024_06_14_adeb8a4d57e8df5fe295g_0041_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( A B C D \\) is a rectangle,\n\nTherefore, \\( A B \\parallel C D \\) and \\( A D \\parallel B C \\).\n\nThus, \\( \\angle A B D = \\angle C D B \\).\n\nFrom the folding, we know: \\( \\angle E B D = \\angle A B E \\) and \\( \\angle F D B = \\angle C D F \\).\n\nTherefore, \\( \\angle E B D = \\angle F D B \\).\n\nHence, \\( B E \\parallel D F \\),\n\nSo, quadrilateral \\( B E D F \\) is a parallelogram.\n\nWhen \\( B E = D E \\), quadrilateral \\( B E D F \\) is a parallelogram,\n\nThus, \\( \\angle E B D = \\angle E D B \\),\n\nAlso, since \\( A D \\parallel B C \\), \\( \\angle E D B = \\angle D B F \\),\n\nTherefore, \\( \\angle E B D = \\angle A B E = \\angle D B F = \\frac{1}{3} \\angle A B C = 30^\\circ \\),\n\nHence, when \\( \\angle D B C = 30^\\circ \\), quadrilateral \\( B E D F \\) becomes a rhombus, and possibility A may hold.\n\nHowever, from the folding method of B, we know \\( M N \\perp B C \\), so \\( B M > B N \\), thus quadrilateral \\( B M D N \\) cannot be a rhombus.\n\nIn summary: Possibility A may hold, while possibility B definitely does not hold,\n\nTherefore, choose B.\n\n[Key Insight] This question examines the properties of folding transformations, the determination of parallelograms, and the properties of rectangles and rhombuses. Folding transformation is a type of symmetry transformation, belonging to axial symmetry, where the shape and size of the figure remain unchanged before and after folding, but the position changes, with corresponding sides and angles being equal." }, { "problem_id": 287, "question": "Four equal-length thin wooden sticks are joined end-to-end to form a quadrilateral $\\mathrm{ABCD}$. By turning this quadrilateral, its shape can be changed. When $\\angle B = 60^\\circ$, as shown in Figure (1), it is measured that $A C = 3$; when $\\angle B = 90^\\circ$, as shown in Figure (2), the length of $\\mathrm{AC}$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $3 \\sqrt{2}$\nB. $2 \\sqrt{3}$\nC. 3\nD. $2 \\sqrt{2}$", "input_image": [ "batch22-2024_06_14_cfc50897275df68cfa35g_0034_1.jpg", "batch22-2024_06_14_cfc50897275df68cfa35g_0034_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in Figure (1), connect $\\mathrm{AC}$,\n\n\n\nFigure (1)\n\n\n\n$\\because \\angle \\mathrm{B}=60^{\\circ}, \\mathrm{AB}=\\mathrm{BC}$,\n\n$\\therefore \\triangle \\mathrm{ABC}$ is an equilateral triangle,\n\n$\\therefore \\mathrm{AC}=\\mathrm{AB}=\\mathrm{BC}=3$,\n\nAs shown in Figure (2), connect $\\mathrm{AC}$,\n\n$\\because \\mathrm{AB}=\\mathrm{BC}=\\mathrm{CD}=\\mathrm{DA}=3, \\angle \\mathrm{B}=90^{\\circ}$,\n\n$\\therefore$ quadrilateral $\\mathrm{ABCD}$ is a square,\n\n$\\therefore A C=\\sqrt{A B^{2}+B C^{2}}=\\sqrt{3^{2}+3^{2}}=3 \\sqrt{2}$.\n\nTherefore, choose: A.\n\n【Key Insight】This question examines the properties and determination of a square, the properties of a rhombus, the Pythagorean theorem, and the determination and properties of an equilateral triangle. The key is to use the determination of an equilateral triangle to confirm the side lengths." }, { "problem_id": 288, "question": "The following figures are composed of parallelograms of the same size according to a certain rule. There are 10 parallelograms in the first figure, 14 in the second, 19 in the third, $\\qquad$. Following this pattern, the number of parallelograms in the seventh figure is ( ) .\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. 40\nB. 44\nC. 47\nD. 49", "input_image": [ "batch22-2024_06_14_d8be910b6d92a90a2fa5g_0012_1.jpg", "batch22-2024_06_14_d8be910b6d92a90a2fa5g_0012_2.jpg", "batch22-2024_06_14_d8be910b6d92a90a2fa5g_0012_3.jpg", "batch22-2024_06_14_d8be910b6d92a90a2fa5g_0012_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: By observing the changes in the figures, we can determine:\n\nIn figure (1), there are a total of \\(7 + 3 = 10\\) parallelograms.\n\nIn figure (2), there are a total of \\(10 + 4 = 14\\) parallelograms.\n\nIn figure (3), there are a total of \\(14 + 5 = 19\\) parallelograms.\n\nIn figure (4), there are a total of \\(19 + 6 = 25\\) parallelograms.\n\nTherefore:\n\nIn figure (5), there are a total of \\(25 + 7 = 32\\) parallelograms.\n\nIn figure (6), there are a total of \\(32 + 8 = 40\\) parallelograms.\n\nIn figure (7), there are a total of \\(40 + 9 = 49\\) parallelograms.\n\nHence, the correct answer is: D.\n\n**Insight:** This question tests the understanding of parallelograms and pattern recognition in figure changes. It is a problem that involves numerical conjecture based on figures, with the key being to deduce the underlying pattern through induction and summarization, and then apply this pattern to solve general problems." }, { "problem_id": 289, "question": "The correct graphical representation of the solution set for the inequality system \n$$\\left\\{\\begin{array}{l}x<2 x+1 \\\\ 2 x-1 \\geq 3(x-1)\\end{array}\\right.$$ \non the number line is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch22-2024_06_14_e70eca755418211ed19fg_0021_1.jpg", "batch22-2024_06_14_e70eca755418211ed19fg_0021_2.jpg", "batch22-2024_06_14_e70eca755418211ed19fg_0021_3.jpg", "batch22-2024_06_14_e70eca755418211ed19fg_0021_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solve the inequality \\( x < 2x + 1 \\), yielding: \\( x > -1 \\).\n\nSolve the inequality \\( 2x - 1 \\geq 3(x - 1) \\), yielding: \\( x \\leq 2 \\).\n\nTherefore, the solution set of the system of inequalities is: \\( -1 < x \\leq 2 \\).\n\nObserving the options, it is clear that option A matches this solution.\n\nHence, the correct choice is A." }, { "problem_id": 290, "question": "In Figure 1, there is a triangular plot of land, and it is planned to divide it into two parts using a fence to plant different plants as shown in Figure 2. The length of the fence $A B$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $2 \\mathrm{~m}$\nB. $3 \\mathrm{~m}$\nC. $4 \\mathrm{~m}$\nD. $1 \\mathrm{~m}$", "input_image": [ "batch22-2024_06_14_e78653ae4415f814bc30g_0033_1.jpg", "batch22-2024_06_14_e78653ae4415f814bc30g_0033_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nFrom the figure, it can be seen that $AB / / DE$.\n\nSince $CB = 3 = \\frac{1}{2} CE$,\n\nTherefore, $B$ is the midpoint of $CE$.\n\nThus, $AB$ is the midline of $\\triangle CDE$.\n\nHence, $AB = \\frac{1}{2} DE = 2 \\mathrm{~m}$.\n\nTherefore, the correct choice is A.\n\n[Key Insight] This question tests the properties of the midline. The key to solving the problem lies in a thorough understanding of the properties of the midline." }, { "problem_id": 291, "question": "Pythagoras's theorem is one of the greatest scientific discoveries in human history. As shown in Figure 1, equilateral triangles are constructed outward along each side of a right-angled triangle; as shown in Figure 2, the smaller equilateral triangles $\\triangle A F G$ and $\\triangle B D E$ are placed inside the largest equilateral triangle $\\triangle A B C$. The lines $D E$ and $F G$ intersect at point $P$, and lines $A P$ and $F E$ are drawn. To find the area of $\\triangle G E C$, which of the following triangle areas is sufficient to know?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\triangle A P G$\nB. $\\triangle A D P$\nC. $\\triangle D F P$\nD. $\\triangle F E G$", "input_image": [ "batch22-2024_06_14_e78653ae4415f814bc30g_0048_1.jpg", "batch22-2024_06_14_e78653ae4415f814bc30g_0048_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we have $S_{\\triangle ABC} = S_{\\triangle AFG} + S_{\\triangle BDE}$, with $FG \\parallel BC$ and $CG \\parallel PE$.\n\n$\\therefore$ Quadrilateral $CEPG$ is a parallelogram,\n\n$\\therefore S_{\\triangle CEG} = \\frac{1}{2} S_{\\text{quadrilateral } ECGP}$,\n\n$\\because S_{\\triangle ABC} = S_{\\triangle AFG} + S_{\\text{quadrilateral } BFPE} + S_{\\text{quadrilateral } ECGP}$,\n\n$\\therefore S_{\\text{quadrilateral } ECGP} = S_{\\triangle DFP}$,\n\n$\\therefore S_{\\triangle CEG} = \\frac{1}{2} S_{\\triangle DFP}$,\n\nTherefore, the correct choice is C.\n\n【Insight】This problem primarily examines the area of figures composed of the sides of a right-angled triangle, the properties and determination of parallelograms. The key to solving the problem lies in correctly understanding the given conditions." }, { "problem_id": 292, "question": "As shown in the figure, the area of the parallelogram paper $A B C D$ is $72 \\mathrm{~cm}^{2}$, and $A D=12 \\mathrm{~cm}$. The parallelogram $A B C D$ can be cut into four triangular paper pieces, A, B, C, D, along its two diagonals. If the triangles A and C are combined to form a symmetrical figure as shown in Figure 2, the sum of the lengths of the two diagonals in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $18 \\mathrm{~cm}$\nB. $20 \\mathrm{~cm}$\nC. $24 \\mathrm{~cm}$\nD. $28 \\mathrm{~cm}$", "input_image": [ "batch22-2024_06_14_e85091efaced6a456a53g_0053_1.jpg", "batch22-2024_06_14_e85091efaced6a456a53g_0053_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Connect $AD$ and $EF$,\n\n\n\nFigure 2\n\nThen, it can be deduced that the diagonal $EF \\perp AD$, and $EF$ is equal in length to the height of the parallelogram,\n\n$\\because$ The area of the parallelogram sheet $ABCD$ is $72 \\mathrm{~cm}^{2}$, and $AD=12 \\mathrm{~cm}$,\n\n$\\therefore BC=AD=12 \\mathrm{~cm}, \\frac{1}{2} EF \\times AD=\\frac{1}{2} \\times 72$,\n\n$\\therefore EF=6 \\mathrm{~cm}$\n\nAlso, $\\because BC=AD=12 \\mathrm{~cm}$,\n\n$\\therefore$ The sum of the diagonals of the quadrilateral in figure 戊 is: $12+6=18 \\mathrm{~cm}$,\n\nTherefore, the answer is: A.\n\n[Insight] This question tests the properties of parallelograms and the symmetry of figures. Mastering the fact that the diagonals of a parallelogram bisect each other is key to solving the problem." }, { "problem_id": 293, "question": "As shown in the figure, in square $A B C D$, $\\angle D A B = 30^\\circ$, $A B = 6$, and $B C = 4$. Point $P$ starts from point $D$ and moves along $D C$ and $C B$ towards endpoint $B$ at a constant speed. Let the distance traveled by point $P$ be $x$, and the area enclosed by the path of point $P$ and segments $A D$, $A P$ be $y$. Which of the following graphs correctly represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch22-2024_06_14_e938ad1d0712c6c47170g_0008_1.jpg", "batch22-2024_06_14_e938ad1d0712c6c47170g_0008_2.jpg", "batch22-2024_06_14_e938ad1d0712c6c47170g_0008_3.jpg", "batch22-2024_06_14_e938ad1d0712c6c47170g_0008_4.jpg", "batch22-2024_06_14_e938ad1d0712c6c47170g_0008_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When point \\( P \\) moves along side \\( DC \\), with \\( 0 \\leq x \\leq 6 \\), as shown in the figure, draw \\( DE \\perp AB \\) at point \\( E \\).\n\n\n\nSince quadrilateral \\( ABCD \\) is a parallelogram,\n\n\\( AD = BC = 4 \\), and \\( CD = AB = 6 \\).\n\nGiven \\( \\angle DAB = 30^\\circ \\) and \\( BC = 4 \\),\n\n\\( DE = \\frac{1}{2} AD = 2 \\),\n\nThus, \\( y = \\frac{1}{2} PD \\cdot DE = \\frac{1}{2} x \\times 2 = x \\).\n\n\n\nWhen point \\( P \\) moves along segment \\( CB \\), with \\( 6 < x \\leq 10 \\), as shown in the figure, draw \\( BF \\perp AD \\), intersecting the extension of \\( AD \\) at point \\( F \\).\n\nGiven \\( \\angle DAB = 30^\\circ \\) and \\( AB = 6 \\),\n\n\\( BF = \\frac{1}{2} AB = 3 \\).\n\nSince \\( CP = x - 6 \\),\n\n\\( y = \\frac{1}{2}(CP + AD) \\cdot BF = \\frac{1}{2} \\times (x - 6 + 4) \\times 3 = \\frac{3}{2}x - 3 \\).\n\n\n\nIn summary, the functional relationship between \\( y \\) and \\( x \\) is:\n\n\\[\ny = \\begin{cases} \nx, & 0 \\leq x \\leq 6 \\\\\n\\frac{3}{2}x - 3, & 6 < x \\leq 10 \n\\end{cases}\n\\]\n\nThe corresponding graph of the function is \\( \\mathrm{A} \\).\n\nTherefore, the correct choice is: A.\n\n**Key Insight:** This problem involves a moving point in a quadrilateral and examines the graph of a function. It tests the properties of parallelograms and the characteristics of 30-degree right triangles. Determining the functional relationship is crucial, and attention must be paid to categorizing different scenarios." }, { "problem_id": 294, "question": "In parallelogram $A B C D$, which of the following pairs of angles can be used as a counterexample to the proposition \"two angles are equal\"?\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch22-2024_06_14_eb32f3a16c255e6590b5g_0030_1.jpg", "batch22-2024_06_14_eb32f3a16c255e6590b5g_0030_2.jpg", "batch22-2024_06_14_eb32f3a16c255e6590b5g_0030_3.jpg", "batch22-2024_06_14_eb32f3a16c255e6590b5g_0030_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a parallelogram,\n\n\\(\\therefore \\angle BAD = \\angle BCD\\), and \\( AD \\parallel BC \\),\n\n\\(\\therefore \\angle 1 = \\angle 2\\), hence options A and D do not fit the context;\n\nSince \\(\\angle 1\\) and \\(\\angle 2\\) are vertical angles,\n\n\\(\\therefore \\angle 1 = \\angle 2\\), hence option B does not fit the context;\n\nIn option C, \\(\\angle 2\\) is an exterior angle of the triangle containing \\(\\angle 1\\) and is not adjacent, thus \\(\\angle 1 \\neq \\angle 2\\), which fits the context;\n\nTherefore, the correct choice is: C.\n\n【Insight】This question tests the properties of parallelograms, the nature of vertical angles, and the properties of parallel lines; mastering the properties of parallelograms is key to solving the problem." }, { "problem_id": 295, "question": "As shown in the figure, in triangle $\\triangle \\mathrm{ABC}$, points $\\mathrm{D}$ and $\\mathrm{E}$ are the midpoints of sides $\\mathrm{AB}$ and $\\mathrm{AC}$, respectively. Triangle $\\triangle \\mathrm{ADE}$ is folded along segment $\\mathrm{DE}$ downward, as shown in Figure 2. Which of the following conclusions about Figure 2 may NOT be true?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. DE// BC\nB. $\\triangle \\mathrm{DBA}$ is an isosceles triangle\nC. Point $\\mathrm{A}$ falls on the midpoint of side $\\mathrm{BC}$\nD. $\\angle \\mathrm{B}+\\angle \\mathrm{C}+\\angle 1=180^{\\circ}$", "input_image": [ "batch23-2024_06_14_234c1c0ec9b3cd707c9ag_0098_1.jpg", "batch23-2024_06_14_234c1c0ec9b3cd707c9ag_0098_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since in triangle \\( \\triangle \\mathrm{ABC} \\), points \\( \\mathrm{D} \\) and \\( \\mathrm{E} \\) are the midpoints of sides \\( \\mathrm{AB} \\) and \\( \\mathrm{AC} \\) respectively,\n\ntherefore \\( \\mathrm{DE} \\) is parallel to \\( \\mathrm{BC} \\);\n\nHence, option A is correct.\n\nSince from the properties of folding, \\( \\mathrm{BD} = \\mathrm{AD} \\),\n\ntherefore \\( \\triangle \\mathrm{DBA} \\) is an isosceles triangle;\n\nHence, option B is correct.\n\nFrom the properties of folding, \\( \\mathrm{AD} = \\mathrm{BD} \\) and \\( \\mathrm{AE} = \\mathrm{EC} \\),\n\nbut it cannot be determined that \\( \\mathrm{AB} = \\mathrm{AC} \\),\n\nhence option C is incorrect.\n\nSince in \\( \\triangle \\mathrm{ABC} \\), \\( \\angle \\mathrm{A} + \\angle \\mathrm{B} + \\angle \\mathrm{C} = 180^{\\circ} \\),\n\nand from the properties of folding, \\( \\angle \\mathrm{A} = \\angle 1 \\),\n\ntherefore \\( \\angle \\mathrm{B} + \\angle \\mathrm{C} + \\angle 1 = 180^{\\circ} \\).\n\nHence, option D is correct.\n\nTherefore, the correct choice is C.\n\n[Highlight] This question examines the folding problem in triangles, which is not difficult. Mastering the properties of folding and the fact that the sum of angles in a triangle is \\( 180^{\\circ} \\) will allow for a smooth solution." }, { "problem_id": 296, "question": "As shown in Figure (1), point $E$ is on side $A D$ of the rectangle $A B C D$. Point $P$ starts from point $B$ and moves along the zigzag path $B-E-D$ until it reaches point $D$, while point $Q$ starts from point $B$ and moves along $B C$ until it reaches point $C$. Both points move at a speed of $0.5 \\mathrm{~cm} / \\mathrm{s}$. Now, $P$ and $Q$ start at the same time. Let the time of motion be $x(s)$, and the area of $\\triangle B P Q$ be $y \\mathrm{~cm}^{2}$. The relationship between $y$ and $x$ is shown in Figure (2). The area of the rectangle $A B C D$ is\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 18\nB. 12\nC. 20\nD. 16", "input_image": [ "batch23-2024_06_14_25ca6f7bb7a468afd8ccg_0077_1.jpg", "batch23-2024_06_14_25ca6f7bb7a468afd8ccg_0077_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the graph, it can be seen that at $10 \\mathrm{~s}$, points $P$ and $E$ coincide, and $B Q = B E = 5 \\mathrm{~cm}$. According to the problem statement, we have:\n\n\\[\n\\frac{1}{2} \\times B Q \\times A B = 7.5,\n\\]\n\n\\[\n\\therefore \\frac{1}{2} \\times 5 \\times A B = 7.5,\n\\]\n\nSolving this, we find $A B = 3 \\mathrm{~cm}$.\n\nSince quadrilateral $A B C D$ is a rectangle,\n\n\\[\n\\angle B A D = 90^{\\circ}\n\\]\n\n\\[\n\\therefore A E = \\sqrt{5^{2} - 3^{2}} = 4 \\mathrm{~cm},\n\\]\n\nFrom the graph, it can be seen that $B E + D E = 7 \\mathrm{~cm}$,\n\n\\[\n\\therefore D E = 2 \\mathrm{~cm},\n\\]\n\n\\[\n\\therefore A D = 6 \\mathrm{~cm},\n\\]\n\nThus, the area of the rectangle is:\n\n\\[\n6 \\times 3 = 18 \\mathrm{~cm}^{2}\n\\]\n\nTherefore, the correct answer is A.\n\n【Key Insight】This problem tests the properties of rectangles, the Pythagorean theorem, and the interpretation of function graphs. Mastering the properties of rectangles and correctly extracting information from function graphs are crucial for solving the problem." }, { "problem_id": 297, "question": "As shown in the figure, $A B C D$ is a rectangular sheet of paper, where $A D = B C = 1$ and $A B = C D = 5$. On the side $A B$ of the rectangle, a point $M$ is taken, and on the side $C D$, a point $N$ is taken. The paper is folded along $M N$ so that $M B$ intersects $D N$ at point $K$, forming $\\triangle M N K$. Which of the following statements is false?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. The minimum value of the crease $M N$ is 1.\nB. The maximum value of the crease $M N$ is $\\sqrt{26}$.\nC. Triangle $M N K$ is an isosceles triangle.\nD. The maximum area of triangle $M N K$ is 1.3.", "input_image": [ "batch23-2024_06_14_2809164ef4e74536d186g_0099_1.jpg", "batch23-2024_06_14_2809164ef4e74536d186g_0099_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: When the crease \\( M N = 1 \\), then \\( M N \\perp A B \\) and \\( B M \\parallel D N \\), which does not match the scenario described in the question. Therefore, option A is correct.\n\nWhen folding along the diagonal \\( A C \\), the crease \\( M N \\) is the longest,\n\\[ M N = A C = \\sqrt{A B^{2} + B C^{2}} = \\sqrt{26}, \\]\nso option B is incorrect.\n\nSince \\( A B C D \\) is a rectangular sheet,\n\\[ A B \\parallel C D, \\]\n\\[ \\angle K N M = \\angle 1. \\]\nFrom the folding, we have:\n\\[ \\angle 1 = \\angle K M N, \\]\n\\[ \\therefore \\angle K N M = \\angle K M N, \\]\n\\[ \\therefore K M = K N, \\]\nand \\( \\triangle K M N \\) is an isosceles triangle, so option C is incorrect.\n\nThere are two scenarios for folding to achieve the maximum area of the triangle:\n\n**Scenario 1:** Fold the rectangular sheet so that point \\( B \\) coincides with point \\( D \\), and point \\( K \\) also coincides with \\( D \\).\n\\[ M K = M B = x, \\]\n\\[ A M = 5 - x. \\]\nBy the Pythagorean theorem:\n\\[ 1^{2} + (5 - x)^{2} = x^{2}, \\]\nsolving gives \\( x = 2.6 \\).\n\\[ \\therefore M D = N D = 2.6. \\]\n\\[ S_{\\triangle M N K} = S_{\\triangle M N D} = \\frac{1}{2} \\times 1 \\times 2.6 = 1.3. \\]\n\n**Scenario 2:** Fold the rectangular sheet along the diagonal \\( A C \\), where the crease is \\( A C \\).\n\\[ M K = A K = C K = x, \\]\n\\[ D K = 5 - x. \\]\nSimilarly, we find:\n\\[ M K = N K = 2.6. \\]\nSince \\( M D = 1 \\),\n\\[ S_{\\triangle M N K} = \\frac{1}{2} \\times 1 \\times 2.6 = 1.3. \\]\nThe maximum area of \\( \\triangle M N K \\) is 1.3, so option D is incorrect.\n\nTherefore, the correct answer is A.\n\n**Key Insight:** This problem tests the properties of rectangles, the application of the Pythagorean theorem, and the identification of isosceles triangles. Understanding \"the properties of symmetry\" is crucial to solving this problem." }, { "problem_id": 298, "question": "As shown in Figure (1), a rectangular paper sheet ($\\mathrm{AD} // \\mathrm{BC}$) is folded along $\\mathrm{EF}$ to form Figure (2). The line segment $\\mathrm{ED}$ intersects $\\mathrm{BC}$ at point $\\mathrm{H}$, and then it is folded along $\\mathrm{HF}$ to form Figure (3). If $\\angle \\mathrm{DEF}$ in Figure (1) is $28^\\circ$, what is the measure of $\\angle \\mathrm{CFE}$ in Figure (3)?\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. $84^{0}$\nB. $96^{0}$\nC. $112^{0}$\nD. $124^{0}$", "input_image": [ "batch23-2024_06_14_28843a4185f44c6dd7b6g_0081_1.jpg", "batch23-2024_06_14_28843a4185f44c6dd7b6g_0081_2.jpg", "batch23-2024_06_14_28843a4185f44c6dd7b6g_0081_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since $\\mathrm{AD} / / \\mathrm{BC}$, it follows that $\\angle \\mathrm{DEF}=\\angle \\mathrm{EFB}=28^{\\circ}$. Given that $\\angle \\mathrm{EFC}+\\angle \\mathrm{EFB}=180^{\\circ}$, we deduce that $\\angle \\mathrm{EFC}=152^{\\circ}$, and consequently, $\\angle \\mathrm{CFE}=152^{\\circ}-28^{\\circ}=124^{\\circ}$.\n\n【Highlight】In transformation by folding (folding problems), equal angles will appear, which is an important point to remember." }, { "problem_id": 299, "question": "As shown in Figure 1, four congruent right-angled triangles form a large square, with a smaller square in the middle. This figure was given by Zhao Shang, a mathematician from the Han Dynasty, when he annotated the \"Zhou Bi Suan Jing\". People call it \"Zhao Shang's Chord Diagram\". In the chord diagram (as shown in Figure 2), connect $A F, D E$, and extend $D E$ to intersect $A F$ at point $K$, then connect $K G$. If $A H = 2 D H = 2 \\sqrt{2}$, then the length of $K G$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\sqrt{3}$\nB. 2\nC. $\\sqrt{5}$\nD. $2 \\sqrt{2}$", "input_image": [ "batch23-2024_06_14_2b7a2e617e420949bb53g_0067_1.jpg", "batch23-2024_06_14_2b7a2e617e420949bb53g_0067_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from point $K$ to $CF$, intersecting the extension of $CF$ at point $M$,\n\n\n\nGiven that $AH = 2DH = 2\\sqrt{2}$, and $AH = DG$,\n\nTherefore, $DH = GH = \\sqrt{2}$,\n\nSince $EFGH$ is a square,\n\nThus, $EF = FG = GH = HE = \\sqrt{2}$, and $AE = AH - HE = 2\\sqrt{2} - \\sqrt{2} = \\sqrt{2}$,\n\nHence, $DH = HE$,\n\nTherefore, $\\angle AEK = \\angle HED = \\angle HDE = 45^\\circ$,\n\nGiven that $\\angle AEB = 90^\\circ$,\n\nThus, $\\angle AEK = \\angle FEK = 45^\\circ$,\n\nTherefore, $AE = EF = \\sqrt{2}$,\n\nHence, $AF = \\sqrt{AE^2 + EF^2} = 2$, and $AK = KF = \\frac{1}{2} AF = \\frac{1}{2} \\times 2 = 1$, with $\\angle AFE = 45^\\circ$,\n\nGiven that $\\angle EFM = 90^\\circ$,\n\nThus, $\\angle MFK = 90^\\circ - \\angle EFK = 45^\\circ$,\n\nAlso, since $KM \\perp CF$,\n\nTherefore, $\\triangle MFK$ is an isosceles right triangle,\n\nHence, $MK = MF = \\frac{\\sqrt{2}}{2} KF = \\frac{\\sqrt{2}}{2}$,\n\nTherefore, in right triangle $\\triangle MGK$, $KG = \\sqrt{MK^2 + MG^2} = \\sqrt{\\left(\\frac{\\sqrt{2}}{2}\\right)^2 + \\left(\\frac{\\sqrt{2}}{2} + \\sqrt{2}\\right)^2} = \\sqrt{5}$.\n\nThus, the answer is: C.\n\n【Key Insight】This problem examines the Pythagorean theorem and the properties of a square, with the key step being the construction of a right triangle." }, { "problem_id": 300, "question": "Xiaoming made a flexible rhombus learning tool using four wooden sticks of equal length. He first adjusted the tool to form the rhombus shown in Figure (1) and measured $\\angle B = 60^\\circ$. Then he adjusted the tool to form a square as shown in Figure (2) and measured the diagonal $AC$ to be $20 \\text{ cm}$. What is the length of the diagonal $AC$ in Figure (1)?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $30 \\text{ cm}$\nB. $20 \\sqrt{2} \\text{ cm}$\nC. $20 \\text{ cm}$\nD. $10 \\sqrt{2} \\text{ cm}$", "input_image": [ "batch23-2024_06_14_34cbf1033997d47727ebg_0020_1.jpg", "batch23-2024_06_14_34cbf1033997d47727ebg_0020_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in Figures (1) and (2), connect $AC$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\nIn Figure (2), since quadrilateral $ABCD$ is a square,\n\n$\\therefore AB = BC$, and $\\angle B = 90^{\\circ}$.\n\nGiven that $AC = 20 \\mathrm{~cm}$,\n\n$\\therefore AB = BC = 10 \\sqrt{2} \\mathrm{~cm}$.\n\nIn Figure (1), since $\\angle B = 60^{\\circ}$ and $BA = BC$,\n\n$\\therefore \\triangle ABC$ is an equilateral triangle,\n\n$\\therefore AC = BC = 10 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This problem primarily combines the determination of an equilateral triangle and the properties of an isosceles triangle to examine calculations related to a rhombus. Mastering the relevant property theorems is crucial for solving this problem." }, { "problem_id": 301, "question": "Given that the length $B C = a$ (where $10 D N$) until a rhombus is formed after the $\\mathrm{n}$-th operation, at which point the operation stops. When $n=3$, the value of $\\mathrm{a}$ is ( )\n\n\n\nFirst Operation\n\nFigure (1)\n\n\n\nSecond Operation\n\nFigure (2)\nA. $\\frac{50}{3}$\nB. 15\nC. $\\frac{40}{3}$\nD. $\\frac{40}{3}$ or $\\frac{50}{3}$", "input_image": [ "batch23-2024_06_14_34cbf1033997d47727ebg_0043_1.jpg", "batch23-2024_06_14_34cbf1033997d47727ebg_0043_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the first operation, we have \\( AF = 10 \\), thus \\( EC = FD = a - 10 \\);\n\nFrom the second operation, we have \\( EM = EC = a - 10 \\), thus \\( MF = 10 - (a - 10) = 20 - a \\);\n\nAfter the third operation, the remaining quadrilateral is a rhombus, so \\( a - 10 - (20 - a) = 20 - a \\), solving this gives \\( a = \\frac{50}{3} \\).\n\nTherefore, the answer is: A.\n\n[Key Insight] This problem mainly examines the folding issues of special quadrilaterals. Mastering the properties of folding, parallelograms, and rhombuses is key to solving the problem.\n\nError Analysis: Medium difficulty. Reasons for incorrect answers: (1) Inability to proficiently grasp and apply the properties of folding; (2) Lack of familiarity with the properties of parallelograms and rhombuses." }, { "problem_id": 302, "question": "To prove that the proposition \"a quadrilateral with equal diagonals is a rectangle\" is a false statement, which of the following figures, as shown, can be used as a counterexample? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch23-2024_06_14_3747c890d3421d4291e7g_0073_1.jpg", "batch23-2024_06_14_3747c890d3421d4291e7g_0073_2.jpg", "batch23-2024_06_14_3747c890d3421d4291e7g_0073_3.jpg", "batch23-2024_06_14_3747c890d3421d4291e7g_0073_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: As shown in Figure $A$, a quadrilateral with equal diagonals is not necessarily a rectangle; it could also be an isosceles trapezoid. Therefore, the correct choice is: A.\n\n[Key Insight] This question tests the ability to judge the validity of propositions. Solving it based on the relevant knowledge points is crucial." }, { "problem_id": 303, "question": "As shown in the figure, in $\\triangle A B C$, $\\angle B A C = 90^\\circ$, $A B = A C$, and $A M$ is the median of $\\triangle A B C$. Point $D$ is on side $B C$ (point $D$ does not coincide with points $B$ or $C$), and line segment $A D$ is drawn. Line segment $A F$ is drawn perpendicular to $A D$ at point $A$, and $F A = D A$. Line segment $B F$ intersects $A M$ at point $E$. If $B D = x$ and $M E = y$, then the approximate graph of $y$ versus $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch23-2024_06_14_3aea59197e45be81d8deg_0087_1.jpg", "batch23-2024_06_14_3aea59197e45be81d8deg_0087_2.jpg", "batch23-2024_06_14_3aea59197e45be81d8deg_0087_3.jpg", "batch23-2024_06_14_3aea59197e45be81d8deg_0087_4.jpg", "batch23-2024_06_14_3aea59197e45be81d8deg_0087_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, connect $CF$,\n\n\n\n$\\because$ In $\\triangle ABC$, $\\angle BAC=90^{\\circ}$, $AB=AC$,\n$\\therefore \\angle ABC=\\angle ACB=45^{\\circ}$.\n\nFrom the problem statement, $\\angle BAD + \\angle DAC = 90^{\\circ}$, $\\angle FAC + \\angle DAC = 90^{\\circ}$,\n\n$\\therefore \\angle BAD = \\angle FAC$.\n\n$\\because AB = AC$, $AD = AF$,\n\n$\\therefore \\triangle BAD \\cong \\triangle CAF$,\n\n$\\therefore BD = CF$, $\\angle ABD = \\angle ACF = \\angle ACB = 45^{\\circ}$,\n\n$\\therefore \\angle BCF = 90^{\\circ}$.\n\n$\\because AB = AC$, $AM$ is the median of $\\triangle ABC$,\n\n$\\therefore \\angle AMC = 90^{\\circ}$,\n\n$\\therefore EM \\parallel CF$,\n\n$\\therefore EM$ is the midline of $\\triangle BCF$,\n\n$\\therefore y = EM = \\frac{1}{2} CF = \\frac{1}{2} BD = \\frac{1}{2} x$,\n\n$\\therefore$ The approximate graph is as shown in option $A$.\n\nTherefore, the answer is: $A$.\n\n【Key Insight】This problem mainly tests the properties and determination of congruent triangles, the determination of parallel lines, and the properties of the midline of a triangle. Mastering these concepts is crucial for solving such problems." }, { "problem_id": 304, "question": "Fold the rectangle $\\mathrm{ABCD}$ in Figure (1) along $\\mathrm{EF}$ to obtain Figure (2). After folding, $\\mathrm{DE}$ and $\\mathrm{BF}$ intersect at point $\\mathrm{P}$. If $\\angle \\mathrm{BPE}=130^{\\circ}$, then the measure of $\\angle \\mathrm{PFE}$ is ( )\n\n\n\nFigure (1)\n\n\nFigure (2)\nA. $60^{\\circ}$\nB. $65^{\\circ}$\nC. $70^{\\circ}$\nD. $75^{\\circ}$", "input_image": [ "batch23-2024_06_14_4b3d919410ee114a92acg_0011_1.jpg", "batch23-2024_06_14_4b3d919410ee114a92acg_0011_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since rectangle $\\mathrm{ABCD}$ has opposite sides $\\mathrm{AD} / / \\mathrm{BC}$,\n\n$\\therefore \\angle \\mathrm{AEP}=180^{\\circ}-\\angle \\mathrm{BPE}=180^{\\circ}-130^{\\circ}=50^{\\circ}$,\n\nBy the property of folding, $\\angle \\mathrm{PEF}=\\frac{1}{2}\\left(180^{\\circ}-\\angle \\mathrm{AEP}\\right)=\\frac{1}{2}\\left(180^{\\circ}-50^{\\circ}\\right)=65^{\\circ}$\n\n$\\therefore \\angle \\mathrm{AEF}=\\angle \\mathrm{AEP}+\\angle \\mathrm{PEF}=115^{\\circ}$\n\n$\\therefore \\angle \\mathrm{PFE}=180^{\\circ}-\\angle \\mathrm{AEF}=180^{\\circ}-115^{\\circ}=65^{\\circ}$\n\nTherefore, choose B.\n\n【Key Insight】This question examines the properties of parallel lines and the transformation of folding. The corresponding angles before and after folding are equal. Remembering this property is key to solving the problem." }, { "problem_id": 305, "question": "As shown in Figure (1), this is a set of Tangram, where the side length of the smallest square is 1. Six pieces are taken to form the shape as shown in Figure (2). A rectangle is constructed along the outer boundary of the shape (the dotted lines). What is the area of this rectangle?\n\n\n\n(1)\n\n\n\n(2)\nA. 35\nB. 35.5\nC. $35 \\sqrt{2}$\nD. $36 \\sqrt{2}$", "input_image": [ "batch23-2024_06_14_5015ba9179d98b9027e0g_0089_1.jpg", "batch23-2024_06_14_5015ba9179d98b9027e0g_0089_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The first step is to calculate the lengths of the corresponding sides of each shape in the original tangram from Figure (1), as shown below:\n\n\n\nThe second step is to determine the lengths of the sides corresponding to the parallelogram in the tangram, as illustrated below:\n\n\n\nThe third step involves calculating the length and width of the rectangle in Figure (2), as shown below:\n\n$\\therefore$ The length of the rectangle is $1+2+\\frac{\\sqrt{2}}{2}+2+1=6+\\frac{2}{2}$, and the width is $1+2-\\frac{\\sqrt{2}}{2}+2+1=6-\\frac{\\sqrt{2}}{2}$.\n\n$\\therefore$ The area of the rectangle is $\\left(6+\\frac{\\sqrt{2}}{2}\\right) \\times\\left(6-\\frac{\\sqrt{2}}{2}\\right)=36-\\frac{1}{2}=35.5$.\n\nTherefore, the correct choice is B.\n\n\n【Key Insight】This problem primarily examines the properties of rectangles, the Pythagorean theorem, and the assembly of shapes in a tangram. Accurately determining the lengths of the sides in each shape is crucial for solving the problem." }, { "problem_id": 306, "question": "Divide the rhombus in Figure 1 into four congruent right-angled triangles along its diagonals, and then assemble these four right-angled triangles into squares as shown in Figure 2 and Figure 3. The area of the rhombus in Figure 1 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. 12\nB. 14\nC. 16\nD. 18", "input_image": [ "batch23-2024_06_14_54c3b9b2253259a41820g_0022_1.jpg", "batch23-2024_06_14_54c3b9b2253259a41820g_0022_2.jpg", "batch23-2024_06_14_54c3b9b2253259a41820g_0022_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Let the two perpendicular sides of the small right triangle in Figures 2 and 3 be $a$ and $b$ respectively. From the data in the figures, we have:\n\n\\[\n\\left\\{\\begin{array}{l}a+b=5 \\\\ b-a=1\\end{array}\\right.\n\\]\n\nSolving the system, we find:\n\n\\[\n\\left\\{\\begin{array}{l}a=2 \\\\ b=3\\end{array}\\right.\n\\]\n\nTherefore, the diagonals of the rhombus are 4 and 6, respectively. Hence, the area of the rhombus is:\n\n\\[\n\\frac{1}{2} \\times 4 \\times 6 = 12.\n\\]\n\nThe correct choice is: A.\n\n[Key Insight] This problem tests the area formula of a rhombus and the method of solving a system of linear equations. Mastering the area formula of a rhombus is crucial for solving this problem." }, { "problem_id": 307, "question": "Cut the rhombus in Figure 1 along its diagonal into four congruent right-angled triangles, and assemble these four right-angled triangles into the squares shown in Figure 2 and Figure 3. The area of the rhombus in Figure 1 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. 10\nB. 12\nC. 24\nD. 25", "input_image": [ "batch23-2024_06_14_54c3b9b2253259a41820g_0032_1.jpg", "batch23-2024_06_14_54c3b9b2253259a41820g_0032_2.jpg", "batch23-2024_06_14_54c3b9b2253259a41820g_0032_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since quadrilateral $\\mathrm{ABCD}$ is a rhombus,\n\n$\\therefore \\mathrm{OA}=\\mathrm{OC}, \\mathrm{OB}+\\mathrm{OD}, \\mathrm{AC} \\perp \\mathrm{BD}$,\n\nLet $\\mathrm{OA}=\\mathrm{x}, \\mathrm{OB}=\\mathrm{y}$,\n\nFrom the problem statement, we have: $\\left\\{\\begin{array}{l}x+y=5 \\\\ x-y=1\\end{array}\\right.$, solving gives: $\\left\\{\\begin{array}{l}x=3 \\\\ y=2\\end{array}\\right.$,\n\n$\\therefore \\mathrm{AC}=2 \\mathrm{OA}=6, \\quad \\mathrm{BD}=2 \\mathrm{OB}=4$,\n\n$\\therefore$ The area of the rhombus $=\\frac{1}{2} A C \\times B D=\\frac{1}{2} \\times 4 \\times 6=12$.\n\nTherefore, choose B.\n\n\n\nUnderstanding the properties, that the diagonals of a rhombus bisect each other perpendicularly and the formula for the area of a rhombus, is key to solving the problem." }, { "problem_id": 308, "question": "For a graph with a fixed position, if all its points are inside or on the edges of a horizontally placed rectangle, and the graph has at least one common point with each side of the rectangle (as shown in Figure 1), then the length of the horizontal side of the rectangle is called the width of the graph, and the length of the vertical side is called the height. As shown in Figure 2, given a rhombus $A B C D$ with side length 1, where side $A B$ is horizontally placed, if the height of the rhombus is $\\frac{2}{3}$ of its width, then the width of the rhombus is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{18}{13}$\nB. $\\frac{13}{9}$\nC. $\\frac{3}{2}$\nD. 2", "input_image": [ "batch23-2024_06_14_586e379e448e55a909b0g_0088_1.jpg", "batch23-2024_06_14_586e379e448e55a909b0g_0088_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Construct a rectangle $E A F C$ on the rhombus as shown in the figure.\n\nLet $A F = x$, then $C F = \\frac{2}{3} x$.\n\nIn the right triangle $\\triangle C B F$, $C B = 1$, and $B F = x - 1$.\n\nBy the Pythagorean theorem: $B C^{2} = B F^{2} + C F^{2}$,\n\n$1^{2} = (x - 1)^{2} + \\left(\\frac{2}{3} x\\right)^{2}$,\n\nSolving gives: $x = \\frac{18}{13}$ or 0 (discarded).\n\nThus, the width of the rhombus is $\\frac{18}{13}$,\n\nTherefore, the correct choice is A.\n\n\n\nFigure 2\n\n[Key Insight] This problem examines new definitions, the properties of rectangles and rhombuses, and the Pythagorean theorem. Understanding the width and height of the rectangle in the new definition is crucial." }, { "problem_id": 309, "question": "Tangram is an ancient Chinese traditional intellectual toy, known as the 'Oriental Magic Board'. A student used the relationship between the side lengths of each board in the tangram (as shown in Figure 1) to form a convex hexagon (as shown in Figure 2). The perimeter of the convex hexagon is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $120 \\mathrm{~cm}$\nB. $120 \\sqrt{2} \\mathrm{~cm}$\nC. $(40 \\sqrt{2}+80) \\mathrm{cm}$\nD. $(80 \\sqrt{2}+40) \\mathrm{cm}$\n\n", "input_image": [ "batch23-2024_06_14_6017bdeaa8de696f18f7g_0035_1.jpg", "batch23-2024_06_14_6017bdeaa8de696f18f7g_0035_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Since the quadrilateral is a square,\n\nTherefore, (1) and (7) have right-angle sides each measuring \\(20 \\sqrt{2} \\mathrm{~cm}\\),\n\n(2) has a short side of \\(10 \\sqrt{2} \\mathrm{~cm}\\),\n\n(3) has a right-angle side of \\(10 \\sqrt{2} \\mathrm{~cm}\\),\n\n(4) has sides measuring \\(10 \\sqrt{2} \\mathrm{~cm}\\),\n\n(5) has a right-angle side of \\(20 \\mathrm{~cm}\\),\n\n(6) has a hypotenuse of \\(20 \\mathrm{~cm}\\).\n\nThus, the perimeter of the convex hexagon is\n\n\\[20 \\sqrt{2} + 20 \\sqrt{2} + 20 + 10 \\sqrt{2} + 10 \\sqrt{2} + 10 \\sqrt{2} + 10 \\sqrt{2} + 20 = (80 \\sqrt{2} + 40) \\mathrm{m}\\]\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This problem primarily examines the properties of a square and the Pythagorean theorem. The key to solving it lies in determining the lengths of each side." }, { "problem_id": 310, "question": "In a square paper strip, $\\angle D E F = 25^\\circ$. Fold the paper strip along $E F$ to form Figure 2, and then fold it again along $B F$ to form Figure 3. In Figure 3,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $105^\\circ$\nB. $120^\\circ$\nC. $130^\\circ$\nD. $145^\\circ$", "input_image": [ "batch23-2024_06_14_6557259aaef3edc96b25g_0089_1.jpg", "batch23-2024_06_14_6557259aaef3edc96b25g_0089_2.jpg", "batch23-2024_06_14_6557259aaef3edc96b25g_0089_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a rectangle,\n\ntherefore \\( AD \\parallel BC \\),\n\nhence \\( \\angle BFE = \\angle DEF = 25^\\circ \\).\n\nFrom the properties of folding, in Figure 2, \\( \\angle EFC = 180^\\circ - \\angle BFE = 155^\\circ \\), and \\( \\angle BFC = \\angle EFC - \\angle BFE = 130^\\circ \\),\n\ntherefore in Figure 3, \\( \\angle CFE = \\angle BFC - \\angle BFE = 105^\\circ \\).\n\nThus, the correct choice is: A.\n\n[Insight] This question examines the transformation of folding and the properties of a rectangle. The key to solving the problem is to find that \\( \\angle CFE = 180^\\circ - 3 \\angle BFE \\). This question is of basic level and not difficult. The key to solving such problems is to identify the equal side and angle relationships based on the folding transformation." }, { "problem_id": 311, "question": "Using the three paper shapes shown in Figure 1 to form the rectangle in Figure 2, an algebraic factorization can be derived. Which of the following is correct?\n\nFigure 1\n\n\nFigure 2\nA. $3 a^{2}+3 a b+b^{2}=(a+b)(b+3 a)$\nB. $3 a^{2}-3 a b+b^{2}=(a-b)(3 a+b)$\n\nC. $3 a^{2}+4 a b+b^{2}=(a+b)(3 a+b) \\quad$ D. $a^{2}+4 a b+3 b^{2}=(a+b)(3 a+b)$", "input_image": [ "batch23-2024_06_14_691dfcfe51d7ef22c152g_0006_1.jpg", "batch23-2024_06_14_691dfcfe51d7ef22c152g_0006_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the given diagram, Figure 2 is composed of 3 squares with an area of \\( a^{2} \\), 4 rectangles with an area of \\( ab \\), and 1 square with an area of \\( b^{2} \\) from Figure 1.\n\nThus, we have:\n\\[ 3a^{2} + 4ab + b^{2} = (a + b)(3a + b), \\]\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the application of factorization, the area of squares, and the area of rectangles. Familiarity with these properties is crucial for solving the problem." }, { "problem_id": 312, "question": "As shown in Figure (1), four rectangles with lengths $\\mathrm{a}$ and widths $\\mathrm{b}$ are arranged as shown in Figure (2), and $\\mathrm{a}=3 \\mathrm{~b}$. Which of the following equations cannot be derived from this figure?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $(a+b)^{2}=4 a b+(a-b)^{2}$\nB. $4 b^{2}+4 a b=(a+b)^{2}$\nC. $(a-b)^{2}=16 b^{2}-4 a b$\nD. $(a-b)^{2}+12 a^{2}=(a+b)^{2}$", "input_image": [ "batch23-2024_06_14_691dfcfe51d7ef22c152g_0021_1.jpg", "batch23-2024_06_14_691dfcfe51d7ef22c152g_0021_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "In Figure (2), the large square has a side length of $(a+b)$ and an area of $(a+b)^{2}$. The small square in the center has a side length of $(a-b)$ and an area of $(a-b)^{2}$.\n\nAccording to the problem, the area of the large square equals the sum of the areas of the four small rectangles and the small square, i.e., $(a+b)^{2} = 4ab + (a-b)^{2}$. Therefore, option A is correct.\n\nGiven that $a = 3b$,\n\nThe area of the small square can be expressed as $4b^{2}$. Thus, the sum of the areas of the four small rectangles and the small square equals the area of the large square, which can be expressed as $4b^{2} + 4ab = (a+b)^{2}$. Therefore, option B is correct.\n\nThe area of the large square can also be expressed as $16b^{2}$. Hence, the area of the large square minus the areas of the four small rectangles equals the area of the small square, which can be expressed as $(a-b)^{2} = 16b^{2} - 4ab$. Therefore, option C is correct.\n\nOnly option D cannot be verified,\n\nThus, the correct choice is: D.\n\n[Key Insight] This problem examines the properties and applications of equations and squares. Deriving algebraic expressions from the given figure is crucial for solving the problem." }, { "problem_id": 313, "question": "In the following figures, each small square grid has a side length of 1. Which figure has a shaded area of $\\frac{5}{2}$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch23-2024_06_14_691dfcfe51d7ef22c152g_0053_1.jpg", "batch23-2024_06_14_691dfcfe51d7ef22c152g_0053_2.jpg", "batch23-2024_06_14_691dfcfe51d7ef22c152g_0053_3.jpg", "batch23-2024_06_14_691dfcfe51d7ef22c152g_0053_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: A. The area of the shaded part is 2, incorrect;\n\nB. The area of the shaded part is 2, incorrect;\n\nC. The area of the shaded part is 2, incorrect;\n\nD. The area of the shaded part is 2.5, correct:\n\nTherefore, choose D.\n[Key Point] This question tests the ability to observe figures. The shaded area can be divided into a triangle and a square to calculate its area." }, { "problem_id": 314, "question": "As shown in Figure 1, cut the rectangular paper $\\mathrm{ABCD}$ and the square paper $\\mathrm{EFGH}$ along their diagonals $\\mathrm{AC}, \\mathrm{EG}$, respectively, and piece them together to form aLMN as shown in Figure 2. If the blank quadrilateral $\\mathrm{OPQR}$ in the middle is a square, and the area of ​​$\\square$ ALMN is 50, then the area of ​​the square $\\mathrm{EFGH}$ is ( )\n\n\nFigure 1\n\n\nFigure 2\nA. 24\nB. 25\nC. 26\nD. 27", "input_image": [ "batch23-2024_06_14_691dfcfe51d7ef22c152g_0084_1.jpg", "batch23-2024_06_14_691dfcfe51d7ef22c152g_0084_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let $\\mathrm{EF}=a$, $\\mathrm{BC}=b$, $\\mathrm{AB}=c$,\n\nthen $\\mathrm{PQ}=a-c$, $\\mathrm{RQ}=b-a$, and $\\mathrm{PQ}=\\mathrm{RQ}$.\n\n$\\therefore a=\\frac{b+c}{2}$,\n\n$\\because$ the area of ALMN is 50,\n\n$\\therefore \\mathrm{bc}+\\mathrm{a}^{2}+(\\mathrm{a}-\\mathrm{c})^{2}=50$,\n\nSubstituting $a=\\frac{b+c}{2}$ and simplifying yields $b+c=10$,\n\n$\\therefore a=5$,\n\n$\\therefore$ the side length of square $\\mathrm{EFGH}$ is 5,\n\n$\\therefore$ the area of square EFGH is 25,\n\nHence, the answer is: B.\n\n[Key Insight] This problem emphasizes the understanding of the properties of squares and rectangles. Mastering these properties is crucial for solving the problem." }, { "problem_id": 315, "question": "Cutting the rhombus in Figure 1 along its diagonals into four triangles, they are assembled without overlap to form the square as shown in Figure 2. If the area of the small square in the middle after assembly is 2, the difference between the longer diagonal and the shorter diagonal of the rhombus is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 2\nB. $\\sqrt{2}$\nC. $2 \\sqrt{2}$\nD. $2 \\sqrt{3}$", "input_image": [ "batch23-2024_06_14_6ca39002b427d61c8718g_0015_1.jpg", "batch23-2024_06_14_6ca39002b427d61c8718g_0015_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the properties of a rhombus, we know that the rhombus in Figure 1, when cut along its diagonals, divides into four congruent right-angled triangles. The longer leg of each right-angled triangle is equal to half the length of the longer diagonal of the rhombus, and the shorter leg is equal to half the length of the shorter diagonal of the rhombus.\n\nSince the area of the small square formed in the center is 2,\n\nThe side length of the small square formed in the center is $\\sqrt{2}$,\n\nTherefore, the difference between the longer and shorter legs of the right-angled triangle is $\\sqrt{2}$,\n\nHence, the difference between the longer and shorter diagonals of the rhombus is $2 \\sqrt{2}$,\n\nThus, the correct choice is: C.\n\n[Key Insight] This question primarily tests the properties of a rhombus. A thorough understanding of these properties is crucial for solving the problem." }, { "problem_id": 316, "question": "Given that $AB = AD$, construct a rhombus $ABCD$ using only a straightedge without markings and a compass. Which of the following methods is correct ( )?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch23-2024_06_14_6ca39002b427d61c8718g_0023_1.jpg", "batch23-2024_06_14_6ca39002b427d61c8718g_0023_2.jpg", "batch23-2024_06_14_6ca39002b427d61c8718g_0023_3.jpg", "batch23-2024_06_14_6ca39002b427d61c8718g_0023_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: From the construction, it can be seen that in option $\\mathrm{C}$, quadrilateral $\\mathrm{ABCD}$ is a rhombus (the reason is that its diagonals bisect each other and are perpendicular).\n\nTherefore, the answer is: C.\n\n[Highlight] This question primarily examines the knowledge points of ruler and compass constructions and the criterion for determining a rhombus. The key to solving this problem lies in the flexible application of the properties of a rhombus." }, { "problem_id": 317, "question": "Master Zhang processed 4 diamond-shaped parts at the request of the customer. Before delivering them to the customer, the 4 parts need to be tested. According to the test results of the parts, the parts that may fail in the picture are ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch23-2024_06_14_6ca39002b427d61c8718g_0066_1.jpg", "batch23-2024_06_14_6ca39002b427d61c8718g_0066_2.jpg", "batch23-2024_06_14_6ca39002b427d61c8718g_0066_3.jpg", "batch23-2024_06_14_6ca39002b427d61c8718g_0066_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Option A: The figure has four equal sides, hence it is a rhombus. This option fits the question's requirement.\n\nOption B: The figure has angles of $120^{\\circ}$ and $60^{\\circ}$ that add up to $180^{\\circ}$, indicating that the opposite sides are parallel. Since these opposite sides are equal and the adjacent sides of the quadrilateral are also equal, it is a rhombus. This option fits the question's requirement.\n\nOption C: The figure has one pair of parallel sides and one pair of equal sides, but this is insufficient to prove it is a rhombus. This option does not fit the question's requirement.\n\nOption D: The figure has supplementary consecutive interior angles, which means both pairs of opposite sides are parallel, and the adjacent sides are equal, making it a rhombus. This option fits the question's requirement.\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the identification of a rhombus. Mastering the relevant knowledge points is crucial for solving the problem." }, { "problem_id": 318, "question": "The Tangram is an outstanding creation of our ancestors, known as the \"Magic Board of the East.\" Xiaoming kept the (1), (2), and (3) blocks of the Tangram (as shown in Figure 1) stationary and rearranged the other blocks to form a \"hoe\" shape (without gaps or overlaps), placing it on a rectangular wooden board (as shown in Figure 2). The perimeter of this wooden board is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $24+2 \\sqrt{2} \\mathrm{~cm}$\nB. $26 \\mathrm{~cm}$\nC. $26 \\sqrt{2} \\mathrm{~cm}$\nD. $(24 \\sqrt{2}+2) \\mathrm{cm}$", "input_image": [ "batch23-2024_06_14_79228479d89c01d830dbg_0026_1.jpg", "batch23-2024_06_14_79228479d89c01d830dbg_0026_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $EN \\perp AP$ at point $N$ and $EM \\perp BF$ at point $M$.\n\nGiven that $AP = 4 \\text{ cm}$,\n\nwe have $AE = EP = 2\\sqrt{2} \\text{ cm}$.\n\nSince $EN \\perp AP$,\n\nit follows that $EN = AN = NP = 2 \\text{ cm}$.\n\nGiven that $EF = \\sqrt{2} + \\sqrt{2} + 2\\sqrt{2} = 4\\sqrt{2} \\text{ cm}$, and $EM \\perp BC$, with $\\angle FEM = \\angle EAP = 45^\\circ$,\n\nwe have $EM = FM = 4 \\text{ cm}$.\n\nGiven that $HF = \\sqrt{2} \\text{ cm}$,\n\nwe have $HC = CF = 1 \\text{ cm}$.\n\nTherefore, $BC = CF + MF + BM = CF + MF + EN = 1 + 4 + 2 = 7 \\text{ cm}$.\n\nAnd $AB = AN + BN = AN + EM = 2 + 4 = 6 \\text{ cm}$.\n\nThus, the perimeter of the wooden board is $2(AB + BC) = 2 \\times (6 + 7) = 26 \\text{ cm}$.\n\nHence, the correct choice is B.\n\n\n\nFigure 1\n\n\n\n[Key Insight] This problem examines the properties of squares and isosceles right triangles; mastering the properties of squares and calculating the lengths of each segment is crucial to solving the problem." }, { "problem_id": 319, "question": "In Figure (1), square $A B C D$ is given, with $A C$ and $B D$ intersecting at point $O$. Point $E$ is the midpoint of $O D$. A moving point $P$ starts from point $E$ and travels along the path $E \\rightarrow O \\rightarrow B \\rightarrow A$ at a speed of 1 unit per second until it reaches point $A$. The length of segment $A P$, denoted by $y$, varies with the motion time $x$, as shown in Figure (2). Determine the length of $A B$.\n\n\n\nFigure (1)\n\n\nA. $4 \\sqrt{2}$\nB. 4\nC. $3 \\sqrt{3}$\nD. $2 \\sqrt{2}$", "input_image": [ "batch23-2024_06_14_79228479d89c01d830dbg_0082_1.jpg", "batch23-2024_06_14_79228479d89c01d830dbg_0082_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "As shown in the figure, connect $\\mathrm{AE}$.\n\nFrom the function graph, it can be seen that $A E=2 \\sqrt{5}$.\n\nLet the side length of square $\\mathrm{ABCD}$ be $4a$, then $A B = A D = 4a$.\nSince quadrilateral $\\mathrm{ABCD}$ is a square,\n\n$\\therefore O A = O D = \\frac{1}{2} B D$, $A C \\perp B D$, and $\\angle B A D = 90^{\\circ}$.\n\n$\\therefore B D = \\sqrt{A B^{2} + A D^{2}} = 4 \\sqrt{2} a$, and $O A = O D = 2 \\sqrt{2} a$.\n\nSince $E$ is the midpoint of $O D$,\n\n$\\therefore O E = \\frac{1}{2} O D = \\sqrt{2} a$.\n\nThen, in right triangle $\\triangle A O E$, by the Pythagorean theorem: $A E = \\sqrt{O A^{2} + O E^{2}} = \\sqrt{10} a$.\n\nTherefore, $\\sqrt{10} a = 2 \\sqrt{5}$.\n\nSolving for $a$, we get $a = \\sqrt{2}$.\n\nThus, $A B = 4 \\sqrt{2}$.\n\nTherefore, the correct choice is: A.\n\n\n\n【Key Insight】This problem examines the properties of squares, the Pythagorean theorem, and function graphs. The key to solving the problem is deducing $A E = 2 \\sqrt{5}$ from the function graph." }, { "problem_id": 320, "question": "A rectangular sheet of paper and a circular sheet of paper are cut as shown in the figure, each resulting in a similar theorem exception figure. $(\\mathrm{AC}=3, \\mathrm{BC}=4, \\mathrm{AB}=5)$, with squares cut out along each side. In Figure 1, the sides $\\mathrm{HI}$ and $\\mathrm{LM}$ and the points $\\mathrm{K}$ and $\\mathrm{J}$ are all on the edges of the rectangular paper. In Figure 2, the center $\\mathrm{O}$ of the circle is at the midpoint of $\\mathrm{AB}$, and points $\\mathrm{H}$ and $\\mathrm{I}$ are on the circle. The ratio of the area of the rectangular paper to the area of the circular paper is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $400: 127 \\pi$\nB. $484: 145 \\pi$\nC. $440: 137 \\pi$\nD. $88: 25 \\pi$", "input_image": [ "batch23-2024_06_14_79228479d89c01d830dbg_0096_1.jpg", "batch23-2024_06_14_79228479d89c01d830dbg_0096_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "As shown in Figure 1, draw a perpendicular line from point B to EF, intersecting at point N.\n\nGiven:\n- AC = 3, BC = 4, AB = 5\n\nTherefore:\n- AC² + BC² = AB²\n- Triangle ABC is a right-angled triangle with ∠ACB = 90°\n\nSince quadrilateral ABJK is a square:\n- AB = BJ, ∠ABJ = 90°\n- ∠BAC + ∠ABC = ∠JBN + ∠ABC = 90°, which implies ∠BAC = ∠JBN\n\nIn triangles ABC and BJN:\n- ∠ACB = ∠BNJ = 90°\n- ∠BAC = ∠JBN\n- AB = BJ\n\nTherefore, triangles ABC and BJN are congruent (AAS):\n- BN = AC = 3\n- JN = BC = 4\n\nSimilarly, it can be proven that triangles JKF and BJN are congruent:\n- FJ = BN = 3\n\nSince quadrilateral BCHI is a square:\n- CH = HI = BI = BC = 4, ∠BIH = 90°\n\nSince quadrilateral DEFG is a rectangle:\n- ∠E = 90°\n- ∠E = ∠BIE = ∠BNE = 90°\n- Therefore, quadrilateral BNEI is a rectangle\n- EN = BI = 4, IE = BN = 3\n\nSimilarly:\n- DH = CM = AC = 3\n- DE = DH + HI + IE = 3 + 4 + 3 = 10\n- EF = FJ + JN + EN = 3 + 4 + 4 = 11\n\nThus, the area of the rectangular board is DE × EF = 10 × 11 = 110.\n\nAs shown in Figure 2, draw a perpendicular line from point O to AH, intersecting at point D, and connect OH. Then, ∠ADO = ∠ACB = 90°.\n\nTherefore, OD is parallel to BC.\n\nSince point O is the midpoint of AB:\n- OD is the midline of triangle ABC\n- CD = ½ AC = 1.5, OD = ½ BC = 2\n- DH = CD + CH = 1.5 + 4 = 5.5\n\nIn right triangle ODH, by the Pythagorean theorem:\n- OH² = OD² + DH² = 2² + (5.5)² = 137/4\n\nThus, the area of the circular board is π × OH² = (137/4)π.\n\nTherefore, the ratio of the areas of the rectangular and circular boards is 110 : (137/4)π = 440 : 137π.\n\nThe correct answer is: C.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Insight】This problem examines the converse of the Pythagorean theorem, properties of squares, triangle congruence theorems, and the midline theorem. The key to solving the problem lies in constructing auxiliary lines to form congruent triangles and right-angled triangles." }, { "problem_id": 321, "question": "As shown in the figure, a rectangle paper with a length of $10 \\mathrm{~cm}$ and a width of $8 \\mathrm{~cm}$ is first folded as shown in Figure A, then folded again as shown in Figure B. Finally, it is cut along the line (dashed line) connecting the midpoints of the adjacent sides of the resulting rectangle, as shown in Figure C. The parts obtained are (1) and (2). The area of the quadrilateral obtained after expanding Figure (1) is ( )\n\n\n\nA\n\n\n\nB\n\nC\nA. $10 \\mathrm{~cm}^{2}$\nB. $20 \\mathrm{~cm}^{2}$\nC. $40 \\mathrm{~cm}^{2}$\nD. $90 \\mathrm{~cm}^{2}$", "input_image": [ "batch23-2024_06_14_8511bb2e904f2163ed85g_0041_1.jpg", "batch23-2024_06_14_8511bb2e904f2163ed85g_0041_2.jpg", "batch23-2024_06_14_8511bb2e904f2163ed85g_0041_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement:\n\nIn Figure C, the obtained rectangle has a length of $5 \\mathrm{~cm}$ and a width of $4 \\mathrm{~cm}$.\n\nSince the endpoints of the dashed line are the midpoints of two adjacent sides of the rectangle,\n\n$\\therefore AC = 4 \\mathrm{~cm}, BD = 5 \\mathrm{~cm}$.\n\nThe quadrilateral obtained by unfolding Figure (1) is the rhombus $ABCD$.\n\n$\\therefore$ The area of the rhombus shown in Figure D is: $\\frac{1}{2} \\times 4 \\times 5 = 10\\left(\\mathrm{~cm}^{2}\\right)$.\n\nTherefore, the correct choice is: A.\n\n\n\nC\n\n\n\nD\n\n【Key Insight】This problem primarily examines the properties of a rhombus and the issue of paper cutting. Determining the lengths of the rhombus's diagonals is the key to solving the problem." }, { "problem_id": 322, "question": "To show that the statement \"a quadrilateral with perpendicular diagonals is a rhombus\" is a false proposition, which of the following figures can serve as a counterexample?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch23-2024_06_14_8c9eaf828dd3b564f406g_0098_1.jpg", "batch23-2024_06_14_8c9eaf828dd3b564f406g_0098_2.jpg", "batch23-2024_06_14_8c9eaf828dd3b564f406g_0098_3.jpg", "batch23-2024_06_14_8c9eaf828dd3b564f406g_0098_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Option A is a rhombus, which satisfies both the condition and the conclusion, thus it does not match the question's requirement;\n\nOption B has a long diagonal that is the perpendicular bisector of the short diagonal, and the quadrilateral has two pairs of adjacent sides equal, but it is not a rhombus. It satisfies the premise but not the conclusion, so it matches the question's requirement;\n\nOption C is a square, which satisfies both the condition and the conclusion, thus it does not match the question's requirement;\n\nOption D is a rectangle, which satisfies neither the condition nor the conclusion, thus it does not match the question's requirement;\n\nTherefore, the correct choice is: B.\n\n[Highlight] This question mainly tests the properties of a rhombus, and mastering these properties is key to solving the problem." }, { "problem_id": 323, "question": "Figure 1 shows a popular refrigerator brand. Figure 2 is a rectangular diagram of its side view. The diagonal $A C = \\frac{3 \\sqrt{53}}{10}$ meters, and the ratio of the height $A B$ to the width $B C$ is $7: 2$. The width $B C$ of the refrigerator is ( )\n\n\n\nWinter 1\n\n\n\nFigure 2\nA. 0.5 meters\nB. 0.6 meters\nC. 0.7 meters\nD. 0.8 meters", "input_image": [ "batch23-2024_06_14_9949fed5d0b749fe1005g_0015_1.jpg", "batch23-2024_06_14_9949fed5d0b749fe1005g_0015_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a rectangle,\n\n\\(\\therefore \\angle ABC = 90^\\circ\\),\nLet \\( AB = 7a \\) meters, then \\( BC = 2a \\) meters,\n\nIn the right triangle \\( \\triangle ABC \\), by the Pythagorean theorem: \\( (7a)^2 + (2a)^2 = \\left(\\frac{3 \\sqrt{53}}{10}\\right)^2 \\),\n\nSolving gives: \\( a = \\frac{3}{10} \\) (the negative value has been discarded),\n\n\\(\\therefore BC = 2a = \\frac{3}{5} = 0.6 \\) meters,\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This problem examines the properties of a rectangle and the Pythagorean theorem. The key to solving the problem lies in deriving the equation based on the Pythagorean theorem." }, { "problem_id": 324, "question": "In triangle $ABC$, $AC = 1$, $AB = \\sqrt{3}$, and $BC = 2$. Using an unmarked straightedge and compass, find a point $D$ on side $BC$ such that $AD = BD$. Which of the following methods is incorrect?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch23-2024_06_14_9d6e7250a85cd2b7c5d0g_0017_1.jpg", "batch23-2024_06_14_9d6e7250a85cd2b7c5d0g_0017_2.jpg", "batch23-2024_06_14_9d6e7250a85cd2b7c5d0g_0017_3.jpg", "batch23-2024_06_14_9d6e7250a85cd2b7c5d0g_0017_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the problem statement, we know that \\( AC^{2} + AB^{2} = 1 + 3 = 4 = BC^{2} \\).\n\nTherefore, \\( \\triangle ABC \\) is a right-angled triangle,\n\nand \\( \\angle BAC = 90^{\\circ} \\).\n\nAccording to the construction steps, the perpendicular bisector of \\( AB \\) is drawn, which implies \\( AD = BD \\), so option A does not fit the context;\n\nAccording to the construction steps, the angle bisector of \\( \\angle BAC \\) is drawn, which implies \\( \\angle BAD = 45^{\\circ} \\), hence \\( AD \\neq BD \\), so option B fits the context;\n\nAccording to the construction steps, \\( AC = CD = 1 \\), meaning point \\( D \\) is the midpoint of \\( BC \\), which implies \\( AD = \\frac{1}{2} BC = BD \\), so option C does not fit the context;\n\nAccording to the construction steps, the perpendicular bisector of \\( BC \\) is drawn, which implies point \\( D \\) is the midpoint of \\( BC \\), hence \\( AD = \\frac{1}{2} BC = BD \\), so option D does not fit the context.\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem mainly tests the knowledge of compass-and-straightedge constructions, the properties of perpendicular bisectors, the definition of angle bisectors, and the determination of isosceles triangles. Mastering the properties and theorems is crucial for solving the problem." }, { "problem_id": 325, "question": "In triangle $\\triangle A B C$ shown in Figure (1), $\\angle A C B = 90^\\circ$, and $A C : B C = 4 : 3$. There is a square on each side of this right-angled triangle. Perform the following operation: from each small square, draw a right-angled triangle with the ratio of its legs as $4: 3$, and then construct a square with the lengths of the legs of the new right-angled triangle as its sides. Figure (2) shows the figure after 1 operation, and Figure (3) shows the figure after 2 operations. If the perimeter of the right-angled triangle in Figure (1) is 12, then the sum of the areas of all the squares in the figure after 10 operations is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA. 225\nB. 250\nC. 275\nD. 300", "input_image": [ "batch23-2024_06_14_9f7e2f0ea59c833a42e4g_0079_1.jpg", "batch23-2024_06_14_9f7e2f0ea59c833a42e4g_0079_2.jpg", "batch23-2024_06_14_9f7e2f0ea59c833a42e4g_0079_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since $\\angle ACB = 90^\\circ$ and $AC:BC = 4:3$,\n\nLet $AC = 4x$, then $BC = 3x$.\n\nAccording to the Pythagorean theorem, $AB = \\sqrt{AC^2 + BC^2} = \\sqrt{(4x)^2 + (3x)^2} = 5x$.\n\nGiven that $3x + 4x + 5x = 12$,\n\nWe find $x = 1$,\n\nThus, $AB = 5$, $BC = 3$, and $AC = 4$.\n\nThe sum of the areas of the squares in Figure (1) is: $3^2 + 4^2 + 5^2 = 25 + 25 = 2 \\times 25 = 50$.\n\nThe sum of the areas of all squares in Figure (2), i.e., after one operation, is:\n\n$3^2 + 4^2 + 3^2 + 4^2 + 5^2 = 25 + 25 + 25 = 25 + 50$.\n\nThe sum of the areas of all squares in Figure (3), i.e., after two operations, is:\n\n$3^2 + 4^2 + 3^2 + 4^2 + 3^2 + 4^2 + 5^2 = 25 + 25 + 25 + 25 = 2 \\times 25 + 50$.\n\nTherefore, the sum of the areas of all squares after $n$ operations is $25n + 50$.\n\nHence, after 10 operations, the sum of the areas of all squares is: $25 \\times 10 + 50 = 300$, so option D is correct.\n\nThe answer is: $25n + 50$.\n\n[Insight] This problem mainly examines the pattern of figures, the properties of right triangles, the Pythagorean theorem, and the properties of squares. The key to solving the problem is understanding the question and flexibly applying the learned knowledge to solve it." }, { "problem_id": 326, "question": "The Pythagorean theorem holds an irreplaceable position in plane geometry. In China's ancient mathematical text \"Zhou Bi Suan Jing,\" it is recorded as \"If the leg (the shorter side) is three, and the hypotenuse (the longer side) is four, then the side opposite the right angle (the hypotenuse) is five.\" As shown in Figure 1, a right-angled triangle $A B C$ with legs of length 1 is constructed using unit squares. The area relationship can be used to verify the Pythagorean theorem. If Figure 1 is \"embedded\" into the rectangle $L M J K$ as shown in Figure 2, what is the area of this rectangle?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 60\nB. 100\nC. 110\nD. 121", "input_image": [ "batch23-2024_06_14_a705467cdbc40a4240d2g_0063_1.jpg", "batch23-2024_06_14_a705467cdbc40a4240d2g_0063_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Extend $AB$ to intersect $KF$ at point $O$, and extend $AC$ to intersect $GM$ at point $P$, as shown in the figure. Then, quadrilateral $OALP$ is a rectangle.\n\n\n\nSince $\\angle CBF = 90^\\circ$,\n\n$\\therefore \\angle ABC + \\angle OBF = 90^\\circ$,\n\nAlso, in right triangle $\\triangle ABC$, $\\angle ABC + \\angle ACB = 90^\\circ$,\n\n$\\therefore \\angle OBF = \\angle ACB$,\n\nIn triangles $\\triangle OBF$ and $\\triangle ACB$,\n\nSince $\\left\\{\\begin{array}{c}\\angle BAC = \\angle BOF \\\\ \\angle ACB = \\angle OBF \\\\ BC = BF\\end{array}\\right.$,\n\n$\\therefore \\triangle OBF \\cong \\triangle ACB$ (by AAS),\n\n$\\therefore AC = OB$,\n\nSimilarly, $\\triangle ACB \\cong \\triangle PGC$,\n\n$\\therefore PC = AB$,\n\n$\\therefore OA = AP$,\n\n$\\therefore$ rectangle $AOLP$ is a square with side length $AO = AB + AC = 3 + 4 = 7$,\n\n$\\therefore KL = 3 + 7 = 10$, and $LM = 4 + 7 = 11$,\n\n$\\therefore$ the area of rectangle $KLMJ$ is $10 \\times 11 = 110$.\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem examines the proof of the Pythagorean theorem, the determination and properties of squares, and the determination and properties of congruent triangles. The key to solving the problem lies in constructing auxiliary lines to prove the congruence of triangles, thereby identifying the square." }, { "problem_id": 327, "question": "As shown in Figure 1, the center pattern of a tablecloth consists of several squares. Xiaoming bought a tablecloth with exactly two square patterns, as shown in Figure 2. If $A B=C E=E F=4$, and points $A, C, E, G$ are on the same straight line, then the length $A G$ of the tablecloth is $(\\quad)$.\n\n\nFigure 1\n\n\nFigure 2\nA. $2 \\sqrt{2}+8$\nB. $8 \\sqrt{2}+4$\nC. $4 \\sqrt{2}+4$\nD. $6 \\sqrt{2}+4$", "input_image": [ "batch23-2024_06_14_a86bfd246eb466e70179g_0063_1.jpg", "batch23-2024_06_14_a86bfd246eb466e70179g_0063_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "As shown in the figure, connect $AC$ and $EG$,\n\n\nSince quadrilaterals $ABCD$ and $EFGH$ are squares,\n\n$\\therefore AB = BC$, $EF = FG$,\n\nGiven that $AB = CE = EF = 4$,\n\n$\\therefore AC = \\sqrt{AB^{2} + BC^{2}} = \\sqrt{2} \\cdot AB = 4\\sqrt{2}$, $EG = \\sqrt{EF^{2} + FG^{2}} = \\sqrt{2} \\cdot EF = 4\\sqrt{2}$,\n\nSince points $A$, $C$, $E$, and $G$ lie on the same straight line,\n\n$\\therefore AG = AC + CE + EG = 4 + 8\\sqrt{2}$,\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem examines the properties of squares and the Pythagorean theorem. Understanding the problem and calculating the diagonals $AC$ and $EG$ are crucial to solving it." }, { "problem_id": 328, "question": "In quadrilateral $A B C D$ as shown in Figure 1, $A B \\parallel C D$, $\\angle B = 90^\\circ$, and $A C = A D$. Point $P$ starts from point $B$ and moves along the broken line $B-A-D-C$ at a constant speed of $a$ units per second. The graph of the area $\\mathrm{S}$ of triangle $B C P$ as a function of the time $t$ (in seconds) during the entire motion is shown in Figure 2. What is the area of quadrilateral $A B C D$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 75\nB. 80\nC. 85\nD. 90", "input_image": [ "batch23-2024_06_14_aa4c93c5d3c13dc155bfg_0021_1.jpg", "batch23-2024_06_14_aa4c93c5d3c13dc155bfg_0021_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From the function graph, it can be seen that when \\( t = 3 \\), point \\( P \\) moves to point \\( A \\); when \\( t = 8 \\), point \\( P \\) moves to point \\( D \\).\n\n\\[\n\\therefore AB = 3a, \\quad AD = (8 - 3)a = 5a.\n\\]\n\\[\n\\because AC = AD,\n\\]\n\\[\n\\therefore AC = 5a.\n\\]\n\\[\n\\because \\angle B = 90^\\circ,\n\\]\n\\[\n\\therefore BC = \\sqrt{AC^2 - AB^2} = 4a.\n\\]\n\\[\n\\because AB \\parallel CD, \\quad \\angle B = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle BCD = 90^\\circ, \\quad \\text{which means} \\quad CD \\perp BC.\n\\]\n\nAs shown in the figure, draw \\( AE \\perp CD \\) at point \\( E \\).\n\n\n\nThen quadrilateral \\( ABCE \\) is a rectangle.\n\n\\[\n\\therefore CE = AB = 3a.\n\\]\n\\[\n\\because AC = AD, \\quad AE \\perp CD,\n\\]\n\\[\n\\therefore CD = 2CE = 6a \\quad \\text{(the three lines of an isosceles triangle coincide)}.\n\\]\n\nFrom the function graph, when point \\( P \\) moves to point \\( D \\), the area of \\( \\triangle BCP \\) is 60.\n\n\\[\n\\therefore \\frac{1}{2} BC \\cdot CD = 60, \\quad \\text{which means} \\quad \\frac{1}{2} \\times 4a \\cdot 6a = 60.\n\\]\n\\[\n\\text{Solving gives} \\quad a^2 = 5.\n\\]\n\\[\n\\therefore \\text{The area of quadrilateral } ABCD \\text{ is} \\quad \\frac{AB + CD}{2} \\cdot BC = \\frac{3a + 6a}{2} \\cdot 4a = 18a^2 = 90.\n\\]\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This problem examines the properties of isosceles triangles (the three lines coincide), the determination and properties of rectangles, and the ability to extract information from function graphs. Understanding the function graph is crucial for solving the problem." }, { "problem_id": 329, "question": "As shown in Figure 1, a square paper $A B C D$ is folded so that $A B$ coincides with $C D$, with the crease being $E F$. As shown in Figure 2, after unfolding, it is folded again so that point $C$ coincides with point $E$, with the crease being $G H$. The corresponding point of point $B$ is point $M$, and $E M$ intersects $A B$ at $N$. If $A D = 8$, then the length of the crease $G H$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $4 \\sqrt{3}$\nB. $4 \\sqrt{2}$\nC. $5 \\sqrt{3}$\nD. $4 \\sqrt{5}$", "input_image": [ "batch23-2024_06_14_c461dda316471c7af465g_0011_1.jpg", "batch23-2024_06_14_c461dda316471c7af465g_0011_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $EC$, and draw $GJ \\perp CD$ at point $J$. Let $EF$ intersect $GH$ at point $Q$.\n\n\n\nSince $\\angle BCD = \\angle ABC = 90^\\circ$,\n\nquadrilateral $BCJG$ is a rectangle.\n\nTherefore, $GJ \\parallel BC$ and $GJ = BC$.\n\nFrom the given conditions: $EF \\perp BC$, and $BC = CD = EF$,\n\nit follows that $EF \\perp GJ$ and $GJ = EF$.\n\nSince points $E$ and $C$ are symmetric with respect to $GH$,\n\n$EC \\perp GH$.\n\nThus, $\\angle EQH + \\angle CEF = \\angle GQF + \\angle HGJ = 90^\\circ$.\n\nSince $\\angle EQH = \\angle GQF$,\n\nit follows that $\\angle CEF = \\angle HGJ$.\n\nIn triangles $\\triangle EFC$ and $\\triangle GJH$, we have:\n\\[\n\\begin{cases}\n\\angle CEF = \\angle HGJ \\\\\nEF = GJ \\\\\n\\angle EFC = \\angle GJH = 90^\\circ\n\\end{cases}\n\\]\nTherefore, $\\triangle EFC \\cong \\triangle GJH$ (ASA).\n\nHence, $GH = EC$.\n\nGiven that $EC = \\sqrt{8^2 + 4^2} = 4\\sqrt{5}$,\n\nit follows that $GH = 4\\sqrt{5}$.\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This problem examines the properties of flip transformations, the properties of squares, the determination and properties of rectangles, the application of the Pythagorean theorem, and the properties of congruent triangles. Mastering the properties of flip transformations is crucial for solving this problem." }, { "problem_id": 330, "question": "As shown in Figure 1, cut along the diagonals $A C$ and $E G$ of the rectangular paper $A B C D$ and the square paper $E F G H$, respectively, and assemble them as shown in Figure 2 to form the quadrilateral $K L M N$. If the blank area in the middle, quadrilateral $O P Q R$, is a square, and the area of $K L M N$ is 50, then the area of the square $E F G H$ is ( )\n\n\nFigure 1\n\n\n\nFigure 2\nA. 25\nB. 26\nC. 27\nD. 28", "input_image": [ "batch23-2024_06_14_c461dda316471c7af465g_0070_1.jpg", "batch23-2024_06_14_c461dda316471c7af465g_0070_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "As shown in the figure, let $P M = P L = N R = K R = a$, and the side length of square $O R Q P$ be $b$.\n\nAccording to the problem: $a^{2} + b^{2} + (a + b)(a - b) = 50$,\n\n$\\therefore a^{2} = 25$,\n\n$\\therefore$ the area of square $E F G H = a^{2} = 25$,\n\nTherefore, the answer is: A.\n\n\n\nFigure 2\n\n【Key Insight】This problem tests knowledge of graphic cutting, properties of rectangles, and properties of squares. The key to solving the problem is to learn to use parameters to construct equations and to apply the concept of combining numbers and shapes to solve problems. It is a challenging multiple-choice question in the high school entrance exam." }, { "problem_id": 331, "question": "Mr. Chen was asked by a client to manufacture 4 rectangular parts, each with a length of $4 \\mathrm{~cm}$ and a width of $3 \\mathrm{~cm}$. Before delivering them to the client, Mr. Chen needs to inspect the 4 parts. Based on the inspection results, which of the parts in the figure might be unqualified?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0031_1.jpg", "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0031_2.jpg", "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0031_3.jpg", "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0031_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "0", "analysis": "Option A: Both pairs of opposite sides are equal.\n\n$\\therefore$ The quadrilateral is a parallelogram.\n\n$\\therefore$ Both pairs of opposite sides are parallel.\n\n$\\because$ One of the interior angles is a right angle.\n\n$\\therefore$ The two adjacent interior angles are both right angles.\n\n$\\therefore$ The quadrilateral is a rectangle.\n\n$\\because$ The measured length is $4 \\mathrm{~cm}$ and the width is $3 \\mathrm{~cm}$.\n\n$\\therefore$ Option A fits the description.\n\nOption B: Three interior angles are right angles.\n\n$\\therefore$ All four angles are right angles, making it a rectangle.\n\n$\\because$ The measured length is $4 \\mathrm{~cm}$ and the width is $3 \\mathrm{~cm}$.\n\n$\\therefore$ Option B fits the description.\n\nOption C: Two opposite angles are right angles, but it cannot be deduced that the other two interior angles are right angles.\n\n$\\therefore$ The quadrilateral may not be a rectangle.\n\n$\\therefore$ Option C does not fit the description.\n\nOption D: Two adjacent interior angles are equal and both are right angles.\n\n$\\therefore$ The two sides measured at $4 \\mathrm{~cm}$ are parallel and equal.\n\n$\\therefore$ The quadrilateral is a rectangle.\n\n$\\because$ The measured length is $4 \\mathrm{~cm}$ and the width is $3 \\mathrm{~cm}$.\n\n$\\therefore$ Option D fits the description.\n\nTherefore, the correct choice is: C.\n\n【Insight】This question tests knowledge of rectangles, parallelograms, and parallel lines. The key to solving it lies in mastering the properties that determine a rectangle, thereby enabling the solution." }, { "problem_id": 332, "question": "Fold a rectangular sheet of paper $M N P Q$ as follows: \nStep 1: Fold a square $M N A B$ at one end of the rectangle using the method shown in Figure 1.\nStep 2: As shown in Figure 2, fold the square along $C D$ into two congruent rectangles, then unfold the paper. \nStep 3: Fold the diagonal $B D$ of the inner rectangle $A B C D$, and fold $B D$ to point $D E$ in Figure 3. \nStep 4: Unfold the paper and fold along the point $E$ to form $E F$, resulting in rectangle $A E F B$. The value of $\\frac{A E}{A B}$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\nA. $\\frac{3}{2}$\nB. $\\frac{5}{2}$\nC. $\\frac{\\sqrt{2}+1}{2}$\nD. $\\frac{\\sqrt{5}-1}{2}$", "input_image": [ "batch23-2024_06_14_d0a1b3041bfff3ed6bd7g_0006_1.jpg", "batch23-2024_06_14_d0a1b3041bfff3ed6bd7g_0006_3.jpg", "batch23-2024_06_14_d0a1b3041bfff3ed6bd7g_0006_2.jpg", "batch23-2024_06_14_d0a1b3041bfff3ed6bd7g_0006_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Folding according to Figure 1 yields the square $M N A B$.\n\nFolding according to Figure 2 gives $A D = B C = \\frac{1}{2} A B$.\n\nThen, $B D = \\sqrt{A B^{2} + A D^{2}} = \\frac{\\sqrt{5}}{2} A B$.\n\nFolding according to Figure 3, we have $D E = B D = \\frac{\\sqrt{5}}{2} A B$.\n\nThus, $A E = D E - A D = \\frac{\\sqrt{5}}{2} A B - \\frac{1}{2} A B = \\frac{\\sqrt{5} - 1}{2} A B$.\n\nTherefore, $\\frac{A E}{A B} = \\frac{\\frac{\\sqrt{5} - 1}{2} A B}{A B} = \\frac{\\sqrt{5} - 1}{2}$.\n\nHence, the correct choice is: D.\n\n【Key Insight】This problem primarily examines the properties of folding, rectangles, squares, and the Pythagorean theorem. Understanding the relationship between corresponding sides after folding is crucial for solving the problem." }, { "problem_id": 333, "question": "In square $A B C D$, the diagonals intersect at point $O$ (as shown in Figure 1). If $\\angle B O C$ rotates clockwise around point $O$, with its sides intersecting sides $A B$ and $B C$ at points $E$ and $F$ (as shown in Figure 2), respectively, and segment $E F$ is connected, then as point $E$ moves from $B$ to $A$, the midpoint $G$ of segment $E F$ traces a path that is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Line segment\nB. Circular arc\nC. Polygonal line\nD. Wavy line", "input_image": [ "batch23-2024_06_14_ddcaf5312fb4fd1f0336g_0039_1.jpg", "batch23-2024_06_14_ddcaf5312fb4fd1f0336g_0039_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Connect $O G$ and $B G$. According to the problem, we know that $\\angle E B F = \\angle E O F = 90^{\\circ}$.\n\n\n\nTherefore, $O G = B G = \\frac{1}{2} E F$.\n\nThus, point $G$ lies on the perpendicular bisector of segment $O B$.\n\nThe path that the midpoint $G$ of segment $E F$ follows is a segment of the perpendicular bisector of $O B$, which is a line segment.\n\nTherefore, the correct answer is: A.\n\n【Insight】This problem tests the determination of the perpendicular bisector of a segment, the property that the median to the hypotenuse of a right triangle is half the hypotenuse, and the properties of a square. Mastering these concepts is key to solving the problem." }, { "problem_id": 334, "question": "Given a rectangular piece of paper, with $\\angle B A F = \\angle B = \\angle C = \\angle D = \\angle F E D = \\angle F = 90^\\circ$, $A B = A F = 2$, and $E F = E D = 1$. There are two cutting and rearrangement schemes, as shown in Figure 1 and 2, where the dashed lines indicate the cuts to be made. Each scheme aims to form a square with the same area as the original paper. Which of the following is true?\n\nScheme A:\n\n\n\nFigure 1 Scheme B:\n\n\n\nFigure 2\nA. Both A and B can be done.\nB. Neither A nor B can be done.\nC. A cannot be done, but B can.\nD. A can be done, but B cannot.\n\n", "input_image": [ "batch23-2024_06_14_ddcaf5312fb4fd1f0336g_0071_1.jpg", "batch23-2024_06_14_ddcaf5312fb4fd1f0336g_0071_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, move $\\triangle A E F$ to position (1), $\\triangle D E H$ to position (2), and quadrilateral $G C H E$ to position (3), and you can obtain a square with the same area as the original;\n\n\n\nAs shown in Figure 2, move $\\triangle A B G$, $\\triangle A H G$, and $\\triangle H G F$ to positions (1), (2), and (3) respectively, and you can obtain a square with the same area as the original;\n\n\n\n$\\therefore$ Both Plan A and Plan B are feasible.\n\nTherefore, the correct choice is: A\n\n【Insight】This question examines the cutting and rearranging of shapes and the properties of squares. The key to solving this problem lies in constructing the figures based on the given information." }, { "problem_id": 335, "question": "By cutting along the dotted lines in Figure (a), two congruent triangles are obtained. These triangles are then rearranged to form the figure in Figure (b). Taking the midpoint $O$ of $B C$, connect $O A, O D$, and $A D$. If $\\angle A C B = 22.5^\\circ$ and $B C = 4$, the perimeter of $\\triangle A O D$ is ( )\n\n\n\n(a)\n\n\nA. 4\nB. $2 \\sqrt{2}$\nC. $4+\\sqrt{2}$\nD. $4+2 \\sqrt{2}$", "input_image": [ "batch23-2024_06_14_e3372b0c06ffa140c062g_0016_1.jpg", "batch23-2024_06_14_e3372b0c06ffa140c062g_0016_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, triangles $\\triangle ABC$ and $\\triangle DBC$ are congruent, and both are right-angled triangles.\n\nSince point $O$ is the midpoint of $BC$,\n\nTherefore, $OA = OD = \\frac{1}{2} BC = BO = CO = 2$,\n\nThus, triangles $\\triangle AOC$ and $\\triangle BOD$ are both isosceles triangles.\n\nGiven that $\\angle ACB = 22.5^\\circ$,\n\nTherefore, $\\angle OAC = 22.5^\\circ$,\n\nHence, $\\angle AOB = \\angle OAC + \\angle ACB = 45^\\circ$,\n\nSimilarly, we can deduce that $\\angle DOC = 45^\\circ$,\n\nTherefore, $\\angle AOD = 180^\\circ - \\angle AOB - \\angle COD = 90^\\circ$,\n\nIn the right-angled triangle $\\triangle AOD$, $AD = \\sqrt{OA^2 + OD^2} = 2\\sqrt{2}$,\n\nThus, the perimeter of $\\triangle AOD$ is $AD + OA + OD = 2\\sqrt{2} + 2 + 2 = 4 + 2\\sqrt{2}$,\n\nTherefore, the correct answer is: D.\n\n[Key Insight] This problem examines the properties of rectangles, the properties of congruent triangles, the midpoint of the hypotenuse in a right-angled triangle, and the Pythagorean theorem. The key to solving the problem lies in proving that $\\triangle AOD$ is an isosceles right-angled triangle." }, { "problem_id": 336, "question": "As shown in the figure, in Figure 1, $A_{1}, B_{1}, C_{1}$ are the midpoints of the sides $B C, C A, A B$ of the equilateral triangle $\\triangle A B C$, in Figure 2, $A_{2}, B_{2}, C_{2}$ are the midpoints of the sides $B_{1} C_{1}, A_{1} C_{1}, A_{1} B_{1}$ of $\\Delta A_{1} B_{1} C_{1}$, ..., following this pattern, the number of rhombuses in the $n$th figure is ( ) in total.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $n^{2}$\nB. $2 n$\nC. $3 n$\nD. $3 n+1$", "input_image": [ "batch23-2024_06_14_e9ac3578399f2856b1b4g_0034_1.jpg", "batch23-2024_06_14_e9ac3578399f2856b1b4g_0034_2.jpg", "batch23-2024_06_14_e9ac3578399f2856b1b4g_0034_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: According to the problem statement,\n\n$\\because A_{1}$, $B_{1}$, and $C_{1}$ are the midpoints of the sides $BC$, $CA$, and $AB$ of the equilateral triangle $\\triangle ABC$, respectively,\n\n$\\therefore A_{1}B_{1} = A_{1}C_{1} = B_{1}C_{1} = A_{1}B = A_{1}C = BC_{1} = AC_{1} = AB_{1} = B_{1}C$,\n\n$\\therefore$ Figure 1 has three rhombuses, which implies that drawing one median line divides the figure into three rhombuses,\n\n$\\therefore$ The number of rhombuses in the $n$th figure is $3n$,\n\nTherefore, the correct answer is C.\n\n[Key Insight] This problem tests the properties of equilateral triangles and the median line theorem. The key to solving it is to recognize that drawing one median line divides the figure into three rhombuses." }, { "problem_id": 337, "question": "A single thread passes through the 4 holes of a button (each hole is passed through only once), and its front view is as shown in the figure. Which of the following 4 figures could be its back view?\n\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch23-2024_06_14_f513d891339227eeea32g_0010_1.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0010_2.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0010_3.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0010_4.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0010_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Observation reveals that there will be two parallel lines on the back, and the thread ends emerge from the diagonal of the button, hence the choice: A.\n\n【Key Point】This question primarily tests the ability of pattern recognition and spatial imagination. The key is to observe and imagine the presence of two parallel lines and the position of the thread ends on the back." }, { "problem_id": 338, "question": "During a mathematics activity class, a math enthusiasts group used a square paper with an area of $100 \\mathrm{~cm}^{2}$ to create a set of Tangram, as shown in Figure 1, and then cooperatively completed the artwork as shown in Figure 2. Calculate the sum of the areas of (1) and (2) in the figure, which is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $12.5 \\mathrm{~cm}^{2}$\nB. $25 \\mathrm{~cm}^{2}$\nC. $37.5 \\mathrm{~cm}^{2}$\nD. $50 \\mathrm{~cm}^{2}$", "input_image": [ "batch23-2024_06_14_f513d891339227eeea32g_0014_1.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0014_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In the diagram, (1) and (2) represent $h$ and $e$ respectively.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nIn the above diagram, half of triangle $\\mathrm{h}$ is congruent to $\\mathrm{b}$, and from the diagram, $b + e + \\frac{1}{2} h$ exactly constitutes a quarter of the square, i.e.,\nthe combined area of (1) and (2) in the diagram is $\\frac{1}{4} \\times 100 \\mathrm{~cm}^{2} = 25 \\mathrm{~cm}^{2}$.\n\nTherefore, the answer is: B.\n\n[Key Insight] This question tests the properties of a square, the ability to formulate algebraic expressions, and the student's ability to interpret graphical information. Mastering these concepts is crucial for solving the problem." }, { "problem_id": 339, "question": "Given that $A C$ is the diagonal of the rectangle $A B C D$, which of the following diagrams definitely shows angles $\\angle 1$ and $\\angle 2$ that are not equal?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch23-2024_06_14_fa808d70992062d50263g_0045_1.jpg", "batch23-2024_06_14_fa808d70992062d50263g_0045_2.jpg", "batch23-2024_06_14_fa808d70992062d50263g_0045_3.jpg", "batch23-2024_06_14_fa808d70992062d50263g_0045_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: In option A, according to the property that vertically opposite angles are equal, $\\angle 1$ and $\\angle 2$ must be equal;\n\nIn options B and C, it is not possible to determine whether $\\angle 1$ and $\\angle 2$ are equal;\n\nIn option D, since $\\angle 1 = \\angle \\mathrm{ACD}$ and $\\angle 2 > \\angle \\mathrm{ACD}$, it follows that $\\angle 2 > \\angle 1$.\n\nTherefore, the correct choice is: D." }, { "problem_id": 340, "question": "As shown in Figure 1, in rectangle $A B C D$, $A D > \\frac{1}{2} A B$. Point $E$ is the midpoint of side $A B$, and point $P$ is a moving point on side $A B$. The line segment $D P$ is drawn. Let the length of $A P$ be $x$, and $P D + P E = y$, where the graph of $y$ as a function of $x$ is shown in Figure 2. Determine the area of rectangle $A B C D$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 15\nB. 24\nC. 35\nD. 36", "input_image": [ "batch23-2024_06_14_fa808d70992062d50263g_0047_1.jpg", "batch23-2024_06_14_fa808d70992062d50263g_0047_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Refer to Figure 2\n\nWhen \\( x = 0 \\), \\( PD + PE = AD + AE = y = 7 \\).\n\nWhen points \\( P \\) and \\( E \\) coincide, \\( PD + PE = DE = y_{\\text{min}} = 5 \\).\n\nLet \\( AD = x \\), then \\( AE = 7 - x \\).\n\nAccording to the Pythagorean theorem, \\( AD^2 + AE^2 = DE^2 \\), which gives \\( x^2 + (7 - x)^2 = 5^2 \\).\n\nSolving the equation yields: \\( x_1 = 3 \\) (discarded as it does not fit the context), \\( x_2 = 4 \\).\n\nTherefore, \\( AD = 4 \\), \\( AE = 3 \\).\n\nHence, \\( AB = 2AE = 6 \\).\n\nThus, the area of rectangle \\( ABCD \\) is \\( S_{ABCD} = AB \\cdot AD = 4 \\times 6 = 24 \\).\n\nThe correct choice is: B\n\n[Key Insight] This problem primarily tests the properties of rectangles and the Pythagorean theorem. Understanding these properties and using the relationship diagram to find the area of the rectangle is crucial for solving the problem." }, { "problem_id": 341, "question": "As shown in the figure, if $a \\| b$, which of the following options can directly use the theorem \"if two parallel lines are intersected by a transversal, the corresponding angles are equal\" to determine that $\\angle 1 = \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch24-2024_06_15_05df2c9ffbad13a48785g_0087_1.jpg", "batch24-2024_06_15_05df2c9ffbad13a48785g_0087_2.jpg", "batch24-2024_06_15_05df2c9ffbad13a48785g_0087_3.jpg", "batch24-2024_06_15_05df2c9ffbad13a48785g_0087_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: A. $\\angle 1, \\angle 2$ are corresponding angles, hence correct;\n\nB. $\\angle 1, \\angle 2$ are alternate interior angles, hence incorrect;\n\nC. $\\angle 1, \\angle 2$ are vertical angles, hence incorrect;\n\nD. $\\angle 1, \\angle 2$ are not corresponding angles, hence incorrect;\n\nTherefore, the correct choice is A.\n\n【Key Point】This question examines the definition of corresponding angles and the properties of parallel lines: (1) Corresponding angles are equal when two lines are parallel, (2) Alternate interior angles are equal when two lines are parallel, (3) Consecutive interior angles are supplementary when two lines are parallel. When applying the properties of parallel lines, it is essential to correctly identify corresponding angles, alternate interior angles, and consecutive interior angles.\n\n## 【Question ID】 2185861722267648" }, { "problem_id": 342, "question": "Given that point $M(2 m-1,1-m)$ is in the fourth quadrant, the correct representation of the range of values for $m$ on the number line is ( )\n\nA.\n\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_1ba61d31fb821579f313g_0021_1.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0021_2.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0021_3.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0021_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the point \\( M(2m - 1, 1 - m) \\) lies in the fourth quadrant,\n\n\\[\n\\begin{cases}\n2m - 1 > 0 \\\\\n1 - m < 0\n\\end{cases}\n\\]\n\nSolving these inequalities gives: \\( m > 1 \\).\n\nTherefore, the range of \\( m \\) is represented on the number line as\n\n\n\nHence, the correct choice is: B\n\n**Key Insight:** This problem primarily examines the characteristics of coordinates in different quadrants of the Cartesian plane and the solution of inequality systems. Mastery of these concepts is crucial for solving such problems." }, { "problem_id": 343, "question": "Given that point $P(a-1, a+2)$ is in the second quadrant of the Cartesian coordinate system, the range of values for $a$ on the number line can be represented as ( ) (shaded part).\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n\n\n##", "input_image": [ "batch24-2024_06_15_1ba61d31fb821579f313g_0029_1.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0029_2.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0029_3.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0029_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the point \\( P(a-1, a+2) \\) lies in the second quadrant of the Cartesian coordinate system, we have \\( a-1 < 0 \\) and \\( a+2 > 0 \\).\n\nSolving these inequalities gives \\( -2 < a < 1 \\).\n\nTherefore, the correct choice is C." }, { "problem_id": 344, "question": "Given that point $P(3-m, m-1)$ is in the fourth quadrant, the range of values for $m$ on the number line can be represented as $(\\quad)$\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch24-2024_06_15_1ba61d31fb821579f313g_0033_1.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0033_2.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0033_3.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0033_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the point $\\mathrm{P}(3-\\mathrm{m}, \\mathrm{m}-1)$ lies in the fourth quadrant, we have\n\n$\\left\\{\\begin{array}{l}3-m>0 \\\\ m-1<0\\end{array}\\right.$,\n\nSolving this, we find $\\mathrm{m}<1$ or $\\mathrm{m}<3$,\n\nThus, the solution set of the inequality system is: $\\mathrm{m}<1$,\n\nRepresented on the number line as:\n\n\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This problem tests the understanding of point coordinates and solving inequality systems. The key to solving it lies in using the fact that in the fourth quadrant, the x-coordinate is positive and the y-coordinate is negative to derive the inequality system." }, { "problem_id": 345, "question": "Given that point $P(2-a, a-3)$ is in the fourth quadrant, the correct representation of the range of values for $a$ on the number line is ( )\nA. \n\nB. \n\nC. \n\nD. \n", "input_image": [ "batch24-2024_06_15_1ba61d31fb821579f313g_0036_1.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0036_2.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0036_3.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0036_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the point \\( P(2 - a, a - 3) \\) lies in the fourth quadrant,\n\n\\[\n\\begin{cases}\n2 - a > 0 \\\\\na - 3 < 0\n\\end{cases}\n\\]\n\nSolving these inequalities gives the range of \\( a \\) as: \\( a < 2 \\).\n\nTherefore, the representation of the inequality on the number line is as shown in option C.\n\nHence, the correct choice is: C.\n\n**Key Insight:** This question tests the ability to represent the solution set of an inequality on a number line. The method for representing the solution set of an inequality on a number line is as follows: \n- For \">\", use an open circle and draw a line to the right.\n- For \"≥\", use a closed circle and draw a line to the right.\n- For \"<\", use an open circle and draw a line to the left.\n- For \"≤\", use a closed circle and draw a line to the left." }, { "problem_id": 346, "question": "If point $P(-2a+2,3-a)$ is located in the second quadrant, the correct representation of the range of values for $a$ on the number line is ( )\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_1ba61d31fb821579f313g_0057_1.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0057_2.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0057_3.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0057_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the point \\( P(-2a + 2, 3 - a) \\) lies in the second quadrant,\n\n\\[\n\\begin{cases}\n-2a + 2 < 0 \\\\\n3 - a > 0\n\\end{cases}\n\\]\n\nSolving the inequalities, we find: \\( 1 < a < 3 \\).\n\nThis solution is represented on the number line as shown in the figure:\n\n\n\nTherefore, the correct choice is: A.\n\n**Key Insight:** This problem tests the ability to solve a system of linear inequalities, represent the solution set on a number line, and understand the coordinate characteristics of points in different quadrants. Mastery of the coordinate features of the second quadrant—where the x-coordinate is negative and the y-coordinate is positive—and the skill to represent inequality solutions on a number line are crucial for solving this problem." }, { "problem_id": 347, "question": "Given that point $P(a, 2-a)$ is in the first quadrant, the correct representation of the range of values for $a$ on the number line is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_1ba61d31fb821579f313g_0065_1.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0065_2.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0065_3.jpg", "batch24-2024_06_15_1ba61d31fb821579f313g_0065_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the point \\( P(a, 2-a) \\) lies in the first quadrant,\n\n\\[\n\\begin{cases}\na > 0 \\\\\n2 - a > 0\n\\end{cases}\n\\]\n\nSolving these inequalities gives: \\( 0 < a < 2 \\).\n\nThe solution set of the inequality system is represented on the number line as shown in the figure:\n\n\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This problem tests the understanding of solving a system of linear inequalities, the coordinates of a point, and representing the solution set of inequalities on a number line. Accurate and proficient calculation is crucial for solving the problem." }, { "problem_id": 348, "question": "Among the following figures, which one can conclude that $AB \\parallel CD$ given $\\angle 1 = \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_5361d9983feca5d1bdc3g_0041_1.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0041_2.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0041_3.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0041_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: \n\nA. Since $\\angle 1 = \\angle 2$ cannot determine that $AB \\parallel CD$, this option does not meet the requirement;\n\nB. Since $\\angle 1 = \\angle 2$ cannot determine that $AB \\parallel CD$, this option does not meet the requirement;\n\nC. Because $\\angle 1 = \\angle 2$, it follows that $AC \\parallel BD$, so this option does not meet the requirement;\n\nD. Because $\\angle 1 = \\angle 2$, it follows that $AB \\parallel CD$, so this option meets the requirement.\nTherefore, the correct choice is D.\n\n[Key Insight] This question tests the determination of parallel lines. Being familiar with the theorem for determining parallel lines is crucial for answering this question correctly." }, { "problem_id": 349, "question": "Among the following options, which one does not allow us to conclude that $A B \\parallel C D$ given $\\angle 1 = \\angle 2$?\nA.\n\nB.\n\nC.\n\nD.\n\n\n##", "input_image": [ "batch24-2024_06_15_5361d9983feca5d1bdc3g_0050_1.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0050_2.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0050_3.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0050_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: \n\nA. Based on the positional relationship of angles $\\angle 1$ and $\\angle 2$, satisfying \"corresponding angles are equal, the two lines are parallel\" can deduce $AB // CD$, which does not meet the requirement of the question.\n\nB. Based on the positional relationship of angles $\\angle 1$ and $\\angle 2$, satisfying \"alternate interior angles are equal, the two lines are parallel\" can deduce $AB // CD$, which does not meet the requirement of the question.\n\nC. Based on the positional relationship of angles $\\angle 1$ and $\\angle 2$, it can be transformed to \"corresponding angles are equal, the two lines are parallel\" which can deduce $AB // CD$, which does not meet the requirement of the question.\n\nD. Based on the positional relationship of angles $\\angle 1$ and $\\angle 2$, it is known that they are vertical angles, which cannot deduce $AB // CD$, thus meeting the requirement of the question. Therefore, the answer is: $D$\n\n[Highlight] This question examines the methods of determining parallel lines: \"if consecutive interior angles are supplementary, the two lines are parallel\", \"if alternate interior angles are equal, the two lines are parallel\", and \"if corresponding angles are equal, the two lines are parallel\". Mastering these methods is key to solving this problem." }, { "problem_id": 350, "question": "Which of the following compass and straightedge constructions for drawing a line parallel to a straight line $l$ through a point $P$ outside $l$ is incorrect?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch24-2024_06_15_5361d9983feca5d1bdc3g_0086_1.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0086_2.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0086_3.jpg", "batch24-2024_06_15_5361d9983feca5d1bdc3g_0086_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "A. This option constructs an angle bisector and an isosceles triangle, which results in a pair of alternate interior angles being equal, thereby proving that the two lines are parallel. Therefore, this option does not meet the requirement of the question;\n\nB. This option constructs an angle equal to a given angle. According to the principle that corresponding angles being equal proves that two lines are parallel, it can be demonstrated that the two lines are parallel. Therefore, this option does not meet the requirement of the question;\n\nC. This option only ensures that two segments are equal in length, which does not guarantee that the two lines are parallel. Therefore, this option meets the requirement of the question;\n\nD. This option constructs an angle equal to a given angle. According to the principle that alternate interior angles being equal proves that two lines are parallel, it can be demonstrated that the two lines are parallel. Therefore, this option does not meet the requirement of the question;\n\nHence, the correct choice is: C.\n\n[Key Insight] This question tests the understanding of compass-and-straightedge constructions and the theorem for determining parallel lines. Mastering the operations of compass-and-straightedge constructions is crucial for solving such problems." }, { "problem_id": 351, "question": "In right triangle $\\triangle A B C$ as shown in Figure 1, $\\angle A C B = 90^\\circ, A C = 1, B C = 2$. Place $\\triangle A B C$ in the Cartesian coordinate system such that point $A$ coincides with the origin and point $C$ is on the positive $x$-axis. Roll $\\triangle A B C$ clockwise as shown in Figure 2 (with no slipping), then after rolling 2021 times, the $x$-coordinate of point $B$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $2020+673 \\sqrt{5}$\nB. $2020+674 \\sqrt{5}$\nC. $2022+673 \\sqrt{5}$\nD. $2022+674 \\sqrt{5}$", "input_image": [ "batch24-2024_06_15_6004c0a91f31a32ead1fg_0005_1.jpg", "batch24-2024_06_15_6004c0a91f31a32ead1fg_0005_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Since $\\angle ACB = 90^\\circ$, $AC = 1$, and $BC = 2$,\n\nthen $AB = \\sqrt{AC^2 + BC^2} = \\sqrt{1^2 + 2^2} = \\sqrt{5}$.\n\nTherefore, the perimeter of the triangle is $1 + 2 + \\sqrt{5} = 3 + \\sqrt{5}$.\n\nGiven that the triangle's rolling pattern repeats every 3 cycles, and $2021 \\div 3 = 673 \\ldots 2$,\n\nafter rolling 2021 times, the horizontal coordinate of point $B$ is $1 + 2 + 673(3 + \\sqrt{5}) = 2022 + 673\\sqrt{5}$.\n\nHence, the correct choice is $C$.\n\n【Key Insight】This problem tests the understanding of numerical patterns in geometric figures. Correctly calculating the cycle length is crucial for solving the problem." }, { "problem_id": 352, "question": "An example that demonstrates that the statement \"equal angles are opposite angles\" is false is ( )\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n", "input_image": [ "batch24-2024_06_15_62e79cb351f5324a756dg_0001_1.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0001_2.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0001_3.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0001_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "In option A, both angles are $30^{\\circ}$, but they are not vertical angles, thus meeting the condition of the question;\n\nIn option B, both angles are $30^{\\circ}$ and they are vertical angles, thus not meeting the condition of the question;\n\nIn options C and D, the two angles are not equal, thus not meeting the condition of the question.\n\nTherefore, the correct choice is: A.\n\n[Highlight] This question tests the judgment of the truth of propositions. A correct proposition is called a true proposition, and an incorrect one is called a false proposition. The key to judging the truth of a proposition lies in being familiar with the properties and theorems in the textbook." }, { "problem_id": 353, "question": "Among the following figures, which one can conclude that $AB // CD$ given that $\\angle 1 = \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_62e79cb351f5324a756dg_0028_1.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0028_2.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0028_3.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0028_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. From $\\angle 1 = \\angle 2$, we cannot conclude that $AB \\parallel CD$. This option does not meet the requirement of the question.\n\nB. From $\\angle 1 = \\angle 2$, we can conclude that $AC \\parallel BD$, but this does not imply that $AB \\parallel CD$. This option does not meet the requirement of the question.\n\nC. From $\\angle 1 = \\angle 2$, we cannot conclude that $AB \\parallel CD$. This option does not meet the requirement of the question.\n\nD. From $\\angle 1 = \\angle 2$, we can conclude that $AB \\parallel CD$. This option meets the requirement of the question.\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question tests the determination of parallel lines. Mastering the relevant theorems is crucial for solving the problem." }, { "problem_id": 354, "question": "Given that $\\angle 1 = \\angle 2$, which of the following figures can conclude that $A B \\parallel C D$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_62e79cb351f5324a756dg_0096_1.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0096_2.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0096_3.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0096_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. $\\angle 1$ and $\\angle 2$ are consecutive interior angles. The fact that $\\angle 1 = \\angle 2$ does not determine that $\\mathrm{AB} \\parallel \\mathrm{CD}$, so this is incorrect.\n\nB. The vertical angle of $\\angle 1$ and $\\angle 2$ are corresponding angles. The fact that $\\angle 1 = \\angle 2$ can determine that $\\mathrm{AB} \\parallel \\mathrm{CD}$, so this is correct.\n\nC. $\\angle 1$ and $\\angle 2$ are alternate interior angles. The fact that $\\angle 1 = \\angle 2$ can determine that $\\mathrm{AC} \\parallel \\mathrm{CD}$, so this is incorrect.\n\nD. $\\angle 1$ and $\\angle 2$ are consecutive interior angles. The fact that $\\angle 1 = \\angle 2$ does not determine that $\\mathrm{AB} \\parallel \\mathrm{CD}$, so this is incorrect.\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question tests the determination of parallel lines. Mastering the theorem for determining parallel lines is crucial for solving the problem." }, { "problem_id": 355, "question": "As shown in Figure (1), two right-angled triangles from a set of triangles are stacked together, where $\\angle C = 90^\\circ, \\angle A = 30^\\circ$, and $\\angle EDC = 45^\\circ$. Triangle $ABC$ is held stationary, while triangle $DCE$ rotates clockwise around point $C$. Figure (2) shows the position of triangle $DCE$ during the rotation process. The rotation stops when points $B, C, E$ are collinear for the first time. Let $\\angle BCD = k \\angle ACE$ (where $k$ is a constant). Given the following four statements:\n\n(1) When $k = 1$, the acute angle formed by the intersection of lines $AB$ and $DE$ is $15^\\circ$;\n\n(2) When $k = 3$, $DE \\parallel BC$;\n\n(3) When $CE \\perp AB$, $k = 2$;\n\n(4) When $CE \\parallel AB$, $k = 5$. The number of correct statements is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch24-2024_06_15_679e02e0c13b71c22c5cg_0020_1.jpg", "batch24-2024_06_15_679e02e0c13b71c22c5cg_0020_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: When the triangle plate $DCE$ rotates at an angle less than 90 degrees, as shown in Figure (2) of the problem, let the lines $AB$ and $ED$ intersect at $F$,\n\n$\\therefore \\angle BCD = \\angle ACB + \\angle ACD$,\n\n$\\therefore \\angle BCD + \\angle ACE = \\angle ACB + \\angle ACE + \\angle ACD = 180^{\\circ}$,\n\nWhen $k=1$, that is, $\\angle BCD = \\angle ACE$, as shown in Figure (1),\n\n$\\therefore \\angle BCD = \\angle ACE = 90^{\\circ}$,\n\n$\\therefore \\angle AFD = \\angle EDC - \\angle A = 15^{\\circ}$;\n\nWhen the triangle plate $DCE$ rotates at an angle greater than $90^{\\circ}$, as shown in Figure (2),\n\n$\\therefore \\angle ACE + \\angle BCD = 360^{\\circ} - \\angle ACB - \\angle ECD = 180^{\\circ}$,\n\n$\\therefore$ when $k=1$, that is, $\\angle BCD = \\angle ACE$,\n\n$\\therefore \\angle BCD = \\angle ACE = 90^{\\circ}$,\n\n$\\therefore$ at this time, $\\triangle ECD$ is in the position of $\\triangle E'CD'$ in the figure,\n\n$\\therefore \\angle AF'E' = \\angle ABE' - \\angle E' = 15^{\\circ}$, thus (1) is correct;\n\n\n\nFigure (1)\n\n\n\nWhen the triangle plate $DCE$ rotates at an angle less than 90 degrees, as shown in Figure (3),\n\nwhen $k=3$, $\\angle BCD = 3 \\angle ACE$,\n\n$\\therefore 3 \\angle ACE + \\angle ACE = 180^{\\circ}$,\n\n$\\therefore \\angle ACE = 45^{\\circ}$,\n\n$\\therefore \\angle BCE = 90^{\\circ} - \\angle ACE = 45^{\\circ}$,\n\n$\\therefore \\angle BCE = \\angle CED$,\n\n$\\therefore DE \\parallel BC$\n\nWhen the triangle plate $DCE$ rotates at an angle greater than $90^{\\circ}$, as shown in Figure (4),\n\nsimilarly, $\\angle ACE = 45^{\\circ}$,\n\n$\\therefore \\angle ACE = \\angle CED$\n\n$\\therefore DE \\parallel AC$,\n\n$\\because AC \\perp BC$,\n\n$\\therefore DE \\perp BC$, thus (2) is incorrect;\n\n\n\nFigure (3)\n\n\n\nAs shown in Figure (5), when $CE \\perp AB$,\n\n$\\because \\angle ABC = 60^{\\circ}$,\n\n$\\therefore \\angle BCE = 30^{\\circ}$,\n$\\therefore \\angle BCD = \\angle BCE + \\angle ECD = 120^{\\circ}, \\angle ACE = \\angle ACB - \\angle BCE = 60^{\\circ}$,\n\n$\\therefore \\angle BCD = 2 \\angle ACE$,\n\n$\\therefore k=2$, thus (3) is correct;\n\n\n\nFigure (5)\n\nSince $\\triangle DCE$ stops rotating clockwise when $B$, $C$, $E$ are collinear,\n\n$\\therefore$ when $CE \\parallel AB$, there is only one case as shown in Figure (6),\n\n\n\nFigure (6)\n\n$\\therefore \\angle ACE = \\angle A = 30^{\\circ}$,\n\n$\\therefore \\angle BCD = 360^{\\circ} - \\angle ACB - \\angle ACE - \\angle ECD = 150^{\\circ}$,\n\n$\\therefore \\angle BCD = 5 \\angle ACE$\n\n$\\therefore k=5$, thus (4) is correct,\n\nTherefore, the answer is: C.\n\n【Key Point】This problem mainly examines the calculation of angles in a triangle plate, the triangle angle sum theorem, the determination of parallel lines, and the correct understanding of the problem is the key to solving it." }, { "problem_id": 356, "question": "Given that $\\angle 1 + \\angle 2 = 180^\\circ$, which of the following figures can deduce that $AB \\parallel CD$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch24-2024_06_15_679e02e0c13b71c22c5cg_0032_1.jpg", "batch24-2024_06_15_679e02e0c13b71c22c5cg_0032_2.jpg", "batch24-2024_06_15_679e02e0c13b71c22c5cg_0032_3.jpg", "batch24-2024_06_15_679e02e0c13b71c22c5cg_0032_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: A. Since $\\angle 1 + \\angle 2 = 180^{\\circ}$, it follows that $AB \\parallel CD$. Therefore, option A is correct.\n\nOptions B, C, and D cannot be deduced from $\\angle 1 + \\angle 2 = 180^{\\circ}$ to conclude that $AB \\parallel CD$. Hence, the correct choice is A.\n\n[Key Insight] This question tests the theorem for determining parallel lines. Mastering the theorem for determining parallel lines is crucial for solving the problem." }, { "problem_id": 357, "question": "Point $C$ is a point on the side of $\\angle B A C$. Using a scaleless straightedge and compass, construct a ray $C D$ parallel to $A B$. Which of the following construction methods is correct ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch24-2024_06_15_679e02e0c13b71c22c5cg_0064_1.jpg", "batch24-2024_06_15_679e02e0c13b71c22c5cg_0064_2.jpg", "batch24-2024_06_15_679e02e0c13b71c22c5cg_0064_3.jpg", "batch24-2024_06_15_679e02e0c13b71c22c5cg_0064_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: According to the method of constructing an angle equal to a given angle using a compass and straightedge, it is clear that option A is correct, therefore the answer is: A.\n\n[Key Point] This question tests the determination of parallel lines and the construction of an angle equal to a given angle using a compass and straightedge. Correctly understanding the problem is crucial for solving it." }, { "problem_id": 358, "question": "While watching a billiards match, we often observe players using the technique of bouncing the cue ball off the table edges to hit the target ball. In this process, the angle between the striking path and the table edge is equal to the angle between the reflected path and the table edge, as shown in Figure 1, where $\\angle 1 = \\angle 2$. As shown in Figure 2, a Cartesian coordinate system $xOy$ is established.\n\nGiven that ball A is located at point $(1,2)$ and ball B is located at point $(6,1)$. Now, hit ball A so that it bounces off the table edge at a whole-number coordinate (where both the x and y coordinates are integers, excluding the pocket positions) and strikes ball B. If ball A can bounce off the table edge at most twice before hitting ball B, then how many whole-number points on the table edge satisfy these conditions?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch24-2024_06_15_6f2e89da18aeebe3371eg_0034_1.jpg", "batch24-2024_06_15_6f2e89da18aeebe3371eg_0034_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Currently, we strike ball $A$ so that it hits the edge of the table at a lattice point (where both the horizontal and vertical coordinates are integers, and the ball cannot rebound at the pocket position). If ball $A$ hits ball $B$ after rebounding off the edge of the table at most twice, then the only lattice point on the table edge that satisfies the condition is $(3,4)$, as shown in the figure.\n\n\n\nTherefore, the answer is: A\n[Key Insight] This question primarily tests the understanding of coordinates and geometry. Mastering the properties of coordinates and geometric figures is crucial for solving the problem." }, { "problem_id": 359, "question": "A school is located $1500 \\mathrm{~m}$ south of and at an angle of $60^{\\circ}$ west from a mall. Which of the following representations is correct ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_6f2e89da18aeebe3371eg_0097_1.jpg", "batch24-2024_06_15_6f2e89da18aeebe3371eg_0097_2.jpg", "batch24-2024_06_15_6f2e89da18aeebe3371eg_0097_3.jpg", "batch24-2024_06_15_6f2e89da18aeebe3371eg_0097_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. A certain shopping mall is located in the direction of 60° west of south from a certain school, and is 1500 meters away from the mall. Therefore, A does not meet the condition;\n\nB. A certain school is located in the direction of 30° west of south from a certain shopping mall, and is 1500 meters away from the mall. Therefore, B does not meet the condition;\n\nC. A certain school is located in the direction of 60° west of south from a certain shopping mall, and is 1500 meters away from the mall. Therefore, C meets the condition;\n\nD. A certain shopping mall is located in the direction of 30° west of south from a certain school, and is 1500 meters away from the mall. Therefore, D does not meet the condition. Hence, the correct choice is C.\n\n[Highlight] This question mainly tests the determination of location, and the key to solving it is to master the representation method of directional angles." }, { "problem_id": 360, "question": "Among the following figures, which one cannot conclude that lines $A B$ are parallel to $C D$ based on $\\angle 1 = \\angle 2$?\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_75de93123d91a70e33b7g_0024_1.jpg", "batch24-2024_06_15_75de93123d91a70e33b7g_0024_2.jpg", "batch24-2024_06_15_75de93123d91a70e33b7g_0024_3.jpg", "batch24-2024_06_15_75de93123d91a70e33b7g_0024_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: A, as shown in the figure,\n\n\n\n$\\because \\angle 1=\\angle 2, \\quad \\angle 3=\\angle 2$,\n\n$\\therefore \\angle 1=\\angle 3$,\n\n$\\therefore AB \\parallel CD$,\n\nTherefore, this option does not meet the requirement;\n\nB, $\\because \\angle 1=\\angle 2$,\n\n$\\therefore AB \\parallel CD$,\n\nTherefore, this option does not meet the requirement;\n\nC, $\\because \\angle 1=\\angle 2$,\n\n$\\therefore AB \\parallel CD$,\n\nTherefore, this option does not meet the requirement;\n\nD, $\\because \\angle 1=\\angle 2$,\n\n$\\therefore AC \\parallel BD$,\n\nTherefore, this option meets the requirement;\n\nHence, the correct choice is: D.\n\n[Key Insight] This question tests the determination of parallel lines. Familiarity with the theorem for determining parallel lines is key to solving the problem." }, { "problem_id": 361, "question": "Given that $A B / / C D$, which of the following figures can lead to $\\angle 1 = \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_89d0ba63ade24546cc6bg_0041_1.jpg", "batch24-2024_06_15_89d0ba63ade24546cc6bg_0041_2.jpg", "batch24-2024_06_15_89d0ba63ade24546cc6bg_0041_3.jpg", "batch24-2024_06_15_89d0ba63ade24546cc6bg_0041_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Since $AB // CD$, it cannot be concluded that $\\angle 1 = \\angle 2$, so this option does not meet the requirement;\n\nB. Since $AB // CD$, it cannot be concluded that $\\angle 1 = \\angle 2$, so this option does not meet the requirement;\n\nC. As shown in the figure,\n\n\n\n$\\because AB // CD$,\n\n$\\therefore \\angle 1 = \\angle 3$,\n\n$\\because \\angle 2 = \\angle 3$,\n\n$\\therefore \\angle 1 = \\angle 2$, so this option meets the requirement;\n\nD. Since $AB // CD$, it cannot be concluded that $\\angle 1 = \\angle 2$, so this option does not meet the requirement;\n\nTherefore, the correct choice is: C\n\n[Key Insight] This question tests the properties of parallel lines. The key to solving it is to thoroughly understand the properties of parallel lines: when two lines are parallel, the alternate interior angles are equal; the corresponding angles are equal; and the consecutive interior angles are supplementary." }, { "problem_id": 362, "question": "Translate the following Chinese mathematical problem into English. If the coordinates of a point $P$ on the circle $\\mathrm{A}$ (with center $A$) are $(m, n)$, what are the coordinates of the corresponding point $P^{\\prime}$ on the translated circle $O$ (with center $O$) as shown in Figure (1) and Figure (2)? \n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $(m+2, n+1)$\nB. $(m-2, n-1)$\nC. $(m-2, n+1)$\nD. $(m+2, n-1)$", "input_image": [ "batch24-2024_06_15_89fdcc8f5269948b3a9fg_0050_1.jpg", "batch24-2024_06_15_89fdcc8f5269948b3a9fg_0050_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the translation rule of point $A$, the movement pattern of this point is to add 2 to the abscissa (x-coordinate) and subtract 1 from the ordinate (y-coordinate). Following this rule, the coordinates of $P^{\\prime}$ are calculated to be $(m+2, n-1)$.\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question examines the relationship between coordinates and graphical transformations. The key to solving this problem lies in identifying the pattern of change in the abscissa and ordinate of each corresponding point based on the known corresponding points." }, { "problem_id": 363, "question": "In quadrilateral $A B C D$ as shown in Figure 1, $A B \\parallel C D$ and $\\angle A D C = 90^\\circ$. Point $P$ starts from point $A$ and moves at a speed of 1 unit per second along the edges in the order $A \\rightarrow B \\rightarrow C \\rightarrow D$. Let the time of $P$'s movement be $t$ seconds, and the area of $\\triangle P A D$ be $S$. The graph of $S$ as a function of $t$ is shown in Figure 2. When point $P$ reaches the midpoint of $B C$, the area of $\\triangle P A D$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 7\nB. 7.5\nC. 8\nD. 8.6", "input_image": [ "batch24-2024_06_15_a50ec6db40562a4758c9g_0003_1.jpg", "batch24-2024_06_15_a50ec6db40562a4758c9g_0003_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, quadrilateral \\( A B C D \\) is a trapezoid.\n\nWhen point \\( P \\) moves from \\( C \\) to \\( D \\), it takes 2 seconds, so \\( C D = 2 \\), and the area of \\( \\triangle A D P \\) is 4.\n\nThus, \\( A D = 4 \\).\n\nFrom the graph, when point \\( P \\) moves to point \\( B \\), the area of \\( \\triangle A D P \\) is 10.\n\nTherefore, \\( A B = 5 \\), and the total movement time is 5 seconds.\n\nHence, the point \\( (5,10) \\) lies on the function graph.\n\nFor the interval \\( 5 < t \\leq 10 \\), let the function be \\( s = k t + b \\).\n\nWe have the system of equations:\n\\[\n\\begin{cases}\n10 = 5k + b \\\\\n4 = 10k + b\n\\end{cases}\n\\]\nSolving this system gives:\n\\[\n\\begin{cases}\nk = -\\frac{6}{5} \\\\\nb = 16\n\\end{cases}\n\\]\nTherefore, for \\( 5 < t \\leq 10 \\), the function is \\( S = -\\frac{6}{5} t + 16 \\).\n\nWhen \\( P \\) reaches the midpoint of \\( B C \\), the time \\( t = 7.5 \\), and \\( S = 7 \\).\n\nThus, the correct answer is: A.\n\n[Key Insight] This problem primarily examines the function graph of a moving point and the formula for the area of a triangle. Utilizing a combination of graphical and algebraic methods is crucial for solving the problem." }, { "problem_id": 364, "question": "Among the following figures, which one can conclude that $AB \\parallel CD$ given $\\angle 1 = \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch24-2024_06_15_ab673a50196ba2332e7fg_0081_1.jpg", "batch24-2024_06_15_ab673a50196ba2332e7fg_0081_2.jpg", "batch24-2024_06_15_ab673a50196ba2332e7fg_0081_3.jpg", "batch24-2024_06_15_ab673a50196ba2332e7fg_0081_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: \n\nA. $\\angle 1$ and $\\angle 2$ are consecutive interior angles. The equality $\\angle 1 = \\angle 2$ does not necessarily imply that $AB \\parallel CD$, so this option is incorrect.\n\nB. $\\angle 1$ and $\\angle 2$ are alternate interior angles. The equality $\\angle 1 = \\angle 2$ implies that $AC \\parallel BD$, so this option is incorrect.\n\nC. $\\angle 1$ and $\\angle 2$ are alternate interior angles. The equality $\\angle 1 = \\angle 2$ implies that $AB \\parallel CD$, so this option is correct.\n\nD. $\\angle 1$ and $\\angle 2$ are opposite angles in a quadrilateral. The equality $\\angle 1 = \\angle 2$ does not imply that $AB \\parallel CD$, so this option is incorrect.\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the understanding of the criteria for determining parallel lines. Familiarity with the theorems related to the determination of parallel lines is crucial for answering this question correctly." }, { "problem_id": 365, "question": "After learning about parallel lines, Xiaomin came up with a new method to draw a parallel line to a given line through a point outside the line by folding a translucent piece of paper (as shown in figures (1) to (4)). From the figures, we can see that Xiaomin's method for drawing parallel lines relies on the following principles:\n\n(1) If two lines are parallel, corresponding angles are equal;\n\n(2) If two lines are parallel, interior angles on opposite sides are equal;\n\n(3) If corresponding angles are equal, the two lines are parallel;\n\n(4) If interior angles on opposite sides are equal, the two lines are parallel.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)\nB. (2)(3)\nC. (3)(4)\nD. (1)(4)", "input_image": [ "batch24-2024_06_15_e078db8917d1060edcc3g_0078_1.jpg", "batch24-2024_06_15_e078db8917d1060edcc3g_0078_2.jpg", "batch24-2024_06_15_e078db8917d1060edcc3g_0078_3.jpg", "batch24-2024_06_15_e078db8917d1060edcc3g_0078_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "1", "analysis": "From the construction process, it can be seen that $\\angle 1 = \\angle 2$, which are alternate interior angles, and $\\angle 1 = \\angle 4$, which are corresponding angles. It can be concluded that Xiaomin's method of drawing parallel lines is based on: (3) if corresponding angles are equal, then the two lines are parallel; (4) if alternate interior angles are equal, then the two lines are parallel.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\n\nTherefore, the correct choice is C." }, { "problem_id": 366, "question": "Among the three-dimensional figures below, which one has a circular front view?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_00e575063edd62623dc1g_0016_1.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0016_2.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0016_3.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0016_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: A's front view is a circle, so A fits the description;\n\nB's front view is a rectangle, so B does not fit the description;\n\nC's front view is a triangle, so C does not fit the description;\n\nD's front view is a square, so D does not fit the description;\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question tests the understanding of the three views of simple geometric shapes. Familiarity with the three views of common geometric shapes is crucial for solving such problems." }, { "problem_id": 367, "question": "In the figure, the front view of the solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_00e575063edd62623dc1g_0017_1.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0017_2.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0017_3.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0017_4.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0017_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Observing the given geometric figure from the front, the resulting diagram is as follows:\n\n\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question tests the knowledge of three-view drawings. The front view is the view obtained by looking at the object from the front. Students may mistakenly choose other options by confusing the three types of views." }, { "problem_id": 368, "question": "Among the main (front) views of the following simple geometric solids, which one is both an axisymmetric and a centrosymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_00e575063edd62623dc1g_0050_1.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0050_2.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0050_3.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0050_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "A. The front view is an isosceles triangle, which is an axisymmetric figure but not a centrally symmetric figure, so it does not meet the requirement;\n\nB. The front view is a circle, which is both an axisymmetric and a centrally symmetric figure, so it meets the requirement;\n\nC. The front view is an equilateral triangle, which is an axisymmetric figure but not a centrally symmetric figure, so it does not meet the requirement;\n\nD. The front view is a regular pentagon, which is an axisymmetric figure but not a centrally symmetric figure, so it does not meet the requirement;\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question tests the understanding of the three views of a geometric solid, the concepts of centrally symmetric and axisymmetric figures. The key to identifying an axisymmetric figure is to find the axis of symmetry; the figure can coincide with itself when folded along the axis. The key to identifying a centrally symmetric figure is to find the center of symmetry; the figure can coincide with itself when rotated 180 degrees around the center." }, { "problem_id": 369, "question": "Among the following geometric solids, which has a left view that is a triangle?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_00e575063edd62623dc1g_0064_1.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0064_2.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0064_3.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0064_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \nA. The left view of this triangular prism is a rectangle, which does not meet the requirement; \n\nB. The left view of the cuboid is a rectangle, which does not meet the requirement; \n\nC. The left view of the cone is a triangle, which meets the requirement; \n\nD. The left view of this cylinder is a rectangle, which does not meet the requirement; \n\nTherefore, the correct answer is: C. \n\n**[Key Point]** This question tests the understanding of the three views of geometric solids. Mastering the definitions is crucial. Note that all visible edges should be represented in the three views." }, { "problem_id": 370, "question": "The three-dimensional shape after cutting off a corner of a cube is shown in the figure. Its front view is ( )\n\n\n\nFrom the front\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_00e575063edd62623dc1g_0084_1.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0084_2.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0084_3.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0084_4.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0084_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: When viewed from the front, both the bottom left and top right are triangles.\n\nA. The option is correct and fits the context;\n\nTherefore, the answer is: A.\n\n[Key Point] This question tests the understanding of three-view drawings: the view directly facing us is called the front view, the surface below the front view is called the horizontal plane, and the right side is called the side view; the view obtained within the front view by observing the object from front to back is called the main view; the view obtained within the horizontal plane by observing the object from top to bottom is called the top view; the view obtained within the side view by observing the object from left to right is called the left view." }, { "problem_id": 371, "question": "The front view of the solid in the figure below is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_00e575063edd62623dc1g_0091_1.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0091_2.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0091_3.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0091_4.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0091_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, the front view of the geometric solid is:\n\n\n\nTherefore, the correct choice is: A.\n\n[Key Point] This question primarily tests the ability to draw the three views of a geometric solid. The knowledge points used are: the front view, side view, and top view are the images obtained by viewing the object from the front, side, and top, respectively." }, { "problem_id": 372, "question": "A geometric solid is constructed from several identical small cubes. The shape viewed from the front and from above is shown in the figure. The maximum number of small cubes used to construct this solid is ( )\n\n\n\nView from the front\n\n\n\nView from above\nA. 4 cubes\nB. 5 cubes\nC. 6 cubes\nD. 7 cubes", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0001_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0001_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, as shown in the figure:\n\n\n\nAt this point, the maximum number of small cubes is: $2 + 2 + 1 + 1 = 6$;\n\nTherefore, the correct choice is C.\n\n[Key Insight] This question tests the ability to determine the number of small cubes based on the three-view drawing. Mastering the use of the top view to determine positions and the front view to determine quantities is crucial for solving the problem." }, { "problem_id": 373, "question": "As shown in the figure, the front view and left view of a simple polyhedron composed of identical small cubes are given. What are the minimum and maximum number of small cubes that make up this polyhedron, respectively?\n\n\n\nFront View\n\n\n\nLeft View\nA. 4, 6\nB. 4, 7\nC. 5, 6\nD. 5, 7", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0003_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0003_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The minimum number of small cubes that make up this geometric figure is $3+1=4$ small cubes, and the maximum is $6+1=7$ small cubes.\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Use the top view to lay the foundation, the front view to build up, and the side view to remove obstructions\" can make it easier to arrive at the answer." }, { "problem_id": 374, "question": "As shown in the figure, these are the shape views of a geometric solid from three directions. Which of the following options represents the shape of this geometric solid?\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from above\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0008_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0008_2.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0008_3.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0008_4.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0008_5.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0008_6.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0008_7.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the top view, options A and D can be excluded.\n\nThen, based on the side view, option B can be excluded.\n\nFinally, it is determined that option C fits the description.\n\nTherefore, choose C.\n\n[Insight] This question tests the ability to reconstruct a geometric figure from its three views. To solve such problems, the elimination method is often more convenient." }, { "problem_id": 375, "question": "A toy model can be regarded as the geometric solid shown in the figure. Which of the following is the front view of the solid?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0009_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0009_2.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0009_3.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0009_4.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0009_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, the following figure can be obtained:\n\n\n\nTherefore, the correct choice is: C.\n\n[Highlight] This question examines the three views of a simple composite object, where the figure obtained from the front view is the main view." }, { "problem_id": 376, "question": "As shown in the figure, which one is the shadow formed by sunlight? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0043_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0043_2.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0043_3.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0043_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The direction of shadows formed under sunlight is the same, so options A and D do not meet the requirement. Observing the images below, it can be seen that option B represents parallel projection, and option C represents central projection.\n\n\n\nB\n\n\n\nC\n\nTherefore, the correct choice is B.\n\n[Insight] This question examines the characteristics and rules of parallel projection. Projection formed by parallel light is called parallel projection. Under parallel light, the length of the shadow of the same object varies at different times; at the same moment, the lengths of the shadows of different objects are proportional to their heights and have the same direction." }, { "problem_id": 377, "question": "The three views of a frustum of a cone are shown in the figure. Based on the data in the figure, the volume of the frustum is ( )\n\n\n\nFront View\n\n\n\nTop View\nA. $12 \\pi \\mathrm{cm}^{3}$\nB. $20 \\pi \\mathrm{cm}^{3}$\nC. $32 \\pi \\mathrm{cm}^{3}$\nD. $48 \\pi \\mathrm{cm}^{3}$", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0061_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0061_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the observation of the three views, it is determined that the radius of the base of the cone is \\(6 \\div 2 = 3 \\, \\text{cm}\\), and the height is \\(4 \\, \\text{cm}\\).\n\nThus, the volume of the cone is calculated as:\n\\[\n\\frac{1}{3} \\pi r^{2} h = \\frac{1}{3} \\pi \\times 3^{2} \\times 4 = 12\\pi \\, \\text{cm}^{3}.\n\\]\n\nTherefore, the correct choice is: A.\n\n**Key Insight:** This question primarily tests the knowledge of three views and the formula for the volume of a cone. Determining the height and the radius of the base circle of the cone from the three views is crucial for solving this problem." }, { "problem_id": 378, "question": "The left view, top view, and related data of a rectangular prism are shown in the figure. Determine the area of its front view ( ).\n\n\n\nLeft View\n\n\n\nTop View\nA. 12\nB. 15\nC. 20\nD. 60", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0083_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0083_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the left view and the top view, we can deduce that the length of the front view is 5 and the width is 3.\n\n$\\therefore$ The area of the front view is $3 \\times 5=15$,\n\nTherefore, the correct choice is B.\n\n【Key Point】This question tests the understanding of the three views and the quantitative relationships between them. Mastering the characteristics of the three views is crucial for solving the problem." }, { "problem_id": 379, "question": "One afternoon, Xiaohong first participated in the school sports meeting's women's $100 \\mathrm{~m}$ race, and after some time, she participated in the women's $400 \\mathrm{~m}$ race. The photographer took two photos from the same position, as shown in the figure. Which of the following statements is correct ( )\n\n\n\nA\n\n\n\nB\n\n\n\nA. Photo B was taken during the $100 \\mathrm{~m}$ race.\n\nB. Photo A was taken during the $100 \\mathrm{~m}$ race.\n\nC. Photo B was taken during the $400 \\mathrm{~m}$ race.\n\nD. It is impossible to determine which race the photos A and B were taken during.", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0099_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0099_2.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0099_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Logic", "image_relavance": "1", "analysis": "Solution: According to the rules of parallel projection, the direction of an object's shadow from morning to evening follows the sequence: west - northwest - north - northeast - east, with the shadow length first shortening and then lengthening. Therefore, photo B corresponds to the 100m event, and photo A corresponds to the 400m event. Hence, the correct choice is A.\n\n[Key Insight] This question tests the understanding of parallel projection. Mastering the rules of parallel projection is crucial for solving the problem." }, { "problem_id": 380, "question": "A geometric solid is constructed of identically sized small cubes. Its top view is shown in the figure, where the number in each small square indicates the number of small cubes at that position. The left view of the solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_103b3c36c8bf95889ea5g_0022_1.jpg", "batch25-2024_06_17_103b3c36c8bf95889ea5g_0022_2.jpg", "batch25-2024_06_17_103b3c36c8bf95889ea5g_0022_3.jpg", "batch25-2024_06_17_103b3c36c8bf95889ea5g_0022_4.jpg", "batch25-2024_06_17_103b3c36c8bf95889ea5g_0022_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the top view, it can be seen that the left view, from left to right, consists of 2, 2, and 3 small squares in sequence.\n\nTherefore, the answer is: D.\n\n[Highlight] This question tests the knowledge of views; the key to solving it lies in thoroughly understanding the properties of the top view and the left view, thereby enabling the solution to be completed." }, { "problem_id": 381, "question": "In the following projections, the orthographic projections are $(\\quad)$\n\n\n\n(1)\n\n\n\n(3)\n\n\n\n(2)\n\n\n\n(4)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch25-2024_06_17_13ac171763bb883f1178g_0025_1.jpg", "batch25-2024_06_17_13ac171763bb883f1178g_0025_2.jpg", "batch25-2024_06_17_13ac171763bb883f1178g_0025_3.jpg", "batch25-2024_06_17_13ac171763bb883f1178g_0025_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "In parallel projection, if the projection rays are perpendicular to the projection plane, then this type of projection is called orthographic projection; according to the concept, the ones that are orthographic projections are: (3), (4).\n\nTherefore, choose B.\n\nKey point: Grasp the concept of orthographic projection." }, { "problem_id": 382, "question": "Among the four figures below, which one is the net of a triangular prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_1a01bcad4bed55b70b27g_0020_1.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0020_2.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0020_3.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0020_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: A. This figure can form a triangular prism, so this option does not meet the requirement of the question;\n\nB. This figure can form a triangular pyramid, so this option meets the requirement of the question;\n\nC. This figure can form a quadrilateral pyramid, so this option does not meet the requirement of the question;\n\nD. It cannot form a three-dimensional figure, so this option does not meet the requirement of the question.\n\nTherefore, the correct choice is: B.\n\n[Highlight] This question mainly examines the development of geometric figures. It is essential to analyze the shape characteristics of the geometric figures to proceed." }, { "problem_id": 383, "question": "Given four types of iron sheets (1)(2)(3)(4) with shapes and dimensions as shown in the figure (units: dm). By selecting two of them, it is possible to make an open-top cylindrical bucket (ignoring the joints). The two iron sheets of the selected types are ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(3)\nB. (1)(4)\nC. (2)(3)\nD. (2)(4)", "input_image": [ "batch25-2024_06_17_1a01bcad4bed55b70b27g_0040_1.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0040_2.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0040_3.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0040_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The length of the lateral (side) expansion of a cylinder equals the circumference of its base. The circumference of a circle with a diameter of \\(2 \\mathrm{dm}\\) is \\(2 \\pi \\mathrm{dm}\\), and the circumference of a circle with a diameter of \\(4 \\mathrm{dm}\\) is \\(4 \\pi \\mathrm{dm}\\). Therefore, options (2) and (3) are appropriate.\n\nThus, the correct choice is C.\n\n[Key Insight] This question tests the characteristics of the lateral expansion of a cylinder. Correctly calculating the circumference of the cylinder's base is crucial for solving the problem." }, { "problem_id": 384, "question": "The unfolding diagram of the cube shown in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_1a01bcad4bed55b70b27g_0088_1.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0088_2.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0088_3.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0088_4.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0088_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "A: Restoring it into a cube,\n\nis the opposite face, which does not match the original geometric shape;\n\nB: Restoring it into a cube, then\n\nis on the right side of the geometric shape, which does not match the original geometric shape;\n\nC: Restoring it into a cube,\n\nis the opposite face, which does not match the original geometric shape;\n\nD: Restoring it into a cube, all features match the original geometric shape;\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question tests the understanding of the unfolded diagram of a geometric shape. Mastering the characteristics of the unfolded diagram of a cube is the key to solving the problem." }, { "problem_id": 385, "question": "As shown in the figure, the geometric solid is part of a certain cylinder, and the cutting plane is a plane. The top view of the solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_26a127df8b63912ac416g_0002_1.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0002_2.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0002_3.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0002_4.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0002_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: The top view of the geometric solid is\n\n\nTherefore, option A is selected.\n\n[Key Point] This question tests the understanding of the three views of a simple composite solid. The view seen from above is the top view, and it is important to note that invisible lines should be represented with dashed lines and must be drawn." }, { "problem_id": 386, "question": "As shown in the figure, a three-dimensional shape is composed of 6 identical cubes. Its front view is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_26a127df8b63912ac416g_0069_1.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0069_2.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0069_3.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0069_4.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0069_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: From the front view, it is easy to determine that the first layer has 3 squares, and the second layer has 1 square located in the middle position.\n\nTherefore, the answer is: B.\n\n[Key Point] This question tests the knowledge of three-view drawings, where the front view is the view obtained by looking at the object from the front." }, { "problem_id": 387, "question": "Among the following four unfolded cube patterns, which one can be folded into the cube shown in the figure?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_35cb0aebc81696508813g_0032_1.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0032_2.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0032_3.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0032_4.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0032_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, in the cube, face $A$, face $B$, and face $C$ are adjacent faces.\n\nFrom the net provided in option $\\mathrm{A}$, it can be seen that face $A$ and face $C$ are opposite faces, thus option $\\mathrm{A}$ does not meet the requirement; from the net provided in option $B$, it can be seen that face $A$, face $B$, and face $C$ are adjacent faces, thus option $B$ meets the requirement;\n\nFrom the net provided in option $C$, it can be seen that face $B$ and face $C$ are opposite faces, thus option $C$ does not meet the requirement; from the net provided in option $D$, it can be seen that face $A$ and face $B$ are opposite faces, thus option $D$ does not meet the requirement;\n\nTherefore, the correct choice is: $B$.\n\n【Insight】This question tests the understanding of the net of a cube, and mastering the characteristics of the cube's net is key to solving the problem." }, { "problem_id": 388, "question": "There is a cube as shown in the figure, which of the following figures is an unfolding diagram of this cube?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_35cb0aebc81696508813g_0061_1.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0061_2.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0061_3.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0061_4.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0061_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \nA. After folding, the three shapes do not share a common vertex, so A does not meet the requirement; \n\nB. After folding, the rectangle and the ordinary triangle are opposite faces, so B does not meet the requirement; \n\nC. After folding, the two triangles are opposite faces, so C does not meet the requirement; \n\nD. After folding, the three shapes share a common vertex, and the green rectangle shares a common edge with the isosceles right triangle, so D meets the requirement. \n\nTherefore, the correct answer is: D. \n\n[Key Point] This question tests the understanding of the surface expansion diagram of a cube. It is an important topic and relatively easy. Mastering the relevant knowledge is key to solving the problem." }, { "problem_id": 389, "question": "An object as shown in the figure is placed on a horizontal table. Its left view is ( )\n\n\n\nView Direction\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0027_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0027_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0027_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0027_4.jpg", "batch25-2024_06_17_493157da588ff6213245g_0027_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: The figure obtained by observing the object from the left side is\n\n\n\nTherefore, the correct choice is: B\n\n[Key Point] This question tests the understanding of the three views of a simple composite object. The view obtained from the left side is the left view. It is important to note that visible lines should be drawn as solid lines, while invisible lines should be drawn as dashed lines." }, { "problem_id": 390, "question": "As shown in the figure, the top view shape of an L-shaped foam plastic is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0034_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0034_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0034_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0034_4.jpg", "batch25-2024_06_17_493157da588ff6213245g_0034_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The top view of the L-shaped foam plastic is as shown in the figure.\n\n\n\nTherefore, choose B.\n\n[Key Point] This question tests the ability to determine the three views of simple geometric shapes. Understanding that the top view is the shape seen from above is crucial for solving the problem." }, { "problem_id": 391, "question": "Among the following geometric solids, which ones have the same front view and top view? ( )\n\n\nCube\n\n\nCone\n\n\nSphere\n\n\nCylinder\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0047_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0047_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0047_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0047_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The main view and top view of a cylinder are a rectangle and a circle, respectively, and they are not the same; the main view and top view of a cone are a triangle and a circle with a center, respectively, and they are not the same; the main view and top view of a sphere are both circles, and they are the same.\n\nThe main view and top view of a cube are both squares, and they are the same.\n\nThere are a total of 2 geometric shapes whose main view and top view are the same.\n\nTherefore, the answer is: B.\n\n[Key Point] Test point: Three views of simple geometric shapes." }, { "problem_id": 392, "question": "Among the following figures, which can be folded into a triangular prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_63d64c842022cf6d6051g_0003_1.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0003_2.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0003_3.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0003_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Option $\\mathrm{A}$ can be folded into a cylinder, option $\\mathrm{B}$ can be folded into a cube, option $\\mathrm{C}$ can be folded into a cone, and option D can be folded into a triangular prism.\n\nTherefore, the correct choice is: D.\n\n[Insight] This question tests the ability to reconstruct a three-dimensional figure from its net, with the key to solving it lying in possessing a certain level of spatial imagination." }, { "problem_id": 393, "question": "Figure 1 is a rectangular prism where $A D=A B=10$ and $A E=6$. $M$ and $N$ are the midpoints of the respective edges. Figure 2 is the unfolded surface of Figure 1. The length of $M N$ in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $11 \\sqrt{2}$\nB. $10 \\sqrt{2}$\nC. 10\nD. 8", "input_image": [ "batch25-2024_06_17_63d64c842022cf6d6051g_0032_1.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0032_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, connect $MN$ and extend the sides of the square to intersect at point $P$;\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nThen, $\\triangle MPN$ is a right-angled triangle.\n\nAccording to the problem, $MP = NP = 5 + 6 = 11$.\n\nBy the Pythagorean theorem, $MN = \\sqrt{11^{2} + 11^{2}} = 11\\sqrt{2}$.\n\nTherefore, the correct answer is A." }, { "problem_id": 394, "question": "As shown in Figure (1), a small triangular prism is cut off from the upper left corner of the cube (points $M$ and $N$ are midpoints of the cube's edges), resulting in the solid shown in Figure (2). When viewed from the front, top, and left sides of the solid in (2), the areas of the visible shapes are denoted as $S_{\\text {front }}$, $S_{\\text {top }}$, and $S_{\\text {left }}$, respectively. Which of the following statements is true?\n\n\n\n(1)\n\n\n\n(2)\n\nA. $S_{\\text {front }}=S_{\\text {top }}=S_{\\text {left }}$\n\nB. $S_{\\text {front }}\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_693961fecc216a55431bg_0091_1.jpg", "batch25-2024_06_17_693961fecc216a55431bg_0091_2.jpg", "batch25-2024_06_17_693961fecc216a55431bg_0091_3.jpg", "batch25-2024_06_17_693961fecc216a55431bg_0091_4.jpg", "batch25-2024_06_17_693961fecc216a55431bg_0091_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: From the \"alternating and $Z$-end opposite faces\" characteristic of the cube's surface net, we can determine that in option A, the \"O face\", \"○ face\", and \"※ face\" all have \"blank\" as their opposite faces, thus option A fits the given condition. In option B, the \"○ face\" and \"※ face\" are opposite, which contradicts the condition, so option B does not fit. In option C, the \"$O$ face\" and \"○ face\" are opposite, which also contradicts the condition, so option C does not fit. In option D, the \"○ face\" and \"※ face\" are opposite, again contradicting the condition, so option D does not fit.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question mainly examines the net of a cube. Understanding the characteristics of a cube's surface net is crucial for correctly answering this question." }, { "problem_id": 396, "question": "In the neighborhood where Xiao Ming lives, there is a straight road with a streetlight in the middle. One night, he walked along this road as shown in the figure. During his walk from point $\\mathrm{A}$ to point $B$, the relationship between the length of his shadow $l$ under the light and the distance he traveled $s$ can be best represented by the graph ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_7b8a6dd337c06631b50cg_0005_1.jpg", "batch25-2024_06_17_7b8a6dd337c06631b50cg_0005_2.jpg", "batch25-2024_06_17_7b8a6dd337c06631b50cg_0005_3.jpg", "batch25-2024_06_17_7b8a6dd337c06631b50cg_0005_4.jpg", "batch25-2024_06_17_7b8a6dd337c06631b50cg_0005_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As he walks from point $A$ to the base of the streetlight, the length of the shadow $l$ gradually decreases. When he walks from the base of the streetlight to point $B$, the length of the shadow $l$ gradually increases. That is, as $S$ gradually increases, $l$ first decreases from a larger value to a smaller one, and then increases from the smaller value back to a larger one.\n\nTherefore, the correct choice is: $A$.\n\n[Key Insight] This question primarily tests the understanding of function graphs and the properties of central projection. Determining the pattern of how $l$ changes with $s$ is crucial to solving the problem." }, { "problem_id": 397, "question": "As shown in the figure, there is a rectangular paper box with a shaded triangle on each of two adjacent faces. Which of the following could be the unfolded view of the box?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_7dd71be433a08193772cg_0018_1.jpg", "batch25-2024_06_17_7dd71be433a08193772cg_0018_2.jpg", "batch25-2024_06_17_7dd71be433a08193772cg_0018_3.jpg", "batch25-2024_06_17_7dd71be433a08193772cg_0018_4.jpg", "batch25-2024_06_17_7dd71be433a08193772cg_0018_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By observing the rectangular prism before it is unfolded, it can be seen that after unfolding, the black triangles will all be located at the upper left corner of the rectangle. Therefore, option A fits the description.\n\nHence, the answer is A.\n\n[Key Insight] This question primarily tests the understanding of the unfolded view of a geometric figure. Determining the position of the triangles based on the geometric figure is the key to solving the problem." }, { "problem_id": 398, "question": "The three views of a cylindrical object are shown in the figure. Based on the data in the figure, the lateral surface area of the cylinder is ( )\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View\nA. $12 \\pi \\mathrm{cm}^{2}$\nB. $15 \\pi \\mathrm{cm}^{2}$\nC. $20 \\pi \\mathrm{cm}^{2}$\nD. $24 \\pi \\mathrm{cm}^{2}$", "input_image": [ "batch25-2024_06_17_802f01be018a9f8ad422g_0003_1.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0003_2.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0003_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By observing the three views, we find that the height of the cone is $4 \\mathrm{~cm}$, and the radius of the base circle is $6 \\div 2=3(\\mathrm{~cm})$.\n\nTherefore, the slant height of the cone $=\\sqrt{3^{2}+4^{2}}=5(\\mathrm{~cm})$.\n\nThus, the lateral surface area of the cone $=\\frac{1}{2} \\times 2 \\pi \\times 3 \\times 5=15 \\pi\\left(\\mathrm{cm}^{2}\\right)$.\n\nHence, the correct choice is: B.\n\n[Insight] This question tests knowledge of three views and the formula for the volume of a cone. The key to solving the problem lies in determining the height of the cone and the radius of the base circle from the three views." }, { "problem_id": 399, "question": "Given a geometric solid as shown in the figure, the left view of the solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_802f01be018a9f8ad422g_0056_1.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0056_2.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0056_3.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0056_4.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0056_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: According to the definition of the left view, the left view of this geometric body is:\n\n\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question tests the ability to determine the left view of a geometric body. Mastering the concept of the left view is crucial for solving this problem." }, { "problem_id": 400, "question": "The top view of a geometric solid built with several identical small cubes is shown in the figure. The numbers in the small squares indicate the number of small cubes at that position. What is the front view of this geometric solid?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_802f01be018a9f8ad422g_0073_1.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0073_2.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0073_3.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0073_4.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0073_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, the leftmost column shows 3 small cubes, the middle column shows 2 small cubes, and the rightmost column shows 2 small cubes.\n\nThus, the front view is:\n\n\n\nTherefore, the correct choice is: B.\n\n[Highlight] This question tests the three-dimensional visualization of geometric shapes, reflecting the core competency of spatial awareness in the mathematics discipline." }, { "problem_id": 401, "question": "Given a stack of instant noodle cans placed on a table, its three views are shown below. How many cans are there in this stack?\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View\nA. 7 cans\nB. 8 cans\nC. 9 cans\nD. 10 cans", "input_image": [ "batch25-2024_06_17_802f01be018a9f8ad422g_0088_1.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0088_2.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0088_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Top View\n\nBased on the comprehensive three-view drawing, there should be 5 buckets of instant noodles at the bottom layer,\n\n2 buckets on the second layer,\n\nand 1 bucket on the third layer,\n\ntotaling to \\(5 + 2 + 1 = 8\\) buckets.\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question aims to test students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills. If you remember the mnemonic \"Top view lays the foundation, front view builds up, side view removes obstructions,\" it becomes easier to arrive at the answer." }, { "problem_id": 402, "question": "As shown in Figure (1), observe a standard six-sided die, where the numbers 1 and 6 are opposite, 2 and 5 are opposite, and 3 and 4 are opposite. If a dot is placed in one of the positions (1), (2), (3), or (4) in Figure (2), and the other three positions are removed, which of the following arrangements can form a cube?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. (1)\nB. (2)\nC. (3)\nD. (4)", "input_image": [ "batch25-2024_06_17_a9cbf53fa0771581ac03g_0089_1.jpg", "batch25-2024_06_17_a9cbf53fa0771581ac03g_0089_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since 1 is opposite to 6, based on the knowledge of the net of a cube, in options (1), (2), (3), and (4), the position of (1) is the opposite face of 6.\n\nTherefore, the option that can form a cube is (1).\n\nHence, the correct choice is A.\n\n[Key Insight] This question tests the unfolding and folding of a cube. You can try folding it yourself to see, or fully utilize spatial imagination to solve it.\n\nError Analysis: This is an easy question. The reason for losing points is: unfamiliarity with the unfolding and folding of a cube." }, { "problem_id": 403, "question": "In the following simple sketch patterns, those that are axisymmetric are ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_b7118e4abd5521e96630g_0048_1.jpg", "batch25-2024_06_17_b7118e4abd5521e96630g_0048_2.jpg", "batch25-2024_06_17_b7118e4abd5521e96630g_0048_3.jpg", "batch25-2024_06_17_b7118e4abd5521e96630g_0048_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. It is not an axisymmetric figure, therefore this option is incorrect;\n\nB. It is an axisymmetric figure, therefore this option is correct;\n\nC. It is not an axisymmetric figure, therefore this option is incorrect;\n\nD. It is not an axisymmetric figure, therefore this option is incorrect.\n\nHence, the correct choice is: B.\n\n[Key Insight] This question tests the concept of axisymmetric figures. The key to identifying an axisymmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide when folded along the axis." }, { "problem_id": 404, "question": "As shown in the figure, it is an unfolded pattern of a cube. Which of the following could be the cube?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_c04268d430149b44ad9ag_0037_1.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0037_2.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0037_3.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0037_4.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0037_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "After folding the net, only option $B$ matches the planar unfolded figure,\n\nTherefore, the answer is: B.\n\n[Key Insight] This question primarily examines the relative positions of patterns on the planar unfolded figure of a cube. Understanding the relative positions of patterns on each face after folding the net is the key to solving the problem." }, { "problem_id": 405, "question": "The following four diagrams represent the shadows cast by two trees in the sunlight at the same time. Which one is it?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_c2b5a48971f4685260d0g_0073_1.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0073_2.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0073_3.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0073_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Sunlight and shadows, at the same moment, the height of the pole and the length of the shadow are directly proportional, and the position of the shadow is in the same direction as the object. It can be concluded that the figure in option B fits the description;\n\nTherefore, the answer is: B\n\n[Highlight] This question tests the understanding of the concept of parallel projection. Mastering the characteristics and properties of parallel projection is the prerequisite for making a correct judgment." }, { "problem_id": 406, "question": "On a wall that is 8 meters long and 3 meters high, there is a window that is 2 meters wide and 1 meter high. Which of the following diagrams could represent this wall?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_c861aa08a70db5000978g_0078_1.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0078_2.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0078_3.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0078_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The problem describes two rectangles, one with a length of 8 and a width of 3, and another with a length of 2 and a width of 1. \n\nConsidering the options provided, only option D satisfies the conditions described in the problem. Therefore, the correct choice is D." }, { "problem_id": 407, "question": "Xiao Zhang took a picture with his phone to get Figure A, and after magnification, he got Figure B. The corresponding segment of segment AB in Figure A in Figure B is ( )\n\n\n\nFigure A\n\n\n\nFigure B\nA. FG\nB. FH\nC. EH\nD. EF", "input_image": [ "batch25-2024_06_17_c861aa08a70db5000978g_0086_1.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0086_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "As shown in the diagram, points $\\mathrm{A}$ and $\\mathrm{E}$ are corresponding vertices,\n\npoints $\\mathrm{B}$ and $\\mathrm{F}$ are corresponding vertices,\n\nand points $\\mathrm{D}$ and $\\mathrm{H}$ are corresponding vertices. Therefore, the corresponding line segment to $\\mathrm{AB}$ in the second diagram is $\\mathrm{EF}$. Hence, the correct choice is D." }, { "problem_id": 408, "question": "The following are the shadows of two buildings at four different times of the day. Arrange them in the correct order of time:\n\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\nA. (3)(1)(4)(2)\nB. (3)(2)(1)(4)\nC. (3)(4)(1)(2)\nD. (2)(4)(1)(3)", "input_image": [ "batch25-2024_06_17_c861aa08a70db5000978g_0091_1.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0091_2.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0091_3.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0091_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "$\\because$ From morning to evening, the direction of an object's shadow points: West → Northwest → North → Northeast → East, and the length of the shadow changes from long to short, then back to long.\n\n$\\therefore$ (1) is Northeast, (2) is East, (3) is West, (4) is Northwest.\n\n$\\therefore$ Arranging them in chronological order gives (3)(4)(1)(2).\n\nTherefore, the correct answer is C.\n\n【Key Point】This question tests the concept of parallel projection. The key to solving it is understanding that sunlight consists of parallel rays and knowing the pattern of how shadows change under sunlight." }, { "problem_id": 409, "question": "A geometric solid is built with small cubes. The shapes viewed from the front, left, and top are shown in the figures. This geometric solid is made of ( ) small cubes.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from above\nA. 6\nB. 5\nC. 4\nD. 3", "input_image": [ "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0088_1.jpg", "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0088_2.jpg", "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0088_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the three-view drawing, it can be determined that there are 2 layers in total. The first layer consists of 4 small squares, and the second layer has 2 small cubes on the left, as shown in the figure.\n\n\n\nTherefore, this geometric shape is composed of 6 small cubes.\n\nHence, the correct choice is: A.\n\n[Key Insight] This question tests the ability to reconstruct a geometric shape from its three-view drawing. Accurate interpretation of the drawing is crucial for solving the problem." }, { "problem_id": 410, "question": "As shown in the figure, there is a small wooden board with a circular hole and a square hole. Which of the following objects can both plug the circular hole and the square hole?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_d78b06d1c38bca83fba0g_0016_1.jpg", "batch25-2024_06_17_d78b06d1c38bca83fba0g_0016_2.jpg", "batch25-2024_06_17_d78b06d1c38bca83fba0g_0016_3.jpg", "batch25-2024_06_17_d78b06d1c38bca83fba0g_0016_4.jpg", "batch25-2024_06_17_d78b06d1c38bca83fba0g_0016_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since it can block both circular and square holes,\n\n$\\therefore$ From the three views of the object, the one that has both circles and squares in its three views can block the small wooden board with circular and square holes.\n\nA. The three views of a cube are all squares, without any circles, so it cannot be option A;\n\nB. When the diameter of a cylinder is equal to its height, its front and side views are squares, and its top view is a circle, possessing both circles and squares, so it can be option B;\n\nC. The front and side views of a cone are triangles, and its top view is a circle, without any squares, so it cannot be option C;\n\nD. The three views of a sphere are all circles, without any squares, so it cannot be option D.\n\nTherefore, choose B.\n\n[Highlight] This question tests the ability of an object to block circular and square holes, essentially examining the views of the object. The key to solving the problem lies in identifying an object with both circles and squares in its three views." }, { "problem_id": 411, "question": "Given the three views of a geometric solid as shown in the figure, determine the volume of the solid.\n\n\n\nFront View\n\n\n\nTop View\n\n\n\nSide View\nA. $\\frac{2 \\sqrt{3}}{3}+\\pi$\nB. $\\frac{2 \\sqrt{3}}{3}+2 \\pi$\nC. $2 \\sqrt{3}+2 \\pi$\nD. $2 \\sqrt{3}+\\pi$", "input_image": [ "batch25-2024_06_17_dc139e69bdca0c6041e1g_0043_1.jpg", "batch25-2024_06_17_dc139e69bdca0c6041e1g_0043_2.jpg", "batch25-2024_06_17_dc139e69bdca0c6041e1g_0043_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, the geometric body is a combination of a triangular prism and a semi-cylinder. The base of the triangular prism is an equilateral triangle with a side length of 2, and its height is 2. The semi-cylinder has a base radius of 1 and a height of 2.\n\nTherefore, the volume \\( V \\) of the geometric body is calculated as:\n\\[\nV = \\frac{1}{2} \\times 2 \\times \\sqrt{3} \\times 2 + \\frac{1}{2} \\times \\pi \\times 1^{2} \\times 2 = 2\\sqrt{3} + \\pi.\n\\]\nHence, the correct answer is: D.\n\n[Key Insight] This problem tests the ability to determine the volume of a geometric body from its three-view drawing. The key to solving the problem lies in identifying the shape of the geometric body and the corresponding geometric measurements from the given data." }, { "problem_id": 412, "question": "The 3D solid shown in the figure is constructed by two identical cubes and a cylinder. Its left view is ( )\n\n\n\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_de12aee7c6c3dcedbe3bg_0063_1.jpg", "batch25-2024_06_17_de12aee7c6c3dcedbe3bg_0063_2.jpg", "batch25-2024_06_17_de12aee7c6c3dcedbe3bg_0063_3.jpg", "batch25-2024_06_17_de12aee7c6c3dcedbe3bg_0063_4.jpg", "batch25-2024_06_17_de12aee7c6c3dcedbe3bg_0063_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since the figure obtained from the left side is the left view,\n\nTherefore, the geometric body, when viewed from the left, has a triangle on the first layer and a small square on the second layer,\n\nHence, the correct choice is: D.\n\n[Key Point] This question tests the understanding of the three views of a simple composite geometric body. The figure seen from the left is the left view, and it's important to note that the left view of a cone is a triangle." }, { "problem_id": 413, "question": "Construct a 3D solid using four identical small cubes, such that when viewed from the front, from the left, and from above, at least two of the resulting shapes are identical. Which of the following four arrangements does not meet this requirement?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0001_1.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0001_2.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0001_3.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0001_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "First, draw the three views, then determine that among the shapes obtained from the front view, the left view, and the top view, at least two of the shapes are the same.\n\n\n\nOnly option D has three views that are all different from each other, so the answer is D.\n\n[Key Point] This question mainly tests the understanding of three views, and spatial imagination is crucial." }, { "problem_id": 414, "question": "In the figures below, which ones can be folded into a prism after folding?\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)(4)\nB. (2)(3)(4)\nC. (1)(2)(3)\nD. (1)(3)(4)", "input_image": [ "batch25-2024_06_17_e5f356909807fe2710f3g_0021_1.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0021_2.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0021_3.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0021_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "(1)(2)(3) can form a prism, (4) when forming a prism, two faces coincide,\n\nTherefore, choose C\n\n[Highlight] This question examines the net of a prism. Understanding the characteristics of a prism and its net is key to solving the problem." }, { "problem_id": 415, "question": "Fold a square piece of paper according to the sequence below, then cut away the small triangle in the upper right corner of the last folded paper along the dotted line. After unfolding the paper, the resulting figure is\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_e5f356909807fe2710f3g_0030_1.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0030_2.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0030_3.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0030_4.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0030_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: It is easy to see that the four small squares cut out are exactly located in the middle of each pair of opposite sides of the original square.\n\nTherefore, the correct choice is A.\n\n[Key Insight] This question primarily examines the problem of paper cutting; the student's spatial imagination and hands-on operation abilities are quite important. When solving such problems, it is essential to cultivate these skills." }, { "problem_id": 416, "question": "As shown in the figure, which of the following figures is not the unfolding diagram of the shape ( )\n\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_e5f356909807fe2710f3g_0064_1.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0064_2.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0064_3.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0064_4.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0064_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, from the planar expansion of the cube, we can deduce:\n\n\n\n$A$ and $C$ are opposite faces, $E$ and $F$ are opposite faces, $B$ and $D$ are opposite faces, and $A$ and $D$ are adjacent faces. Moreover, the vertex of the blank figure on face $B$ lies on the edge of the blank figure on faces $E$ and $F$. Therefore, in option $\\mathrm{D}$ of the cube, the vertex of the blank figure on face $B$ does not lie on the edge of the blank figure on face $E$ or $F$, so option D does not meet the requirement.\n\nThus, the correct choice is D.\n\n【Key Point】This question tests the understanding of the surface expansion of a cube. Mastering \"the characteristics of opposite and adjacent faces in the cube's expansion diagram\" is crucial to solving this problem." }, { "problem_id": 417, "question": "A cube's faces are labeled with numbers $1, 2, 3$, and negative numbers are placed on their opposite faces so that the sum of any two opposite faces is 0. After cutting the cube along certain edges, the following shapes are obtained. Among these shapes, the one where $x$ corresponds to the number -3 is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_e5f356909807fe2710f3g_0091_1.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0091_2.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0091_3.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0091_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: A. $x=-3$\n\nB. $x=-2$\n\nC. $x=-2$\n\nD. $x=-2$\n\nTherefore, the correct answer is: A\n\n[Key Point] This question primarily examines the text on opposite faces of a cube. Pay attention to the spatial representation of the cube and approach the problem by analyzing the opposite faces to derive and solve the question." }, { "problem_id": 418, "question": "As shown in the figure, it is a horizontally placed fully enclosed object. Which of the following is its top view?\n\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_e768c97899b57b6ada19g_0001_1.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0001_2.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0001_3.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0001_4.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0001_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the observation above, we can obtain:\n\n\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question examines the concept of three-view drawings of simple geometric shapes. The key point here is to consider that the edges visible in the top view are represented by solid lines." }, { "problem_id": 419, "question": "As shown in the figure, the left view of the solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_e768c97899b57b6ada19g_0041_1.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0041_2.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0041_3.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0041_4.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0041_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The left view of this geometric figure is a rectangle, with a hidden line represented by a dashed line, as shown in the figure:\n\n\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question tests the understanding of three-view drawings. Having spatial imagination is crucial for solving the problem, and it's important to note that invisible lines should be depicted with dashed lines." }, { "problem_id": 420, "question": "As shown in the figure, the left view of the given solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_e768c97899b57b6ada19g_0092_1.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0092_2.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0092_3.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0092_4.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0092_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: Its left view is\n\n\n\nTherefore, the correct choice is: B.\n\n[Key Point] This question tests the understanding of the left view of a simple geometric solid. The key to solving the problem lies in understanding the definition of the left view: \"The view obtained from the left side is the left view.\" It is important to note that lines that are not visible but exist should be drawn as dashed lines." }, { "problem_id": 421, "question": "The top view of the solid shown in the figure is ( )\n\n\n\nFront View\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0068_1.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0068_2.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0068_3.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0068_4.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0068_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "From above, the interior of the rectangle has two perpendicular dashed lines,\n\nTherefore, the answer is: D.\n\n[Highlight] This question tests the knowledge of three-view drawings, note: the view obtained from above is the top view." }, { "problem_id": 422, "question": "Among the following geometric solids, which one has a front view and a top view that are both rectangles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0096_1.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0096_2.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0096_3.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0096_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. The front view is a rectangle and the top view is a triangle, therefore it is incorrect;\n\nB. Both the front view and the top view are rectangles, therefore it is correct;\n\nC. The front view is a rectangle and the top view is a circle, therefore it is incorrect;\n\nD. The front view is a triangle and the top view is a rectangle with diagonals, therefore it is incorrect;\n\nHence, the correct choice is: B.\n\n[Key Insight] This question tests the understanding of the three views of simple geometric shapes. Mastering the three views of common geometric shapes is crucial for solving such problems." }, { "problem_id": 423, "question": "In Figure 1 and Figure 2, both are cubes. In Figure 3, a geometric solid composed of several identical cubes is shown from the front and the side. After studying these figures, Xiaomin comes to the following conclusions:\n\n(1) If the surface of the cube in Figure 1 is cut along certain edges to form a flat shape, 7 edges need to be cut;\n\n(2) When a plane cuts through the cube in Figure 1 from different directions, the resulting cross-section can be a triangle, quadrilateral, pentagon, or hexagon;\n\n(3) When a plane cuts through the cube in Figure 1, it results in the triangle $A B C$ in the cross-section of Figure 2, where $\\angle A B C = 45^\\circ$;\n\n(4) In Figure 3, the minimum number of cubes needed to build the geometric solid is $a$, and the maximum number is $b$. Then $a + b = 19$.\n\nHow many of these conclusions are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFront View\n\n\n\nSide View\n\nFigure 3\nA. 2\nB. 3\nC. 4\nD. 5", "input_image": [ "batch25-2024_06_17_f3b68e01e9dc18006b97g_0024_1.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0024_2.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0024_3.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0024_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: (1) If the surface of the cube in Figure 1 is cut along certain edges and unfolded into a planar figure, it is necessary to cut 7 edges; this is correct because a cube has 6 faces and 12 edges, and to unfold it into a planar figure, 5 edges must remain connected, so at least \\(12-5=7\\) edges need to be cut.\n\n(2) Using a plane to intersect the cube in Figure 1 from different directions, the resulting cross-section could be a triangle, quadrilateral, pentagon, or hexagon; this is correct because when a plane intersects a cube, it can intersect up to six faces to form a hexagon, and at least three faces to form a triangle.\n\n(3) Using a plane to intersect the cube in Figure 1 to obtain Figure 2, the angle \\(\\angle ABC\\) in the triangular cross-section \\(ABC\\) is \\(45^\\circ\\); this is incorrect because \\(\\triangle ABC\\) is an equilateral triangle, so \\(\\angle ABC=60^\\circ\\).\n\n(4) As shown in Figure 3, the minimum number of cubes needed to construct the geometry is \\(a\\), and the maximum is \\(b\\), then \\(a+b=19\\). This is incorrect; it should be \\(a=6\\), \\(b=11\\), so \\(a+b=17\\).\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question primarily examines the properties of the cube's net, the cross-sections of a cube, and the three-view drawings of simple composite geometries. Understanding that a planar figure must have 5 connected edges based on the properties of the net is crucial for solving the problem." }, { "problem_id": 424, "question": "As shown in the figure, this is the three views of a three-dimensional solid composed of two rectangular prisms. According to the dimensions marked in the figure (in units of $\\mathrm{mm}$), calculate the surface area of this solid in $\\mathrm{mm}^{2}$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View\nA. 200\nB. 280\nC. 350\nD. None of the above answers are correct", "input_image": [ "batch25-2024_06_17_f3b68e01e9dc18006b97g_0033_1.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0033_2.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0033_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the three-view drawing, we can determine that the upper cuboid has a length of $4 \\mathrm{~mm}$, a height of $4 \\mathrm{~mm}$, and a width of $2 \\mathrm{~mm}$. The lower cuboid has a length of $8 \\mathrm{~mm}$, a width of $6 \\mathrm{~mm}$, and a height of $2 \\mathrm{~mm}$.\n\nTherefore, the surface area of the three-dimensional figure is:\n\n$4 \\times 4 \\times 2 + 4 \\times 2 \\times 2 + 4 \\times 2 \\times 2 + 6 \\times 2 \\times 2 + 8 \\times 2 \\times 2 + 6 \\times 8 \\times 2 - 4 \\times 2 \\times 2 = 200\\left(\\mathrm{~mm}^{2}\\right)$.\n\nHence, the correct choice is: A.\n\n[Key Insight] This problem primarily tests the ability to interpret a three-dimensional figure from its three-view drawing and to calculate the surface area of the figure. The key to solving the problem lies in accurately identifying the length, width, and height of the cuboids from the drawing." }, { "problem_id": 425, "question": "As shown in the figure, the rectangular prism is composed of the four small geometric solids $A, B, C, D$ below. Which geometric solid corresponds to the fourth part in the figure?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_f3b68e01e9dc18006b97g_0065_1.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0065_2.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0065_3.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0065_4.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0065_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the cuboid and the geometric shapes corresponding to the first, second, and third parts, it can be inferred that,\n\nThe geometric shape corresponding to the fourth part has one cube in a row and three cubes in another row, with the single cube positioned in the middle of the three cubes at the back.\n\nTherefore, the correct choice is: A.\n\n[Highlight] This question tests the understanding of three-dimensional shapes, and the key to solving it lies in correctly analyzing the positions of the cubes." }, { "problem_id": 426, "question": "The three views of the geometric solid in the figure are ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_fb7a7b941e483c077eaeg_0013_1.jpg", "batch25-2024_06_17_fb7a7b941e483c077eaeg_0013_2.jpg", "batch25-2024_06_17_fb7a7b941e483c077eaeg_0013_3.jpg", "batch25-2024_06_17_fb7a7b941e483c077eaeg_0013_4.jpg", "batch25-2024_06_17_fb7a7b941e483c077eaeg_0013_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "From the geometric figure, it can be seen that the three views of this geometric figure are:\n\n\n\nTherefore, choose C.\n\n[Key Point] This question tests the understanding of the three views of a simple geometric figure. Mastering the orientation and drawing method of the three views is crucial for solving the problem. Note that edges that actually exist but are not blocked by other edges and are not visible from the viewing direction should be represented by dashed lines." }, { "problem_id": 427, "question": "Among the four figures below, which one cannot be obtained by rotating $\\triangle A B C$ to form $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_0aab922c48aded34ec53g_0032_1.jpg", "batch26-2024_06_17_0aab922c48aded34ec53g_0032_2.jpg", "batch26-2024_06_17_0aab922c48aded34ec53g_0032_3.jpg", "batch26-2024_06_17_0aab922c48aded34ec53g_0032_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "In option $\\mathrm{A}$, $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ can be considered as $\\triangle A B C$ rotated $180^{\\circ}$ around the midpoint of $C C^{\\prime}$, which does not meet the requirement;\n\nIn option $\\mathrm{B}$, $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ can be considered as $\\triangle A B C$ rotated $180^{\\circ}$ around point $C$, which does not meet the requirement;\n\nIn option $\\mathrm{C}$, $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ can be considered as $\\triangle A B C$ rotated around point $O$, which does not meet the requirement;\n\nIn option $\\mathrm{D}$, $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ can be considered as $\\triangle A B C$ obtained through axial symmetry, which meets the requirement;\n\nTherefore, the correct choice is: D.\n\n【Key Point】This question tests the concept of rotation, and the key to solving it is to identify the center of rotation." }, { "problem_id": 428, "question": "Place a pair of right-angled triangles (as shown in Figure (a)), where $\\angle A C B = \\angle D E C = 90^\\circ, \\angle A = 45^\\circ, \\angle D = 30^\\circ$, with the hypotenuse $A B = 6 \\sqrt{2}$ cm, and $D C = 8 \\sqrt{2}$ cm. Rotate triangle $D C E$ clockwise around point $C$ by $15^\\circ$ to form $\\triangle D_{I} C E_{I}$ (as shown in Figure (b)). At this time, $A B$ intersects $C D_{I}$ at point $O$ and $D_{I} E_{l}$ at point $F$. The length of segment $A D_{I}$ is ( )\n\n\n\n(a)\n\n\n\n(b)\nA. $5 \\sqrt{3}$ cm\nB. $5 \\sqrt{2}$ cm\nC. $17 \\sqrt{2}$ cm\nD. $2 \\sqrt{17}$ cm", "input_image": [ "batch26-2024_06_17_13af33e16678a18085deg_0100_1.jpg", "batch26-2024_06_17_13af33e16678a18085deg_0100_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since the rotation angle is $15^{\\circ}$,\n\n$\\therefore \\angle O C B=60^{\\circ}-15^{\\circ}=45^{\\circ}$,\n\n$\\therefore \\angle C O B=180^{\\circ}-45^{\\circ}-45^{\\circ}=90^{\\circ}$,\n\n$\\therefore C D_{1} \\perp A B$,\n\nAlso, since $\\angle D=30^{\\circ}$\n\n$\\therefore A O=C O=\\frac{1}{2} A B=\\frac{1}{2} \\times 6 \\sqrt{2}=3 \\sqrt{2} \\quad(\\mathrm{~cm})$,\n\n$\\therefore O D_{I}=D C-C O=8 \\sqrt{2}-3 \\sqrt{2}=5 \\sqrt{2} \\quad(\\mathrm{~cm})$,\n\nIn the right triangle $\\triangle A D_{l} O$, by the Pythagorean theorem, $A D_{l}=\\sqrt{A O^{2}+O D_{1}^{2}}=\\sqrt{(3 \\sqrt{2})^{2}+(5 \\sqrt{2})^{2}}=2 \\sqrt{17}(\\mathrm{~cm})$; therefore, the answer is: D.\n\n【Key Insight】This problem examines the properties of rotation, the Pythagorean theorem, the properties of a right triangle with a $30^{\\circ}$ angle, and the properties of an isosceles right triangle. Mastering the Pythagorean theorem is crucial for solving this problem." }, { "problem_id": 429, "question": "The following figure that is a central symmetric figure but not an axis symmetric figure is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_177e0089b0dcdd52b64bg_0100_1.jpg", "batch26-2024_06_17_177e0089b0dcdd52b64bg_0100_2.jpg", "batch26-2024_06_17_177e0089b0dcdd52b64bg_0100_3.jpg", "batch26-2024_06_17_177e0089b0dcdd52b64bg_0100_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. It is a centrally symmetric figure but not an axially symmetric figure; therefore, this option is correct.\n\nB. It is both a centrally symmetric figure and an axially symmetric figure; therefore, this option is incorrect.\n\nC. It is both a centrally symmetric figure and an axially symmetric figure; therefore, this option is incorrect.\n\nD. It is not a centrally symmetric figure but is an axially symmetric figure; therefore, this option is incorrect.\n\nThus, the correct choice is A.\n\n[Key Point] This question tests the concepts of centrally symmetric and axially symmetric figures: The key to identifying an axially symmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide upon folding along the axis; a centrally symmetric figure requires finding the center of symmetry, where the figure coincides with its original form upon a 180-degree rotation." }, { "problem_id": 430, "question": "Place a pair of right-angled triangles as shown in Figure 1, where $\\angle A C B = \\angle D E C = 90^\\circ$, $\\angle A = 45^\\circ$, and $\\angle D = 30^\\circ$. The hypotenuse $A B$ is 6 units, and $C D$ is 8 units. Rotate triangle $D C E$ clockwise around point $C$ by $15^\\circ$ to obtain $\\triangle D_{I} C E_{I}$ (as shown in Figure 2). At this time, $A B$ intersects $C D_{1}$ at point $O$. The length of segment $A D_{1}$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 4\nB. $3 \\sqrt{2}$\nC. 6\nD. $\\sqrt{34}$", "input_image": [ "batch26-2024_06_17_19263ae180636aead5d2g_0072_1.jpg", "batch26-2024_06_17_19263ae180636aead5d2g_0072_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the problem statement, it is easy to know: $\\angle CAB = 45^{\\circ}, \\angle ACD = 30^{\\circ}$.\n\nIf the rotation angle is $15^{\\circ}$, then $\\angle ACO = 30^{\\circ} + 15^{\\circ} = 45^{\\circ}$.\n\n$\\therefore \\angle AOC = 180^{\\circ} - \\angle ACO - \\angle CAO = 90^{\\circ}$\n\n$\\therefore$ According to the Three Lines Coincidence Theorem, point $O$ is the midpoint of $AB$.\n\nIn the isosceles right triangle $\\triangle ABC$, $AB = 6$, so $AO = OC = 3$.\n\nIn the right triangle $\\triangle AOD_{I}$, $OD_{I} = CD_{1} - OC = 5$,\n\nBy the Pythagorean theorem, $AD_{1} = \\sqrt{AO^{2} + OD^{2}} = \\sqrt{34}$.\n\nTherefore, the answer is D.\n\n[Key Insight] This problem mainly examines the properties of rotation, the Three Lines Coincidence Theorem in isosceles triangles, and the Pythagorean theorem. The key to solving the problem lies in mastering and applying these related concepts." }, { "problem_id": 431, "question": "Two identical right-angled triangles are placed as shown in Figure (1) (points $A, B, D$ are on the same straight line), where $\\angle B = 60^\\circ$ and $\\angle C = 30^\\circ$. Triangle $A B C$ is rotated clockwise around the right-angled vertex $A$ to form triangle $A F G$, and $A G$ intersects $D E$ at point $H$ (as shown in Figure 2). Let the rotation angle be $\\beta\\left(0^\\circ < \\beta < 90^\\circ\\right)$. When triangle $A D H$ is an isosceles triangle, the measure of the rotation angle $\\beta$ is ( )\n\n\n\n(1)\n\n\n\n(2)\nA. $20^\\circ$\nB. $20^\\circ$ or $60^\\circ$\nC. $15^\\circ$ or $60^\\circ$\nD. $60^\\circ$", "input_image": [ "batch26-2024_06_17_1efe5c1b1fdcfa56fa64g_0016_1.jpg", "batch26-2024_06_17_1efe5c1b1fdcfa56fa64g_0016_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since \\(\\angle B = 60^\\circ\\) and \\(\\angle C = 30^\\circ\\),\n\n\\(\\therefore \\angle D = \\angle G = 30^\\circ\\).\n\nBecause \\(\\triangle ABC\\) is rotated clockwise around the right angle vertex \\(A\\) to obtain \\(\\triangle AFG\\),\n\n\\(\\therefore \\angle CAG = \\beta\\).\n\nWhen \\(HA = HD\\), then \\(\\angle HAD = \\angle D = 30^\\circ\\),\n\n\\(\\therefore \\angle CAG = 90^\\circ - \\angle HAD = 60^\\circ\\), that is, \\(\\beta = 60^\\circ\\);\n\nWhen \\(DA = DH\\),\n\n\\(\\therefore \\angle AHD = \\angle DHA\\),\n\n\\(\\because \\angle D = 30^\\circ\\),\n\n\\(\\therefore \\angle HAD = \\frac{1}{2}(180^\\circ - 30^\\circ) = 75^\\circ\\),\n\n\\(\\therefore \\angle CAG = 90^\\circ - 75^\\circ = 15^\\circ\\), that is, \\(\\beta = 15^\\circ\\);\n\nIn summary, the rotation angle \\(\\beta\\) is either \\(15^\\circ\\) or \\(60^\\circ\\).\n\nTherefore, the correct answer is C.\n\n[Key Insight] This problem mainly examines the properties of rotation, the properties of isosceles triangles, and the triangle angle sum theorem. Accurate calculation is the key to solving the problem." }, { "problem_id": 432, "question": "Examine the figure below and, according to the pattern, draw the appropriate figures in the two \"field\" grids pointed to by the arrows. Which of the following is correct?\n\n\n\n\n\nA\n\n\n\nB\n\n\n\nC\n\n\n\nD\nA. A\nB. B\nC. C\nD. D", "input_image": [ "batch26-2024_06_17_201bada2148f276a9b39g_0058_1.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0058_2.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0058_3.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0058_4.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0058_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorics", "image_relavance": "1", "analysis": "From the diagram, it can be seen that the line passing through the top vertex of the white triangle and perpendicular to the base is the axis of symmetry. Therefore, the center line of the \"田\" grid indicated by the second arrow is the axis of symmetry, and it is the white triangle. The black dot is located in the top-right box indicated by the first arrow.\n\nThus, the correct choice is B." }, { "problem_id": 433, "question": "As shown in Figure (1), there is a $4 \\times 4$ square grid, where two square cells have already been painted black. Please paint two more cells black in such a way that the resulting pattern is axisymmetric. Two patterns that are identical after rotation are considered the same. In Figure (2), the two diagrams are considered the same. How many different patterns can be obtained?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 6\nB. 7\nC. 8\nD. 9", "input_image": [ "batch26-2024_06_17_201bada2148f276a9b39g_0062_1.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0062_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Question Analysis: As shown in the figure, there are a total of 8 types.\n\n\nTherefore, choose C\n\nKey Point: Axial Symmetry Figures" }, { "problem_id": 434, "question": "In the figure, the image of the graph in the grid paper rotated clockwise around point $\\mathrm{O}$ by $90^{\\circ}$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_201bada2148f276a9b39g_0067_1.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0067_2.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0067_3.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0067_4.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0067_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "According to the properties of rotation, the figure obtained by rotating $90^{\\circ}$ clockwise around point $O$ is\n\n\n\nTherefore, choose B.\n\n[Highlight] This question tests the properties of rotation. Before and after the rotation transformation, the corresponding line segments and angles remain equal, and the size and shape of the figure do not change." }, { "problem_id": 435, "question": "The transformation from figure A to figure B, which of the following statements is completely correct ( )\n\n(1)\n\n\n(2)\n\n\n(3)\n\nA. (1) Reflection, (2) Rotation, (3) Translation\n\nB. (1) Reflection, (2) Translation, (3) Rotation\n\nC. (1) Translation, (2) Reflection, (3) Rotation\n\nD. (1) Translation, (2) Rotation, (3) Reflection", "input_image": [ "batch26-2024_06_17_25561aac9e7e1a7fd0ebg_0081_1.jpg", "batch26-2024_06_17_25561aac9e7e1a7fd0ebg_0081_2.jpg", "batch26-2024_06_17_25561aac9e7e1a7fd0ebg_0081_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, we can see that:\n\n(1) Folding figure A will yield figure B;\n\n(2) Rotating figure A clockwise by $90^{\\circ}$ around its right angle vertex will yield figure B;\n\n(3) Translating figure A to the right and then upwards will yield figure B.\n\nTherefore, the transformations involved are: (1) folding, (2) rotation, (3) translation.\n\nHence, the correct choice is: A.\n\n[Key Insight] This question tests the understanding of common geometric transformation types. Accurately interpreting the diagram and determining the transformation process from figure A to figure B is crucial for solving the problem." }, { "problem_id": 436, "question": "After rotating the figure about the center of the circle by $180^\\circ$, the resulting pattern is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_4ff1ec270366d069efb8g_0068_1.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0068_2.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0068_3.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0068_4.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0068_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "The pattern obtained by rotating it $180^{\\circ}$ around the center of the circle is:\n\nTherefore, the correct choice is: D.\n\n[Highlight] This question tests the use of rotation in designing patterns, which is a fundamental and relatively easy concept. Mastering the relevant knowledge is key to solving the problem." }, { "problem_id": 437, "question": "As shown in the figure, in a grid, after rotating Rt $\\triangle A O B$ clockwise around point $B$ by $90^{\\circ}$, we obtain Rt $\\triangle A^{\\prime} O^{\\prime} B$. Which of the following four diagrams is correct ( )\n\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_4ff1ec270366d069efb8g_0098_1.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0098_2.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0098_3.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0098_4.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0098_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "A. Rt $\\triangle A^{\\prime} O^{\\prime} B$ is obtained by reflecting Rt $\\triangle A O B$ over the line that passes through point $\\mathrm{B}$ and is perpendicular to $\\mathrm{OB}$, so option A does not meet the requirement;\n\nB. Rt $\\triangle A^{\\prime} O^{\\prime} B$ is obtained by rotating Rt $\\triangle A O B$ clockwise by $90^{\\circ}$ around point $B$, so option B meets the requirement;\n\nC. The corresponding points of Rt $\\triangle A^{\\prime} O^{\\prime} B$ and Rt $\\triangle A O B$ have changed, so option C does not meet the requirement;\n\nD. Rt $\\triangle A O B$ is obtained by rotating Rt $\\triangle A O B$ counterclockwise by $90^{\\circ}$ around point $B$, so option D does not meet the requirement.\n\nTherefore, the correct answer is: B.\n\n[Key Insight] This question tests the understanding of rotational transformations. The key to solving the problem is to determine the direction and degree of rotation." }, { "problem_id": 438, "question": "A bamboo weaving is first folded around the straight line $\\mathrm{MN}$ as shown in the figure, flipping it $180^{\\circ}$, and then it is rotated counterclockwise by $90^{\\circ}$. The resulting bamboo weaving is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_7271d973a6f7af6646c5g_0018_1.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0018_2.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0018_3.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0018_4.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0018_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Question Analysis: First, flip it $180^{\\circ}$ around the straight line MN as shown in the figure, then rotate it $90^{\\circ}$ counterclockwise. The resulting bamboo strip weave is $\\mathrm{B}$, therefore choose $\\mathrm{B}$.\n\nExam Focus: Designing patterns using rotation." }, { "problem_id": 439, "question": "Among the following four options, which figure cannot be obtained by rotating or translating the figure shown below?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_7271d973a6f7af6646c5g_0035_1.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0035_2.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0035_3.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0035_4.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0035_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "A. By rotating the figure clockwise by $90^{\\circ}$, it can be obtained; therefore, this option does not meet the requirement;\n\nB. By rotating the figure counterclockwise by $90^{\\circ}$, it can be obtained; therefore, this option does not meet the requirement;\n\nC. It cannot be obtained by rotating or translating the given figure; therefore, this option meets the requirement;\n\nD. By rotating the figure clockwise by $180^{\\circ}$, it can be obtained; therefore, this option does not meet the requirement;\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question examines the concept of rotation. Rotation is a transformation that turns a figure around a fixed point by a certain angle. Thus, rotation must have a center of rotation and an angle of rotation, and the figure before and after rotation must coincide. This is the key to determining rotation." }, { "problem_id": 440, "question": "In the four triangles shown in the figure, which one cannot be obtained by rotating or translating $\\triangle A B C$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_78c82601763351be2b36g_0061_1.jpg", "batch26-2024_06_17_78c82601763351be2b36g_0061_2.jpg", "batch26-2024_06_17_78c82601763351be2b36g_0061_3.jpg", "batch26-2024_06_17_78c82601763351be2b36g_0061_4.jpg", "batch26-2024_06_17_78c82601763351be2b36g_0061_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: Option A can be obtained through rotation, which does not meet the requirement of the question;\n\nOption B is obtained through axial symmetry and cannot be achieved by rotation or translation, thus it meets the requirement of the question;\n\nOption C can be obtained through rotation and translation, which does not meet the requirement of the question;\n\nOption D can be obtained through rotation and translation, which does not meet the requirement of the question.\n\nTherefore, the correct choice is: B\n\n[Key Insight] This question tests the understanding of geometric transformations. The key to solving it lies in being proficient with geometric transformation graphics, establishing a sense of space, and accurately interpreting the diagrams." }, { "problem_id": 441, "question": "Place a pair of right-angled triangles as shown in Figure (1), where $\\angle A C B = \\angle D E C = 90^\\circ$, $\\angle A = 45^\\circ$, and $\\angle D = 30^\\circ$. The hypotenuse $A B$ is 6 units long, and $D C$ is 7 units long. Rotate triangle $D C E$ clockwise around point $C$ by $15^\\circ$ to obtain $\\triangle D_1 C E_1$ (as shown in Figure (2)). When $A B$ intersects $C D_1$ at point $O$, find the length of segment $A D_1$.\n\n\n\n(1)\n\n\n\n(2)\nA. 6\nB. 5\nC. 4\nD. $\\sqrt{31}$", "input_image": [ "batch26-2024_06_17_9b08795585671626fbc8g_0016_1.jpg", "batch26-2024_06_17_9b08795585671626fbc8g_0016_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since \\(\\angle ACB = \\angle DEC = 90^\\circ\\),\n\nTherefore, \\(DE \\parallel AC\\).\n\nThus, \\(\\angle ACD = \\angle D = 30^\\circ\\).\n\nFrom the property of rotation, we know that \\(CD = CD_1 = 7\\), and the rotation angle is \\(15^\\circ\\).\n\nTherefore, \\(\\angle ACO = 30^\\circ + 15^\\circ = 45^\\circ\\).\n\nHence, \\(\\angle ACO = \\angle BCO = 45^\\circ\\).\n\nSince \\(\\triangle ABC\\) is an isosceles right triangle,\n\nTherefore, \\(CO\\) is the perpendicular bisector of \\(AB\\).\n\nThus, \\(OA = OC = 3\\).\n\nHence, \\(OD_1 = CD_1 - OC = 4\\).\n\nIn the right triangle \\(\\triangle AOD_1\\), by the Pythagorean theorem, we find \\(AD_1 = \\sqrt{OA^2 + OD_1^2} = \\sqrt{3^2 + 4^2} = 5\\).\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This problem tests the properties of isosceles right triangles, the properties of rotation, the determination and properties of parallel lines, and the Pythagorean theorem. Proficiency in applying these concepts is crucial for solving the problem." }, { "problem_id": 442, "question": "In right triangle $\\triangle A B C$ shown in Figure 1, $\\angle A = 90^\\circ$, $A B = A C$, points $D$ and $E$ are on sides $A B$ and $A C$ respectively, with $A D = A E$. The segment $D C$ is drawn, and points $M$, $P$, and $N$ are the midpoints of segments $D E$, $D C$, and $B C$ respectively. Triangle $\\triangle A D E$ is free to rotate around point $A$ in the plane (as shown in Figure 2). If $A D = 4$ and $A B = 10$, what is the maximum area of triangle $\\triangle P M N$?\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\nA. $\\frac{49}{4}$\nB. 18\nC. $\\frac{49}{2}$\nD. $\\frac{25}{2}$", "input_image": [ "batch26-2024_06_17_9b08795585671626fbc8g_0026_1.jpg", "batch26-2024_06_17_9b08795585671626fbc8g_0026_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nBy the properties of rotation, we have: $\\angle BAD = \\angle CAE$.\n\nIn triangles $\\triangle ABD$ and $\\triangle ACE$,\n\n$$\n\\begin{aligned}\n& \\left\\{\\begin{array}{l}\nAB = AC \\\\\n\\angle BAD = \\angle CAE \\\\\nAD = AE\n\\end{array}\\right. \\\\\n& \\therefore \\triangle ABD \\cong \\triangle ACE \\quad (\\text{SAS}) \\\\\n& \\therefore \\angle ABD = \\angle ACE, \\quad BD = CE\n\\end{aligned}\n$$\n\nSince points $M$, $P$, and $N$ are the midpoints of $DE$, $DC$, and $BC$ respectively,\n\n$$\n\\begin{aligned}\n& \\therefore PN = \\frac{1}{2} BD, \\quad PM = \\frac{1}{2} CE, \\quad PM \\parallel CE, \\quad PN \\parallel BD \\\\\n& \\therefore PM = PN \\\\\n& \\therefore \\triangle PMN \\text{ is an isosceles triangle}\n\\end{aligned}\n$$\n\nSince $PM \\parallel CE$,\n\n$$\n\\angle DPM = \\angle DCE\n$$\n\nSince $PN \\parallel BD$,\n\n$$\n\\angle PNC = \\angle DBC\n$$\n\nSince $\\angle DPN = \\angle DCB + \\angle PNC = \\angle DCB + \\angle DBC$,\n\n$$\n\\begin{aligned}\n\\angle MPN &= \\angle DPM + \\angle DPN \\\\\n&= \\angle DCE + \\angle DCB + \\angle DBC \\\\\n&= \\angle BCE + \\angle DBC \\\\\n&= \\angle ACB + \\angle ACE + \\angle DBC \\\\\n&= \\angle ACB + \\angle ABD + \\angle DBC \\\\\n&= \\angle ACB + \\angle ABC\n\\end{aligned}\n$$\n\nSince $\\angle BAC = 90^\\circ$,\n\n$$\n\\angle ACB + \\angle ABC = 90^\\circ\n$$\n\nThus,\n\n$$\n\\angle MPN = 90^\\circ\n$$\n\nTherefore, $\\triangle PMN$ is an isosceles right triangle.\n\nHence,\n\n$$\nPM = PN = \\frac{1}{2} BD\n$$\n\nWhen $PM$ is maximized, the area of $\\triangle PMN$ is maximized, which occurs when $BD$ is maximized.\n\nTherefore, when point $D$ lies on the extension of $BA$, $BD$ is maximized.\n\nThus,\n\n$$\nBD = AB + AD = 14\n$$\n\nHence,\n\n$$\nPM = 7\n$$\n\nTherefore, the maximum area of $\\triangle PMN$ is:\n\n$$\nS_{\\triangle PMN \\text{ (max)}} = \\frac{1}{2} PM^2 = \\frac{1}{2} \\times 7^2 = \\frac{49}{2}\n$$\n\n**Answer:** C\n\n**Key Insight:** This problem is a comprehensive geometric transformation question, primarily testing the properties of the midsegment theorem, the properties of isosceles right triangles, the criteria for congruent triangles, and the properties of right triangles. The key is to recognize that $\\triangle ABD \\cong \\triangle ACE$ and that the area of $\\triangle PMN$ is maximized when $MN$ is maximized." }, { "problem_id": 443, "question": "As shown in Figure (1), two congruent triangles have their right-angle vertices and one of their right-angled sides overlapping. Triangle $\\triangle \\mathrm{ACB}$ is rotated clockwise around point $\\mathrm{C}$ to the position of $\\triangle A^{\\prime} C B^{\\prime}$, where $A^{\\prime} C$ intersects the straight line $\\mathrm{AD}$ at point $\\mathrm{E}$, and $A^{\\prime} B^{\\prime}$ intersects the straight lines $\\mathrm{AD}$ and $\\mathrm{AC}$ at points F and G, respectively. In Figure (2), how many pairs of congruent triangles are there?\n\n\n\n(1)\n\n\n\n(2)\nA. 5 pairs\nB. 4 pairs\nC. 3 pairs\nD. 2 pairs", "input_image": [ "batch26-2024_06_17_9b08795585671626fbc8g_0049_1.jpg", "batch26-2024_06_17_9b08795585671626fbc8g_0049_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Question Analysis: Based on the properties of rotation and the criteria for congruent triangles, we have\n\n$\\triangle A^{\\prime} C B^{\\prime} \\cong \\triangle \\mathrm{ACD}, \\quad \\Delta G B^{\\prime} C \\cong \\triangle \\mathrm{FDC}, \\quad \\triangle A^{\\prime} C A \\cong \\triangle \\mathrm{ACE}, \\quad \\triangle A^{\\prime} E F \\cong \\triangle \\mathrm{AGF}$.\n\nA total of 4 pairs. Therefore, the correct choice is B." }, { "problem_id": 444, "question": "As shown in the figure, in the Cartesian coordinate system, point $A$ has coordinates $(-\\sqrt{3}, 1)$. Point $B$ is a moving point on the $x$-axis. An equilateral triangle $A B C$ is constructed with $A B$ as a side. When point $C(x, y)$ is in the first quadrant, which of the following graphs can represent the function relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_9b892babf4cf5cef7a9ag_0037_1.jpg", "batch26-2024_06_17_9b892babf4cf5cef7a9ag_0037_2.jpg", "batch26-2024_06_17_9b892babf4cf5cef7a9ag_0037_3.jpg", "batch26-2024_06_17_9b892babf4cf5cef7a9ag_0037_4.jpg", "batch26-2024_06_17_9b892babf4cf5cef7a9ag_0037_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "On the $y$-axis, intercept $OD = 2$, draw $CF \\perp y$-axis at point $F$, connect $AD$, $CD$, $OA$, and draw $AP \\perp OB$ at $P$,\n\n\n\nSince the coordinates of point $A$ are $(-\\sqrt{3}, 1)$,\n\n$\\therefore OP = \\sqrt{3}$, $AP = 1$\n\n$\\therefore OA = \\sqrt{AP^{2} + OP^{2}} = \\sqrt{4} = 2$,\n\n$\\therefore \\sin \\angle AOP = \\frac{AP}{AO} = \\frac{1}{2}$,\n\n$\\therefore \\angle AOP = 30^{\\circ}$,\n\n$\\therefore \\angle AOD = 60^{\\circ}$,\n\n$\\therefore \\triangle AOD$ is an equilateral triangle,\n\n$\\therefore AO = AD$,\n\nSince $\\triangle ABC$ is an equilateral triangle,\n\n$\\therefore AB = AC$, $\\angle CAB = \\angle OAD = 60^{\\circ}$,\n\n$\\therefore \\angle CAD = \\angle OAB$,\n\n$\\therefore \\triangle ADC \\cong \\triangle AOB$,\n\n$\\therefore \\angle ADC = \\angle AOB = 150^{\\circ}$,\nSince $\\angle ADF = 120^{\\circ}$,\n\n$\\therefore \\angle CDF = 30^{\\circ}$,\n\n$\\therefore DF = \\sqrt{3} CF$,\n\n$\\therefore y - 2 = \\sqrt{3} x$, which is $y = \\sqrt{3} x + 2$.\n\nAlso, since $x > 0$, the functional relationship between $y$ and $x$ can be represented by option A.\n\nTherefore, choose A." }, { "problem_id": 445, "question": "In right triangle $\\triangle A B C$ as shown in Figure 1, $A C = B C$ and $\\angle C = 90^\\circ$. Point $D$ is the midpoint of side $A B$. $\\angle E D F = 90^\\circ$, and this angle is rotated around point $D$. Its sides intersect the lines containing $A C$ and $C B$ at points $E$ and $F$. There are four conclusions:\n\n(1) $C E = B F$;\n(2) $\\angle D E C + \\angle D F C = 180^\\circ$;\n(3) $E F^2 = 2 D E^2$;\n(4) As shown in Figure 2, when points $E$ and $F$ are on the extensions of $A C$ and $C B$, $S_{\\triangle D E F} - S_{\\triangle C E F} = \\frac{1}{2} S_{\\triangle A B C}$.\n\nWhich of these conclusions hold true during the rotation process?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. (1)(2)\nB. (2)(3)\nC. (1)(2)(3)\nD. (1)(3)(4)", "input_image": [ "batch26-2024_06_17_a6032b2eb5938f68b987g_0023_1.jpg", "batch26-2024_06_17_a6032b2eb5938f68b987g_0023_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:** As shown in the figure, connect $DC$.\n\n\n\nSince $AC = BC$, $\\angle ACB = 90^\\circ$, and $D$ is the midpoint of $AB$,\n\n$\\therefore \\angle B = 45^\\circ$, $\\angle DCE = \\frac{1}{2} \\angle ACB = 45^\\circ$, $CD \\perp AB$, and $CD = \\frac{1}{2} AB = BD$.\n\n$\\therefore \\angle DCE = \\angle B$, and $\\angle CDB = 90^\\circ$.\n\nSince $\\angle EDF = 90^\\circ$,\n\n$\\therefore \\angle CDE = \\angle BDF$.\n\nIn triangles $\\triangle CDE$ and $\\triangle BDF$,\n\n$$\n\\begin{aligned}\n& \\left\\{\\begin{array}{l}\n\\angle CDE = \\angle BDF \\\\\nCD = BD \\\\\n\\angle DCE = \\angle B\n\\end{array},\\right. \\\\\n& \\therefore \\triangle CDE \\cong \\triangle BDF (\\text{ASA}), \\\\\n& \\therefore CE = BF, \\angle BFD = \\angle CED, \\quad DE = DF, \\\\\n& \\therefore \\angle BFD + \\angle DFC = 180^\\circ = \\angle CED + \\angle DFC,\n\\end{aligned}\n$$\n\nAs shown in the figure, when points $E$ and $F$ lie on the extensions of $AC$ and $CB$ respectively, connect $CD$.\n\n\n\nSimilarly, it can be proven that $\\triangle DEC \\cong \\triangle DFB$.\n\n$\\therefore DE = DF$, and $\\angle DEC = \\angle DFC$, hence (1) is correct; (2) is incorrect.\n\nWhen $E$ and $F$ lie on $AC$ and $BC$ respectively,\n\n$\\because \\angle BDC = 90^\\circ$,\n\n$\\therefore \\angle BDF + \\angle CDF = \\angle CDE + \\angle CDF = 90^\\circ$,\n\n$\\therefore \\angle EDF = 90^\\circ$,\n\n$\\therefore EF^2 = DE^2 + DF^2 = 2DE^2$.\n\nWhen $E$ and $F$ lie on the extensions of $AC$ and $CB$ respectively, similarly, $EF^2 = DE^2 + DF^2 = 2DE^2$, hence (3) is correct.\n\nAs shown in the figure, connect $CD$.\n\n\n\nSimilarly, it can be proven that $\\triangle DEC \\cong \\triangle DFB$, and $\\angle DCE = \\angle DBF = 135^\\circ$.\n\n$\\therefore S_{\\triangle DEF} = S_{\\triangle CFE} + S_{\\triangle DBC} = S_{\\triangle CFE} + \\frac{1}{2} S_{\\triangle ABC}$.\n\n$\\therefore S_{\\triangle DEF} - S_{\\triangle CFE} = \\frac{1}{2} S_{\\triangle ABC}$. Hence, (4) is correct.\n\nTherefore, the correct answer is: D.\n\n**Key Insight:** This problem examines the properties of rotation, the criteria and properties of congruent triangles, the application of the Pythagorean theorem, the properties of the median to the hypotenuse of a right triangle, and the properties of isosceles right triangles. The key to solving the problem lies in the flexible application of these properties." }, { "problem_id": 446, "question": "During the Wei and Jin Dynasties, the mathematician Liu Hui used differently colored counting rods (small stick-shaped counting tools) to represent positive and negative numbers (white for positive, gray for negative) in his book \"The Nine Chapters on the Mathematical Art.\" Figure 1 represents the calculation process of $(+21)+(-32)=-11$, then Figure 2 represents the calculation process of ( )\n\n\nFigure 1\n\n\nFigure 2\nA. $(+23)+(-11)=12$\nB. $(-32)+(+11)=-21$\nC. $(-23)+(-11)=-12$\nD. $(-23)+(+11)=-12$", "input_image": [ "batch26-2024_06_17_ae8eeb8d1f5f2c348fd7g_0032_1.jpg", "batch26-2024_06_17_ae8eeb8d1f5f2c348fd7g_0032_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: According to the problem statement,\n\nThe calculation process represented in Figure 2 is: $(-23)+(+11)=-12$;\n\nTherefore, the correct choice is: D\n\n[Key Insight] This question primarily tests the understanding of positive and negative numbers, as well as general mathematical knowledge. It is a reading comprehension type of problem, where understanding the meaning of the diagram and applying it proficiently is crucial for solving the problem." }, { "problem_id": 447, "question": "As shown in Figure 1, right triangle $ABC$ is rotated counterclockwise around point $A$ by $180^\\circ$. During this process, the corresponding points of $A, B, C$ are $A, B', C$, respectively. Connect $B'C$. Let the rotation angle be $x^\\circ (0 \\leq x \\leq 180)$, and $y = B'C^2$. The graph of the function relationship between $y$ and $x$ is shown in Figure 2. When $x = 150^\\circ$, the value of $y$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 4\nB. 3\nC. $\\sqrt{3}$\nD. $\\sqrt{13}$", "input_image": [ "batch26-2024_06_17_b9382cd200010fae7901g_0091_1.jpg", "batch26-2024_06_17_b9382cd200010fae7901g_0091_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from point \\( B^{\\prime} \\) to \\( AC \\), intersecting at point \\( D \\), as shown in the figure below.\n\n\n\nFrom the problem statement, when \\( x = 0^{\\circ} \\), \\( y = 5 \\),\n\n\\[\n\\therefore BC = \\sqrt{5}.\n\\]\n\nWhen \\( x = 90^{\\circ} \\), \\( y = 1 \\), then \\( AC = AB + 1 \\).\n\nLet \\( AB = a \\), then \\( AC = a + 1 \\).\n\nSince \\( \\triangle ABC \\) is a right-angled triangle, by the Pythagorean theorem, we have \\( AB^{2} + AC^{2} = BC^{2} \\),\n\n\\[\na^{2} + (a + 1)^{2} = (\\sqrt{5})^{2}.\n\\]\n\nSolving this equation gives \\( a = 1 \\) or \\( a = -2 \\) (discarded),\n\n\\[\n\\therefore AB = a = 1, \\quad AC = a + 1 = 2.\n\\]\n\nWhen \\( x = 150^{\\circ} \\), \\( \\angle B^{\\prime}AB = 150^{\\circ} \\), \\( \\angle B^{\\prime}AD = \\angle B^{\\prime}AB - \\angle BAC = 60^{\\circ} \\),\n\n\\[\n\\angle AB^{\\prime}D = 90^{\\circ} - \\angle B^{\\prime}AD = 30^{\\circ}.\n\\]\n\n\\[\n\\therefore AD = \\frac{1}{2} AB^{\\prime} = \\frac{1}{2}, \\quad B^{\\prime}D = \\sqrt{AB^{\\prime 2} - AD^{2}} = \\frac{\\sqrt{3}}{2}, \\quad CD = AC - AD = \\frac{3}{2}.\n\\]\n\n\\[\n\\therefore y = B^{\\prime}C^{2} = B^{\\prime}D^{2} + CD^{2} = \\left(\\frac{\\sqrt{3}}{2}\\right)^{2} + \\left(\\frac{3}{2}\\right)^{2} = 3.\n\\]\n\nTherefore, the correct answer is: B.\n\n【Insight】This problem examines knowledge of rotational transformations, the Pythagorean theorem, and function graphs. The key to solving it lies in constructing equations using parameters." }, { "problem_id": 448, "question": "On the number line, points $A$, $B$, and $C$ represent numbers $a$, $1$, and $C$, respectively, and $|c-1|-|a-1|=|a-c|$. Which of the following options correctly represents the positions of points $A$, $B$, and $C$ on the number line?\n\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_d232904ecb09bec2020eg_0015_1.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0015_2.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0015_3.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0015_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "A. $|c-1|-|a-1|=c-1-a+1=c-a$, and $|a-c|=c-a$, which satisfies $|c-1|-|a-1|=|a-c|$, so it is correct;\n\nB. $|c-1|-|a-1|=1-c-a+1=2-c-a$, and $|a-c|=a-c$, which does not satisfy $|c-1|-|a-1|=|a-c|$, so it is incorrect;\n\nC. $|c-1|-|a-1|=1-c+a-1=a-c$, and $|a-c|=c-a$, which does not satisfy $|c-1|-|a-1|=|a-c|$, so it is incorrect;\n\nD. $|c-1|-|a-1|=c-1-a+1=c-a$, and $|a-c|=a-c$, which does not satisfy $|c-1|-|a-1|=|a-c|$, so it is incorrect.\n\nTherefore, the correct answer is: A.\n\n【Key Point】This question examines the concept of absolute values and number lines. Mastering the definition and properties of absolute values is crucial for solving such problems." }, { "problem_id": 449, "question": "As shown in Figure (1), a small cube is placed horizontally. Figures (2) and (3) are formed by stacking such small cubes. Following this pattern, the total number of small cubes in the seventh stacked figure should be ( ) cubes.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. 91\nB. 66\nC. 25\nD. 120", "input_image": [ "batch26-2024_06_17_d232904ecb09bec2020eg_0017_1.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0017_2.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0017_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "From the problem statement, we can deduce:\n\n- In Figure (1), there is \\(1 \\times 1 = 1\\) small cube.\n- In Figure (2), there are \\(1 \\times 2 + 4 \\times 1 = 6\\) small cubes.\n- In Figure (3), there are \\(1 \\times 3 + 4 \\times 2 + 4 \\times 1 = 15\\) small cubes.\n\nFollowing this pattern, the total number of small cubes in the seventh stacked figure should be 91.\n\nTherefore, the correct answer is A.\n\n**Key Insight:** This problem tests the student's ability to generalize from specific cases. Note that in the seventh stacked figure, the total number of small cubes is calculated as:\n\\[1 \\times 7 + 4 \\times 6 + 4 \\times 5 + 4 \\times 4 + 4 \\times 3 + 4 \\times 2 + 4 \\times 1 = 7 + 4 \\times (6 + 5 + 4 + 3 + 2 + 1) = 91\\]" }, { "problem_id": 450, "question": "During the battle against the COVID-19 pandemic, Huaxi Rongzhu Garden community implemented an identity recognition system as shown in Figure (1). Figure (2) represents the identification pattern for a certain resident. The gray small squares indicate 1, and the white small squares indicate 0. The result of the calculation $a \\times 2^{3}+b \\times 2^{2}+c \\times 2^{1}+d \\times 2^{0}$, where $a, b, c, d$ are the digits in the first row from left to right, represents the building number for that resident's house. For example, in Figure (2), the first row digits from left to right are $0,1,0,1$. Calculating this gives $0 \\times 2^{3}+1 \\times 2^{2}+0 \\times 2^{1}+1 \\times 2^{0}=5$, which means this resident is in Building 5. If Xiaomin's family lives in Building 11, the identification pattern for their house would be ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_d232904ecb09bec2020eg_0088_1.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0088_2.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0088_3.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0088_4.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0088_5.jpg", "batch26-2024_06_17_d232904ecb09bec2020eg_0088_6.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution:\n\nA. The numbers in the first row from left to right are $1, 0, 0, 1$. Since $1 \\times 2^{3} + 0 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0} = 9$,\n\n$\\therefore$ This owner is a resident of building 9, which does not match the requirement;\n\nB. The numbers in the first row from left to right are $1, 0, 1, 1$. Since $1 \\times 2^{3} + 0 \\times 2^{2} + 1 \\times 2^{1} + 1 \\times 2^{0} = 11$, $\\therefore$ This owner is a resident of building 11, which matches the requirement;\n\nC. The numbers in the first row from left to right are $0, 1, 0, 1$. Since $0 \\times 2^{3} + 1 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0} = 5$, $\\therefore$ This owner is a resident of building 5, which does not match the requirement;\n\nD. The numbers in the first row from left to right are $1, 1, 0, 1$. Since $1 \\times 2^{3} + 1 \\times 2^{2} + 0 \\times 2^{1} + 1 \\times 2^{0} = 13$, $\\therefore$ This owner is a resident of building 13, which does not match the requirement;\n\nTherefore, the correct answer is: B.\n\n[Key Insight] This question tests the practical application of mixed operations with rational numbers. Correctly understanding the problem and formulating the equation is key to solving it." }, { "problem_id": 451, "question": "As shown in the figure, in $\\triangle \\mathrm{ABC}$, $\\mathrm{AB}=\\mathrm{AC}=2$, $\\angle \\mathrm{B}=30^{\\circ}$, and $\\triangle \\mathrm{ABC}$ is rotated counterclockwise around point $\\mathrm{A}$ by an angle $\\alpha\\left(0<\\alpha<120^{\\circ}\\right)$ to form $\\triangle A B^{\\prime} C^{\\prime}$. The line segments $B^{\\prime} C^{\\prime}$ intersect $\\mathrm{BC}$ and $\\mathrm{AC}$ at points $\\mathrm{D}$ and $\\mathrm{E}$, respectively. Let $C D+D E=x$, and the area of $\\triangle A E C^{\\prime}$ be $y$. The graph of $y$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_d55f8a59b09ae1e2fc61g_0007_1.jpg", "batch26-2024_06_17_d55f8a59b09ae1e2fc61g_0007_2.jpg", "batch26-2024_06_17_d55f8a59b09ae1e2fc61g_0007_3.jpg", "batch26-2024_06_17_d55f8a59b09ae1e2fc61g_0007_4.jpg", "batch26-2024_06_17_d55f8a59b09ae1e2fc61g_0007_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Connect $\\mathrm{B}^{\\prime} \\mathrm{C}$, and draw $\\mathrm{AH} \\perp \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}$, with the foot of the perpendicular at $\\mathrm{H}$.\n\nSince $\\mathrm{AB}=\\mathrm{AC}$ and $\\angle \\mathrm{B}=30^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{C}=\\angle \\mathrm{B}=30^{\\circ}$.\n\nSince $\\triangle \\mathrm{ABC}$ is rotated counterclockwise around point $\\mathrm{A}$ by an angle $\\alpha$ ($0<\\alpha<120^{\\circ}$) to obtain $\\triangle A B^{\\prime} C^{\\prime}$,\n\nTherefore, $\\mathrm{AB}^{\\prime}=\\mathrm{AB}=\\mathrm{AC}=\\mathrm{AC}^{\\prime}=2$, and $\\angle \\mathrm{AB}^{\\prime} \\mathrm{C}^{\\prime}=\\angle \\mathrm{C}^{\\prime}=30^{\\circ}$.\n\nThus, $\\mathrm{AH}=\\frac{1}{2} \\mathrm{AC}^{\\prime}=1$,\n\nAnd $\\mathrm{C}^{\\prime} \\mathrm{H}=\\sqrt{A C^{\\prime 2}-A H^{2}}=\\sqrt{3}$,\n\nTherefore, $\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}=2 \\mathrm{C}^{\\prime} \\mathrm{H}=2 \\sqrt{3}$.\n\nSince $\\mathrm{AB}^{\\prime}=\\mathrm{AC}$,\n\nTherefore, $\\angle \\mathrm{AB}^{\\prime} \\mathrm{C}=\\angle \\mathrm{ACB}^{\\prime}$.\n\nSince $\\angle \\mathrm{AB}^{\\prime} \\mathrm{D}=\\angle \\mathrm{ACD}=30^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{AB}^{\\prime} \\mathrm{C}-\\angle \\mathrm{AB}^{\\prime} \\mathrm{D}=\\angle \\mathrm{ACB}^{\\prime}-\\angle \\mathrm{ACD}$,\n\nWhich means $\\angle \\mathrm{DB}^{\\prime} \\mathrm{C}=\\angle \\mathrm{DCB}^{\\prime}$,\n\nThus, $\\mathrm{B}^{\\prime} \\mathrm{D}=\\mathrm{CD}$.\n\nSince $\\mathrm{CD}+\\mathrm{DE}=\\mathrm{x}$,\n\nTherefore, $\\mathrm{B}^{\\prime} \\mathrm{D}+\\mathrm{DE}=\\mathrm{x}$, which means $\\mathrm{B}^{\\prime} \\mathrm{E}=\\mathrm{x}$.\n\nThus, $\\mathrm{C}^{\\prime} \\mathrm{E}=\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}-\\mathrm{B}^{\\prime} \\mathrm{E}=2 \\sqrt{3}-\\mathrm{x}$.\n\nTherefore, $\\mathrm{y}=\\frac{1}{2} C^{\\prime} E \\cdot A H=\\frac{1}{2} \\times(2 \\sqrt{3}-\\mathrm{x}) \\times 1=-\\frac{1}{2} x+\\sqrt{3}$.\n\nObserving that only the graph of option B fits the description,\n\nTherefore, choose B.\n\n\n\n【Insight】This problem tests comprehensive geometric knowledge, involving properties of rotation, determination and properties of isosceles triangles, the Pythagorean theorem, and the application of linear functions. Correctly adding auxiliary lines and proficiently mastering and flexibly applying relevant knowledge are key to solving the problem." }, { "problem_id": 452, "question": "Among the following figures, which one is a central symmetric figure? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_d5e27fbd06bdf2e2fd56g_0095_1.jpg", "batch26-2024_06_17_d5e27fbd06bdf2e2fd56g_0095_2.jpg", "batch26-2024_06_17_d5e27fbd06bdf2e2fd56g_0095_3.jpg", "batch26-2024_06_17_d5e27fbd06bdf2e2fd56g_0095_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: A is a centrally symmetric figure, which fits the question's requirement;\n\nB is not a centrally symmetric figure, thus does not meet the requirement;\n\nC is not a centrally symmetric figure, thus does not meet the requirement;\n\nD is not a centrally symmetric figure, thus does not meet the requirement.\n\nTherefore, the correct choice is: A.\n\n[Key Point] This question primarily tests the recognition of centrally symmetric figures. Correctly understanding the definition of a centrally symmetric figure is crucial for solving the problem." }, { "problem_id": 453, "question": "In Figure (1), $\\triangle \\mathrm{ABC}$ and $\\triangle \\mathrm{ADE}$ are both isosceles right triangles, with $\\angle \\mathrm{ACB}$ and $\\angle \\mathrm{D}$ both being right angles. Point $C$ is on $\\mathrm{AE}$. After $\\triangle \\mathrm{ABC}$ is rotated counterclockwise around point $\\mathrm{A}$, it coincides with $\\triangle \\mathrm{AED}$. Then, the Figure (1) is taken as the \"base figure\" and rotated counterclockwise around point A to obtain Figure (2). The angles of these two rotations are ( )\n\n\n\n(1)\n\n\n\n(2)\nA. $45^\\circ, 90^\\circ$\nB. $90^\\circ, 45^\\circ$\nC. $60^\\circ, 30^\\circ$\nD. $30^\\circ, 60^\\circ$", "input_image": [ "batch26-2024_06_17_d641fb53fe07ff163a07g_0062_1.jpg", "batch26-2024_06_17_d641fb53fe07ff163a07g_0062_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since both $\\triangle ABC$ and $\\triangle ADE$ are isosceles right triangles, $\\therefore \\angle CAB = \\angle DAE = 45^{\\circ}$.\n\nMoreover, since $\\triangle ABC$ can coincide with $\\triangle AED$ by rotating $x$ degrees counterclockwise around point $A$, $\\therefore$ the center of rotation is point $A$, and the rotation angle is $45^{\\circ}$, i.e., $x = 45$. If we take Figure (1) as the \"basic figure\" and rotate it $y$ degrees counterclockwise around point $A$ to obtain Figure (2), then $y = 45 + 45 = 90$, hence the correct choice is A." }, { "problem_id": 454, "question": "Congruent triangles are also called similar triangles. In a plane, similar triangles are divided into true similar triangles and mirror similar triangles. Suppose $\\triangle A B C$ and $\\triangle A_{1} B_{1} C_{1}$ are similar triangles, with points $A$ corresponding to $A_{1}$, $B$ to $B_{1}$, and $C$ to $C_{1}$. When moving along the boundaries $A \\rightarrow B \\rightarrow C \\rightarrow A$ and $A_{1} \\rightarrow B_{1} \\rightarrow C_{1} \\rightarrow A_{1}$, if the directions of motion are the same, they are called true similar triangles (as shown in Figure 1). If the directions of motion are opposite, they are called mirror similar triangles (as shown in Figure 2). Two true similar triangles can be superimposed by translation or rotation in the plane, while for two mirror similar triangles to overlap, one must be flipped $180^\\circ$. Among the following combinations of similar triangles, which are mirror similar triangles?\n\n\nFigure (1)\n\n\nFigure (2)\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_e775ac44b0bd009d76dag_0085_1.jpg", "batch26-2024_06_17_e775ac44b0bd009d76dag_0085_2.jpg", "batch26-2024_06_17_e775ac44b0bd009d76dag_0085_3.jpg", "batch26-2024_06_17_e775ac44b0bd009d76dag_0085_4.jpg", "batch26-2024_06_17_e775ac44b0bd009d76dag_0085_5.jpg", "batch26-2024_06_17_e775ac44b0bd009d76dag_0085_6.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the problem statement, we understand the characteristics of truly congruent triangles and mirror congruent triangles. It can be determined that to make the two triangles in group C coincide, one of them must be flipped by 180 degrees; whereas, the congruent triangles in other groups can be made to coincide through translation or rotation within the plane.\n\nTherefore, the correct choice is: C.\n\n[Insight] This question tests the understanding of graphical transformations such as translation, rotation, and symmetry, as well as the student's reading comprehension and spatial imagination skills, which are quite flexible. Carefully reading the problem and thoroughly understanding the requirements are key to correctly solving this problem." }, { "problem_id": 455, "question": "Mr. Zhang presented a problem in class: As shown in Figure (1), in quadrilateral $A B C D$, $A B=A D$, $\\angle A=\\angle C=90^{\\circ}$, given that $S_{\\text {quadrilateral } A B C D}=9$ and $C D=2$, find the length of $A D$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\nXiao Ming's approach: Draw $A E \\perp B C$ at point $E$ as shown in Figure (2), rotate $\\triangle A B E$ counterclockwise around point $A$ by $90^{\\circ}$ to get $\\triangle A D F$, converting the irregular quadrilateral into a square for solution. Using Xiao Ming's method, solve the problem: In right triangle $A B C$ as shown in Figure (3), $A C=B C$, $\\angle A C B=90^{\\circ}$, point $M$ is inside $\\triangle A B C$, and $A M=\\sqrt{5}$, $B M=3$, $C M=\\sqrt{2}$. The area of the shaded region is $(\\quad)$\n\nA. $\\sqrt{5}$\nB. $\\sqrt{5}+1$\nC. $\\sqrt{5}+2$\nD. $2 \\sqrt{5}+2$\n\n##", "input_image": [ "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0046_1.jpg", "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0046_2.jpg", "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0046_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, rotate $\\triangle BCM$ counterclockwise by $90^{\\circ}$ around point $C$ to obtain $\\triangle ACN$, then connect $MN$. Thus, $\\angle MCN = 90^{\\circ}$, $CN = CM = \\sqrt{2}$, and $AN = BM = 3$.\n\n\n\nTherefore, $MN = \\sqrt{CM^{2} + CN^{2}} = 2$.\n\nSince $AM = \\sqrt{5}$,\n\nThen, $AM^{2} + MN^{2} = (\\sqrt{5})^{2} + 2^{2} = 9 = AN^{2}$,\n\nHence, $\\angle AMN = 90^{\\circ}$,\n\nThus, the area $S_{\\text{shadow}} = S_{\\triangle BCM} + S_{\\triangle ACM} = S_{\\triangle ACN} + S_{\\triangle ACM} = S_{\\triangle CMN} + S_{\\triangle AMN} = \\frac{1}{2} \\times (\\sqrt{2})^{2} + \\frac{1}{2} \\times \\sqrt{5} \\times 2 = \\sqrt{5} + 1$.\n\nTherefore, the answer is: B\n\n[Key Insight] This problem primarily examines the rotation of shapes, the Pythagorean theorem, and its converse. Mastering the properties of shape rotation and the Pythagorean theorem and its converse is crucial for solving the problem." }, { "problem_id": 456, "question": "As shown in the figure, $\\triangle A B C$ is transformed into $\\triangle A B^{\\prime} C^{\\prime}$ by rotation or reflection. Among the following options, which one is $\\triangle A B C$ rotated counterclockwise around point $A$ by $60^{\\circ}$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0092_1.jpg", "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0092_2.jpg", "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0092_3.jpg", "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0092_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Option $\\mathrm{A}$ represents rotating $\\triangle ABC$ counterclockwise around point $A$ by $90^{\\circ}$ to obtain $\\triangle AB^{\\prime}C^{\\prime}$, so $\\mathrm{A}$ does not meet the requirement;\n\nOption $\\mathrm{B}$ represents reflecting $\\triangle ABC$ over a certain line to obtain $\\triangle AB^{\\prime}C^{\\prime}$, so B does not meet the requirement;\n\nOption $\\mathrm{C}$ represents reflecting $\\triangle ABC$ over a certain line to obtain $\\triangle AB^{\\prime}C^{\\prime}$, so $\\mathrm{C}$ does not meet the requirement;\n\nOption D represents rotating $\\triangle ABC$ counterclockwise around point $A$ by $60^{\\circ}$ to obtain $\\triangle AB^{\\prime}C^{\\prime}$, so D meets the requirement;\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question tests the concepts of reflection and rotation transformations. Understanding the characteristics of reflection and rotation transformations is key to solving the problem." }, { "problem_id": 457, "question": "In $\\triangle A B C$ as shown, $\\angle A C B = 90^\\circ$. Using a compass and straightedge, a perpendicular bisector $C P$ of the hypotenuse is constructed. Which of the following methods is definitely correct? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_04279b26e178d3a91c87g_0041_1.jpg", "batch28-2024_06_17_04279b26e178d3a91c87g_0041_2.jpg", "batch28-2024_06_17_04279b26e178d3a91c87g_0041_3.jpg", "batch28-2024_06_17_04279b26e178d3a91c87g_0041_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. From the construction, it can be seen that \\( CP = BC \\), which does not satisfy the condition that point \\( P \\) is the midpoint of \\( AB \\). Therefore, it does not meet the requirement of the problem.\n\nB. From the construction, it can be seen that \\( BP = BC \\), which does not satisfy the condition that point \\( P \\) is the midpoint of \\( AB \\). Therefore, it does not meet the requirement of the problem.\n\nC. From the construction, it can be seen that point \\( P \\) is the midpoint of \\( AB \\). Therefore, it meets the requirement of the problem.\n\nD. From the construction, it can be seen that \\( CP \\) bisects \\( \\angle ACB \\). Therefore, it does not meet the requirement of the problem.\n\nThus, the correct answer is C.\n\n**Key Insight**: This problem primarily tests the theorem of the median to the hypotenuse of a right triangle and the construction of the perpendicular bisector of a segment, as well as the construction of an angle bisector. Mastering the use of a compass and straightedge for these constructions is crucial for solving the problem." }, { "problem_id": 458, "question": "As shown in the figure, point $P$ is a moving point on the side of the rhombus $A B C D$. It starts from point $A$ and moves along the path $A \\rightarrow B \\rightarrow C \\rightarrow D$ at a uniform speed until it reaches point $D$. Let the area of $\\triangle P A D$ be $y$, and the time of movement of point $P$ be $x$. The graph of $y$ as a function of $x$ would approximately look like $(\\quad)$.\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_2af8f22c2462c71c8b73g_0002_1.jpg", "batch28-2024_06_17_2af8f22c2462c71c8b73g_0002_2.jpg", "batch28-2024_06_17_2af8f22c2462c71c8b73g_0002_3.jpg", "batch28-2024_06_17_2af8f22c2462c71c8b73g_0002_4.jpg", "batch28-2024_06_17_2af8f22c2462c71c8b73g_0002_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: There are three cases to consider:\n\n(1) When point \\( P \\) is on side \\( AB \\), as shown in Figure 1,\n\n\n\nFigure 1\n\nLet the height of the rhombus be \\( h \\),\n\n\\[ y = \\frac{1}{2} \\cdot AP \\cdot h, \\]\n\nSince \\( AP \\) increases as \\( x \\) increases, and \\( h \\) remains constant,\n\n\\[ y \\] has a linear relationship with \\( x \\),\n\nTherefore, options C and D are incorrect;\n\n(2) When \\( P \\) is on side \\( BC \\), as shown in Figure 2,\n\n\n\nFigure 2\n\n\\[ y = \\frac{1}{2} \\cdot AD \\cdot h, \\]\n\nBoth \\( AD \\) and \\( h \\) remain unchanged,\n\nThus, during this process, \\( y \\) remains constant,\n\nTherefore, option A is incorrect;\n\n(3) When \\( P \\) is on side \\( CD \\), as shown in Figure 3,\n\n\n\nFigure 3\n\n\\[ y = \\frac{1}{2} \\cdot PD \\cdot h, \\]\n\nSince \\( PD \\) decreases as \\( x \\) increases, and \\( h \\) remains constant,\n\n\\[ y \\] decreases as \\( x \\) increases,\n\nGiven that point \\( P \\) moves uniformly from point \\( A \\) along the path \\( A \\rightarrow B \\rightarrow C \\rightarrow D \\) to point \\( D \\),\n\nThe time \\( P \\) spends moving on each of the three segments is the same,\n\nTherefore, option B is correct;\n\nHence, the correct answer is: B.\n\n【Key Insight】This problem examines the function graph of a moving point and the properties of a rhombus. The key to solving it lies in applying the classification discussion method based on the different positions of point \\( P \\), and deriving the expression for the area of \\( \\triangle PAD \\) in three segments." }, { "problem_id": 459, "question": "In the figure below, (1) is a work of artist M. C. Escher, who combined mathematics and painting to create a three-dimensional effect on a flat surface. (2) is a rhombus. By cutting off a smaller rhombus with a side length half that of the original, we obtain (3), which is then used to tile to form (4). After coloring (4), tiling it again results in (1). The measure of $\\angle A B C$ in figure (4) is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\nA. $30^{\\circ}$\nB. $45^{\\circ}$\nC. $60^{\\circ}$\nD. $75^{\\circ}$", "input_image": [ "batch28-2024_06_17_2af8f22c2462c71c8b73g_0004_1.jpg", "batch28-2024_06_17_2af8f22c2462c71c8b73g_0004_2.jpg", "batch28-2024_06_17_2af8f22c2462c71c8b73g_0004_3.jpg", "batch28-2024_06_17_2af8f22c2462c71c8b73g_0004_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\n$\\because \\angle B A D=\\angle B A E=\\angle D A E$,\n\nand $\\because \\angle B A D+\\angle B A E+\\angle D A E=360^{\\circ}$,\n\n$\\therefore \\angle B A D=\\angle B A E=\\angle D A E=120^{\\circ}$,\n\n$\\because B C / / A D$,\n\n$\\therefore \\angle A B C+\\angle B A D=180^{\\circ}$,\n\n$\\therefore \\angle A B C=180^{\\circ}-120^{\\circ}=60^{\\circ}$, hence option C is correct.\n\nTherefore, the answer is: C.\n\n【Key Insight】This question primarily examines the properties of a rhombus and the student's ability to read and understand the problem. Understanding the problem, accurately interpreting the diagram, and determining the measure of $\\angle B A D$ are crucial to solving the problem." }, { "problem_id": 460, "question": "Given the figure $X$ and the line $l$ in the diagram, the reflection of figure $X$ across line $l$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch28-2024_06_17_87fa3b0abb7db750811bg_0049_1.jpg", "batch28-2024_06_17_87fa3b0abb7db750811bg_0049_2.jpg", "batch28-2024_06_17_87fa3b0abb7db750811bg_0049_3.jpg", "batch28-2024_06_17_87fa3b0abb7db750811bg_0049_4.jpg", "batch28-2024_06_17_87fa3b0abb7db750811bg_0049_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: The axisymmetric figure of the given shape is\n\n\n\nTherefore, the correct choice is: $C$.\n\n[Key Insight] This question tests the concept of axisymmetric figures. The key to identifying an axisymmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide when folded along the axis." }, { "problem_id": 461, "question": "As shown in the figure, a cube has not been completely unfolded. If one more edge is cut, which of the following plane unfoldings is impossible?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_c8bd428e774ea485a3ebg_0006_1.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0006_2.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0006_3.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0006_4.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0006_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Cutting along the back and bottom surfaces yields $\\mathrm{C}$, cutting along the back and right surfaces yields $\\mathrm{A}$, and cutting along the bottom and right surfaces yields $\\mathrm{D}$. Therefore, the planar development cannot be B.\n\nHence, the correct choice is B.\n\n[Key Insight] This question examines the surface development of a cube. A cube has a total of 11 distinct surface developments, and it is important to consider different scenarios in the discussion." }, { "problem_id": 462, "question": "Among the four figures below, which figure can represent the same angle using all three methods of $\\angle 1, \\angle \\mathrm{ABC}, \\angle \\mathrm{B}$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_c8bd428e774ea485a3ebg_0016_1.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0016_2.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0016_3.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0016_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The figure that can represent the same angle using all three notations $\\angle 1$, $\\angle \\mathrm{ABC}$, and $\\angle \\mathrm{B}$ simultaneously can only be B. Therefore, the correct choice is B.\n\n[Highlight] This question tests the understanding of angle representation methods." }, { "problem_id": 463, "question": "Using two identical $30^\\circ$-angled triangles to draw parallel lines, the following drawing of lines $a$ and $b$ that may not be parallel is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_1342a555d4b736320128g_0003_1.jpg", "batch29-2024_06_14_1342a555d4b736320128g_0003_2.jpg", "batch29-2024_06_14_1342a555d4b736320128g_0003_3.jpg", "batch29-2024_06_14_1342a555d4b736320128g_0003_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. According to the equality of corresponding angles, the two lines are parallel, hence $a / / b$; therefore, it does not meet the requirement of the question;\n\nB. According to the equality of alternate interior angles, the two lines are parallel, hence $a / / b$; therefore, it does not meet the requirement of the question;\n\nC. The drawn lines $a$ and $b$ are not necessarily parallel; therefore, it meets the requirement of the question;\n\nD. According to the equality of alternate interior angles, the two lines are parallel, hence $a / / b$; therefore, it does not meet the requirement of the question; thus, choose C.\n\n[Key Point] This question tests the determination of parallel lines. Mastering the theorem for determining parallel lines is crucial for solving the problem." }, { "problem_id": 464, "question": "In the figures below, which pair of angles $\\angle 1$ and $\\angle 2$ are vertical angles?\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_1342a555d4b736320128g_0025_1.jpg", "batch29-2024_06_14_1342a555d4b736320128g_0025_2.jpg", "batch29-2024_06_14_1342a555d4b736320128g_0025_3.jpg", "batch29-2024_06_14_1342a555d4b736320128g_0025_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: \n\nA. The vertices of $\\angle 1$ and $\\angle 2$ are not the same, so they are not vertical angles. This option does not meet the requirement.\n\nB. One side of $\\angle 1$ and $\\angle 2$ is not the reverse extension of the other, so they are not vertical angles. This option does not meet the requirement.\n\nC. $\\angle 1$ and $\\angle 2$ are vertical angles, so this option meets the requirement.\n\nD. One side of $\\angle 1$ and $\\angle 2$ is not the reverse extension of the other, so they are not vertical angles. This option does not meet the requirement.\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the judgment of vertical angles. Vertical angles share a common vertex, and the sides of one angle are the reverse extensions of the sides of the other angle. Two angles with this positional relationship are called vertical angles. The key to solving the problem is to thoroughly understand the definition and make correct judgments." }, { "problem_id": 465, "question": "In the figures below, which pair of angles, $\\angle 1$ and $\\angle 2$, are adjacent supplementary angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_17194f2d3b6db9da2c2cg_0017_1.jpg", "batch29-2024_06_14_17194f2d3b6db9da2c2cg_0017_2.jpg", "batch29-2024_06_14_17194f2d3b6db9da2c2cg_0017_3.jpg", "batch29-2024_06_14_17194f2d3b6db9da2c2cg_0017_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: \nA. The other side does not have a reverse extension line, so it is not an adjacent supplementary angle. Therefore, A does not meet the requirement of the question. \n\nB. It is an adjacent supplementary angle, so B meets the requirement of the question. \n\nC. It is not an angle formed by the intersection of two straight lines, so C does not meet the requirement of the question. \n\nD. It is not an angle formed by the intersection of two straight lines, so D does not meet the requirement of the question. \n\nThus, the correct answer is B. \n\n**Key Insight:** This question tests the definition of adjacent supplementary angles. Correctly grasp the definition: two angles are adjacent supplementary angles only if they share a common side, and the other sides of the angles are reverse extensions of each other. \n\n## 2. Fill-in-the-Blank Questions" }, { "problem_id": 466, "question": "As shown in the figure, in the Cartesian coordinate system, quadrilateral $A B C D$ is a rhombus. The coordinates of point $B$ are $(0,4)$, and the coordinates of point $D$ are $(8 \\sqrt{3}, 4)$. Points $M$ and $N$ are two moving points. Point $M$ starts from point $B$ and moves along $B A$ at a constant speed of 2 units per second until it reaches point $A$, where it stops. At the same time, point $N$ starts from point $B$ and moves along the broken line $B C \\rightarrow C D$ at a constant speed of 4 units per second until it reaches point $D$, where it stops. If one point stops moving, the other point also stops moving. Let the time for points $M$ and $N$ to move be $x$, and the area of $\\triangle B M N$ be $y$. Which of the following graphs approximately represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch29-2024_06_14_24cd39b7ca7ae27652ccg_0027_1.jpg", "batch29-2024_06_14_24cd39b7ca7ae27652ccg_0027_2.jpg", "batch29-2024_06_14_24cd39b7ca7ae27652ccg_0027_3.jpg", "batch29-2024_06_14_24cd39b7ca7ae27652ccg_0027_4.jpg", "batch29-2024_06_14_24cd39b7ca7ae27652ccg_0027_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When \\(0 < x \\leqslant 2\\), as shown in Figure 1:\n\n\n\nFigure 1\n\nConnect \\(\\mathrm{BD}\\) and \\(\\mathrm{AC}\\), intersecting at point \\(\\mathrm{O}^{\\prime}\\). Connect \\(\\mathrm{NM}\\), and draw \\(\\mathrm{CP} \\perp \\mathrm{AB}\\) from point \\(\\mathrm{C}\\), with the foot of the perpendicular at point \\(\\mathrm{P}\\).\n\n\\(\\therefore \\angle \\mathrm{CPB} = 90^{\\circ}\\).\n\nSince quadrilateral \\(\\mathrm{ABCD}\\) is a rhombus, with point \\(\\mathrm{B}\\) at coordinates \\((0,4)\\) and point \\(\\mathrm{D}\\) at \\((8 \\sqrt{3}, 4)\\),\n\n\\(\\therefore \\mathrm{BO}^{\\prime} = 4 \\sqrt{3}\\), \\(\\mathrm{CO}^{\\prime} = 4\\),\n\n\\(\\therefore \\mathrm{BC} = \\mathrm{AB} = \\sqrt{O^{\\prime} B^{2} + O^{\\prime} C^{2}} = 8\\).\n\nSince \\(\\mathrm{AC} = 8\\),\n\n\\(\\therefore \\triangle \\mathrm{ABC}\\) is an equilateral triangle,\n\n\\(\\therefore \\angle \\mathrm{ABC} = 60^{\\circ}\\).\n\n\\(\\therefore \\mathrm{CP} = \\mathrm{BC} \\times \\sin 60^{\\circ} = 8 \\times \\frac{\\sqrt{3}}{2} = 4 \\sqrt{3}\\), and \\(\\mathrm{BP} = 4\\).\n\n\\(\\mathrm{BN} = 4x\\), \\(\\mathrm{BM} = 2x\\),\n\n\\(\\frac{BM}{BP} = \\frac{2x}{4} = \\frac{x}{2}\\), \\(\\frac{BN}{BC} = \\frac{x}{2}\\),\n\n\\(\\therefore \\frac{BM}{BP} = \\frac{BN}{BC}\\).\n\nAlso, since \\(\\angle \\mathrm{NBM} = \\angle \\mathrm{CBP}\\),\n\n\\(\\therefore \\triangle \\mathrm{NBM} \\sim \\triangle \\mathrm{CBP}\\),\n\n\\(\\therefore \\angle \\mathrm{NMB} = \\angle \\mathrm{CPB} = 90^{\\circ}\\),\n\n\\(\\therefore S_{\\triangle CBP} = \\frac{1}{2} \\times BP \\times CP = \\frac{1}{2} \\times 4 \\times 4 \\sqrt{3} = 8 \\sqrt{3}\\);\n\n\\(\\therefore \\frac{S_{\\triangle NBM}}{S_{\\triangle CBP}} = \\left(\\frac{BN}{BC}\\right)^{2}\\),\n\ni.e., \\(\\mathrm{y} = S_{\\triangle NBM} = S_{\\triangle CBP} \\times \\left(\\frac{BN}{BC}\\right)^{2} = 8 \\sqrt{3} \\times \\left(\\frac{x}{2}\\right)^{2} = 2 \\sqrt{3} x^{2}\\).\n\nWhen \\(2 < x \\leqslant 4\\), draw \\(NE \\perp AB\\), with the foot of the perpendicular at \\(E\\).\n\n\n\nFigure 2\n\nSince quadrilateral \\(\\mathrm{ABCD}\\) is a rhombus,\n\n\\(\\therefore \\mathrm{AB} \\parallel \\mathrm{CD}\\),\n\n\\(\\therefore \\mathrm{NE} = \\mathrm{CP} = 4 \\sqrt{3}\\),\n\n\\(\\mathrm{BM} = 2x\\),\n\n\\(\\therefore \\mathrm{y} = \\frac{1}{2} \\times BM \\times NE = \\frac{1}{2} \\cdot 2x \\cdot 4 \\sqrt{3} = 4 \\sqrt{3} x\\);\n\nTherefore, the answer is D.\n\n【Key Point】This problem mainly examines the function graph of moving point problems. Mastering the function graph of moving point problems is the key to solving the problem." }, { "problem_id": 467, "question": "In the following four figures, which pair of angles, $\\angle 1$ and $\\angle 2$, are interior alternate angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_280938cab17da31e4f9eg_0015_1.jpg", "batch29-2024_06_14_280938cab17da31e4f9eg_0015_2.jpg", "batch29-2024_06_14_280938cab17da31e4f9eg_0015_3.jpg", "batch29-2024_06_14_280938cab17da31e4f9eg_0015_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "In option A, $\\angle 1$ and $\\angle 2$ are not alternate interior angles, so it is incorrect;\n\nIn option B, $\\angle 1$ and $\\angle 2$ are alternate interior angles, so it is correct;\n\nIn option C, $\\angle 1$ and $\\angle 2$ are not alternate interior angles, so it is incorrect;\n\nIn option D, $\\angle 1$ and $\\angle 2$ are not alternate interior angles, so it is incorrect;\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question mainly tests the understanding of alternate interior angles. The key to solving the problem is to distinguish between corresponding angles, alternate interior angles, and consecutive interior angles." }, { "problem_id": 468, "question": "In the following figures, which pair of angles $\\angle 1$ and $\\angle 2$ are not corresponding angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_280938cab17da31e4f9eg_0077_1.jpg", "batch29-2024_06_14_280938cab17da31e4f9eg_0077_2.jpg", "batch29-2024_06_14_280938cab17da31e4f9eg_0077_3.jpg", "batch29-2024_06_14_280938cab17da31e4f9eg_0077_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Because\n\n\nTherefore, the answer is: B.\n\n[Highlight] This question tests the definition of corresponding angles: two angles located on the same side of two lines and on the same side of a transversal. Mastering the definition is key to solving the problem." }, { "problem_id": 469, "question": "As shown in the figure, among these quadrilaterals, the one where sides $\\mathrm{AB}$ are not parallel to $\\mathrm{CD}$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0018_1.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0018_2.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0018_3.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0018_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "A is a parallelogram, from which it can be deduced that the opposite sides $\\mathrm{AB}$ and $\\mathrm{CD}$ are parallel;\n\nB is a trapezoid, from which it can be deduced that the top base $\\mathrm{AB}$ is parallel to the bottom base $\\mathrm{CD}$;\n\nC is a square, from which it can be deduced that the opposite sides $\\mathrm{AB}$ and $\\mathrm{CD}$ are parallel;\n\nD is a general quadrilateral, where $\\mathrm{AB}$ is not parallel to $\\mathrm{CD}$.\n\nTherefore, the correct choice is: D." }, { "problem_id": 470, "question": "The correct method for drawing a perpendicular segment from point P to segment AB is ( )\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0035_1.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0035_2.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0035_3.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0035_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: The distance from a point to a line refers to the length of the perpendicular segment drawn from a point outside the line to the known line. The perpendicular segment must form a right angle, hence the correct answer is D." }, { "problem_id": 471, "question": "In the following figures, which pair of angles, $\\angle 1$ and $\\angle 2$, are interior alternate angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0094_1.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0094_2.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0094_3.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0094_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: By observing the figure, it is noted that only in option B are the two angles located between two straight lines and on opposite sides of another straight line, hence $\\angle 1$ and $\\angle 2$ are alternate interior angles. In the other options, $\\angle 1$ and $\\angle 2$ do not form a pair of alternate interior angles.\n\nTherefore, the correct choice is: B." }, { "problem_id": 472, "question": "A math club organized a hands-on activity and designed the three shapes shown in the figure. They plan to create corresponding structures using wire according to the shapes. Which of the following statements best describes the relationship between the lengths of the wire used for each design?\n\n\nA\n\n\nB\n\n\nC\nA. The wire used for the A design is the longest.\nB. The wire used for the B design is the longest.\nC. The wire used for the C design is the longest.\nD. The wire used for all three designs is the same length.", "input_image": [ "batch29-2024_06_14_4cdc77ea69d91e19c2beg_0067_1.jpg", "batch29-2024_06_14_4cdc77ea69d91e19c2beg_0067_2.jpg", "batch29-2024_06_14_4cdc77ea69d91e19c2beg_0067_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Question Analysis:**\n\n**Solution:** From the diagram, we can deduce the following:\n\n- The length of wire used by **A** is: $2a + 2b$,\n- The length of wire used by **B** is: $2a + 2b$,\n- The length of wire used by **C** is: $2a + 2b$.\n\nTherefore, the length of wire used in all three schemes is the same.\n\nHence, the correct answer is **D**.\n\n**Exam Point:** Translational phenomena in everyday life." }, { "problem_id": 473, "question": "A rectangle is divided into three right-angled triangles as shown in Figure 1. The larger two triangular paper pieces are arranged in two ways as shown in Figure 2 (1) and (2). Let the area of the shaded region in (1) be $S_{1}$, and the area of the shaded region in (2) be $S_{2}$. When $S_{2}=S_{1}$, the ratio of the longer to the shorter sides of the rectangle is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2 (1)\n\n\n\nFigure 2 (2)\nA. 2\nB. $\\sqrt{2}$\nC. $\\frac{4}{3}$\nD. $\\sqrt{3}$", "input_image": [ "batch29-2024_06_14_5f55591362ac80f6b8e2g_0024_1.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0024_2.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0024_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of the rectangle be \\( AB = a \\), the width be \\( BC = b \\), and \\( DE = c \\).\n\n\n\nFigure 1\n\nSince quadrilateral \\( \\mathrm{ABCD} \\) is a rectangle,\n\n\\[ S_{\\triangle ABC} = S_{\\triangle CDA} \\]\n\nThat is,\n\n\\[ \\frac{1}{2} a b = \\frac{1}{2} \\sqrt{a^{2}+b^{2}} \\cdot c \\]\n\nThus,\n\n\\[ c = \\frac{a b}{\\sqrt{a^{2}+b^{2}}} \\]\n\nIn (1), \\( \\triangle DEF \\) is an isosceles triangle,\n\n\n\nFigure 2(1)\n\nTherefore, its height is\n\n\\[ \\frac{1}{2} DE \\cdot \\tan \\angle CDE = \\frac{a b}{2 \\sqrt{a^{2}+b^{2}}} \\cdot \\frac{a}{b} = \\frac{a^{2}}{2 \\sqrt{a^{2}+b^{2}}} \\]\n\nThus,\n\n\\[ S_{1} = \\frac{1}{2} \\cdot \\frac{a^{2}}{2 \\sqrt{a^{2}+b^{2}}} \\cdot \\frac{a b}{\\sqrt{a^{2}+b^{2}}} \\]\n\nIn (2), the area of the shaded part is the area of the smallest triangle in the figure,\n\nThe smallest triangle is similar to the largest triangle, and the area ratio is the square of the similarity ratio,\n\nThus,\n\n\\[ S_{2} = \\frac{1}{2} a b \\cdot \\left( \\frac{b}{\\sqrt{a^{2}+b^{2}}} \\right)^{2} \\]\n\nWhen \\( S_{2} = S_{1} \\), that is,\n\n\\[ \\frac{1}{2} a b \\cdot \\left( \\frac{b}{\\sqrt{a^{2}+b^{2}}} \\right)^{2} = \\frac{1}{2} \\cdot \\frac{a^{2}}{2 \\sqrt{a^{2}+b^{2}}} \\cdot \\frac{a b}{\\sqrt{a^{2}+b^{2}}} \\]\n\nSimplifying, we get\n\n\\[ \\frac{a}{b} = \\sqrt{2} \\]\n\nTherefore, the answer is: B.\n\n【Highlight】This question tests the understanding of solving right triangles, the determination and properties of similarity, etc. Proficiency in applying various theorems is key to solving the problem." }, { "problem_id": 474, "question": "In rhombus $A B C D$ as shown in Figure (1), $\\angle A = 120^\\circ$, point $E$ is the midpoint of side $B C$, and point $F$ is a moving point on the diagonal $B D$. Let the length of $F D$ be $x$ and the sum of the lengths of $E F$ and $C F$ be $y$. Figure (2) shows the graph of $y$ as a function of $x$. Point $P$ is the lowest point on the graph. The coordinates of the right endpoint $Q$ of the function graph are ( ).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $(6,4 \\sqrt{3})$\nB. $(4 \\sqrt{3}, 3 \\sqrt{3})$\nC. $(4 \\sqrt{3}, 6)$\nD. $(6,3 \\sqrt{3})$", "input_image": [ "batch29-2024_06_14_5f55591362ac80f6b8e2g_0069_1.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0069_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Connect $AF$, as shown in the figure,\n\n\n\n$\\because$ In rhombus $ABCD$, point $A$ and point $C$ are symmetric with respect to $BD$,\n\n$\\therefore AF = CF$,\n\n$\\therefore y = EF + CF = EF + AF$,\n\nWhen points $A$, $F$, and $E$ are collinear, $y$ reaches its minimum value, which is the length of segment $AE$,\n\nFrom Figure (2), we know that at this point $x = 4$, i.e., $F_{1}D = 4$, and in the rhombus, point $E$ is the midpoint of side $BC$,\n\nIt is easy to deduce that $AE \\perp BC$, $EA \\perp AD$,\n\n$\\because \\angle A = 120^{\\circ}$, $AB = AD$,\n\n$\\therefore \\angle ADB = 30^{\\circ}$,\n\n$\\therefore BC = AD = F_{1}D \\cdot \\cos \\angle ADB = 2 \\sqrt{3}$,\n\n$\\because AD / / BC$,\n$\\therefore \\triangle ADF_{1} \\sim \\triangle EBF_{1}$,\n\n$\\therefore \\frac{F_{1}D}{F_{1}B} = \\frac{AD}{BE} = 2$,\n\n$\\therefore F_{1}B = \\frac{1}{2} F_{1}D = 2$, $BE = \\frac{1}{2} BC = \\sqrt{3}$,\n\n$\\therefore BD = F_{1}B + F_{1}D = 6$,\n\nWhen point $F$ coincides with point $B$, $x$ reaches its maximum value of $6$, and $y = EF + CF = EB + CB = \\sqrt{3} + 2 \\sqrt{3} = 3 \\sqrt{3}$.\n\n$\\therefore$ The coordinates of point $Q$ are $(6, 3 \\sqrt{3})$,\n\nTherefore, the correct choice is D.\n\n【Key Insight】This problem examines the function image of moving points, the properties of a rhombus, and the solution of right triangles. The key to solving this problem is to understand the given conditions, identify the necessary conditions for the problem, and use a combination of numerical and graphical methods to find the solution." }, { "problem_id": 475, "question": "As shown in the figure, the sides of the rhombus $A B C D$ are $5 \\mathrm{~cm}$ long, and $\\sin A=\\frac{4}{5}$. Point $P$ starts from point $A$ and moves along the broken line $A B-B C-C D$ at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$ until it reaches point $D$. Point $Q$ also starts from point $A$ at the same time and moves along $A D$ at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$ until it reaches point $D$. Let the area of $\\triangle A P Q$ be $y\\left(\\mathrm{~cm}^{2}\\right)$ when point $P$ has traveled for $x$ (s). Which of the following graphs best represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_5f55591362ac80f6b8e2g_0071_1.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0071_2.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0071_3.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0071_4.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0071_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the side length of rhombus \\( A B C D \\) is \\( 5 \\mathrm{~cm} \\), and the speeds of points \\( P \\) and \\( Q \\) are both \\( 1 \\mathrm{~cm/s} \\),\n\nWhen \\( 0 \\leq x \\leq 5 \\), both points \\( Q \\) and \\( P \\) are moving, and the area \\( y \\) is given by:\n\\[\ny = \\frac{x \\cdot \\frac{4}{5} x}{2} = \\frac{2}{5} x^{2},\n\\]\nThus, options \\( A \\) and \\( D \\) are incorrect.\n\nWhen \\( 5 < x \\leq 10 \\), point \\( Q \\) stops while point \\( P \\) continues to move, and the height remains constant. The area \\( y \\) is:\n\\[\ny = \\frac{5 \\times 5 \\times \\frac{4}{5}}{2} = 10.\n\\]\n\nWhen \\( 10 < x \\leq 15 \\), point \\( Q \\) remains stationary while point \\( P \\) moves, and the area \\( y \\) is:\n\\[\ny = \\frac{5 \\times (5 \\times 3 - x) \\times \\frac{4}{5}}{2} = 30 - 2x.\n\\]\n\nTherefore, option \\( B \\) is incorrect, and option \\( C \\) is correct.\n\nThe correct choice is: \\( C \\).\n\n【Key Insight】This problem examines knowledge points such as trigonometric functions and properties of a rhombus. The key to solving the problem lies in discussing the positions of moving points on different sides, deriving the corresponding functional relationships, and then making the correct judgment." }, { "problem_id": 476, "question": "As shown in Figure 1, rhombus paper sheet $A B C D$ has side length $2$ and $\\angle A B C = 60^\\circ$. As shown in Figure 2, fold $\\angle A B C$ and $\\angle A D C$ so that the vertices of the two angles overlap at a point $P$ on the diagonal $B D$. $E F$ and $G H$ are the creases. Let $B E = x (0 < x < 2)$. The following statements are given: \n(1) When $x = 1$, the length of $D P$ is $\\sqrt{3}$.\n(2) The value of $E F + G H$ changes with $x$.\n(3) The maximum area of hexagon $A E F C H G$ is $3 \\sqrt{3}$.\n(4) The perimeter of hexagon $A E F C H G$ remains constant.\n\nWhich of the statements are correct? \n\n\n\nFigure 1\n\n\n\nFigure 2\nA. (1)(2)\nB. (1)(4)\nC. (2)(3)(4)\nD. (1)(3)(4)", "input_image": [ "batch29-2024_06_14_5f55591362ac80f6b8e2g_0088_1.jpg", "batch29-2024_06_14_5f55591362ac80f6b8e2g_0088_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since the side length of rhombus \\( A B C D \\) is 2,\n\n\\(\\therefore A B = B C = 2\\),\n\nSince \\( \\angle A B C = 60^{\\circ} \\),\n\n\\(\\therefore A C = A B = 2\\), \\( B D = 2 \\sqrt{3} \\),\n\nFrom the folding, \\( \\triangle B E F \\) is an equilateral triangle,\n\nWhen \\( x = 1 \\), then \\( B E = 1 \\), \\( B M = B E \\sin 60^{\\circ} = \\frac{\\sqrt{3}}{2} \\),\n\nFrom the folding, \\( B P = 2 \\times \\frac{\\sqrt{3}}{2} = \\sqrt{3} \\), \\( D P = \\sqrt{3} \\), so (1) is correct,\n\n\n\nFigure 1\n\nSince \\( B E = x \\),\n\n\\(\\therefore A E = 2 - x \\),\n\nSince \\( \\triangle B E F \\) is an equilateral triangle,\n\n\\(\\therefore E F = x \\),\n\n\\(\\therefore B M = \\sqrt{3} E M = \\sqrt{3} \\times \\frac{1}{2} E F = \\frac{\\sqrt{3}}{2} x \\),\n\n\\(\\therefore B P = 2 B M = \\sqrt{3} x \\),\n\n\\(\\therefore D P = B D - B P = 2 \\sqrt{3} - \\sqrt{3} x \\),\n\n\\(\\therefore D N = \\frac{1}{2} D P = \\frac{1}{2}(2 \\sqrt{3} - \\sqrt{3} x) \\),\n\n\\( G N = \\frac{D N}{\\tan 60^{\\circ}} = \\frac{1}{2}(2 - x) \\)\n\n\\(\\therefore G H = 2 G N = 2 - x \\),\n\n\\(\\therefore E F + G H = 2 = A C \\), so (2) is incorrect;\n\nThe area of \\( \\triangle B E F \\) is: \\( \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} x \\times x = \\frac{\\sqrt{3} x^{2}}{4} \\),\n\nThe area of \\( \\triangle G H D \\) is: \\( \\frac{1}{2} \\times \\frac{1}{2}(2 \\sqrt{3} - \\sqrt{3} x) \\times (2 - x) = \\frac{\\sqrt{3}(2 - x)^{2}}{4} \\),\n\nThe area of hexagon \\( A E F C H G \\) \\( = S_{\\text{yellow}} A A B C D - S_{\\triangle} B E F - S_{\\triangle} D G H \\)\n\n\\( = \\frac{1}{2} \\times 2 \\times 2 \\sqrt{3} - \\frac{\\sqrt{3}(2 - x)^{2}}{4} - \\frac{\\sqrt{3} x^{2}}{4} \\)\n\\( = -\\frac{\\sqrt{3}}{2}(x - 1)^{2} + \\frac{3 \\sqrt{3}}{2} \\),\n\n\\(\\therefore\\) When \\( x = 1 \\), the area of hexagon \\( A E F C H G \\) reaches its maximum of \\( \\frac{3 \\sqrt{3}}{2} \\), so (3) is incorrect,\n\nThe perimeter of hexagon \\( A E F C H G \\) \\( = A E + E F + F C + C H + H G + A G \\)\n\n\\( = A E + B E + F C + B F + D G + A G \\)\n\n\\( = A B + C B + D A \\)\n\n\\( = 6 \\) is a constant,\n\nSo (4) is correct, i.e., the correct statements are (1) and (4),\n\nTherefore, the answer is: B.\n\n【Highlight】This problem is a comprehensive question on quadrilaterals, mainly testing the properties of rhombuses, the determination and properties of equilateral triangles, the area formula of triangles, the area formula of rhombuses, solving right triangles, and the key to solving this problem is expressing the relevant segment lengths in terms of \\( x \\)." }, { "problem_id": 477, "question": "As shown in the figure, in right triangle $\\triangle A B C$, $\\angle A C B = 90^\\circ$, $A C = B C = 2\\sqrt{2}$, and $C D \\perp A B$ at point $D$. Point $P$ starts from point $A$ and moves along the path $A \\rightarrow D \\rightarrow C$ until it reaches point $C$. A perpendicular $P E$ is drawn from point $P$ to $A C$ at point $E$, and a perpendicular $P F$ is drawn from point $P$ to $B C$ at point $F$. Let the distance traveled by point $P$ be $x$, and the area of quadrilateral $C E P F$ be $y$. The graph that best represents the functional relationship between $y$ and $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_615670bb7fcc0025dfefg_0040_1.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0040_2.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0040_3.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0040_4.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0040_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since $\\angle ACB = 90^\\circ$ and $AC = BC = 2\\sqrt{2}$,\n\nTherefore, $\\angle A = 45^\\circ$ and $AB = 4$.\n\nMoreover, since $CD \\perp AB$,\n\nThus, $AD = BD = CD = 2$, and $\\angle ACD = \\angle BCD = 45^\\circ$.\n\nSince $PE \\perp AC$ and $PF \\perp BC$,\n\nTherefore, quadrilateral $CEPF$ is a rectangle.\n\nI. When point $P$ is on segment $AD$, i.e., $0 < x \\leq 2$, as shown in Solution Diagram 1,\n\n\n\n## Solution Diagram 1\n\nThus, $AE = PE = AP \\cdot \\sin A = \\frac{\\sqrt{2}}{2} x$,\n\nTherefore, $CE = 2\\sqrt{2} - \\frac{\\sqrt{2}}{2} x$,\n\nHence, the area of quadrilateral $CEPF$ is $y = \\frac{\\sqrt{2}}{2} x\\left(2\\sqrt{2} - \\frac{\\sqrt{2}}{2} x\\right) = -\\frac{1}{2} x^{2} + 2x$. The graph of this function is a parabola,\n\nOpening downward, so options C and D are incorrect;\n\nII. When point $P$ is on segment $CD$, i.e., $2 < x \\leq 4$, as shown in Solution Diagram 2:\n\n\n\n## Solution Diagram 2\n\nAccording to the problem, $CP = 4 - x$,\n\nSince $\\angle ACD = \\angle BCD = 45^\\circ$ and $PE \\perp AC$,\n\nThus, $CE = PE = CP \\times \\sin \\angle ECP$,\n\nTherefore, $CE = PE = (4 - x) \\sin 45^\\circ = \\frac{\\sqrt{2}}{2}(4 - x)$,\n\nHence, the area of quadrilateral $CEPF$ is $y = \\left[\\frac{\\sqrt{2}}{2}(4 - x)\\right]^{2} = \\frac{1}{2} x^{2} - 4x + 8$. The graph of this function is a parabola,\n\nOpening upward, so option B is incorrect;\n\nTherefore, the correct answer is: A.\n\n【Insight】This problem examines the function graph of a moving point. The key to solving it is to write the function's analytical expression in segments and analyze it with the aid of graphical representation." }, { "problem_id": 478, "question": "As shown in Figure 1, point P starts from vertex A of the right-angled triangle ABC and moves along the path A → C → B at a constant speed until it reaches point B. Line segment PQ is drawn perpendicular to AB, with Q as the point of intersection. Let the distance traveled by point P be x, and the length of PQ be y. If the relationship between y and x is as shown in Figure 2, when x = 6, what is the length of PQ, which is $($  )?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 1\nB. $\\frac{4}{5}$\nC. $\\frac{3}{5}$\nD. $\\frac{2}{5}$", "input_image": [ "batch29-2024_06_14_615670bb7fcc0025dfefg_0082_1.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0082_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 2, we know that $\\mathrm{AC}=3$ and $\\mathrm{BC}=4$.\n\nIn the right triangle $\\triangle ABC$, $AB=\\sqrt{AC^{2}+BC^{2}}=\\sqrt{9+16}=5$.\n\nTherefore, $\\sin B=\\frac{AC}{AB}=\\frac{3}{5}$.\n\nWhen $\\mathrm{x}=6$, point $\\mathrm{P}$ is on $\\mathrm{BC}$, as shown in the figure below:\n\n\n\nIn the right triangle $\\triangle PQB$, $PQ=BP \\sin B=(7-x) \\times \\frac{3}{5}$.\n\nWhen $\\mathrm{x}=6$, $PQ=\\frac{3}{5}$.\n\nHence, the correct choice is: C.\n\n【Key Insight】This problem examines the issue of moving point images, involving knowledge such as solving right triangles. The key to such problems is to clearly understand the correspondence between the images and the figures at different time periods, and then solve accordingly." }, { "problem_id": 479, "question": "As shown in the figure, in rhombus $A B C D$, $A B=4 \\mathrm{~cm}$ and $\\angle A=60^{\\circ}$. Point $P$ starts from point $A$ and moves along $A \\rightarrow D \\rightarrow C$ at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$ until it reaches point $C$. At the same time, point $Q$ starts from point $A$ and moves along $A B$ at the same speed until it reaches point $B$. If the area of $\\triangle A P Q$ is $s$ (square centimeters) and the time of movement is $t(\\mathrm{~s})$, then the graph that best represents the functional relationship between $S$ and $t$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch29-2024_06_14_615670bb7fcc0025dfefg_0089_1.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0089_2.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0089_3.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0089_4.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0089_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem, when \\(0 < t \\leq 4\\), \\(\\triangle APQ\\) is an equilateral triangle with \\(AP = AQ = t\\). Drawing \\(PE \\perp AB\\) at point \\(E\\), we have \\(PE = AP \\cdot \\sin 60^\\circ = \\frac{\\sqrt{3}}{2} t\\).\n\nTherefore, the area \\(s\\) of \\(\\triangle APQ\\) is:\n\\[\ns = \\frac{1}{2} \\cdot PE \\cdot AQ = \\frac{1}{2} \\cdot t \\cdot \\frac{\\sqrt{3}}{2} t = \\frac{\\sqrt{3}}{4} t^2.\n\\]\nThus, options A and B can be excluded.\n\nWhen point \\(Q\\) reaches point \\(B\\), point \\(P\\) reaches point \\(D\\), at which time \\(t = 4\\), and the area \\(s\\) reaches its maximum value:\n\\[\ns = \\frac{\\sqrt{3}}{4} \\times 4^2 = 4\\sqrt{3}.\n\\]\nFor \\(4 < t \\leq 8\\), point \\(Q\\) remains stationary at point \\(B\\), while point \\(P\\) moves along \\(CD\\). The area \\(s\\) of \\(\\triangle APQ\\) remains constant, so option C can be excluded. Therefore, the graph that best represents the relationship between \\(S\\) and \\(t\\) is option D.\n\nHence, the correct choice is: D.\n\n\n\n**Insight:** This problem examines the properties of a rhombus, the determination and properties of an equilateral triangle, solving right triangles, the sine of the special angle \\(60^\\circ\\), the area of a triangle, and the graph of a quadratic function. Carefully analyzing the problem, understanding the movement paths of points \\(P\\) and \\(Q\\), and deriving the functional relationship between \\(s\\) and \\(t\\) are key to solving the problem." }, { "problem_id": 480, "question": "In a $5 \\times 5$ grid paper, the position of figure $N$ after translation is shown in Figure (2) as per Figure (1). Which of the following is the correct method of translation?\n\n(1)\n\n\n\n(2)\n\n\nA. Move down 1 grid, then move left 1 grid\nB. Move down 1 grid, then move left 2 grids\nC. Move down 2 grids, then move left 1 grid\nD. Move down 2 grids, then move left 2 grids", "input_image": [ "batch29-2024_06_14_85c79faaf97b4a25229eg_0025_1.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0025_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the concept of translation, the figure first moves down by 2 units, then moves left by 1 unit, or first moves left by 1 unit, then moves down by 2 units. Considering the options, only option C fits.\n\nTherefore, the answer is: C\n\n[Highlight] This question tests the basic concept and rules of translation, which is a relatively simple geometric transformation. The key is to observe and compare the positions of the object before and after the translation." }, { "problem_id": 481, "question": "Among the options below, which pair of angles, $\\angle 1$ and $\\angle 2$, are adjacent supplementary angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_85c79faaf97b4a25229eg_0087_1.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0087_2.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0087_3.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0087_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Options A and B: Angles $\\angle 1$ and $\\angle 2$ do not share a common vertex and are not adjacent, hence they are not adjacent supplementary angles.\n\nOption C: Angles $\\angle 1$ and $\\angle 2$ are not supplementary, so they are not adjacent supplementary angles.\n\nOption D: Angles $\\angle 1$ and $\\angle 2$ are supplementary and adjacent, making them adjacent supplementary angles. Therefore, the correct choice is D.\n\n[Key Insight] This question tests the concept of adjacent supplementary angles: two angles that share a common side, with their non-common sides forming a straight line. Such angles are supplementary and adjacent, and are known as adjacent supplementary angles." }, { "problem_id": 482, "question": "As shown in Figure 1, point $P$ is a moving point on the side of $\\triangle ABC$, traveling along the path $A \\rightarrow C \\rightarrow B$. Point $P$ draws a perpendicular $P D$ to side $AB$, with the foot of the perpendicular being point $D$. Let $AD = x$ and the area of $\\triangle APD$ be $y$. Figure 2 shows the graph of $y$ as a function of $x$. Based on the quantitative relationships in the graph, calculate the perimeter of $\\triangle ABC$ as ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $14+\\sqrt{3}$\nB. 15\nC. $9+3 \\sqrt{3}$\nD. $7+2 \\sqrt{5}$", "input_image": [ "batch29-2024_06_14_8680915691a88ef3ed6fg_0080_1.jpg", "batch29-2024_06_14_8680915691a88ef3ed6fg_0080_2.jpg", "batch29-2024_06_14_8680915691a88ef3ed6fg_0080_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "From the graph, it can be seen that the coordinates of the inflection point of the function graph are $(4,6)$.\n\nCombining with Figure 3, when point $\\mathrm{P}$ moves to point $\\mathrm{C}$, $\\mathrm{y}$ reaches its maximum value of 6.\n\nIt can be deduced that: $\\mathrm{y}=\\frac{1}{2} \\mathrm{AD} \\cdot \\mathrm{CD}$, and substituting the data gives $\\mathrm{CD}=3$.\n\nIn the right triangle $\\triangle \\mathrm{ADC}$, $\\mathrm{AC}=\\sqrt{4^{2}+3^{2}}=5$.\n\nWhen point $\\mathrm{D}$ moves to point $\\mathrm{B}$, the function value is 0, hence $\\mathrm{AB}=4+\\sqrt{3}$.\n\n$\\therefore \\mathrm{BD}=4+\\sqrt{3}-4=\\sqrt{3}$.\n\nIn the right triangle $\\triangle \\mathrm{BDC}$, $\\mathrm{CD}=3$, $\\mathrm{BD}=\\sqrt{3}$, and $\\tan B=\\sqrt{3}$.\n\nThis gives $\\angle B=60^{\\circ}$, and from $B D=B C \\cdot \\cos 60^{\\circ}$, we get $B C=2 \\sqrt{3}$.\n\n$\\therefore$ The perimeter of $\\triangle \\mathrm{ABC}$ is: $5+2 \\sqrt{3}+4+\\sqrt{3}=9+3 \\sqrt{3}$.\n\nTherefore, the correct choice is: C.\n\n【Key Insight】This question examines the function graph of a moving point problem. The key to solving it is to understand the problem clearly and use a combination of numerical and graphical methods to find the solution." }, { "problem_id": 483, "question": "Cut the rectangle in Figure 1, which has a perimeter of 32, into squares 1, 2, 3, 4, and a rectangle 5, and arrange them as shown in Figure 2, which fits into a larger rectangle with a perimeter of 48. The perimeter of the unshaded area in the larger rectangle is ( ).\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\nA. 16\nB. 24\nC. 30\nD. 40", "input_image": [ "batch29-2024_06_14_95d6af0392df862fb97dg_0073_1.jpg", "batch29-2024_06_14_95d6af0392df862fb97dg_0073_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Let the side length of square 1 be \\( x \\), and the side length of square 2 be \\( y \\). Then, the side length of square 3 is \\( x + y \\), the side length of square 4 is \\( 2x + y \\), the length of rectangle 5 is \\( 3x + y \\), and its width is \\( y - x \\).\n\nFrom the perimeter of the rectangle in Figure 1 being 32, we have:\n\\[\ny + 2(x + y) + (2x + y) = 16\n\\]\nSolving this, we get:\n\\[\nx + y = 4\n\\]\n\nAs shown in the figure,\n\nSince the perimeter of the rectangle in Figure 2 is 48,\n\\[\nAB + 2(x + y) + 2x + y + y - x = 24\n\\]\nThus,\n\\[\nAB = 24 - 3x - 4y\n\\]\n\nBy translation, the perimeter of the uncovered shaded region is equal to the perimeter of quadrilateral \\( ABCD \\),\n\\[\n2(AB + AD) = 2(24 - 3x - 4y + x + y + 2x + y + y - x) = 2(24 - x - y) = 48 - 2(x + y) = 48 - 8 = 40\n\\]\nTherefore, the correct answer is: D.\n\n\n\n(Figure 2)\n\n【Insight】This problem tests the application of integer addition and subtraction, the properties of translation, and the use of translation properties to transform irregular shapes into regular shapes for solving. The key to solving the problem is to set up unknowns and express each segment with algebraic expressions to find the solution." }, { "problem_id": 484, "question": "Given right triangle $\\triangle ABC$ where $\\angle C = 90^\\circ, \\angle B = 30^\\circ$, and $BC = 3$, which of the following diagrams depict right triangles that are congruent to $\\triangle ABC$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_b9af2d17c1bfa2bebf76g_0006_1.jpg", "batch29-2024_06_14_b9af2d17c1bfa2bebf76g_0006_2.jpg", "batch29-2024_06_14_b9af2d17c1bfa2bebf76g_0006_3.jpg", "batch29-2024_06_14_b9af2d17c1bfa2bebf76g_0006_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Since in triangle $\\triangle \\mathrm{ABC}$, $\\angle \\mathrm{C}=90^{\\circ}$ and $\\angle \\mathrm{B}=30^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{A}=90^{\\circ}-30^{\\circ}=60^{\\circ}$.\n\nGiven that $\\mathrm{BC}=3$, which is exactly the side opposite the $60^{\\circ}$ angle,\n\nThus, the corresponding sides and angles of the triangle in option B are exactly equal,\n\nTherefore, option B meets the condition. The triangles in the other options do not.\n\nHence, choose B.\n\n[Highlight] This question examines the method of determining the congruence of right-angled triangles; the methods to determine if two right-angled triangles are congruent include: SSS, SAS, AAS, ASA, and HL. When solving, combine the given conditions with the congruence determination methods to verify one by one." }, { "problem_id": 485, "question": "Using the method shown in the figure, a line parallel to a given straight line through a point outside the line can be constructed. The following justifications are given:\n\n(1) Same-side interior angles are equal, so the two lines are parallel, (2) Interior angles on the same side of the transversal are equal, so the two lines are parallel, (3) Consecutive interior angles are supplementary, so the two lines are parallel, (4) If two lines are parallel, same-side interior angles are equal. Among these, the number of reasonable justifications is ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch29-2024_06_14_c753650a642e029cdb2bg_0061_1.jpg", "batch29-2024_06_14_c753650a642e029cdb2bg_0061_2.jpg", "batch29-2024_06_14_c753650a642e029cdb2bg_0061_3.jpg", "batch29-2024_06_14_c753650a642e029cdb2bg_0061_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: As shown in the figure, from the operation in diagram (2), we know that \\( PE \\perp CD \\), so \\( \\angle PEC = \\angle PED = 90^\\circ \\). From the operation in diagram (3), we know that \\( AB \\perp PE \\), so \\( \\angle APE = \\angle BPE = 90^\\circ \\). Therefore,\n\n\\[\n\\angle PEC = \\angle PED = \\angle APE = \\angle BPE = 90^\\circ,\n\\]\n\nso we can determine that \\( AB \\parallel CD \\) based on conclusions (2), (3), or (4).\n\nHence, the correct answer is C.\n\n\n\nDiagram (1)\n\n\n\nDiagram (2)\n\n\n\nDiagram (3)\n\n\n\nDiagram (4)\n\n【Key Point】This question examines the determination of parallel lines: if corresponding angles are equal, the lines are parallel; if alternate interior angles are equal, the lines are parallel; if consecutive interior angles are supplementary, the lines are parallel; if two lines are both perpendicular to the same line, then the two lines are parallel." }, { "problem_id": 486, "question": "Among the following figures, which one can be obtained by translating $\\triangle A B C$ to get $\\triangle D E F$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_e7f29cc027e50efb8665g_0062_1.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0062_2.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0062_3.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0062_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, it can be seen that only option A, $\\triangle ABC$, when translated, can result in $\\triangle DEF$.\n\nTherefore, the answer is: A.\n\n[Key Point] This question tests the fundamental properties of translation: (1) Translation does not alter the shape and size of the figure; (2) After translation, the corresponding connected segments are parallel and equal in length, the corresponding segments are parallel and equal, and the corresponding angles are equal." }, { "problem_id": 487, "question": "Among the following figures, the length of segment $A D$ represents the distance from point $A$ to the straight line $B C$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_e7f29cc027e50efb8665g_0098_1.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0098_2.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0098_3.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0098_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Since $AD$ is not perpendicular to $BC$, the length of segment $AD$ cannot represent the distance from point $A$ to line $BC$, which does not meet the requirement;\n\nB. If $AD$ is perpendicular to $BC$ at $D$, then the length of segment $AD$ represents the distance from point $A$ to line $BC$, which meets the requirement;\n\nC. Since $AD$ is not perpendicular to $BC$, the length of segment $AD$ cannot represent the distance from point $A$ to line $BC$, which does not meet the requirement;\n\nD. Since $AD$ is not perpendicular to $BC$, the length of segment $AD$ cannot represent the distance from point $A$ to line $BC$, which does not meet the requirement.\n\nTherefore, the correct choice is: B.\n\n[Insight] This question tests the definition of the distance from a point to a line. Note that the perpendicular segment drawn from a point outside the line to the line is called the perpendicular segment." }, { "problem_id": 488, "question": "As shown in the figure, in rectangle $A B C D$, $A B = 4$ and $B C = 3$. Point $E$ starts from the midpoint of $B C$ and moves counterclockwise along the sides of the rectangle until it reaches the midpoint of $A D$. Let the distance traveled by point $E$ be $x$ and the area of triangle $A B E$ be $y$. Which of the following graphs correctly represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch2-2024_06_13_690b95b9aaba78d1bb26g_0003_1.jpg", "batch2-2024_06_13_690b95b9aaba78d1bb26g_0003_2.jpg", "batch2-2024_06_13_690b95b9aaba78d1bb26g_0003_3.jpg", "batch2-2024_06_13_690b95b9aaba78d1bb26g_0003_4.jpg", "batch2-2024_06_13_690b95b9aaba78d1bb26g_0003_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As point $E$ moves along $BC$, the area of the triangle continuously increases. Initially, the area $=\\frac{1}{2} AB \\cdot \\frac{1}{2} BC=\\frac{1}{2} \\times 4 \\times \\frac{1}{2} \\times 3=3$;\n\nThe maximum area $=\\frac{1}{2} \\times AB \\cdot BC=\\frac{1}{2} \\times 4 \\times 3=6$;\n\nWhen point $E$ moves along $DC$, the area of the triangle remains constant at 6.\n\nWhen point $E$ moves along $AD$, the area of the triangle continuously decreases, reducing to an area of 3.\n\nObserving the four options, option D fits the scenario,\n\nTherefore, the answer is: D.\n\n【Key Insight】This problem primarily examines the functional image of a moving point. Determining the changes in the area of $\\triangle ABE$ as point $E$ moves along $BC$, $DC$, and $AD$ is crucial to solving the problem." }, { "problem_id": 489, "question": "As shown in the figure, in $\\triangle A B C$, $\\angle A = 20^\\circ$. Point $D$ is on side $A C$ (as shown in Figure 1). First, $\\triangle A B D$ is folded along $B D$, so that point $A$ lands at point $A^\\prime$, and $A^\\prime B$ intersects $A C$ at point $E$ (as shown in Figure 2). Then, $\\triangle B C E$ is folded along $B E$, and point $C$ lands at point $C^\\prime$ on $B D$, where $\\angle C^\\prime E B = 66^\\circ$ (as shown in Figure 3). The measure of $\\angle A B C$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $66^\\circ$\nB. $23^\\circ$\nC. $46^\\circ$\nD. $69^\\circ$", "input_image": [ "batch2-2024_06_13_690b95b9aaba78d1bb26g_0005_1.jpg", "batch2-2024_06_13_690b95b9aaba78d1bb26g_0005_2.jpg", "batch2-2024_06_13_690b95b9aaba78d1bb26g_0005_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we have $\\angle ABC' = \\angle C'BE = \\angle EBC = \\frac{1}{3} \\angle ABC$, and $\\angle C'EB = \\angle CEB = 66^\\circ$. Let $\\angle ABC = x$, then $\\angle ABC' = \\angle C'BE = \\angle EBC = \\frac{1}{3}x$.\n\nSince the sum of the interior angles of a triangle is $180^\\circ$,\n\nin $\\triangle ABC$, $\\angle A + \\angle ABC = 180^\\circ - \\angle C$, which means $20^\\circ + x = 180^\\circ - \\angle C$;\n\nin $\\triangle BCE$, $\\angle CEB + \\angle CBE = 180^\\circ - \\angle C$, which means $66^\\circ + \\frac{1}{3}x = 180^\\circ - \\angle C$;\n\nTherefore, $20^\\circ + x = 66^\\circ + \\frac{1}{3}x$,\n\nSolving for $x$, we get: $x = 69^\\circ$,\n\nHence, the answer is: D.\n\n【Key Insight】This problem examines the equality of corresponding angles after a fold and utilizes the fact that the sum of the interior angles of a triangle is $180^\\circ$. Setting up an unknown and establishing an equation is key to solving the problem. The challenge here is that $\\angle C$ is a common angle in both triangles, leading to the formulation of the equation for solution." }, { "problem_id": 490, "question": "Among the following figures, the shaded areas that are equal in size are $\\qquad$ .\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch2-2024_06_13_6cc23543e4e8a4f1b736g_0089_1.jpg", "batch2-2024_06_13_6cc23543e4e8a4f1b736g_0089_2.jpg", "batch2-2024_06_13_6cc23543e4e8a4f1b736g_0089_3.jpg", "batch2-2024_06_13_6cc23543e4e8a4f1b736g_0089_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: The area formula for a triangle is: base × height ÷ 2; the area formula for a parallelogram is: base × height.\n\nAssuming the side length of each small square is 1, and the area of the shaded part is S,\n\nthen: A. S = 1 × 1 = 1,\n\nB. S = 2 × 2 ÷ 2 = 2,\n\nC. S = 1 × 1 ÷ 2 = 0.5,\n\nD. S = 1 × 1 ÷ 2 = 0.5\n\nTherefore, the shaded areas that are equal are C and D,\n\nso the answer is CD.\n\n[Key Insight] This question mainly examines the area calculation of parallelograms and triangles, which is quite fundamental." }, { "problem_id": 491, "question": "As shown in the figure, a standard pair of right-angled triangles is placed in four different positions. Which of the following arrangements satisfies that $\\angle \\alpha$ and $\\angle \\beta$ are complementary angles? ( \n\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch30-2024_06_17_006ee311fce70e2fc37ag_0060_1.jpg", "batch30-2024_06_17_006ee311fce70e2fc37ag_0060_2.jpg", "batch30-2024_06_17_006ee311fce70e2fc37ag_0060_3.jpg", "batch30-2024_06_17_006ee311fce70e2fc37ag_0060_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. $\\angle \\alpha + \\angle \\beta = 90^{\\circ}$, therefore incorrect;\n\nB. $\\angle \\alpha = \\angle \\beta$, therefore incorrect;\n\nC. $\\angle \\alpha + \\angle \\beta = 270^{\\circ}$, therefore incorrect;\n\nD. $\\angle \\alpha + \\angle \\beta = 180^{\\circ}$, therefore correct.\n\nHence, the correct choice is: D.\n\n[Key Insight] This question examines the concepts of complementary and supplementary angles. If the sum of two angles equals $90^{\\circ}$, they are said to be complementary. If the sum equals $180^{\\circ}$, they are said to be supplementary." }, { "problem_id": 492, "question": "Among the following figures, which one has segment $P Q$ representing the distance from point $P$ to the line $l$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch30-2024_06_17_1ef92c6bf4715abd5e91g_0005_1.jpg", "batch30-2024_06_17_1ef92c6bf4715abd5e91g_0005_2.jpg", "batch30-2024_06_17_1ef92c6bf4715abd5e91g_0005_3.jpg", "batch30-2024_06_17_1ef92c6bf4715abd5e91g_0005_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: In figures A, B, and C, the line segment \\( PQ \\) is not perpendicular to the line \\( l \\), so the length of segment \\( PQ \\) cannot represent the distance from point \\( P \\) to line \\( l \\).\n\nIn figure D, the line segment \\( PQ \\) is perpendicular to the line \\( l \\), with the foot of the perpendicular at point \\( Q \\). Therefore, the length of segment \\( PQ \\) can represent the distance from point \\( P \\) to line \\( l \\).\n\nHence, the correct choice is: D\n\n[Key Insight] This question tests the concept of the distance from a point to a line. Correctly understanding the concept of the distance from a point to a line is crucial for solving the problem." }, { "problem_id": 493, "question": "In the four figures shown below, the number of figures in which $\\angle 1$ and $\\angle 2$ are adjacent angles is ( )\n\n\n\n\n\nA. 0\nB. 1\nC. 2\nD. 3", "input_image": [ "batch30-2024_06_17_3b48496b0942f6c6c5bag_0088_1.jpg", "batch30-2024_06_17_3b48496b0942f6c6c5bag_0088_2.jpg", "batch30-2024_06_17_3b48496b0942f6c6c5bag_0088_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Vertical Angles: Two angles that share a common vertex and whose sides are opposite extensions of each other. Among the given options, only the third figure satisfies this condition.\n\nTherefore, the correct choice is: B\n\n[Key Insight] This question tests the understanding of the definition of vertical angles. The key to solving this problem lies in the ability to proficiently identify vertical angles. Vertical angles are defined as two angles that share a common vertex and whose sides are opposite extensions of each other." }, { "problem_id": 494, "question": "The following statements are given ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n(1) As shown in Figure (1), by straightening the curved river $B C A$ into a straight path $B A$, the journey can be shortened:\n\n(2) As shown in Figure (2), when water is diverted to the pond $C$, a point $D$ can be found on the canal bank $A B$ such that $C D \\perp A B$. Digging a ditch along $C D$ will result in the shortest ditch;\n\n(3) In Figure (3), Cars A and B start simultaneously from roads $A C$ and $B C$ to head for City $C$. If both cars have the same speed, Car A will arrive at City $C$ first.\n\nThe statements that apply the property \"the shortest distance between two points is a straight line\" are ( )\nA. (1)(2)\nB. (1)(3)\nC. (2)(3)\nD. (1)(2)(3)", "input_image": [ "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0020_1.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0020_2.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0020_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "(1) As shown in Figure (1), changing the curved river path BCA to a straight path BA can shorten the sailing distance based on the principle that the shortest distance between two points is a straight line.\n\n(2) As shown in Figure (2), to channel water into the pool $\\mathrm{C}$, a point $\\mathrm{D}$ can be found on the bank $\\mathrm{AB}$ such that $\\mathrm{CD} \\perp \\mathrm{AB}$. Digging a ditch along $\\mathrm{CD}$ ensures the shortest ditch length, based on the principle that the perpendicular segment is the shortest.\n\n(3) As shown in Figure (3), two cars, Car A and Car B, start simultaneously from roads $\\mathrm{AC}$ and $\\mathrm{BC}$ respectively, heading towards city C. If both cars travel at the same speed, Car A will arrive at city C first, based on the principle that the perpendicular segment is the shortest.\n\nTherefore, the correct answer is C.\n\n[Key Insight] This question primarily tests the properties of perpendicular segments, emphasizing the importance of understanding the fundamental theorem." }, { "problem_id": 495, "question": "As shown in the figure, a rectangular paper is folded along the dotted lines in Figure (1) to obtain Figure (2). The angle formed by the fold line and one side of the rectangle, $\\angle 1$, is $55^\\circ$. The paper is then folded again along the dotted lines in Figure (2) to obtain Figure (3). What is the measure of $\\angle 2$?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA. $20^\\circ$\nB. $25^\\circ$\nC. $30^\\circ$\nD. $35^\\circ$", "input_image": [ "batch30-2024_06_17_5a24cb0314474b337ca9g_0098_1.jpg", "batch30-2024_06_17_5a24cb0314474b337ca9g_0098_2.jpg", "batch30-2024_06_17_5a24cb0314474b337ca9g_0098_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, the angles are labeled,\n\n\n\nFrom the properties of folding, we have: $\\angle 5=\\angle 1=55^{\\circ}, \\angle 3=\\angle 4$,\n\n$\\therefore \\angle 3=\\angle 4=\\frac{1}{2}\\left(180^{\\circ}-\\angle 1-\\angle 5\\right)=35^{\\circ}$,\n\n$\\because$ The opposite sides of a rectangle are parallel,\n\n$\\therefore \\angle 2=\\angle 3=35^{\\circ}$,\n\nTherefore, the correct choice is D.\n\n【Highlight】This question tests the properties of parallel lines and the properties of folding. Remembering these properties is key to solving the problem." }, { "problem_id": 496, "question": "In the following figures, $\\angle 1$ and $\\angle 2$ are not necessarily equal in:\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch30-2024_06_17_6e31a288e444fbe80462g_0060_1.jpg", "batch30-2024_06_17_6e31a288e444fbe80462g_0060_2.jpg", "batch30-2024_06_17_6e31a288e444fbe80462g_0060_3.jpg", "batch30-2024_06_17_6e31a288e444fbe80462g_0060_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. According to the property of vertically opposite angles, $\\angle 1 = \\angle 2$, so this option does not meet the requirement of the question.\n\nB. Since $\\angle 1 + \\angle 2 = 90^{\\circ}$, $\\angle 1$ and $\\angle 2$ are not necessarily equal, which makes this option meet the requirement of the question.\n\nC. Based on the properties of parallel lines, $\\angle 1 = \\angle 2$, so this option does not meet the requirement of the question.\n\nD. According to the property that the complements of equal angles are equal, $\\angle 1 = \\angle 2$, so this option does not meet the requirement of the question.\n\nTherefore, the correct choice is: B.\n\n[Insight] This question tests the knowledge of the properties of parallel lines, vertically opposite angles, and the equality of the complements of equal angles. The key to solving the problem lies in mastering the fundamental concepts, which are commonly tested in middle school exams." }, { "problem_id": 497, "question": "As shown in Figure (1), in $\\triangle A B C$, $\\angle A = 42^\\circ$, and $\\angle B = 73^\\circ$. As shown in Figure (2), the side $C B$ of Figure (1) is rotated counterclockwise around point $C$ for a full revolution until it returns to its original position, forming segment $C B^{\\prime}$. During the entire rotation, if $C B^{\\prime} \\parallel A B$, then the measure of $\\angle B C B^{\\prime}$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $73^\\circ$\nB. $107^\\circ$\nC. $73^\\circ$ or $107^\\circ$\nD. $42^\\circ$ or $107^\\circ$", "input_image": [ "batch30-2024_06_17_86e71f2eee792fb62b01g_0007_1.jpg", "batch30-2024_06_17_86e71f2eee792fb62b01g_0007_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nSince \\( CB' \\parallel AB \\),\n\nTherefore, \\( \\angle B'CB + \\angle B = 180^\\circ \\).\n\nGiven that \\( \\angle B = 73^\\circ \\),\n\nThus, \\( \\angle BCB' = 180^\\circ - \\angle B = 180^\\circ - 73^\\circ = 107^\\circ \\).\n\nAs shown in the figure,\n\n\n\nSince \\( CB' \\parallel AB \\),\n\nTherefore, \\( \\angle BCB' = \\angle B \\).\n\nGiven that \\( \\angle B = 73^\\circ \\),\n\nThus, \\( \\angle BCB' = 73^\\circ \\).\n\nIn summary: \\( \\angle BCB' = 107^\\circ \\) or \\( 73^\\circ \\).\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This problem primarily examines the properties of rotation and the properties of parallel lines. The key to solving the problem lies in applying the concept of classification and discussion." }, { "problem_id": 498, "question": "As shown in Figure 1, when light is incident obliquely from air to a transparent liquid, refraction occurs, satisfying the angle of incidence $\\angle 1$ and the angle of refraction $\\angle 2$ ratio of $3: 2$. As shown in Figure 2, on the same plane, two light rays are incident obliquely from air to the liquid at the same time. The angles between the two incident light rays and the horizontal liquid surface are $\\alpha, \\beta$ respectively, and the angle between the two refracted light rays in the liquid is $\\gamma$,\nThen the quantitative relationship between $\\alpha, \\beta, \\gamma$ is ( $)$\n\n\nFigure 1\n\n\nFigure 2\nA. $\\frac{2}{3}(\\alpha+\\beta)=\\gamma$\nB. $\\frac{2}{3}(\\alpha+\\beta)=120^{\\circ}-\\gamma$\nC. $\\alpha+\\beta=\\gamma$\nD. $\\alpha+\\beta+\\gamma=180^{\\circ}$", "input_image": [ "batch30-2024_06_17_b2c6b659c4565a065270g_0036_1.jpg", "batch30-2024_06_17_b2c6b659c4565a065270g_0036_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in the figure: Draw perpendiculars to the horizontal line through points $B$, $D$, and $F$, then $PC // DE // QG$.\n\n\n\n$\\therefore \\angle BDF = \\angle BDE + \\angle FDE = \\angle DBC + \\angle DFG$. Given that $\\angle DBC = \\frac{2}{3} \\angle ABP = \\frac{2}{3}(90^\\circ - \\alpha)$.\n\n$\\angle DFG = \\frac{2}{3} \\angle HFQ = \\frac{2}{3}(90^\\circ - \\beta)$.\n\n$\\therefore \\angle BDF = \\frac{2}{3}(90^\\circ - \\alpha) + \\frac{2}{3}(90^\\circ - \\beta)$.\n\n$= \\frac{2}{3}(180^\\circ - \\alpha - \\beta)$.\n\nThat is: $\\gamma = 120^\\circ - \\frac{2}{3}(\\alpha + \\beta)$.\n\nThus, $\\frac{2}{3}(\\alpha + \\beta) = 120^\\circ - \\gamma$.\n\nTherefore, choose B.\n\n【Key Insight】This question tests the properties of parallel lines and optical principles. Understanding the problem and mastering the properties of parallel lines are crucial." }, { "problem_id": 499, "question": "In rectangle $A B C D$ as shown in Figure (1), point $E$ is on $A D$, and $\\angle A E B = 60^\\circ$. The figure is folded along the creases $B E$ and $C E$ and pressed flat, as shown in Figure (2). If $\\angle A E D = 10^\\circ$, what is the measure of $\\angle D E C$?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $25^\\circ$\nB. $30^\\circ$\nC. $35^\\circ$\nD. $40^\\circ$", "input_image": [ "batch30-2024_06_17_ca1f21fdad6cd0e86c22g_0023_1.jpg", "batch30-2024_06_17_ca1f21fdad6cd0e86c22g_0023_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the folding, we can deduce that $BE$ bisects $\\angle A'EA$ and $CE$ bisects $\\angle DED'$.\n\nGiven that $\\angle AEB = 60^\\circ$,\n\nTherefore, $\\angle A'EA = 2 \\angle AEB = 120^\\circ$.\n\nGiven that $\\angle AED = 10^\\circ$,\n\nThus, $\\angle E D = 180^\\circ - 120^\\circ + 10^\\circ = 70^\\circ$,\n\nHence, $\\angle DEC = \\frac{1}{2} \\times 70^\\circ = 35^\\circ$.\n\nTherefore, the correct answer is C.\n\n[Key Insight] This problem tests the understanding of angle sum and difference relationships and the properties of symmetry. The key to solving this problem lies in using the properties of folding to determine that $BE$ bisects $\\angle A'EA$ and $CE$ bisects $\\angle DED'$." }, { "problem_id": 500, "question": "In the following diagrams, $\\angle 1$ and $\\angle 2$ are definitely complementary angles in which figure(s)?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch30-2024_06_17_f66b39c27dd26482c43dg_0011_1.jpg", "batch30-2024_06_17_f66b39c27dd26482c43dg_0011_2.jpg", "batch30-2024_06_17_f66b39c27dd26482c43dg_0011_3.jpg", "batch30-2024_06_17_f66b39c27dd26482c43dg_0011_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "A. $\\angle 1$ and $\\angle 2$ are vertical angles, hence $\\angle 1 = \\angle 2$, which does not meet the condition;\n\nB. $\\angle 1 + \\angle 2 = 180^{\\circ}$, meaning $\\angle 1$ and $\\angle 2$ are supplementary, which meets the condition;\n\nC. $\\angle 1 + \\angle 2 = 90^{\\circ}$, meaning $\\angle 1$ and $\\angle 2$ are complementary, which does not meet the condition;\n\nD. Only when the two lines are parallel does $\\angle 1 + \\angle 2 = 180^{\\circ}$, making $\\angle 1$ and $\\angle 2$ supplementary, which does not meet the condition. Therefore, the correct choice is B.\n\n【Key Point】This question mainly examines the definitions of supplementary and complementary angles, adjacent supplementary angles, vertical angles, and the properties of parallel lines. Familiarity with the definition of supplementary angles is crucial for answering this question." }, { "problem_id": 501, "question": "As shown in the figure, YY folded a \"parallel line to a given line through a point outside the line\" (i.e., $b / / a$) using a square piece of paper. The steps are as follows, with the basis for each step being:\n\n\n\nA\n\n\n\nB\n\n\n\nC\n\n\n\nD\n\nA. There is exactly one line parallel to a given line through a point outside the line.\nB. Two lines parallel to the same line are parallel to each other.\nC. If two lines are parallel, the interior angles on the same side are supplementary.\nD. If corresponding angles are equal, the two lines are parallel.", "input_image": [ "batch31-2024_06_14_5cb09016da4908896958g_0007_1.jpg", "batch31-2024_06_14_5cb09016da4908896958g_0007_2.jpg", "batch31-2024_06_14_5cb09016da4908896958g_0007_3.jpg", "batch31-2024_06_14_5cb09016da4908896958g_0007_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: After the first fold, the positional relationship between the resulting crease $AB$ and the straight line $m$ is perpendicular;\n\nUnfolding the square paper and performing the second fold, the positional relationship between the resulting crease $CD$ and the first crease is perpendicular;\n\n\n\n$\\because AB \\perp m, \\quad CD \\perp m$,\n\n$\\therefore \\angle 1=\\angle 2=\\angle 3=\\angle 4=90^{\\circ}$,\n\n$\\because \\angle 3=\\angle 1$,\n\n$\\therefore AB / / CD$ (corresponding angles are equal, the two lines are parallel).\n\nTherefore, the answer is: D.\n\n【Key Insight】This question mainly tests the understanding of the parallel line criterion and the transformation of folding. The key is to master the parallel line criterion theorem." }, { "problem_id": 502, "question": "As shown in Figure 1, a strip of paper with $A D / / B C$ is given. Fold and flatten the paper as shown, first along $E F$, and then along $B F$. If in Figure 3, $\\angle C F E = 24^\\circ$, what is the measure of $\\angle A E F$ in Figure 2?\n\n\n(Figure 1)\n\n\n\n(Figure 2)\n\n\n\n(Figure 3)\nA. $112^\\circ$\nB. $68^\\circ$\nC. $48^\\circ$\nD. $136^\\circ$", "input_image": [ "batch31-2024_06_14_5cb09016da4908896958g_0036_1.jpg", "batch31-2024_06_14_5cb09016da4908896958g_0036_2.jpg", "batch31-2024_06_14_5cb09016da4908896958g_0036_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to Figure 2, it is known that $\\angle BFE$ has been folded twice, which means $2 \\angle BFE + \\angle BFC = 180^{\\circ}$, and $\\angle BFE - \\angle BFC = \\angle CFE = 24^{\\circ}$.\n\nFrom Figure 3, it is understood that $\\angle BFE$ has been folded three times but is short by $\\angle CFE$.\n\nTherefore, $\\angle BFE = \\frac{1}{3}(180^{\\circ} + 24^{\\circ}) = 68^{\\circ}$.\n\nSince $AE \\parallel BF$,\n\n$\\angle AEF = 180^{\\circ} - \\angle BFE = 112^{\\circ}$.\n\nHence, the correct choice is: A.\n\n[Key Insight] This problem tests the properties of parallel lines, the nature of folding, and the calculation of angles. The key to solving the problem lies in calculating the degree of $\\angle BFE$." }, { "problem_id": 503, "question": "As shown in the figure, $\\triangle A B C$ is cut along $D E$ into a small triangle $A D E$ and a quadrilateral $D^{\\prime} E^{\\prime} C B$. If $D E \\parallel B C$, and the lengths of the sides of the quadrilateral $D^{\\prime} E^{\\prime} C B$ are as shown in the figure, then the small triangle $A D E$ should be ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch31-2024_06_14_f8afee3bcb06c08cc378g_0093_1.jpg", "batch31-2024_06_14_f8afee3bcb06c08cc378g_0093_2.jpg", "batch31-2024_06_14_f8afee3bcb06c08cc378g_0093_3.jpg", "batch31-2024_06_14_f8afee3bcb06c08cc378g_0093_4.jpg", "batch31-2024_06_14_f8afee3bcb06c08cc378g_0093_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Let \\( AD = x \\), \\( AE = y \\).\n\nSince \\( DE \\parallel BC \\),\n\n\\(\\triangle ADE \\sim \\triangle ABC\\),\n\\(\\therefore \\frac{AD}{AB} = \\frac{AE}{AC} = \\frac{DE}{BC}\\),\n\n\\(\\therefore \\frac{x}{x + 12} = \\frac{y}{y + 16} = \\frac{6}{14}\\),\n\n\\(\\therefore x = 9\\), \\( y = 12\\),\n\nTherefore, the correct choice is \\( C \\).\n\n【Highlight】This problem examines the properties of parallel lines, the determination and properties of similar triangles, and other related knowledge. The key to solving the problem lies in mastering the fundamental concepts, which are commonly tested in high school entrance exams." }, { "problem_id": 504, "question": "Given segments $a, b, c$, if we are to construct segment $x$ such that $a : b = c : x$, which of the following constructions is correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_078d0a0e18bf66a6fa5dg_0067_1.jpg", "batch32-2024_06_14_078d0a0e18bf66a6fa5dg_0067_2.jpg", "batch32-2024_06_14_078d0a0e18bf66a6fa5dg_0067_3.jpg", "batch32-2024_06_14_078d0a0e18bf66a6fa5dg_0067_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "$A 、 a: b=x: c$ does not match the given $a: b=c: x$, so option $A$ is incorrect;\n\n$B 、 a: b=c: x$ matches the given $a: b=c: x$, so option $B$ is correct;\n\n$C 、 a: c=x: b$ does not match the given $a: b=c: x$, so option $C$ is incorrect;\n\n$D 、 a: x=b: c$ does not match the given $a: b=c: x$, so option $D$ is incorrect;\n\nTherefore, the correct choice is: $B$.\n\n【Highlight】This question examines the theorem of proportional segments divided by parallel lines and complex constructions. It is essential to understand that a line parallel to one side of a triangle that intersects the other two sides (or their extensions) divides those sides proportionally." }, { "problem_id": 505, "question": "Given segments $a, b, c$, construct segment $x$ such that $b x = a c$. Which of the following methods is correct ( )?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0009_1.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0009_2.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0009_3.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0009_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: \n\nA. $\\frac{x}{c} = \\frac{a}{b}$, which implies $b x = a c$. However, since $x$ is an unknown line segment and cannot be drawn, it does not meet the requirement of the problem.\n\nB. $\\frac{x}{b} = \\frac{a}{c}$, which implies $a b = c x$. This does not meet the requirement of the problem.\n\nC. $\\frac{a}{b} = \\frac{c}{x}$, which implies $a x = b c$. This does not meet the requirement of the problem.\n\nD. $\\frac{c}{x} = \\frac{b}{a}$, which implies $a c = b x$. This meets the requirement of the problem.\n\nTherefore, the correct choice is: D\n\n[Key Insight] This problem mainly tests the concept of proportional segments divided by parallel lines. The key to solving it is to understand that \"when two lines are intersected by a set of parallel lines (at least three), the lengths of the corresponding segments intercepted are proportional.\"" }, { "problem_id": 506, "question": "When a general triangle, equilateral triangle, square, and rectangle are each expanded outward by 1 unit along their respective sides, the resulting geometric figures are not necessarily similar to the original figures in which option?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0031_1.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0031_2.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0031_3.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0031_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \nA: The shapes are identical, satisfying the definition of similar shapes, and the corresponding angles are equal. Therefore, the triangles are similar, so option A does not meet the requirement. \n\nB: The shapes are identical, satisfying the definition of similar shapes, so option B does not meet the requirement. \n\nC: The shapes are identical, satisfying the definition of similar shapes, so option C does not meet the requirement. \n\nD: Although the two rectangles have corresponding angles that are equal, their corresponding sides are not necessarily proportional. Therefore, option D meets the requirement. Hence, the correct choice is D. \n\n**Key Point**: This question tests the judgment of similar polygons. Familiarity with the concept of polygon similarity is crucial." }, { "problem_id": 507, "question": "Given segments $a, b, c$, construct segment $x$ such that $a b = c x$. Which of the following constructions ( $A B \\parallel C D$) is correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0085_1.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0085_2.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0085_3.jpg", "batch32-2024_06_14_0a2e4ed0bb716236d4e8g_0085_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "In option A, since \\( AB \\parallel CD \\), we have: \\(\\frac{a}{c} = \\frac{b}{x}\\), which implies \\( a x = b c \\), and this does not match the intended condition;\n\nIn option B, since \\( AB \\parallel CD \\), we have: \\(\\frac{a}{b} = \\frac{c}{x}\\), which implies \\( a x = b c \\), and this does not match the intended condition;\n\nIn option C, since \\( AB \\parallel CD \\), we have: \\(\\frac{a}{x} = \\frac{b}{c}\\), which implies \\( a c = b x \\), and this does not match the intended condition;\n\nIn option D,\n\nSince \\( AB \\parallel CD \\),\n\nTherefore, \\(\\frac{a}{c} = \\frac{x}{b}\\)\n\nThus, \\( a b = c x \\)\n\nHence, option D matches the intended condition.\n\nTherefore, the correct choice is D.\n\n[Note] This question tests the ability to perform complex geometric constructions and the theorem of proportional segments divided by parallel lines. The key to solving the problem lies in understanding the given conditions and flexibly applying the learned knowledge." }, { "problem_id": 508, "question": "Given segments $a, b, c$, the incorrect method for constructing segment $x$ such that $a : b = c : x$ is ( )\nA.\nB.\nC.\nD.\n\n##", "input_image": [ "batch32-2024_06_14_1aa9edfeb22f70b3d1f7g_0095_1.jpg", "batch32-2024_06_14_1aa9edfeb22f70b3d1f7g_0095_2.jpg", "batch32-2024_06_14_1aa9edfeb22f70b3d1f7g_0095_3.jpg", "batch32-2024_06_14_1aa9edfeb22f70b3d1f7g_0095_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: \n\nA. Draw $AC \\parallel BD$,\n\nthen $a: b = c: x$. The construction in this option is correct, which does not match the question's requirement;\n\nB. Draw $AC \\parallel BD$,\n\nthen $a: b = x: c$. The construction in this option is incorrect, which matches the question's requirement;\n\nC. Draw $AC \\parallel BD$,\n\nthen $a: b = c: x$. The construction in this option is correct, which does not match the question's requirement;\n\nD. Draw $AC \\parallel BD$,\n\nthen $a: b = c: x$. The construction in this option is correct, which does not match the question's requirement;\n\nTherefore, the correct choice is: B.\n\n[Key Insight] The question mainly tests the property of proportional segments divided by parallel lines. Mastering this property is crucial for solving the problem." }, { "problem_id": 509, "question": "As shown in Figure (1), there is an irregular pattern (the shaded part) on a flat ground. Kaidian, a student, wants to know the area of this pattern. He took the following approach: he used a rectangle with an area of $200 \\mathrm{~cm}^{2}$ to enclose the irregular pattern, then randomly threw small balls into the rectangular area at appropriate positions, and recorded the number of times the balls landed on the irregular pattern (balls landing on the boundaries or outside the rectangular area were not considered valid trials). He plotted the results of several valid trials in the form of a line graph, as shown in Figure (2). Based on this, he estimates the area of the irregular pattern to be approximately ( ).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $50 \\mathrm{~cm}^{2}$\nB. $55 \\mathrm{~cm}^{2}$\nC. $60 \\mathrm{~cm}^{2}$\nD. $110 \\mathrm{~cm}^{2}$", "input_image": [ "batch32-2024_06_14_1c0b16b6a4661c2e717dg_0002_1.jpg", "batch32-2024_06_14_1c0b16b6a4661c2e717dg_0002_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Statistics", "image_relavance": "1", "analysis": "Solution: Assume the area of the irregular shape is \\( x \\mathrm{~cm}^{2} \\).\n\nGiven: The area of the rectangle is \\( 200 \\mathrm{~cm}^{2} \\).\n\nAccording to the geometric probability formula, the probability that the ball lands on the irregular shape is: \\( \\frac{x}{200} \\).\n\nWhen the number of trials for event \\( A \\) is sufficiently large, i.e., the sample size is large enough, the frequency can be used as an estimate for the probability of event \\( A \\) occurring. Therefore, from the line graph, the probability that the ball lands on the irregular shape is approximately 0.55.\n\nThus, we have: \\( \\frac{x}{200} = 0.55 \\).\n\nSolving for \\( x \\): \\( x = 110 \\).\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This question tests geometric probability and the use of frequency to estimate probability. The key to solving the problem is understanding the context and determining that the probability of the ball landing within the irregular shape is approximately 0.55." }, { "problem_id": 510, "question": "As shown in Figure 1, a fair spinner is divided into 10 equal sections, labeled $1, 2, 3, 4, 5, 6, 7, 8, 9,$ and $10$. Xiaoye performs a frequency estimation probability experiment by spinning the spinner. The number the spinner lands on after stopping is considered the outcome of the experiment. Figure 2 shows Xiaoye's recorded results. Which of the following describes the experiment Xiaoye recorded?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Spinning the spinner and landing on an even number.\nB. Spinning the spinner and landing on a number divisible by 3.\nC. Spinning the spinner and landing on a number greater than 6.\nD. Spinning the spinner and landing on a number divisible by 5.", "input_image": [ "batch32-2024_06_14_1c0b16b6a4661c2e717dg_0005_1.jpg", "batch32-2024_06_14_1c0b16b6a4661c2e717dg_0005_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Statistics", "image_relavance": "1", "analysis": "Solution: From Figure 2, it can be seen that when the number of rotations is 600, the frequency is 0.3, so the probability of this event is approximately 0.3.\n\nA. After spinning the turntable, the probability of landing on an even number is 0.5, which does not match the given condition;\n\nB. After spinning the turntable, the probability of landing on a number divisible by 3. There are 10 numbers on the turntable, among which the numbers divisible by 3 are 3, 6, and 9, totaling 3 numbers. The probability is 0.3, which matches the given condition;\n\nC. After spinning the turntable, the probability of landing on a number greater than 6. There are 10 numbers on the turntable, among which the numbers greater than 6 are 7, 8, 9, and 10, totaling 4 numbers. The probability is 0.4, which does not match the given condition;\n\nD. After spinning the turntable, the probability of landing on a number divisible by 5. There are 10 numbers on the turntable, among which the numbers divisible by 5 are 5 and 10, totaling 2 numbers. The probability is 0.2, which does not match the given condition;\n\nTherefore, the correct answer is B.\n\n[Key Insight] This question tests the use of frequency to estimate probability. Mastering the method of using frequency to estimate probability is key to solving the problem." }, { "problem_id": 511, "question": "Zhou Dynasty mathematician Shang Gao summarized a method of measuring the height of an object using a \"square\" (as shown in Figure 1): Place the square's sides as shown in Figure 2, with the observer's eye at point A and the line of sight passing through point C. Let the observer's standing position be point B, and measure $BG = x$ (meters). Then, the height of the object $EG$ can be calculated. Given $BG = x (\\text{m}), EG = y (\\text{m})$, if $a = 30 \\text{ cm}, b = 60 \\text{ cm}$, and $AB = 1.6 \\text{ m}$, the function expression for $y$ in terms of $x$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $y=\\frac{1}{2} x$\nB. $y=\\frac{1}{2} x+1.6$\nC. $y=2 x+1.6$\nD. $y=\\frac{1800}{x}+1.6$", "input_image": [ "batch32-2024_06_14_20d9f7a744747cd4199bg_0011_1.jpg", "batch32-2024_06_14_20d9f7a744747cd4199bg_0011_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, quadrilateral $A B G F$ is a rectangle,\n\n$\\therefore A F=B G=x \\mathrm{~m}, F G=A B=1.6 \\mathrm{~m}$,\n\n$\\because E G=y \\mathrm{~m}$,\n\n$\\therefore E F=E G-F G=(y-1.6) \\mathrm{m}$,\nAlso, since $C D \\perp A F, E F \\perp A F$,\n\n$\\therefore C D \\| E F$,\n\n$\\therefore \\triangle A E F \\sim \\triangle A C D$,\n\n$\\therefore \\frac{E F}{C D}=\\frac{A F}{A D}$,\n\n$\\because C D=a=30 \\mathrm{~cm}=0.3 \\mathrm{~m}, A D=b=60 \\mathrm{~cm}=0.6 \\mathrm{~m}$,\n\n$\\therefore \\frac{y-1.6}{0.3}=\\frac{x}{0.6}$,\n\nSimplifying, we get: $y=\\frac{1}{2} x+1.6$,\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem examines the determination and properties of rectangles, the determination and properties of similar triangles, and the geometric application of linear functions. Mastering the determination and properties of similar triangles is crucial for solving the problem." }, { "problem_id": 512, "question": "As shown in Figure 1, point $\\mathrm{E}$ is on side $\\mathrm{AD}$ of rectangle $\\mathrm{ABCD}$. Points $\\mathrm{P}$ and $\\mathrm{Q}$ start from point $\\mathrm{B}$ simultaneously. Point $\\mathrm{P}$ moves along $\\mathrm{BE} \\rightarrow \\mathrm{ED} \\rightarrow \\mathrm{DC}$ until it reaches point $\\mathrm{C}$, and point $\\mathrm{Q}$ moves along $\\mathrm{BC}$ until it reaches point $\\mathrm{C}$. Both points move at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$. Let the area of $\\triangle \\mathrm{BPQ}$ be $\\mathrm{ycm}$ when $\\mathrm{P}$ and $\\mathrm{Q}$ have been moving for $\\mathrm{t}$ seconds. The graph of the function relationship between $\\mathrm{y}$ and $\\mathrm{t}$ is shown in Figure 2 (curve $\\mathrm{OM}$ is a part of a parabola). The following conclusions are given: (1) $A D=B E=5 \\mathrm{~cm}$; (2) when $0\n\nFigure (1)\n\n\n\nFigure (2)\nA. 4\nB. 3\nC. 2\nD. 1", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0007_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0007_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "According to Figure (2), when point P reaches point E, point Q reaches point C.\n\nSince both points P and Q move at a speed of 1 cm per second,\n\n\n\nTherefore, BC = BE = 5 cm. Hence, AD = BE = 5, so conclusion (1) is correct.\n\nAs shown in Figure 1, drawing a perpendicular PF from point P to BC at point F,\n\nGiven that the area of triangle BPQ remains constant at 10, we can deduce that AB = 4,\n\nSince AD is parallel to BC, angle AEB equals angle PBF.\n\nThus, sin angle PBF = sin angle AEB = AB / BE = 4/5.\n\nTherefore, PF = PB * sin angle PBF = (4/5)t.\n\nThus, when 0 < t ≤ 5, y = (1/2) * BQ * PF = (1/2) * t * (4/5)t = (2/5)t². So conclusion (2) is correct.\n\nFrom the constant area between 5 to 7 seconds, we can deduce that ED = 2,\n\nWhen point P moves to point C, the area becomes 0, and the distance traveled by point P is BE + ED + DC = 11, so the coordinates of point H are (11,0).\n\nAssuming the equation of line NH is y = kx + b,\n\nSubstituting the coordinates of point H(11,0) and point N(7,10) gives: \n\\[\n\\begin{cases}\n11k + b = 0 \\\\\n7k + b = 10\n\\end{cases}\n\\]\nSolving these gives: \n\\[\nk = -\\frac{5}{2}, \\quad b = \\frac{55}{2}\n\\]\n\nTherefore, the equation of line NH is: y = -\\(\\frac{5}{2}\\)t + \\(\\frac{55}{2}\\). Thus, conclusion (3) is incorrect.\n\nAs shown in Figure 2, when triangle ABE is similar to triangle QBP, point P is on DC,\n\n\n\nSince tan angle PBQ = tan angle ABE = 3/4, then PQ / BQ = 3/4, which means (11 - t)/5 = 3/4.\n\nSolving this gives: t = 29/4. So conclusion (4) is correct.\n\nIn summary, conclusions (1), (2), and (4) are correct, totaling 3. Therefore, the correct choice is B.\n\nKey points: Function graphs of moving point problems, double moving point problems, properties of rectangles, definitions of acute angle trigonometric functions, application of the method of undetermined coefficients, relationship between the coordinates of points on a curve and equations, properties of similar triangles, application of classification thinking." }, { "problem_id": 513, "question": "In parallelogram $A B C D$ as shown, the diagonals $A C$ and $B D$ intersect at point $O$, with $A C = 6$ and $B D = 8$. Point $P$ is arbitrarily on diagonal $B D$, and line $E F$ is drawn parallel to $A C$ through point $P$, intersecting the parallelogram's sides at points $E$ and $F$. Let $B P = x$ and $E F = y$. The graph that can roughly represent the relationship between $y$ and $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0016_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0016_2.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0016_3.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0016_4.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0016_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "When \\(0 \\leq x \\leq 4\\),\n\nSince \\(BO\\) is the median of \\(\\triangle ABC\\) and \\(EF \\parallel AC\\),\n\nTherefore, \\(BP\\) is the median of \\(\\triangle BEF\\), and \\(\\triangle BEF \\sim \\triangle BAC\\),\n\nThus, \\(\\frac{BP}{BO} = \\frac{EF}{AC}\\), which means \\(\\frac{x}{4} = \\frac{y}{6}\\), solving gives \\(y = \\frac{3}{2} x\\),\n\nSimilarly, when \\(4 < x \\leq 8\\), \\(y = \\frac{3}{2}(8 - x)\\).\n\nTherefore, the correct choice is D.\n\n【Key Point】This question tests the function graph of moving point problems, with the key to solving it lying in the use of triangle similarity." }, { "problem_id": 514, "question": "In $\\triangle A B C$ as shown, $\\angle C = 90^\\circ, \\angle B = 30^\\circ, A B = 10 \\text{ cm}$. Points $P$ and $Q$ start simultaneously from point $A$. Point $P$ moves at a speed of $2 \\text{ cm/s}$ along $A \\rightarrow B \\rightarrow C$, and point $Q$ moves at a speed of $1 \\text{ cm/s}$ along $A \\rightarrow C \\rightarrow B$. They stop after meeting. During this process, if the distance between points $P$ and $Q$, denoted as $P Q = y$, the relationship between $y$ and time $t$ can be best represented by the graph ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0020_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0020_2.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0020_3.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0020_4.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0020_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: In triangle \\( \\triangle ABC \\), \\( \\angle C = 90^\\circ \\), \\( \\angle B = 30^\\circ \\), and \\( AB = 10 \\).\n\nTherefore, \\( AC = 5 \\), and \\( \\frac{AC}{AB} = \\frac{1}{2} \\).\n\nI. When \\( 0 \\leq t \\leq 5 \\), point \\( P \\) is on \\( AB \\), and point \\( Q \\) is on \\( AC \\). According to the problem, \\( AP = 2t \\), and \\( AQ = t \\).\n\nFrom the given conditions: \\( \\frac{AQ}{AP} = \\frac{1}{2} \\).\n\nSince \\( \\angle A = \\angle A \\), it follows that \\( \\triangle APQ \\sim \\triangle ABC \\).\n\nThus, \\( \\angle AQP = \\angle C = 90^\\circ \\), and \\( PQ = \\sqrt{3} t \\).\n\nII. When \\( t > 5 \\), points \\( P \\) and \\( Q \\) are on \\( BC \\). According to the problem, the distance traveled by \\( P \\) is \\( 2t \\), and the distance traveled by \\( Q \\) is \\( t \\).\n\nTherefore, \\( PQ = 15 + 5\\sqrt{3} - 3t \\).\n\nHence, the correct answer is: A.\n\n[Key Insight] This problem primarily examines the functional relationship in the context of moving points. Correctly understanding that the length of \\( PQ \\) is a linear function of time and deriving the functional relationship is crucial to solving the problem." }, { "problem_id": 515, "question": "As shown in the figure, in rhombus $A B C D$, the diagonals $A C$ and $B D$ are drawn, intersecting at point $O$. Given $A C = 4$ and $B D = 2$. Point $C$ is the midpoint of the side $E F$ of the rectangle $E F G H$, where $E F = B D$ and $F G = A C$. Line $l$ passes through points $A$ and $C$. Rhombus $A B C D$ moves to the right along line $l$ at a speed of 1 unit per second until point $C$ reaches the side $G H$. Which of the following graphs approximately represents the relationship between the perimeter $y$ of the rhombus when it is inside the rectangle and the time $t$ of the movement?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0024_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0024_2.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0024_3.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0024_4.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0024_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let $AC$ intersect $EF$ at point $R$, then $CR = t$.\n\nSince in rhombus $ABCD$, $AC = 4$ and $BC = 2$,\n\n$\\therefore OD = 1$, and $CO = 2$,\n\n$\\therefore CD = \\sqrt{1^2 + 2^2} = \\sqrt{5}$.\n\nAs shown in Figure (1), when $BD$ is on the left side of $EF$, let $EF$ intersect $CD$ and $CB$ at points $M$ and $N$ respectively.\n\nSince $MN \\parallel DB$,\n\n$\\therefore \\triangle CMR \\sim \\triangle CDO$,\n\n$\\therefore \\frac{CR}{CO} = \\frac{CM}{CD}$, that is, $\\frac{t}{2} = \\frac{CM}{\\sqrt{5}}$,\n\n$\\therefore CM = \\frac{\\sqrt{5}}{2} t$,\n\n$\\therefore y = 2CM = \\sqrt{5} t$.\n\nAs shown in Figure (2), when $BD$ is on the right side of $EF$, let $EF$ intersect $AD$ and $AB$ at points $P$ and $Q$ respectively, and $AR = 4 - t$.\n\nSince $PQ \\parallel DB$,\n\n$\\therefore \\triangle APR \\sim \\triangle ADO$,\n\n$\\therefore \\frac{AR}{AO} = \\frac{AP}{AD}$, that is, $\\frac{4 - t}{2} = \\frac{AP}{\\sqrt{5}}$,\n\n$\\therefore AP = \\frac{\\sqrt{5}}{2}(4 - t)$,\n\n$\\therefore PD + CD = AD + CD - AP = 2\\sqrt{5} - \\frac{\\sqrt{5}}{2}(4 - t) = \\frac{\\sqrt{5}}{2} t$,\n\n$\\therefore y = 2(PD + CD) = \\sqrt{5} t$.\n\nIn summary, the functional relationship between the perimeter $y$ of the rhombus entering the rectangle and the time $t$ of movement is $y = \\sqrt{5} t$ ($0 \\leq t \\leq 4$).\n\nTherefore, the correct answer is: B\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n【Key Insight】This problem examines the properties of translation and the properties of similar triangles to derive a linear function. The key to solving the problem is to express the value of $CE$ or $DP$ using an algebraic expression containing $t$." }, { "problem_id": 516, "question": "As shown in Figure (1), in rectangle $A B C D$, when the right-angled triangle ruler $M P N$ with its right-angle vertex $P$ moving along $B C$, the right-angled side $M P$ always passes through point $A$. Let the other right-angled side $P N$ intersect $C D$ at point $Q$. During the movement, the length of segment $B P$ is $x$, and the length of segment $C Q$ is $y$. The relationship between $y$ and $x$ is shown in Figure (2). The length of $A B$ is $($ ).\n\n\n\n(1)\n\n\n\n(2)\nA. 2.25\nB. 3\nC. 4\nD. 6", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0059_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0059_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( AB = m \\).\n\nGiven that the maximum value of \\( x \\) from the function graph is 6,\n\n\\[\n\\therefore BC = 6.\n\\]\n\nSince quadrilateral \\( ABCD \\) is a rectangle,\n\n\\[\n\\therefore \\angle B = \\angle C = 90^\\circ.\n\\]\n\nMoreover, because of the right-angled triangle, \\( \\angle APQ = 90^\\circ \\),\n\n\\[\n\\therefore \\angle APB + \\angle BAP = \\angle APB + \\angle QPC = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle BAP = \\angle CPQ.\n\\]\n\nThus, triangles \\( \\triangle ABP \\) and \\( \\triangle PCQ \\) are similar,\n\n\\[\n\\therefore \\frac{AB}{PC} = \\frac{BP}{CQ}, \\quad \\text{that is} \\quad \\frac{m}{6 - x} = \\frac{x}{y},\n\\]\n\n\\[\n\\therefore y = \\frac{1}{m} x(6 - x).\n\\]\n\nWhen \\( x = 3 \\), the function reaches its maximum value of 2.25,\n\n\\[\n\\therefore y = \\frac{1}{m} \\times 3(6 - 3) = 2.25,\n\\]\n\n\\[\n\\therefore m = \\frac{9}{2.25} = 4.\n\\]\n\nTherefore, the correct choice is: \\( C \\).\n\n**Key Insight:** This problem tests the properties of rectangles, the characteristics of right-angled triangles, the criteria and properties of similar triangles, and the ability to interpret information from the graph of a quadratic function. Correctly understanding the graph information and flexibly applying the learned knowledge is crucial for solving the problem." }, { "problem_id": 517, "question": "In triangle $ABC$, the side length $AB=2$, the area is 1, and line $PQ$ is parallel to $BC$, intersecting $AB$ and $AC$ at $P$ and $Q$ respectively. If $AP=t$ and the area of triangle $APQ$ is $S$, the graph of $S$ as a function of $t$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0062_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0062_2.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0062_3.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0062_4.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0062_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( PQ \\parallel BC \\),\n\n\\[\n\\angle AQP = \\angle C, \\quad \\angle APQ = \\angle B\n\\]\n\n\\[\n\\therefore \\triangle APQ \\sim \\triangle ABC,\n\\]\n\n\\[\n\\therefore \\frac{S_{\\triangle AQP}}{S_{\\triangle ACB}} = \\frac{AP^2}{AB^2} = \\left(\\frac{AP}{AB}\\right)^2\n\\]\n\n\\[\n\\therefore S = \\left(\\frac{AP}{AB}\\right)^2,\n\\]\n\n\\[\n\\therefore \\left(\\frac{t}{2}\\right)^2 = S,\n\\]\n\n\\[\n\\therefore S = \\frac{1}{4} t^2, \\quad 0 \\leq t \\leq 2,\n\\]\n\nConsidering the graph of the quadratic function, the corresponding graph is \\( B \\).\n\nTherefore, the correct choice is: \\( B \\).\n\n【Key Insight】This problem examines the determination and properties of similar triangles, the expression of a quadratic function, and the graph of a quadratic function. Mastering the determination and properties of similar triangles, the expression of a quadratic function, and the graph of a quadratic function is crucial for solving the problem." }, { "problem_id": 518, "question": "As shown in the figure, in rhombus $A B C D$, the diagonals $A C$ and $B D$ intersect at point $O$, with $A C = 6$ and $B D = 8$. Point $P$ starts at point $B$ and moves along the sides $A B$ and $A D$ of the rhombus in the direction $B \\rightarrow A \\rightarrow D$, stopping at point $D$. Point $P^{\\prime}$ is the reflection of point $P$ across $B D$, and line segment $P P^{\\prime}$ intersects $B D$ at point $M$. If $B M = x (0 < x < 8)$, the area of triangle $D P P^{\\prime}$ is $y$. Which of the following graphs correctly represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0072_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0072_2.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0072_3.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0072_4.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0072_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a rhombus,\n\nTherefore, \\( AB = BC = CD = DA \\), \\( OA = \\frac{1}{2} AC = 3 \\), \\( OB = \\frac{1}{2} BD = 4 \\), and \\( AC \\perp BD \\).\n\n(1) When \\( BM \\leq 4 \\),\n\nSince point \\( P' \\) is symmetric to point \\( P \\) with respect to \\( BD \\),\nTherefore, \\( P'P \\perp BD \\),\n\nThus, \\( P'P \\parallel AC \\),\n\nHence, \\( \\triangle P'BP \\sim \\triangle CBA \\),\n\nTherefore, \\( \\frac{PP'}{AC} = \\frac{BM}{OB} \\), which means \\( \\frac{PP'}{6} = \\frac{x}{4} \\),\n\nSo, \\( PP' = \\frac{3}{2}x \\),\n\nSince \\( DM = 8 - x \\),\n\nTherefore, the area of \\( \\triangle DPP' \\) is \\( y = \\frac{1}{2} PP' \\cdot DM = \\frac{1}{2} \\times \\frac{3}{2}x(8 - x) = -\\frac{3}{4}x^2 + 6x \\);\n\nThus, the function graph of \\( y \\) versus \\( x \\) is a parabola opening downward, passing through (0, 0) and (4, 12);\n\n(2) When \\( BM \\geq 4 \\), as shown in the figure:\n\n\n\nSimilarly, \\( \\triangle P'DP \\sim \\triangle CDA \\),\n\nTherefore, \\( \\frac{PP'}{AC} = \\frac{DM}{OD} \\), which means \\( \\frac{PP'}{6} = \\frac{8 - x}{4} \\),\n\nSo, \\( PP' = \\frac{3}{2}(8 - x) \\),\n\nTherefore, the area of \\( \\triangle DPP' \\) is \\( y = \\frac{1}{2} PP' \\cdot DM = \\frac{1}{2} \\times \\frac{3}{2} (8 - x)^2 = \\frac{3}{4} (8 - x)^2 \\);\n\nThus, the function graph of \\( y \\) versus \\( x \\) is a parabola opening upward, passing through (4, 12) and (8, 0); In summary, the function graph of \\( y \\) versus \\( x \\) is approximately:\n\n\n\nTherefore, the correct choice is: D.\n\n【Key Insight】This problem examines the function graph of moving point problems, the properties of rhombuses, the determination and properties of similar triangles, the calculation of triangle areas, and the application of quadratic functions; mastering the properties of rhombuses and deriving the quadratic function expression based on the given conditions are crucial for solving the problem." }, { "problem_id": 519, "question": "In triangle $ABC$, $\\angle A = 90^\\circ, AB = AC = 1$, and $BF \\perp BC$. Point $D$ starts from point $A$ and moves along segment $AB$ at a constant speed of 1 unit per second, while point $E$ starts from point $B$ and moves along ray $BF$ at a constant speed of $\\sqrt{2}$ units per second. Both points stop moving when point $D$ coincides with point $B$. Connect $DE$, $EC$, and $CD$. Let the time it takes for point $D$ to move be $x$ seconds, and the area of triangle $CDE$ be $y$. Which of the following graphs approximately represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0079_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0079_2.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0079_3.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0079_4.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0079_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, let $BC$ intersect $DE$ at point $P$, and draw $DQ \\perp BC$ at point $Q$, then $\\angle BQD = 90^\\circ$.\n\n\n\nAccording to the problem, we have: $AD = x$, $BE = \\sqrt{2}x$, thus $BD = 1 - x$.\n\nSince $\\angle A = 90^\\circ$ and $AB = AC = 1$,\n\n$\\therefore \\angle ABC = 45^\\circ$, and $BC = \\sqrt{2}$.\n\nTherefore, $\\triangle BDQ$ is an isosceles right triangle,\n\n$\\therefore BQ = DQ = \\frac{\\sqrt{2}}{2} BD = \\frac{\\sqrt{2}}{2}(1 - x)$.\n\nSince $BF \\perp BC$,\n\n$\\therefore BF \\parallel DQ$,\n\n$\\therefore \\triangle BEP \\sim \\triangle QDP$,\n\n$\\therefore \\frac{BE}{DQ} = \\frac{BP}{PQ}$, that is, $\\frac{\\sqrt{2}x}{\\frac{\\sqrt{2}}{2}(1 - x)} = \\frac{BP}{\\frac{\\sqrt{2}}{2}(1 - x) - BP}$.\n\nSolving this, we get: $BP = \\frac{\\sqrt{2}x(1 - x)}{1 + x}$.\n\n$\\therefore PC = \\sqrt{2} - \\frac{\\sqrt{2}x(1 - x)}{1 + x} = \\frac{\\sqrt{2}x^2 + \\sqrt{2}}{1 + x}$.\n\n$\\therefore y = S_{\\triangle CDP} + S_{\\triangle CEP} = \\frac{1}{2}\\left(\\frac{\\sqrt{2}}{2}(1 - x) + \\sqrt{2}x\\right) \\times \\frac{\\sqrt{2}x^2 + \\sqrt{2}}{1 + x} = \\frac{1}{2}x^2 + \\frac{1}{2}$.\n\nSince $x \\geq 0$,\n\n$\\therefore$ the graph of this function is a segment of a parabola located on and to the right of the $y$-axis.\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem mainly examines the moving point problem in function graphs and the graph of quadratic functions. The key to solving the problem is to derive the function's analytical expression based on the given conditions." }, { "problem_id": 520, "question": "As shown in the figure, in rectangle $A B C D$, $A B = 2$ and $B C = 4$. Point $M$ starts from point $A$ and moves at a constant speed of 1 unit per second along the path $A \\rightarrow B \\rightarrow C$. A perpendicular line is drawn from point $M$ to diagonal $A C$, with the foot of the perpendicular being point $N$. Let the time of motion be $t$ seconds, and the area of triangle $A M N$ be $S$. Which of the following graphs can roughly represent the functional relationship between $S$ and $t$? $(\\quad)$\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0081_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0081_2.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0081_3.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0081_4.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0081_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: In the right triangle $\\triangle ABC$, $AC = \\sqrt{AB^2 + BC^2} = \\sqrt{2^2 + 4^2} = 2\\sqrt{5}$.\n\nThe area of $\\triangle ABC$ is $S_{\\triangle ABC} = \\frac{1}{2} \\times BC \\times AB = \\frac{1}{2} \\times 4 \\times 2 = 4$.\n\nWhen $0 < t \\leq 2$, point $M$ is on $AB$, and from the problem, $AM = t$.\n\nSince $\\angle MAN = \\angle CAB$ and $\\angle ANM = \\angle ABC = 90^\\circ$,\n\n$\\triangle AMN \\sim \\triangle ACB$,\n\nThus, $\\frac{S_{\\triangle AMN}}{S_{\\triangle ACB}} = \\left(\\frac{AM}{AC}\\right)^2$, which means $\\frac{S}{4} = \\left(\\frac{t}{2\\sqrt{5}}\\right)^2$, so $S = \\frac{1}{5}t^2$.\n\nWhen $2 < t < 6$, point $M$ is on $BC$, as shown in the figure, and $CM = 6 - t$.\n\nSince $\\angle MCN = \\angle ACB$ and $\\angle CNM = \\angle CBA = 90^\\circ$,\n\n$\\triangle CMN \\sim \\triangle CAB$,\n\nThus, $\\frac{CM}{AC} = \\frac{CN}{BC} = \\frac{MN}{AB}$, which means $\\frac{6 - t}{2\\sqrt{5}} = \\frac{CN}{4} = \\frac{MN}{2}$,\n\nSo, $MN = \\frac{6 - t}{\\sqrt{5}}$, $CN = \\frac{12 - 2t}{\\sqrt{5}}$,\n\nTherefore, $AN = AC - CN = 2\\sqrt{5} - \\frac{12 - 2t}{\\sqrt{5}} = \\frac{2t - 2}{\\sqrt{5}}$,\n\nThus, $S = \\frac{1}{2} \\times AN \\times MN = \\frac{1}{2} \\times \\frac{2t - 2}{\\sqrt{5}} \\times \\frac{6 - t}{\\sqrt{5}} = -\\frac{1}{5}t^2 + \\frac{7}{5}t - \\frac{6}{5}$.\n\nIn summary, $S = \\begin{cases} \n\\frac{1}{5}t^2, & 0 < t \\leq 2, \\\\ \n-\\frac{1}{5}t^2 + \\frac{7}{5}t - \\frac{6}{5}, & 2 < t < 6.\n\\end{cases}$\n\nTherefore, the option that reflects the functional relationship between $S$ and $t$ is option B.\n\nAnswer: B.\n\n[Highlight] This problem examines the practical application of quadratic functions in geometry, similar triangles, and the Pythagorean theorem. The key to solving the problem is deriving the function expression and analyzing the options based on the graph of the quadratic function." }, { "problem_id": 521, "question": "The two triangles in each group of figures below satisfy $\\triangle A B C \\sim \\triangle D E F$. Which pair of triangles is not a pair of similar figures by dilation?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_28c345e53f655ea57a93g_0023_1.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0023_2.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0023_3.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0023_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: Two similar polygons whose corresponding vertices' connecting lines intersect at a single point are called homothetic figures.\n\nBased on the concept of homothetic figures, among the three figures A, C, and D, any two are homothetic figures;\n\nThe two figures in B do not conform to the concept of homothetic figures as their corresponding sides are not parallel, hence they are not homothetic figures.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question primarily examines the concept of homothety, noting that homothety and similarity are related yet distinct. Similarity only requires that two figures have exactly the same shape, whereas homothety additionally requires that the connecting lines of corresponding points intersect at a single point." }, { "problem_id": 522, "question": "The following pairs of figures are not similar figures ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_28c345e53f655ea57a93g_0045_1.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0045_2.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0045_3.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0045_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: In options $A$, $B$, and $C$, both figures are homothetic (similar) figures. The center of homothety in $A$ is point $C$,\n\n\n\nin $B$ it is point $O$,\n\n\n\nand in $C$ it is also point $O$.\n\n\n\nOnly in option $D$ do the connecting lines of corresponding vertices not intersect at a single point, and the corresponding sides are not parallel. Therefore, $D$ is not a homothetic image. Hence, the correct choice is $D$.\n\n【Key Insight】This question tests the concept of homothety transformation. The key to solving it is understanding that two figures are homothetic not only if they are similar but also if the lines connecting corresponding vertices intersect at a single point and the corresponding sides are parallel." }, { "problem_id": 523, "question": "As shown in the figure, there are several ways to draw a triangle similar to $\\triangle A B C$. Among these, the correct ones are $($ )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch32-2024_06_14_28c345e53f655ea57a93g_0046_1.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0046_2.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0046_3.jpg", "batch32-2024_06_14_28c345e53f655ea57a93g_0046_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the method of drawing similar figures, all four figures are similar to $\\triangle ABC$.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question primarily examines the concept of similarity transformation. Understanding the definition of similar figures is crucial for solving the problem." }, { "problem_id": 524, "question": "As shown in the figure, there is a pair of triangles, $\\triangle A D E$ and $\\triangle A B C$, where $A D = A B$. The vertex $A$ of both triangles coincides. $\\triangle A D E$ is rotated around its vertex $A$. Among the following four positions, which one does not have similar triangles?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_3c613934e88ba6f21b85g_0010_1.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0010_2.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0010_3.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0010_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Option B, since $\\angle BCA = \\angle AED = 90^\\circ$ and $\\angle CAF = \\angle DAE$,\n\n$\\therefore \\triangle ACF \\sim \\triangle AED$, hence option B does not meet the requirement;\n\nOption C, as shown in the figure, let the intersection of $CD$ and $AE$ be point $O$,\n\n\n\nSince $\\angle ACO = \\angle DEO = 90^\\circ$ and $\\angle AOC = \\angle DOE$,\n\n$\\therefore \\triangle ACO \\sim \\triangle DEO$, hence option C does not meet the requirement;\n\nOption D, since $AD = AB$, $\\angle B = 30^\\circ$, and $\\angle D = 45^\\circ$,\n\n$\\therefore AC = \\frac{1}{2} AB$, and $AE = \\frac{\\sqrt{2}}{2} AD = \\frac{\\sqrt{2}}{2} AB$,\n\n$\\therefore CE = \\sqrt{AE^2 - AC^2} = \\frac{1}{2} AB$,\n\n$\\therefore AC = EC$,\n\n$\\therefore \\angle CAE = 45^\\circ = \\angle EAD$,\n\nAlso, since $\\angle C = \\angle AED = 90^\\circ$,\n\n$\\therefore \\triangle ACE \\sim \\triangle AED$, hence option D does not meet the requirement,\n\nTherefore, the correct choice is: A.\n\n【Key Insight】This question tests the determination of similar triangles and the properties of right-angled triangles. Mastering the method of determining similar triangles is key to solving the problem." }, { "problem_id": 525, "question": "As shown in the figure, in $\\triangle A B C$, $\\angle A = 78^\\circ, A B = 4, A C = 6$. The triangle $\\triangle A B C$ is cut along the lines in the figure. Among the following four methods of cutting, which of the following methods result in a shaded similar triangle to the original triangle?\n\n\n\nFigure 1\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)\nB. (3)(4)\nC. (1)(2)(3)(4)\nD. (1)(2)(4)", "input_image": [ "batch32-2024_06_14_3c613934e88ba6f21b85g_0025_1.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0025_2.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0025_3.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0025_4.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0025_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: (1) The triangle in the shaded area has two angles equal to those of the original triangle; thus, it can be determined that the two triangles are similar. (2) The triangle in the shaded area has two angles equal to those of the original triangle; thus, it can be determined that the two triangles are similar.\n\n(3) Since the length of $BC$ is not provided, the relationship $\\frac{2}{AB} = \\frac{3}{BC}$ cannot be established; therefore, it cannot be determined that the two triangles are similar.\n\n(4) The lengths of the two sides of the triangle in the shaded area are: $AB - 1 = 3$ and $AC - 4 = 2$; based on $\\frac{2}{4} = \\frac{3}{6}$ and $\\angle A = \\angle A$, it can be determined that the two triangles are similar.\n\nTherefore, the correct answer is: D.\n\n[Key Insight] This question tests the determination of similar triangles; mastering the method of determining similar triangles is key to solving the problem." }, { "problem_id": 526, "question": "Given that in $\\triangle A B C$, $\\angle A = 78^\\circ, A B = 4, A C = 6$, which of the following options has a shaded triangle that is not similar to the original $\\triangle A B C$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_3c613934e88ba6f21b85g_0037_1.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0037_2.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0037_3.jpg", "batch32-2024_06_14_3c613934e88ba6f21b85g_0037_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Since two triangles with two corresponding angles equal are similar, it can be proven that the shaded triangle is similar to the original $\\triangle ABC$. Therefore, option A does not meet the requirement of the question;\n\nB. Although the corresponding sides are proportional, the included angles are not necessarily equal. Hence, it cannot be proven that the shaded triangle is similar to the original $\\triangle ABC$, making option B the correct choice;\n\nC. Similar to option A, since two triangles with two corresponding angles equal are similar, it can be proven that the shaded triangle is similar to the original $\\triangle ABC$. Therefore, option C does not meet the requirement of the question;\n\nD. Two triangles are similar if the ratios of two corresponding sides are equal and the included angles are equal. Therefore, option D does not meet the requirement of the question. Thus, the correct choice is B.\n\n[Key Insight] This question tests the understanding of the criteria for determining similar triangles. Mastering the methods to determine similar triangles is crucial for solving such problems." }, { "problem_id": 527, "question": "Given segments $m$ and $n$, construct segment $x$ such that $x = \\frac{b^2}{2a}$. Which of the following constructions is correct?\n\nA.\n\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_3faf974290235a6e5797g_0062_1.jpg", "batch32-2024_06_14_3faf974290235a6e5797g_0062_2.jpg", "batch32-2024_06_14_3faf974290235a6e5797g_0062_3.jpg", "batch32-2024_06_14_3faf974290235a6e5797g_0062_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: \n\nA. According to the problem, we have: $\\frac{x}{2 a}=\\frac{b}{b}$, which implies $x=2 a$. Therefore, this option does not meet the requirement of the problem;\n\nB. According to the problem, we have: $\\frac{2 a}{x}=\\frac{b}{b}$, which implies $x=2 a$. Therefore, this option does not meet the requirement of the problem;\n\nC. According to the problem, we have: $\\frac{x}{b}=\\frac{b}{2 a}$, which implies $x=\\frac{b^{2}}{2 a}$. Therefore, this option meets the requirement of the problem;\n\nD. According to the problem, we have: $\\frac{b}{b}=\\frac{x}{2 a}$, which implies $x=2 a$. Therefore, this option does not meet the requirement of the problem;\n\nHence, the correct choice is: C\n\n[Key Insight] This problem mainly tests the proportional division of segments by parallel lines and complex constructions. It is crucial to understand that a line parallel to one side of a triangle that intersects the other two sides (or their extensions) divides those sides proportionally." }, { "problem_id": 528, "question": "We know that the centroid of $\\triangle A B C$ is the intersection point $G$ of the three medians $A D, B E, C F$. As shown in Figure 1, $\\frac{G D}{A D} = \\frac{G E}{B E} = \\frac{G F}{C F} = \\frac{1}{3}$. In Figure 2, for the right-angled triangle $A B C$ with $\\angle C = 90^\\circ$, $A C = 4$, and $B C = 8$, $A B C$ is rotated around its centroid $G$. The corresponding points of $A, B, C$ are $A_1, B_1, C_1$. The value closest to the maximum possible length of $C A_1$ is ( )\n\n\n\nFigure 1\n\n\nA. 5.5\nB. 6.5\nC. 7.5\nD. 8.5", "input_image": [ "batch32-2024_06_14_4a07680e90202dde6dedg_0080_1.jpg", "batch32-2024_06_14_4a07680e90202dde6dedg_0080_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Connect $AG$ and extend it to intersect $BC$ at point $E$,\n\n\n\nFigure 2\n\n$\\because$ Point $G$ is the centroid of $\\triangle ABC$, and $BC=8$,\n\n$\\therefore CE=4$, and $AG:AE=2:3$,\n\n$\\therefore AE=\\sqrt{AC^{2}+CE^{2}}=\\sqrt{4^{2}+4^{2}}=4 \\sqrt{2}$,\n\n$\\therefore AG=\\frac{2}{3} AE=\\frac{2}{3} \\times 4 \\sqrt{2}=\\frac{8 \\sqrt{2}}{3}$,\n\nWith point $G$ as the center and $OG$ as the radius, draw a circle. Connect $CG$ and extend it to intersect $\\odot G$ at point $F$, and intersect $AB$ at point $H$. At this point, the length of $CF$ is the maximum value of $CA_{1}$,\n\n$\\because AC=4, \\quad BC=8$,\n\n$\\therefore AB=4 \\sqrt{5}$,\n\n$\\because$ Point $G$ is the centroid of $\\triangle ABC$, and $\\angle ACB=90^{\\circ}$,\n\n$\\therefore CH=\\frac{1}{2} AB=2 \\sqrt{5}$,\n\n$\\therefore CG=\\frac{2}{3} CH=\\frac{2}{3} \\times 2 \\sqrt{5}=\\frac{4 \\sqrt{5}}{3}$,\n\n$\\therefore CF=CG+GF=CG+GA=\\frac{4 \\sqrt{5}}{3}+\\frac{8 \\sqrt{2}}{3} \\approx 6.75$,\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem examines the centroid of a triangle and the properties of rotation. The key to solving the problem lies in determining the position of point $A_{I}$ when it is farthest from point $C$ using the properties of rotation." }, { "problem_id": 529, "question": "As shown in the figure, in $\\triangle A B C$, $\\angle A = 78^\\circ, A B = 6, A C = 9$. By cutting along the dashed lines in the figure, which of the following shaded triangles is not similar to the original triangle $\\triangle A B C$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_5a8449461015d69a47d9g_0018_1.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0018_2.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0018_3.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0018_4.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0018_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "A. There are two sides with corresponding proportional lengths, but the included angles are not equal, so the two triangles are not similar, which fits the question's requirement.\n\nB. $\\frac{6-3}{9-7}=\\frac{3}{2}, \\frac{9}{6}=\\frac{3}{2}, \\angle A=\\angle A$. The two triangles have two sides with corresponding proportional lengths and equal included angles, so the two triangles are similar, which does not fit the question's requirement.\n\nC. The triangle in the shaded part has two angles equal to those of the original triangle, so the two triangles are similar, which does not fit the question's requirement.\n\nD. The triangle in the shaded part has two angles equal to those of the original triangle, so the two triangles are similar, which does not fit the question's requirement. Therefore, the correct choice is: A.\n\n[Key Insight] This question tests the determination of similar triangles. If two angles of one triangle are equal to two angles of another triangle, the two triangles are similar; if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, the two triangles are similar; if all three sides of one triangle are proportional to all three sides of another triangle, the two triangles are similar." }, { "problem_id": 530, "question": "Question: As shown in Figure 1, in rectangle paper $A B C D$, $A B = 2$ and $B C = 3$. The task is to cut the rectangle into two pieces without overlapping or leaving gaps, and then join to form a square. Students A and B determined the side length of the square to be $\\sqrt{6}$ based on the fact that the area remains unchanged before and after cutting and pasting, and designed the following plans.\n\n A: As shown in Figure 2, find point $E$ on $A D$ such that $B E = \\sqrt{6}$, and draw $\\angle D C F = \\angle A E B$, which intersects $B E$ at point $F$ to complete the cut;\n\n B: As shown in Figure 3, find point $F$ on $A D$ such that $C F = \\sqrt{6}$. Draw a circle with $B C$ as the diameter, which intersects $C F$ at point $E$. Connect $B E$ to complete the cut. Which of the following conclusions is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. Neither A nor B's cutting is correct\nB. Both A and B's cutting are correct\nC. B's cutting is correct, and in Figure 3, $A F = \\sqrt{2}$\nD. A's cutting is correct, and in Figure 2, $A E = 2 \\sqrt{2}$", "input_image": [ "batch32-2024_06_14_5a8449461015d69a47d9g_0082_1.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0082_2.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0082_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The side length of the formed square is: $\\sqrt{2 \\times 3}=\\sqrt{6}$.\n\nFor Part A: As shown in the figure\n\n\n\n$\\because \\angle D C F=\\angle A E B$,\n\n$\\because \\angle A E B+\\angle D E F=180^{\\circ}$,\n\n$\\therefore \\angle D C F+\\angle D E F=180^{\\circ}$,\n\n$\\therefore \\angle D+\\angle E F C=180^{\\circ}$,\n\n$\\therefore \\angle E F C=90^{\\circ}=\\angle A$,\n\n$\\therefore \\triangle B A E \\sim \\triangle C F B$,\n\n$\\therefore \\frac{B A}{C F}=\\frac{B E}{B C}$\n\n$\\therefore \\frac{2}{C F}=\\frac{\\sqrt{6}}{3}$\n\nSolving gives: $C F=\\sqrt{6}$\n\n$\\therefore C F=B E$\n\n$\\therefore$ Translating $\\triangle A B E$ to $\\triangle C D M$, and translating $\\triangle C B F$ to $\\triangle M E N$, yields the square $C F N M$.\n\nFor Part B: $\\because B C$ is the diameter of the circle,\n\n$\\therefore \\angle B E C=90^{\\circ}$,\n\nSimilarly, we can derive $\\triangle B E C \\sim \\triangle C D F, C F=B E=\\sqrt{6}$,\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem examines the properties of rectangles, the determination of squares, the properties and determination of similar triangles, translation, the fact that the angle subtended by the diameter is a right angle, and proving $C F=B E=\\sqrt{6}$ is key to solving the problem." }, { "problem_id": 531, "question": "As shown in the figure, in rectangle $A B C D$, $B C = 4$, $A B = 2$. Point $P$ starts from point $A$ and moves along the broken line $A-D-C$. A perpendicular line is drawn from point $P$ to the diagonal $A C$, intersecting the broken line $A B C$ at point $Q$. Let the distance traveled by point $P$ be $x$, and the area of triangle $A P Q$ be $y$. The graph of $y$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_5a8449461015d69a47d9g_0096_1.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0096_2.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0096_3.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0096_4.jpg", "batch32-2024_06_14_5a8449461015d69a47d9g_0096_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, when \\(0 < x < 1\\), point \\(P\\) is on side \\(AB\\).\n\n\\[\n\\because \\angle DAC + \\angle BAC = \\angle AQP + \\angle BAC = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle DAC = \\angle AQP,\n\\]\n\\[\n\\because \\angle ADC = \\angle QAP = 90^\\circ,\n\\]\n\\[\n\\therefore \\triangle ADC \\sim \\triangle QAP,\n\\]\n\\[\n\\therefore \\frac{AP}{CD} = \\frac{AQ}{AD}, \\text{ i.e., } \\frac{x}{2} = \\frac{AQ}{4},\n\\]\n\\[\n\\therefore AQ = 2x,\n\\]\n\\[\n\\therefore S_{\\triangle APQ} = \\frac{1}{2} AP \\cdot AQ = \\frac{1}{2} \\times 2x \\cdot x = x^2,\n\\]\n\\[\n\\therefore \\text{The graph is an upward-opening parabola.}\n\\]\n\n\n\nAs shown in the figure, when \\(1 \\leq x \\leq 4\\), point \\(P\\) is on side \\(AD\\), and point \\(Q\\) is on side \\(BC\\).\n\nIn \\(\\triangle APQ\\), the height corresponding to side \\(AP\\) is 2, so \\(S_{\\triangle APQ} = \\frac{1}{2} \\times 2 \\times AP = x\\).\n\nAt this time, the area of \\(\\triangle APQ\\) increases as \\(AP\\) increases and is a linear function of \\(x\\).\n\n\\[\n\\therefore \\text{When } 1 \\leq x \\leq 4, \\text{ the graph is a line segment.}\n\\]\n\n\n\nAs shown in the figure, when \\(4 < x < 6\\), point \\(P\\) is on side \\(CD\\), and point \\(Q\\) is on side \\(BC\\).\n\n\n\nIn rectangle \\(ABCD\\), \\(\\angle ADC = \\angle BCD = 90^\\circ\\),\n\n\\[\n\\therefore \\angle CAD + \\angle ACD = 90^\\circ,\n\\]\n\\[\n\\because PQ \\perp AC,\n\\]\n\\[\n\\therefore \\angle CPQ + \\angle ACD = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle CAD = \\angle CPQ,\n\\]\n\\[\n\\therefore \\triangle ACD \\sim \\triangle PQC,\n\\]\n\\[\n\\therefore \\frac{AD}{CP} = \\frac{CD}{CQ}, \\text{ i.e., } \\frac{4}{4 + 2 - x} = \\frac{2}{CQ},\n\\]\n\\[\n\\therefore CQ = 3 - \\frac{1}{2}x,\n\\]\n\\[\n\\therefore BQ = BC - CQ = 1 + \\frac{1}{2}x,\n\\]\n\\[\n\\therefore S_{\\triangle APQ} = 2 \\times 4 - \\frac{1}{2} \\times 2 \\times \\left(1 + \\frac{1}{2}x\\right) - \\frac{1}{2} \\times 4 \\times (x - 4) - \\frac{1}{2} \\times (6 - x)\\left(3 - \\frac{1}{2}x\\right)\n\\]\n\\[\n= -\\frac{1}{4}x^2 + \\frac{1}{2}x + 6\n\\]\n\\[\n\\therefore \\text{When } 4 < x < 6, \\text{ the graph is a downward-opening parabola.}\n\\]\n\nTherefore, the correct answer is: C\n\n【Highlight】This problem examines the function image of a moving point, the determination and properties of similar triangles, and the image of a quadratic function. The key to solving the problem is to express the area of \\(\\triangle APQ\\) based on the movement of point \\(P\\)." }, { "problem_id": 532, "question": "As shown in the figure, in rhombus $A B C D$, the diagonals $A C$ and $B D$ intersect at point $O$. Given $A C = 4$ and $B D = 8$. Point $N$ moves along $B O$. Through point $N$, a line $E F$ is drawn parallel to $A C$, intersecting $A B$ at $E$ and $B C$ at $F$. By folding $\\triangle B E F$ along $E F$, we obtain $\\triangle E F G$. If $O N = x$ and the area of the overlapping part between $\\triangle E F G$ and $\\triangle A B C$ is $y$, which of the following graphs correctly represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_616739ac8cb65fd7d225g_0028_1.jpg", "batch32-2024_06_14_616739ac8cb65fd7d225g_0028_2.jpg", "batch32-2024_06_14_616739ac8cb65fd7d225g_0028_3.jpg", "batch32-2024_06_14_616739ac8cb65fd7d225g_0028_4.jpg", "batch32-2024_06_14_616739ac8cb65fd7d225g_0028_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a rhombus,\n\n\\[\n\\therefore BD \\perp AC, \\quad BO = DO = \\frac{1}{2} BD = 4,\n\\]\n\n\\[\n\\therefore \\angle AOB = \\angle AOD = 90^\\circ;\n\\]\n\nWe discuss the problem in two cases:\n\n(1) When point \\( G \\) is to the left of point \\( O \\) after folding (as shown in Figure (1)), i.e., \\( 2 \\leq x \\leq 4 \\),\n\n\n\nFigure (1)\n\nSince \\( EF \\parallel AC \\),\n\n\\[\n\\therefore \\angle BEF = \\angle BAC, \\quad \\angle BFE = \\angle BCA,\n\\]\n\n\\[\n\\therefore \\triangle BEF \\sim \\triangle BAC,\n\\]\n\n\\[\n\\therefore \\frac{BN}{EF} = \\frac{BO}{AC} = 1, \\quad \\text{which implies} \\quad BN = EF = 4 - x,\n\\]\n\nSince \\( EF \\parallel AC \\),\n\n\\[\n\\therefore \\angle BNE = \\angle AOB = 90^\\circ,\n\\]\n\nAfter folding, the area of the overlapping part is:\n\n\\[\ny = S_{\\triangle EFG} = S_{\\triangle BEF} = \\frac{1}{2}(4 - x)^2 = \\frac{1}{2}x^2 - 4x + 8 \\quad (2 \\leq x \\leq 4);\n\\]\n\n(2) When point \\( G \\) is to the right of point \\( O \\) after folding (as shown in Figure (2)), i.e., \\( 0 \\leq x \\leq 2 \\),\n\n\n\nFigure (2)\n\nAfter folding, the overlapping part \\( y = S_{\\text{trapezoid } HIEF} \\),\n\n\\[\n\\because ON = x, \\quad BN = 4 - x, \\quad GN = BN = 4 - x,\n\\]\n\n\\[\n\\therefore OG = 4 - 2x,\n\\]\n\nSince \\( EF \\parallel AC \\),\n\nSimilarly, we have \\( \\triangle GHI \\sim \\triangle GEF \\),\n\n\\[\n\\therefore HI = OG = 4 - 2x,\n\\]\n\n\\[\n\\therefore y = \\frac{1}{2}[(4 - x) + (4 - 2x)] \\cdot x = 4x - \\frac{3}{2}x^2 \\quad (0 \\leq x \\leq 2),\n\\]\n\nIn summary,\n\n\\[\ny = \\begin{cases}\n4x - \\frac{3}{2}x^2 & (0 \\leq x \\leq 2) \\\\\n\\frac{1}{2}x^2 - 4x + 8 & (2 < x \\leq 4)\n\\end{cases},\n\\]\n\nTherefore, option A is correct.\n\nAnswer: A.\n\n【Key Insight】This problem mainly examines the properties of a rhombus, the determination and properties of similar triangles, the properties of parallel lines, and the calculation of triangle areas. The key to solving the problem is to consider different cases and derive the functional relationship between \\( y \\) and \\( x \\)." }, { "problem_id": 533, "question": "In triangle $A B C$ shown in Figure 1, $\\angle B = 36^\\circ$. Point $P$ moves from point $A$ along the broken line $A \\rightarrow B \\rightarrow C$ at a constant speed until it reaches point $C$. If the speed of point $P$ is $1 \\mathrm{~cm} / \\mathrm{s}$, and the time of movement of point $P$ is denoted as $t(s)$, with the length of $A P$ being $y(\\mathrm{~cm})$, the graph of $y$ against $t$ is shown in Figure 2. The value of $t$ when $A P$ bisects $\\angle B A C$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $2 \\sqrt{3}+2$\nB. $2 \\sqrt{3}-2$\nC. $2 \\sqrt{5}+2$\nD. $2 \\sqrt{5}-2$", "input_image": [ "batch32-2024_06_14_616739ac8cb65fd7d225g_0048_1.jpg", "batch32-2024_06_14_616739ac8cb65fd7d225g_0048_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw the angle bisector $AP$ of $\\angle BAC$ intersecting $BC$ at point $P$. From Figure 2, we know that $AB = BC = 4$.\n\n\n\nSince $\\angle B = 36^\\circ$ and $AB = BC$,\n\n$\\therefore \\angle BAC = \\angle C = 72^\\circ$.\n\nSince $AP$ bisects $\\angle BAC$,\n\n$\\therefore \\angle BAP = \\angle PAC = \\angle B = 36^\\circ$.\n\n$\\therefore AP = BP$, and $\\angle APC = \\angle B + \\angle BAP = 72^\\circ = \\angle C$.\n\n$\\therefore AP = AC = BP$.\n\nSince $\\angle PAC = \\angle B$ and $\\angle C = \\angle C$,\n\n$\\therefore \\triangle APC \\sim \\triangle BAC$.\n\n$\\therefore \\frac{AP}{BA} = \\frac{PC}{AC}$.\n\n$\\therefore AP \\cdot AC = AB \\cdot PC$.\n\n$\\therefore AP^{2} = AB \\cdot PC = 4(4 - AP)$.\n\nSolving this, we get $AP = 2\\sqrt{5} - 2$ or $AP = -2\\sqrt{5} - 2$ (discarded).\n\n$\\therefore BP = 2\\sqrt{5} - 2$.\n\n$\\therefore t = \\frac{2\\sqrt{5} - 2 + 4}{1} = 2\\sqrt{5} + 2$.\n\nTherefore, the correct answer is C.\n\n【Key Insight】This problem tests the determination and properties of similar triangles, the triangle angle sum theorem, and the determination of isosceles triangles. The key to solving the problem is proving that $\\triangle APC \\sim \\triangle BAC$." }, { "problem_id": 534, "question": "As shown in Figures 1, 2, and 3, they respectively represent the routes of individuals A, B, and C from location A to location B. It is known that A's route is: A → C → B;\n\n B's route is: A → D → E → F → B, where E is the midpoint of AB;\n\n C's route is: A → I → J → K → B, where J is on AB, and AJ > JB.\n\nIf the symbol $[\\rightarrow]$ represents [move straight ahead], based on the data in Figures 1, 2, and 3, determine the order of the lengths of the routes traveled by the three individuals as $(\\quad$ ）\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. A $=$ B= C\nB. A $<$ B $<$ C\nC. B< C $<$ A\nD. C $<$ B< A", "input_image": [ "batch32-2024_06_14_6ab249b0777bef36876ag_0085_1.jpg", "batch32-2024_06_14_6ab249b0777bef36876ag_0085_2.jpg", "batch32-2024_06_14_6ab249b0777bef36876ag_0085_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Analysis: From the degree measures of the angles, it can be deduced that the corresponding sides of the two triangles in figures 2 and 3 are parallel. Therefore, the triangles in figures 2 and 3 are similar to the triangle in figure 1. Moreover, the triangle in figure 2 is congruent, and the triangle in figure 3 is similar.\n\nDetailed Explanation: Based on the above analysis, in figure 2, we have $AE = BE$, $AD = EF$, and $DE = BE$.\n\nSince $AE = BE = \\frac{1}{2} AB$, it follows that $AD = EF = \\frac{1}{2} AC$, and $DE = BE = \\frac{1}{2} BC$. Therefore, triangle A is equal to triangle B.\n\nIn figure 3 and figure 1, the three triangles are similar, so $\\frac{JK}{AI} = \\frac{JB}{AJ} = \\frac{BK}{IJ}$, and $\\frac{AI}{AC} = \\frac{AJ}{AB} = \\frac{IJ}{BC}$.\n\nSince $AJ + BJ = AB$, it follows that $AI + JK = AC$, and $IJ + BK = BC$.\n\nTherefore, triangle A is equal to triangle C. Hence, triangle A equals triangle B equals triangle C.\n\nThus, the correct choice is A.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nKey Point: This question tests the properties of parallelograms. The key to solving this problem lies in using the translation of similar triangles to determine the relationships between the line segments." }, { "problem_id": 535, "question": "As shown in the figure, in $\\triangle A B C$, $A C=B C=8$, and $\\angle A C B=90^{\\circ}$. Points $D$ and $E$ are the midpoints of $A C$ and $B C$ respectively. Point $P$ moves from point $A$ towards point $D$, and point $Q$ is on segment $D E$ such that $D Q=D P$. The line segment $C Q$ is drawn, and from point $Q$, line segment $Q F$ is drawn perpendicular to $C Q$ intersecting $A B$ at point $F$. Let the distance traveled by point $P$ be $x$, and the area of $\\triangle C Q F$ be $y$. The graph that best represents the relationship between $y$ and $x$ is\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_6bcdbda069c7776d8801g_0076_1.jpg", "batch32-2024_06_14_6bcdbda069c7776d8801g_0076_2.jpg", "batch32-2024_06_14_6bcdbda069c7776d8801g_0076_3.jpg", "batch32-2024_06_14_6bcdbda069c7776d8801g_0076_4.jpg", "batch32-2024_06_14_6bcdbda069c7776d8801g_0076_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Draw a perpendicular from point \\( F \\) to \\( BC \\) at point \\( N \\), and extend \\( NF \\) to intersect the extension of \\( DE \\) at point \\( M \\), as shown in the figure.\n\n\n\nSince points \\( D \\) and \\( E \\) are the midpoints of \\( AC \\) and \\( BC \\) respectively,\n\n\\[ DE = \\frac{1}{2} BC = 4, \\quad DE \\parallel BC. \\]\n\nSince \\( FN \\perp BC \\),\n\n\\[ MN \\perp DE. \\]\n\nGiven that \\( \\angle ACB = 90^\\circ \\),\n\nquadrilateral \\( CDMN \\) is a rectangle,\n\n\\[ MN = CD = \\frac{1}{2} BC = 4. \\]\n\nGiven \\( AC = BC = 8 \\) and \\( \\angle ACB = 90^\\circ \\),\n\n\\[ \\angle B = 45^\\circ. \\]\n\nSince \\( FN \\perp BC \\),\n\n\\[ \\angle NFB = 45^\\circ, \\]\n\n\\[ \\angle EFM = \\angle NFB = 45^\\circ. \\]\n\nThus, triangle \\( MEF \\) is an isosceles right triangle,\n\n\\[ ME = MF. \\]\n\nLet \\( ME = MF = m \\).\n\nGiven \\( PA = x \\), then \\( DP = 4 - x \\).\n\nSince \\( DQ = DP \\),\n\n\\[ DQ = 4 - x, \\]\n\n\\[ QE = DE - DQ = 4 - (4 - x) = x. \\]\n\nSince \\( QF \\perp CQ \\),\n\n\\[ \\angle DQC + \\angle MQF = 90^\\circ. \\]\n\nGiven \\( \\angle DQC + \\angle DCQ = 90^\\circ \\),\n\n\\[ \\angle DCQ = \\angle MQF. \\]\n\nSince \\( \\angle CDQ = \\angle QMF = 90^\\circ \\),\n\n\\[ \\triangle DCQ \\sim \\triangle MQF, \\]\n\n\\[ \\frac{CD}{DQ} = \\frac{MQ}{MF}, \\]\n\n\\[ \\frac{4}{4 - x} = \\frac{m + x}{m}. \\]\n\nSolving gives \\( m = 4 - x \\),\n\n\\[ MF = 4 - x. \\]\n\nThus,\n\n\\[ FN = MN - MF = x. \\]\n\nThe area \\( S_{\\triangle CQF} \\) is given by:\n\n\\[ S_{\\triangle CQF} = S_{\\text{trapezoid } CDEB} - S_{\\triangle CDQ} - S_{\\triangle QEF} - S_{\\triangle BCF}, \\]\n\n\\[ y = \\frac{1}{2}(DE + BC) \\cdot CD - \\frac{1}{2} \\times CD \\cdot DQ - \\frac{1}{2} \\times QE \\cdot MF - \\frac{1}{2} \\times BC \\cdot NF \\]\n\n\\[ = \\frac{1}{2} \\times (4 + 8) \\times 4 - \\frac{1}{2} \\times 4(4 - x) - \\frac{1}{2} \\times x(4 - x) - \\frac{1}{2} \\times 8x \\]\n\n\\[ = 24 - 8 + 2x - 2x + \\frac{1}{2}x^2 - 4x \\]\n\n\\[ = \\frac{1}{2}x^2 + 4x + 16 \\]\n\n\\[ = \\frac{1}{2}(x - 4)^2 + 8. \\]\n\nSince \\( \\frac{1}{2} > 0 \\),\n\nthe parabola opens upwards with its vertex at \\( (4, 8) \\).\n\nGiven the range of \\( x \\) as \\( 0 \\leq x \\leq 4 \\),\n\nwhen \\( x = 0 \\), \\( y = 16 \\), and when \\( x = 4 \\), \\( y = 8 \\).\n\nThus, the function \\( y \\) versus \\( x \\) is part of the parabola \\( y = \\frac{1}{2}(x - 4)^2 + 8 \\) with endpoints at \\( (4, 8) \\) and \\( (0, 16) \\). Therefore, the correct choice is: C.\n\n【Highlight】This problem examines the function image of a moving point. It primarily tests the understanding of the median of a triangle, the properties of rectangles, the properties of isosceles triangles, the properties of similar triangles, and the properties of quadratic functions. This problem is comprehensive and is a typical high school entrance exam challenge. The key to solving it lies in accurately determining the function's expression." }, { "problem_id": 536, "question": "As shown in Figure 1, this is a figure studied by the Qing Dynasty mathematician Li Zhixuan in his book \"Ji He Yi Jian Ji\". After studying this figure, Xiao Yuan designed Figure 2. Extend the side $B C$ of the square $A B C D$ to point $M$, and construct a rectangle $A B M N$. Draw a semicircle $O$ with $B M$ as the diameter, which intersects $C D$ at point $E$. Construct a square $C E F G$ with $C E$ as a side, where point $G$ lies on $B C$. Let the areas of the square $A B C D$, the square $C E F G$, and the rectangle $C M N D$ be $S_{1}, S_{2}, S_{3}$, respectively. Then $\\frac{S_{1}}{S_{2}+S_{3}} = ($ ) \n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{3+\\sqrt{5}}{4}$\nB. $\\frac{1+\\sqrt{5}}{2}$\nC. $\\frac{3+\\sqrt{2}}{4}$\nD. $\\frac{1+\\sqrt{2}}{2}$", "input_image": [ "batch32-2024_06_14_6c3de9971dd3c9d355f4g_0098_1.jpg", "batch32-2024_06_14_6c3de9971dd3c9d355f4g_0098_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "**Solution:**\n\nConnect \\( BF \\), \\( ME \\), and \\( BE \\) as shown in the figure.\n\n\n\nFigure 2\n\nSince \\( EF \\parallel BM \\),\n\n\\[\n\\overparen{BF} = \\overparen{ME},\n\\]\n\n\\[\nBF = ME.\n\\]\n\nGiven that \\( \\angle BGF = \\angle MCE = 90^\\circ \\) and \\( GF = CE \\),\n\n\\[\n\\text{Rt} \\triangle BGF \\cong \\text{Rt} \\triangle MCE \\quad (\\text{HL}),\n\\]\n\n\\[\nBG = CM.\n\\]\n\nSince \\( BM \\) is the diameter of circle \\( O \\),\n\n\\[\n\\angle BEM = 90^\\circ,\n\\]\n\n\\[\n\\angle CEM + \\angle CEB = \\angle CEM + \\angle CME = 90^\\circ,\n\\]\n\n\\[\n\\angle CEB = \\angle CME.\n\\]\n\nGiven that \\( \\angle BCE = \\angle ECM = 90^\\circ \\),\n\n\\[\n\\triangle BCE \\sim \\triangle ECM,\n\\]\n\n\\[\n\\frac{CE}{CB} = \\frac{CM}{CE}, \\quad \\text{which implies} \\quad CE^2 = CB \\cdot CM.\n\\]\n\nLet the side length of square \\( ABCD \\) be \\( a \\), and the side length of square \\( CEFG \\) be \\( b \\). Let \\( BG = CM = c \\).\n\n\\[\n\\begin{cases}\nb = a - c \\\\\nb^2 = a c\n\\end{cases},\n\\]\n\n\\[\n(a - c)^2 = a c,\n\\]\n\nSimplifying, we get:\n\n\\[\na^2 + c^2 = 3 a c,\n\\]\n\n\\[\n\\frac{a}{c} + \\frac{c}{a} = 3.\n\\]\n\nThus,\n\n\\[\n\\frac{a}{c} = \\frac{3 + \\sqrt{5}}{2} \\quad \\text{or} \\quad \\frac{a}{c} = \\frac{3 - \\sqrt{5}}{2}.\n\\]\n\nSince \\( a > c \\),\n\n\\[\n\\frac{a}{c} = \\frac{3 - \\sqrt{5}}{2} \\quad \\text{is discarded}.\n\\]\n\nTherefore,\n\n\\[\n\\frac{S_1}{S_2 + S_3} = \\frac{a^2}{b^2 + a c} = \\frac{a^2}{a c + a c} = \\frac{a}{2 c} = \\frac{3 + \\sqrt{5}}{4}.\n\\]\n\nHence, the correct answer is: **A**.\n\n**Key Points:** This problem primarily examines the properties of squares, basic properties of circles, the inscribed angle theorem, and the properties and criteria for congruent and similar triangles. The key is to derive the quantitative relationship between the side lengths of the squares and rectangles." }, { "problem_id": 537, "question": "As shown in Figure 1, point $E$ is on side $A D$ of the rectangle $A B C D$. Points $P$ and $Q$ start simultaneously from point $B$. Point $P$ moves at a speed of $1 \\mathrm{~cm}$ per second along the broken line $B E-E D-D C$ until it reaches point $C$, and point $Q$ moves at a speed of $2 \\mathrm{~cm}$ per second along $B C$ until it reaches point $C$. Let $y \\mathrm{~cm}^{2}$ be the area of $\\triangle B P Q$ when $P$ and $Q$ have been moving for $t$ seconds. Given that the graph of the function $y$ versus $t$ is shown in Figure 2 (where curve $O G$ is a part of a parabola, and the rest of the graph consists of line segments), the following conclusions are drawn:\n\n(1) For $0\n\nFigure 1\n\n\n\nFigure 2\nA. 2\nB. 3\nC. 4\nD. 5", "input_image": [ "batch32-2024_06_14_78b5e3e9fbbbfbfddf38g_0001_1.jpg", "batch32-2024_06_14_78b5e3e9fbbbfbfddf38g_0001_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: By observing the graph, we can determine that $AD = BC = 5 \\times 2 = 10$, $BE = 1 \\times 10 = 10$, $ED = 4 \\times 1 = 4$, and $AE = 10 - 4 = 6$.\n\nIn the right triangle $\\triangle ABE$, $AB = \\sqrt{BE^2 - AE^2} = \\sqrt{10^2 - 6^2} = 8$.\n\nThus, $AD = BE = 10$ and $CD = AB = 8$.\n\nAs shown in Figure (1), we draw $PM \\perp BC$ at point $M$.\n\n\n\nFigure (1)\n\nSince $\\angle ABC = \\angle BMP = \\angle A = 90^\\circ$,\n\n$\\angle ABM + \\angle BMP = 180^\\circ$,\n\nTherefore, $AB \\parallel MP$,\n\n$\\angle ABE = \\angle BPM$,\n\nThus, $\\triangle ABE \\sim \\triangle MPB$,\n\n$\\frac{PB}{BE} = \\frac{PM}{AB}$,\n\n$\\frac{t}{10} = \\frac{PM}{8}$,\n\n$PM = \\frac{4}{5}t$.\n\nWhen $0 < t \\leq 5$, the area of $\\triangle BPQ$ is $y = \\frac{1}{2} \\cdot BQ \\cdot PM = \\frac{1}{2} \\cdot 2 \\cdot \\frac{4}{5}t = \\frac{4}{5}t^2$, so statement (1) is correct.\n\nWhen $t = 6$ seconds, $BP = 6$, and point $Q$ is at point $C$, so $BQ = BC = 10$.\n\nSince $AD \\parallel BC$,\n\n$\\angle AEB = \\angle PBQ$,\n\n$AE = BP = 6$, $BE = BQ = 10$,\n\nThus, $\\triangle ABE \\cong \\triangle PQB$, so statement (2) is correct.\n\nSince $AE = 6$, $BE = 10$,\n\nIn the right triangle $\\triangle BAE$,\n\n$\\cos \\angle AEB = \\frac{AE}{BE} = \\frac{6}{10} = \\frac{3}{5}$,\n\n$\\angle CBE = \\angle AEB$,\n\nThus, $\\cos \\angle CBE = \\cos \\angle AEB = \\frac{3}{5}$, so statement (3) is incorrect.\n\nWhen $t = \\frac{29}{2}$ seconds, point $P$ is on side $DC$, and point $Q$ coincides with point $C$, as shown in the figure:\n\n\n\nAt this time, $DP = 1 \\times \\frac{29}{2} - 1 \\times 14 = \\frac{1}{2}$,\n\n$PC = 8 - \\frac{1}{2} = \\frac{15}{2}$,\n\n$\\frac{BQ}{PQ} = \\frac{BC}{PC} = \\frac{10}{\\frac{15}{2}} = \\frac{4}{3}$, $\\frac{AB}{BE} = \\frac{8}{6} = \\frac{4}{3}$, $\\angle A = \\angle Q = 90^\\circ$,\n\nThus, $\\triangle ABE \\sim \\triangle QBP$, so statement (4) is correct.\n\nThe total movement time of point $P$ is $\\frac{10 + 4 + 8}{1} = 22$,\n\nThus, the coordinates of point $F$ are $(22, 0)$.\n\nLet the equation of line $NF$ be $y = kx + b$. Substituting $(14, 40)$ and $(22, 0)$ into the equation:\n\n$\\left\\{\\begin{array}{l}14k + b = 40 \\\\ 22k + b = 0\\end{array}\\right.$,\n\nSolving gives $\\left\\{\\begin{array}{l}k = -5 \\\\ b = 110\\end{array}\\right.$,\n\nThus, the equation of line $NF$ is $y = -5x + 110$, so statement (5) is incorrect.\n\nIn summary, statements (1), (2), and (4) are correct, totaling 3 correct statements, so option B is correct.\n\nAnswer: B.\n\n【Key Insight】This problem examines the determination and properties of similar triangles, the Pythagorean theorem, the function graph of moving points, and the derivation of a linear function's equation. The key to solving the problem lies in determining from Figure (2) that point $P$ reaches point $E$ in $10$ seconds and point $Q$ reaches point $C$ in $5$ seconds." }, { "problem_id": 538, "question": "In the following figures, which one is not an example of similar figures?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_96c7356882fd2c57db1dg_0078_1.jpg", "batch32-2024_06_14_96c7356882fd2c57db1dg_0078_2.jpg", "batch32-2024_06_14_96c7356882fd2c57db1dg_0078_3.jpg", "batch32-2024_06_14_96c7356882fd2c57db1dg_0078_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Two similar polygons whose corresponding vertices' connecting lines intersect at a single point are called homothetic figures.\n\nAccording to the concept of homothetic figures, two of the three figures A, B, and C are homothetic figures;\n\nThe two figures in D do not conform to the concept of homothetic figures, as their corresponding vertices cannot intersect at a single point, hence they are not homothetic figures.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question primarily examines the understanding of homothetic figures, noting that homothety and similarity are related yet distinct concepts. Similarity merely requires that two figures have identical shapes, whereas homothety, in addition to similarity, requires that the connecting lines of corresponding points intersect at a single point." }, { "problem_id": 539, "question": "As shown in the figure, in triangle paper $A B C$, $A B = 9, A C = 6, B C = 12$. Among the shaded triangles cut along the dotted lines, the one that is similar to $\\triangle A B C$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_a755eb3b7bc6c9b3a430g_0050_1.jpg", "batch32-2024_06_14_a755eb3b7bc6c9b3a430g_0050_2.jpg", "batch32-2024_06_14_a755eb3b7bc6c9b3a430g_0050_3.jpg", "batch32-2024_06_14_a755eb3b7bc6c9b3a430g_0050_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In the triangle piece $ABC$, $AB=9$, $AC=6$, $BC=12$.\n\nA. Since $\\frac{6}{BC}=\\frac{6}{12}=\\frac{1}{2}$, and the corresponding side $\\frac{AB}{BC}=\\frac{9}{12}=\\frac{3}{4}$, with $\\frac{1}{2} \\neq \\frac{3}{4}$, the shaded triangle cut along the dotted line is not similar to $\\triangle ABC$, hence this option is incorrect;\n\nB. Since $\\frac{4}{AC}=\\frac{4}{6}=\\frac{2}{3}$, and the corresponding side $\\frac{AC}{AB}=\\frac{6}{9}=\\frac{2}{3}$, with $\\angle A=\\angle A$, the shaded triangle cut along the dotted line is similar to $\\triangle ABC$, hence this option is correct;\n\nC. Since $\\frac{4}{AB}=\\frac{4}{9}$, and the corresponding side $\\frac{AB}{BC}=\\frac{9}{12}=\\frac{3}{4}$, meaning $\\frac{4}{9} \\neq \\frac{3}{4}$, the shaded triangle cut along the dotted line is not similar to $\\triangle ABC$, hence this option is incorrect;\n\nD. Since $\\frac{4}{6}=\\frac{2}{3}$, and the corresponding side $\\frac{AC}{BC}=\\frac{6}{12}=\\frac{1}{2}$, with $\\frac{2}{3} \\neq \\frac{1}{2}$, the shaded triangle cut along the dotted line is not similar to $\\triangle ABC$, hence this option is incorrect;\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This question primarily tests the understanding of the criteria for similar triangles. The key to solving it lies in correctly using the condition that two triangles are similar if the ratios of two corresponding sides are equal and the included angles are equal." }, { "problem_id": 540, "question": "Archimedes was one of the greatest mathematicians of ancient Greece, and he discovered the Archimedes chord theorem using the figure in Figure 1. As shown in Figure 2, given that $B C$ is the diameter of $\\odot O$, $A B$ is a chord $(B C > A B)$, point $M$ is a point on $\\overline{A B C}$, $M D$ is perpendicular to $B C$ at point $D$, and $M D$ extended intersects chord $A B$ at point $E$, with $B M = \\sqrt{6}$ and $A B = 4$, the length of $A E$ is $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{5}{2}$\nB. $\\frac{9}{4}$\nC. $\\frac{12}{5}$\nD. $\\frac{13}{5}$", "input_image": [ "batch32-2024_06_14_ac7ad2afd712c43f62f7g_0028_1.jpg", "batch32-2024_06_14_ac7ad2afd712c43f62f7g_0028_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nExtend $\\mathrm{ME}$ and let it intersect the circle at point $\\mathrm{F}$. Connect $\\mathrm{BF}$ and $\\mathrm{AF}$ as shown in the figure.\n\n\n\nSince $BC$ is the diameter of circle $\\mathrm{O}$, and $\\mathrm{MD} \\perp \\mathrm{BC}$ at point $\\mathrm{D}$,\n\nwe have $\\mathrm{MB} = \\mathrm{FB} = \\sqrt{6}$, and $\\angle \\mathrm{BMF} = \\angle \\mathrm{BFM}$.\n\nMoreover, $\\angle \\mathrm{BMF} = \\angle \\mathrm{FAB}$,\n\nso $\\angle \\mathrm{BFM} = \\angle \\mathrm{FAB}$.\n\nThus, $\\angle \\mathrm{BFE} = \\angle \\mathrm{FAB}$.\n\nSince $\\angle \\mathrm{EBF} = \\angle \\mathrm{FBA}$,\n\nwe have $\\triangle \\mathrm{BFA} \\sim \\triangle \\mathrm{BEF}$.\n\nTherefore, $\\frac{BF}{AB} = \\frac{BE}{BF}$,\n\nwhich gives $\\frac{\\sqrt{6}}{4} = \\frac{BE}{\\sqrt{6}}$.\n\nSolving for $\\mathrm{BE}$, we get $\\mathrm{BE} = \\frac{3}{2}$.\n\nThus, $\\mathrm{AE} = 4 - \\frac{3}{2} = \\frac{5}{2}$.\n\nThe correct answer is: **A**.\n\n**Key Insight:** This problem tests the understanding of the **Perpendicular Diameter Theorem** and the **properties of similar triangles**. The key to solving it lies in accurately constructing the auxiliary lines and identifying the similar triangles." }, { "problem_id": 541, "question": "In the square $A B C D$ with side length 10, point $E$ is on the extension of $C B$, and line segment $E D$ intersects $A B$ at point $F$. Given that $A F = x$ (where $2 \\leq x \\leq 8$) and $E C = y$, which of the following graphs approximately represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_ac7ad2afd712c43f62f7g_0039_1.jpg", "batch32-2024_06_14_ac7ad2afd712c43f62f7g_0039_2.jpg", "batch32-2024_06_14_ac7ad2afd712c43f62f7g_0039_3.jpg", "batch32-2024_06_14_ac7ad2afd712c43f62f7g_0039_4.jpg", "batch32-2024_06_14_ac7ad2afd712c43f62f7g_0039_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem, we know that \\( BF = 10 - x \\) and \\( BE = y - 10 \\).\n\nSince quadrilateral \\( ABCD \\) is a square, \\( AD \\parallel BC \\).\n\nThus, triangles \\( EFB \\) and \\( EDC \\) are similar.\n\nTherefore, \\( \\frac{BF}{DC} = \\frac{BE}{EC} \\), which means \\( \\frac{10 - x}{10} = \\frac{y - 10}{y} \\).\n\nSolving this, we get \\( y = \\frac{100}{x} \\) where \\( 2 \\leq x \\leq 8 \\). The graph of this function is a part of a hyperbola located in the first quadrant.\n\nThe graphs of options A and D are parts of straight lines, option B is part of a parabola, and option C is part of a hyperbola.\n\nHence, the correct choice is C.\n\n【Key Insight】This problem tests the understanding of function graphs related to moving points. Familiarity with the relevant properties is crucial for solving it." }, { "problem_id": 542, "question": "Given that quadrilateral $A B C D$ is a rhombus with $\\angle B = 60^\\circ$, place a $30^\\circ$-angled triangle as shown on the rhombus. Which of the following pairs of shaded triangles are similar?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_aff4f783cee54a536a7eg_0024_1.jpg", "batch32-2024_06_14_aff4f783cee54a536a7eg_0024_2.jpg", "batch32-2024_06_14_aff4f783cee54a536a7eg_0024_3.jpg", "batch32-2024_06_14_aff4f783cee54a536a7eg_0024_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "As shown in the diagram, since quadrilateral \\( ABCD \\) is a rhombus,\n\n\\(\\therefore AB = BC\\).\n\nSince \\(\\angle B = 60^\\circ\\),\n\n\\(\\therefore \\triangle ABC\\) is an equilateral triangle.\n\n\\(\\therefore \\angle B = \\angle ACB = 60^\\circ\\).\n\nSince \\(\\angle AEF = 60^\\circ\\),\n\n\\(\\therefore \\angle AEB + \\angle FEC = 120^\\circ\\)\n\n\\(\\therefore \\angle EFC + \\angle FEC = 120^\\circ\\),\n\n\\(\\therefore \\angle AEB = \\angle EFC\\).\n\n\\(\\therefore \\triangle AEB \\sim \\triangle EFC\\).\n\nThe other three options do not meet the conditions for similarity.\n\nTherefore, the correct choice is: A.\n\n\n\n【Key Insight】This question tests the properties of a rhombus, the properties of a right triangle, the determination and properties of an equilateral triangle, and the determination of similar triangles. Mastering these properties and applying them to solve problems is crucial.\n\nError Analysis: Medium difficulty. The reasons for losing points are: 1. Not mastering the properties of a right triangle with a \\(30^\\circ\\) angle; 2. Not mastering the properties of a rhombus." }, { "problem_id": 543, "question": "As shown in Figure (1), extend the sides of a regular hexagon to form a regular hexagram $A F B D C E$, whose area is 1. Taking the midpoints of the sides of $\\triangle A B C$ and $\\triangle D E F$, connect them to form a regular hexagram $A_{1} F_{1} B_{1} D_{1} C_{1} E_{1}$, as shown in the shaded area of Figure (2). Taking the midpoints of the sides of $\\triangle A_{1} B_{1} C_{1}$ and $\\triangle D_{1} E_{1} F_{1}$, connect them to form a regular hexagram $A_{2} F_{2} B_{2} D_{2} C_{2} E_{2}$, as shown in the shaded area of Figure (3), and so on... What is the area of the regular hexagram $A_{4} F_{4} B_{4} D_{4} C_{4} E_{4}$?\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA. $\\frac{1}{16}$\nB. $\\frac{1}{64}$\nC. $\\frac{1}{128}$\nD. $\\frac{1}{256}$", "input_image": [ "batch32-2024_06_14_c63a15acc5199478291dg_0026_1.jpg", "batch32-2024_06_14_c63a15acc5199478291dg_0026_2.jpg", "batch32-2024_06_14_c63a15acc5199478291dg_0026_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Since points $\\mathrm{A}_{1}$, $\\mathrm{F}_{1}$, $\\mathrm{B}_{1}$, $\\mathrm{D}_{1}$, $\\mathrm{C}_{1}$, and $\\mathrm{E}_{1}$ are the midpoints of the sides of triangles $\\triangle \\mathrm{ABC}$ and $\\triangle \\mathrm{DEF}$ respectively,\n\nTherefore, the regular hexagram $\\mathrm{AFBDCE}$ is similar to the regular hexagram $\\mathrm{A}_{1}\\mathrm{F}_{1}\\mathrm{B}_{1}\\mathrm{D}_{1}\\mathrm{C}_{1}\\mathrm{E}_{1}$ with a similarity ratio of $2:1$.\n\nGiven that the area of the regular hexagram $\\mathrm{AFBDCE}$ is 1,\n\nThus, the area of the regular hexagram $\\mathrm{A}_{1}\\mathrm{F}_{1}\\mathrm{B}_{1}\\mathrm{D}_{1}\\mathrm{C}_{1}\\mathrm{E}_{1}$ is $\\frac{1}{4}$.\n\nSimilarly, the area of the second hexagram is: $\\frac{1}{4^{2}}=\\frac{1}{16}$,\nThe area of the third hexagram is: $\\frac{1}{4^{3}}=\\frac{1}{64}$,\n\nThe area of the fourth hexagram is: $\\frac{1}{4^{4}}=\\frac{1}{256}$,\n\nTherefore, the correct choice is D.\n\n【Key Insight】This question tests the properties of similar polygons and the Midsegment Theorem of triangles. The key to solving this problem is understanding that the ratio of the areas of similar polygons is equal to the square of the similarity ratio." }, { "problem_id": 544, "question": "A type of paper clip is shown in Figure 1. Figure 2 is a diagram of the closed position, and Figure 3 is a diagram of the open position (at this time, $A B \\parallel C D$). The relevant data is shown in the figure (units: $\\mathrm{cm}$). From the closed position in Figure 2 to the open position in Figure 3, the distance between points $B$ and $D$ decreases by $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $2 \\mathrm{~cm}$\nB. $3 \\mathrm{~cm}$\nC. $4 \\mathrm{~cm}$\nD. $5 \\mathrm{~cm}$", "input_image": [ "batch32-2024_06_14_e0c87936841f97183ff9g_0072_1.jpg", "batch32-2024_06_14_e0c87936841f97183ff9g_0072_2.jpg", "batch32-2024_06_14_e0c87936841f97183ff9g_0072_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Connect $BD$ as shown in the figure:\n\n\n\nFrom the given conditions, $\\frac{AE}{AB}=\\frac{AF}{AD}$, and $\\angle A=\\angle A$,\n\n$\\therefore \\triangle AEF \\sim \\triangle ABD$,\n\n$\\therefore \\frac{AE}{AB}=\\frac{EF}{BD}$,\n\n$\\therefore \\frac{2}{5}=\\frac{2}{BD}$,\n\n$\\therefore BD=5 \\mathrm{~cm}$,\n\n$\\therefore$ The distance between points $B$ and $D$ has decreased by $5-2=3(\\mathrm{~cm})$,\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem examines the application of similar triangles, and the key to solving it lies in correctly identifying the geometric relationships in the figure." }, { "problem_id": 545, "question": "As shown in the figure, in right triangle $ABC$, $\\angle C = 90^\\circ, AC = 3, BC = 4$. Line $l$ is perpendicular to $AB$. Line $l$ is translated along the $AB$ direction from point $A$ to point $B$. If line $l$ intersects $AB$ at point $P$ and $AC$ (or $BC$) at point $Q$, and $AP = x, CQ = y$, then which of the following graphs approximately represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_e0c87936841f97183ff9g_0084_1.jpg", "batch32-2024_06_14_e0c87936841f97183ff9g_0084_2.jpg", "batch32-2024_06_14_e0c87936841f97183ff9g_0084_3.jpg", "batch32-2024_06_14_e0c87936841f97183ff9g_0084_4.jpg", "batch32-2024_06_14_e0c87936841f97183ff9g_0084_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since $\\angle C=90^{\\circ}$, $AC=3$, and $BC=4$,\n\n$\\therefore AB=\\sqrt{AC^{2}+BC^{2}}=5$.\n\nCase Analysis: (1) When point $Q$ is on side $AC$, as shown in the figure,\n\n\n\nSince $PQ \\perp AB$,\n\n$\\therefore \\angle APQ=90^{\\circ}=\\angle C$.\n\nAlso, since $\\angle A=\\angle A$,\n\n$\\therefore \\triangle APQ \\sim \\triangle ACB$,\n\n$\\therefore \\frac{AP}{AC}=\\frac{AQ}{AB}$,\n\n$\\therefore AQ=\\frac{AP \\cdot AB}{AC}=\\frac{5}{3} x$,\n\n$\\therefore y=CQ=AC-AQ=3-\\frac{5}{3} x$,\n\n$\\therefore$ the graph of the function relationship of $y$ with respect to $x$ is a decreasing line segment;\n\n(2) When point $Q$ is on side $BC$, as shown in the figure,\n\n\n\nSince $PQ \\perp AB$,\n\n$\\therefore \\angle BPQ=\\angle C=90^{\\circ}$.\n\nAlso, since $\\angle B=\\angle B$,\n\n$\\therefore \\triangle BPQ \\sim \\triangle BCA$,\n$\\therefore \\frac{BP}{BC}=\\frac{BQ}{AB}$,\n\n$\\therefore BQ=\\frac{AB \\cdot BP}{BC}$.\n\nSince $BP=AB-AP=5-x$,\n\n$\\therefore BQ=\\frac{5(5-x)}{4}$,\n\n$\\therefore y=CQ=BC-BQ=4-\\frac{5(5-x)}{4}=\\frac{5}{4} x-\\frac{9}{4}$,\n\n$\\therefore$ the graph of the function relationship of $y$ with respect to $x$ is an increasing line segment,\n\nTherefore, the correct choice is B.\n\n【Key Insight】This problem examines the function graph of moving point problems, the determination and properties of similar triangles, and other knowledge. It is a challenging multiple-choice question in the high school entrance exam, with the key to solving it being to find the functional relationship between $y$ and $x$." }, { "problem_id": 546, "question": "The three views of a geometric solid are shown as follows, then which of the following solids could it be ( )\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_4c8122bf8583225dfc55g_0091_1.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0091_2.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0091_3.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0091_4.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0091_5.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0091_6.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0091_7.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: Based on the planar figures of the front view, left view, and top view, it can be determined that the geometric shape is A.\n\nTherefore, the answer is: A" }, { "problem_id": 547, "question": "As shown in the figure, a pair of right-angled triangles are placed in different positions. Which of the following arrangements makes $\\angle \\alpha$ and $\\angle \\beta$ both acute and equal?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\nA. Figure (1)\nB. Figure (2)\nC. Figure (3)\nD. Figure (4)", "input_image": [ "batch33-2024_06_14_5d8a6f6d844d1168bb3bg_0010_1.jpg", "batch33-2024_06_14_5d8a6f6d844d1168bb3bg_0010_2.jpg", "batch33-2024_06_14_5d8a6f6d844d1168bb3bg_0010_3.jpg", "batch33-2024_06_14_5d8a6f6d844d1168bb3bg_0010_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "In Figure (1), $\\angle \\alpha + \\angle \\beta = 90^{\\circ}$, and the two angles are not equal; in Figure (2), based on the fact that the complementary angles of the same angle are equal, it can be deduced that $\\angle \\alpha$ and $\\angle \\beta$ are equal and both are acute angles; in Figure (3), although the two angles are equal, they are both obtuse angles; in Figure (4), $\\angle \\alpha + \\angle \\beta = 180^{\\circ}$, and the two angles are not equal.\n\nTherefore, choose B.\n\n【Key Point】This question examines the relationship between angles in a triangle, with the key being to master the relevant fundamental knowledge." }, { "problem_id": 548, "question": "As shown in the figure, MN is the diameter of the circular base of a cylindrical oil drum, and NP is the height of the drum. An ant is on the side of the drum and wants to travel from P to M and then back to P along the shortest path. If the side of the drum is cut along NP, the resulting expanded viewof the side is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_629ecf96069e663c619eg_0080_1.jpg", "batch33-2024_06_14_629ecf96069e663c619eg_0080_2.jpg", "batch33-2024_06_14_629ecf96069e663c619eg_0080_3.jpg", "batch33-2024_06_14_629ecf96069e663c619eg_0080_4.jpg", "batch33-2024_06_14_629ecf96069e663c619eg_0080_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Since the unfolded surface of a cylinder is a rectangle, the unfolding of $MP$ should result in two straight lines with a common point $M$.\n\nTherefore, the correct choice is $A$.\n\n【Key Insight】This question primarily tests the understanding of the unfolded diagram of a cylinder, the principle that the shortest distance between two points is a straight line, and the student's spatial thinking ability." }, { "problem_id": 549, "question": "Unfolding the given square as shown, which of the following could be the flat unfolded pattern?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_7fcc9a9aa722b575bb21g_0013_1.jpg", "batch33-2024_06_14_7fcc9a9aa722b575bb21g_0013_2.jpg", "batch33-2024_06_14_7fcc9a9aa722b575bb21g_0013_3.jpg", "batch33-2024_06_14_7fcc9a9aa722b575bb21g_0013_4.jpg", "batch33-2024_06_14_7fcc9a9aa722b575bb21g_0013_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: By unfolding the square and labeling the vertices, we obtain the following diagram:\n\n\n\nHere, $C_{1}$ connects to $\\mathrm{C}$, $B_{1}$ connects to $\\mathrm{B}$, $D_{1}$ connects to $\\mathrm{D}$, $A_{1}$ connects to $\\mathrm{A}$, $B_{1}^{\\prime}$ connects to $B^{\\prime}$, and $D_{1}^{\\prime}$ connects to $D^{\\prime}$.\n\nThus, the correct option is B.\n\nTherefore, the answer is: B.\n\n【Key Insight】This question tests the understanding of the surface development of a cube and spatial imagination. A common mistake is the inaccurate visualization of the positions of the related figures, leading to incorrect choices. To solve such problems, it is advisable to physically manipulate the shapes, which can help clarify the spatial relationships." }, { "problem_id": 550, "question": "There are two squares with side lengths of 1 and 2, respectively. They are placed together, and each is to be cut along the dotted lines and rearranged to form a square with the same area as the original. Two cutting and rearrangement schemes are shown in the figure, then ( )\n\n\n\nScheme One\n\n\n\nScheme Two\nA. Both Scheme One and Scheme Two are possible.\nB. Neither Scheme One nor Scheme Two is possible.\nC. Scheme One is not possible, but Scheme Two is.\nD. Scheme One is possible, but Scheme Two is not.", "input_image": [ "batch33-2024_06_14_93f12ebaa8a40c7bd1aeg_0059_1.jpg", "batch33-2024_06_14_93f12ebaa8a40c7bd1aeg_0059_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, the method involves cutting and rearranging,\n\n\nOption 1\n\nOption 2\n\nBoth Option 1 and Option 2 can be rearranged into a square with the same area as the original,\n\nTherefore, the answer is: A.\n\n【Key Point】This question tests the ability to cut and rearrange shapes, with the key to solving it lying in ensuring that the area of the rearranged square is equal to that of the original shape." }, { "problem_id": 551, "question": "As shown in the figure, a square paper with sides of $16 \\mathrm{~cm}$ is cut into a smaller square at each corner, and the remaining part is folded into an open-top rectangular box. When the side length of the squares cut from the corners changes from $2 \\mathrm{~cm}$ to $4 \\mathrm{~cm}$, the volume of the paper box ( ) $\\mathrm{cm}^{3}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. Decreased by 32\nB. Decreased by 80\nC. Increased by 32\nD. Increased by 80", "input_image": [ "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0013_1.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0013_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem, when the side length of the cut-out square is \\(2 \\mathrm{~cm}\\), the volume of the cuboid is \\((16-2 \\times 2)^{2} \\times 2 = 288 \\mathrm{~cm}^{3}\\).\n\nWhen the side length of the cut-out square is \\(4 \\mathrm{~cm}\\), the volume of the cuboid is \\((16-2 \\times 4)^{2} \\times 4 = 256 \\mathrm{~cm}^{3}\\).\n\nThe difference in volume is \\(288 - 256 = 32 \\mathrm{~cm}^{3}\\).\n\nTherefore, when the side length of the cut-out square increases from \\(2 \\mathrm{~cm}\\) to \\(4 \\mathrm{~cm}\\), the volume of the cuboid paper box decreases by \\(32 \\mathrm{~cm}^{3}\\). Hence, the correct answer is: A.\n\n【Key Insight】This problem tests the practical application of mixed operations with rational numbers. Understanding the problem and correctly formulating the expression based on the volume formula for a cuboid is crucial for solving it." }, { "problem_id": 552, "question": "The following diagrams represent the unfolding of a cylinder. Which of the following is (are) correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0090_1.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0090_2.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0090_3.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0090_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "As shown in the figure, the diameter of the base of the cylinder is 6, and the height is 4.\n\nTherefore, the circumference of the base of the cylinder (which is the length of the rectangle obtained by unfolding the side of the cylinder) is:\n\n$6 \\times 3.14 = 18.84$,\n\nHence, the correct choice is: C.\n\n【Key Point】This question tests the understanding of the unfolded diagram of a cylinder. Comprehending the relationship between each part of the unfolded diagram and the cylinder itself is essential for correct calculations." }, { "problem_id": 553, "question": "Below are Xiaoming's descriptions of four geometric figures: (1) Figure 1: Line $E F$ passes through point $C$; (2) Figure 2: Point $A$ is outside line $l$; (3) Figure 3: Ray $O P$ bisects $\\angle A O B$; (4) Figure 4: Lines $A B, C D$ intersect at point $O$. Which of the following are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\nA. (1)(2)(3)\nB. (1)(2)(4)\nC. (1)(3)(4)\nD. (2)(3)(4)", "input_image": [ "batch33-2024_06_14_c720ff8294c46ce65b03g_0074_1.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0074_2.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0074_3.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0074_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "(1) Figure 1: The line $EF$ passes through point $C$, correct. \n(2) Figure 2: Point $A$ lies outside the line $l$, correct. \n(3) Figure 3: The ray $OP$ bisects $\\angle AOB$, incorrect. \n(4) Figure 4: The lines $AB$ and $CD$ intersect at point $O$, correct. Therefore, the answer is B. \n\n**Key Insight**: This question primarily tests the fundamental properties of plane geometry. The key to solving it lies in understanding the relationships between points and lines, as well as the relationships between lines themselves." }, { "problem_id": 554, "question": "Among the four figures below, which one can represent the same angle using $\\angle 1, \\angle A O B, \\angle O$ in three different ways?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_cab5fd474189d082ffe5g_0045_1.jpg", "batch33-2024_06_14_cab5fd474189d082ffe5g_0045_2.jpg", "batch33-2024_06_14_cab5fd474189d082ffe5g_0045_3.jpg", "batch33-2024_06_14_cab5fd474189d082ffe5g_0045_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: \n\nA. Can be represented as: $\\angle 1, \\angle A O B$, but cannot be represented as $\\angle O$, which does not meet the requirement;\n\nB. Can be represented as: $\\angle 1, \\angle A O B$, but cannot be represented as $\\angle O$, which does not meet the requirement;\n\nC. Can be represented as: $\\angle 1, \\angle A O B, \\angle O$, which meets the requirement;\n\nD. Can be represented as: $\\angle 1, \\angle A O B$, but cannot be represented as $\\angle O$, which does not meet the requirement;\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question primarily tests the methods of representing angles. Mastering the three correct ways to denote angles is crucial for solving such problems." }, { "problem_id": 555, "question": "Xiaoming uses a roller, as shown in the figure below, to paint the wall from left to right. Which of the following plane figures best represents the area painted by the roller?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0051_1.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0051_2.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0051_3.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0051_4.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0051_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the pattern obtained by the glue roller, the small black square is located in the middle of the far left. As the glue roller moves from left to right, the first impression it leaves is the pattern with a small black square in the middle.\n\nTherefore, the correct choice is: A.\n\n[Key Point] This question tests the understanding of planar figures. The key to solving the problem lies in recognizing the sequence in which the patterns on the right, as seen by the glue roller, appear first on the far left." }, { "problem_id": 556, "question": "The Rubik's Cube is a well-known mathematical game consisting of seven solid shapes. In the 17th shape shown, as viewed from the front, how many blocks have the same shape as ( )?\n\n\nRubik's Cube\n\n\n1st shape\n\n\n2nd shape\n\n\n3rd shape\n\n\n4th shape\n\n\n5th shape\n\n\n6th shape\n\n\n7th shape\nA. 2\nB. 3\nC. 4\nD. 5", "input_image": [ "batch33-2024_06_14_dc09671304abd41736e6g_0055_1.jpg", "batch33-2024_06_14_dc09671304abd41736e6g_0055_2.jpg", "batch33-2024_06_14_dc09671304abd41736e6g_0055_3.jpg", "batch33-2024_06_14_dc09671304abd41736e6g_0055_4.jpg", "batch33-2024_06_14_dc09671304abd41736e6g_0055_5.jpg", "batch33-2024_06_14_dc09671304abd41736e6g_0055_6.jpg", "batch33-2024_06_14_dc09671304abd41736e6g_0055_7.jpg", "batch33-2024_06_14_dc09671304abd41736e6g_0055_8.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the front view, the shapes of numbers 1, 6, and 7 are identical,\n\nTherefore, choose B.\n\n[Highlight] This question tests the understanding and application of three-view drawings and spatial imagination. One can analyze the number of layers from top to bottom and left to right on the main view." }, { "problem_id": 557, "question": "The diagram in the figure is the unfolding pattern of a ( ) cube.\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_fa30bd0163a7a41365acg_0058_1.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0058_2.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0058_3.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0058_4.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0058_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\nTherefore, the correct choice is: B.\n\n【Key Point】The key to solving this problem lies in understanding the positional relationships of the patterns on the cube's net when it is folded into a cube." }, { "problem_id": 558, "question": "As shown in the figure, two right-angled triangles $A B C$ and $D E F$ with hypotenuses of length 6 are placed as depicted, where $\\angle A C B = \\angle D E F = 30^\\circ$, $B C$ and $E F$ lie on a straight line, and points $C$ and $E$ coincide. With triangle $D E F$ fixed, triangle $A B C$ is translated along the straight line $l$ to the right until point $B$ coincides with point $F$. During this process, let the distance that point $B$ moves be $x$, and the area of the overlapping part of the two triangles be $y$. The graph of $y$ against $x$ would approximately look like ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch34-2024_06_17_010d88a8bf3f2e00a466g_0012_1.jpg", "batch34-2024_06_17_010d88a8bf3f2e00a466g_0012_2.jpg", "batch34-2024_06_17_010d88a8bf3f2e00a466g_0012_3.jpg", "batch34-2024_06_17_010d88a8bf3f2e00a466g_0012_4.jpg", "batch34-2024_06_17_010d88a8bf3f2e00a466g_0012_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: When \\(0 \\leq x \\leq 6\\), as shown in Figure (1), according to the problem statement,\n\n\\(CE = x\\),\n\nSince \\(\\angle ACB = \\angle DEF = 30^\\circ\\),\n\nThe height of \\(\\triangle EGC\\) is: \\(h = \\frac{1}{2} EC \\tan 30^\\circ = \\frac{\\sqrt{3}}{6} x\\),\n\nThus, \\(y = \\frac{1}{2} x h = \\frac{1}{2} x \\times \\frac{\\sqrt{3}}{6} x = \\frac{\\sqrt{3}}{12} x^{2} \\quad (0 \\leq x \\leq 6)\\), so the images A and C fit the description;\n\n\n\nFigure (1)\n\nWhen \\(6 \\leq x \\leq 9\\), as shown in Figure (2), according to the problem statement,\n\nSince \\(\\angle ACB = \\angle DEF = 30^\\circ\\),\n\n\\(\\angle ABC = \\angle DFE = 60^\\circ\\),\n\n\\(\\angle AMP = \\angle DNP = 30^\\circ\\),\n\nThus, \\(AM = DN = 9 - x\\), \\(NF = FC = x - 6\\),\n\n\\(y = S_{\\triangle ABC} - S_{\\triangle APM} - S_{\\triangle NCF} = \\frac{1}{2} \\times 3 \\times 3 \\sqrt{3} - \\frac{1}{2} \\times (9 - x) \\times \\frac{9 - x}{\\sqrt{3}} \\frac{1}{2} \\times \\frac{x - 6}{2} \\times \\sqrt{3}(x - 6)\\)\n\n\\(= -\\frac{5}{12} \\sqrt{3} x^{2} + 6 \\sqrt{3} x - 18 \\sqrt{3}\\), so the images A and C fit the description;\n\n\n\nFigure (2)\n\nWhen \\(9 \\leq x \\leq 12\\), as shown in Figure (3), according to the problem statement,\n\n\\(BF = 12 - x\\),\n\nSince \\(\\angle ACB = \\angle DEF = 30^\\circ\\),\n\n\\(\\angle QFB = \\angle QBF = 60^\\circ\\),\n\nThe height of \\(\\triangle QFB\\) is: \\(h = \\frac{1}{2} BF \\tan 60^\\circ = \\frac{1}{2}(12 - x) \\times \\sqrt{3} = 6 \\sqrt{3} - \\frac{\\sqrt{3}}{2} x\\),\n\nThus, \\(y = \\frac{1}{2}(12 - x) \\times \\left(6 \\sqrt{3} - \\frac{\\sqrt{3}}{2} x\\right) = \\frac{\\sqrt{3}}{4}(12 - x)^{2}\\), so only option A fits the description;\n\n\n\nFigure (3)\n\nTherefore, choose A;\n\n【Key Insight】This problem tests the properties of quadratic function graphs, solving right triangles, and the properties of isosceles triangles. The key to solving it is to classify and discuss the representation of the analytical expression." }, { "problem_id": 559, "question": "After folding a square piece of paper as shown in (1) and (2) in the figure, then cutting along the dotted line in (3), and finally unfolding the paper piece in (4), the resulting pattern should be one of the following patterns ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch35-2024_06_17_0f1da1946bcafdc7bf2fg_0058_1.jpg", "batch35-2024_06_17_0f1da1946bcafdc7bf2fg_0058_2.jpg", "batch35-2024_06_17_0f1da1946bcafdc7bf2fg_0058_3.jpg", "batch35-2024_06_17_0f1da1946bcafdc7bf2fg_0058_4.jpg", "batch35-2024_06_17_0f1da1946bcafdc7bf2fg_0058_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Question Analysis: Strictly follow the sequence in the diagram: fold to the right, then fold upwards, cut out a small rectangle from the midpoint of the top side of the square, and cut out another small rectangle from the bottom left corner of the square. Unfold to arrive at the conclusion.\n\nTherefore, choose B.\n\nExam Focus: Paper-cutting problem." }, { "problem_id": 560, "question": "Paper cutting is a traditional folk art in China. As shown in Figure (1), (2), a piece of paper is folded twice, and then cut along the dotted lines in Figure (3). After opening and flattening the paper in Figure (4), the resulting pattern should be ( )\n\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_1.jpg", "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_2.jpg", "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_3.jpg", "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_4.jpg", "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_5.jpg", "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_6.jpg", "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_7.jpg", "batch35-2024_06_17_39d6ca47a7505c7d39e8g_0021_8.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "After restoration, only B fits the question's requirement,\n\nTherefore, choose B.\n\n【Key Point】This question mainly tests the problem of paper cutting. The key to solving this question lies in accurately analyzing the method of folding the paper and the position of the cut, which allows for an intuitive answer." }, { "problem_id": 561, "question": "As shown in the figure, take a thin square piece of paper, fold it along the diagonal to form an isosceles right triangle. Then fold it again along the altitude of the base. Repeat this process, and finally cut along the curved line. Discard the smaller part, and when you unfold and lay it flat, the resulting shape should be ( )\n\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n\n\n##", "input_image": [ "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0028_1.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0028_2.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0028_3.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0028_4.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0028_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem statement, the resulting figure is option A.\n\nTherefore, the answer is: A.\n\n[Key Insight] This question tests paper-cutting problems and spatial imagination skills. The key to solving it lies in hands-on practice." }, { "problem_id": 562, "question": "A square piece of paper is folded twice and a small square is cut out as shown in the figure. The unfolded shape is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0052_1.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0052_2.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0052_3.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0052_4.jpg", "batch35-2024_06_17_8001d1fc55eb3c50e07bg_0052_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "After the first fold, there is a rectangle in the middle, eliminating options B and C;\n\nAfter the second fold, the rectangle lies on the crease, eliminating option D;\n\nTherefore, the correct choice is: A.\n\n【Key Point】This question tests students' spatial thinking ability and hands-on operation skills. The key is to solve it by understanding the folding of planar figures and the symmetry of the unfolded figures." }, { "problem_id": 563, "question": "Given $\\angle O$ and a fixed point $P$ inside it, find points $A$ and $B$ on the sides of $\\angle O$ such that the perimeter of $\\triangle PAB$ is minimized. The correct option is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch35-2024_06_17_8ac35b52de720b07595ag_0074_1.jpg", "batch35-2024_06_17_8ac35b52de720b07595ag_0074_2.jpg", "batch35-2024_06_17_8ac35b52de720b07595ag_0074_3.jpg", "batch35-2024_06_17_8ac35b52de720b07595ag_0074_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "In the diagram, the perimeter of the triangle $= \\mathrm{AP} + \\mathrm{BP} + \\mathrm{AB} = \\mathrm{P}_{1}\\mathrm{A} + \\mathrm{AB} + \\mathrm{BP}_{2} = \\mathrm{P}_{1}\\mathrm{P}_{2}$, which is a straight line segment, hence it is the minimum. The other three options do not represent the minimum perimeter.\n\nTherefore, choose D.\n【Key Insight】This question primarily tests the properties of symmetry and the definition of perimeter. The key to solving the problem lies in understanding the properties of symmetry." }, { "problem_id": 564, "question": "As shown in Figure 2, the figure is abstracted from the clock in Figure 1. Given that triangle $A B C$ is an equilateral triangle with $\\angle A = 60^\\circ$, the hour hand $O P$ points to point A exactly at $12:00$. If the hour hand $O P$ is parallel to one side of triangle $A B C$, the time the hour hand points to cannot be ( )\n\n\n\nClock Figure 1\n\n\n\nAbstracted Figure 2\nA. $1:00$\nB. $3:00$\nC. $5:00$\nD. $8:00$", "input_image": [ "batch36-2024_06_17_7d94fe8f4bba43bf3f09g_0034_1.jpg", "batch36-2024_06_17_7d94fe8f4bba43bf3f09g_0034_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, it is necessary to consider three scenarios, as illustrated below:\n\n\n\nFigure 2(1)\n\n\n\nFigure 2(2)\n\n\n\nFigure 2(3)\n\nWhen \\( O P \\parallel A B \\), as shown in Figure 2(1), the corresponding times are 1:00 or 7:00;\n\nWhen \\( O P \\parallel A C \\), as shown in Figure 2(2), the corresponding times are 5:00 or 11:00;\n\nWhen \\( O P \\parallel B C \\), as shown in Figure 2(3), the corresponding times are 3:00 or 9:00;\n\nTherefore, the correct answer is: D.\n\n[Key Insight] This problem primarily tests the concept of categorical discussion and the understanding of clock mechanics. Identifying each scenario is the key to solving the problem." }, { "problem_id": 565, "question": "A reconnaissance plane observes target $R$ at an elevation angle of $30^\\circ$ from point $P$. After flying east for 2 minutes, it reaches point $Q$, where the elevation angle to target $R$ is $45^\\circ$. Which of the following diagrams meets the criteria? ( ).\nA.\n\n\nB.\n\n\nC.\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_06b63790d29ee4769a3eg_0092_1.jpg", "batch37-2024_06_14_06b63790d29ee4769a3eg_0092_2.jpg", "batch37-2024_06_14_06b63790d29ee4769a3eg_0092_3.jpg", "batch37-2024_06_14_06b63790d29ee4769a3eg_0092_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "According to the problem, the schematic diagram is obtained.\n\n\n\nTherefore, the answer is: A.\n\n【Key Point】This question examines the practical application of solving right-angled triangles in the context of angle of depression problems. Correctly understanding the definition of the angle of depression is crucial for solving the problem." }, { "problem_id": 566, "question": "As shown in the figure, in right triangle $A B C$, $C D$ is the altitude on the hypotenuse $A B$. The two triangles $\\triangle A C D$ and $\\triangle B C D$ are arranged in the following three ways as shown in Figure (1), Figure (2), and Figure (3). Let the areas of the shaded regions in these three figures be $S_{1}, S_{2},$ and $S_{3}$, respectively. If $S_{1}=S_{2}$, what is the relationship between $S_{1}$ and $S_{3}$?\n\n\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. $S_{1}=1.5 S_{3}$\nB. $S_{1}=2 S_{3}$\nC. $S_{1}=3 S_{3}$\nD. $S_{1}=3.5 S_{3}$", "input_image": [ "batch37-2024_06_14_06b63790d29ee4769a3eg_0095_1.jpg", "batch37-2024_06_14_06b63790d29ee4769a3eg_0095_2.jpg", "batch37-2024_06_14_06b63790d29ee4769a3eg_0095_3.jpg", "batch37-2024_06_14_06b63790d29ee4769a3eg_0095_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (2), draw a perpendicular line from point $F$ to $BD$, intersecting $BD$ at point $E$,\n\n\n\n(2)\n\n$$\n\\begin{aligned}\n& S_{1}=S_{\\triangle B C D}-S_{\\triangle A C D^{\\prime}} \\\\\n& =\\frac{1}{2} \\times B D \\times C D-\\frac{1}{2} \\times A D \\times C D \\\\\n& =\\frac{1}{2} \\times\\left(B D-A D^{\\prime}\\right) \\times C D \\\\\n& S_{2}=\\frac{1}{2} C^{\\prime} D^{\\prime} \\times F E, \\\\\n& \\because S_{1}=S_{2}, \\\\\n& \\therefore E F=B D-A D=D D \\\\\n& \\therefore S_{1}=\\frac{1}{2} \\times C D \\times D D^{\\prime}, \\\\\n& S_{3}=\\frac{1}{2} B D^{\\prime} \\times D^{\\prime} K,\n\\end{aligned}\n$$\n\nFrom the given figure and the properties of the altitude in a right-angled triangle, we know:\n\n$C^{\\prime} K=1.5 D^{\\prime} K, D D^{\\prime}=0.6 B D$,\n\n$C^{\\prime} K+D^{\\prime} K=C^{\\prime} D^{\\prime}=2.5 D^{\\prime} K$,\n\n$\\therefore S 1=\\frac{1}{2} \\times 2.5 D^{\\prime} K \\times 0.6 B^{\\prime} D$,\n\n$\\therefore S_{1}=1.5 S_{3}$.\n\nTherefore, the answer is: $A$.\n\n【Key Insight】This question tests the application of the triangle area formula. The key to solving the problem lies in finding the area expressions within their respective figures, utilizing the given equal relationships, and combining the provided figure and the properties of the altitude in a right-angled triangle to determine the relationship between $S_{1}$ and $S_{3}$." }, { "problem_id": 567, "question": "As shown in Figure 1, in square $A B C D$, point $E$ is the midpoint of $A B$, and $\\angle C = 60^\\circ$. Point $F$ is a moving point that starts from point $B$ and moves along the edges of square $A B C D$ in the order $B \\rightarrow C \\rightarrow D \\rightarrow A$. Let the distance traveled by point $F$ be $x$, and the area of $\\triangle A E F$ be $y$. The graph in Figure 2 shows the function relationship between $y$ and $x$. When $x = 8$, the area of $\\triangle A E F$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 18\nB. 9\nC. $\\frac{9 \\sqrt{3}}{2}$\nD. $9 \\sqrt{3}$", "input_image": [ "batch37-2024_06_14_130a72fe01bab086139fg_0033_1.jpg", "batch37-2024_06_14_130a72fe01bab086139fg_0033_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw $BG \\perp CD$ at point $G$.\n\nFrom Figure 2, we know that $BC = CD = 6$, and when $x = 8$, point $F$ lies on $CD$.\n\n$\\therefore$ Quadrilateral $ABCD$ is a parallelogram,\n\n$\\therefore AB = CD = 6$,\n\n$\\therefore AE = \\frac{1}{2} AB = 3$.\n\nIn right triangle $\\triangle BCG$, $BG = BC \\sin \\angle C = 6 \\sin 60^\\circ = 6 \\times \\frac{\\sqrt{3}}{2} = 3\\sqrt{3}$,\n\n$\\therefore S_{\\triangle AEF} = \\frac{1}{2} \\times AE \\times GB = \\frac{1}{2} \\times 3 \\times 3\\sqrt{3} = \\frac{9\\sqrt{3}}{2}$.\n\nTherefore, the answer is: C.\n\n\n\n【Key Insight】This problem examines the properties of parallelograms and the application of trigonometric functions. It is a comprehensive problem involving moving points and function graphs. The key to solving it lies in extracting information about the lengths of line segments from the function graph." }, { "problem_id": 568, "question": "As shown in Figure (1), point $E$ is on side $A D$ of rectangle $A B C D$. Points $P$ and $Q$ start simultaneously from point $B$. Point $P$ moves along the broken line $B E - E D - D C$ until it reaches point $C$, and point $Q$ moves along $B C$ until it reaches point $C$. Both points move at a speed of $1 \\mathrm{~cm} /$ second. Let $y \\mathrm{~cm}^{2}$ be the area of $\\triangle B P Q$ when $P$ and $Q$ have been moving for $t$ seconds. The graph of the function $y$ versus $t$ is shown in Figure (2) (curve $O M$ is a part of a parabola). Which of the following conclusions is false?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\nA. $A D=B E=5 \\mathrm{~cm}$\n\nB. $\\cos \\angle A B E=\\frac{3}{5}$\n\nC. When $0\n\nFigure (1)\n\n\n\nFigure (2)\n\nSince the speeds of points \\( P \\) and \\( Q \\) are both \\( 1 \\, \\text{cm/s} \\),\n\n\\[ BC = BE = 5, \\]\n\n\\[ AD = BE = 5, \\] thus option A is correct and does not meet the question's requirement.\n\nAccording to Figure (2), the change from \\( M \\) to \\( N \\) takes 2 seconds,\n\n\\[ ED = 2, \\]\n\n\\[ AE = AD - ED = 5 - 2 = 3, \\]\n\nIn right triangle \\( \\triangle ABE \\), \\( AB = \\sqrt{BE^2 - AE^2} = \\sqrt{5^2 - 3^2} = 4 \\),\n\n\\[ \\cos \\angle ABE = \\frac{AB}{BE} = \\frac{4}{5}, \\] thus option B is incorrect and meets the question's requirement.\n\nAs shown in Figure (1), draw \\( PF \\perp BC \\) at point \\( F \\),\n\nSince \\( AD \\parallel BC \\),\n\n\n\n\\[ \\angle AEB = \\angle PBF, \\]\n\n\\[ \\sin \\angle PBF = \\sin \\angle AEB = \\frac{AB}{BE} = \\frac{4}{5}, \\]\n\n\\[ PF = PB \\sin \\angle PBF = \\frac{4}{5} t, \\]\n\nThus, when \\( 0 < t \\leq 5 \\), \\( y = \\frac{1}{2} BQ \\cdot PF = \\frac{1}{2} t \\cdot \\frac{4}{5} t = \\frac{2}{5} t^2 \\), so option C is correct and does not meet the question's requirement.\n\nWhen \\( t = \\frac{29}{4} \\) seconds, point \\( P \\) is on \\( CD \\), and at this time, \\( PD = \\frac{29}{4} - BE - ED = \\frac{29}{4} - 5 - 2 = \\frac{1}{4} \\),\n\n\\[ PQ = CD - PD = 4 - \\frac{1}{4} = \\frac{15}{4}, \\]\n\nSince \\( \\frac{AB}{AE} = \\frac{4}{3}, \\quad \\frac{BQ}{PQ} = \\frac{4}{3} \\),\n\n\\[ \\frac{AB}{AE} = \\frac{BQ}{PQ}, \\]\n\nAnd since \\( \\angle A = \\angle Q = 90^\\circ \\),\n\n\\[ \\triangle ABE \\sim \\triangle QBP, \\] thus option D is correct and does not meet the question's requirement.\n\nTherefore, the correct choice is B.\n\n【Key Insight】This problem examines the function graph of moving point problems, the determination of similar triangles, and the solution of right triangles. The key to solving the problem, and the breakthrough point, is determining from Figure (2) that when point \\( P \\) reaches point \\( E \\), point \\( Q \\) reaches point \\( C \\)." }, { "problem_id": 569, "question": "As shown in Figure 1, it is the 'Pythagorean tree' drawn by mathematician Pythagoras based on the Pythagorean theorem. As shown in Figure 2, in Rt $\\triangle A B C$, $\\angle B A C=90^{\\circ}$, use its three sides as the sides to make squares outwards, extend $EC, D B$ to intersect $GF, AH$ at points $N, K$ respectively, connect $KN$ to intersect $AG$ at point $M$, if $\\frac{S_{1}}{S_{2}}=\\frac{9}{16}$, then $\\tan \\angle A C B$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{1}{2}$\nB. $\\frac{2}{3}$\nC. $\\frac{3}{4}$\nD. $\\frac{5}{12}$", "input_image": [ "batch37-2024_06_14_75fe8f1a75ca44f27540g_0001_1.jpg", "batch37-2024_06_14_75fe8f1a75ca44f27540g_0001_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilaterals \\( A B C D \\), \\( A C F G \\), and \\( A B J H \\) are squares,\n\n\\[\n\\angle B C N = \\angle A C F = 90^\\circ, \\quad \\angle B A C = \\angle N F C = 90^\\circ, \\quad \\text{and} \\quad A C = C F.\n\\]\n\nThus,\n\n\\[\n\\angle A C B = \\angle F C N.\n\\]\n\nTherefore,\n\n\\[\n\\triangle A B C \\cong \\triangle F N C,\n\\]\n\nwhich implies\n\n\\[\nN F = A B.\n\\]\n\nLet \\( A C = 7a \\) and \\( A B = 7b \\). Then,\n\n\\[\n\\tan \\angle A C B = \\frac{b}{a}.\n\\]\n\nHence,\n\n\\[\nG N = G F - N F = 7a - 7b.\n\\]\n\nSince \\( A C \\parallel G F \\),\n\n\\[\n\\triangle A K M \\sim \\triangle G N M.\n\\]\n\nGiven that\n\n\\[\n\\frac{S_{1}}{S_{2}} = \\frac{9}{16} = \\left(\\frac{G M}{A M}\\right)^2,\n\\]\n\nwe have\n\n\\[\n\\frac{A K}{G N} = \\frac{A M}{G M} = \\frac{4}{3}.\n\\]\n\nSince\n\n\\[\n\\angle K B C = \\angle B A C = 90^\\circ,\n\\]\n\nit follows that\n\n\\[\n\\angle A C B + \\angle A B C = \\angle A B C + \\angle A B K.\n\\]\n\nThus,\n\n\\[\n\\angle A C B = \\angle A B K.\n\\]\n\nTherefore,\n\n\\[\n\\tan \\angle A C B = \\frac{A B}{A C} = \\tan \\angle A B K = \\frac{A K}{A B}.\n\\]\n\nHence,\n\n\\[\nA K = \\frac{A B^2}{A C} = \\frac{(7b)^2}{7a} = \\frac{7b^2}{a}.\n\\]\n\nThus,\n\n\\[\n\\frac{G N}{A K} = \\frac{7a - 7b}{\\frac{7b^2}{a}} = \\frac{3}{4}.\n\\]\n\nThis simplifies to\n\n\\[\n4a^2 - 4ab = 3b^2.\n\\]\n\nSince \\( a \\neq 0 \\),\n\n\\[\n4 - 4\\frac{b}{a} = 3\\left(\\frac{b}{a}\\right)^2.\n\\]\n\nThis can be rewritten as\n\n\\[\n3\\left(\\frac{b}{a}\\right)^2 + 4\\left(\\frac{b}{a}\\right) - 4 = 0.\n\\]\n\nSolving this quadratic equation yields\n\n\\[\n\\frac{b}{a} = \\frac{2}{3} \\quad \\text{or} \\quad \\frac{b}{a} = -2 \\quad (\\text{discarded}).\n\\]\n\nTherefore,\n\n\\[\n\\tan \\angle A C B = \\frac{2}{3}.\n\\]\n\nHence, the correct answer is **B**.\n\n**Key Insight:** This problem examines the properties of squares, the properties and determination of similar triangles, solving quadratic equations, and the tangent function. Mastering the properties and determination of similar triangles is crucial for solving this problem." }, { "problem_id": 570, "question": "A rectangle is divided into three right-angled triangles as shown in Figure 1. The two larger right-angled triangle paper pieces are placed in the ways depicted in Figure 2 (1) and (2). Let the area of the shaded region in (1) be $S_{1}$ and the area of the shaded region in (2) be $S_{2}$. When $S_{1}=S_{2}$, the ratio of the two sides of the rectangle is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2 (1)\n\n\n\nFigure 2 (2)\nA. $\\frac{3}{2}$\nB. $\\frac{4}{3}$\nC. $\\sqrt{2}$\nD. $\\sqrt{3}$", "input_image": [ "batch37-2024_06_14_9c2045d9977bd2288d98g_0010_1.jpg", "batch37-2024_06_14_9c2045d9977bd2288d98g_0010_2.jpg", "batch37-2024_06_14_9c2045d9977bd2288d98g_0010_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in Figure 1, let the length of the rectangle be $\\mathrm{AB}=\\mathrm{a}$ and the width be $\\mathrm{BC}=\\mathrm{b}$. According to the problem, we have:\n\n$\\mathrm{AC}=\\sqrt{a^{2}+b^{2}}, \\quad \\mathrm{DE}=\\frac{a b}{\\sqrt{a^{2}+b^{2}}}, \\tan \\angle \\mathrm{BAC}=\\frac{b}{a}$.\n\nAs shown in Figure 2(1), $\\angle \\mathrm{BCA}=\\angle \\mathrm{FHC}$,\n\n$\\therefore \\mathrm{FH}=\\mathrm{FC}$.\n\nDraw $\\mathrm{FG} \\perp \\mathrm{BC}$ through point $\\mathrm{F}$, with the foot of the perpendicular at $\\mathrm{G}$,\n\n$\\therefore \\angle \\mathrm{BAC}=\\angle \\mathrm{GFC}, \\mathrm{HG}=\\mathrm{GC}=\\frac{1}{2} \\mathrm{DE}=\\frac{a b}{2 \\sqrt{a^{2}+b^{2}}}$.\n\nSince $\\tan \\angle \\mathrm{BAC}=\\tan \\angle \\mathrm{GFC}=\\frac{G C}{F G}$,\n\n$\\therefore \\mathrm{FG}=\\frac{a b}{2 \\sqrt{a^{2}+b^{2}}} \\times \\frac{a}{b}$.\n\nThus, $S_{1}=\\frac{1}{2} \\mathrm{HC} \\times \\mathrm{FG}=\\frac{1}{2} \\times \\frac{a b}{2 \\sqrt{a^{2}+b^{2}}} \\times \\frac{a}{b} \\times \\frac{a b}{\\sqrt{a^{2}+b^{2}}}=\\frac{a^{3} b}{4\\left(a^{2}+b^{2}\\right)}$.\n\n\n\nFigure 1\n\n\n\nFigure 2(1)\n\n\n\nFigure 2(2)\n\nAs shown in Figure 2(2), $\\mathrm{AC}=\\sqrt{a^{2}+b^{2}}, \\mathrm{MN}=\\mathrm{a}$.\n\nSince $\\angle \\mathrm{BAC}=\\angle \\mathrm{BMN}$,\n\n$\\therefore \\mathrm{MN} / / \\mathrm{AC}$.\n\nThus, $\\triangle \\mathrm{BMN} \\sim \\triangle \\mathrm{BAC}$,\n\n$\\therefore S_{\\triangle B M N}: S_{\\triangle A B C}=(M N: A C)^{2}$.\n\nTherefore, $S_{\\triangle B M N}=\\frac{a^{3} b}{2\\left(a^{2}+b^{2}\\right)}$.\n\nThus, $S_{2}=\\frac{1}{2} \\mathrm{ab}-\\frac{a^{3} b}{2\\left(a^{2}+b^{2}\\right)}=\\frac{a b^{3}}{2\\left(a^{2}+b^{2}\\right)}$.\n\nSince $S_{1}=S_{2}$,\n\n$\\therefore \\frac{a^{3} b}{4\\left(a^{2}+b^{2}\\right)}=\\frac{a b^{3}}{2\\left(a^{2}+b^{2}\\right)}$.\n\nThus, $\\mathrm{a}^{2}=2 b^{2}$,\n\n$\\therefore \\mathrm{a}: \\mathrm{b}=\\sqrt{2}$.\n\nTherefore, the answer is C.\n\n【Key Insight】This problem tests the understanding of solving right triangles, the properties of similar triangles, and the application of various theorems. Mastery of these theorems is crucial for solving the problem." }, { "problem_id": 571, "question": "As shown in Figure 1, it is a folding table composed of the bracket $A D$ and $B C$ and the tabletop. As shown in Figure 2, it is given that $O A=O B=O C=O D=20 \\sqrt{3} \\mathrm{~cm}$ and $\\angle C O D=60^{\\circ}$. The distance from point $A$ to the ground (the plane of $C D$) is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $30 \\sqrt{3} \\mathrm{~cm}$\nB. $60 \\sqrt{3} \\mathrm{~cm}$\nC. $40 \\mathrm{~cm}$\nD. $60 \\mathrm{~cm}$", "input_image": [ "batch37-2024_06_14_aa09a85f660683fabf7cg_0023_1.jpg", "batch37-2024_06_14_aa09a85f660683fabf7cg_0023_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "As shown in the figure, connect $CD$, and draw $OF \\perp CD$ through point $O$, intersecting at point $F$. Extend $FO$ to intersect $AB$ at point $E$.\n\n\n\nFrom the problem statement, we know that $AB \\parallel CD$,\n\n$\\therefore OE \\perp AB$,\n\n$\\because OA = OB = OC = OD = 20\\sqrt{3}$, and $\\angle COD = \\angle AOB = 60^\\circ$,\n\n$\\therefore \\triangle COD$ and $\\triangle AOB$ are equilateral triangles, and $\\triangle COD \\cong \\triangle AOB$,\n\n$\\therefore \\angle COF = 30^\\circ$, and $OF = OE$,\n\n$\\therefore OF = OC \\cdot \\cos \\angle COF = 20\\sqrt{3} \\times \\frac{\\sqrt{3}}{2} = 30 \\text{ cm}$,\n\n$\\therefore EF = 2OF = 60 \\text{ cm}$,\n\nThus, the distance from point $A$ to the ground is $60 \\text{ cm}$.\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This problem examines the trigonometric functions of special angles, the determination and properties of equilateral triangles. The key to solving the problem lies in correctly drawing the auxiliary lines." }, { "problem_id": 572, "question": "As shown in Figure 1, in a plane, a fixed point $O$ is selected, along with a directed ray $O x$, and a unit length is chosen. Then, the position of any point $M$ in the plane can be determined by the length $m$ of $O M$ and the angle $\\theta$ in degrees of $\\angle M O x$. The ordered pair $(m, \\theta)$ is called the \"polar coordinates\" of point $M$, and this system of coordinates is called a \"polar coordinate system\".\n\nApplication: In the polar coordinate system of Figure 2, if $\\odot A$ is tangent to $O x$ at point $B$, with $O B = 2$, and ray $O A$ intersects $\\odot A$ at points $C$ and $D$, with $B C$ drawn and $\\angle O B C = 30^\\circ$, then the polar coordinates of point $D$ should be recorded as ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\left(\\frac{4 \\sqrt{3}}{3}, 30^\\circ\\right)$\nB. $\\left(\\frac{2 \\sqrt{3}}{3}, 60^\\circ\\right)$\nC. $\\left(\\sqrt{3}, 60^\\circ\\right)$\nD. $\\left(2 \\sqrt{3}, 30^\\circ\\right)$", "input_image": [ "batch37-2024_06_14_aa09a85f660683fabf7cg_0031_1.jpg", "batch37-2024_06_14_aa09a85f660683fabf7cg_0031_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $AB$,\n\n\n\nSince $\\odot A$ is tangent to $O x$,\n\n$\\therefore AB \\perp OB$,\n\nGiven $\\angle OBC = 30^{\\circ}$,\n\n$\\therefore \\angle ABC = 60^{\\circ}$,\n\nSince $AB = AC$,\n\n$\\therefore \\triangle ABC$ is an equilateral triangle,\n\n$\\therefore \\angle BAC = 60^{\\circ}$,\n\n$\\therefore \\cos \\angle BAC = \\frac{AB}{OA}$, and $\\angle AOB = 30^{\\circ}$,\n\n$\\therefore OA = 2AB$\n\nBy the Pythagorean theorem: $OA^{2} - AB^{2} = OB^{2}$, which is $(2AB)^{2} - AB^{2} = 2^{2}$,\n\nSolving gives: $AB = \\frac{2\\sqrt{3}}{3}$,\n\n$\\therefore OD = 2\\sqrt{3}$,\n\n$\\therefore$ The polar coordinates of point $D$ should be recorded as $\\left(2\\sqrt{3}, 30^{\\circ}\\right)$,\n\nTherefore, the answer is: D.\n\n【Key Insight】This problem examines the properties of tangents and the definition of polar coordinates. Understanding that the tangent to a circle is perpendicular to the radius at the point of contact is crucial for solving the problem." }, { "problem_id": 573, "question": "In the isosceles trapezoid $A B C D$, $A D = D C = B C = 4$ and $A B = 8$. Point $E$ moves along the path $A \\rightarrow D \\rightarrow C \\rightarrow B$ and point $F$ moves along the path $A \\rightarrow B \\rightarrow C$, both with a speed of 1 unit per second. The movement stops when the two points meet. The graph of the area $y$ of $\\triangle A E F$ versus the time $x$ in seconds is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch37-2024_06_14_b93c19209f895a7f6de9g_0047_1.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0047_2.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0047_3.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0047_4.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0047_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: As shown in the figure, draw line $DM$ through point $D$ such that $DM // BC$, intersecting $AB$ at point $M$.\n\n\n\nSince quadrilateral $ABCD$ is an isosceles trapezoid,\n\n$\\therefore AB // DC$, and $\\angle A = \\angle B$.\n\nGiven that $AB // DC$ and $DM // BC$,\n\n$\\therefore$ quadrilateral $MBCD$ is a parallelogram.\n\n$\\therefore BM = CD = 4$, and $DM = BC = 4$.\n\nAlso, since $AB = 8$,\n\n$\\therefore AM = AB - BM = 4$.\n\n$\\therefore AM = AD = DM$,\n\n$\\therefore \\triangle ADM$ is an equilateral triangle.\n\n$\\therefore \\angle B = \\angle A = 60^{\\circ}$.\n\nSince point $E$ moves along $A \\rightarrow D \\rightarrow C \\rightarrow B$ and point $F$ moves along $A \\rightarrow B \\rightarrow C$, both at a speed of 1 unit per second, and $\\angle A = 60^{\\circ}$,\n\n$\\therefore$ when point $E$ is on side $AD$, $\\triangle AEF$ is an equilateral triangle.\n\nGiven that $AD = DC = BC = 4$,\n\n$\\therefore$ when $0 \\leqslant x \\leqslant 4$, $AE = AF = x$, and the area of $\\triangle AEF$ is $y = \\frac{1}{2} x \\cdot x \\cdot \\sin 60^{\\circ} = \\frac{\\sqrt{3}}{4} x^{2}$.\n\nWhen $4 < x \\leqslant 8$, as shown in Figure 1, $AF = x$. Draw $DG \\perp AB$ at $G$, then $DG = 4 \\sin 60^{\\circ} = 2 \\sqrt{3}$.\n\n\n\nFigure 1\n\n$\\therefore$ the area of $\\triangle AEF$ is $y = \\frac{1}{2} AF \\cdot DG = \\frac{1}{2} x \\times 4 \\times \\frac{\\sqrt{3}}{2} = \\sqrt{3} x$.\n\nWhen $8 < x \\leqslant 10$, as shown in Figure 2, $CE = x - 8$, and $BF = x - 8$.\n\nThen $EF = 4 - (x - 8) - (x - 8) = 20 - 2x$.\n\n\n\nFigure 2\n\nDraw $DG \\perp AB$ and $CH \\perp AB$, and connect $AC$.\n\n$\\therefore AG = BH = 4 \\times \\cos 60^{\\circ} = 2$, and $GH = DC = 4$.\n\n$\\therefore AH = 2 + 4 = 6$, and $CH = DG = 2 \\sqrt{3}$.\n\n$\\therefore$ by the Pythagorean theorem, $AC = \\sqrt{AH^{2} + CH^{2}} = \\sqrt{6^{2} + (2 \\sqrt{3})^{2}} = 4 \\sqrt{3}$.\n\nSince $AC^{2} + BC^{2} = 48 + 16 = 64 = AB^{2}$,\n\n$\\therefore \\angle ACB = 90^{\\circ}$.\n\n$\\therefore$ the area of $\\triangle AEF$ is $y = \\frac{1}{2} EF \\cdot AC = 2 \\sqrt{3}(20 - 2x)$.\n\n$\\therefore$ at this point, $y$ is a linear function of $x$, and option A is correct.\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This problem examines the function graph of moving points, the properties of isosceles trapezoids, the determination and properties of equilateral triangles, and the solution of right triangles. The key to solving the problem lies in applying the concept of combining numbers and shapes and deriving the function's analytical expression in segments." }, { "problem_id": 574, "question": "As shown in the figure, $P$ is a moving point on the side of regular hexagon $A B C D E F$. Point $P$ starts from point $D$ and moves along the sides of the hexagon at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$ in a counterclockwise direction until it reaches point $C$. Let the time of point $P$'s movement be $x(\\mathrm{~s})$, and the area of the triangle with vertices $P, C$, and $D$ be $y\\left(\\mathrm{~cm}^{2}\\right)$. Which of the following graphs can roughly represent the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_b93c19209f895a7f6de9g_0094_1.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0094_2.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0094_3.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0094_4.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0094_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\nLet the side length of the regular hexagon \\( A B C D E F \\) be 1. When point \\( P \\) lies on side \\( D E \\):\n\n- Draw \\( P H \\perp C D \\) at point \\( H \\). Given that \\( \\angle C D P = 120^\\circ \\) and \\( P D = x \\),\n- It follows that \\( \\angle P D H = 60^\\circ \\),\n- Therefore, \\( P H = P D \\cdot \\sin 60^\\circ = \\frac{\\sqrt{3}}{2} x \\),\n- Hence, \\( y = \\frac{1}{2} \\cdot C D \\cdot P H = \\frac{1}{2} \\cdot 1 \\cdot \\frac{\\sqrt{3}}{2} x = \\frac{\\sqrt{3}}{4} x \\).\n\nWhen \\( P \\) lies on side \\( E F \\):\n\n- Extend \\( C D \\) and \\( F E \\) to meet at point \\( M \\). Draw \\( P Q \\perp C D \\) at point \\( Q \\),\n- Similarly, \\( \\angle C D E = \\angle F E D = 120^\\circ \\),\n- Thus, \\( \\angle E D M = \\angle D E M = 60^\\circ \\), making \\( \\triangle D E M \\) an equilateral triangle,\n- Therefore, \\( \\angle E M D = 60^\\circ \\), \\( E M = E D = 1 \\), and \\( P M = P E + E M = P E + E D = x \\),\n- Hence, \\( P Q = P M \\cdot \\sin 60^\\circ = \\frac{\\sqrt{3}}{2} x \\),\n- So, \\( y = \\frac{1}{2} \\cdot C D \\cdot P Q = \\frac{1}{2} \\cdot 1 \\cdot \\frac{\\sqrt{3}}{2} x = \\frac{\\sqrt{3}}{4} x \\).\n\nWhen \\( P \\) lies on side \\( A F \\):\n\n- Connect \\( A C \\) and \\( C F \\),\n- By the properties of a regular hexagon, \\( \\angle A B C = \\angle B A F = \\angle A F E = 120^\\circ \\), and \\( B A = B C \\),\n- Therefore, \\( \\angle B A C = \\frac{1}{2} (180^\\circ - 120^\\circ) = 30^\\circ \\), and \\( \\angle C A F = 120^\\circ - 30^\\circ = 90^\\circ \\),\n- By the symmetry of the regular hexagon, \\( \\angle A F C = \\frac{1}{2} \\angle A F E = 60^\\circ \\), and \\( A F = 1 \\),\n- Thus, \\( A C = A F \\cdot \\tan 60^\\circ = \\sqrt{3} \\),\n- Hence, \\( y = \\frac{1}{2} \\cdot C D \\cdot A C = \\frac{1}{2} \\cdot 1 \\cdot \\sqrt{3} = \\frac{\\sqrt{3}}{2} \\).\n\nBy the symmetry of the regular hexagon:\n\n- The graph of \\( P \\) on side \\( A B \\) is symmetric to the graph of \\( P \\) on side \\( E F \\),\n- The graph of \\( P \\) on side \\( B C \\) is symmetric to the graph of \\( P \\) on side \\( D E \\).\n\nTherefore, the correct answer is **A**.\n\n**Key Insight:** This problem examines the function graph of a moving point, the application of trigonometric functions, and the properties of regular polygons. Clear classification and discussion are crucial to solving this problem." }, { "problem_id": 575, "question": "As shown in Figure (1), it is a paperboard of a hexagram, where each acute angle is $60^{\\circ}$, each obtuse angle is $120^{\\circ}$, and all sides are equal. Now, the paperboard is cut as shown in Figure (2) and pieced together seamlessly without overlap to form a rectangle $A B C D$. If the area of the hexagram paperboard is $9 \\sqrt{3} \\mathrm{~cm}^{2}$, then the perimeter of rectangle $A B C D$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $18 \\sqrt{2} \\mathrm{~cm}$\nB. $4 \\sqrt{6} \\mathrm{~cm}$\nC. $(6 \\sqrt{3}+6) \\mathrm{cm}$\nD. $(2 \\sqrt{3}+6 \\sqrt{2}) \\mathrm{cm}$", "input_image": [ "batch37-2024_06_14_d25d6761981b932bcfbcg_0019_1.jpg", "batch37-2024_06_14_d25d6761981b932bcfbcg_0019_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular line $EF \\perp AB$ at point $F$.\n\nSince all six acute angles are $60^{\\circ}$ and all six obtuse angles are $120^{\\circ}$,\n\nlet $AE = x \\text{ cm}$, then $AD = 3x$.\n\nGiven that $\\angle AEB = 120^{\\circ}$,\n\nit follows that $\\angle EAB = 30^{\\circ}$.\n\nTherefore, $AB = 2AF = 2x \\cdot \\cos 30^{\\circ} = \\sqrt{3}x$.\n\nSince the area of the hexagonal star board is $9\\sqrt{3} \\text{ cm}^2$,\n\nwe have $AB \\cdot AD = 9\\sqrt{3}$, which means $\\sqrt{3}x \\cdot 3x = 9\\sqrt{3}$,\n\nsolving for $x$ gives $x = \\sqrt{3}$.\n\nThus, $AD = 3\\sqrt{3}$ and $AB = 3$.\n\nTherefore, the perimeter of rectangle $ABCD$ is $2(3\\sqrt{3} + 3) = (6\\sqrt{3} + 6) \\text{ cm}$.\n\nThe correct choice is: C.\n\n\n\n【Key Insight】This problem primarily tests knowledge of trigonometric values of special angles and is of medium difficulty. Common reasons for losing points include: 1. Lack of spatial imagination and unfamiliarity with the cutting and assembling of shapes; 2. Confusion over the trigonometric values of special angles." }, { "problem_id": 576, "question": "The barrier at the exit of a garage is shown in the figure. Point $A$ is the pivot point of the barrier, and point $E$ is the connection point of the two segments of the barrier. When a vehicle passes, the barrier $A E F$ can only rise to the position shown in Figure 2, as illustrated in Figure 3 (the width of the barrier is ignored). In the figure, $A B \\perp B C, E F / / B C, \\angle A E F=143^\\circ, A B=1.18$ meters, and $A E=1.2$ meters. What is the appropriate height limit sign for vehicles entering this underground garage? (Reference data: $\\sin 37^\\circ \\approx 0.60$, $\\cos 37^\\circ \\approx 0.80, \\tan 37^\\circ \\approx 0.75$)\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_d25d6761981b932bcfbcg_0031_1.jpg", "batch37-2024_06_14_d25d6761981b932bcfbcg_0031_2.jpg", "batch37-2024_06_14_d25d6761981b932bcfbcg_0031_3.jpg", "batch37-2024_06_14_d25d6761981b932bcfbcg_0031_4.jpg", "batch37-2024_06_14_d25d6761981b932bcfbcg_0031_5.jpg", "batch37-2024_06_14_d25d6761981b932bcfbcg_0031_6.jpg", "batch37-2024_06_14_d25d6761981b932bcfbcg_0031_7.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, extend $BA$ and $FE$ to intersect at point $D$.\n\n\n\nSince $AB \\perp BC$ and $EF \\parallel BC$,\n\nTherefore, $BD \\perp DF$, which means $\\angle ADE = 90^{\\circ}$.\n\nGiven that $\\angle AEF = 143^{\\circ}$,\n\nThus, $\\angle AED = 37^{\\circ}$.\n\nIn the right triangle $\\triangle ADE$,\n\nSince $\\sin \\angle AED = \\frac{AD}{AE}$ and $AE = 1.2$ meters,\n\nTherefore, $AD = AE \\cdot \\sin \\angle AED = 1.2 \\times \\sin 37^{\\circ} \\approx 0.72$ meters,\n\nThen, $BD = AB + AD = 1.18 + 0.72 = 1.9$ meters.\n\nHence, the correct choice is: A.\n\n【Key Insight】This problem tests the application of solving right triangles. The key to solving it is to construct a right triangle based on the given conditions and to be proficient in the concept of the sine function." }, { "problem_id": 577, "question": "As shown in the figure, the sides of the rhombus $A B C D$ are $4 \\mathrm{~cm}$, and $\\angle A = 60^{\\circ}$. Points $E$ and $F$ are on the sides of the rhombus $A B C D$. Starting from point $A$ simultaneously, they move along $A \\rightarrow B \\rightarrow C$ and $A \\rightarrow D \\rightarrow C$ at a speed of $1 \\mathrm{~cm}$ per second, respectively, and stop when they reach point $C$. The area of the region swept by segment $E F$ is denoted as $y\\left(\\mathrm{~cm}^{2}\\right)$, and the time of motion is denoted as $x(\\mathrm{~s})$. The graph that can roughly represent the functional relationship between $y$ and $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_dbf632bd78317adc6b59g_0081_1.jpg", "batch37-2024_06_14_dbf632bd78317adc6b59g_0081_2.jpg", "batch37-2024_06_14_dbf632bd78317adc6b59g_0081_3.jpg", "batch37-2024_06_14_dbf632bd78317adc6b59g_0081_4.jpg", "batch37-2024_06_14_dbf632bd78317adc6b59g_0081_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Analysis: When \\(0 \\leq x \\leq 4\\), draw \\(FM \\perp AB\\) at point \\(M\\) from point \\(F\\), as shown in Figure 1.\n\n\\[\n\\therefore AF = AE = x, \\angle A = 60^\\circ,\n\\]\n\nthen \\(FM = AF \\cdot \\sin A = \\frac{\\sqrt{3}}{2} x\\),\n\n\\[\n\\therefore \\text{The area } y \\text{ swept by segment } EF \\text{ is } \\frac{1}{2} x \\cdot \\frac{\\sqrt{3}}{2} x = \\frac{\\sqrt{3}}{4} x^{2}.\n\\]\n\nThe graph is a part of a parabola opening upwards, located to the right of the \\(y\\)-axis.\n\nWhen \\(4 < x \\leq 8\\),\n\nas shown in Figure 2, draw \\(FN \\perp BC\\) at point \\(N\\) from point \\(F\\), then \\(CE = CF = 8 - x\\),\n\n\\[\n\\therefore FN = \\frac{\\sqrt{3}}{2}(8 - x),\n\\]\n\\[\n\\therefore \\text{The area } y \\text{ swept by segment } EF \\text{ is } 4 \\times 2 \\sqrt{3} - \\frac{1}{2} \\times (8 - x) \\times \\frac{\\sqrt{3}}{2}(8 - x) = 8 \\sqrt{3} - \\frac{\\sqrt{3}}{4}(8 - x)^{2}.\n\\]\n\nThe graph is a part of a parabola opening downwards, located to the left of the symmetry axis line \\(x = 8\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This question examines the function graph of moving point problems, the properties of a rhombus, solving right triangles, and the properties of quadratic function graphs. Mastering the properties of quadratic function graphs is key to solving the problem." }, { "problem_id": 578, "question": "As shown in the figure, equilateral triangle $\\triangle A B C$ has a side length of $6$. Point $P$ moves along $C \\rightarrow B \\rightarrow A$ at a speed of 2 units per second, and point $Q$ moves along $B \\rightarrow A \\rightarrow C$ at the same speed. The area of $\\triangle B P Q$ is $y$. The graph of the function $y$ in terms of the movement time $x$ seconds is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0017_1.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0017_2.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0017_3.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0017_4.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0017_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem, we have: $PC = BQ = 2x$.\n\nAs shown in the figure, when $0 \\leq x \\leq 3$, $BP = 6 - 2x$. Draw $QD \\perp BC$ at point $D$.\n\n\n\nSince $\\triangle ABC$ is an equilateral triangle,\n\n$\\angle B = 60^\\circ$, and $\\sin 60^\\circ = \\frac{QD}{BQ}$,\n\nThus, $QD = 2x \\cdot \\frac{\\sqrt{3}}{2} = \\sqrt{3}x$,\n\nTherefore, $y = S_{\\triangle BPQ} = \\frac{1}{2} PB \\cdot QD = \\frac{1}{2}(6 - 2x) \\cdot \\sqrt{3}x = -\\sqrt{3}x^2 + 3\\sqrt{3}x$.\n\nWhen $3 \\leq x \\leq 6$, $BP = AQ = 2x - 6$. Draw $QE \\perp AB$ at point $E$.\n\n\n\nThus, $\\angle A = 60^\\circ$, and $\\sin 60^\\circ = \\frac{EQ}{AQ}$,\n\nTherefore, $EQ = \\frac{\\sqrt{3}}{2} \\cdot (2x - 6) = \\sqrt{3}(x - 3)$,\n\nHence, $y = S_{\\triangle BPQ} = \\frac{1}{2}(2x - 6) \\cdot \\sqrt{3}(x - 3) = \\sqrt{3}(x - 3)^2$.\n\nIn summary, when $0 \\leq x \\leq 3$, $y = -\\sqrt{3}x^2 + 3\\sqrt{3}x$, where $y$ is a quadratic function of $x$ opening downward; when $3 \\leq x \\leq 6$, $y = \\sqrt{3}(x - 3)^2$, where $y$ is a quadratic function of $x$ opening upward.\n\nTherefore, the correct choice is: C.\n\n【Key Insight】This problem examines the movement of two moving points. It is crucial to determine the distance and position of the moving points. By using a combination of numerical and graphical methods, the area's expression at different stages can be derived to make the correct judgment." }, { "problem_id": 579, "question": "In triangle $ABC$ as shown, $\\angle ABC = 90^\\circ$, $\\angle ACB = 30^\\circ$, and $AB = 2$. $BD$ is the median of side $AC$. The triangle $BCD$ is translated along ray $CB$ at a rate of $\\sqrt{3}$ units per second. The translated triangle is denoted as $\\triangle B_1 C_1 D_1$. Let the area of the overlapping part between $\\triangle B_1 C_1 D_1$ and $\\triangle ABD$ be $y$, and the translation time be $x$. When point $C_1$ coincides with point $B$, the motion of $\\triangle B_1 C_1 D_1$ stops. Which of the following graphs accurately represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD. $\\frac{\\sqrt{3}}{3}$ (", "input_image": [ "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0021_1.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0021_2.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0021_3.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0021_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Let $BD$ intersect $C_{1}D_{1}$ at point $E$, as shown in the figure,\n\n\n\nWhen $x=1$, it translates by $\\sqrt{3}$ units in length, i.e., $CC_{1}=\\sqrt{3}$.\n\nSince $\\angle ABC=90^{\\circ}$, $\\angle ACB=30^{\\circ}$, and $AB=2$,\n\nTherefore, $AC=2AB=4$, and $BC=\\frac{AB}{\\tan 30^{\\circ}}=\\frac{2}{\\frac{\\sqrt{3}}{3}}=2\\sqrt{3}$.\n\nThus, $BC_{1}=CC_{1}=\\sqrt{3}$, and the area of $\\triangle ABC$ is $S_{\\triangle ABC}=\\frac{1}{2}AB \\cdot BC=\\frac{1}{2} \\times 2 \\times 2\\sqrt{3}=2\\sqrt{3}$.\n\nTherefore, the area of $\\triangle ABD$ is $S_{\\triangle ABD}=\\frac{1}{2}S_{\\triangle ABC}=\\frac{1}{2} \\times 2\\sqrt{3}=\\sqrt{3}$.\n\nSince in $\\triangle ABC$, $\\angle ABC=90^{\\circ}$ is the median on side $AC$,\n\nTherefore, $BD=\\frac{1}{2}AC=AD=CD=2$.\n\nThus, $\\triangle BCD$ and $\\triangle B_{1}C_{1}D_{1}$ are isosceles triangles.\n\nSince $\\triangle BCD$ is translated along the direction of ray $CB$ to form $\\triangle B_{1}C_{1}D_{1}$,\n\nTherefore, $AC \\parallel C_{1}D_{1}$.\n\nSince $BC_{1}=CC_{1}=\\sqrt{3}$,\n\nTherefore, $D_{1}E$ is the midline of $\\triangle ABD$,\n\nThus, $D_{1}E=\\frac{1}{2}AD$.\n\nTherefore, the area of $\\triangle BD_{1}E$ is $S_{\\triangle BD_{1}E}=\\frac{1}{4}S_{\\triangle ABD}=\\frac{1}{4} \\times \\sqrt{3}=\\frac{\\sqrt{3}}{4}$.\n\nThat is, when $x=1$, $S_{\\triangle BD_{1}E}=\\frac{1}{4}S_{\\triangle ABD}=\\frac{\\sqrt{3}}{4}$, hence options C and D are incorrect.\n\nWhen $x>1$, $AB$ intersects $C_{1}D_{1}$ at point $H$, as shown in the figure:\n\n\n\nSince $AC \\parallel C_{1}D_{1}$,\n\nTherefore, $\\triangle BHE \\sim \\triangle BAD$, and $\\frac{BC_{1}}{BC}=\\frac{BE}{BD}$.\n\nThat is, $S_{\\triangle BHE}=\\left(\\frac{BE}{BD}\\right)^{2}S_{\\triangle ABD}$.\n\nAccording to the speed of motion: $C_{1}C=\\sqrt{3}x$, i.e., $BC_{1}=BC-CC_{1}=2\\sqrt{3}-\\sqrt{3}x$.\n\nTherefore, $\\frac{BC_{1}}{BC}=\\frac{2\\sqrt{3}-\\sqrt{3}x}{2\\sqrt{3}}$.\n\nThat is, $\\frac{BC_{1}}{BC}=\\frac{BE}{BD}=\\frac{2\\sqrt{3}-\\sqrt{3}x}{2\\sqrt{3}}$.\n\nTherefore, $S_{\\triangle BHE}=\\left(\\frac{2\\sqrt{3}-\\sqrt{3}x}{2\\sqrt{3}}\\right)^{2}S_{\\triangle ABD}$.\n\n$=\\left(\\frac{2\\sqrt{3}-\\sqrt{3}x}{2\\sqrt{3}}\\right)^{2} \\times \\sqrt{3}$.\n\n$=\\frac{\\sqrt{3}}{4}x^{2}-\\sqrt{3}x+\\sqrt{3}$.\n\nIt can be seen that when $x>1$, $S_{\\triangle BHE}=\\frac{\\sqrt{3}}{4}x^{2}-\\sqrt{3}x+\\sqrt{3}$, and the function graph is an upward-opening parabola, so option A is correct, and option B, being a linear function, is incorrect.\n\nTherefore, the answer is: A.\n\n【Insight】This problem is a comprehensive question on similarity, examining the midline theorem, the properties of similar triangles, the definition of trigonometric functions, the properties of right triangles, and the graphs of quadratic functions. The key to solving the problem lies in skillfully using the method of special points for graphical analysis." }, { "problem_id": 580, "question": "As shown in the figure, the sides of the rhombus $A B C D$ are $2$ units long, and $\\angle A = 60^\\circ$. Points $P$ and $Q$ start simultaneously from points $B$ and $C$, respectively, moving along the ray $B C$ to the right at the same speed. A perpendicular $Q H$ is drawn from point $Q$ to $B D$, with the foot of the perpendicular being $H$. The line segment $P H$ is drawn. Let the distance traveled by point $P$ be $x$ (where $0 < x \\leq 2$), and the area of $\\triangle B P H$ be $S$. Which of the following graphs can roughly represent the functional relationship between $S$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0048_1.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0048_2.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0048_3.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0048_4.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0048_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the side length of rhombus \\( A B C D \\) is 2 and \\( \\angle A = 60^\\circ \\),\n\nTherefore, \\( \\angle D B C = \\frac{1}{2} \\angle A B C = \\frac{1}{2} \\left(180^\\circ - \\angle A\\right) = 60^\\circ \\), and \\( B C = 2 \\).\n\nFrom the problem statement: \\( B P = C Q = x \\),\n\nThus, \\( B Q = B C + C Q = 2 + x \\).\n\nSince \\( Q H \\perp B D \\), \\( \\angle B Q H = 90^\\circ - \\angle D B C = 30^\\circ \\),\n\nTherefore, \\( B H = \\frac{1}{2} B Q = 1 + \\frac{1}{2} x \\).\n\nAs shown in the figure, draw \\( H G \\perp B C \\) at point \\( G \\),\n\n\n\nThus, \\( H G = B H \\cdot \\sin \\angle D B C = \\frac{\\sqrt{3}}{2} B H = \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{3}}{4} x \\),\n\nTherefore, \\( S = \\frac{1}{2} B P \\cdot H G = \\frac{\\sqrt{3}}{8} x^{2} + \\frac{\\sqrt{3}}{4} x \\) (for \\( 0 < x \\leq 2 \\)),\n\nHence, the correct choice is: A.\n\n【Key Insight】This problem examines the geometric application of quadratic functions, the properties of a rhombus, and solving right triangles. Mastering the properties of a rhombus is crucial for solving this problem." }, { "problem_id": 581, "question": "As shown in the figure, the side length of the square $A B C D$ is 2. Point $E$ is any point on the side $B C$ (not coinciding with points $B$ or $C$). The line segment $A E$ is drawn, and the bisector of $\\angle B A E$ intersects $B C$ at point $P$. A perpendicular line $P F$ is drawn from point $P$ to $A E$ at point $F$, and the bisector of $\\angle F P E$ intersects $D C$ at point $Q$. Let $P F = x$ and $C Q = y$. The graph of $y$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0087_1.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0087_2.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0087_3.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0087_4.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0087_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since the angle bisector of $\\angle BAE$ intersects $BC$ at point $P$, with $PB \\perp AB$ and $PF \\perp AE$,\n\n$\\therefore BP = PF = x$,\n\nSince $\\angle BAP = \\angle FAP$, $\\angle ABP = \\angle AFP = 90^{\\circ}$, and $PB = PF$,\n\n$\\therefore \\triangle ABP \\cong \\triangle AFP$ (by AAS),\n\n$\\therefore \\angle APB = \\angle APF$,\n\nSince $PQ$ bisects $\\angle FPC$, it follows that $\\angle FPQ = \\angle CPQ$,\n\nSince $\\angle APB + \\angle APF + \\angle FPQ + \\angle CPQ = 180^{\\circ}$,\n\n$\\therefore \\angle APF + \\angle QPF = 90^{\\circ}$, which implies $AP \\perp PQ$,\n\nSince $\\angle APB + \\angle QPC = 90^{\\circ}$ and $\\angle QPC + \\angle PQC = 90^{\\circ}$,\n\n$\\therefore \\angle APB = \\angle PQC$,\n\n$\\therefore \\tan \\angle APB = \\tan \\angle PQC$, hence $\\frac{AB}{BP} = \\frac{PC}{QC}$,\n\n$\\therefore \\frac{2}{x} = \\frac{2 - x}{y}$,\n\n$\\therefore y = -\\frac{1}{2}x(x - 2)$,\n\nTherefore, the correct choice is: C.\n\n【Highlight】This problem examines the function graph of a moving point, involving knowledge of quadratic functions, triangle congruence, and solving right triangles. The key to solving this problem is determining that $PA$ and $PQ$ are perpendicular to each other." }, { "problem_id": 582, "question": "As shown in the figure, in rhombus $A B C D$, $\\angle A B C = 120^\\circ$ and $A B = 2$. Point $P$ starts from point $A$ and moves at a speed of 2 units per second along the broken line $A D \\rightarrow D C$ to point $C$. At the same time, point $Q$ also starts from point $A$ and moves at a speed of $\\sqrt{3}$ units per second along $A C$ to point $C$. When one point stops moving, the other point stops as well. Let the area of triangle $A P Q$ be $y$, and the time of movement be $x$ seconds. Which of the following graphs can roughly represent the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0091_1.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0091_2.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0091_3.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0091_4.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0091_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Since \\(\\angle ABC = 120^\\circ\\),\n\n\\(\\angle DAB = 60^\\circ\\), thus \\(\\angle DAQ = 30^\\circ\\).\n\nWhen point \\(P\\) moves on \\(AD\\), as shown in the figure below,\n\n\n\nDraw \\(QH \\perp AD\\) through point \\(Q\\).\n\nFrom the given conditions: \\(AP = 2t\\), \\(AQ = \\sqrt{3}t\\), \\(\\angle HAQ = 30^\\circ\\),\n\nthen \\(y = \\frac{1}{2} \\times AP \\times HQ = \\frac{1}{2} \\times 2t \\times AQ \\times \\sin \\angle HAQ = \\frac{1}{2} \\times 2t \\times \\sqrt{3}t \\times \\frac{1}{2} = \\frac{\\sqrt{3}}{2} t^{2}\\), which is an upward-opening parabola.\n\nWhen point \\(P\\) moves on \\(CD\\), similarly, we get \\(y = -\\frac{\\sqrt{3}}{2} t(t-2)\\), which is a downward-opening parabola. Therefore, the correct answer is A.\n\n【Key Insight】This problem mainly tests the comprehensive understanding of function graphs. Accurate analysis and judgment are crucial to solving the problem." }, { "problem_id": 583, "question": "Examine the following compass and straightedge construction traces. Which one shows that segment $A D$ is the angle bisector of $\\triangle A B C$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch3-2024_06_14_fa34f5860d3cd15f347eg_0005_1.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0005_2.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0005_3.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0005_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: For option A, based on the construction marks, AD is the angle bisector of ∠CAB, so option A fits the description;\n\nFor option B, based on the construction marks, AD is the altitude on side BC of triangle ABC, so option B does not fit the description;\n\nFor option C, based on the construction marks, AD is the median of triangle ABC, so option C does not fit the description;\n\nFor option D, based on the construction marks, AD is the altitude on side BC of triangle ABC, so option D does not fit the description. Therefore, the correct choice is A.\n\n[Key Insight] This question tests the basic construction skills of drawing the angle bisector, median, and altitude of a triangle. Mastering the basic construction methods is crucial for solving this problem." }, { "problem_id": 584, "question": "Given the three methods of constructing an isosceles triangle using a ruler and compass as shown in the figure:\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\nCarefully observe the construction trace, which of the following known conditions correspond to the correct sequence of construction methods?\n\n(1) Given the base and the height of the base of the isosceles triangle;\n\n(2) Given the base and the side of the isosceles triangle;\n\n(3) Given the base and one base angle of the isosceles triangle.\n\nA. (1)(2)(3)\nB. (2)(1)(3)\nC. (3)(1)(2)\nD. (2)(3)(1)", "input_image": [ "batch3-2024_06_14_fa34f5860d3cd15f347eg_0019_1.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0019_2.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0019_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: The construction basis for figure (1) is \"(2) given the base and the legs of an isosceles triangle\";\n\nThe construction basis for figure (2) is \"(1) given the base and the height from the base of an isosceles triangle\";\n\nThe construction basis for figure (3) is \"(3) given the base and one base angle of an isosceles triangle\".\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question mainly tests the compass-and-straightedge construction of an isosceles triangle. Mastering the properties of isosceles triangles and the methods of construction is key to solving the problem." }, { "problem_id": 585, "question": "The following patterns are axisymmetric figures ( ).\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch3-2024_06_14_fa34f5860d3cd15f347eg_0026_1.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0026_2.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0026_3.jpg", "batch3-2024_06_14_fa34f5860d3cd15f347eg_0026_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "An axisymmetric figure is one where the two parts of the figure can coincide when folded along the axis of symmetry. Figures A, B, and C do not meet this condition. Only figure D, when folded along a certain line (the axis of symmetry), allows the two parts of the figure to coincide.\n\nTherefore, the correct choice is D." }, { "problem_id": 586, "question": "While exploring the proof that \"the sum of the interior angles of a triangle is $180^\\circ$\", the students in the comprehensive practice group drew the following four auxiliary lines, as shown in the figure. Among them, the ones that can prove \"the sum of the interior angles of $\\triangle A B C$ is $180^\\circ$\" are ( )\n\n\n\n(1) Draw $E F \\parallel A B$ through point $C$\n\n\n\n(2) Extend $A C$ to point $F$, and draw $C E \\parallel A B$ through point $C$\n\n\n\n(3) Draw $C D \\perp A B$ at point $D$\n\n\n\n(4) Draw $D E \\parallel B C$ and $D F \\parallel A C$ through a point $D$ on $A B$\n\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch5-2024_06_14_50e1bab738255386ff18g_0095_1.jpg", "batch5-2024_06_14_50e1bab738255386ff18g_0095_2.jpg", "batch5-2024_06_14_50e1bab738255386ff18g_0095_3.jpg", "batch5-2024_06_14_50e1bab738255386ff18g_0095_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: \n\n(1). Since \\( EF \\parallel AB \\), it follows that \\( \\angle ECA = \\angle A \\) and \\( \\angle FCB = \\angle B \\). Given that \\( \\angle ECA + \\angle ACB + \\angle FCB = 180^\\circ \\), we deduce that \\( \\angle A + \\angle ACB + \\angle B = 180^\\circ \\), which satisfies the condition.\n\n(2). Since \\( CE \\parallel AB \\), it follows that \\( \\angle A = \\angle FCE \\) and \\( \\angle B = \\angle BCE \\). Given that \\( \\angle FCE + \\angle ECB + \\angle ACB = 180^\\circ \\), we deduce that \\( \\angle A + \\angle B + \\angle ACB = 180^\\circ \\), which satisfies the condition.\n\n(3). Since \\( CD \\perp AB \\) at point \\( D \\), it follows that \\( \\angle ADC = \\angle CDB = 90^\\circ \\). However, this does not prove that the sum of the angles in the triangle is \\( 180^\\circ \\), so it does not satisfy the condition.\n\n(4). Since \\( DF \\parallel AC \\), it follows that \\( \\angle EDF = \\angle AED \\) and \\( \\angle A = \\angle FDB \\). Since \\( ED \\parallel BC \\), it follows that \\( \\angle EDA = \\angle B \\) and \\( \\angle C = \\angle AED \\), hence \\( \\angle C = \\angle EDF \\). Given that \\( \\angle ADE + \\angle EDF + \\angle FDB = 180^\\circ \\), we deduce that \\( \\angle B + \\angle A + \\angle C = 180^\\circ \\), which satisfies the condition.\n\nIn total, conditions (1), (2), and (4) satisfy the requirement.\n\nTherefore, the correct choice is: C.\n\n【Key Insight】This problem primarily examines the proof of the triangle angle sum theorem. Mastering the concept of transformation and the definition of a straight angle is crucial for solving this problem." }, { "problem_id": 587, "question": "Pythagorean theorem is an important theorem in geometry, and there is a record in the ancient Chinese mathematical book \"Zhou Bi Suan Jing\" that states, \"If the leg (the shorter side) is three, and the side (the longer side) is four, then the hypotenuse (the side opposite the right angle) is five.\" As shown in Figure 1, a figure is constructed using equal small squares and right-angled triangles, which can be used to verify the Pythagorean theorem through their area relationships. Figure 2 is obtained by placing Figure 1 inside a rectangle, where $\\angle B A C = 90^\\circ, A B = 6, B C = 10$. Points $D, E, F, G, H, I$ are on the edges of the rectangle $K L M J$. The area of the rectangle $K L M J$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 420\nB. 440\nC. 430\nD. 410", "input_image": [ "batch5-2024_06_14_77ce3e8cf88e61cf1a7dg_0087_1.jpg", "batch5-2024_06_14_77ce3e8cf88e61cf1a7dg_0087_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, extend $AB$ to intersect $KL$ at $P$, and extend $AC$ to intersect $LM$ at $Q$.\n\nFrom the given conditions, we have $\\angle BAC = \\angle BPF = \\angle FBC = 90^\\circ$, and $BC = BF$.\n\nTherefore, $\\angle ABC + \\angle ACB = 90^\\circ = \\angle PBF + \\angle ABC$,\n\nHence, $\\angle ACB = \\angle PBF$,\n\nThus, $\\triangle ABC \\cong \\triangle PFB$ (by AAS).\n\nSimilarly, it can be proven that $\\triangle ABC \\cong \\triangle QCG$ (by AAS).\n\nTherefore, $PB = AC = 8$, and $CQ = AB = 6$.\n\nSince Figure 2 is obtained by placing Figure 1 inside a rectangle,\n\nWe have $IP = 8 + 6 + 8 = 22$, and $DQ = 6 + 8 + 6 = 20$.\n\nThus, the area of rectangle $KLMJ$ is $22 \\times 20 = 440$.\n\nThe correct answer is: B.\n\n\n\nFigure 2\n\n【Key Insight】This problem examines the proof of the Pythagorean theorem, the properties and determination of congruent triangles. The key to solving the problem lies in constructing auxiliary lines to form congruent triangles and determining the lengths of the adjacent sides of the rectangle, which is also the challenging part of this problem." }, { "problem_id": 588, "question": "As shown in the figure, there is a square frame in the workshop. Xiaohua finds it easy to deform. Which of the following reinforcement plans is the best?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0031_1.jpg", "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0031_2.jpg", "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0031_3.jpg", "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0031_4.jpg", "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0031_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the stability of the triangle, it can be concluded that option $D$ is the best reinforcement plan.\n\nTherefore, the answer is: D.\n\n[Key Point] This question primarily examines the stability of triangles. The key to solving it lies in understanding that once the lengths of the three sides of a triangle are determined, the shape and size of the triangle are uniquely determined, hence the triangle possesses stability." }, { "problem_id": 589, "question": "Given the conditions in the following figures, find the congruent triangles.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1) and (2)\nB. (2) and (3)\nC. (1) and (3)\nD. (1) and (4)", "input_image": [ "batch5-2024_06_14_7ad1f93a1f17b69ef356g_0038_1.jpg", "batch5-2024_06_14_7ad1f93a1f17b69ef356g_0038_2.jpg", "batch5-2024_06_14_7ad1f93a1f17b69ef356g_0038_3.jpg", "batch5-2024_06_14_7ad1f93a1f17b69ef356g_0038_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Since (1) and (4) satisfy the SAS (Side-Angle-Side) condition,\n\nTherefore, triangles (1) and (4) are congruent;\n\nHence, choose D.\n\n【Key Point】This question tests the method of determining triangle congruence. Mastering the criteria for congruence is crucial for solving such problems." }, { "problem_id": 590, "question": "Given a right-angled triangle, squares are constructed externally on each side (as shown in Figure 1). Then, the two smaller squares are arranged inside the largest square as shown in Figure 2. If the area of the shaded region is known, then it is definitely possible to find out ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. The area of quadrilateral $A B C D$\nB. The area of quadrilateral $D C E G$\nC. The area of quadrilateral $H G F P$\nD. The area of $\\triangle G E F$", "input_image": [ "batch5-2024_06_14_ad6e7e10493b0d750827g_0082_1.jpg", "batch5-2024_06_14_ad6e7e10493b0d750827g_0082_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the area of quadrilateral \\( A B C D \\) be \\( S_{1} \\), the area of quadrilateral \\( D C E G \\) be \\( S_{2} \\), the area of triangle \\( G E F \\) be \\( S_{3} \\), and the area of quadrilateral \\( H G F P \\) be \\( S_{4} \\). Let the area of the large square be \\( c \\), the area of the medium square be \\( b \\), and the area of the small square be \\( a \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nSince \\( S_{4} + S_{\\text{shaded}} = \\frac{1}{2}(c - a) \\), \\( S_{3} + S_{4} = \\frac{1}{2} b \\), and \\( b = c - a \\),\n\nit follows that \\( S_{4} + S_{\\text{shaded}} = S_{3} + S_{4} \\).\n\nTherefore, \\( S_{3} = S_{\\text{shaded}} \\).\n\nThis means that if the area of the shaded part in the figure is known, the area of triangle \\( G E F \\) can be determined.\n\nHence, the correct choice is: D.\n\n【Key Insight】This problem examines the Pythagorean theorem and the operations of algebraic expressions. Mastering the Pythagorean theorem is crucial for solving the problem." }, { "problem_id": 591, "question": "To facilitate green travel for citizens, a city launched a shared bike service. Figure (1) shows a real image of a shared bike placed on a horizontal ground, and Figure (2) is a diagram of the bike, where $A B, C D$ are both parallel to the ground $l$, $\\angle B C D = 60^\\circ$, and $\\angle B A C = 55^\\circ$. When $\\angle M A C$ is ( ) degrees, $A M$ is parallel to $C B$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 55\nB. 65\nC. 75\nD. 80", "input_image": [ "batch5-2024_06_14_ad6e7e10493b0d750827g_0084_1.jpg", "batch5-2024_06_14_ad6e7e10493b0d750827g_0084_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AB \\parallel l \\) and \\( CD \\parallel l \\),\nit follows that \\( AB \\parallel CD \\).\n\nGiven that \\( \\angle BCD = 60^\\circ \\),\nwe have \\( \\angle ABC = \\angle BCD = 60^\\circ \\).\n\nGiven that \\( \\angle BAC = 55^\\circ \\),\nwe can find \\( \\angle ACB \\) as follows:\n\\[\n\\angle ACB = 180^\\circ - \\angle BAC - \\angle ABC = 180^\\circ - 55^\\circ - 60^\\circ = 65^\\circ.\n\\]\n\nFor \\( AM \\) to be parallel to \\( CB \\), the condition \\( \\angle ACB = \\angle MAC \\) must be satisfied.\nTherefore, \\( \\angle MAC = 65^\\circ \\), which confirms that option B is correct.\n\nThe correct answer is: B.\n\n[Key Insight] This problem primarily tests the understanding of the criteria and properties of parallel lines. Familiarity with the theorems related to the determination and properties of parallel lines is crucial for solving such problems." }, { "problem_id": 592, "question": "Among the following everyday objects, which one does not make use of the stability of a triangle?\nA.\n\n\nSolar water heater\nB.\n\n\nCollapsible clothes hanger\nC.\n\n\nTripod\nD.\n\n\nBasketball hoop", "input_image": [ "batch5-2024_06_14_ad7ed33db7a371745ab4g_0054_1.jpg", "batch5-2024_06_14_ad7ed33db7a371745ab4g_0054_2.jpg", "batch5-2024_06_14_ad7ed33db7a371745ab4g_0054_3.jpg", "batch5-2024_06_14_ad7ed33db7a371745ab4g_0054_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "A. The bracket of a solar water heater is triangular, which has the stability of a triangle and does not meet the requirement of the question;\n\nB. The clothes rack is quadrilateral and does not have the stability of a triangle, thus meeting the requirement of the question;\n\nC. The tripod is triangular, which has the stability of a triangle and does not meet the requirement of the question;\n\nD. The basketball stand is triangular, which has the stability of a triangle and does not meet the requirement of the question; therefore, the correct choice is B.\n\n[Key Point] This question tests the stability of triangles. Mastering the stability of triangles is crucial for solving the problem." }, { "problem_id": 593, "question": "In the following figures, which one correctly draws the altitude from vertex $A$ to side $A C$ in $\\triangle A B C$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_ad7ed33db7a371745ab4g_0100_1.jpg", "batch5-2024_06_14_ad7ed33db7a371745ab4g_0100_2.jpg", "batch5-2024_06_14_ad7ed33db7a371745ab4g_0100_3.jpg", "batch5-2024_06_14_ad7ed33db7a371745ab4g_0100_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: In triangle \\( \\triangle ABC \\), the altitude on side \\( AC \\) is the perpendicular line segment from vertex \\( B \\) to the line containing \\( AC \\). According to the definition, the correct option is D.\n\nTherefore, the answer is: D.\n\n**Key Point:** This question tests the understanding of the definition of an altitude in a triangle. Grasping the definition is crucial.\n\nPage 56 of 56 in the exam paper." }, { "problem_id": 594, "question": "As shown in the figure, in Figure (1) there are 3 triangles with height $M N$, in Figure (2) there are 10 triangles with height $M N$. In Figure (3) there are triangles with height $M N$, $\\ldots$, and so on. The number of triangles with height $M N$ in Figure (6) is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA. 55\nB. 78\nC. 96\nD. 105", "input_image": [ "batch5-2024_06_14_cff2fd86c2bf8908338bg_0089_1.jpg", "batch5-2024_06_14_cff2fd86c2bf8908338bg_0089_2.jpg", "batch5-2024_06_14_cff2fd86c2bf8908338bg_0089_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In the first figure, there are \\(1 + 2 = 3\\) triangles;\n\nIn the second figure, there are \\(1 + 2 + 3 + 4 = 10\\) triangles;\n\nIn the third figure, there are \\(1 + 2 + 3 + 4 + 5 + 6 = 21\\) triangles;\n\nIn the \\(n\\)th figure, the number of triangles is given by \\(1 + 2 + 3 + 4 + 5 + \\ldots + 2n = n(2n + 1)\\).\n\nTherefore, the number of triangles in the sixth figure is \\(1 + 2 + 3 + \\ldots + 12 = 6 \\times 13 = 78\\),\n\nHence, the correct answer is: B.\n\n[Key Insight] This problem primarily examines the pattern of changes in the figures. The key to solving it lies in deriving the relationship for the number of triangles in the \\(n\\)th figure." }, { "problem_id": 595, "question": "In Figure (1) and (2), $\\angle A = 42^\\circ, \\angle 1 = \\angle 2, \\angle 3 = \\angle 4$. The sum of $\\angle O_1 + \\angle O_2$ is ( )\n\n\n\n(1)\n\n\n\n(2)\nA. 111\nB. 174\nC. 153\nD. 132", "input_image": [ "batch5-2024_06_14_d212e25dd7dcee36ccd8g_0080_1.jpg", "batch5-2024_06_14_d212e25dd7dcee36ccd8g_0080_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since in (1) and (2), $\\angle A=42^{\\circ}$, $\\angle 1=\\angle 2$, and $\\angle 3=\\angle 4$,\n\nTherefore, in (1), $\\angle 2 + \\angle 4 = \\frac{1}{2}(\\angle 1 + \\angle 2 + \\angle 3 + \\angle 4) = \\frac{1}{2} \\times (180^{\\circ} - 42^{\\circ}) = 69^{\\circ}$,\n\nHence, $\\angle O_{I} = 180^{\\circ} - 69^{\\circ} = 111^{\\circ}$;\n\nIn (2), $\\angle O_{2} = \\angle 4 - \\angle 2 = \\frac{1}{2}[(\\angle 3 + \\angle 4) - (\\angle 1 + \\angle 2)] = \\frac{1}{2} \\angle A = 21^{\\circ}$;\n\nThus, $\\angle O_{1} + \\angle O_{2} = 111^{\\circ} + 21^{\\circ} = 132^{\\circ}$,\n\nTherefore, the correct choice is: D.\n\n【Key Insight】This problem examines the triangle angle sum theorem, the properties of exterior angles, the concept of supplementary angles, and their applications. Mastering and applying the triangle angle sum theorem and the properties of exterior angles are crucial for solving this problem." }, { "problem_id": 596, "question": "Given that the area of triangle $ABC$ is $a$, as shown in Figure (1), if sides $BC$ and $AC$ are divided into two equal parts, $BE_1$ and $AD_1$ intersect at point $O$, and the area of triangle $AOB$ is denoted as $S_1$; as shown in Figure (2), if sides $BC$ and $AC$ are divided into three equal parts, $BE_1$ and $AD_1$ intersect at point $O$, and the area of triangle $AOB$ is denoted as $S_2$; and so on. If $S_5 = \\frac{3}{11}$, then the value of $a$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA. 1\nB. 2\nC. 6\nD. 3", "input_image": [ "batch5-2024_06_14_e4559f258117b7931eb3g_0090_1.jpg", "batch5-2024_06_14_e4559f258117b7931eb3g_0090_2.jpg", "batch5-2024_06_14_e4559f258117b7931eb3g_0090_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In Figure (1), connect $OC$,\n\n\n\nFigure (1)\n\n$\\because AE_{1}=CE_{1}, \\quad BD_{1}=CD_{1}$,\n\n$\\therefore S_{\\triangle OAE_{1}}=S_{\\triangle OCE_{1}}, S_{\\triangle OBD_{1}}=S_{\\triangle OCD_{1}}, S_{\\triangle ABE_{1}}=S_{\\triangle ABD_{1}}=\\frac{1}{2} S_{\\triangle ABC}=\\frac{1}{2} a$,\n\n$\\because S_{\\triangle OAE_{1}}=S_{\\triangle ABE_{1}}-S_{\\triangle OAB}, \\quad S_{\\triangle OBD_{1}}=S_{\\triangle ABD_{1}}-S_{\\triangle OAB}$,\n\n$\\therefore S_{\\triangle OAE_{1}}=S_{\\triangle OBD_{1}}$,\n\n$\\therefore S_{\\triangle OAE_{1}}=S_{\\triangle OCE_{1}}=S_{\\triangle OBD_{1}}=S_{\\triangle OCD_{1}}$,\n\nLet $S_{\\triangle OAE_{1}}=S_{\\triangle OCE_{1}}=S_{\\triangle OBD_{1}}=S_{\\triangle OCD_{1}}=x$, then\n\n$\\left\\{\\begin{array}{l}S_{1}+x=\\frac{1}{2} a \\\\ S_{1}+4x=a\\end{array}\\right.$\n\nSolving gives $S_{1}=\\frac{1}{3} a$;\n\nIn Figure (2), connect $OE_{2}$, $OC$, $OD_{2}$,\n\n\n\nFigure (2)\n\nThen $S_{\\triangle ABE_{1}}=S_{\\triangle ABD_{1}}=\\frac{1}{3} a, S_{\\triangle OAE_{1}}=S_{\\triangle OE_{1}E_{2}}=S_{\\triangle OCE_{2}}=S_{\\triangle OBD_{1}}=S_{\\triangle OD_{1}D_{2}}=S_{\\triangle OCD_{2}}$,\n\nLet $S_{\\triangle OAE_{1}}=S_{\\triangle OE_{1}E_{2}}=S_{\\triangle OCE_{2}}=S_{\\triangle OBD_{1}}=S_{\\triangle OD_{1}D_{2}}=S_{\\triangle OCD_{2}}=x$, then\n\n$\\left\\{\\begin{array}{l}S_{2}+x=\\frac{1}{3} a \\\\ S_{3}+6x=a\\end{array}\\right.$\n\nSolving gives $S_{2}=\\frac{1}{5} a$;\n\nIn Figure (3), connect $OE_{2}$, $OE_{3}$, $OC$, $OD_{2}$, $OD_{3}$,\n\n\n\nFigure (3)\n\nThen $S_{\\triangle ABE_{1}}=S_{\\triangle ABD_{1}}=\\frac{1}{4} a, S_{\\triangle OAE_{1}}=S_{\\triangle OE_{1}E_{2}}=S_{\\triangle OE_{2}E_{3}}=S_{\\triangle OCE_{3}}=S_{\\triangle OBD_{1}}=S_{\\triangle OD_{1}D_{2}}=S_{\\triangle OD_{2}D_{3}}=S_{\\triangle OCD_{3}}$,\n\nLet $S_{\\triangle OAE_{1}}=S_{\\triangle OE_{1}E_{2}}=S_{\\triangle OE_{2}E_{3}}=S_{\\triangle OCE_{3}}=S_{\\triangle OBD_{1}}=S_{\\triangle OD_{1}D_{2}}=S_{\\triangle OD_{2}D_{3}}=S_{\\triangle OCD_{3}}=x$, then\n\n$\\left\\{\\begin{array}{l}S_{3}+x=\\frac{1}{4} a \\\\ S_{3}+8x=a\\end{array}\\right.$\n\nSolving gives $S_{3}=\\frac{1}{7} a$,\n\nFrom this, it can be seen that $S_{n}=\\frac{1}{2n+1} a$,\n\n$\\because S_{5}=\\frac{3}{11}$,\n\n$\\therefore \\frac{1}{2 \\times 5+1} a=\\frac{3}{11}$,\n\nSolving gives $a=3$.\n\nTherefore, the answer is: D\n 【Key Point】This problem examines the area formula of triangles, with the key being to find the area of each figure by setting up and solving a system of equations, thereby identifying the pattern." }, { "problem_id": 597, "question": "A water level alarm device was designed by a merchant, and its circuit diagram is shown in Figure 1. Here, the constant resistor $R_{1} = 10 \\Omega$, and $R_{2}$ is a pressure-sensitive resistor. It is wrapped in an insulating film and placed in a hard, concave insulating box, which is then placed at the bottom of the water tank. The contact surface is horizontal, and the area $S$ it covers is $0.01 \\mathrm{~m}^{2}$. The resistance of the pressure-sensitive resistor $R_{2}$ varies with the liquid pressure $F$ as shown in Figure 2 (the deeper the water, the greater the pressure). The voltage source maintains a constant $6 \\mathrm{~V}$, and the alarm (resistance ignored) starts to sound when the current in the circuit is $0.3 \\mathrm{~A}$. The relationship between water pressure and depth is shown in Figure 3 (reference formula: $I = \\frac{U}{R}, F = pS, 1000 \\mathrm{~Pa} = 1 \\mathrm{kPa}$). Which of the following statements is NOT correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nA. When the water tank is empty $(h = 0 \\mathrm{~m})$, the pressure $p$ is $0 \\mathrm{kPa}$\n\nB. When the alarm just starts to sound, the force $F$ exerted on the water tank is $40 \\mathrm{~N}$\n\nC. When the alarm just starts to sound, the depth $h$ of the water in the tank is $0.8 \\mathrm{~m}$\n\nD. To make the alarm sound when the water depth is $1 \\mathrm{~m}$, the resistance of the constant resistor $R_{1}$ should be $12 \\Omega$\n\n##", "input_image": [ "batch6-2024_06_14_03c0f2dd13de177f4c24g_0055_1.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0055_2.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0055_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "A. From Figure 3, when \\( h = 0 \\), \\( p = 0 \\), so this statement is correct;\n\nB. When the alarm just starts to sound, \\( 0.3 = \\frac{6}{10 + R_{2}} \\), solving gives \\( R_{2} = 10 \\Omega \\). From Figure 2, we can find: \\( R_{2} = \\frac{800}{F} \\), therefore \\( 10 = \\frac{800}{F} \\), solving gives \\( F = 80 \\mathrm{~N} \\), so this statement is incorrect;\n\nC. When the alarm just starts to sound, from above \\( F = 80 N \\), then \\( 80 = p \\times 0.01 \\), therefore \\( p = 8 \\mathrm{kPa} \\). From Figure 3, \\( p = 10h \\), \\( 8 = 10h \\), solving gives: \\( h = 0.8 \\), so this statement is correct;\n\nD. When the alarm just starts to sound: \\( 0.3 = \\frac{6}{R_{1} + R_{2}} \\), therefore \\( R_{1} + R_{2} = 20 \\Omega \\). When \\( h = 1 \\), \\( p = 10 \\times 1 = 10 \\mathrm{kPa} \\), therefore \\( F = 10000 \\times 0.01 = 100 \\mathrm{~N} \\), \\( R_{2} = \\frac{800}{100} = 8 \\Omega \\), therefore \\( R_{1} = 20 - 8 = 12 \\Omega \\), so this statement is correct.\n\nTherefore, the correct choice is: B.\n\n【Key Point】This question cross-examines the practical application of inverse proportion functions and linear functions across disciplines. Understanding the actual meaning of each variable is key to solving the problem." }, { "problem_id": 598, "question": "In the graph of the inverse proportion function $y = \\frac{6}{x}$, which of the following shaded areas does not equal 6?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch6-2024_06_14_03c0f2dd13de177f4c24g_0093_1.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0093_2.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0093_3.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0093_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: \n\nA. The area of the shaded part is \\( x y = k = 6 \\), which does not meet the requirement;\n\nB. The area of the shaded part is \\( \\frac{1}{2}(x y + x y + x y) = \\frac{1}{2} \\times 3 x y = \\frac{3}{2} k = 9 \\), which meets the requirement;\n\nC. The area of the shaded part is \\( \\frac{1}{2} \\times 2 x \\cdot y = x y = k = 6 \\), which does not meet the requirement;\n\nD. The area of the shaded part is \\( \\frac{1}{2} x \\cdot 2 y = x y = k = 6 \\), which does not meet the requirement.\n\nTherefore, the correct answer is: B.\n\n【Key Insight】The problem primarily examines the geometric significance of \\( k \\) in the inverse proportional function \\( y = \\frac{k}{x} \\). Utilizing the combination of numerical and graphical analysis is crucial for solving this problem." }, { "problem_id": 599, "question": "\"Nine Chapters on the Mathematical Art\" is a classic mathematical treatise compiled in the early Eastern Han Dynasty in China. In its \"Equations\" chapter, systems of linear equations were represented by counting rods arranged in rows. The counting rod diagrams in the \"Nine Chapters\" were arranged vertically, but for ease of viewing, we will transpose them into horizontal rows, as shown in Figure 1 and Figure 2. The numbers of counting rods listed from left to right in each row of the diagrams represent the coefficients of the unknowns $x, y$ and the corresponding constant terms. The counting rod diagram in Figure 1 can be expressed in the familiar form of a system of equations as $\\left\\{\\begin{array}{l}3 x+2 y=19 \\\\ x+4 y=23\\end{array}\\right.$, and the system of equations represented by the counting rod diagram in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\left\\{\\begin{array}{l}2 x+y=11 \\\\ 4 x+3 y=27\\end{array}\\right.$\nB. $\\left\\{\\begin{array}{l}2 x+y=1 \\\\ 4 x+3 y=7\\end{array}\\right.$\nC. $\\left\\{\\begin{array}{l}2 x+y=27 \\\\ 4 x+3 y=11\\end{array}\\right.$\nD. $\\left\\{\\begin{array}{l}2 y+x=11 \\\\ 4 y+3 x=27\\end{array}\\right.$", "input_image": [ "batch6-2024_06_14_575997d6e283bc8cc5f6g_0043_1.jpg", "batch6-2024_06_14_575997d6e283bc8cc5f6g_0043_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: The system of equations represented by the counting rods diagram shown in Figure 2 is:\n\n\\[\n\\left\\{\n\\begin{array}{l}\n2x + y = 11 \\\\\n4x + 3y = 27\n\\end{array}\n\\right.\n\\]\n\nTherefore, the correct choice is: **A**.\n\n**Key Point**: This question tests the application of a system of linear equations in two variables. Understanding the problem and interpreting the method of representing the system of equations in Figure 1 is crucial for solving it." }, { "problem_id": 600, "question": "Xiaoming, Xiaoying, and Xiaoliang play a dart game. Each of them throws at the target 5 times, as shown in the figure. The rule is that scoring the same ring earns the same points. If Xiaoming scores 21 points and Xiaoliang scores 17 points, then Xiaoying's score is ( )\n\n\nXiaoming\n\n\nXiaoliang\n\n\nXiaoying\nA. 19 points\nB. 20 points\nC. 21 points\nD. 22 points", "input_image": [ "batch6-2024_06_14_575997d6e283bc8cc5f6g_0067_1.jpg", "batch6-2024_06_14_575997d6e283bc8cc5f6g_0067_2.jpg", "batch6-2024_06_14_575997d6e283bc8cc5f6g_0067_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let scoring in the outer ring be worth \\( x \\) points, and scoring in the inner ring be worth \\( y \\) points. According to the problem, we have the system of equations:\n\n\\[\n\\left\\{\n\\begin{array}{c}\n2x + 3y = 21 \\\\\n4x + y = 17\n\\end{array}\n\\right.\n\\]\n\nSolving the system, we find:\n\n\\[\n\\left\\{\n\\begin{array}{l}\nx = 3 \\\\\ny = 5\n\\end{array}\n\\right.\n\\]\n\nThus, \\( 3x + 2y = 3 \\times 3 + 2 \\times 5 = 19 \\) points.\n\nTherefore, Xiaoying's score is 19 points.\n\nThe correct choice is: A.\n\n**Key Insight:** This problem tests the application of a system of linear equations. Understanding the problem and identifying the correct relationships is crucial for solving it." }, { "problem_id": 601, "question": "As shown in Figure (1), on the first balance, the mass of weight $A$ equals the sum of the masses of weights $B$ and $C$. As shown in Figure (2), on the second balance, the combined mass of weights $A$ and $B$ equals the mass of three weights $C$. Please determine: how many weights $C$ are equivalent in mass to 1 weight $A$?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch6-2024_06_14_575997d6e283bc8cc5f6g_0079_1.jpg", "batch6-2024_06_14_575997d6e283bc8cc5f6g_0079_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the masses of weights \\( A \\), \\( B \\), and \\( C \\) be \\( x \\), \\( y \\), and \\( z \\) respectively. According to the problem, we have:\n\n\\[\n\\left\\{\n\\begin{array}{l}\nx = y + z \\quad (1) \\\\\nx + y = 3z \\quad (2)\n\\end{array}\n\\right.\n\\]\n\nAdding equations (1) and (2), we get:\n\n\\[\n2x = 4z,\n\\]\n\n\\[\nx = 2z.\n\\]\n\nThis means that the mass of one weight \\( A \\) is equal to the mass of two weights \\( C \\).\n\nTherefore, the correct answer is B.\n\n**Key Point:** This problem requires correctly setting up a system of equations based on the balance scale and then using addition or subtraction to eliminate variables." }, { "problem_id": 602, "question": "Utilizing two identical rectangular blocks to measure the height of a table, first arrange them as shown in Figure (1), then switch the positions of the blocks and arrange them as shown in Figure (2). Given the measurements as depicted, the height of the table is ( )\n\n\n\n(1)\n\n\n\n(2)\nA. $73 \\mathrm{~cm}$\nB. $74 \\mathrm{~cm}$\nC. $75 \\mathrm{~cm}$\nD. $76 \\mathrm{~cm}$", "input_image": [ "batch6-2024_06_14_575997d6e283bc8cc5f6g_0088_1.jpg", "batch6-2024_06_14_575997d6e283bc8cc5f6g_0088_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the height of the table be \\( h \\mathrm{~cm} \\), the length of the first rectangular block be \\( x \\mathrm{~cm} \\), and the width of the second rectangular block be \\( y \\mathrm{~cm} \\). From the first figure, the height of the table can be expressed as:\n\\[ h - y + x = 80 \\]\n\nFrom the second figure, the height of the table can be expressed as:\n\\[ h - x + y = 72 \\]\n\nAdding the two equations together, we get:\n\\[ (h - y + x) + (h - x + y) = 152 \\]\n\nSolving the equation, we find:\n\\[ h = 76 \\mathrm{~cm} \\]\n\nTherefore, the correct answer is **D**.\n\n**Key Insight**: This problem primarily tests the application of equation-solving skills, holistic thinking, and the ability to interpret graphical information. The key is to understand the meaning of the figures and identify the relationships they represent." }, { "problem_id": 603, "question": "In the \"Equations\" chapter of the ancient Chinese mathematical text \"The Nine Chapters on the Mathematical Art,\" systems of linear equations were represented by arrangements of counting rods. As shown in Figure 1, the counting rods in each row, from left to right, represent the coefficients of the unknowns $x$ and $y$ and the corresponding constant terms. Expressing the counting rod diagram in Figure 1 in the familiar form of a system of equations gives us $\\left\\{\\begin{array}{l}x+4 y=10, \\\\ 6 x+11 y=34 .\\end{array}\\right.$ Similarly, the system of equations representing the counting rod diagram in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\left\\{\\begin{array}{l}2 x+y=7 \\\\ x+3 y=11\\end{array}\\right.$\nB. $\\left\\{\\begin{array}{l}2 x+y=12 \\\\ x+3 y=6\\end{array}\\right.$\nC. $\\left\\{\\begin{array}{l}2 x+y=12 \\\\ x+3 y=11\\end{array}\\right.$\nD. $\\left\\{\\begin{array}{l}2 x+y=7 \\\\ x+3 y=6\\end{array}\\right.$", "input_image": [ "batch6-2024_06_14_8fa8123c9845349638ffg_0002_1.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0002_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the representation method of counting rods shown in Figure 1,\n\nwe can derive the system of equations represented by the counting rods in Figure 2: $\\left\\{\\begin{array}{l}2 x+y=7 \\\\ x+3 y=11\\end{array}\\right.$\n\nTherefore, the correct choice is: A.\n\n【Key Point】This question tests the application of a system of linear equations in two variables. Observing the diagram and correctly setting up the system of linear equations is the key to solving the problem." }, { "problem_id": 604, "question": "Given the figure, there are 360 rectangular sheets and 140 square sheets as shown in Figure 1. These are used to make the A and B types of rectangular, open-top boxes as shown in Figure 2, using all the sheets. How many A-type and B-type boxes can be made? If we denote the number of square sheets used for A-type boxes as $x$ and the number of square sheets used for B-type boxes as $y$, which of the following systems of equations is correct?\n\n\n\nSheet of Paper\n\n\n\nVertical Box\n\n\n\nHorizontal Box\n\nFigure 2\nA. $\\left\\{\\begin{array}{l}x+y=140 \\\\ 4 x+\\frac{3}{2} y=360\\end{array}\\right.$\nB. $\\left\\{\\begin{array}{l}x+y=140 \\\\ 4 x+3 y=360\\end{array}\\right.$\nC. $\\left\\{\\begin{array}{l}x+2 y=140 \\\\ 4 x+\\frac{3}{2} y=360\\end{array}\\right.$\nD. $\\left\\{\\begin{array}{l}x+2 y=140 \\\\ 4 x+3 y=360\\end{array}\\right.$", "input_image": [ "batch6-2024_06_14_8fa8123c9845349638ffg_0020_1.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0020_2.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0020_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the number of square sheets used to make type $A$ boxes be $x$, and the number of square sheets used to make type $B$ boxes be $y$. Then, the number of type $A$ boxes that can be made is $x$, and the number of type $B$ boxes is $\\frac{y}{2}$. According to the problem, we have the following system of equations:\n\n\\[\n\\left\\{\n\\begin{array}{l}\nx + y = 140 \\\\\n4x + \\frac{3}{2}y = 360\n\\end{array}\n\\right.\n\\]\n\nThus, option A is correct.\n\nTherefore, the answer is: A.\n\n**Key Insight:** This problem primarily tests the application of a system of linear equations. The key to solving it lies in identifying the relationships given in the problem." }, { "problem_id": 605, "question": "A large square and four identical small squares are arranged in two ways as shown in Figure 1 and Figure 2. The length of the side of the large square is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{a+b}{2}$\nB. $\\frac{a-b}{2}$\nC. $\\frac{a+b}{4}$\nD. $\\frac{a-b}{4}$", "input_image": [ "batch6-2024_06_14_8fa8123c9845349638ffg_0045_1.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0045_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the side length of the large square be \\( x_{1} \\), and the side length of the small square be \\( x_{2} \\). From figures (1) and (2), we can set up the following system of equations:\n\n\\[\n\\left\\{\n\\begin{array}{l}\nx_{1} + 2x_{2} = a \\\\\nx_{1} - 2x_{2} = b\n\\end{array}\n\\right.\n\\]\n\nSolving the system, we obtain:\n\n\\[\n\\left\\{\n\\begin{array}{l}\nx_{1} = \\frac{a + b}{2} \\\\\nx_{2} = \\frac{a - b}{4}\n\\end{array}\n\\right.\n\\]\n\nTherefore, the side length of the large square is \\( \\frac{a + b}{2} \\), so option A is correct.\n\nThe correct choice is: A.\n\n【Key Insight】This problem examines the application of a system of linear equations with two variables. The key is to correctly understand the problem and identify the relationships within the problem, expressing the side lengths of the large and small squares accordingly." }, { "problem_id": 606, "question": "Magic squares are ancient mathematical problems. The earliest magic square, known as the \"Palace Grid,\" was recorded in China's ancient book, \"The Book of Lu.\" To fill a magic square, 9 numbers must be placed in the empty cells such that the sum of each horizontal row, each vertical column, and both diagonals is equal. For example, Figure (1) is a magic square. Figure (2) is an incomplete magic square. What is the value of $x - y$?\n\n\n(1)\n\n\n\n(2)\nA. 0\nB. -4\nC. -10\nD. 32", "input_image": [ "batch6-2024_06_14_8fa8123c9845349638ffg_0052_1.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0052_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let the number in the middle be $a$, and the first number in the third row be $b$,\n\n\n\n(2)\n\n\\[\n\\left\\{\n\\begin{array}{l}\na + b + x = y + a + 2 \\quad (1) \\\\\n10 + 6 + x = 6 + y + b \\quad (2)\n\\end{array}\n\\right.\n\\]\n\nFrom equation (1), we get $x - y = 2 - b$.\n\nFrom equation (2), we get $x - y = b - 10$.\n\nTherefore, $2 - b = b - 10$,\n\nSolving for $b$, we get: $b = 6$.\n\nThus, $x - y = 2 - 6 = -4$.\n\nHence, the correct choice is: B.\n\n【Key Insight】This problem tests the application of a system of linear equations with two variables. The key to solving it lies in setting up the system of equations based on the given conditions." }, { "problem_id": 607, "question": "There are several identical jigsaw puzzle pieces, each shaped as shown in Figure 1, and when arranged horizontally, they can fit tightly into a row, with the bottom edge aligning with a straight line. When four pieces are arranged tightly into a row, the length is $23 \\text{ cm}$, as shown in Figure 2. When ten pieces are arranged tightly into a row, the length is $56 \\text{ cm}$, as shown in Figure 3. Given that the lengths of the two parts in Figure 1 are $a \\text{ cm}$ and $b \\text{ cm}$, respectively, which of the following statements is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nA. According to the problem, $4(a+b)=23$\n\nB. The length of one puzzle piece is $5.75 \\text{ cm}$\n\nC. When arranging the pieces tightly into a row, each additional piece increases the total length by $6.5 \\text{ cm}$\n\nD. When arranging $n$ pieces tightly into a row, the total length is $(5.5 n+1) \\text{ cm}$", "input_image": [ "batch6-2024_06_14_8fa8123c9845349638ffg_0069_1.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0069_2.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0069_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: \n\n$\\because$ When 4 puzzle pieces are tightly arranged in a row, the total length is $23 \\mathrm{~cm}$,\n\n$\\therefore 4a + b = 23$ (1), hence option A is incorrect and does not meet the requirement;\n\n$\\because$ When 10 puzzle pieces are tightly arranged in a row, the total length is $56 \\mathrm{~cm}$,\n\n$\\therefore 10a + b = 56$ (2),\n\nFrom (1) and (2), we can derive $a = 5.5$ and $b = 1$,\n\n$\\therefore$ The length of one puzzle piece is $6.5 \\mathrm{~cm}$, hence option B is incorrect and does not meet the requirement;\n\nWhen puzzle pieces are tightly arranged in a row, each additional piece increases the total length by $5.5 \\mathrm{~cm}$, hence option C is incorrect and does not meet the requirement;\n\nWhen $n$ puzzle pieces are tightly arranged in a row, the total length is $(5.5n + 1) \\mathrm{cm}$, hence option D is correct and meets the requirement.\n\nTherefore, the correct choice is: D.\n\n【Key Insight】This problem tests the ability to set up algebraic expressions and apply systems of linear equations in practical scenarios. The key to solving the problem lies in understanding the given information and determining the values of $a$ and $b$." }, { "problem_id": 608, "question": "Surround a square using an appropriate number of identical rectangular paper pieces. With 4 rectangular paper pieces, a square is formed as shown in Figure (1), with the shaded area being 81. With 8 rectangular paper pieces, a square is formed as shown in Figure (2), with the shaded area being 64. With 12 rectangular paper pieces, a square is formed as shown in Figure (3), with the shaded area being ( ).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\nA. 48\nB. 36\nC. 50\nD. 49", "input_image": [ "batch6-2024_06_14_8fa8123c9845349638ffg_0072_1.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0072_2.jpg", "batch6-2024_06_14_8fa8123c9845349638ffg_0072_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: In Figure (1), the shaded area is 81, and the side length is 9. In Figure (2), the shaded area is 64, and the side length is 8. Let the length of the rectangle be \\( a \\) and the width be \\( b \\). According to the problem, we have the following system of equations:\n\n\\[\n\\left\\{\n\\begin{array}{l}\na - b = 9 \\\\\na - 2b = 8\n\\end{array}\n\\right.\n\\]\n\nSolving the system, we find:\n\n\\[\n\\left\\{\n\\begin{array}{l}\na = 10 \\\\\nb = 1\n\\end{array}\n\\right.\n\\]\n\nTherefore, the shaded area in Figure (3) is:\n\n\\[\n(a - 3b)^2 = (10 - 3)^2 = 49\n\\]\n\nThus, the correct answer is: D.\n\n**Key Insight:** This problem tests the application of a system of linear equations. The key to solving it lies in setting up the correct system based on the given conditions." }, { "problem_id": 609, "question": "As shown in the figure, if the number of lattice points (points with integer coordinates) strictly inside the enclosed region (excluding the boundary) formed by the parabola $y = -x^2 + 3$ and the $x$-axis is $k$, then the graph of the inverse proportion function $y = \\frac{k}{x}(x > 0)$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch6-2024_06_14_97ec6f90182b40a82bfeg_0023_1.jpg", "batch6-2024_06_14_97ec6f90182b40a82bfeg_0023_2.jpg", "batch6-2024_06_14_97ec6f90182b40a82bfeg_0023_3.jpg", "batch6-2024_06_14_97ec6f90182b40a82bfeg_0023_4.jpg", "batch6-2024_06_14_97ec6f90182b40a82bfeg_0023_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Question Analysis: Since the number of integer points within the closed region is 4, it follows that $\\mathrm{k}=4$. Therefore, the correct answer is $\\mathrm{D}$. Key Points: Graphs of quadratic functions, Graphs of inverse proportional functions." }, { "problem_id": 610, "question": "A new operation is defined as follows: $p \\oplus q = \\left\\{\\begin{array}{c}\\frac{p}{q}, \\text{ if } q > 0 \\\\ -\\frac{p}{q}, \\text{ if } q < 0\\end{array}\\right.$, for example: $2 \\oplus 3 = \\frac{2}{3}, 2 \\oplus (-3) = -\\frac{2}{3}$. The graph of the function $y = 4 \\oplus x (x \\neq 0)$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch6-2024_06_14_d6212742b424b34c5ef2g_0001_1.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0001_2.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0001_3.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0001_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem, we have \n\n\\[ y = \\begin{cases} \n\\frac{4}{x} & (x > 0) \\\\\n-\\frac{4}{x} & (x < 0)\n\\end{cases} \\]\n\nThe graph of \\( y = \\frac{4}{x} \\) for \\( x > 0 \\) is located in the first quadrant, and the graph of \\( y = \\frac{4}{x} \\) for \\( x < 0 \\) is located in the second quadrant.\n\nTherefore, the correct choice is: D.\n\n**Key Insight:** This problem primarily tests the understanding of the graph and properties of inverse proportional functions. Grasping the new definition is crucial for solving the problem." }, { "problem_id": 611, "question": "As shown in the figure, in the rhombus $A B C D$, $A C$ and $B D$ intersect at point $O$, $A C = 4$, and $B D = 2$. Point $N$ is the midpoint of $C D$. Point $P$ starts from point $A$ and moves along the path $A-O-B-C$. A perpendicular line $P Q$ is drawn from point $P$ to $A C$, intersecting the sides of the rhombus at $Q$ (point $Q$ is above point $P$). The lines $P N$ and $Q N$ are connected. The movement of point $P$ stops when point $Q$ coincides with point $N$. Let the area of triangle $P Q N$ be $y$, and the distance traveled by point $P$ be $x$. Which of the following graphs can roughly represent the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch6-2024_06_14_d6212742b424b34c5ef2g_0007_1.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0007_2.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0007_3.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0007_4.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0007_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a rhombus, with \\( AC = 4 \\) and \\( BD = 2 \\),\n\n\\[\n\\therefore AO = CO = 2, \\quad BO = DO = 1.\n\\]\n\n(1) When point \\( P \\) is on \\( AO \\),\n\n\\[\n\\because PQ \\perp AC, \\quad DO \\perp AC,\n\\]\n\\[\n\\therefore PQ \\parallel DO,\n\\]\n\\[\n\\therefore \\triangle APQ \\sim \\triangle ADO,\n\\]\n\\[\n\\therefore \\frac{PQ}{DO} = \\frac{AP}{AO}.\n\\]\n\\[\n\\because AO = 2, \\quad AP = x,\n\\]\n\\[\n\\therefore \\frac{PQ}{1} = \\frac{x}{2}, \\quad \\text{solving gives: } PQ = \\frac{x}{2}.\n\\]\n\nDraw \\( MN \\perp OC \\) at point \\( M \\),\n\n\\[\n\\because MN \\perp OC, \\quad \\text{and point } N \\text{ is the midpoint of } CD,\n\\]\n\\[\n\\therefore OM = \\frac{1}{2} OC = 1,\n\\]\n\\[\n\\therefore PM = PO + OM = 2 - x + 1 = 3 - x,\n\\]\n\\[\n\\therefore y = \\frac{1}{2} PQ \\cdot PM = \\frac{1}{2} \\times \\frac{x}{2} \\times (3 - x) = \\frac{1}{4}x^2 + \\frac{3}{4} \\quad (0 \\leq x \\leq 2).\n\\]\n\n(2) When point \\( P \\) is on \\( OB \\),\n\n\\[\n\\because AO = 2, \\quad AO + OP = x,\n\\]\n\\[\n\\therefore OP = x - 2,\n\\]\n\\[\n\\therefore DP = DO + OP = 1 + (x - 2) = x - 1,\n\\]\n\\[\n\\therefore y = \\frac{1}{2} PQ \\cdot OM = \\frac{1}{2} \\times (x - 1) \\times 2 = x - 1 \\quad (2 \\leq x \\leq 3).\n\\]\n\n(3) When point \\( P \\) is on \\( BC \\),\n\n\\[\n\\because AO + BO = 3, \\quad AO + BO + BP = x,\n\\]\n\\[\n\\therefore BP = x - 3.\n\\]\n\nBy the Pythagorean theorem:\n\n\\[\nBC = \\sqrt{BO^2 + CO^2} = \\sqrt{5},\n\\]\n\\[\n\\therefore CP = BC - BP = \\sqrt{5} - (x - 3) = \\sqrt{5} + 3 - x.\n\\]\n\\[\n\\because PQ \\perp AC, \\quad BD \\perp AC,\n\\]\n\\[\n\\therefore PQ \\parallel BD,\n\\]\n\\[\n\\therefore \\triangle CPQ \\sim \\triangle CBD,\n\\]\n\\[\n\\therefore \\frac{PQ}{BD} = \\frac{CP}{BC}, \\quad \\text{that is, } \\frac{PQ}{2} = \\frac{\\sqrt{5} + 3 - x}{\\sqrt{5}},\n\\]\n\\[\n\\text{Solving gives: } PQ = \\frac{10 + 6\\sqrt{5}}{5} - \\frac{2\\sqrt{5}}{5}x = 2 + \\frac{6\\sqrt{5}}{5} - \\frac{2\\sqrt{5}}{5}x.\n\\]\n\\[\n\\because \\frac{OE}{OC} = \\frac{BP}{BC}, \\quad \\text{that is, } \\frac{OE}{2} = \\frac{x - 3}{\\sqrt{5}},\n\\]\n\\[\n\\text{Solving gives: } OE = \\frac{2\\sqrt{5}x - 6\\sqrt{5}}{5},\n\\]\n\\[\n\\therefore NF = 1 - \\frac{2\\sqrt{5}x - 6\\sqrt{5}}{5},\n\\]\n\\[\n\\therefore y = \\frac{1}{2} PQ \\cdot NF = \\frac{1}{2} \\times \\left(2 + \\frac{6\\sqrt{5}}{5} - \\frac{2\\sqrt{5}}{5}x\\right) \\left(1 - \\frac{2\\sqrt{5}x - 6\\sqrt{5}}{5}\\right) = -\\frac{1}{50}(10 + 12\\sqrt{5} - 2\\sqrt{5}x)(5 + 6\\sqrt{5} - 2\\sqrt{5}x) \\quad (3 \\leq x \\leq 4).\n\\]\n\nIn summary:\n\n- When \\( 0 \\leq x \\leq 2 \\), \\( y \\) is a quadratic function of \\( x \\), opening downward.\n- When \\( 2 \\leq x \\leq 3 \\), \\( y \\) is a linear function of \\( x \\).\n- When \\( 3 \\leq x \\leq 4 \\), \\( y \\) is a quadratic function of \\( x \\), opening upward.\n\nTherefore, the correct answer is: B.\n\n**Key Insight:** This problem primarily examines the properties of a rhombus, the determination and properties of similar triangles, and the graphs of quadratic and linear functions. The key to solving the problem lies in correctly understanding the scenario and performing a case-by-case analysis to derive the expressions." }, { "problem_id": 612, "question": "In square $A B C D$, point $E$ is a point on side $A B$, with $A B = 1$. Connect $C E$, and draw $D F \\perp C E$ at point $F$. Let $C E = x$, $D F = y$, and the graph of the function relationship between $y$ and $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch6-2024_06_14_fc2841d7b09b8762476ag_0100_1.jpg", "batch6-2024_06_14_fc2841d7b09b8762476ag_0100_2.jpg", "batch6-2024_06_14_fc2841d7b09b8762476ag_0100_3.jpg", "batch6-2024_06_14_fc2841d7b09b8762476ag_0100_4.jpg", "batch6-2024_06_14_fc2841d7b09b8762476ag_0100_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: In square $\\mathrm{ABCD}$, $\\mathrm{AB}=1$,\n\n\n\n$\\therefore \\mathrm{BC}=\\mathrm{CD}=1, \\angle \\mathrm{ABC}=90^{\\circ}, \\mathrm{AB} / / \\mathrm{CD}$,\n\n$\\therefore \\angle \\mathrm{BEC}=\\angle \\mathrm{FCD}$,\n\n$\\because \\mathrm{DF} \\perp \\mathrm{CE}$,\n\n$\\therefore \\angle \\mathrm{CFD}=\\angle \\mathrm{EBC}=90^{\\circ}$,\n\n$\\therefore \\triangle \\mathrm{BCE} \\sim \\triangle \\mathrm{FDC}$,\n\n$\\therefore \\frac{C E}{D C}=\\frac{B C}{F D}$ which means $\\frac{x}{1}=\\frac{1}{y}$,\n\n$\\therefore y=\\frac{1}{x}(1 \\leq x \\leq \\sqrt{2})$,\n\nFrom the above, it can be concluded that the graph of the function of $\\mathrm{y}$ with respect to $\\mathrm{x}$ is a hyperbola in the first quadrant.\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem mainly examines the ability to derive a function expression based on practical problems and determine the function graph, the properties of a square, the properties and determination of similar triangles. The key is to derive the expression of $\\mathrm{y}$ with respect to $\\mathrm{x}$ based on the given conditions." }, { "problem_id": 613, "question": "Given that in the square $A B C D$, point $P$ is a moving point on the diagonal $B D$. Lines parallel to $C D$ and $A D$ are drawn through $P$ and intersect the sides of the square $A B C D$ at $E$, $F$, $M$, and $N$. If $B P = x$, and the area of the shaded region is $y$, then the graph of the relationship between $y$ and $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch6-2024_06_14_fe3bda29f9b5de8dac4dg_0002_1.jpg", "batch6-2024_06_14_fe3bda29f9b5de8dac4dg_0002_2.jpg", "batch6-2024_06_14_fe3bda29f9b5de8dac4dg_0002_3.jpg", "batch6-2024_06_14_fe3bda29f9b5de8dac4dg_0002_4.jpg", "batch6-2024_06_14_fe3bda29f9b5de8dac4dg_0002_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Let the side length of square \\( A B C D \\) be \\( a \\).\n\nSince quadrilateral \\( A B C D \\) is a square,\n\nTherefore, \\( A B = B C \\), \\( \\angle M B P = \\angle F B P = 45^{\\circ} \\), \\( A B \\parallel C D \\), and \\( B C \\parallel A D \\).\n\nSince lines parallel to \\( C D \\) and \\( A D \\) are drawn through point \\( P \\), intersecting the sides of square \\( A B C D \\) at points \\( E \\), \\( F \\), \\( M \\), and \\( N \\),\n\nTherefore, quadrilaterals \\( A M P E \\) and \\( F C N P \\) are rectangles,\n\nAnd quadrilaterals \\( B F P M \\) and \\( E P N D \\) are squares.\n\nThus, \\( B F = F P = P M = B M = \\frac{\\sqrt{2}}{2} B P = \\frac{\\sqrt{2}}{2} x \\),\n\nAnd \\( E P = P N = N D = D E = a - \\frac{\\sqrt{2}}{2} x \\).\n\nTherefore, the area of the shaded region \\( S_{\\text{shaded}} = S_{\\text{square } A B C D} - S_{\\text{square } B F P M} - S_{\\text{square } E P N D} \\)\n\n\\[\n= a^{2} - \\left(\\frac{\\sqrt{2}}{2} x\\right)^{2} - \\left(a - \\frac{\\sqrt{2}}{2} x\\right)^{2}\n\\]\n\n\\[\n= a^{2} - \\frac{1}{2} x^{2} - a^{2} + \\sqrt{2} a x - \\frac{1}{2} x^{2}\n\\]\n\n\\[\n= -x^{2} + \\sqrt{2} a x\n\\]\n\nThus, \\( y = -x^{2} + \\sqrt{2} a x \\) (where \\( x \\geq 0 \\), \\( y \\geq 0 \\)).\n\nTherefore, the graph of this function is a downward-opening parabola.\n\nHence, the correct choice is: D.\n\n【Key Insight】This problem examines the determination and properties of squares, the derivation of function expressions, and the relationship between geometric problems and quadratic functions. Accurately deriving the function expression is crucial to solving this problem." }, { "problem_id": 614, "question": "As shown in the figure, in an isosceles right triangle $A B C$, $\\angle B = 90^\\circ$ and $A C = 4$. In the square $A D E F$, $A F = 2$, and points $F$, $A$, and $C$ are collinear. The square $A D E F$ is translated along the ray $F A$ until point $F$ coincides with point $C$. If the distance traveled by point $F$ is $x$, and the area of the overlapping part between the two figures during the translation is $y$, which of the following graphs correctly represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_8db9b77b9d3b77f76112g_0069_1.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0069_2.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0069_3.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0069_4.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0069_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, when \\(0 \\leq x \\leq 2\\), \\(y = \\frac{1}{2} x^{2}\\);\n\n\n\nAs shown in the figure, when \\(2 < x < 4\\), \\(y = \\frac{1}{2} \\times 4 \\times 2 - \\frac{1}{2}(x-2)^{2} - \\frac{1}{2}(4-x)^{2} = -(x-3)^{2} + 3\\)\n\n\n\nAs shown in the figure, when \\(0 \\leq x \\leq 2\\), \\(y = \\frac{1}{2}[2 - (x - 4)]^{2} = \\frac{1}{2}(6 - x)^{2}\\)\n\n\n\nThus, the graph of the function corresponds to option B.\n\nTherefore, the correct choice is B.\n\n【Key Point】This question examines the function graph of a moving point problem, and the key to solving it lies in the mathematical concept of categorical discussion." }, { "problem_id": 615, "question": "Given the function $y=(x-m)(x-n)$ (where $m\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0014_1.jpg", "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0014_2.jpg", "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0014_3.jpg", "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0014_4.jpg", "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0014_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let \\( y = 0 \\), then \\( (x - m)(x - n) = 0 \\).\n\nSolving this, we get: \\( x_1 = m \\), \\( x_2 = n \\).\n\nTherefore, the points where the quadratic function intersects the \\( x \\)-axis are \\( (m, 0) \\) and \\( (n, 0) \\).\n\nSince \\( m < n \\), and given \\( m < -1 \\) and \\( 0 < n < 1 \\), the function \\( y = n x + m \\) passes through the first, third, and fourth quadrants, and its intersection with the \\( y \\)-axis is below the point \\( (0, -1) \\). Hence, the correct choice is: D.\n\n**Key Insight:** This problem primarily examines the graphs and properties of quadratic and linear functions. A thorough understanding of the graphs and properties of these functions is crucial for solving the problem." }, { "problem_id": 616, "question": "The graphs of the functions $y = \\frac{k}{x}$ and $y = kx^2 - k$ (where $k \\neq 0$) in the same Cartesian coordinate system are approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_da13b7672084f31d9ca1g_0014_1.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0014_2.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0014_3.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0014_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Discuss in two cases:\n\n(1) When \\( k < 0 \\), the inverse proportional function \\( y = \\frac{k}{x} \\) is located in the second and fourth quadrants, while the quadratic function \\( y = kx^{2} - k \\) opens downward and intersects the \\( y \\)-axis above the origin. Therefore, options B, C, and D do not meet the requirements, and option A is correct.\n\n(2) When \\( k > 0 \\), the inverse proportional function \\( y = \\frac{k}{x} \\) is located in the first and third quadrants, while the quadratic function \\( y = kx^{2} - k \\) opens upward and intersects the \\( y \\)-axis below the origin. Therefore, options A, B, C, and D do not meet the requirements.\n\nThus, the correct answer is: A.\n\n**Key Insight:** This question primarily examines the graphs of quadratic and inverse proportional functions. The general steps to solve such problems are: (1) Determine whether the value of \\( k \\) is contradictory based on the characteristics of the graph; (2) Check whether the intersection of the parabola with the \\( y \\)-axis meets the requirements based on the graph of the quadratic function." }, { "problem_id": 617, "question": "As shown in the figure, in rectangle $A B C D$, $A B = 2$ and $B C = 4$. Point $P$ is a moving point on side $B C$ (point $P$ does not coincide with points $B$ or $C$). Now, $\\triangle A B P$ is folded along the straight line $A P$, so that point $B$ lands at point $B'$. The bisector of $\\angle B' P C$ intersects $C D$ at point $E$. Let $B P = x$ and $C E = y$. Which of the following graphs approximately represents the relationship between $y$ and $x$ as a function?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_da13b7672084f31d9ca1g_0037_1.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0037_2.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0037_3.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0037_4.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0037_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Since triangle ABP is folded along the line AP to obtain triangle AB'P,\n\nTherefore, angle APB equals angle APB',\n\nSince PE bisects angle B'PC,\n\nTherefore, angle B'PE equals angle CPE,\n\nTherefore, angle APB' plus angle EPB' equals half of 180 degrees, which is 90 degrees,\n\nSince angle C is 90 degrees,\n\nTherefore, angle CPE plus angle CEP equals 90 degrees,\n\nTherefore, angle APB equals angle CEP,\n\nSince angle B equals angle C, both being 90 degrees,\n\nTherefore, triangle ABP is similar to triangle PCE,\n\nTherefore, the ratio of AB to PC equals the ratio of PB to EC,\n\nSince BP equals x, CE equals y, and in rectangle ABCD, AB equals 2, BC equals 4,\n\nTherefore, PC equals 4 minus x,\nTherefore, the ratio of 2 to (4 - x) equals the ratio of x to y,\n\nTherefore, y equals half of x times (4 - x), which is negative half of x squared plus 2x.\n\nTherefore, the graph of this function is a parabola opening downward.\n\nHence, the correct choice is: D.\n\n[Key Insight] This problem examines the function graph of a moving point, and mastering the properties of folding and the determination and properties of similar triangles is key to solving this problem." }, { "problem_id": 618, "question": "As shown in the figure, the graphs of the functions $y = a x^{2} (a \\neq 0)$ and $y = -a x + b (a \\neq 0)$ in the same coordinate system could be\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch7-2024_06_14_dd9b6a481087dff4e73bg_0061_1.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0061_2.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0061_3.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0061_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "When \\( a > 0 \\), the quadratic function opens upwards, and the coefficient of the linear term in the linear function \\( -a < 0 \\). The graph will pass through the second and fourth quadrants, so option A does not meet the requirement, while option D does.\n\nWhen \\( a < 0 \\), the quadratic function opens downwards, and the coefficient of the linear term in the linear function \\( -a > 0 \\). The graph will pass through the first and third quadrants, so options B and C do not meet the requirement.\n\nTherefore, the correct choice is D.\n\n【Key Insight】This question examines the combination of linear and quadratic function graphs. The key to solving the problem lies in the integration of numerical and graphical analysis." }, { "problem_id": 619, "question": "As shown in the figure, the graphs of the functions $y = ax^2 - 2x + 1$ and $y = ax - a$ (where $a$ is a constant and $a \\neq 0$) in the same Cartesian coordinate system could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_dd9b6a481087dff4e73bg_0079_1.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0079_2.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0079_3.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0079_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "A. From the graph of the linear function \\( y = a x - a \\), we can deduce that \\( a < 0 \\). In this case, the graph of the quadratic function \\( y = a x^{2} - 2 x + 1 \\) should open downward. Therefore, the option is incorrect;\n\nB. From the graph of the linear function \\( y = a x - a \\), we can deduce that \\( a > 0 \\). In this case, the graph of the quadratic function \\( y = a x^{2} - 2 x + 1 \\) should open upward, and the axis of symmetry \\( x = -\\frac{-2}{2 a} > 0 \\). Therefore, the option is correct;\n\nC. From the graph of the linear function \\( y = a x - a \\), we can deduce that \\( a > 0 \\). In this case, the graph of the quadratic function \\( y = a x^{2} - 2 x + 1 \\) should open upward, and the axis of symmetry \\( x = -\\frac{-2}{2 a} > 0 \\), intersecting the positive half-axis of \\( x \\). Therefore, the option is incorrect;\n\nD. From the graph of the linear function \\( y = a x - a \\), we can deduce that \\( a > 0 \\). In this case, the graph of the quadratic function \\( y = a x^{2} - 2 x + 1 \\) should open upward. Therefore, the option is incorrect.\n\nHence, the correct choice is: B.\n\n【Key Insight】This question examines the graphs of quadratic and linear functions. The key to solving it lies in remembering the quadrant in which the linear function \\( y = a x - a \\) lies under different conditions, as well as mastering the relevant properties of quadratic functions: the direction of opening, the axis of symmetry, the vertex coordinates, etc." }, { "problem_id": 620, "question": "As shown in the figure, in right triangle $\\triangle A B C$, $\\angle A = 90^\\circ, A C = A B = 4$. Point $D$ starts from point $A$ and moves along segment $A B$ at a speed of 1 unit per second. The movement stops when point $D$ coincides with point $B$. A square $A D E F$ is constructed above $A D$ with $A D$ as a side. If the time of movement is $x$ seconds $(0 < x \\leq 4)$, the area of the overlapping part between the square $A D E F$ and $\\triangle A B C$ is $y$. Which of the following graphs can roughly represent the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_ec1e735fb15846e30563g_0037_1.jpg", "batch7-2024_06_14_ec1e735fb15846e30563g_0037_2.jpg", "batch7-2024_06_14_ec1e735fb15846e30563g_0037_3.jpg", "batch7-2024_06_14_ec1e735fb15846e30563g_0037_4.jpg", "batch7-2024_06_14_ec1e735fb15846e30563g_0037_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When \\(0 < x \\leq 2\\), \\(y = x^{2}\\).\n\nThe graph of the function in this interval is the part of \\(y = x^{2}\\) where \\(0 < x \\leq 2\\), with the vertex at \\((2, 4)\\).\n\nTherefore, options \\(C\\) and \\(D\\) do not apply.\n\nWhen \\(2 < x \\leq 4\\), as shown in the figure:\n\n\n\nSince in the right triangle \\(\\triangle ABC\\), \\(\\angle A = 90^{\\circ}\\), and \\(AC = AB = 4\\),\n\\(\\angle B = \\angle C = 45^{\\circ}\\).\n\nSince quadrilateral \\(ADEF\\) is a square,\n\n\\(AD = ED = EF = x\\), and \\(\\angle EDB = \\angle E = \\angle CFE = 90^{\\circ}\\),\n\n\\(\\angle CMF = \\angle EMN = \\angle ENM = \\angle BND = 45^{\\circ}\\),\n\n\\(BD = DN = AB - AD = 4 - x\\), and \\(EM = EN = ED - DN = x - (4 - x) = 2x - 4\\),\n\nThus, \\(y = S_{\\text{square } ADEF} - S_{\\triangle EMN}\\)\n\n\\(= x^{2} - \\frac{1}{2}(2x - 4)^{2}\\)\n\n\\(= -x^{2} + 8x - 8\\)\n\n\\(= -(x - 4)^{2} + 8\\)\n\nThe graph of the function in this interval is the part of \\(y = -(x - 4)^{2} + 8\\) where \\(2 < x \\leq 4\\), with the vertex at \\((4, 8)\\).\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem examines the function graph of a moving point and the derivation of a quadratic function's equation. The key to solving this problem lies in adopting a classified discussion approach." }, { "problem_id": 621, "question": "Sketch the graphs of $y_{1}=3 x^{2}, y_{2}=-3 x^{2}, y_{3}=\\frac{1}{2} x^{2}$ in the same coordinate system. Which of the following is correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0037_1.jpg", "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0037_2.jpg", "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0037_3.jpg", "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0037_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: When \\( x = 1 \\), the corresponding points on the graphs of \\( y_{1} \\), \\( y_{2} \\), and \\( y_{3} \\) are \\( (1, 3) \\), \\( (1, -3) \\), and \\( \\left(1, \\frac{1}{2}\\right) \\), respectively. It can be observed that two of these points lie in the first quadrant and one lies in the fourth quadrant, thus eliminating options B and C.\n\nWithin the first quadrant, the corresponding point of \\( y_{1} \\), \\( (1, 3) \\), is above the corresponding point of \\( y_{3} \\), \\( \\left(1, \\frac{1}{2}\\right) \\), which eliminates option A.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question examines the relationship between the graph of a quadratic function and its coefficient \\( a \\). For the quadratic function \\( y = a x^{2} \\), if the coefficient \\( a \\) is positive, the parabola opens upwards; if \\( a \\) is negative, the parabola opens downwards. The larger the absolute value of \\( a \\), the narrower the parabola's opening." }, { "problem_id": 622, "question": "As shown in the figure, in the square $A B C D$, $A B=4$. Point $P$ starts from point $A$ and moves along the path $A \\rightarrow B \\rightarrow C$ towards the endpoint $C$. Connect $D P$, and draw the perpendicular bisector $M N$ of $D P$ which intersects the sides of the square $A B C D$ at points $M$ and $N$. Let the distance traveled by point $P$ be $x$, and the area of $\\triangle P M N$ be $y$. Which of the following graphs can roughly represent the relationship between $y$ and $x$ as a function?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0070_1.jpg", "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0070_2.jpg", "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0070_3.jpg", "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0070_4.jpg", "batch8-2024_06_14_1b6e75fcf14707b6c65ag_0070_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: (1) As shown in the figure, when \\(0 \\leq x \\leq 4\\), point \\(P\\) is on \\(AB\\). We draw \\(NE \\perp AD\\) at point \\(E\\), and let \\(MN\\) intersect \\(PD\\) at point \\(F\\).\n\n\n\nThus, \\(NE = DC = AD\\),\n\nthen \\(PD = \\sqrt{PA^{2} + AD^{2}} = \\sqrt{x^{2} + 4^{2}} = \\sqrt{x^{2} + 16}\\),\n\nand since \\(MN\\) is the perpendicular bisector of \\(PD\\),\n\n\\(PF = \\frac{1}{2} PD = \\frac{1}{2} \\sqrt{x^{2} + 16}\\),\n\n\\(\\angle MDF + \\angle FMD = \\angle MNE + \\angle FME = 90^{\\circ}\\),\n\n\\(\\angle MNE = \\angle PDA\\),\n\nIn triangles \\(MNE\\) and \\(PDA\\),\n\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle A = \\angle NEM \\\\\nAD = EN \\\\\n\\angle PDA = \\angle MNE\n\\end{array}\n\\right.\n\\]\n\n\\(\\triangle APD \\cong \\triangle EMN\\)\n\nTherefore, \\(PD = MN = \\sqrt{x^{2} + 16}\\),\n\n\\(y = \\frac{1}{2} MN \\cdot PF = \\frac{1}{2} \\sqrt{x^{2} + 16} \\cdot \\frac{1}{2} \\sqrt{x^{2} + 16} = \\frac{1}{4} x^{2} + 4\\),\n\n(2) As shown in the figure, when \\(4 < x \\leq 8\\), point \\(P\\) is on \\(BC\\),\n\n\n\nWe draw \\(NE \\perp CD\\) at point \\(E\\), and let \\(MN\\) intersect \\(PD\\) at point \\(F\\),\n\nthen \\(PD = \\sqrt{PC^{2} + CD^{2}} = \\sqrt{(8 - x)^{2} + 16}\\),\n\n\\(PF = \\frac{1}{2} \\sqrt{(8 - x)^{2} + 16}\\)\n\nUsing the method from (1), we get\n\n\\(MN = \\sqrt{(8 - x)^{2} + 16}\\),\n\n\\(y = \\frac{1}{2} \\sqrt{(8 - x)^{2} + 16} \\cdot \\frac{1}{2} \\sqrt{(8 - x)^{2} + 16} = \\frac{1}{4}(x - 8)^{2} + 4\\),\n\nThus, \\(y = \\left\\{\n\\begin{array}{l}\n\\frac{1}{4} x^{2} + 4 \\quad (0 \\leq x \\leq 4) \\\\\n\\frac{1}{4}(x - 8)^{2} + 4 \\quad (4 < x \\leq 8)\n\\end{array}\n\\right.\\)\n\nTherefore, the correct choice is A.\n\n【Key Insight】This problem examines piecewise functions. The key to solving it is to determine the range of the independent variable based on the position of point \\(P\\) and derive the function's expression accordingly." }, { "problem_id": 623, "question": "As shown in the figure, there is an equilateral triangle $\\triangle A B C$ with side length 2 and an equilateral triangle $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ with side length 1. Their sides $B^{\\prime} C^{\\prime}$ and $B C$ lie on the same straight line $l$. Initially, point $C^{\\prime}$ coincides with point $B$, and $\\triangle A B C$ remains fixed. Then $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is translated from left to right along the straight line $l$, stopping when it exits $\\triangle A B C$ (when point $B^{\\prime}$ coincides with point $C$). Let the distance that $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ translates be $x$, and the area of the overlapping part of the two triangles be $y$. The graph of $y$ as a function of $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_3014dab89f701789b5aeg_0087_1.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0087_2.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0087_3.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0087_4.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0087_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: (1) As shown in the figure, during the movement of the equilateral triangle $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ to $\\triangle A_{1}^{\\prime} B_{1}^{\\prime} C_{1}^{\\prime}$, that is, for $0 \\leq x \\leq 1$, let the intersection of $A B$ and $A^{\\prime} C^{\\prime}$ be point $\\mathrm{P}$, and draw $P Q \\perp B C$ to point $\\mathrm{Q}$.\n\nSince the translation distance of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is $x$, then $B Q=\\frac{1}{2} x$.\n\nAccording to the problem, $\\angle P B Q=60^{\\circ}$.\n\nTherefore, in the right triangle $\\triangle P B Q$, $P Q=\\sqrt{3} B Q=\\frac{\\sqrt{3}}{2} x$.\n\nThus, the area of the overlapping part of the two triangles during this process is $y=\\frac{1}{2} x \\cdot \\frac{\\sqrt{3}}{2} x=\\frac{\\sqrt{3}}{4} x^{2}$.\n\nTherefore, when $x=1$, $y=\\frac{\\sqrt{3}}{4}$;\n\n\n\n(2) As shown in the figure, during the movement of the equilateral triangle $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ from $\\triangle A_{1}^{\\prime} B_{1}^{\\prime} C_{1}^{\\prime}$ to $\\triangle A_{2}^{\\prime} B_{2}^{\\prime} C_{2}^{\\prime}$, that is, for $1\n\n(3) As shown in the figure, during the movement of the equilateral triangle $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ from $\\triangle A_{2}^{\\prime} B_{2}^{\\prime} C_{2}^{\\prime}$ to $\\triangle A_{3}^{\\prime} B_{3}^{\\prime} C_{3}^{\\prime}$, that is, for $2\n\n\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem examines the area of graphic translation and quadratic functions. Understanding the translation process in the problem, flexibly classifying and discussing, and accurately solving the quadratic function expression are the keys to solving the problem." }, { "problem_id": 624, "question": "A small ball is thrown vertically upwards from the ground. The relationship between the height $h$ (in meters, $m$) of the ball and the time $t$ (in seconds, $s$) of its motion is given by $h = -5t^2 + 30t$ (for $0 \\leq t \\leq 6$). Which graph represents the relationship between the height of the ball and the time of its motion?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_39f616e1ffa2dea341d7g_0005_1.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0005_2.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0005_3.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0005_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: The equation given is \\( h = -5t^{2} + 30t = -5(t - 3)^{2} + 45 \\) for \\( 0 \\leq t \\leq 6 \\).\n\nSetting \\( h = 0 \\), we have:\n\\[\n-5t^{2} + 30t = 0\n\\]\nSolving this equation yields:\n\\[\nt = 0 \\text{ or } t = 6\n\\]\nTherefore, the graph representing the height of the ball as a function of time is a downward-opening parabola segment above the \\( t \\)-axis, with its vertex at the point \\( (3, 45) \\).\n\nThus, the graph corresponding to option B fits the description.\n\nThe correct choice is: B\n\n[Key Insight] This problem primarily tests the understanding of the graph and properties of a parabola. Mastering the graph and properties of a parabola is crucial for solving such problems." }, { "problem_id": 625, "question": "As shown in the figure, the side length of square $A B C D$ is 10. Four congruent circles are drawn with the vertices $A, B, C,$ and $D$ as centers. If the radius of each circle is $x$, where $0 < x \\leq 5$, the area of the shaded region is $y$. Which of the following graphs approximately represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_39f616e1ffa2dea341d7g_0053_1.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0053_2.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0053_3.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0053_4.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0053_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Since the sum of the internal angles of a regular hexagon is $360^{\\circ}$, the four shaded parts form a circle with radius $x$.\n\nTherefore, $y = \\pi x^{2}$ for $0 < x \\leq 5$,\nWhen $x = 5$, $y = 25\\pi$.\n\nHence, the correct choice is D.\n\n[Key Insight] This problem primarily tests the application of quadratic functions. The key to solving it lies in establishing the functional relationship between $y$ and $x$ based on the given conditions." }, { "problem_id": 626, "question": "As shown in Figure 7, there is a square and an equilateral triangle, both with sides of length 1, where one side of each shape is on the same horizontal line. The triangle moves horizontally to the right at a constant speed across the square. Let the time it takes for the triangle to fully cross the square be $t$, and the area of the overlapping part (the blank area) between the square and the triangle be $S$. The approximate graph of the function $S$ with respect to $t$ should be ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_39f616e1ffa2dea341d7g_0096_1.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0096_2.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0096_3.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0096_4.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0096_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Given a square and an equilateral triangle, both with side lengths of 1, sharing one side on the same horizontal line, the triangle moves uniformly from left to right through the square. The time taken to pass through is \\( t \\), and the area of overlap between the square and the triangle is \\( S \\) (the blank part).\n\nTherefore, the function \\( S \\) with respect to \\( t \\) should have the following approximate graph: Before the triangle enters the square, the blank area gradually increases. When \\( 0 \\leq t \\leq \\frac{1}{2} \\),\n\n\\[\nS = \\frac{1}{2} \\times t \\times \\sqrt{3} t = \\frac{\\sqrt{3}}{2} t^{2},\n\\]\n\nwhen \\( \\frac{1}{2} < t \\leq 1 \\),\n\n\\[\nS = \\frac{1}{2} \\times 1 \\times \\frac{\\sqrt{3}}{2} - \\frac{1}{2} \\times (1 - t) \\times \\sqrt{3} (1 - t) = -\\frac{\\sqrt{3}}{2} t^{2} + \\sqrt{3} t - \\frac{\\sqrt{3}}{4},\n\\]\n\nwhen \\( 1 < t \\leq \\frac{3}{2} \\),\n\n\\[\nS = \\frac{1}{2} \\times 1 \\times \\frac{\\sqrt{3}}{2} - \\frac{1}{2} \\times (t - 1) \\times \\sqrt{3} (t - 1) = -\\frac{\\sqrt{3}}{2} t^{2} + \\sqrt{3} t - \\frac{\\sqrt{3}}{4},\n\\]\n\nand when \\( \\frac{3}{2} < t \\leq 2 \\),\n\n\\[\nS = \\frac{1}{2} \\times (2 - t) \\times \\sqrt{3} (2 - t) = \\frac{\\sqrt{3}}{2} t^{2} - 4 \\sqrt{3} t + 2 \\sqrt{3}.\n\\]\n\nThus, \\( S \\) and \\( t \\) have a quadratic function relationship. Therefore, only option \\( D \\) meets the requirements. Hence, the answer is \\( D \\)." }, { "problem_id": 627, "question": "(2018. Kaifeng Second Simulation) As shown in the figure, in the Cartesian coordinate system, points $A(0,1)$ and $B(\\sqrt{3}, 0)$ are given. A rhombus $A B C D$ is constructed with $A B$ as a side, and point $D$ is on the $y$-axis. The rhombus $A B C D$ slides along the ray $A B$ at a speed of 2 units per second until vertex $D$ falls on the $x$-axis. Let $S$ denote the area of the part of the rhombus that is below the $x$-axis. Which of the following graphs represents the relationship between $S$ and the sliding time $t$?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_4b79afec46b2311e7644g_0080_1.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0080_2.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0080_3.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0080_4.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0080_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Given points \\( A(0,1) \\) and \\( B(\\sqrt{3}, 0) \\),\n\n\\[\n\\therefore OA = 1, \\quad OB = \\sqrt{3},\n\\]\n\n\\[\n\\therefore AB = \\sqrt{OA^{2} + OB^{2}} = \\sqrt{1^{2} + (\\sqrt{3})^{2}} = 2,\n\\]\n\n\\[\n\\because \\tan \\angle BAO = \\frac{OB}{OA} = \\frac{\\sqrt{3}}{1} = \\sqrt{3},\n\\]\n\n\\[\n\\therefore \\angle BAO = 60^{\\circ},\n\\]\n\n\\[\n\\therefore \\text{The height of rhombus } ABCD \\text{ is } 2 \\times \\frac{\\sqrt{3}}{2} = \\sqrt{3},\n\\]\n\n\\[\n\\because \\text{Rhombus } ABCD \\text{ slides along ray } AB \\text{ at a speed of 2 units per second},\n\\]\n\n\\[\n\\therefore \\text{The sliding speed along the } y\\text{-axis is } 1,\n\\]\n\n\\[\n\\text{and the sliding speed along the } x\\text{-axis is } \\sqrt{3},\n\\]\n\n(1) When point \\( A \\) is above the \\( x \\)-axis, the part below the \\( x \\)-axis is a triangle,\n\n\\[\n\\text{Area } S = \\frac{1}{2} \\cdot t \\cdot \\sqrt{3} t = \\frac{\\sqrt{3}}{2} t^{2},\n\\]\n\n(2) When point \\( A \\) is below the \\( x \\)-axis and point \\( C \\) is above the \\( x \\)-axis, the part below the \\( x \\)-axis is a trapezoid,\n\n\\[\n\\text{Area } S = \\frac{1}{2}[t + (t - 1) \\cdot 1] \\times \\sqrt{3} = \\sqrt{3} t - \\frac{\\sqrt{3}}{2},\n\\]\n\n(3) When point \\( C \\) is below the \\( x \\)-axis,\n\n\\[\n\\text{The part below the } x\\text{-axis is the area of the rhombus minus the area of the triangle above the } x\\text{-axis},\n\\]\n\n\\[\nS = 2 \\times \\sqrt{3} - \\frac{1}{2}(3 - t) \\cdot \\frac{\\sqrt{3}}{2}(6 - 2t) = 2\\sqrt{3} - \\frac{\\sqrt{3}}{2}(3 - t)^{2},\n\\]\n\nAfter reviewing all the options, only option A matches the graph.\n\nTherefore, the answer is: A.\n\n**Key Insight:** This problem examines the function graph of a moving point, utilizing the properties of a rhombus and solving right triangles. The key to solving the problem lies in dividing the scenario into three segments, determining the shape of the part below the \\( x \\)-axis, and deriving the corresponding functional relationships." }, { "problem_id": 628, "question": "As shown in Figure (1), \"The Gate of the East\" integrates traditional culture with modern architecture through simple geometric curves, maximizing the inheritance of Suzhou's historical and cultural heritage. As shown in Figure (2), the inner curve of the \"door\" is parabolic in shape. It is known that its base width is 80 meters and its height is 200 meters. The horizontal width at a height of 150 meters from the ground (i.e., the length of $C D$) is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 40 meters\nB. 30 meters\nC. 25 meters\nD. 20 meters", "input_image": [ "batch8-2024_06_14_abd5efc71c86223e46b3g_0004_1.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0004_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Establish a plane rectangular coordinate system with the straight line at the bottom as the $x$-axis and the perpendicular bisector of the line segment $CD$ as the $y$-axis:\n\n\n\n$\\therefore A(-40,0), \\quad B(40,0), \\quad E(0,200)$\n\nLet the equation of the parabola be $y=a(x+40)(x-40)$. Substituting $E(0,200)$ into the equation, we get:\n\n$200=a(0+40)(0-40)$,\n\nSolving for $a$: $a=-\\frac{1}{8}$,\n\n$\\therefore$ The equation of the parabola is $y=-\\frac{1}{8} x^{2}+200$,\n\nSubstituting $y=150$ into the equation: $-\\frac{1}{8} x^{2}+200=150$,\n\nSolving for $x$: $x= \\pm 20$,\n\n$\\therefore C(-20,150), \\quad D(20,150)$,\n\n$\\therefore CD=40 \\mathrm{~m}$,\n\nTherefore, the correct choice is: A\n\n【Key Insight】This problem examines the application of quadratic functions in practical problems. The key to solving it lies in combining numerical and graphical methods and mastering the method of undetermined coefficients." }, { "problem_id": 629, "question": "As shown in the figure, lines $a$ and $b$ are both perpendicular to line $l$, with the feet of the perpendiculars being $E$ and $F$, respectively, where $EF = 1$. Square $A B C D$ has a side length of $\\sqrt{2}$, and its diagonal $A C$ lies on line $l$. Point $C$ is located at point $E$. The square $A B C D$ is translated to the right along line $l$ until point $A$ coincides with point $F$. Let the distance that point $C$ moves be $x$, and the area of the part of the square $A B C D$ between lines $a$ and $b$ (the shaded area) be $y$. The graph of the function $y$ as a function of $x$ would approximately look like ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_abd5efc71c86223e46b3g_0012_1.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0012_2.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0012_3.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0012_4.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0012_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: (1) When \\(0 \\leq x \\leq 1\\), as shown in Figure 1,\n\n\n\nFigure 1\n\nLet the translated square intersect line \\(a\\) at points \\(G\\) and \\(H\\),\n\nthen \\(EC = x\\), and \\(\\triangle GHC\\) is an isosceles right triangle, so \\(GH = 2x\\),\n\nthus \\(y = S_{\\triangle HGC} = \\frac{1}{2} \\times EC \\cdot GH = \\frac{1}{2} \\cdot x \\cdot 2x = x^{2}\\), which is an upward-opening parabola;\n\n(2) When \\(1 < x \\leq 2\\), as shown in Figure 2,\n\n\n\nFigure 2\n\nLet the translated square intersect line \\(b\\) at points \\(M\\) and \\(N\\), and line \\(a\\) at points \\(G\\) and \\(H\\),\n\nthen \\(\\triangle A'GH\\) and \\(\\triangle MNC'\\) are both isosceles right triangles,\n\nthus \\(y = S_{\\text{square }} ABCD - (S_{\\triangle A'GH} + S_{\\triangle MNC'})\\)\n\n\\(= (\\sqrt{2})^{2} - \\frac{1}{2}[(2 - x)(2 - x) \\times 2 - 2 \\times (x - 1)(x - 1)]\\)\n\n\\(= -2x^{2} + 6x - 3\\);\n\nThis function is a downward-opening parabola;\n\n(3) When \\(2 < x \\leq 3\\), similarly, we get: \\(y = (3 - x) \\times 2(3 - x) \\times \\frac{1}{2} = x^{2} - 6x + 9\\),\n\nThis function is an upward-opening parabola;\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem examines the function graph of a moving point, involving properties of squares and isosceles right triangles. Correctly classifying the problem based on the graph is key to solving it." }, { "problem_id": 630, "question": "As shown in the figure, in equilateral triangle $\\triangle A B C$, $A B = 2$. Point $D$ starts from point $A$ and moves along the broken line $A-C-B$ at a speed of 2 units per second. A perpendicular line is drawn from point $D$ to $A B$, with the foot of the perpendicular being point $E$. Let the time for point $D$ to move be $x$ seconds, and the area of $\\triangle A D E$ be $y$ (when $A, D, E$ are collinear, set $y = 0$). Which of the following graphs roughly represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_abd5efc71c86223e46b3g_0037_1.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0037_2.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0037_3.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0037_4.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0037_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When point \\( D \\) is on \\( AC \\), i.e., \\( 0 \\leq x \\leq 1 \\), as shown in the figure,\n\n\n\nFrom the movement of point \\( D \\), we know that \\( AD = 2x \\).\n\nSince \\( \\triangle ABC \\) is an equilateral triangle,\n\n\\( \\angle A = 60^\\circ \\).\n\nSince \\( DE \\perp AB \\),\n\n\\( \\angle AED = 90^\\circ \\),\n\n\\( \\angle ADE = 30^\\circ \\),\n\nThus, \\( AD = x \\), \\( DE = \\sqrt{3}x \\),\n\nTherefore, \\( y = \\frac{1}{2} AE \\cdot DE = \\frac{1}{2} x \\cdot \\sqrt{3}x = \\frac{\\sqrt{3}}{2}x^2 \\).\n\nThe graph of this function is a parabola opening upwards, symmetric about the \\( y \\)-axis and passing through the origin. In the range \\( 0 \\leq x \\leq 1 \\), \\( y \\) increases as \\( x \\) increases, reaching a maximum value of \\( \\frac{\\sqrt{3}}{2} \\) when \\( x = 1 \\).\n\nWhen point \\( D \\) is on \\( BC \\), i.e., \\( 1 \\leq x \\leq 2 \\), as shown in the figure,\n\n\n\nFrom the movement of point \\( D \\), we know that \\( AC + CD = 2x \\),\n\nThus, \\( BD = 4 - 2x \\).\n\nSince \\( \\triangle ABC \\) is an equilateral triangle,\n\n\\( \\angle B = 60^\\circ \\).\n\nSince \\( DE \\perp AB \\),\n\n\\( \\angle DEB = 90^\\circ \\),\n\n\\( \\angle BDE = 30^\\circ \\),\n\nThus, \\( BE = 2 - x \\), \\( AE = x \\),\n\n\\( DE = \\sqrt{3}(2 - x) \\),\n\nTherefore, \\( y = \\frac{1}{2} AE \\cdot DE = \\frac{1}{2} x \\cdot \\sqrt{3}(2 - x) = -\\frac{\\sqrt{3}}{2}x^2 + \\sqrt{3}x \\).\n\nThe graph of this function is a parabola opening downwards, reaching a maximum value of \\( \\frac{\\sqrt{3}}{2} \\) when \\( x = 1 \\). In the range \\( 1 \\leq x \\leq 2 \\), \\( y \\) decreases as \\( x \\) increases.\n\nIn summary, graph \\( C \\) satisfies the conditions.\n\nTherefore, the answer is: C.\n\n【Key Insight】This problem examines the function graph of a moving point, the properties of quadratic functions, the properties of equilateral triangles, and the properties of right triangles with a 30-degree angle. The key to solving the problem lies in combining graphical analysis with the ability to write the relevant function expressions accurately." }, { "problem_id": 631, "question": "As shown in the figure, the side length of the square $A B C D$ is $8 \\mathrm{~cm}$. Point $E$ starts from point $C$ and moves along side $C B$ towards point $B$ at a speed of $2 \\mathrm{~cm} / \\mathrm{s}$, while point $F$ starts from point $C$ and moves along side $C D$ towards point $D$ at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$. Let the time be $x(\\mathrm{~s})$, and when $x>0$, a rectangle $C F H E$ is constructed with sides $C E$ and $C F$. Let the area of the remaining part of the square $A B C D$ after removing the rectangle $C F H E$ be $y\\left(\\mathrm{~cm}^{2}\\right)$. The graph of the general relationship between $y$ and $x$ is ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_abd5efc71c86223e46b3g_0075_1.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0075_2.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0075_3.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0075_4.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0075_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem, $CE = 2x \\text{ cm}, CF = x \\text{ cm}$.\n\n$\\therefore DF = (8 - x) \\text{ cm}$.\n\nWhen $0 \\leq x \\leq 2$,\n\n$\\therefore y = 64 - 2x \\cdot x = 64 - 2x^{2}$,\n\nWhen $2 < x \\leq 4$,\n\nthen $y = 8(8 - x) = -8x + 64$,\n\n$\\therefore y = \\left\\{\\begin{array}{l}-2x^{2} + 64 \\quad (\\begin{array}{lll}0 & x & 2\\end{array}) \\\\ -8x + 64 \\quad (2 < x \\leq 4\\end{array}\\right);$\n\nTherefore, the graph that fits the problem is D,\n\nHence, choose D.\n\n【Key Insight】This problem examines the application of the area formulas for squares and rectangles, the use of quadratic and linear function expressions, the use of piecewise functions, and the determination of the maximum and minimum values of linear and quadratic functions. The key to solving this problem is to first find the function expressions." }, { "problem_id": 632, "question": "As shown in the figure, in square $A B C D$, $A B=4 \\text{ cm}$. Point $E$ starts from point $A$ and moves along the broken line $A B-B C$ until it reaches point $C$. A line $E F$ is drawn perpendicular to $A E$ passing through point $E$ and intersecting $C D$ at point $F$. Let the distance traveled by point $E$ be $x \\text{ cm}$ and $D F=y \\text{ cm}$. The graph of the approximate relationship between $y$ and $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_abd5efc71c86223e46b3g_0077_1.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0077_2.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0077_3.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0077_4.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0077_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Given that \\( AB = BC = 4 \\),\n\nWhen point \\( E \\) is on \\( AB \\), as shown in Figure 1, i.e., \\( 0 < x \\leq 4 \\),\n\n\n\nFigure 1\n\nAt this time, \\( DF = AE = x \\),\n\nTherefore, when \\( 0 < x \\leq 4 \\), the functional relationship is: \\( y = x \\).\n\nWhen \\( E \\) is on \\( BC \\), as shown in Figure 2, i.e., \\( 4 < x \\leq 8 \\),\n\n\n\nFigure 2\n\nSince \\( EF \\perp AE \\),\n\nTherefore, \\( \\triangle ABE \\sim \\triangle ECF \\),\n\nThus, \\( \\frac{AB}{BE} = \\frac{CE}{FC} \\),\n\nSo, \\( \\frac{4}{x-4} = \\frac{8-x}{FC} \\),\n\nTherefore, \\( CF = -\\frac{1}{4}x^2 + 3x - 8 \\),\n\nHence, \\( y = 4 - \\left(-\\frac{1}{4}x^2 + 3x - 8\\right) = \\frac{1}{4}x^2 - 3x + 12 \\).\n\nFrom this, we can conclude: when \\( 0 < x \\leq 4 \\), the functional relationship is \\( y = x \\); when \\( 4 < x \\leq 8 \\), the functional relationship is\n\n\\( y = \\frac{1}{4}x^2 - 3x + 12 \\).\n\nTherefore, the correct answer is: A\n\n【Highlight】This question examines the problem of moving point images, involving knowledge of quadratic functions, linear functions, and solving right triangles. The key to such problems is: to clarify the correspondence between the images and the figures in different time periods, and then to solve accordingly." }, { "problem_id": 633, "question": "In the rectangle $A B C D$ shown, $A B=2 \\text{ cm}, B C=4 \\sqrt{3} \\text{ cm}$, and $E$ is the midpoint of $A D$. The segments $B E$ and $C E$ are drawn. Point $P$ starts from point $B$ and moves along the direction of $B C$ at a speed of $\\sqrt{3} \\text{ cm/s}$ until it reaches point $C$. At the same time, point $Q$ starts from point $B$ and moves along the direction of $B E-E C$ at a speed of $1 \\text{ cm/s}$ until it reaches point $C$. If the area of $\\triangle B P Q$ is $y(\\text{cm}^{2})$ and the time of motion is $x$ (s), which of the following graphs best represents the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_abd5efc71c86223e46b3g_0086_1.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0086_2.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0086_3.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0086_4.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0086_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since \\( E \\) is the midpoint of \\( AD \\),\n\n\\[\n\\therefore AE = \\frac{1}{2} AD = 2\\sqrt{3}.\n\\]\n\nIn the right triangle \\( \\triangle ABE \\),\n\n\\[\nBE = \\sqrt{AB^2 + AE^2} = \\sqrt{2^2 + (2\\sqrt{3})^2} = 4.\n\\]\n\nSimilarly,\n\n\\[\nCE = 4.\n\\]\n\nThe area of \\( \\triangle BCE \\) is\n\n\\[\nS_{\\triangle BCE} = \\frac{1}{2} \\times 4\\sqrt{3} \\times 2 = 4\\sqrt{3}.\n\\]\n\n(1) When \\( 0 < x \\leq 4 \\), point \\( P \\) is on \\( BC \\) and point \\( Q \\) is on \\( BE \\), with \\( BP = \\sqrt{3}x \\) and \\( BQ = x \\) (as shown in Figure 1).\n\n\n\nFigure 1\n\nSince the heights of the triangles are the same, we have:\n\n\\[\ny = S_{\\triangle BPQ} = \\frac{x}{4} S_{\\triangle BPE} = \\frac{x}{4} \\times \\frac{\\sqrt{3}x}{4\\sqrt{3}} S_{\\triangle BCE} = \\frac{x}{4} \\times \\frac{\\sqrt{3}x}{4\\sqrt{3}} \\times 4\\sqrt{3} = \\frac{\\sqrt{3}}{4} x^2.\n\\]\n\nThe graph of the function \\( y = \\frac{\\sqrt{3}}{4} x^2 \\) is a parabola opening upwards, thus eliminating options A and C.\n\n(2) When \\( 4 < x < 8 \\), point \\( P \\) coincides with point \\( C \\), and point \\( Q \\) is on \\( EC \\), with \\( CQ = 8 - x \\) (as shown in Figure 2).\n\n\n\nFigure 2\n\n\\[\ny = S_{\\triangle BPQ} = \\frac{8 - x}{4} S_{\\triangle BCE} = \\frac{8 - x}{4} \\times 4\\sqrt{3} = 8\\sqrt{3} - \\sqrt{3}x.\n\\]\n\nThe graph of the function \\( y = 8\\sqrt{3} - \\sqrt{3}x \\) is a straight line, thus eliminating option B.\n\nTherefore, the correct choice is D.\n\n【Key Insight】This problem examines the function graph of a moving point problem. The key to solving this problem lies in determining the different expressions for the area of the triangle based on the different positions of the moving points \\( P \\) and \\( Q \\). The solution involves the application of the classification discussion method and the establishment of functional relationships." }, { "problem_id": 634, "question": "Given the quadratic function $y = -2 x^{2} + b x + c$ (where $b < 0$ and $c$ is a constant), which of the following graphs could represent the function?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_bba33c093999874a2a33g_0065_1.jpg", "batch8-2024_06_14_bba33c093999874a2a33g_0065_2.jpg", "batch8-2024_06_14_bba33c093999874a2a33g_0065_3.jpg", "batch8-2024_06_14_bba33c093999874a2a33g_0065_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Given the quadratic function \\( y = -2x^{2} + bx + c \\) (where \\( b < 0 \\) and \\( c \\) is a constant).\n\nSince the coefficient \\( a = -2 < 0 \\), the parabola opens downward.\n\nGiven that \\( b < 0 \\), the axis of symmetry \\( x = -\\frac{b}{2 \\times (-2)} \\) is negative, meaning the axis of symmetry lies to the left of the y-axis.\n\nOnly option D satisfies these conditions.\n\nTherefore, the correct choice is: D.\n\n**Key Insight:** The problem primarily tests the basic properties and graph of a quadratic function. Mastering the fundamental properties of quadratic functions is crucial for solving such problems." }, { "problem_id": 635, "question": "As shown in the figure, the perimeter of the rectangle $A B C D$ is $28 \\mathrm{~cm}$, and $A B$ is $2 \\mathrm{~cm}$ longer than $B C$. If point $P$ starts from point $A$ and moves at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$ along the path $A \\rightarrow D \\rightarrow C$ at a constant speed, and point $Q$ starts from point $A$ and moves at a speed of $2 \\mathrm{~cm} / \\mathrm{s}$ along the path $A \\rightarrow B \\rightarrow C$ at a constant speed, both points stop moving when one of them reaches point $C$. If the time of movement is denoted as $t(s)$ and the area of triangle $A P Q$ is denoted as $S\\left(\\mathrm{~cm}^{2}\\right)$, the approximate graph of the function relationship between $S\\left(\\mathrm{~cm}^{2}\\right)$ and $t(s)$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_be77b09795d7e25ae1bag_0010_1.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0010_2.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0010_3.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0010_4.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0010_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: From the problem statement, we have \\(2AB + 2BC = 28\\) and \\(AB = BC + 2\\),\n\nSolving these equations gives \\(AB = 8\\) and \\(BC = 6\\), which implies \\(AD = 6\\).\n\n(1) When \\(0 \\leq t \\leq 4\\), point \\(Q\\) is on side \\(AB\\) and point \\(P\\) is on side \\(AD\\), as shown in Figure 1.\n\n\n\nFigure 1\n\nThe area of triangle \\(APQ\\) is given by:\n\\[\nS_{\\triangle APQ} = \\frac{1}{2} \\cdot AP \\cdot AQ = \\frac{1}{2} \\cdot t \\cdot 2t = t^2\n\\]\nThe graph of this function is an upward-opening parabola, so options B and C are incorrect.\n\n(2) When \\(4 < t \\leq 6\\), point \\(Q\\) is on side \\(BC\\) and point \\(P\\) is on side \\(AD\\), as shown in Figure 2.\n\n\n\nFigure 2\n\nThe area of triangle \\(APQ\\) is given by:\n\\[\nS_{\\triangle APQ} = \\frac{1}{2} \\cdot AP \\cdot AB = \\frac{1}{2} \\cdot t \\cdot 8 = 4t\n\\]\nThe graph of this function is a straight line, so option D is incorrect.\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This problem examines the function graph of a moving point problem. By determining the different positions of points \\(P\\) and \\(Q\\), the area of the triangle changes. The key to solving this problem lies in using a case-by-case approach to derive the functional relationship between \\(S\\) and \\(t\\)." }, { "problem_id": 636, "question": "As shown in the figure, point $C$ is a moving point on the semicircle with center $O$ and diameter $A B$ (point $C$ does not coincide with points $A$ or $B$), where $A B = 4$. Let the length of chord $A C$ be $x$ and the area of triangle $A B C$ be $y$. Which of the following graphs approximately represents the relationship between $y$ and $x$ as a function?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_cd22dcb38e13acb480a0g_0011_1.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0011_2.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0011_3.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0011_4.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0011_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since \\( AB = 4 \\) and \\( AC = x \\),\n\n\\[ BC = \\sqrt{AB^2 - AC^2} = \\sqrt{16 - x^2} \\]\n\nThus, the area of triangle \\( ABC \\) is:\n\n\\[ S_{\\triangle ABC} = \\frac{1}{2} BC \\cdot AC = \\frac{1}{2} x \\sqrt{16 - x^2} \\]\n\nSince this function is neither quadratic nor linear,\n\nWe can eliminate options A and C.\n\nGiven that \\( AB \\) is a fixed value, the area of \\( \\triangle ABC \\) is maximized when \\( OC \\) is perpendicular to \\( AB \\). At this point, \\( AC = 2\\sqrt{2} \\), meaning when \\( x = 2\\sqrt{2} \\), \\( y \\) reaches its maximum. Therefore, option D is also eliminated.\n\nHence, the correct choice is B.\n\n**Key Insight:** This problem examines the function graph of a moving point. The key to solving it lies in establishing the functional relationship and applying the concept of combining numbers with shapes." }, { "problem_id": 637, "question": "In a large square flowerbed in a school campus, as shown in Figure A, it is composed of four small squares with sides of 3 meters, and the planting scheme for each small square is the same. One of the small squares, $A B C D$, is shown in Figure B, where $D G = 1$ meter, $A E = A F = x$ meters. Flowers are planted in the pentagon region $E F B C G$. The graph of the function $y$, representing the area of the large square flowerbed planted with flowers, versus $x$, is approximately ( )\n\nFigure A\n Figure B\nA\n\n\n\nB\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_cd22dcb38e13acb480a0g_0059_1.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0059_2.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0059_3.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0059_4.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0059_5.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0059_6.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: The area of triangle \\( \\triangle AEF \\) is \\( S_{\\triangle AEF} = \\frac{1}{2} \\times AE \\times AF = \\frac{1}{2} x^{2} \\).\n\nThe area of triangle \\( \\triangle DEG \\) is \\( S_{\\triangle DEG} = \\frac{1}{2} \\times DG \\times DE = \\frac{1}{2} \\times 1 \\times (3 - x) = \\frac{3 - x}{2} \\).\n\nThe area of the pentagon \\( EFBCG \\) is \\( S_{\\text{pentagon } EFBCG} = S_{\\text{square } ABCD} - S_{\\triangle AEF} - S_{\\triangle DEG} = 9 - \\frac{1}{2} x^{2} - \\frac{3 - x}{2} = \\frac{1}{2} x^{2} + \\frac{1}{2} x + \\frac{15}{2} \\).\n\nThus, \\( y = 4 \\times \\left(-\\frac{1}{2} x^{2} + \\frac{1}{2} x + \\frac{15}{2}\\right) = -2x^{2} + 2x + 30 \\).\n\nSince \\( AE < AD \\), it follows that \\( x < 3 \\).\n\nIn summary, the function is \\( y = -2x^{2} + 2x + 30 \\) for \\( 0 < x < 3 \\).\n\nTherefore, the correct choice is: A.\n\n【Key Insight】This problem examines the function graph of a moving point, and the key to solving it lies in determining the functional relationship between \\( y \\) and \\( x \\)." }, { "problem_id": 638, "question": "As shown in the figure, in the square $A B C D$, $A B=3 \\mathrm{~cm}$. Point $M$ starts from point $A$ and moves along $A B$ at a speed of $1 \\mathrm{~cm}$ per second. At the same time, point $N$ starts from point $D$ and moves along the path $D C-C B$ at a speed of $2 \\mathrm{~cm}$ per second. The motion stops when point $N$ reaches point $B$. Let the area of $\\triangle A M N$ be $y\\left(\\mathrm{~cm}^{2}\\right)$, and the time of motion be $x$ (seconds). Which of the following graphs can roughly represent the functional relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_cd22dcb38e13acb480a0g_0070_1.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0070_2.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0070_3.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0070_4.jpg", "batch8-2024_06_14_cd22dcb38e13acb480a0g_0070_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since point $N$ starts from point $D$ and moves along the polyline $D C-C B$ at a speed of $2 \\mathrm{~cm}$ per second, the movement stops when it reaches point $B$.\n\nTherefore, the time for $N$ to reach $C$ is: $x=3 \\div 2=1.5$ seconds.\n\nWe can divide the problem into two parts:\n\n(1) When $0 \\leqslant x \\leqslant 1.5$, as shown in Figure 1, $N$ is on $D C$.\n\nThe area of $\\triangle A M N$ is: $S_{\\triangle A M N}=y=\\frac{1}{2} A M \\cdot A D=\\frac{1}{2} x \\times 3=\\frac{3}{2} x\\left(\\mathrm{~cm}^{2}\\right)$.\n\n(2) When $1.5 < x \\leqslant 3$, as shown in Figure 2, $N$ is on $B C$.\n\n\n\nFigure 2\n\nThus, $D C + C N = 2x$,\n\nSo, $B N = 6 - 2x$,\n\nTherefore, the area of $\\triangle A M N$ is: $S_{\\triangle A M N}=y=\\frac{1}{2} A M \\cdot B N=\\frac{1}{2} x(6-2 x)=\\left(-x^{2}+3 x\\right)\\left(\\mathrm{cm}^{2}\\right)$.\n\nHence, the correct choice is: A.\n\n\n\nFigure 1\n\n【Insight】This problem examines the function image problem of moving points; the key to solving this problem is to obtain the corresponding functional relationship based on different ranges of the independent variable." }, { "problem_id": 639, "question": "As shown in the figure, in the Cartesian coordinate system, quadrilateral $A B C D$ is a rhombus, with $A B$ parallel to the $x$-axis. The coordinates of point $B$ are $(4,1)$. $\\angle B A D = 60^\\circ$. A perpendicular line $l$ to the $x$-axis starts from the $y$-axis and moves along the positive direction of the $x$-axis at a speed of 1 unit per second. Let line $l$ intersect the sides of rhombus $A B C D$ at points $M$ and $N$ (point $N$ is above point $M$), and connect $O M$ and $O N$. If the area of triangle $O M N$ is $S$, and the time of the movement of line $l$ is $t$ seconds $(0 \\leq t \\leq 6)$, then the graph of $S$ as a function of $t$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_e35fe70123f4447b467cg_0051_1.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0051_2.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0051_3.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0051_4.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0051_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem, we know that \\( AB = AD = CD = BC = 4 \\).\n\nSince \\( \\angle BAD = 60^\\circ \\),\n\nWhen the line \\( l \\) passes through point \\( D \\), the movement time is 2,\n\nTherefore, the horizontal coordinate of \\( C \\) is 6.\n\nAs shown in Figure 1, when \\( 0 \\leq t \\leq 2 \\),\n\n\n\nFigure 1\n\nSince \\( l \\) is parallel to the y-axis,\n\nThe area \\( S_{\\triangle AMN} = S_{\\triangle OMN} = S \\),\n\nSince \\( AM = t \\), and \\( \\angle BAD = 60^\\circ \\),\n\nTherefore, \\( MN = \\sqrt{3} t \\),\n\nThus, \\( S = \\frac{1}{2} \\times t \\times \\sqrt{3} t = \\frac{\\sqrt{3}}{2} t^{2} \\);\n\nThe graph is a segment of a parabola passing through the origin and opening upwards;\n\nAs shown in Figure 2, when \\( 2 < t \\leq 4 \\), \\( MN \\) is of constant length,\n\n\n\nFigure 2\n\nSince \\( AD = 4 \\), and \\( \\angle BAD = 60^\\circ \\),\n\nTherefore, \\( MN = 2 \\sqrt{3} \\),\n\nThus, \\( S = \\frac{1}{2} \\times t \\times 2 \\sqrt{3} = \\sqrt{3} t \\);\n\nThe graph is a segment of a direct proportion function passing through the origin;\n\nSince the proportionality coefficient 2 of \\( y = 2x \\) is greater than \\( \\sqrt{3} \\),\n\nTherefore, the slope of the area segment is steeper than that of \\( y = 2x \\);\n\nAs shown in Figure 3, when \\( 4 < t \\leq 6 \\),\n\n\n\nFigure 3\n\nSince \\( BC = 4 \\), and \\( \\angle CBG = 60^\\circ \\),\n\nTherefore, \\( BG = 2 \\), \\( CG = 2 \\sqrt{3} \\),\n\nThus, \\( B(4,1) \\), \\( C(6,2 \\sqrt{3}+1) \\),\n\nTherefore, the system of equations is:\n\n\\[\n\\begin{cases}\n4k + b = 1 \\\\\n6k + b = 2 \\sqrt{3} + 1\n\\end{cases}\n\\]\n\nSolving gives:\n\n\\[\n\\begin{cases}\nk = \\sqrt{3} \\\\\nb = 1 - 4 \\sqrt{3}\n\\end{cases}\n\\]\n\nTherefore, the equation of the line is \\( y = \\sqrt{3} x + 1 - 4 \\sqrt{3} \\),\n\nThus, the coordinates of \\( N \\) are \\( (t, 2 \\sqrt{3} + 1) \\), and the coordinates of \\( M \\) are \\( (t, \\sqrt{3} t + 1 - 4 \\sqrt{3}) \\),\n\nTherefore, \\( MN = 2 \\sqrt{3} + 1 - (\\sqrt{3} t + 1 - 4 \\sqrt{3}) = -\\sqrt{3} t + 6 \\sqrt{3} \\),\n\nThus, \\( S = \\frac{1}{2} \\times t \\times (-\\sqrt{3} t + 6 \\sqrt{3}) = -\\frac{\\sqrt{3} t^{2}}{2} + 3 \\sqrt{3} t \\)\n\nThe graph is a segment of a parabola opening downwards;\n\nTherefore, the correct choice is: \\( C \\).\n\n【Key Insight】This problem mainly tests the understanding and mastery of functional images of moving points, the Pythagorean theorem, the area of triangles, the graphs of quadratic functions, the graphs of direct proportion functions, the properties of right triangles with a 30-degree angle, and the properties of rhombuses. The key to solving this problem is the ability to perform calculations based on these properties, using the mathematical concept of classification and discussion." }, { "problem_id": 640, "question": "As shown in the figure, in right triangle $\\triangle A B C$, $\\angle A C B = 90^\\circ, A C = 3 \\text{ cm}, B C = 4 \\text{ cm}$. Point $M$ starts from point $C$ and moves along the path $C \\rightarrow A \\rightarrow B$ at a speed of $1 \\text{ cm/s}$. Point $N$ starts from point $B$ and moves along the path $B \\rightarrow C$ at a speed of $\\frac{1}{2} \\text{ cm/s}$. Both points $M$ and $N$ start at the same time. When point $M$ reaches point $B$, point $N$ stops moving. Connect points $M$ and $N$. Let the distance traveled by point $M$ be $x \\text{ cm}$, and the area of triangle $\\triangle C M N$ be $y \\text{ cm}^2$. The graph of $y$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_e35fe70123f4447b467cg_0082_1.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0082_2.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0082_3.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0082_4.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0082_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "In the right triangle \\( \\triangle ABC \\), \\( \\angle ACB = 90^\\circ \\), \\( AC = 3 \\, \\text{cm} \\), and \\( BC = 4 \\, \\text{cm} \\).\n\nTherefore, \\( AB = \\sqrt{AC^2 + BC^2} = 5 \\, \\text{cm} \\).\n\nThus, \\( AC + AB = 3 + 5 = 8 \\, \\text{cm} \\), and \\( \\sin B = \\frac{AC}{AB} = \\frac{3}{5} \\).\n\nAccording to the problem, \\( CN = \\left(4 - \\frac{1}{2}x\\right) \\, \\text{cm} \\), and \\( CM = x \\, \\text{cm} \\).\n\nWhen \\( 0 \\leq x \\leq 3 \\), that is, when point \\( M \\) moves along \\( AC \\),\n\n\\[\ny = \\frac{1}{2} \\times CN \\times CM = \\frac{1}{2}\\left(4 - \\frac{1}{2}x\\right) \\times x = -\\frac{1}{4}x^2 + 2x.\n\\]\n\nThis represents a part of a downward-opening parabola, so options C and D are excluded.\n\nWhen \\( 3 < x \\leq 8 \\), as shown in the figure, draw \\( MD \\perp BC \\) at point \\( D \\).\n\n\n\nThen, \\( MD = MB \\cdot \\sin B = \\frac{3}{5} MB \\).\n\nSince \\( MB = AC + AB - x = (8 - x) \\, \\text{cm} \\),\n\n\\[\nMD = \\frac{3}{5}(8 - x) \\, \\text{cm}.\n\\]\n\nTherefore,\n\n\\[\ny = \\frac{1}{2} CN \\cdot MD = \\frac{1}{2} \\times \\left(4 - \\frac{1}{2}x\\right) \\times \\frac{3}{5}(8 - x) = \\frac{3}{20}x^2 - \\frac{12}{5}x + \\frac{48}{5}.\n\\]\n\nThis represents a part of an upward-opening parabola, so option B is excluded.\n\nHence, the correct choice is: A.\n\n【Key Insight】This problem examines the function image of a moving point, the properties of quadratic function graphs, and solving right triangles. Mastering the properties of quadratic function graphs is crucial for solving the problem. Note that two different cases must be considered." }, { "problem_id": 641, "question": "As shown in Figure 1, given $\\mathrm{AB}=\\mathrm{AC}$, point $\\mathrm{D}$ is a point on the bisector of $\\angle \\mathrm{BAC}$. Connect $\\mathrm{BD}$ and $\\mathrm{CD}$. As shown in Figure 2, given $\\mathrm{AB}=\\mathrm{AC}$, points $\\mathrm{D}$ and $\\mathrm{E}$ are two points on the bisector of $\\angle \\mathrm{BAC}$. Connect $\\mathrm{BD}$, $\\mathrm{CD}$, $\\mathrm{BE}$, and $\\mathrm{CE}$. As shown in Figure 3, given $\\mathrm{AB}=\\mathrm{AC}$, points $\\mathrm{D}$, $\\mathrm{E}$, and $\\mathrm{F}$ are three points on the bisector of $\\angle \\mathrm{BAC}$. Connect $\\mathrm{CD}$, $\\mathrm{BE}$, $\\mathrm{CE}$, $\\mathrm{BF}$, and $\\mathrm{CF}$; ..., following this pattern, the number of pairs of congruent triangles in the 200th figure is ( )\n\n\nFigure 1\n\n\nFigure 2\n\n\nFigure 3\nA. 200 pairs\nB. 399 pairs\nC. 603 pairs\nD. 20100 pairs", "input_image": [ "batch9-2024_05_23_972e6822ae9b8cf26e8bg_0040_1.jpg", "batch9-2024_05_23_972e6822ae9b8cf26e8bg_0040_2.jpg", "batch9-2024_05_23_972e6822ae9b8cf26e8bg_0040_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: In the first figure, there is $\\frac{1}{2} \\times 2 \\times 1=1$ pair of congruent triangles;\n\nIn the second figure, there are $\\frac{1}{2} \\times 3 \\times 2=3$ pairs of congruent triangles;\n\nIn the third figure, there are $\\frac{1}{2} \\times 4 \\times 3=6$ pairs of congruent triangles;\n\nThus, in the $\\mathrm{n}$th figure, the number of pairs of congruent triangles is $\\frac{1}{2} \\times(\\mathrm{n}+1) \\times \\mathrm{n}=\\frac{(n+1) n}{2}$.\n\nTherefore, the 200th figure has $\\frac{1}{2} \\times 201 \\times 200=20100$ pairs of congruent triangles.\n\nHence, the correct answer is D.\n\n【Key Insight】This problem primarily tests the identification of congruent triangles. The key to solving it lies in identifying the underlying pattern." }, { "problem_id": 642, "question": "As shown in the figure, from an interior point $O$ of $\\triangle A B C$, $\\triangle A B C$ is cut into three triangles (as shown in Figure 1), with sides $A B, B C, A C$ lying on the same line, and point $O$ falls on the line $M N$ (as shown in Figure 2), where line $M N$ is parallel to $A C$. Then point $O$ is the\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. intersection of the three angle bisectors\n\nB. intersection of the three altitudes\n\nC. intersection of the three medians\n\nD. intersection of the three perpendicular bisectors of the sides", "input_image": [ "batch9-2024_05_23_9fc6cdbf563e40e02c6fg_0026_1.jpg", "batch9-2024_05_23_9fc6cdbf563e40e02c6fg_0026_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: Since the line \\( MN \\) is parallel to \\( AC \\),\n\nBy the properties of parallel lines, the distance from point \\( O \\) to \\( BC \\), the distance from \\( O \\) to \\( AC \\), and the distance from \\( O \\) to \\( BA \\) are equal. Therefore, the distances from point \\( O \\) to the three sides are equal.\n\nThus, point \\( O \\) is the intersection point of the three angle bisectors of the triangle.\n\nHence, the correct choice is A.\n\n[Key Insight] This question tests the properties of angle bisectors. Understanding the properties of angle bisectors is crucial for solving the problem." }, { "problem_id": 643, "question": "Given that $A B=A C$. As shown in Figure 1, $D$ and $E$ are points on the bisectors of $\\angle B A C$, and segments $B D$, $C D$, $B E$, and $C E$ are drawn; in Figure 2, $D$, $E$, and $F$ are points on the bisectors of $\\angle B A C$, and segments $B D$, $C D$, $B E$, $C E$, $B F$, and $C F$ are drawn; in Figure 3, $D$, $E$, $F$, and $G$ are points on the bisectors of $\\angle B A C$, and segments $B D$, $C D$, $B E$, $C E$, $B F$, $C F$, $B G$, and $C G$ are drawn, and so on. According to this pattern, the number of pairs of congruent triangles in the 17th figure is $(\\quad)$.\n\n\nFigure 1\n\n\nFigure 2\n\n\nFigure 3\nA. 17\nB. 54\nC. 153\nD. 171", "input_image": [ "batch9-2024_05_23_a168fda90a2d3ac802e0g_0013_1.jpg", "batch9-2024_05_23_a168fda90a2d3ac802e0g_0013_2.jpg", "batch9-2024_05_23_a168fda90a2d3ac802e0g_0013_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In Figure 1, when there are 2 points $D$ and $E$, there are $1+2=3$ pairs of congruent triangles;\n\nIn Figure 2, when there are 3 points $D$, $E$, and $F$, there are $1+2+3=6$ pairs of congruent triangles;\n\nIn Figure 3, when there are 4 points, there are $1+2+3+4=10$ pairs of congruent triangles;\n\nIn Figure $n$, when there are $(n+1)$ points, there are $\\frac{(n+1)(n+2)}{2}$ pairs of congruent triangles.\n\nWhen $n=17$, the number of pairs of congruent triangles is $\\frac{18 \\times 19}{2}=171$.\n\nTherefore, the correct answer is: D.\n\n[Key Insight] This question examines the pattern of graphical transformations and the determination of congruent triangles. Identifying the pattern of graphical transformations is the key to solving the problem." }, { "problem_id": 644, "question": "In a math class, the teacher presented the following problem:\nAs shown in Figure 1, $\\angle B = \\angle C = 90^\\circ$, point $E$ is the midpoint of $B C$, and $D E$ bisects $\\angle A D C$. Prove that $A B + C D = A D$. Xiaoming thought about it this way: To prove $A B + C D = A D$, it suffices to find a point $F$ on $A D$ such that $A F = A B$ and $D F = C D$. As shown in Figure 2, after some consideration, Xiaoming came up with the following three ways to add auxiliary lines.\n\n(1) Draw $E F \\perp A D$ through point $E$ and let it intersect $A D$ at point $F$;\n\n(2) Draw $E F = E C$ and let it intersect $A D$ at point $F$;\n\n(3) Take a point $F$ on $A D$ such that $D F = D C$, and then connect $E F$.\n\nOf the three methods of adding auxiliary lines, which ones can be used to prove \" $A B + C D = A D$ \"? ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. (1)(2)\nB. (1)(3)\nC. (2)(3)\nD. (1)(2)(3)", "input_image": [ "batch9-2024_05_23_f909a2d1d2e438a81155g_0076_1.jpg", "batch9-2024_05_23_f909a2d1d2e438a81155g_0076_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: \n\n(1) As shown in Figure 1, draw $EF \\perp AD$, with the foot of the perpendicular at point $F$.\n\n\n\nFigure 1\n\nIt follows that $\\angle DFE = 90^\\circ$,\n\nthen $\\angle DFE = \\angle C$,\n\nSince $DE$ bisects $\\angle ADC$,\n\n$\\therefore \\angle FDE = \\angle CDE$,\n\nIn triangles $\\triangle DCE$ and $\\triangle DFE$,\n\n$\\left\\{\\begin{array}{l}\\angle C = \\angle DFE \\\\ \\angle CDE = \\angle FDE, \\\\ DE = DE\\end{array}\\right.$\n\n$\\therefore \\triangle DEF \\cong \\triangle DCE$ (by AAS)\n\n$\\therefore CE = EF, \\quad DC = DF, \\quad \\angle CED = \\angle FED$,\n\nSince $E$ is the midpoint of $BC$,\n\n$\\therefore CE = EB$,\n\n$\\therefore EF = EB$,\n\nIn right triangles $\\triangle ABE$ and $\\triangle AFE$,\n\n$\\left\\{\\begin{array}{l}BE = FE \\\\ AE = AE\\end{array}\\right.$,\n\n$\\therefore \\mathrm{Rt} \\triangle AFE \\cong \\mathrm{Rt} \\triangle ABE$ (by HL)\n\n$\\therefore AF = AB$\n\n$\\therefore AD = AF + DF = AB + CD$.\n\n(2) As shown in Figure 2, draw $EF = EC$, intersecting $AD$ at point $F$;\n\n\n\nFigure 2\n\nSince $EF = EC, \\quad DE = DE, \\quad \\angle FDE = \\angle CDE$,\n\n$\\therefore$ According to SSA, we cannot prove $\\triangle DEF \\cong \\triangle DCE$,\n\n$\\therefore$ This method of adding auxiliary lines cannot prove the conclusion $AD = AB + CD$.\n\n(3) As shown in Figure 3, take a point $F$ on $AD$ such that $DF = DC$, and connect $EF$,\n\n\n\nFigure 3\n\nIn triangles $\\triangle DCE$ and $\\triangle DFE$,\n\n$\\left\\{\\begin{array}{l}DC = DF \\\\ \\angle CDE = \\angle FDE, \\\\ DE = DE\\end{array}\\right.$\n\n$\\therefore \\triangle DEF \\cong \\triangle DCE$ (by SAS);\n\n$\\therefore CE = EF, \\angle ECD = \\angle EFD = 90^\\circ$,\n\nSince $E$ is the midpoint of $BC$,\n\n$\\therefore CE = EB$,\n\n$\\therefore EF = EB$,\n\nIn right triangles $\\triangle ABE$ and $\\triangle AFE$,\n\n$\\left\\{\\begin{array}{l}BE = FE \\\\ AE = AE\\end{array}\\right.$,\n$\\therefore \\mathrm{Rt} \\triangle AFE \\cong \\mathrm{Rt} \\triangle ABE$ (by HL)\n\n$\\therefore AF = AB$\n\n$\\therefore AD = AF + DF = AB + CD$.\n\nTherefore, the correct choice is: $B$.\n\n【Highlight】This question examines the determination and properties of congruent triangles, the properties of angle bisectors, and the key to this problem is the appropriate addition of auxiliary lines to construct congruent triangles." }, { "problem_id": 645, "question": "A polyhedron made up of several small cubes is shown as the bottom left figure, its left view is ( )\n\n\n\nView from the front\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_f5b8b25a104af99401e0g_0076_1.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0076_2.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0076_3.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0076_4.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0076_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "From the left side view: the far right consists of two layers of small squares, the middle layer has one layer of small squares, and the left side has one layer of small squares, therefore choose\n\nB." }, { "problem_id": 646, "question": "In the diagrams shown, which one is not an unfolding pattern of a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_0ce863b43d59865f0f74g_0078_1.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0078_2.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0078_3.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0078_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: There are 11 forms of the planar development of a cube, and $D$ is not one of them.\n\nTherefore, choose $D$.\n\n[Key Insight] This question examines the planar development of geometric solids, which is a basic and relatively simple concept." }, { "problem_id": 647, "question": "Figure 1 shows the unfolded surface of a small cube. The small cube is flipped from the position shown in Figure 2 to the 1st, 2nd, and 3rd grids in sequence. At this time, the word on the face of the small cube facing up is ( ).\n\n\n\nFigure 1\n\n\nA. 信\nB. 国\nC. 友\nD. 善", "input_image": [ "batch33-2024_06_14_0ce863b43d59865f0f74g_0053_1.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0053_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The first flip places \"Sincerity\" on the bottom, the second flip places \"Love\" on the bottom, and the third flip places \"Country\" on the bottom. \"Trust\" is opposite to \"Country.\"\n\nTherefore, choose A.\n\n[Highlight] This question examines the text on opposite faces of a cube. Two faces separated by one face are opposites. Paying attention to the sequence of flips to determine which face is on the bottom each time is key to solving the problem." }, { "problem_id": 648, "question": "In the following groups of figures, which pairs of figures form an axis of symmetry?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_84572c67c65ffa3dc0f5g_0012_1.jpg", "batch14-2024_06_15_84572c67c65ffa3dc0f5g_0012_2.jpg", "batch14-2024_06_15_84572c67c65ffa3dc0f5g_0012_3.jpg", "batch14-2024_06_15_84572c67c65ffa3dc0f5g_0012_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "According to the concept of axisymmetric figures, the figure that is axisymmetric is: A.\n\nTherefore, choose A.\n[Highlight] This question tests the understanding of axisymmetric figures. The key to axisymmetry is to find the axis of symmetry, where the images on both sides can coincide when folded." }, { "problem_id": 649, "question": "Among the following figures, which one belongs to the category of prisms?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_f5b8b25a104af99401e0g_0022_1.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0022_2.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0022_3.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0022_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of a prism, only option $\\mathrm{C}$ fits the definition of a prism.\n\nTherefore, the answer is C." }, { "problem_id": 650, "question": "A geometric solid is composed of several identical small cubes. Its top view and side view are shown in the figure. The maximum number of small cubes needed to form this solid is ( )\n\n\n\nTop View\n\n\n\nSide View\nA. 4\nB. 5\nC. 6\nD. 7", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0020_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0020_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view and the side view, the maximum distribution of small cubes required for this geometric figure is as shown in the following diagram:\n\n\n\nTop View\n\n[Key Insight] This problem primarily tests the ability to determine a geometric figure from its three views. Mastering the mnemonic \"Top view lays the foundation, front view builds wildly, side view removes violations\" is crucial for solving the problem." }, { "problem_id": 651, "question": "Form a cube from the unfolded net of the cube shown in Figure 1 (with words facing outward), and then roll the cube according to Figure 2, moving to the first, second, third, and fourth positions in sequence. At this time, the word on the top face of the cube is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 富\nB. 强\nC. 文\nD. 民", "input_image": [ "batch10-2024_06_14_da259e8c8bb047f1aad6g_0097_1.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0097_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Question Analysis: From Figure 1, it can be deduced that \"富\" (wealth) is opposite to \"文\" (culture); \"强\" (strength) is opposite to \"主\" (master); \"民\" (people) is opposite to \"明\" (brightness). From Figure 2, when the small cube flips from its position in Figure 2 to the 4th square, \"文\" is on the bottom. Therefore, the character on the top face of the small cube at this moment is \"富\".\n\nThus, the correct answer is A." }, { "problem_id": 652, "question": "Xiaoming and Xiaohua used wire to make a staircase model in their handicraft class, as shown in the figure. Then, the amount of wire they used is\n\n\n\nXiaoming\n\n\n\nXiaohua\nA. Xiaohua used more.\nB. Xiaoming used more.\nC. They both used the same amount.\nD. It's uncertain who used more.", "input_image": [ "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0074_1.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0074_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since both shapes can be transformed into two rectangles with equal side lengths through translation, both individuals used the same amount. Therefore, the correct choice is C.\n\nKey Point: This question primarily tests the properties of translation. Mastering the properties of translation is crucial for solving the problem." }, { "problem_id": 653, "question": "As shown in the figure, the front view of the geometric solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_da259e8c8bb047f1aad6g_0099_1.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0099_2.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0099_3.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0099_4.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0099_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "According to the definition of the three-view drawing, the front view consists of a rectangle with equal length and width at the top and bottom, so option B is chosen." }, { "problem_id": 654, "question": "As shown in the figure, these are the front view and top view of a geometric solid. The name of the solid is\n\n\nFront View\n\n\nTop View\nA. Cone\nB. Cylinder\nC. Sphere\nD. Cuboid", "input_image": [ "batch25-2024_06_17_f3b68e01e9dc18006b97g_0013_1.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0013_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, since the main view is a triangle, options B, C, and D can be directly ruled out.\n\nA. The cross-section of a cone is an isosceles triangle, with the main view being an isosceles triangle and the top view being a circle;\n\nB. The main view of a cylinder is a rectangle, and the top view is a circle;\n\nC. Both the main view and the top view of a sphere are circles;\n\nD. Both the main view and the top view of a rectangular prism are rectangles;\n\nTherefore, the correct answer is A.\n\n[Key Insight] This question tests the ability to determine the geometric shape from its three views. Understanding the characteristics of various geometric shapes is crucial for solving such problems." }, { "problem_id": 655, "question": "The shape viewed from the front and the side of a geometric solid made of several identical small cubes is shown in the figure. The minimum number of small cubes needed to build this solid is ( )\n\n\n\nView from the front\n\n\n\nView from the side\nA. 4\nB. 5\nC. 6\nD. 8", "input_image": [ "batch25-2024_06_17_dc139e69bdca0c6041e1g_0085_1.jpg", "batch25-2024_06_17_dc139e69bdca0c6041e1g_0085_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By carefully observing the front view and the left view of the object, it can be determined that the lower part of the geometric body must consist of at least three small cubes, and the upper part must have at least one small cube. Therefore, the geometric body is composed of a minimum of 4 small cubes.\n\nHence, the correct choice is: A.\n\n[Highlight] This question tests the knowledge of three-view drawings. The front view is obtained by looking at the object from the front, and the left view is obtained by looking at the object from the left. It examines the students' ability to observe carefully and is considered a basic level question." }, { "problem_id": 656, "question": "As shown in the figure, the front view of the geometric solid is 【】\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0038_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0038_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0038_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0038_4.jpg", "batch25-2024_06_17_493157da588ff6213245g_0038_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "When viewed from the front, the object appears as an isosceles trapezoid. Therefore, choose $\\mathrm{C}$." }, { "problem_id": 657, "question": "Which of the following options can be obtained by translating the figure below? ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_85c79faaf97b4a25229eg_0056_1.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0056_2.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0056_3.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0056_4.jpg", "batch29-2024_06_14_85c79faaf97b4a25229eg_0056_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "The one that can be obtained by translating the left figure is: Option C.\n\nTherefore, choose C.\n\n[Highlight] Examine the properties of translation, mastering the properties of translation is the key to solving the problem." }, { "problem_id": 658, "question": "Which of the following figures cannot be folded into a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_f529096e40f8fe228a22g_0024_1.jpg", "batch25-2024_06_17_f529096e40f8fe228a22g_0024_2.jpg", "batch25-2024_06_17_f529096e40f8fe228a22g_0024_3.jpg", "batch25-2024_06_17_f529096e40f8fe228a22g_0024_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Options A, C, and D can all form a cube, while option B is missing one face when assembled.\n\nTherefore, choose B." }, { "problem_id": 659, "question": "Which of the following diagrams is not an unfolded net of a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_f529096e40f8fe228a22g_0029_1.jpg", "batch25-2024_06_17_f529096e40f8fe228a22g_0029_2.jpg", "batch25-2024_06_17_f529096e40f8fe228a22g_0029_3.jpg", "batch25-2024_06_17_f529096e40f8fe228a22g_0029_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "B, C, D are all unfolded views of a cube;\n\nA cannot be folded into a cube.\n\nTherefore, the correct answer to this question is: A\n\n[Highlight] This question mainly examines several scenarios of the unfolded views of a cube, with moderate difficulty." }, { "problem_id": 660, "question": "Three congruent equilateral triangles are arranged as shown in Figure 1. Now, add another equilateral triangle of the same size so that the four equilateral triangles form a centrally symmetric figure (as shown in Figure 2). The position where the additional equilateral triangle is placed is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. (1)\nB. (2)\nC. (3)\nD. (4)", "input_image": [ "batch14-2024_06_15_86080ee49a80ebd25483g_0010_1.jpg", "batch14-2024_06_15_86080ee49a80ebd25483g_0010_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, by adding the equilateral triangle (4), a centrally symmetric figure can be obtained. Therefore, the correct choice is: D.\n\n[Highlight] This question tests the concept of a centrally symmetric figure. Mastering the definition of a centrally symmetric figure is crucial for solving the problem." }, { "problem_id": 661, "question": "Among the following patterns, which can be designed using translation? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0064_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0064_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0064_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0064_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Question Analysis: \n\nA. Utilizes central symmetry in its design, which does not meet the requirement;\n\nB and C utilize axial symmetry in their designs, which also do not meet the requirement;\n\nD. Utilizes translation in its design, which aligns with the requirement.\n\nTherefore, the correct choice is D." }, { "problem_id": 662, "question": "As shown in the figure, this is part of a cartoon drawn by Xiaoming. Which of the following options represents the other part of the cartoon that is symmetric about line l?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch35-2024_06_17_0e7e7b1770ce0b2f7c3ag_0022_1.jpg", "batch35-2024_06_17_0e7e7b1770ce0b2f7c3ag_0022_2.jpg", "batch35-2024_06_17_0e7e7b1770ce0b2f7c3ag_0022_3.jpg", "batch35-2024_06_17_0e7e7b1770ce0b2f7c3ag_0022_4.jpg", "batch35-2024_06_17_0e7e7b1770ce0b2f7c3ag_0022_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "According to the definition of symmetry, the other part of the figure that is symmetric about the line $l$ should be B.\n\nTherefore, choose B." }, { "problem_id": 663, "question": "As shown in the figure, the front view and left view of a geometric solid composed of some identical small cubes are given. Which of the following numbers of small cubes cannot be used to form this solid?\n\n\n\nFront View\n\n\n\nLeft View\nA. 3\nB. 4\nC. 5\nD. 6", "input_image": [ "batch24-2024_06_15_75432e11b1fecd000fb8g_0004_1.jpg", "batch24-2024_06_15_75432e11b1fecd000fb8g_0004_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Test Question Analysis: Based on the front view and the side view, the first layer of the cube has 2, 3, or 4 small cubes, and the second layer has 1 small cube. Therefore, the total number of small cubes that make up this geometric figure is 3, 4, or 5. It is impossible to have 6 small cubes. Hence, the correct answer is D.\n\nExam Point: Determining the geometric figure from its three views." }, { "problem_id": 664, "question": "As shown in the figure, the top view of the given 3D solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0004_1.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0004_2.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0004_3.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0004_4.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0004_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "From the diagram, it can be seen that the top view of this geometric solid is\n\n\n\nTherefore, the correct choice is A.\n\n[Key Insight] This question primarily tests the judgment of the top view. The key to solving it lies in understanding the definition of a top view." }, { "problem_id": 665, "question": "Among the following geometric solids, which is a pyramid?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_f557af24e20d9104e53fg_0016_1.jpg", "batch33-2024_06_14_f557af24e20d9104e53fg_0016_2.jpg", "batch33-2024_06_14_f557af24e20d9104e53fg_0016_3.jpg", "batch33-2024_06_14_f557af24e20d9104e53fg_0016_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: The geometric shapes in the options are as follows: A. Cylinder; B. Cone; C. Quadrilateral Pyramid; D. Sphere; Therefore, the correct choice is: C.\n\n[Key Point] This question tests the recognition of geometric shapes, and mastering the characteristics of each geometric shape is the key to solving the problem." }, { "problem_id": 666, "question": "Among the following quadrilaterals, which ones have equal diagonals?\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch23-2024_06_14_f513d891339227eeea32g_0012_1.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0012_2.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0012_3.jpg", "batch23-2024_06_14_f513d891339227eeea32g_0012_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Slightly; briefly; roughly; to omit; to delete; strategy; plan; scheme; to capture; to seize. (Note: The translation of \"略\" depends on the context in which it is used. The above are some common translations.)" }, { "problem_id": 667, "question": "The figure shows the three views of a geometric solid composed of several identical small cubes. The number of small cubes used to build this solid is\n\n\n\nFront View\n\n\n\nTop View\n\n\n\nLeft View\nA. 5\nB. 6\nC. 7\nD. 8", "input_image": [ "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0007_1.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0007_2.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0007_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the front view, it can be seen that there are 4 small cubes on the front. From the left view, it can be observed that there should be at least one small cube in front of these four small cubes. Finally, from the top view, it is known that there is a small cube in front of one of the small cubes in the middle of the four small cubes. Therefore, there are a total of 5 cubes. Choose A." }, { "problem_id": 668, "question": "Construct a geometric solid using a number of small cubes such that its left view and top view are as shown in the figure. The maximum number of small cubes used to build the solid is ( )\n\n\n\nLeft View\n\n\n\nTop View\nA. 15 cubes\nB. 14 cubes\nC. 13 cubes\nD. 12 cubes", "input_image": [ "batch5-2024_06_14_43010c41ad60ab0a5523g_0065_1.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0065_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By synthesizing the left view and the top view, the maximum number of small cubes on the bottom layer is \\(3 + 2 + 1 = 6\\), on the second layer is \\(2 + 2 + 1 = 5\\), and on the third layer is \\(1 + 1 + 1 = 3\\).\n\nTherefore, the maximum number of small cubes in the constructed geometric figure is \\(6 + 5 + 3 = 14\\), so the correct choice is B." }, { "problem_id": 669, "question": "The top view of the solid in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_aa8ac36ee4741b0da18ag_0027_1.jpg", "batch25-2024_06_17_aa8ac36ee4741b0da18ag_0027_2.jpg", "batch25-2024_06_17_aa8ac36ee4741b0da18ag_0027_3.jpg", "batch25-2024_06_17_aa8ac36ee4741b0da18ag_0027_4.jpg", "batch25-2024_06_17_aa8ac36ee4741b0da18ag_0027_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "From the characteristics of the three views, it is known that B is the top view of the geometric body.\n\nTherefore, choose B." }, { "problem_id": 670, "question": "During a mathematics activity class, the \"Wisdom Group\" designed to create a rectangular box using a large rectangle, as shown in the figure. The dimensions of the box are required to be $4, 3, 1$ in length, width, and height, respectively. Then the length of this large rectangle is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 14\nB. 10\nC. 8\nD. 7", "input_image": [ "batch25-2024_06_17_1a01bcad4bed55b70b27g_0093_1.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0093_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: From Figure (1), it can be seen that the length of the large rectangle is \\(1 + 4 + 1 + 4 = 10\\).\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question tests the understanding of geometric figure expansions. The key to solving it lies in having spatial imagination skills." }, { "problem_id": 671, "question": "Which of the following geometric figures is a hexagonal prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_9e6988a33fdf375f8541g_0022_1.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0022_2.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0022_3.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0022_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "According to the definition of a hexagonal prism, the base is a hexagon, and only option $\\mathrm{D}$ fits this definition.\n\nTherefore, the answer is D.\n\n[Key Insight] This question tests the definition of a hexagonal prism. It is a basic question, and a thorough understanding of the definition of a hexagonal prism is crucial for determining the correct answer." }, { "problem_id": 672, "question": "The three views of a geometric solid are shown in the figure. Determine the shape of the solid.\n\n\n\nFront View\n\n\n\nTop View\nA. Triangular Prism\nB. Cylindrical\nC. Cuboid\nD. Cubic", "input_image": [ "batch25-2024_06_17_f3b68e01e9dc18006b97g_0010_1.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0010_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since both the front view and the left view are rectangles, the geometric body can be determined as a prism. Furthermore, based on the top view being a triangle, the geometric body is identified as a triangular prism.\n\nTherefore, the correct choice is A.\n\n[Key Insight] This question tests the ability to determine a three-dimensional shape based on its three views. Mastering the three views of common three-dimensional shapes is crucial for solving this problem." }, { "problem_id": 673, "question": "The shape of the view from the front and the top of a 3D structure made of identical small cubes is shown in the figure. The maximum number of small cubes used to build this structure is ( )\n\n\n\nView from the front\n\n\n\nView from above\nA. 9 cubes\nB. 8 cubes\nC. 7 cubes\nD. 6 cubes", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0046_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0046_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By combining the views from above and from the front, we can deduce that the upper layer has at most 3 small cubes and at least 2, while the lower layer must have 4.\n\nTherefore, the total number of small cubes that could form this geometric figure is either 7 or 6.\n\nThus, the maximum number of small cubes needed to construct this geometric figure is 7.\n\nHence, the correct choice is: C.\n\n[Insight] This question tests the students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills." }, { "problem_id": 674, "question": "The figure below is a 3D structure made of 7 identical small cubes. What is the left view of this structure?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_9e565a461277c7b1b67cg_0077_1.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0077_2.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0077_3.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0077_4.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0077_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "From the left side, the number of small squares in the two columns from left to right can be seen as: 3, 1. Therefore, the correct choice is A." }, { "problem_id": 675, "question": "As shown in the figure, a factory is designing a type of geometric solid made up of five identical small cubes. Which of the following is the top view of the solid?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_f5b8b25a104af99401e0g_0077_1.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0077_2.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0077_3.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0077_4.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0077_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: From the top view, it is easy to see that there are 3 squares on the upper layer and one square in the middle of the lower layer. Therefore, the correct choice is A." }, { "problem_id": 676, "question": "The following pairs of figures (solid lines) are not similar figures:\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_a9bda9fa88e05a01666cg_0074_1.jpg", "batch32-2024_06_14_a9bda9fa88e05a01666cg_0074_2.jpg", "batch32-2024_06_14_a9bda9fa88e05a01666cg_0074_3.jpg", "batch32-2024_06_14_a9bda9fa88e05a01666cg_0074_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Options A, B, and D all have a homothetic center, and the images are similar. Option C has congruent images. Therefore, choose C." }, { "problem_id": 677, "question": "The shaded part of the figure is a sector of a circle for ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_8b41935007d2325ece3bg_0098_1.jpg", "batch20-2024_05_23_8b41935007d2325ece3bg_0098_2.jpg", "batch20-2024_05_23_8b41935007d2325ece3bg_0098_3.jpg", "batch20-2024_05_23_8b41935007d2325ece3bg_0098_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem statement, the shaded part of the figure in option D is a sector.\n\nTherefore, the answer is: D.\n\n[Key Point] This question primarily tests the definition of a sector. Mastering the definition of a sector is crucial for solving the problem." }, { "problem_id": 678, "question": "The top view of the following solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0090_1.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0090_2.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0090_3.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0090_4.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0090_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Find the shape obtained from the top view, ensuring that all visible edges are represented in the top view. \nSolution: From the top view, this geometric figure has only one layer and consists of 3 small squares, so the correct choice is A." }, { "problem_id": 679, "question": "A geometric solid is composed of several identical cubes. Its front view and top view are shown in the figure. The maximum number of cubes in this solid is ( )\n\n\n\nFront View\n\n\n\nTop View\nA. 4\nB. 5\nC. 6\nD. 7", "input_image": [ "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0003_1.jpg", "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0003_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By combining the front view and the top view, it can be determined that the upper left layer can have a maximum of 2 cubes, the lower left layer can also have a maximum of 2 cubes, and the right side has only one layer with a maximum of 1 cube.\n\nTherefore, the maximum number of small cubes in the figure is 5.\n\nHence, the correct choice is: B.\n\n[Key Insight] This question primarily tests the ability to deduce the geometry from three-view drawings, assessing students' mastery and flexible application of three-view concepts, as well as their spatial imagination skills." }, { "problem_id": 680, "question": "Among the following figures, which one does not have stability?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_033ee52ad856509d52d9g_0044_1.jpg", "batch5-2024_06_14_033ee52ad856509d52d9g_0044_2.jpg", "batch5-2024_06_14_033ee52ad856509d52d9g_0044_3.jpg", "batch5-2024_06_14_033ee52ad856509d52d9g_0044_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The triangle has stability.\n\nA. Has stability;\n\nB. Has stability;\n\nC. Has stability;\n\nD. Does not have stability;\n\nTherefore, choose D.\n\n[Key Point] This question tests the stability of a triangle. Mastering this knowledge point is crucial for solving the problem." }, { "problem_id": 681, "question": "Which of the following objects gives us the image of a cylinder ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_9e6988a33fdf375f8541g_0017_1.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0017_2.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0017_3.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0017_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Since the top and bottom surfaces of a cylinder are circles,\n\nTherefore, option C is chosen.\n\n[Key Point] This question tests the recognition of a cylinder, which is a simple problem. Familiarity with the definition of three-dimensional shapes is key to solving it." }, { "problem_id": 682, "question": "As shown in the figure, the front view and left view of a geometric solid composed of some identical small cubes are given. What is the minimum number of small cubes used to build this geometric solid?\n\n\n\nFront View\n\n\n\nLeft View\nA. 3\nB. 4\nC. 5\nD. 6", "input_image": [ "batch25-2024_06_17_802f01be018a9f8ad422g_0026_1.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0026_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the front view provided in the problem, we can see that the small cubes are arranged in two columns, with the left column having one layer and the right column having two layers.\n\nFrom the side view, it is evident that the small cubes are arranged in two columns, with both the left and right columns having two layers each. The minimum number of layers is as shown in the top view,\n\n\n\nTop view\n\nTherefore, the minimum number of small cubes in the figure is 5.\n\nHence, the correct choice is: C.\n\n[Insight] This question primarily tests the students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills. If one masters the mnemonic \"Top view lays the foundation, front view builds up, side view removes the excess,\" it becomes easier to arrive at the answer." }, { "problem_id": 683, "question": "The top view of the solid shown in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0080_1.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0080_2.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0080_3.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0080_4.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0080_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: When viewed from above, it is a triangle.\n\nTherefore, the correct choice is: $A$.\n\n[Key Insight] This question examines the three views of a simple composite solid. The key to solving the problem lies in recognizing that the view from above is the top view." }, { "problem_id": 684, "question": "As shown in the figure, which of the following is the front view of the object?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_00e575063edd62623dc1g_0038_1.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0038_2.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0038_3.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0038_4.jpg", "batch25-2024_06_17_00e575063edd62623dc1g_0038_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The figure obtained from the front view is a rectangle.\n\nTherefore, the correct choice is: B.\n\n[Highlight] This question tests the knowledge of the three views of a simple composite solid. The figure obtained from the front view is the main view." }, { "problem_id": 685, "question": "The front view of the geometric solid in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0008_1.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0008_2.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0008_3.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0008_4.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0008_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The front view of the figure is:\n\n\n\nTherefore, the correct choice is: B.\n\n【Key Point】This question tests the understanding of the three views of a simple composite solid. The view obtained from the front is the main view." }, { "problem_id": 686, "question": "Examine the following figures, which of them are centro-symmetric?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch15-2024_06_15_ddd1d533d60a0f1ae3ecg_0009_1.jpg", "batch15-2024_06_15_ddd1d533d60a0f1ae3ecg_0009_2.jpg", "batch15-2024_06_15_ddd1d533d60a0f1ae3ecg_0009_3.jpg", "batch15-2024_06_15_ddd1d533d60a0f1ae3ecg_0009_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: When the figure in option D is rotated by $180^{\\circ}$ around a certain point, it can completely coincide with the original figure. Therefore, this figure is centrally symmetric.\n\nThus, the correct choice is: D." }, { "problem_id": 687, "question": "As shown in the figure below, compared to the triangle in (1), the change in the triangle in (2) is ( )\n\n\n(1)\n\n\nA. A left translation by 3 units\nB. A left translation by 2 units\nC. An upward translation by 3 units\nD. An upward translation by 1 unit", "input_image": [ "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0024_1.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0024_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From Figure (1) to Figure (2), the point $(1,1)$ is translated to the point $(-2,1)$,\n\nand the point $(3,1)$ is translated to the point $(0,1)$. Both points are shifted 3 units to the left.\n\n$\\therefore$ The rule for the translation of the figure is: Shift 3 units to the left.\n\nTherefore, the correct choice is A." }, { "problem_id": 688, "question": "The left view of the solid shown in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_c2b5a48971f4685260d0g_0052_1.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0052_2.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0052_3.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0052_4.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0052_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the problem statement, we can deduce that,\n\n\n\nTherefore, the correct choice is C.\n\n[Key Insight] This question tests the understanding of the left view in the three-view drawing of a solid figure: the view obtained by looking at the solid figure from the left side." }, { "problem_id": 689, "question": "In the figure, （）is not an unfolded net of a cube.\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_9e6988a33fdf375f8541g_0095_1.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0095_2.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0095_3.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0095_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Option D is not the unfolded diagram of a cube,\n\nTherefore, choose D." }, { "problem_id": 690, "question": "Among the four figures below, which one is a central symmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0037_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0037_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0037_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0037_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: A is a centrally symmetric figure, hence the option is correct;\n\nB is not a centrally symmetric figure, hence the option is incorrect;\n\nC is not a centrally symmetric figure, hence the option is incorrect;\n\nD is not a centrally symmetric figure, hence the option is incorrect.\n\nTherefore, choose A.\n\n[Key Point] This question tests the understanding of centrally symmetric figures." }, { "problem_id": 691, "question": "In the figure, there are straight lines, rays, and line segments. Which of the following can intersect?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n", "input_image": [ "batch33-2024_06_14_1241c422ecb863904091g_0094_1.jpg", "batch33-2024_06_14_1241c422ecb863904091g_0094_2.jpg", "batch33-2024_06_14_1241c422ecb863904091g_0094_3.jpg", "batch33-2024_06_14_1241c422ecb863904091g_0094_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The figure that can intersect is the one in option B.\n\nTherefore, the answer is: B.\n\n[Key Point] This question examines the properties of lines, rays, and segments; understanding the extensibility of these three is crucial." }, { "problem_id": 692, "question": "After unfolding the four cubes in the options, the unfolded net that is different from the one shown in the figure is ( )\n\n\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0089_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0089_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0089_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0089_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0089_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By observing the figure, it can be seen that when the four cubes in the options are unfolded, the resulting planar development that differs from the above development is option B." }, { "problem_id": 693, "question": "In the figures below, which one is the net of the triangular prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_2eea509d719aacf4fdebg_0048_1.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0048_2.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0048_3.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0048_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The planar development of a triangular prism consists of two triangular bases and three rectangles in between, therefore the correct choice is A." }, { "problem_id": 694, "question": "As shown in the figure, the geometric solid is composed of a cube and a cylinder. Its top view is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_43010c41ad60ab0a5523g_0044_1.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0044_2.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0044_3.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0044_4.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0044_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The top view of a cone is a point and a circle, and since the cone is tangent to the cube, the correct choice is D." }, { "problem_id": 695, "question": "As shown in the figure, these are three views of a three-dimensional structure composed of some identical small cubes. How many small cubes are used to form this three-dimensional structure? ( )\n\n\n\nLeft View\n\n\n\nFront View\n\n\n\nTop View\nA. 6\nB. 7\nC. 9\nD. 8", "input_image": [ "batch5-2024_06_14_43010c41ad60ab0a5523g_0012_1.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0012_2.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0012_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Question Analysis: From the top view, it is easy to determine that the bottom layer consists of 6 cubes, and the second layer has 2 cubes. Therefore, the total number of cubes is $6 + 2 = 8$.\n\nThus, the correct answer is D.\n\nKey Point: Determining the geometry from the three views." }, { "problem_id": 696, "question": "The top view of the solid shown in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_e768c97899b57b6ada19g_0028_1.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0028_2.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0028_3.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0028_4.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0028_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: According to the definition of a top view, the top view of this geometric figure is\n\n\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the understanding of the top view. Mastering the definition of a top view is crucial for solving the problem." }, { "problem_id": 697, "question": "The solid figure is composed of two small cubes and a cylinder. Its front view is \n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_682dd3cab3079f1333ccg_0041_1.jpg", "batch25-2024_06_17_682dd3cab3079f1333ccg_0041_2.jpg", "batch25-2024_06_17_682dd3cab3079f1333ccg_0041_3.jpg", "batch25-2024_06_17_682dd3cab3079f1333ccg_0041_4.jpg", "batch25-2024_06_17_682dd3cab3079f1333ccg_0041_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The front view is the view obtained from the front. When looking from the front, the cone on top appears as a triangle, and the two cubes below appear as two squares. Therefore, the correct choice is B." }, { "problem_id": 698, "question": "Among the following figures, those that are axisymmetric are ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_7b577704b9b4a50255f1g_0074_1.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0074_2.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0074_3.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0074_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Question Analysis: According to the definition of an axisymmetric figure, it can be determined that D is an axisymmetric figure.\n\nTherefore, the answer is D." }, { "problem_id": 699, "question": "On the table is a geometric solid composed of several identical small cubes. Its front view and left view are shown in the figure. The maximum number of small cubes that can make up this solid is ( )\n\n\n\n(Front View)\n\n\n\n(Left View)\nA. 10 cubes\nB. 11 cubes\nC. 12 cubes\nD. 13 cubes", "input_image": [ "batch25-2024_06_17_83c3aa5e6600cb504863g_0018_1.jpg", "batch25-2024_06_17_83c3aa5e6600cb504863g_0018_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The bottom layer can have a maximum of 9 cubes, and the second layer can have a maximum of 4 cubes. Therefore, the total number of small cubes that make up this geometric figure can be at most 13.\n\nHence, the answer is: D.\n\n[Key Insight] This question tests the knowledge of determining the geometry from three-view drawings. The key to solving this problem lies in using the principle of \"the front view covers extensively, and the side view removes the excess\" to find the maximum number of cubes required." }, { "problem_id": 700, "question": "The following figures are not unfolded patterns of a cube:\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_1.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_2.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_3.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem statement,\n\n\n\ncannot be assembled into a cube,\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question primarily tests the understanding of the net of a cube. Mastering the net of a cube is crucial for solving such problems." }, { "problem_id": 701, "question": "As shown in the figure, it is a three-way connector for a water pipe, and its left view is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_26a127df8b63912ac416g_0074_1.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0074_2.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0074_3.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0074_4.jpg", "batch25-2024_06_17_26a127df8b63912ac416g_0074_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: Its left view consists of a circle at the bottom and a rectangle on top, with the bottom side of the rectangle connected to the cylinder below.\n\nTherefore, the correct choice is: B." }, { "problem_id": 702, "question": "The perspective view of a geometric solid composed of 5 identical cubes is shown as ( \n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0048_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0048_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0048_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0048_4.jpg", "batch25-2024_06_17_493157da588ff6213245g_0048_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: On the left side, a small square is visible; in the middle, one can be seen at both the top and bottom, and on the right side, there is only one. Therefore, choose A." }, { "problem_id": 703, "question": "In the figure below, what Xiaoming sees from above is ( ):\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0056_1.jpg", "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0056_2.jpg", "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0056_3.jpg", "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0056_4.jpg", "batch25-2024_06_17_cd2fbf97f9f5f7e75de7g_0056_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: From the top view, the figure seen is:\n\n\n\nTherefore, choose A.\n\n[Highlight] This question tests the three-view representation of a simple composite solid, primarily aiming to cultivate students' observational skills and spatial imagination." }, { "problem_id": 704, "question": "Which of the following figures cannot form a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0015_1.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0015_2.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0015_3.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0015_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Options A, B, and D can all form a cube, while option C, when folded, results in two squares overlapping.\n\nTherefore, choose C." }, { "problem_id": 705, "question": "In the four geometric solids shown in the figure, how many have a circular top view?\n\n\nCone\n\n\nCuboid\n\n\nCylinder\n\n\nSphere\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch33-2024_06_14_c7449420eec16015971ag_0072_1.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0072_2.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0072_3.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0072_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: When viewed from above, the shapes of a cylinder and a sphere are circles, therefore option B fits the question's requirement; hence, select B." }, { "problem_id": 706, "question": "Among the following figures, which one is the lateral surface unfolding of a triangular prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_4c8122bf8583225dfc55g_0040_1.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0040_2.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0040_3.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0040_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The lateral expansion diagram of a triangular prism is a rectangle composed of three small rectangles.\n\nTherefore, choose A." }, { "problem_id": 707, "question": "As shown in the figure, the left view of the geometric solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_e768c97899b57b6ada19g_0049_1.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0049_2.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0049_3.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0049_4.jpg", "batch25-2024_06_17_e768c97899b57b6ada19g_0049_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, the left view of the geometric solid is:\n\n\n\nTherefore, the correct choice is: C.\n\n[Highlight] This question tests the understanding of the three views of a simple composite solid. The view obtained from the left side is the left view." }, { "problem_id": 708, "question": "There are two diagrams below, both representing classifications of triangles. Which of the following is correct ( )\n\n\n\n(1)\n\n\n\n(2)\nA. (1) is correct, (2) is incorrect.\nB. (2) is correct, (1) is incorrect.\nC. Both (1) and (2) are incorrect.\nD. Both (1) and (2) are correct.", "input_image": [ "batch5-2024_06_14_e4559f258117b7931eb3g_0058_1.jpg", "batch5-2024_06_14_e4559f258117b7931eb3g_0058_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "An isosceles triangle includes an equilateral triangle, therefore the classification in (1) is incorrect; the classification of the triangle in figure (2) is correct. Hence, the answer is: B.\n\n【Key Point】The examination focuses on the classification of triangles, with the key to solving the problem being mastering the classification method. According to the equality of sides, triangles are classified into: scalene triangles and isosceles triangles, noting that isosceles triangles include equilateral triangles." }, { "problem_id": 709, "question": "By translating the isosceles right triangle shown, the resulting pattern is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_0170c451960296f01a9cg_0006_1.jpg", "batch14-2024_06_15_0170c451960296f01a9cg_0006_2.jpg", "batch14-2024_06_15_0170c451960296f01a9cg_0006_3.jpg", "batch14-2024_06_15_0170c451960296f01a9cg_0006_4.jpg", "batch14-2024_06_15_0170c451960296f01a9cg_0006_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "From the properties of translation, it can be determined that only option B can be obtained through translation, therefore the answer is B." }, { "problem_id": 710, "question": "As shown in the figure, it is a figure composed of two identical cylindrical shapes. Its top view is 【】\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_c7449420eec16015971ag_0003_1.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0003_2.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0003_3.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0003_4.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0003_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Find the shape obtained from the top view: A standing cylinder viewed from above is a circle, while a lying cylinder is a rectangle, and\n\nthe rectangle is located to the right of the circle. Therefore, choose C." }, { "problem_id": 711, "question": "Which of the following figures has stability?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_da253f0356f25e28dec0g_0059_1.jpg", "batch5-2024_06_14_da253f0356f25e28dec0g_0059_2.jpg", "batch5-2024_06_14_da253f0356f25e28dec0g_0059_3.jpg", "batch5-2024_06_14_da253f0356f25e28dec0g_0059_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Among the options, only option D is composed of triangles, hence it possesses stability.\n\nTherefore, the answer is: D.\n\n[Key Point] This question examines the stability of triangles, which is a fundamental concept that requires memorization. The key is to answer based on the principle that triangles have stability." }, { "problem_id": 712, "question": "Among the following figures, which one is obtained by a translation transformation from the given figure ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0036_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0036_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0036_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0036_4.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0036_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "By observing the figure, it can be seen that pattern $\\mathrm{C}$ can be obtained through translation.\n\nTherefore, the answer is C." }, { "problem_id": 713, "question": "Which of the following diagrams can be folded into a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0012_1.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0012_2.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0012_3.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0012_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "According to the net of the cube\n\n\nit can be determined that $\\mathrm{A}$, $\\mathrm{B}$, and $\\mathrm{C}$ cannot form a cube, only $\\mathrm{D}$ can form a cube.\n\nTherefore, the correct choice is D." }, { "problem_id": 714, "question": "A geometric solid is placed as shown in Figure 1. Figure 2 could be its ( )\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\nA. Front view\nB. Left view\nC. Top view\nD. Cannot be determined", "input_image": [ "batch25-2024_06_17_f3b68e01e9dc18006b97g_0053_1.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0053_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: The shape seen from the left side of Figure 1 is the same as that of Figure 2, therefore Figure 2 could be its left view, so option B is correct.\n\nThus, the answer is: B.\n\n[Highlight] This question mainly tests the understanding of three-view drawings. The key to solving it lies in mastering the definitions of the three views: the front view is the shape seen from the front, the left view is the shape seen from the left, and the top view is the shape seen from above." }, { "problem_id": 715, "question": "Among the following figures, which one is the unfolded surface of a prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_0ce863b43d59865f0f74g_0013_1.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0013_2.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0013_3.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0013_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Based on the characteristics of the prism's net: it consists of four rectangular lateral faces and two non-continuous top and bottom bases, we can determine that the correct answer is B.\n\n【Highlight】This question tests the understanding of the net of a prism. A thorough comprehension of the prism is crucial for solving this problem." }, { "problem_id": 716, "question": "In the diagrams below, which one represents an unfolded net of a cube?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0035_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0035_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0035_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0035_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "According to the unfolded diagram of the cube, when A, B, and D are folded together, there will be overlapping areas. Only C can form a cube, so the correct choice is C.\n\n## [Key Point] This question tests the understanding of the unfolded diagram of a cube." }, { "problem_id": 717, "question": "In the following figures, which one does not show the correct shadow of the object?\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_13ac171763bb883f1178g_0078_1.jpg", "batch25-2024_06_17_13ac171763bb883f1178g_0078_2.jpg", "batch25-2024_06_17_13ac171763bb883f1178g_0078_3.jpg", "batch25-2024_06_17_13ac171763bb883f1178g_0078_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Because sunlight is a bundle of parallel rays, option B is incorrect.\n\nTherefore, choose B.\n\n[Highlight] This question examines the difference between parallel projection and central projection. Sunlight consists of parallel rays, whereas light from a lamp does not consist of parallel rays." }, { "problem_id": 718, "question": "The shape of a section passing through the vertex of a cone might be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_452fbe550559079ef90fg_0052_1.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0052_2.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0052_3.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0052_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Question Analysis: The shape of the cross-section passing through the apex of the cone may resemble the figure in option B, therefore the answer is B." }, { "problem_id": 719, "question": "Among the following geometric figures, which one is not an axisymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_712dae00cc969aad55f7g_0028_1.jpg", "batch5-2024_06_14_712dae00cc969aad55f7g_0028_2.jpg", "batch5-2024_06_14_712dae00cc969aad55f7g_0028_3.jpg", "batch5-2024_06_14_712dae00cc969aad55f7g_0028_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "A parallelogram is a centrally symmetric figure but not an axisymmetric figure.\n\nCircles, isosceles triangles, and equilateral triangles are all axisymmetric figures.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question tests the recognition of axisymmetric figures. Understanding the definition of an axisymmetric figure is crucial for solving the problem." }, { "problem_id": 720, "question": "The front view and left view of a geometric solid made of identical small cubes are shown in the figure. How many different arrangements of the small cubes are there to form this solid?\n\n\n\nFront View\n\n\n\nLeft View\nA. 6\nB. 5\nC. 4\nD. 3", "input_image": [ "batch25-2024_06_17_83c3aa5e6600cb504863g_0014_1.jpg", "batch25-2024_06_17_83c3aa5e6600cb504863g_0014_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the front view and the side view, the small cubes that make up the geometric body can be arranged in the following different ways:\n\n\n\n1\n\n\n\n2\n\n\n\n3\n\n\n\n4\n\n\n\n5\n\nThere are a total of 5 different arrangement methods,\n\nTherefore, the answer is: B.\n\n[Key Insight] This question tests the understanding of three-view drawings. The key to solving the problem lies in comprehending the definition of three-view drawings. For general three-dimensional figures, careful observation and imagination are required before drawing their diagrams." }, { "problem_id": 721, "question": "When viewed from different angles, which of the following is the perspective view of a teapot? \n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_4c8122bf8583225dfc55g_0096_1.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0096_2.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0096_3.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0096_4.jpg", "batch33-2024_06_14_4c8122bf8583225dfc55g_0096_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "The top view is the planar figure seen from above, and it is also the orthographic projection on the horizontal projection plane. It is easy to determine that option A is correct." }, { "problem_id": 722, "question": "As shown in the figure, which of the following pairs of angles are vertical angles? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0055_1.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0055_2.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0055_3.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0055_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: According to the definition of vertical angles, only figure B satisfies the definition of vertical angles.\n\nTherefore, the correct choice is B." }, { "problem_id": 723, "question": "In Figure 1 and Figure 2, all the squares are congruent. When the square in Figure 1 is placed in one of the positions (1)(2)(3)(4) in Figure 2, the arrangement that forms a centrally symmetric figure is ( )\n\n\n\nFigure 1\n\n\nA. (1)(2)\nB. (2)(3)\nC. (3)(4)\nD. (2)(4)", "input_image": [ "batch26-2024_06_17_83481c83c3572e9930e6g_0064_1.jpg", "batch26-2024_06_17_83481c83c3572e9930e6g_0064_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Place the square from Figure 1 in one of the positions (1), (2), (3), or (4) in Figure 2. The positions where the resulting figure is centrally symmetric are: (3) and (4).\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question primarily tests the understanding of the definition of a centrally symmetric figure. Mastering the definition of a centrally symmetric figure is crucial for solving this problem." }, { "problem_id": 724, "question": "As shown in the figure, which of the following is not a central symmetric figure? ( )\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_844a73fd19ba25585b19g_0041_1.jpg", "batch26-2024_06_17_844a73fd19ba25585b19g_0041_2.jpg", "batch26-2024_06_17_844a73fd19ba25585b19g_0041_3.jpg", "batch26-2024_06_17_844a73fd19ba25585b19g_0041_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "A centrally symmetric figure can coincide with the original figure after being rotated by $180^{\\circ}$ around the center of symmetry. Only the figure in option B does not meet this condition.\n\nTherefore, choose B." }, { "problem_id": 725, "question": "A right-angled triangle rotated around one of its legs forms a geometric solid that could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_43010c41ad60ab0a5523g_0100_1.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0100_2.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0100_3.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0100_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "A right triangle rotated around one of its right-angled sides for one full revolution forms a geometric shape known as a cone,\n\nTherefore, the correct choice is D." }, { "problem_id": 726, "question": "Among the following groups of figures, which group is congruent?\nA. \nB. \nC. \nD. \n\nA. (A)\nB. (B)\nC. (C)\nD. (D)", "input_image": [ "batch9-2024_05_23_500b579d4482fc00b365g_0060_1.jpg", "batch9-2024_05_23_500b579d4482fc00b365g_0060_2.jpg", "batch9-2024_05_23_500b579d4482fc00b365g_0060_3.jpg", "batch9-2024_05_23_500b579d4482fc00b365g_0060_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Congruent figures are two shapes that can completely coincide with each other, from which it can be deduced that only option D fits the description. Therefore, D is chosen." }, { "problem_id": 727, "question": "Among the following figures, which one has stability? ( )\n\n\nA\n\n\nB\n\n\nC\n\n\nD\nA. A\nB. B\nC. C\nD. D", "input_image": [ "batch5-2024_06_14_033ee52ad856509d52d9g_0004_1.jpg", "batch5-2024_06_14_033ee52ad856509d52d9g_0004_2.jpg", "batch5-2024_06_14_033ee52ad856509d52d9g_0004_3.jpg", "batch5-2024_06_14_033ee52ad856509d52d9g_0004_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Because the shapes in option $\\mathrm{C}$ are all composed of triangles, and triangles possess stability, the shapes in option $\\mathrm{C}$ are stable. Therefore, option C is chosen." }, { "problem_id": 728, "question": "Among the following figures, which one is a right-angled triangle?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_78c5686324495f47dbfbg_0078_1.jpg", "batch5-2024_06_14_78c5686324495f47dbfbg_0078_2.jpg", "batch5-2024_06_14_78c5686324495f47dbfbg_0078_3.jpg", "batch5-2024_06_14_78c5686324495f47dbfbg_0078_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Slightly; briefly; roughly; to omit; to delete; strategy; plan; to capture; to seize. (Note: The translation of \"略\" depends on the context in which it is used. The above are some common translations.)" }, { "problem_id": 729, "question": "The plane unfolding of the following geometric figure ( ) is shown in the figure.\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_2eea509d719aacf4fdebg_0059_1.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0059_2.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0059_3.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0059_4.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0059_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Using the process of elimination, all elements in $\\mathrm{A}$ are rectangles, $\\mathrm{C}$ contains triangles, and $\\mathrm{D}$ includes circles. Therefore, the only option left is $\\mathrm{B}$." }, { "problem_id": 730, "question": "As shown in the figure, which of the following figures are all part of a cylinder?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_f5b8b25a104af99401e0g_0025_1.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0025_2.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0025_3.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0025_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. There is a triangular pyramid, which does not meet the requirement;\n\nB. There is an irregular polyhedron, which does not meet the requirement;\n\nC. They are respectively a cylinder and two quadrangular prisms;\n\nD. There is a truncated cone, which does not meet the requirement.\n\nTherefore, the correct answer is: C." }, { "problem_id": 731, "question": "As shown in Figure 1, this is the unfolded side view of a small cube. The small cube is flipped from the position shown in Figure 2 to the 1st, 2nd, and 3rd grids in sequence. At this time, the word on the top face of the small cube is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 共\nB. 建\nC. 实\nD. 外", "input_image": [ "batch25-2024_06_17_e5f356909807fe2710f3g_0016_1.jpg", "batch25-2024_06_17_e5f356909807fe2710f3g_0016_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the principle of \"alternate and $Z$-end faces are opposite\" in the surface unfolding diagram of a cube, the faces labeled \"Gong\" and \"Wai\" are opposite, the faces labeled \"Jian\" and \"Li\" are opposite, and the faces labeled \"Mei\" and \"Shi\" are opposite.\n\nAfter the first flip, \"Mei\" is on the bottom.\n\nAfter the second flip, \"Gong\" is on the bottom.\n\nAfter the third flip, the face on top is \"Jian\".\n\nTherefore, the correct answer is: B.\n\n[Key Point] This question tests the topic: Text on opposite faces of a cube. The key to solving it lies in understanding the surface unfolding diagram of a cube." }, { "problem_id": 732, "question": "Among the four three-dimensional figures below, which one has a circular front view?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0082_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0082_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0082_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0082_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Since the front view of a cylinder is a rectangle, the front view of a cone is an isosceles triangle, the front view of a sphere is a circle, and the front view of a cube is a square, it follows that the geometric solid whose front view is a circle is the sphere. Therefore, the correct choice is B.\n\nKey Point: Three views of simple geometric solids." }, { "problem_id": 733, "question": "As shown in Figure 1, it is the unfolded view of a side of a cube. The small cube moves from its position in Figure 2 to the 1st, 2nd, and 3rd grids successively. At this time, the word on the top face of the small cube is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 和\nB. 谐\nC. 社\nD. 会", "input_image": [ "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0034_1.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0034_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From Figure 1, it can be deduced that \"建\" corresponds to \"谐\"; \"和\" corresponds to \"社\"; and \"构\" corresponds to \"会\".\n\nFrom Figure 2, when the small cube flips from its position in Figure 2 to the third grid, \"构\" is on the bottom, which means the character on the top face of the small cube at this time is \"会\".\n\nTherefore, the correct choice is D." }, { "problem_id": 734, "question": "Among the solid-line figures below, which one is a semicircle?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0001_1.jpg", "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0001_2.jpg", "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0001_3.jpg", "batch19-2024_05_24_1a013bf8f56fa69ae2ffg_0001_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "A semicircle is the arc subtended by the diameter, but it does not include the diameter itself.\n\nTherefore, option $B$ is selected.\n\n【Key Point】This question primarily tests the basic properties of a circle, and the solution is based on a clear understanding of the definition of a semicircle." }, { "problem_id": 735, "question": "Among the following figures, which ones have opposite angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0045_1.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0045_2.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0045_3.jpg", "batch30-2024_06_17_4f8e209ac7e6ceea3279g_0045_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Based on the criterion that opposite angles are formed by the intersection of two straight lines, figures A, B, and D are not formed by the intersection of two straight lines, hence they are incorrect; figure C is formed by the intersection of two straight lines, therefore it is correct." }, { "problem_id": 736, "question": "Which of the following is an unfolding of a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_c04268d430149b44ad9ag_0100_1.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0100_2.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0100_3.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0100_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Based on the types of cube net diagrams \"141\", \"132\", \"33\", and \"222\", only option C satisfies the condition.\n\nTherefore, choose C.\n\n[Highlight] This question tests the understanding of cube net diagrams through unfolding and folding. Using common mnemonics can help in quickly determining the correct answer." }, { "problem_id": 737, "question": "As shown in the figure, after the surface of the cylinder is unfolded, the resulting flat shape is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_63d64c842022cf6d6051g_0034_1.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0034_2.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0034_3.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0034_4.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0034_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The lateral surface of a cylinder, when unfolded, forms a rectangular plane figure, with the top and bottom bases being two circles. Therefore, the correct choice is B." }, { "problem_id": 738, "question": "The three views and corresponding edge lengths of a geometric solid are shown in the figure. The area of the left view is ( )\n\n\n\nFront View\n\n\n\nLeft View\n\n\nA. 15\nB. 30\nC. 45\nD. 62", "input_image": [ "batch25-2024_06_17_f3b68e01e9dc18006b97g_0052_1.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0052_2.jpg", "batch25-2024_06_17_f3b68e01e9dc18006b97g_0052_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: By observing the figure, we can determine that the geometric shape is a rectangular prism with a length of 3, a width of 3, and a height of 5. The area of the left view is $3 \\times 5=15$.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question primarily tests the understanding of three-view drawings. Mastering the three views of geometric shapes is crucial for solving such problems." }, { "problem_id": 739, "question": "Among the following figures, which one's lateral surface is the net of a cylinder?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0065_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0065_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0065_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0065_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The lateral development of a cone is a smooth curved surface, which is a sector.\n\nTherefore, the correct choice is: A.\n\n[Key Point] This question examines the development of geometric solids, where the lateral development of a cone is a sector." }, { "problem_id": 740, "question": "The following figures are congruent figures ( )\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n", "input_image": [ "batch9-2024_05_23_fba44ff85c69ee94778dg_0067_1.jpg", "batch9-2024_05_23_fba44ff85c69ee94778dg_0067_2.jpg", "batch9-2024_05_23_fba44ff85c69ee94778dg_0067_3.jpg", "batch9-2024_05_23_fba44ff85c69ee94778dg_0067_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "A. The two figures are similar, incorrect;\n\nB. The two figures are congruent, correct;\n\nC. The two figures are similar, incorrect;\n\nD. The two figures are not congruent, incorrect;\n\nTherefore, choose B." }, { "problem_id": 741, "question": "As shown in the figure, in Figure 1 and Figure 2, the size relationship between the angles $\\alpha$ and $\\beta$ when the scissors are open is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\alpha>\\beta$\nB. $\\alpha<\\beta$\nC. $\\alpha=\\beta$\nD. Cannot be determined", "input_image": [ "batch32-2024_06_14_c63a15acc5199478291dg_0030_1.jpg", "batch32-2024_06_14_c63a15acc5199478291dg_0030_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, it can be seen that the two pairs of scissors are similar figures.\n\n$\\therefore \\alpha=\\beta$,\n\nTherefore, the correct choice is C.\n\n[Key Insight] This question tests the properties of similar figures, which is a basic concept. The key to solving the problem lies in determining that the two figures are similar." }, { "problem_id": 742, "question": "As shown in the figure, Xiaohua designed a brain teaser game: Six coins are arranged in a triangle (as shown in Figure 1). What is the minimum number of coins that need to be moved to form the circular arrangement shown in Figure 2?\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch26-2024_06_17_a2a41eecc90827e9e711g_0012_1.jpg", "batch26-2024_06_17_a2a41eecc90827e9e711g_0012_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the characteristics of a ring, each circle has exactly two adjacent circles. Therefore, it is necessary to move the coins labeled 1 and 2 from Figure 1 to the positions labeled 1 and 2 in Figure 2. This means that a minimum of 2 coins need to be moved to arrange them into the ring shape shown in Figure 2, as illustrated below.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\n\nThus, the correct answer is B.\n\n[Key Insight] This question tests the ability to explore patterns in graphical arrangements. The key to solving this problem lies in identifying the pattern based on the characteristics of a ring." }, { "problem_id": 743, "question": "As shown in the figure, when the left figure is folded, which of the following three-dimensional shapes on the right will it become ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0094_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0094_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0094_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0094_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0094_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Question Analysis: The adjacent face to the circular face is a rectangle, and the rectangle does not point towards the circle.\nTherefore, the correct choice is B." }, { "problem_id": 744, "question": "Using 10 sticks to form the pattern shown in Figure 1, please move 3 sticks to form the pattern shown in Figure 2. How many ways are there to move the sticks?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 1 way\nB. 2 ways\nC. 3 ways\nD. 4 ways", "input_image": [ "batch14-2024_06_15_00a109ed8135f00b577eg_0056_1.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0056_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (2) and Figure (3):\n(3)\n(5)\n(6)\n(7)\n(8)\n(9)\n(10)\n\n\nFigure 2\n\n\nFigure 3\n\nTherefore, there are two translation methods: (2)(4)(6) or (1)(8)(" }, { "problem_id": 745, "question": "Among the following figures, which one can be translated to form the entire figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0054_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0054_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0054_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0054_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Question Analysis: The entire figure is obtained by translating one of the shapes, which is option A, therefore the answer is A." }, { "problem_id": 746, "question": "The approximate graph of the inverse proportional function $y = \\frac{k}{x} (k < 0)$ is.\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_84512915d0d2c440e574g_0079_1.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0079_2.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0079_3.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0079_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "The graph of the inverse proportional function \\( y = \\frac{k}{x} \\) (where \\( k < 0 \\)) should be in the second and fourth quadrants. Therefore, the correct choice is B." }, { "problem_id": 747, "question": "Among the following patterns, which one is not an axisymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_7b577704b9b4a50255f1g_0050_1.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0050_2.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0050_3.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0050_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Question Analysis: Option A is not an axisymmetric figure, while options B, C, and D are all axisymmetric figures.\n\nTherefore, the correct choice is A.\n\nKey Point: An axisymmetric figure is one that can be completely overlapped with its mirror image when folded along a certain straight line." }, { "problem_id": 748, "question": "The following statement is incorrect ():\n\n A\n\n B\n A. A plane intersecting a sphere will always produce a circular cross-section.\nB. A plane intersecting a cube can produce a pentagonal cross-section.\nC. The cross-section of a prism cannot be a circle.\nD. In the figures A and B, only B can be folded into a cube.", "input_image": [ "batch33-2024_06_14_452fbe550559079ef90fg_0060_1.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0060_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Options A, B, and C are all correct. In option D, both Jia and Yi can be folded into a cube. Therefore, choose D." }, { "problem_id": 749, "question": "As shown in Figure 1, one of the four sides of a regular tetrahedron (with a square base and equal edges) is cut open, resulting in Figure 2. The four sides that could have been cut open are ( )\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\nA. PA, PB, AD, BC\nB. PD, DC, BC, AB\nC. PA, AD, PC, BC\nD. PA, PB, PC, AD", "input_image": [ "batch33-2024_06_14_9e6988a33fdf375f8541g_0080_1.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0080_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the unfolding characteristics of the prism, the four edges that could have been cut are possibly $PA, PB, AD, BC$.\n\nTherefore, the correct choice is A." }, { "problem_id": 750, "question": "As shown in the figure, after rotating Figure 1 around a certain point, we obtain Figure 2. Among the following rotation methods, which one satisfies the condition?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Clockwise rotation by $90^{\\circ}$\nB. Counterclockwise rotation by $90^{\\circ}$\nC. Clockwise rotation by $45^{\\circ}$\nD. Counterclockwise rotation by $45^{\\circ}$", "input_image": [ "batch26-2024_06_17_4ff1ec270366d069efb8g_0072_1.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0072_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the properties of rotation, we know that rotating Figure 1 clockwise by $90^{\\circ}$ will yield Figure 2.\n\nTherefore, the correct choice is: A\n[Key Insight] This question tests the properties of rotation; remembering the relevant knowledge points is crucial for solving the problem." }, { "problem_id": 751, "question": "In the figures below, which of the angles is a central angle of a circle?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_dd341a611d5d139e3378g_0005_1.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0005_2.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0005_3.jpg", "batch19-2024_05_24_dd341a611d5d139e3378g_0005_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Central Angle: An angle whose vertex lies on the circle and whose sides both intersect the circle is called a central angle. Observing the four options, only the angle in option B satisfies the definition.\n\nTherefore, the answer is: B.\n\n[Highlight] This question tests the understanding of central angles; mastering the definition is key to solving the problem." }, { "problem_id": 752, "question": "As shown in the figure, which of the following is not stable?\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_01becfa046433bf11e2fg_0092_1.jpg", "batch5-2024_06_14_01becfa046433bf11e2fg_0092_2.jpg", "batch5-2024_06_14_01becfa046433bf11e2fg_0092_3.jpg", "batch5-2024_06_14_01becfa046433bf11e2fg_0092_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Logic", "image_relavance": "0", "analysis": "Solution: A, C, and D are composed of multiple triangles, which provide stability. B is a quadrilateral and does not have stability. Therefore, the correct choice is B." }, { "problem_id": 753, "question": "As shown in the figure, in the following figures, those with vertical angles are ( )\n\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\nA. (1)\nB. (1)(2)\nC. (2)(4)\nD. (2)(3)", "input_image": [ "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0094_1.jpg", "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0094_2.jpg", "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0094_3.jpg", "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0094_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: According to the diagram, only figures (2) and (4) have vertical angles.\n\nTherefore, the correct choice is C." }, { "problem_id": 754, "question": "The front view and left view of a rectangular prism are shown in the figure. Determine the volume of the rectangular prism ( ).\n\n\n\nFront View\n\n\n\nLeft View\nA. 18\nB. 24\nC. 36\nD. 48", "input_image": [ "batch25-2024_06_17_c2b5a48971f4685260d0g_0083_1.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0083_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, the length of the cuboid is 4.\n\nFrom the side view, the height of the cuboid is 3, and the width is 2.\n\nThus, the volume of this cuboid is \\(3 \\times 2 \\times 4 = 24\\).\n\nTherefore, the volume of this cuboid is 24.\n\nHence, the correct choice is B.\n\n[Key Insight] This question tests the understanding of the three views of a cuboid. The key to solving this problem lies in the flexible application of the knowledge learned." }, { "problem_id": 755, "question": "The following groups of figures can be obtained from each other by translation ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_42999df1f8dd14bc18ffg_0022_1.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0022_2.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0022_3.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0022_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: By observing the figure, it can be seen that pattern $C$ can be obtained through translation.\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the understanding of the translation of figures. Mastering the properties of translation is crucial for solving the problem." }, { "problem_id": 756, "question": "As shown in the figure, the 3D geometric structure is made up of five small cubes. What is seen from its front view is ( ).\n\n\n\nFront View\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n", "input_image": [ "batch33-2024_06_14_c7449420eec16015971ag_0062_1.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0062_2.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0062_3.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0062_4.jpg", "batch33-2024_06_14_c7449420eec16015971ag_0062_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: When viewed from the front, the number of squares from left to right in the three columns is: 1, 1, 2. Therefore, the correct answer is C." }, { "problem_id": 757, "question": "Which of the following figures is a prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_9e6988a33fdf375f8541g_0036_1.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0036_2.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0036_3.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0036_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "A. It is a prism, hence incorrect;\n\nB. Correct:\n\nC. It is a sphere, hence incorrect;\n\nD. It is a cylinder, hence incorrect;\n\nTherefore, choose B.\n\n【Key Point】This question tests the understanding of solid geometry, with the key to solving it lying in the recognition of shapes." }, { "problem_id": 758, "question": "As shown in the figure, a three-dimensional figure is formed by six identical small cubes. If it changes from Figure (1) to Figure (2), which of the following remains unchanged ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. Front view\nB. Side view\nC. Top view\nD. Side view and Top view", "input_image": [ "batch25-2024_06_17_de12aee7c6c3dcedbe3bg_0019_1.jpg", "batch25-2024_06_17_de12aee7c6c3dcedbe3bg_0019_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "The main view consists of three squares on the first layer and one square on the left side of the second layer, hence option A is correct; therefore, the answer is: A.\n\n【Highlight】This question tests the knowledge of three-view drawings, where the main view is the view obtained by looking at the object from the front." }, { "problem_id": 759, "question": "As shown in the figure, these are the shadows of a tree and a flagpole at the same moment in the school. If the height of the tree is 3 meters and its shadow is 1.2 meters long, and the height of the flagpole is 5 meters, then the length of its shadow is ( ).\n\n\n\nFlagpole and Shadow\n\n\nA. 4 meters\nB. 2 meters\nC. 1.8 meters\nD. 3.6 meters", "input_image": [ "batch32-2024_06_14_b37cc082ac7e3fd7d5d9g_0009_1.jpg", "batch32-2024_06_14_b37cc082ac7e3fd7d5d9g_0009_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: Let the length of the flagpole's shadow be \\( x \\). According to the problem, the two figures are similar, so we have the proportion:\n\n\\[\n\\frac{3}{5} = \\frac{1.2}{x}\n\\]\n\nSolving for \\( x \\), we get:\n\n\\[\nx = 2 \\text{ meters}\n\\]\n\nAfter verification, \\( x = 2 \\) is indeed the solution to the original equation.\n\nTherefore, the correct choice is: **B**" }, { "problem_id": 760, "question": "The front view of the geometric solid shown in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0079_1.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0079_2.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0079_3.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0079_4.jpg", "batch25-2024_06_17_04aeb9c470b3dffc7fcbg_0079_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: The shape seen from the front corresponds to the figure in option B, hence B is correct.\n\nTherefore, the answer is: B.\n\n[Highlight] This question primarily tests the understanding of the three views of common geometric solids, with the front view being the shape seen from the front of the object." }, { "problem_id": 761, "question": "In the figures below, which pair of angles, $\\angle 1$ and $\\angle 2$, are adjacent supplementary angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch30-2024_06_17_94cae917628bb2f1b9fbg_0047_1.jpg", "batch30-2024_06_17_94cae917628bb2f1b9fbg_0047_2.jpg", "batch30-2024_06_17_94cae917628bb2f1b9fbg_0047_3.jpg", "batch30-2024_06_17_94cae917628bb2f1b9fbg_0047_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of adjacent supplementary angles, only the angles in diagram D are adjacent supplementary angles; the others are not.\n\nTherefore, the correct choice is D." }, { "problem_id": 762, "question": "Fold the plane figure in the diagram into a cube. The correct cube should be\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_9e6988a33fdf375f8541g_0086_1.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0086_2.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0086_3.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0086_4.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0086_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By observing the figure, it can be seen that the two faces with circled patterns are opposite each other, so options A and B are incorrect;\n\nIn option C, the position of the triangle is incorrect.\n\nTherefore, the correct choice is D." }, { "problem_id": 763, "question": "As shown in the figure, after cutting off a corner of the small cube and unfolding it, which of the following is the correct net?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_452fbe550559079ef90fg_0061_1.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0061_2.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0061_3.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0061_4.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0061_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Only $\\mathrm{D}$ can be restored, so choose $\\mathrm{D}$." }, { "problem_id": 764, "question": "The following image is formed by translating which of the patterns below ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_52ffba5930bb3584c0fcg_0038_1.jpg", "batch14-2024_06_15_52ffba5930bb3584c0fcg_0038_2.jpg", "batch14-2024_06_15_52ffba5930bb3584c0fcg_0038_3.jpg", "batch14-2024_06_15_52ffba5930bb3584c0fcg_0038_4.jpg", "batch14-2024_06_15_52ffba5930bb3584c0fcg_0038_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the properties of translation, the figure in option B conforms to the given condition.\n\nTherefore, the answer is: B.\n\n[Key Insight] This question examines the concept of translation in real-life scenarios. Understanding the properties of translation is essential for making correct judgments." }, { "problem_id": 765, "question": "A geometric solid is composed of several identical small cubes. When viewed from above and from the left, as shown in the figures, the maximum number of small cubes used to construct the solid is ( )\n\n\n\nView from above\n\n\n\nView from the left\nA. 8 cubes\nB. 10 cubes\nC. 12 cubes\nD. 13 cubes", "input_image": [ "batch25-2024_06_17_c2b5a48971f4685260d0g_0003_1.jpg", "batch25-2024_06_17_c2b5a48971f4685260d0g_0003_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we have the following diagram:\n\n| 3 | 3 | 3 |\n| :--- | :--- | :--- |\n| | 2 | 2 |\n\nAt this point, the maximum number of small cubes is: $3+3+3+2+2=13$;\n\nTherefore, the correct choice is: D.\n\n【Key Point】This question tests the ability to determine the number of small cubes based on the three-view drawing. Mastering the use of the top view to determine positions and the side view to determine quantities is crucial for solving the problem." }, { "problem_id": 766, "question": "As shown in the figure, the two triangles are congruent. The measure of angle $\\alpha$ is ( )\n\n\n\nb\n\n\nA. $58^{\\circ}$\nB. $72^{\\circ}$\nC. $50^{\\circ}$\nD. $60^{\\circ}$", "input_image": [ "batch10-2024_06_14_234bba88fa4a56aa2b46g_0023_1.jpg", "batch10-2024_06_14_234bba88fa4a56aa2b46g_0023_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since the two triangles are congruent,\n\ntherefore, $\\alpha=50^{\\circ}$.\n\nThus, the correct choice is C.\n\n[Key Insight] This question tests the properties of congruent triangles. Remembering these properties and accurately identifying the corresponding angles in the diagram are crucial for solving the problem." }, { "problem_id": 767, "question": "Which of the following figures is the side view unfolded of a square pyramid? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_c04268d430149b44ad9ag_0097_1.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0097_2.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0097_3.jpg", "batch25-2024_06_17_c04268d430149b44ad9ag_0097_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The side of a quadrilateral pyramid is a triangle, which indicates that option B is correct.\n\nTherefore, the answer is: B.\n\n[Key Point] This question tests the development diagram of a three-dimensional figure. Noting that the side of a quadrilateral pyramid is a triangle is the key to solving the problem." }, { "problem_id": 768, "question": "Among the three-dimensional figures below, which one has a sector as its lateral surface when unfolded?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_63d64c842022cf6d6051g_0056_1.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0056_2.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0056_3.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0056_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Based on the characteristics of a cone, the lateral development is a sector of a circle, which is a cone.\n\nTherefore, choose B." }, { "problem_id": 769, "question": "Among the following figures, which one is not a central symmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_844a73fd19ba25585b19g_0055_1.jpg", "batch26-2024_06_17_844a73fd19ba25585b19g_0055_2.jpg", "batch26-2024_06_17_844a73fd19ba25585b19g_0055_3.jpg", "batch26-2024_06_17_844a73fd19ba25585b19g_0055_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "According to the concept of centrally symmetric figures, A, B, and C are all centrally symmetric figures; whereas D is not a centrally symmetric figure.\nTherefore, D is the correct choice." }, { "problem_id": 770, "question": "Among the following four patterns, which one is not a central symmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0017_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0017_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0017_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0017_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the concept of centrally symmetric figures, figure B is not a centrally symmetric figure.\n\nTherefore, the answer is: B." }, { "problem_id": 771, "question": "The unfolding of a cylinder could be one of the following figures ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_1a01bcad4bed55b70b27g_0058_1.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0058_2.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0058_3.jpg", "batch25-2024_06_17_1a01bcad4bed55b70b27g_0058_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: The net of a cone consists of a sector and a circle.\n\nAs shown in the figure:\n\n\n\nTherefore, the correct answer is: C\n\n[Key Point] This question primarily tests the understanding of the net of simple geometric shapes, and the problem is relatively straightforward." }, { "problem_id": 772, "question": "By rotating the figure below around the straight line $l$, the resulting figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_452fbe550559079ef90fg_0029_1.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0029_2.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0029_3.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0029_4.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0029_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Question Analysis: The figure formed by rotating the given shape around the straight line for one full rotation is the figure shown in option D.\n\nTherefore, the answer is D.\n\nKey Points: Points, Lines, Surfaces, and Solids." }, { "problem_id": 773, "question": "In a $5 \\times 5$ grid, the position of the figure $N$ after translation is shown in Figure 2, as depicted in Figure 1. Which of the following methods for translating figure $N$ is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Move down 2 cells\nB. Move down 3 cells\nC. Move up 2 cells\nD. Move up 3 cells", "input_image": [ "batch14-2024_06_15_42999df1f8dd14bc18ffg_0045_1.jpg", "batch14-2024_06_15_42999df1f8dd14bc18ffg_0045_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: By observing the figures, it can be seen that from Figure 1 to Figure 2, figure $N$ can be moved downward by 3 units.\n\nTherefore, the correct choice is: B\n\n[Key Insight] This question primarily examines the basic concept and rules of translation, which is a relatively simple geometric transformation. The key to solving the problem lies in observing and comparing the positions of the figures before and after the translation." }, { "problem_id": 774, "question": "Among the following car logos, which can be obtained by translating some basic geometric shapes?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_90e5fb0e42c896d6f42eg_0051_1.jpg", "batch14-2024_06_15_90e5fb0e42c896d6f42eg_0051_2.jpg", "batch14-2024_06_15_90e5fb0e42c896d6f42eg_0051_3.jpg", "batch14-2024_06_15_90e5fb0e42c896d6f42eg_0051_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the properties of translation transformations, option A satisfies the condition.\n\nTherefore, the answer is: A.\n\n[Key Insight] This question examines the properties of translation. The key to solving the problem lies in a thorough understanding and application of these properties." }, { "problem_id": 775, "question": "As shown in the figure, the ( ) view of (1) and (2) is the same.\n\n\n\n(1)\n\n\n\n(2)\nA. Front view\nB. Left view\nC. Top view\nD. Left view, Top view", "input_image": [ "batch33-2024_06_14_9499ad9aa1a29e158e4fg_0049_1.jpg", "batch33-2024_06_14_9499ad9aa1a29e158e4fg_0049_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: First, when viewed from the front, the front views of the two are different.\n\nWhen viewed from the left, the left views of both consist of two squares arranged vertically, which are identical.\n\nWhen viewed from above, the top views of both consist of four squares arranged horizontally, which are also identical.\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question tests the understanding of the three views of a geometric solid. Remembering the definitions of the front view, left view, and top view is crucial for solving the problem." }, { "problem_id": 776, "question": "As shown in the figure, how can Figure 2 be obtained from Figure 1 ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Similarity transformation\nB. Rotation\nC. Reflection\nD. Translation", "input_image": [ "batch14-2024_06_15_00a109ed8135f00b577eg_0077_1.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0077_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Figure 2 is obtained by translating Figure 1 a certain distance to the right.\n\nTherefore, the correct choice is: D.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n[Key Point] This question primarily examines the translation of figures. The key to solving the problem lies in proficiently applying the properties of translation to make a judgment." }, { "problem_id": 777, "question": "Among the following figures, the one that is completely consistent with the pattern in the left figure is\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_7271d973a6f7af6646c5g_0026_1.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0026_2.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0026_3.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0026_4.jpg", "batch26-2024_06_17_7271d973a6f7af6646c5g_0026_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Two figures that can completely coincide are called congruent figures. The pattern does not match those in A, C, or D, but only matches the pattern in B. Therefore, choose B." }, { "problem_id": 778, "question": "As shown in the figure, it is a bordered template. Which of the following \"basic patterns\" is obtained by a single translation inside it ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0021_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0021_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0021_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0021_4.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0021_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: It is obtained by translating the second \"basic pattern\" once, hence the answer is B." }, { "problem_id": 779, "question": "The lateral surface expanded viewof a geometric solid is shown in the figure. Determine the base of the solid ( ).\n\n\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n", "input_image": [ "batch33-2024_06_14_9499ad9aa1a29e158e4fg_0094_1.jpg", "batch33-2024_06_14_9499ad9aa1a29e158e4fg_0094_2.jpg", "batch33-2024_06_14_9499ad9aa1a29e158e4fg_0094_3.jpg", "batch33-2024_06_14_9499ad9aa1a29e158e4fg_0094_4.jpg", "batch33-2024_06_14_9499ad9aa1a29e158e4fg_0094_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Based on the development diagram, the geometric shape is deduced to be a quadrangular prism with a quadrilateral base.\n\nTherefore, choose B.\n\n[Key Insight] Assessment focus: Three-view drawing of geometric shapes." }, { "problem_id": 780, "question": "Figure 1 consists of two identical rectangles. If the area of the shaded region in Figure 1 is 10, then the area of the shaded region in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 40\nB. 30\nC. 20\nD. 10", "input_image": [ "batch14-2024_06_15_82623ec71bb7bc51c88bg_0019_1.jpg", "batch14-2024_06_15_82623ec71bb7bc51c88bg_0019_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Question Analysis: According to the area formula of a triangle and the fact that the distance between two parallel lines is equal everywhere, it can be concluded that: In Figure 1, the sum of the areas of the two triangles is half the area of the rectangle; in Figure 2, the shaded area is also half the area of the rectangle, which is 10.\n\nTherefore, the correct answer is D.\n\nKey Point: The area of triangles with equal bases and equal heights." }, { "problem_id": 781, "question": "The square with the heart symbol \"$\\nabla$\" in Figure 1 forms part of the surface of the cube in Figure 2. In Figure 1, the square with the heart symbol \"$\\nabla$\" is on the face:\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $C D H E$\nB. $B C E F$\nC. $A B F G$\nD. $A D H G$", "input_image": [ "batch25-2024_06_17_63d64c842022cf6d6051g_0100_1.jpg", "batch25-2024_06_17_63d64c842022cf6d6051g_0100_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the red heart \"$\\checkmark$\" symbol in Figure 1, it can be seen that it is adjacent to the equilateral triangle. When folded into a cube, it becomes face $CDHE$ of the cube.\nTherefore, the correct choice is: A.\n\n[Key Point] Test Point: Folding a net into a geometric solid." }, { "problem_id": 782, "question": "Regarding the classification of triangles, there are two methods, as shown in Figure 1 (Method A) and Figure 2 (Method B). Then ( )\n\n\n\nMethod A\n\n\n\nMethod B\nA. Both Method A and Method B are correct\nB. Both Method A and Method B are incorrect\nC. Method A is incorrect, and Method B is correct\nD. Method A is correct, and Method B is incorrect", "input_image": [ "batch5-2024_06_14_85a87c4eb891206e633bg_0053_1.jpg", "batch5-2024_06_14_85a87c4eb891206e633bg_0053_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Method A is correct, and the correct classification for Method B should be:\n\n\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question tests the classification of triangles. The key to solving it lies in being familiar with the classification criteria for triangles, and it is easy to overlook that isosceles triangles include equilateral triangles." }, { "problem_id": 783, "question": "Which of the following figures is not an axisymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch35-2024_06_17_a5af53d889bd5704e3e0g_0007_1.jpg", "batch35-2024_06_17_a5af53d889bd5704e3e0g_0007_2.jpg", "batch35-2024_06_17_a5af53d889bd5704e3e0g_0007_3.jpg", "batch35-2024_06_17_a5af53d889bd5704e3e0g_0007_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "According to the definition of an axisymmetric figure, it is easy to obtain D." }, { "problem_id": 784, "question": "As shown in the figure, the figure rotates around the dotted line for a week, forming a geometric solid which is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_d6e7887b4e796c8b833cg_0040_1.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0040_2.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0040_3.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0040_4.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0040_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: The right triangle above forms a cone when rotated around its axis, and the rectangle below forms a cylinder when rotated around its axis.\n\nTherefore, the resulting shape is a combination of a cone and a cylinder.\n\nHence, the correct choice is B." }, { "problem_id": 785, "question": "The cross-section of the solid in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0074_1.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0074_2.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0074_3.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0074_4.jpg", "batch5-2024_06_14_346d3efaab3dcbfdd6e3g_0074_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Question Analysis: As can be seen from the diagram, the cross-section is a rectangle.\n\nTherefore, the correct choice is B." }, { "problem_id": 786, "question": "As shown in the figure, Xiaohua designed a brain teaser game: Six coins are arranged in a triangle (as shown in Figure 1). What is the minimum number of coins that need to be moved to form the circular arrangement shown in Figure 2?\n\n\n\n(figure 1)\n\n\n\n(figure 2)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch26-2024_06_17_201bada2148f276a9b39g_0093_1.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0093_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the characteristics of a circular arrangement, each circle has exactly two adjacent circles. Therefore, it is necessary to move the coins labeled 1 and 2 from Figure 1 to the positions labeled 1 and 2 in Figure 2. This means that a minimum of 2 coins need to be moved to arrange them into the circular pattern shown in Figure 2, as illustrated below:\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\n\nThus, the correct answer is B.\n\n[Key Insight] This question tests the ability to explore patterns in graphical arrangements. Understanding the characteristics of a circular arrangement is crucial to solving this problem." }, { "problem_id": 787, "question": "In the figure below, which pair of rays $O A$ and $O B$ represent the same ray?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_c720ff8294c46ce65b03g_0068_1.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0068_2.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0068_3.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0068_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: A ray is represented by two uppercase letters, with the letter denoting the endpoint written first. According to the definition, the correct choice for this question is B." }, { "problem_id": 788, "question": "Among the following graphs, which one represents the graph of an inverse proportion function?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_664d6e156f235987779ag_0048_1.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0048_2.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0048_3.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0048_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "A. The straight line passes through the origin, hence it is the graph of a direct proportion function;\n\nB. It is the graph of a linear function;\n\nC. It is a parabola, the graph of a quadratic function;\n\nD. It is a hyperbola, the graph of an inverse proportion function.\n\nTherefore, the correct choice is D." }, { "problem_id": 789, "question": "As shown in Figure ( A), there are four cards. If only one of the cards is rotated $180^{\\circ}$ to obtain Figure ( B), which card was rotated?\n\n\n\n( A)\n\n\n\n( B)\nA. The first card\nB. The second card\nC. The third card\nD. The fourth card", "input_image": [ "batch26-2024_06_17_4ff1ec270366d069efb8g_0043_1.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0043_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Upon careful observation of the diagram, it becomes evident that option D is the correct choice.\n\nTherefore, the answer is D." }, { "problem_id": 790, "question": "In the figure shown, which figure is both axisymmetric and centrosymmetric?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_83481c83c3572e9930e6g_0062_1.jpg", "batch26-2024_06_17_83481c83c3572e9930e6g_0062_2.jpg", "batch26-2024_06_17_83481c83c3572e9930e6g_0062_3.jpg", "batch26-2024_06_17_83481c83c3572e9930e6g_0062_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "According to the concepts of axisymmetric and centrally symmetric figures, options A, B, and C are both axisymmetric and centrally symmetric figures, while option D is an axisymmetric figure but not a centrally symmetric one. Therefore, the correct choice is D." }, { "problem_id": 791, "question": "As shown in Figure 1, there is a $2 \\times 5$ rectangular grid. Using the $1 \\times 2$ black rectangles as depicted in Figure 2 (allowing only one type), there are $($ different ways to fill the grid.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 7\nB. 8\nC. 9\nD. 10", "input_image": [ "batch14-2024_06_15_637191b7c30eeed99c12g_0078_1.jpg", "batch14-2024_06_15_637191b7c30eeed99c12g_0078_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\nThere are $1 + 4 + 3 = 8$ possibilities,\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question tests the ability to cut and rearrange shapes. The key to solving it lies in understanding the properties of the shapes and being able to think of different combinations. It is best to try them out practically." }, { "problem_id": 792, "question": "In the figures below, which one is not a possible net of a cylinder?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_9e6988a33fdf375f8541g_0096_1.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0096_2.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0096_3.jpg", "batch33-2024_06_14_9e6988a33fdf375f8541g_0096_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "In the lateral expansion diagram of a cone, the base circle should connect with the arc of the sector, not with the radius of the sector. Therefore, option A is selected." }, { "problem_id": 793, "question": "By the law of reflection of light, the incident ray and the reflected ray are symmetric about the normal (Figure 1). In Figure 2, the light ray enters from point $P$ and after reflection from the mirror $E F$, the point it passes through is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Point $\\mathrm{A}$\nB. Point $B$\nC. Point $C$\nD. Point $D$", "input_image": [ "batch14-2024_06_15_c672595237947d6cada1g_0007_1.jpg", "batch14-2024_06_15_c672595237947d6cada1g_0007_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Complete Figure 2 based on the properties of straight lines and draw the normal line $O K$, as shown in the following diagram:\n\n\n\nFigure 2\n\nFrom the diagram, it can be seen that $O B$ is the reflected ray,\n\nTherefore, the correct choice is: B.\n\n[Highlight] This question mainly examines the properties of axial symmetry and the method of drawing perpendicular lines. The key to completing the light ray is to derive equal angles based on the properties of axial symmetry." }, { "problem_id": 794, "question": "As shown in the figure, which of the following is the front view of the solid? \n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_c8bd428e774ea485a3ebg_0005_1.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0005_2.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0005_3.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0005_4.jpg", "batch28-2024_06_17_c8bd428e774ea485a3ebg_0005_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: The figure seen from the front of the geometric solid is\n\n\n\nTherefore, the correct choice is: D.\n\n[Highlight] This question tests the ability to view a geometric solid from different perspectives, and the problem is relatively simple." }, { "problem_id": 795, "question": "As shown in the figure, which of the line segments or rays can intersect?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_c720ff8294c46ce65b03g_0094_1.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0094_2.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0094_3.jpg", "batch33-2024_06_14_c720ff8294c46ce65b03g_0094_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Among the lines and rays in the figure, those that can intersect are\n\n\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question tests the understanding of lines, rays, and segments. Mastering the properties of segments and rays is crucial for solving this problem." }, { "problem_id": 796, "question": "After folding the plane figure shown into a cube, it could be ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0008_1.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0008_2.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0008_3.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0008_4.jpg", "batch33-2024_06_14_21d2414c9e0a3cf386ddg_0008_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Question Analysis: \n\nA. The two blank faces are adjacent, which does not match the original cube; \n\nB. It matches the original cube; \n\nD. After folding, the two solid circle patterns are adjacent, which does not match the original cube. \n\nTherefore, the correct answer is B." }, { "problem_id": 797, "question": "As shown in Figure 1, given segments $a$ and $b$, the segment $A B$ in Figure 2 represents ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $a-b$\nB. $a+b$\nC. $2 a-b$\nD. $a-2 b$", "input_image": [ "batch10-2024_06_14_55584d21b09f596e3cf6g_0092_1.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0092_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: The length of segment $AB$ in Figure 2 is $2a - b$,\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question tests the understanding of the distance between two points, and correctly interpreting the diagram is crucial for solving the problem." }, { "problem_id": 798, "question": "As shown in the figure, the top view of the workpiece is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_43010c41ad60ab0a5523g_0075_1.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0075_2.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0075_3.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0075_4.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0075_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: When viewed from above, it appears as concentric circles, with the outer circle being a solid line and the inner circle being a dashed line.\n\nTherefore, the correct choice is: B." }, { "problem_id": 799, "question": "Viewing the solid in Figure 1 from a certain direction, if the resulting view is Figure 2, then this direction is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Above\nB. Left\nC. Above or Front\nD. Left or Front", "input_image": [ "batch25-2024_06_17_756a44c095a36b650569g_0053_1.jpg", "batch25-2024_06_17_756a44c095a36b650569g_0053_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: According to the definition of three-view drawings, the view obtained from either the left or the front side is\n\n\n\nThe view obtained from the top is\n\n\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question tests the understanding of three-view drawings of simple composite solids. Mastering the definition of three-view drawings is crucial for solving the problem." }, { "problem_id": 800, "question": "As shown in the figure, which of the following is the correct front view of the component?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_3b4acb0b083caafa0f44g_0054_1.jpg", "batch25-2024_06_17_3b4acb0b083caafa0f44g_0054_2.jpg", "batch25-2024_06_17_3b4acb0b083caafa0f44g_0054_3.jpg", "batch25-2024_06_17_3b4acb0b083caafa0f44g_0054_4.jpg", "batch25-2024_06_17_3b4acb0b083caafa0f44g_0054_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "The main view is obtained by looking from the front, resulting in\n\n\n\nTherefore, the answer is: D\n\n[Highlight] Assessment knowledge point: Three views. Understand the significance of the three views and pay attention to the angle of observation." }, { "problem_id": 801, "question": "(2017・Yancheng) Among the following figures, which ones are axisymmetric?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch28-2024_06_17_7b577704b9b4a50255f1g_0068_1.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0068_2.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0068_3.jpg", "batch28-2024_06_17_7b577704b9b4a50255f1g_0068_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Question Analysis: The figure of D is folded along the central line, and the parts on both sides of the line can coincide, therefore D is the correct choice." }, { "problem_id": 802, "question": "The incorrect perspective drawing is ( )\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_c861aa08a70db5000978g_0076_1.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0076_2.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0076_3.jpg", "batch25-2024_06_17_c861aa08a70db5000978g_0076_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "The sun's rays are parallel and cannot intersect, therefore option B is incorrect." }, { "problem_id": 803, "question": "In triangle $EFG$, $\\angle G = 90^\\circ$, $EG = FG = 2\\sqrt{2}$. Square $ABCD$ has a side length of $1$, and $AD$ is on the same line as $EF$. Point $A$ coincides with point $E$. Now, square $ABCD$ moves along the direction of $EF$ at a constant speed of $1$ unit per second. The area $S$ of the overlapping part between square $ABCD$ and triangle $EFG$ and the function graph of $S$ with respect to the movement time $t$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_59ecf865bdf1009ec380g_0007_1.jpg", "batch8-2024_06_14_59ecf865bdf1009ec380g_0007_2.jpg", "batch8-2024_06_14_59ecf865bdf1009ec380g_0007_3.jpg", "batch8-2024_06_14_59ecf865bdf1009ec380g_0007_4.jpg", "batch8-2024_06_14_59ecf865bdf1009ec380g_0007_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since \\( EG = FG = 2\\sqrt{2} \\) and \\( \\angle G = 90^\\circ \\),\n\nby the Pythagorean theorem, we have \\( EF = 5 \\).\n\n(1) When \\( 0 \\leq t \\leq 1 \\), as shown in Figure 1,\n\nthen \\( AE = t = AH \\),\n\n\n\nFigure 1\n\n\\( S = \\frac{1}{2} \\times AE \\times AH = \\frac{1}{2} t^{2} \\), the function is an upward-opening parabola, and when \\( t = 1 \\), \\( S = \\frac{1}{2} \\);\n\n(2) When \\( 1 < t \\leq 2 \\), as shown in Figure 2, let \\( EG \\) intersect \\( CD \\) at point \\( H \\), and \\( BC \\) intersect \\( EG \\) at point \\( G \\),\n\n\n\nFigure 2\n\nthen \\( ED = AE - AD = t - 1 = HD \\), so \\( CH = CD - HD = 2 - t = CG \\),\n\n\\( S = S_{\\text{square }} ABCD - S_{\\triangle} CGH = 1 - \\frac{1}{2} \\times CH \\times CG = 1 - \\frac{1}{2}(2 - t)^{2} \\), the function is a downward-opening parabola, and when \\( t = 2 \\), \\( S = 1 \\);\n\n(3) When \\( 2 < t \\leq 3 \\), as shown in Figure 3,\n\n\n\nFigure 3\n\n\\( S = S_{\\text{square }} ABCD = 1 \\),\n\n(4) When \\( 3 < t \\leq 4 \\), as shown in Figure 4, let \\( AB \\) and \\( BC \\) intersect \\( FG \\) at points \\( N \\) and \\( M \\) respectively,\n\n\n\nFigure 4\n\nthen \\( AF = 4 - t = AN \\)\n\n\\(\\therefore BN = BM = AB - AN = 1 - (4 - t) = t - 3 \\)\n\n\\(\\therefore S = S_{\\text{square }} ABCD - S_{\\triangle} BMN = 1 - \\frac{1}{2} \\times BM \\times BN = 1 - \\frac{1}{2}(t - 3)^{2} \\)\n\nThe function is a downward-opening parabola, and when \\( t = 4 \\), \\( S = \\frac{1}{2} \\)\n\nTherefore, the answer is: C.\n\n【Key Insight】This question examines the graph and properties of quadratic functions. Understanding the pattern of the overlapping parts of the square and triangle during the movement and then classifying the discussion is the key and difficulty in solving the problem." }, { "problem_id": 804, "question": "As shown in the figure, $\\triangle A B C$ is an equilateral triangle with $A B=6 \\mathrm{~cm}$. Point $M$ starts from point $C$ and moves along $C B$ at a constant speed of $1 \\mathrm{~cm} / \\mathrm{s}$ until it reaches point $B$. At the same time, point $N$ starts from point $C$ and moves along ray $C A$ at a constant speed of $2 \\mathrm{~cm} / \\mathrm{s}$. When point $M$ stops, point $N$ also stops. A line $M P$ parallel to $C A$ is drawn through point $M$ intersecting $A B$ at point $P$. The segments $M N$ and $N P$ are drawn, and the triangle $\\triangle M N P$ is reflected about the line $M P$ to form $\\triangle M N^{\\prime} P$. Let the time of motion be $t \\mathrm{~s}$, and the area of the overlapping part of $\\triangle M N^{\\prime} P$ and $\\triangle B M P$ be $S \\mathrm{~cm}^{2}$. The approximate graph of the function relationship between $S$ and $t$ is ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_b93c19209f895a7f6de9g_0020_1.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0020_2.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0020_3.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0020_4.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0020_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: As shown in Figure 1, when point $N^{\\prime}$ falls on $AB$, take the midpoint $T$ of $CN$ and connect $MT$.\n\n\n\nFigure 1\n\n$\\because CM = t, \\quad CN = 2t, \\quad CT = TN$\n\n$\\therefore CT = TN = t$,\n\n$\\because \\triangle ABC$ is an equilateral triangle,\n$\\therefore \\angle C = \\angle A = 60^{\\circ}$,\n\n$\\therefore \\triangle MCT$ is an equilateral triangle,\n\n$\\therefore TM = TC = TN$,\n\n$\\therefore \\angle CMN = 90^{\\circ}$,\n\n$\\because MP \\parallel AC$,\n\n$\\therefore \\angle BPM = \\angle A = \\angle MPN = 60^{\\circ}, \\angle BMP = \\angle C = 60^{\\circ}, \\angle C + \\angle CMP = 180^{\\circ}$,\n\n$\\therefore \\angle CMP = 120^{\\circ}, \\triangle BMP$ is an equilateral triangle,\n\n$\\therefore BM = MP$,\n\n$\\because \\angle CMP + \\angle MPN = 180^{\\circ}$,\n\n$\\therefore CM \\parallel PN$,\n\n$\\because MP \\parallel CN$,\n\n$\\therefore$ quadrilateral $CMPN$ is a parallelogram,\n\n$\\therefore PM = CN = BM = 2t$,\n\n$\\therefore 3t = 6$,\n\n$\\therefore t = 2$,\n\nAs shown in Figure 2, when $0 < t \\leq 2$, draw $MK \\perp AC$ at point $K$, then $MK = CM \\cdot \\sin 60^{\\circ} = \\frac{\\sqrt{3}}{2} t$,\n\n\n\nFigure 2\n\n$\\therefore S = \\frac{1}{2} \\cdot (6 - t) \\cdot \\frac{\\sqrt{3}}{2} t = -\\frac{\\sqrt{3}}{4} t^{2} + \\frac{\\sqrt{3}}{2} t$.\n\nAs shown in Figure 3, when $2 < t \\leq 3$, $S = \\frac{1}{2} \\times \\frac{\\sqrt{3}}{4} (6 - t)^{2}$,\n\n\n\nFigure 3\n\nObserving the graph, it can be seen that option $A$ fits the description,\n\nTherefore, the answer is: $A$.\n\n【Insight】This problem examines the properties of moving points, equilateral triangles, and quadratic functions. The key to solving the problem lies in using a categorical discussion approach, making it a challenging multiple-choice question in the high school entrance exam." }, { "problem_id": 805, "question": "As shown in Figures (1), (2), (3), and (4), a paper triangle $A B C$ with equal sides $A B=A C$ is folded twice. First, $A B$ is folded onto $A C$, with the crease being $A D$. The paper is then flattened. Next, point $A$ is folded onto point $C$, with the crease being $E F$. The paper is flattened again, and the lines $A D$ and $E F$ intersect at point $G$. If $A B=A C=5 \\mathrm{~cm}$ and $B C=6 \\mathrm{~cm}$, what is the length of $D G$? ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\nA. $\\frac{3}{4} \\mathrm{~cm}$\nB. $\\frac{7}{8} \\mathrm{~cm}$\nC. $1 \\mathrm{~cm}$\nD. $\\frac{7}{6} \\mathrm{~cm}$", "input_image": [ "batch37-2024_06_14_130a72fe01bab086139fg_0047_1.jpg", "batch37-2024_06_14_130a72fe01bab086139fg_0047_2.jpg", "batch37-2024_06_14_130a72fe01bab086139fg_0047_3.jpg", "batch37-2024_06_14_130a72fe01bab086139fg_0047_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the properties of folding, we have:\n$AD$ is the perpendicular bisector of $BC$, and $EF$ is the perpendicular bisector of $AC$.\n\n$\\therefore CD=3 \\mathrm{~cm}, CF=\\frac{5}{2} \\mathrm{~cm}$.\n\nIn right triangle $\\triangle ADC$, $AD=\\sqrt{AC^{2}-DC^{2}}=4 \\mathrm{~cm}$.\n\n$\\therefore \\sin \\angle CAD=\\frac{CD}{AC}=\\frac{3}{5}, \\tan \\angle CAD=\\frac{CD}{AD}=\\frac{3}{4}$.\n\nSince $\\angle C + \\angle CAD = 90^{\\circ}$ and $\\angle C + \\angle CEF = 90^{\\circ}$,\n\n$\\therefore \\angle CAD = \\angle CEF$.\n\nIn right triangle $\\triangle CFE$, $CE = CF \\div \\sin \\angle CEF = \\frac{5}{2} \\div \\frac{3}{5} = \\frac{25}{6} \\mathrm{~cm}$.\n\n$\\therefore DE = CE - DC = \\frac{25}{6} - 3 = \\frac{7}{6} \\mathrm{~cm}$.\n\nIn right triangle $\\triangle EDG$, $DG = DE \\cdot \\tan \\angle DEG = \\frac{7}{6} \\times \\frac{3}{4} = \\frac{7}{8} \\mathrm{~cm}$.\n\nTherefore, the answer is: B.\n【Key Insight】This problem examines the properties of folding and solving right triangles; mastering the sine and tangent trigonometric functions is crucial for solving it." }, { "problem_id": 806, "question": "As shown in Figure 1, a square paper sheet $A B C D$ has a side length of 2. Fold $\\angle B$ and $\\angle D$ so that the vertices of the two right angles overlap at a point $P$ on the diagonal $B D$. $E F$ and $G H$ are the creases (as shown in Figure 2). Given that $A E = x (0 < x < 2)$, the following statements are made: (1) When $x = \\frac{3}{2}$, $E F + A B > A C$; (2) The perimeter of hexagon $A E F C H G$ is a constant value; (3) The area of hexagon $A E F C H G$ is a constant value. Which of these statements are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. (1)(2)\nB. (1)(3)\nC. (2)\nD. (2)(3)", "input_image": [ "batch22-2024_06_14_77881459dbacbd0f1d19g_0030_1.jpg", "batch22-2024_06_14_77881459dbacbd0f1d19g_0030_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Because of folding,\n\nTherefore, \\( \\mathrm{BE} = \\mathrm{EP} \\), \\( \\mathrm{BF} = \\mathrm{PF} \\), and \\( \\angle \\mathrm{ABC} = \\angle \\mathrm{EPF} = 90^\\circ \\).\n\nSince \\( \\mathrm{BD} \\) bisects \\( \\angle \\mathrm{ABC} \\) and \\( \\mathrm{EF} \\) is the perpendicular bisector of \\( \\mathrm{BP} \\),\n\nTherefore, \\( \\mathrm{BE} = \\mathrm{BF} \\),\n\nThus, quadrilateral \\( \\mathrm{BEPF} \\) is a rhombus, and \\( \\angle \\mathrm{EBF} = 90^\\circ \\),\n\nTherefore, quadrilateral \\( \\mathrm{BEPF} \\) is a square.\n\nSimilarly, quadrilateral \\( \\mathrm{PGDH} \\) is a square.\n\nTherefore, \\( \\angle \\mathrm{AGP} = 90^\\circ \\), \\( \\angle \\mathrm{AEP} = 90^\\circ \\),\n\nThus, quadrilateral \\( \\mathrm{AEPG} \\) is a rectangle.\n\nSimilarly, quadrilateral \\( \\mathrm{CFPH} \\) is a rectangle.\n\nTherefore, \\( \\mathrm{AE} = \\mathrm{PG} = \\mathrm{GD} = \\mathrm{DH} = \\mathrm{PH} = \\mathrm{FC} \\), and \\( \\mathrm{BE} = \\mathrm{BF} = \\mathrm{EP} = \\mathrm{PF} = \\mathrm{AG} = \\mathrm{CH} \\).\n\nWhen \\( x = \\frac{3}{2} \\), then \\( \\mathrm{BE} = \\frac{1}{2} \\),\n\nTherefore, \\( EF = \\frac{\\sqrt{2}}{2} \\),\n\nThus, \\( \\mathrm{AB} + \\mathrm{EF} = 2 + \\frac{\\sqrt{2}}{2} \\),\n\nSince \\( \\mathrm{AB} = \\mathrm{BC} = 2 \\),\n\nTherefore, \\( \\mathrm{AC} = 2\\sqrt{2} \\),\n\nThus, \\( \\mathrm{AB} + \\mathrm{EF} < \\mathrm{AC} \\),\n\nHence, statement (1) is incorrect.\n\nSince the perimeter of hexagon \\( \\mathrm{AEFCHG} \\) is \\( \\mathrm{AE} + \\mathrm{AG} + \\mathrm{CH} + \\mathrm{CF} + \\mathrm{EF} + \\mathrm{GH} = \\mathrm{AE} + \\mathrm{BE} + \\mathrm{CF} + \\mathrm{BF} + \\sqrt{2} \\mathrm{BE} + \\sqrt{2} \\mathrm{AE} \\),\n\nTherefore, the perimeter of hexagon \\( \\mathrm{AEFCHG} \\) is \\( \\mathrm{AB} + \\mathrm{BC} + \\sqrt{2} (\\mathrm{AE} + \\mathrm{BE}) = 4 + 2\\sqrt{2} \\), which is a constant value.\n\nHence, statement (2) is correct.\n\nSince the area of hexagon \\( \\mathrm{AEFCHG} \\) is \\( 2 \\times 2 - \\frac{1}{2} \\mathrm{BE}^2 - \\frac{1}{2} \\mathrm{GD}^2 = 4 - \\frac{1}{2} (\\mathrm{EP}^2 + \\mathrm{AE}^2) = 4 - \\frac{1}{2} \\mathrm{EG}^2 \\),\n\nTherefore, the area of hexagon \\( \\mathrm{AEFCHG} \\) is not a constant value.\n\nHence, statement (3) is incorrect.\n\nTherefore, the correct choice is:\n【Highlight】This question examines folding problems, the properties and determination of squares, and finding the relationships between line segments is key to solving this problem." }, { "problem_id": 807, "question": "As shown in the figure, rectangle $O A B C$ has its vertices $A$ and $C$ on the $x$-axis and $y$-axis, respectively. $O A = 4$ and $O C = 3$. Line $m: y = -\\frac{3}{4} x$ starts from the origin $O$ and moves along the positive direction of the $x$-axis at a speed of 1 unit per second. Let line $m$ intersect the sides of rectangle $O A B C$ at points $M$ and $N$. The time that line $m$ moves is $t$ (seconds). Let the area of triangle $O M N$ be $S$. Which of the following graphs approximately represents the functional relationship between $S$ and $t$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_0d247365a88bfbce6c4bg_0019_1.jpg", "batch32-2024_06_14_0d247365a88bfbce6c4bg_0019_2.jpg", "batch32-2024_06_14_0d247365a88bfbce6c4bg_0019_3.jpg", "batch32-2024_06_14_0d247365a88bfbce6c4bg_0019_4.jpg", "batch32-2024_06_14_0d247365a88bfbce6c4bg_0019_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in Figure 1, when \\(0 < t \\leq 4\\),\n\n\n\nSince \\(MN \\parallel CA\\),\n\nTherefore, \\(OM : OA = ON : OC\\),\n\nThus, \\(OM : ON = OA : OC = 4 : 3\\),\n\nHence, \\(OM = t\\), \\(ON = \\frac{3}{4} t\\),\n\nTherefore, \\(y = \\frac{1}{2} \\cdot OM \\cdot ON = \\frac{3}{8} t^{2}\\).\n\nAs shown in Figure 2, when \\(4 < t \\leq 8\\),\n\n\n\nFigure 2\n\n\\(y = S_{\\triangle EOF} - S_{\\triangle EON} - S_{\\triangle OFM} = \\frac{3}{8} t^{2} - \\frac{1}{2} \\cdot \\frac{3}{4} t \\cdot (t - 4) - \\frac{1}{2} \\cdot t \\cdot \\frac{3}{4} (t - 4) = -\\frac{3}{8} t^{2} + 3 t\\).\n\nIn summary, \\(y = \\left\\{\\begin{array}{l}\\frac{3}{8} t^{2} \\quad (0 < t \\leqslant 4) \\\\ -\\frac{3}{8} t^{2} + 3 t \\quad (4 < t \\leqslant 8)\\end{array}\\right.\\).\n\nTherefore, the correct choice is \\(D\\).\n\n【Key Insight】This question examines the function image of moving points, properties of rectangles, and the area of triangles. The key to solving the problem lies in learning to classify and discuss, deriving the piecewise function's analytical expression, which is a common type of question in middle school exams." }, { "problem_id": 808, "question": "In ancient China, the mathematician Zhao Shang created a diagram to prove the Pythagorean Theorem, known as \"Zhao Shang's String Diagram\" (as shown in Figure 1). Figure 2 is derived from the string diagram, formed by joining two congruent right-angled triangles. Let the areas of the squares ABCD, EFGH, and MNKT be denoted as $S_1$, $S_2$, and $S_3$, respectively. If $S_1 + S_2 + S_3 = 10$, what is the value of $S_2$?\n\n\nFigure 1\n\n\n\nFigure 2\nA. 5\nB. $\\frac{10}{3}$\nC. $\\frac{25}{4}$\nD. 4", "input_image": [ "batch12-2024_06_15_8372d55620e8759965f8g_0054_1.jpg", "batch12-2024_06_15_8372d55620e8759965f8g_0054_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Let the area of quadrilateral MNKT be \\( \\mathrm{x} \\), and the area of each of the other eight congruent triangles be \\( \\mathrm{y} \\).\n\nGiven that the areas of squares \\( \\mathrm{ABCD} \\), \\( \\mathrm{EFGH} \\), and MNKT are \\( \\mathrm{S}_{1} \\), \\( \\mathrm{S}_{2} \\), and \\( \\mathrm{S}_{3} \\) respectively, and \\( \\mathrm{S}_{1} + \\mathrm{S}_{2} + \\mathrm{S}_{3} = 10 \\),\n\nwe can derive:\n\\[\n\\mathrm{S}_{1} = 8\\mathrm{y} + \\mathrm{x}, \\quad \\mathrm{S}_{2} = 4\\mathrm{y} + \\mathrm{x}, \\quad \\mathrm{S}_{3} = \\mathrm{x}.\n\\]\n\nTherefore:\n\\[\n\\mathrm{S}_{1} + \\mathrm{S}_{2} + \\mathrm{S}_{3} = 3\\mathrm{x} + 12\\mathrm{y} = 10,\n\\]\nwhich simplifies to:\n\\[\n\\mathrm{x} + 4\\mathrm{y} = \\frac{10}{3}.\n\\]\n\nThus:\n\\[\nS_{2} = \\mathrm{x} + 4\\mathrm{y} = \\frac{10}{3}.\n\\]\n\nThe correct answer is **B**.\n\n**Key Insight**: This problem tests the understanding of the geometric background of the Pythagorean theorem. The key to solving it lies in analyzing the area relationships between the squares and triangles." }, { "problem_id": 809, "question": "Fold and flatten a rectangular sheet of paper with a length of 2 and a width of $a$ (where $a$ is greater than 1 and less than 2) as shown in Figure (1) and cut out a square with a side length equal to the width of the rectangle, which is called the first operation. Then, fold and flatten the remaining rectangle as shown in Figure (2) and cut out a square with a side length equal to the width of the rectangle at that time, which is called the second operation. Repeat this process... If after the $\\mathrm{n}^{\\text {th }}$ operation, the remaining rectangle is a square, the operation terminates. When $\\mathrm{n}=3$, the value of $a$ is $(\\quad)$.\n\n\n\n(1)\n\n\n\n(2)\nA. 1.8 or 1.5\nB. 1.5 or 1.2\nC. 1.5\nD. 1.2", "input_image": [ "batch35-2024_06_17_5a4127c6911cd24b9845g_0086_1.jpg", "batch35-2024_06_17_5a4127c6911cd24b9845g_0086_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: In the first operation, the side length of the cut-out square is \\( a \\), and the remaining rectangle has length and width of \\( a \\) and \\( 2a \\) respectively. Given that \\( 1 < a < 2 \\), it follows that \\( a > 2 - a \\). In the second operation, the side length of the cut-out square is \\( 2 - a \\), so the remaining rectangle has sides of \\( 2 - a \\) and \\( a - (2 - a) = 2a - 2 \\).\n\n(1) When \\( 2a - 2 < 2 - a \\), i.e., \\( a < \\frac{4}{3} \\), then in the third operation, the side length of the cut-out square is \\( 2a - 2 \\), and the remaining rectangle has sides of \\( 2a - 2 \\) and \\( (2 - a) - (2a - 2) = 4 - 3a \\). Setting \\( 2a - 2 = 4 - 3a \\) gives \\( a = 1.2 \\).\n\n(2) When \\( 2a - 2 > 2 - a \\), i.e., \\( a > \\frac{4}{3} \\), then in the third operation, the side length of the cut-out square is \\( 2 - a \\), and the remaining rectangle has sides of \\( 2 - a \\) and \\( (2a - 2) - (2 - a) = 3a - 4 \\). Setting \\( 2 - a = 3a - 4 \\) gives \\( a = 1.5 \\).\n\nTherefore, the correct answer is: B.\n\n**Key Insight:** This problem tests knowledge of numerical patterns, geometric patterns, and linear equations, incorporating the method of classification and discussion. It is comprehensive and moderately challenging. Careful analysis to identify the pattern and mastery of the relevant knowledge are crucial for solving the problem." }, { "problem_id": 810, "question": "As shown in the figure, the side length of regular hexagon $A B C D E F$ is $6 \\mathrm{~cm}$. Point $P$ is a moving point on the diagonal $B E$. A line $l$ perpendicular to $B E$ is drawn through point $P$. Point $P$ moves from point $B$ at a constant speed of $1 \\mathrm{~cm} / \\mathrm{s}$ to point $E$. Let the area of the region of the regular hexagon swept by line $l$ be $S \\mathrm{~cm}^{2}$, and the time of point $P$'s movement be $t(\\mathrm{~s})$. Which of the following graphs approximately represents the functional relationship between $S$ and $t$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_3eec3ce0215c5cefc697g_0038_1.jpg", "batch19-2024_05_24_3eec3ce0215c5cefc697g_0038_2.jpg", "batch19-2024_05_24_3eec3ce0215c5cefc697g_0038_3.jpg", "batch19-2024_05_24_3eec3ce0215c5cefc697g_0038_4.jpg", "batch19-2024_05_24_3eec3ce0215c5cefc697g_0038_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "As shown in the figure, when \\(0 \\leq t \\leq 3\\):\n\nSince hexagon \\(ABCDEF\\) is a regular hexagon,\n\n\\(\\angle ABE = 60^\\circ\\),\n\nGiven \\(BP = t\\), and \\(\\tan \\angle ABE = PG : BP\\),\n\nThus, \\(PG = \\sqrt{3} t\\),\n\nTherefore, \\(S = \\frac{1}{2} BP \\times GH = \\frac{1}{2} t \\times \\sqrt{3} t \\times 2 = \\sqrt{3} t^{2}\\), which is a parabola opening upwards,\n\nHence, neither \\(A\\) nor \\(B\\) fits the context;\n\nAs shown in the figure, when \\(3 < t \\leq 9\\):\n\nSince hexagon \\(ABCDEF\\) is a regular hexagon,\n\n\\(\\angle ABE = 60^\\circ\\), \\(AC = 6 \\sqrt{3}\\), and \\(AM = t - 3\\),\n\nThus, \\(S = S_{ABC} + AC \\times AM = (t - 3) \\times 6 \\sqrt{3} + 9 \\sqrt{3} = 6 \\sqrt{3} t - 9 \\sqrt{3}\\), which is a line segment sloping from high on the right to low on the left;\n\nHence, both \\(C\\) and \\(D\\) fit the context;\n\nAs shown in the figure, when \\(9 < t \\leq 12\\):\n\nSince hexagon \\(ABCDEF\\) is a regular hexagon,\n\n\\(\\angle QEP = 60^\\circ\\), and \\(PE = 12 - t\\),\n\nGiven \\(PE = 12 - t\\), and \\(\\tan \\angle QEP = PQ : PE\\),\n\nThus, \\(PQ = \\sqrt{3}(12 - t)\\),\n\nTherefore, \\(S = S_{\\text{hexagon } ABCDEF} - \\frac{1}{2} PE \\times QR = \\frac{1}{2} \\times 2 \\times (12 + 6) \\times 3 \\sqrt{3} - \\frac{1}{2} \\times \\sqrt{3} \\times (12 - t) \\times (12 - t) \\times 2\\)\n\n\\(= -\\sqrt{3} t^{2} + 24 \\sqrt{3} t - 90 \\sqrt{3}\\), hence the graph is a parabola opening downwards,\n\nThus, \\(C\\) fits the context, but \\(D\\) does not;\n\nTherefore, the correct choice is \\(C\\).\n\n【Key Insight】This question examines the properties of a regular hexagon, the analytical expression and graph of a quadratic function, the method of calculating area by partitioning shapes, and trigonometric functions of special angles. Mastery of classification thinking and accurate determination of area expressions are crucial for solving the problem." }, { "problem_id": 811, "question": "As shown in the figure, in quadrilateral $A B C D$, $A D \\parallel B C, \\angle A=45^\\circ, \\angle C=90^\\circ, A D=4 \\text{ cm}, C D=3 \\text{ cm}$. Points $M$ and $N$ start simultaneously from point $A$. Point $M$ moves at a speed of $\\sqrt{2} \\text{ cm/s}$ along $A B$ towards endpoint $B$, while point $N$ moves at a speed of $2 \\text{ cm/s}$ along the polyline $A D-D C$ towards endpoint $C$. Let the time of point $N$'s movement be $t \\text{ s}$, and the area of triangle $A M N$ be $S \\text{ cm}^2$. Which of the following graphs can roughly represent the function relationship between $S$ and $t$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_be77b09795d7e25ae1bag_0052_1.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0052_2.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0052_3.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0052_4.jpg", "batch8-2024_06_14_be77b09795d7e25ae1bag_0052_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: $\\angle \\mathrm{A}=45^{\\circ}$, $\\mathrm{CD}=3 \\mathrm{~cm}$,\n\n$\\mathrm{AB}=\\sqrt{3^{2}+3^{2}}=3 \\sqrt{2} \\mathrm{~cm}$\n\n$\\therefore \\mathrm{M}$ takes 3 seconds to move from $\\mathrm{A}$ to $\\mathrm{B}$, $\\mathrm{N}$ takes 2 seconds to move from $\\mathrm{A}$ to $\\mathrm{D}$, and 3.5 seconds to reach $\\mathrm{C}$.\n\nWe will discuss the following three scenarios:\n\n(1) When $N$ is on $A D$, i.e., $0\n\nFigure 1\n\nDraw $\\mathrm{ME} \\perp \\mathrm{AD}$ at $\\mathrm{E}$,\n\nWe know $\\mathrm{AN}=2 \\mathrm{t}$, $\\mathrm{AM}=\\sqrt{2} t$,\n\n$\\therefore \\mathrm{EM}=\\mathrm{t}$,\n\n$\\therefore s=\\frac{1}{2} A N \\cdot M E=\\frac{1}{2} \\times 2 t \\cdot t=t^{2}$\n\nThus, this segment of the graph is an upward-opening parabola;\n\n(2) When $N$ is on $C D$ and $M$ has not reached $B$, i.e., $2\n\nFigure 2\n\nDraw $\\mathrm{MF} \\perp \\mathrm{CD}$ at $\\mathrm{F}$, extend $\\mathrm{AB}$ and $\\mathrm{DC}$ to intersect at $\\mathrm{O}$,\n\nWe know $\\mathrm{DN}=2 \\mathrm{t}-4$, $\\mathrm{AM}=\\sqrt{2} t$, $\\mathrm{OD}=4$, $\\mathrm{OA}=4 \\sqrt{2}$,\n\n$\\therefore \\mathrm{ON}=4-\\mathrm{DN}=8-2 \\mathrm{t}$, $\\mathrm{OM}=4 \\sqrt{2}-\\sqrt{2} t$,\n\n$\\therefore \\mathrm{MF}=4-\\mathrm{t}$,\n\n$\\therefore s_{\\triangle O A D}=\\frac{1}{2} O D \\cdot A D=\\frac{1}{2} \\times 4 \\times 4=8$,\n\n$s_{\\triangle N A D}=\\frac{1}{2} N D \\cdot A D=\\frac{1}{2} \\times(2 t-4) \\times 4=4 t-8$,\n\n$s_{\\triangle O M N}=\\frac{1}{2} O N \\cdot M F=\\frac{1}{2} \\times(8-2 t) \\cdot(4-t)=(4-t)^{2}$,\n\n$\\therefore s=8-(4 t-8)-(4-t)^{2}=-t^{2}+4 t$,\n\nThus, this segment of the graph is a downward-opening parabola;\n\n(3) When $\\mathrm{N}$ is on $\\mathrm{CD}$ and $\\mathrm{M}$ coincides with $\\mathrm{B}$, i.e., $3 \\leq t \\leq 3.5$, as shown in Figure 3,\n\n\n\nFigure 3\n\nWe know $\\mathrm{BC}=1$, $\\mathrm{DN}=2 \\mathrm{t}-4$,\n\n$\\therefore \\mathrm{CN}=3-\\mathrm{DN}=7-2 \\mathrm{t}$,\n\n$\\therefore s_{A B C D}=\\frac{1}{2}(B C+A D) \\cdot C D=\\frac{1}{2} \\times 5 \\times 3=\\frac{15}{2}$,\n\n$s_{\\triangle N A D}=\\frac{1}{2} N D \\cdot A D=\\frac{1}{2} \\times(2 t-4) \\times 4=4 t-8$,\n\n$s_{\\triangle B C N}=\\frac{1}{2} B C \\cdot C N=\\frac{1}{2} \\times 1 \\times(7-2 t)=\\frac{7}{2}-t$,\n\n$\\therefore s=\\frac{15}{2}-(4 t-8)-\\left(\\frac{7}{2}-t\\right)=12-3 t$,\n\nThus, this segment of the graph is a declining line segment;\n\nIn summary, the answer is B.\n\n【Highlight】This problem examines the function graph of a moving point: the function graph is a typical combination of numbers and shapes, and the application of graphs is extensive. By viewing the graph to obtain information, it can not only solve practical problems in life but also improve the ability to analyze and solve problems. The key to solving this problem is to use the idea of classification discussion to find the functional relationship between $\\mathrm{S}$ and $\\mathrm{t}$." }, { "problem_id": 812, "question": "The median of a triangle divides it into two parts of equal area. If two triangles have the same height, then their area ratio is equal to the ratio of their corresponding bases. In Figure (1), $\\triangle M B C$, point $M$ is on segment $B C$. Therefore, $\\frac{\\triangle A B M}{\\triangle A C M} = \\frac{B M}{C M}$. In Figure (2), $\\triangle A B C$, point $M$ is on segment $B C$ and $B M = \\frac{1}{4} B C$, and point $N$ is the midpoint of segment $A C$. If the area of $\\triangle A B C$ is 1, then the area of $\\triangle A D N$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $\\frac{3}{20}$\nB. $\\frac{3}{10}$\nC. $\\frac{3}{8}$\nD. $\\frac{9}{20}$", "input_image": [ "batch5-2024_06_14_8dccab626ba7b956511ag_0017_1.jpg", "batch5-2024_06_14_8dccab626ba7b956511ag_0017_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Connect $CD$, as shown in the figure:\n\nSince $N$ is the midpoint of $AC$,\n\n$\\therefore \\frac{S_{\\triangle ADN}}{S_{\\triangle CDN}}=\\frac{AN}{CN}=1$,\n\n$\\therefore S_{\\triangle ADN}=S_{\\triangle CDN}$,\n\nSimilarly: $S_{\\triangle ABN}=S_{\\triangle CBN}$,\n\nLet $S_{\\triangle ADN}=S_{\\triangle CDN}=a$,\n\nSince the area of $\\triangle ABC$ is 1,\n\n$\\therefore S_{\\triangle ABN}=S_{\\triangle CBN}=\\frac{1}{2}$,\n\n$\\therefore S_{\\triangle BCD}=S_{\\triangle ABD}=\\frac{1}{2}-a$,\n\nSince $BM=\\frac{1}{4} BC$,\n\n$\\therefore \\frac{BM}{CM}=\\frac{1}{3}$,\n\n$\\therefore \\frac{S_{\\triangle BDM}}{S_{\\triangle CDM}}=\\frac{BM}{CM}=\\frac{1}{3}, \\frac{S_{\\triangle ABM}}{S_{\\triangle ACM}}=\\frac{BM}{CM}=\\frac{1}{3}$,\n\n$\\therefore S_{\\triangle CDM}=3 S_{\\triangle BDM}, S_{\\triangle ACM}=3 S_{\\triangle ABM}$,\n\n$\\therefore S_{\\triangle CDM}=\\frac{3}{4} S_{\\triangle BCD}=\\frac{3}{4} \\times\\left(\\frac{1}{2}-a\\right)=\\frac{3}{8}-\\frac{3}{4} a, S_{\\triangle ACM}=\\frac{3}{4} S_{\\triangle ABC}=\\frac{3}{4}$,\n\nSince $S_{\\triangle ACM}=S_{\\text{quadrilateral }} CM DN+S_{\\triangle ADN}=S_{\\triangle CDM}+S_{\\triangle CDN}+S_{\\triangle ADN}$,\n\nThat is: $\\frac{3}{4}=\\frac{3}{8}-\\frac{3}{4} a+a+a$,\n\nSolving gives: $a=\\frac{3}{10}$,\n\n$\\therefore S_{\\triangle ADN}=\\frac{3}{10}$,\n\nTherefore, the answer is: B.\n\n\n\n【Key Insight】This question examines the properties of medians and the area of triangles. Mastering the properties of medians in triangles is crucial for solving the problem." }, { "problem_id": 813, "question": "As shown in the figure, given rhombus $A B C D$ where $A B = 5$ and $B D = 6$. Point $P$ starts from point $A$ and moves at a speed of 1 unit per second along the path $A \\rightarrow B \\rightarrow O$. Let the time when point $P$ is moving be $t$, and $O P^2$ be $y$. The graph of $y$ against $t$ would approximately look like ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_ae6064ab4e0558310798g_0021_1.jpg", "batch32-2024_06_14_ae6064ab4e0558310798g_0021_2.jpg", "batch32-2024_06_14_ae6064ab4e0558310798g_0021_3.jpg", "batch32-2024_06_14_ae6064ab4e0558310798g_0021_4.jpg", "batch32-2024_06_14_ae6064ab4e0558310798g_0021_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: In rhombus \\( A B C D \\), \\( A B = 5 \\), \\( B D = 6 \\),\n\n\\(\\therefore A C \\perp B D\\), \\( O B = O D = \\frac{1}{2} B D = 3 \\),\n\n\\(\\therefore\\) In right triangle \\( \\triangle O A B \\), \\( O A = \\sqrt{A B^{2} - O B^{2}} = \\sqrt{5^{2} - 3^{2}} = 4 \\),\n\nWhen point \\( P \\) moves on \\( A B \\), as shown in the figure, draw \\( P M \\perp O B \\) at point \\( M \\),\n\n\n\n\\(\\therefore \\triangle B P M \\sim \\triangle B A O \\), and point \\( P \\) starts from point \\( A \\), moving at a speed of 1 unit length per second. Let the movement time of point \\( P \\) be \\( t \\),\n\n\\(\\therefore A P = t \\), \\( B P = 5 - t \\),\n\\(\\therefore \\frac{B P}{B A} = \\frac{P M}{A O} = \\frac{B M}{B O} \\), that is \\( \\frac{5 - t}{5} = \\frac{P M}{4} = \\frac{B M}{3} \\),\n\n\\(\\therefore P M = 4 - \\frac{4}{5} t \\), \\( B M = 3 - \\frac{3}{5} t \\), then \\( O M = O B - B M = 3 - \\left(3 - \\frac{3}{5} t\\right) = \\frac{3}{5} t \\),\n\n\\(\\therefore O P^{2} = P M^{2} + O M^{2} = \\left(4 - \\frac{4}{5} t\\right)^{2} + \\left(\\frac{3}{5} t\\right)^{2} = t^{2} - \\frac{32}{5} t + 16 \\) (\\( 0 \\leq t \\leq 5 \\)),\n\nWhen \\( P \\) moves on \\( B O \\), as shown in the figure,\n\n\n\n\\( O P^{2} = (A B + O B - t)^{2} = (8 - t)^{2} = t^{2} - 16 t + 64 \\) (\\( 5 < t \\leq 8 \\)),\n\nTherefore, the answer is: A.\n\n【Key Insight】This problem mainly examines the relationship between moving points and function graphs, the determination and properties of similar triangles. Mastering the properties of rhombuses and the determination and properties of similar triangles is key to solving the problem." }, { "problem_id": 814, "question": "As shown in Figure (1), in $\\triangle A B C$, $\\angle A = 42^{\\circ}$, and side $B C$ rotates clockwise around point $C$ for a full revolution back to its original position. During the rotation (Figure (2)), when $\\angle A C B^{\\prime} = ($____$)$, $C B^{\\prime} \\parallel A B$.\n\n\n\n(1)\n\n\n\n(2)\nA. $42^{\\circ}$\nB. $138^{\\circ}$\nC. $42^{\\circ}$ or $138^{\\circ}$\nD. $42^{\\circ}$ or $128^{\\circ}$", "input_image": [ "batch24-2024_06_15_a22509bc0ead1b6c010ag_0092_1.jpg", "batch24-2024_06_15_a22509bc0ead1b6c010ag_0092_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (2)(1),\n\n\n\nWinter (2)(1)\n\nWhen $\\angle A C B^{\\prime}=42^{\\circ}$,\n\nSince $\\angle A=42^{\\circ}$,\n\nTherefore, $\\angle A C B^{\\prime}=\\angle A$.\n\nThus, $C B^{\\prime} \\parallel A B$.\n\nAs shown in Figure (2)(2),\n\n\n\nFigure (2)(2)\n\nWhen $\\angle A C B^{\\prime}=138^{\\circ}$,\n\nSince $\\angle A=42^{\\circ}$,\n\nTherefore, $\\angle A C B^{\\prime}+\\angle A=138^{\\circ}+42^{\\circ}=180^{\\circ}$.\n\nThus, $C B^{\\prime} \\parallel A B$.\n\nIn summary, when $\\angle A C B^{\\prime}=42^{\\circ}$ or $\\angle A C B^{\\prime}=138^{\\circ}$, $C B^{\\prime} \\parallel A B$.\n\nTherefore, the correct answer is: C\n\n[Key Insight] This question examines the determination of parallel lines and the mathematical concept of classification and discussion. The key to solving the problem lies in classifying and discussing the different positions of $C B^{\\prime}$ during its rotation." }, { "problem_id": 815, "question": "After learning the chapter \"Cartesian Coordinate System\" in Grade 7, a school organized a \"Describing Positions\" math learning activity. The teacher printed out a map of the current Grade 7 schools in the region for the students, as shown in Figure 1. Xiaoming, a student, received the map and first created a grid with 1 unit length as the side, as shown in Figure 2. Then, he established a Cartesian coordinate system, recording Huabei School's coordinates as $(1,3)$ and Tanghekou Middle School's coordinates as $\\left(\\frac{1}{2}, 11\\right)$.\n\n\n\nMap 1\n\n\n\nMap 2\n\nThe students discussed Xiaoming's approach and Map 2, and the following were conclusions made by some students:\n\n(1) Xiaoming established the Cartesian coordinate system with Hongzhong as the origin, the positive direction of the $x$-axis as the east, and the positive direction of the $y$-axis as the north;\n\n(2) The schools located in the fourth quadrant are No. 2 High School, No. 3 High School, No. 5 High School, Zhangzhong High School, Beifang High School, and Yangsong High School;\n\n(3) The coordinates of Qiazi Middle School and Yangsong High School are $(-2,-1.5),(1.5,-2)$;\n\n(4) Tanghekou Middle School is 30 degrees west of north from 101 Huairou Branch School;\n\n(5) If Bohai Middle School's coordinates are $(-0.5,1.5)$ and Jiuduohu River Middle School's coordinates are $(-1,-0.5)$, then Huabei School's coordinates are $(5,2)$.\n\nOf these conclusions, which are definitely correct?\nA. (1)(3)(5)\nB. (2)(4)(5)\nC. (1)(2)(4)\nD. (3)(4)(5)", "input_image": [ "batch24-2024_06_15_6004c0a91f31a32ead1fg_0038_1.jpg", "batch24-2024_06_15_6004c0a91f31a32ead1fg_0038_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Since the coordinates of Huaibei School are recorded as $(1,3)$, and those of Tanghekou Middle School as $\\left(\\frac{1}{2}, 11\\right)$,\n\nTherefore, the student Xiaoming established a plane rectangular coordinate system with Red Center as the origin, the positive east direction as the positive $x$-axis, and the positive north direction as the positive $y$-axis, making statement (1) correct;\n\nFrom the diagram, the schools located in the fourth quadrant are Wuzhong, Zhangzhong, Beifang Middle School, and Yangsong Middle School, making statement (2) incorrect;\n\nThe coordinates of Qiaozi Middle School and Yangsong Middle School are (-2, -1.5) and (1.5, -2) respectively, making statement (3) correct;\n\nTanghekou Middle School is not located to the north-west at a 30-degree angle from the 101 Huairou Branch School, making statement (4) incorrect;\n\nIf the coordinates of Bohai Middle School are $(-0.5,1.5)$ and those of Jiuduhe Middle School are $(-1,-0.5)$, then the coordinates of Huaibei School are $(5, 2)$, making statement (5) correct;\n\nThe conclusions that are definitely correct are (1), (3), and (5),\nTherefore, the answer is: A.\n\n[Key Insight] This question tests the method of determining a plane coordinate system from point coordinates and the method of determining point coordinates. Establishing a plane rectangular coordinate system is key to solving the problem." }, { "problem_id": 816, "question": "As shown in Figure (1), a folding chair is depicted. Figure (2) illustrates the side profile of the chair when it is unfolded, where the legs $A B$ and $C D$ have equal lengths, and $O$ is their midpoint. To make the folding chair both comfortable and sturdy, the manufacturer designs the unfolded chair's height to be $32 \\mathrm{~cm}$, and $\\angle D O B = 100^{\\circ}$. What should be the length of the leg $A B$ (result rounded to $0.1 \\mathrm{~cm}$)? Use the reference data: $\\sin 50^{\\circ} = \\cos 40^{\\circ} \\approx 0.77, \\sin 40^{\\circ} = \\cos 50^{\\circ} \\approx 0.64, \\tan 40^{\\circ} \\approx 0.84, \\tan 50^{\\circ} \\approx 1.19$.\n\n\n\n(1)\n\n\n\n(2)\nA. $38.1 \\mathrm{~cm}$\nB. $49.8 \\mathrm{~cm}$\nC. $41.6 \\mathrm{~cm}$\nD. $45.3 \\mathrm{~cm}$", "input_image": [ "batch37-2024_06_14_ffd632d0a4dcfebd1fb9g_0041_1.jpg", "batch37-2024_06_14_ffd632d0a4dcfebd1fb9g_0041_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Connect $\\mathrm{BD}$.\n\n\n\nFigure 1\n\nFigure 2\n\nAccording to the problem, $\\mathrm{OA}=\\mathrm{OB}=\\mathrm{OC}=\\mathrm{OD}$.\n\nSince $\\angle \\mathrm{DOB}=100^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{ADO}=50^{\\circ}$, $\\angle \\mathrm{OAD}=\\angle \\mathrm{ODB}=40^{\\circ}$,\n\nThus, $\\angle \\mathrm{ADB}=90^{\\circ}$.\n\nAlso, since $\\mathrm{BD}=32$,\n\nTherefore, $\\mathrm{AB}=32 \\div \\sin 50^{\\circ} \\approx 41.6(\\mathrm{~cm})$.\n\nHence, the correct choice is C.\n\n【Key Point】This problem focuses on using trigonometric relationships in a right triangle to calculate an unknown side." }, { "problem_id": 817, "question": "Xiaojin cuts a square piece of paper as shown in Figure 1, removing the small square numbered 4, and assembles it into a rectangle as shown in Figure 2. If it is given that $A B = 9$ and $B C = 16$, then the perimeter of the shape numbered 3 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $22+\\sqrt{33}$\nB. $22+\\sqrt{34}$\nC. $23+\\sqrt{33}$\nD. $23+\\sqrt{34}$", "input_image": [ "batch37-2024_06_14_b93c19209f895a7f6de9g_0083_1.jpg", "batch37-2024_06_14_b93c19209f895a7f6de9g_0083_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, according to the problem statement: (Figure 1) is a square, (Figure 2) is a rectangle, and figure 4 is a small square.\n\n$\\therefore AB = CD, AD = BC, AD \\parallel BC, \\angle ABC = \\angle AFG = 90^{\\circ}$,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nLet $CE = x$, and $AB = 9$, $BC = 16$,\n\n$\\therefore BE = 16 - x$,\n\nCombining the related information from (Figure 1) and (Figure 2), we get:\n\n$FG = 7 - x = DF, AF = 9 + x$,\n\n$\\because AD \\parallel BC$,\n\n$\\therefore \\triangle FAG \\sim \\triangle AEB, \\tan \\angle FAG = \\tan \\angle AEB$,\n$\\therefore \\frac{9}{16 - x} = \\frac{7 - x}{9 + x}$,\n\nSimplifying, we get: $x^{2} - 32x + 31 = 0$,\n\nSolving, we find: $x_{1} = 1, x_{2} = 31$,\n\nAfter verification: $x = 31$ does not fit the problem's context, so we take $x = 1$,\n\n$\\therefore FG = DF = 7 - 1 = 6, CE = 1, AF = 9 + 1 = 10$\n\nExtend $FG$ to intersect $BC$ at $M$, then $FG \\perp BC$, and quadrilateral $ABMF$ is a rectangle,\n\n$\\therefore FM = AB = 9, BM = AF = 10, GM = FM - FG = 9 - 6 = 3, ME = 16 - 1 - 10 = 5$,\n\n$\\therefore GE = \\sqrt{GM^{2} + ME^{2}} = \\sqrt{34}$,\n\nTherefore, the perimeter of figure 3 is: $FG + GE + CE + CD + DF = 6 + \\sqrt{34} + 1 + 9 + 6 = 22 + \\sqrt{34}$,\n\nHence, the correct choice is B.\n\n【Key Insight】This problem tests the determination and properties of rectangles, the properties of squares, the application of acute trigonometric functions, and the application of quadratic equations. The key to solving this problem lies in deducing the equality of sides in the figures from the related information in (Figure 1) and (Figure 2)." }, { "problem_id": 818, "question": "(2019. Zhoukou Second Simulation) As shown in Figure 1, point $E$ is on side $A D$ of the rectangle $A B C D$. Point $P$ moves along the broken line $B E - E D - D C$ from point $B$ to point $C$, and point $Q$ moves along the segment $B C$ from point $B$ to point $C$. Both points move at a speed of $2 \\mathrm{~cm} / \\mathrm{s}$. If $P$ and $Q$ start moving simultaneously, let the time be $t(s)$ and the area of $\\triangle B P Q$ be $y\\left(\\mathrm{~cm}^{2}\\right)$. Given the graph of the function relationship between $y$ and $t$ as shown in Figure 2, what is the value of $\\frac{C D}{B E}$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{\\sqrt{5}}{3}$\nB. $\\frac{\\sqrt{3}}{2}$\nC. $\\frac{\\sqrt{5}}{6}$\nD. $\\frac{\\sqrt{7}}{4}$", "input_image": [ "batch8-2024_06_14_4b79afec46b2311e7644g_0079_1.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0079_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 2, it can be observed that at $\\mathrm{t}=8$, point $\\mathrm{P}$ coincides with point $\\mathrm{E}$, and point $\\mathrm{Q}$ coincides with point $\\mathrm{C}$; at $\\mathrm{t}=10$, point $\\mathrm{P}$ coincides with point $\\mathrm{D}$.\n\nGiven that the speed of point $\\mathrm{P}$ is $2 \\mathrm{~cm/s}$,\n\nTherefore, $\\mathrm{DE}=4$, and $\\mathrm{BE}=\\mathrm{BC}=16$.\n\nSince the area of triangle $\\mathrm{BCD}$ is $\\frac{1}{2} \\mathrm{BC} \\cdot \\mathrm{CD}=32 \\sqrt{7}$,\n\nIt follows that $C D=4 \\sqrt{7}$.\n\nThus, $\\frac{C D}{B E}=\\frac{4 \\sqrt{7}}{16}=\\frac{\\sqrt{7}}{4}$.\n\nHence, the correct answer is: D.\n\n【Key Insight】This problem examines the function image of a moving point, involving knowledge of quadratic functions, linear functions, etc. The key to solving such problems is to understand the correspondence between the image and the figure at different time intervals, and then to proceed with the solution." }, { "problem_id": 819, "question": "As shown in Figure 1, a pair of right-angled triangle rulers is stacked, with the $45^\\circ$ triangle $A D E$ fixed and the $30^\\circ$ triangle $A B C$ rotating clockwise around vertex $A$ so that at least one pair of sides of the two triangles becomes parallel, as shown in Figure 2. When $\\angle B A D = 15^\\circ$, $B C \\parallel D E$. The other possible values of $\\angle B A D (0^\\circ < \\angle B A D < 180^\\circ)$ that satisfy this condition are\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $60^\\circ, 115^\\circ, 135^\\circ$\nB. $45^\\circ, 60^\\circ, 105^\\circ, 135^\\circ$\nC. $15^\\circ, 30^\\circ, 45^\\circ, 135^\\circ$\nD. $45^\\circ, 60^\\circ, 30^\\circ, 15^\\circ$", "input_image": [ "batch30-2024_06_17_d6d9810f3c0b0bb1c454g_0082_1.jpg", "batch30-2024_06_17_d6d9810f3c0b0bb1c454g_0082_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure\n\n\n\nWhen $AC // DE$, $\\angle BAD = \\angle DAE = 45^\\circ$;\n\nWhen $BC // AD$, $\\angle DAB = \\angle B = 60^\\circ$;\n\nWhen $BC // AE$, since $\\angle EAB = \\angle B = 60^\\circ$,\n\n$\\therefore \\angle BAD = \\angle DAE + \\angle EAB = 45^\\circ + 60^\\circ = 105^\\circ$;\n\nWhen $AB // DE$, since $\\angle E = \\angle EAB = 90^\\circ$,\n\n$\\therefore \\angle BAD = \\angle DAE + \\angle EAB = 45^\\circ + 90^\\circ = 135^\\circ$.\n\nTherefore, choose B.\n\n[Key Insight] This question examines the determination and properties of parallel lines. Drawing the figure according to the problem statement and utilizing the properties of parallel lines and the characteristics of a right triangle are crucial for solving this problem." }, { "problem_id": 820, "question": "As shown in Figure 1, in quadrilateral $A B C D$, $A D \\parallel B C$, and $\\angle B = 30^\\circ$. Line $l$ is perpendicular to $A B$. When line $l$ slides along ray $B C$ from point $B$ to the right, it intersects the sides of quadrilateral $A B C D$ at points $E$ and $F$. Let the distance line $l$ slides to the right be $x$, and the length of segment $E F$ be $y$. The function relationship between $y$ and $x$ is shown in Figure 2. Determine the area of the quadrilateral\n\n\n\nFigure 1\n\n\n\nFigure ?\nA. $10+2 \\sqrt{3}$\nB. $12+2 \\sqrt{3}$\nC. $10+\\frac{5 \\sqrt{3}}{2}$\nD. $12+\\frac{5 \\sqrt{3}}{2}$", "input_image": [ "batch37-2024_06_14_130a72fe01bab086139fg_0007_1.jpg", "batch37-2024_06_14_130a72fe01bab086139fg_0007_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\angle B=30^{\\circ}$ and the line $l \\perp A B$,\n\nTherefore, $B E=2 E F$.\n\nFrom the diagram, we can derive:\n\n$$\nA B=4 \\cos 30^{\\circ}=4 \\times \\frac{\\sqrt{3}}{2}=2 \\sqrt{3}, \\quad A O=\\frac{1}{2} B O=2=D M, \\quad B C=5, \\quad A D=7-4=3,\n$$\n\nFrom the image, we also have: $A N=5-4=1, N D=C M=7-5=2$,\n\nSince $\\angle B=30^{\\circ}$ and $E F \\perp A B$,\n\nTherefore, $\\angle C M D=60^{\\circ}$,\n\nMoreover, since $D M=M C=2$,\n\nThus, $\\triangle D M C$ is an equilateral triangle,\n\nTherefore, $D C=D M=2$,\n\nHence, the perimeter of quadrilateral $A B C D$ is: $A B+B C+A D+C D=2 \\sqrt{3}+5+3+2=10+2 \\sqrt{3}$, so the correct choice is: A.\n\n\n\nFigure 1\n\n【Insight】This problem examines the function graph of moving point problems and the application of trigonometric functions of acute angles. The key to solving this problem is to understand the question clearly and use a combination of numerical and graphical methods to find the solution." }, { "problem_id": 821, "question": "The Tangram, originating in the pre-Qin period of China, is a puzzle consisting of seven geometric pieces that can be arranged to form various shapes. It evolved from the dissection of squares described in the ancient Chinese mathematical text \"The Zhou Bi Suan Jing.\" In the 19th century, it spread overseas and was called \"Tangram\" (meaning \"puzzle from China\"). As shown in Figure 2, the \"Ip Man Kick\" pattern is formed using a Tangram created from a 4x4 square. The area of the \"leg\" that is raised (the shaded region) in the figure is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 3\nB. $\\frac{7}{2}$\nC. 2\nD. $\\frac{5}{2}$", "input_image": [ "batch33-2024_06_14_7fcc9a9aa722b575bb21g_0056_1.jpg", "batch33-2024_06_14_7fcc9a9aa722b575bb21g_0056_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure below, the tangram made by dividing a square with a side length of 4 consists of the following shapes:\n\n(1) An isosceles right triangle with a leg length of \\(2 \\sqrt{2}\\),\n\n\n\n(2) An isosceles right triangle with a leg length of 2,\n\n\n\n(3) An isosceles right triangle with a leg length of \\(\\sqrt{2}\\),\n\n\n\n(4) A square with a side length of \\(\\sqrt{2}\\),\n\n\n\n(5) A parallelogram with side lengths of 2 and \\(\\sqrt{2}\\), and vertex angles of \\(45^{\\circ}\\) and \\(135^{\\circ}\\),\n\n\n\nAccording to Figure 2, the raised \"leg\" (i.e., the shaded part) in the figure is composed of an isosceles right triangle with a leg length of \\(\\sqrt{2}\\) and a parallelogram with side lengths of 2 and \\(\\sqrt{2}\\), and vertex angles of \\(45^{\\circ}\\) and \\(135^{\\circ}\\),\n\nAs shown in the figure below,\n\n\n\nAccording to the properties of a parallelogram, the height of the parallelogram with vertex angles of \\(45^{\\circ}\\) and \\(135^{\\circ}\\) is \\(D B\\), and \\(D B = \\sqrt{2}\\),\n\n\\(\\therefore\\) The area of an isosceles right triangle with a leg length of \\(\\sqrt{2}\\) is: \\(\\frac{1}{2} \\times \\sqrt{2} \\times \\sqrt{2} = 1\\),\n\nThe area of the parallelogram with vertex angles of \\(45^{\\circ}\\) and \\(135^{\\circ}\\) is: \\(\\sqrt{2} \\times \\sqrt{2} = 2\\),\n\n\\(\\therefore\\) The area of the shaded part is: \\(1 + 2 = 3\\),\n\nTherefore, the correct answer is: A.\n\n【Key Point】This question examines the composition and area calculation of shapes in a tangram. Familiarity with the classification of shapes in a tangram is key to solving the problem." }, { "problem_id": 822, "question": "As shown in the figure, the side length of the square $A B C D$ is $1$. Points $E, F, G,$ and $H$ are on each side (not coinciding with $A, B, C,$ or $D$), and $A E = B F = C G = D H$. Let the area of the small square $E F G H$ be $S$, and the length of $A E$ be $x$. The graph of the function $S$ in terms of $x$ would approximately look like ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_4b79afec46b2311e7644g_0068_1.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0068_2.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0068_3.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0068_4.jpg", "batch8-2024_06_14_4b79afec46b2311e7644g_0068_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since quadrilaterals $\\mathrm{ABCD}$ and $\\mathrm{EFGH}$ are squares,\n\n$\\angle \\mathrm{A}=\\angle \\mathrm{B}=90^{\\circ}$, and $\\mathrm{EH}=\\mathrm{EF}$.\n\nAlso, $\\mathrm{AE}=\\mathrm{BF}$.\n\nTherefore, $\\triangle \\mathrm{AEH} \\cong \\triangle \\mathrm{BFE}$ (by the HL theorem).\n\nGiven that $\\mathrm{AE}=\\mathrm{x}$, it follows that $\\mathrm{AH}=\\mathrm{BE}=1-\\mathrm{x}$.\n\nThus, $\\mathrm{s}=\\mathrm{EH}^{2}=\\mathrm{AE}^{2}+\\mathrm{AH}^{2}=\\mathrm{x}^{2}+(1-\\mathrm{x})^{2}$.\n\nHence, $s=2x^{2}-2x+1=2\\left(x-\\frac{1}{2}\\right)^{2}+\\frac{1}{2}$.\n\nTherefore, when $\\mathrm{x}=\\frac{1}{2}$, that is, when $\\mathrm{E}$ is at the midpoint of $\\mathrm{AB}$, $\\mathrm{s}$ reaches its minimum value of $\\frac{1}{2}$.\n\nThe graph is a parabola opening upwards, with the vertex at $\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$.\n\nThus, the correct choice is: A.\n\n【Key Insight】The focus of this problem is on the graph of the function, and the key to solving it lies in establishing the function's equation. This is a basic problem." }, { "problem_id": 823, "question": "Pythagorean theorem is one of the most important theorems in junior high school mathematics. As shown in Figure 1, squares are constructed outwardly on each side of a right-angled triangle, and the two smaller squares are arranged inside the largest square as shown in Figure 2. Let the area of the quadrilateral $A B C D$ be $S_{1}$, the area of the quadrilateral $D C E I$ be $S_{2}$, the area of the quadrilateral $C E F G$ be $S_{3}$, and the area of the quadrilateral $V E H I$ be $S_{4}$. If the area of the shaded region in the figure is known, then it is definitely possible to find ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $S_{1}$\nB. $S_{2}$\nC. $S_{3}$\nD. $S_{4}$", "input_image": [ "batch12-2024_06_15_ed1911a30eded62729c0g_0092_1.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0092_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let the area of the large square be $c$, the area of the medium square be $b$, and the area of the small square be $a$,\n\n\n\n$\\because S_{\\text{shaded}} + S_{\\text{quadrilateral } 1HPQ} = \\frac{1}{2}(c - a)$, and $S_{4} + S_{\\text{quadrilateral } 1HPQ} = \\frac{1}{2} b$\n\nSolving gives $S_{\\text{shaded}} + \\frac{1}{2} b - S_{4} = \\frac{1}{2}(c - a)$, which means $S_{\\text{shaded}} - S_{4} = \\frac{1}{2}(c - a) - \\frac{1}{2} b$\n\n$\\because c = a + b$,\n\n$\\therefore S_{\\text{shaded}} - S_{4} = \\frac{1}{2}(c - a) - \\frac{1}{2} b = 0$,\n$\\therefore S_{4} = S_{\\text{shaded}}$,\n\n$\\therefore$ Knowing the area of the shaded part in the figure, we can definitely determine $S_{4}$.\n\nTherefore, the answer is: D.\n\n【Key Point】This problem tests the Pythagorean theorem and the operations of algebraic expressions. Mastering the Pythagorean theorem is crucial for solving this problem." }, { "problem_id": 824, "question": "As shown in the figure, in right triangle $\\triangle A B C$, $\\angle C = 90^\\circ$, $\\angle A B C = 30^\\circ$, and $A B = 4$. A moving point $P$ starts from point $C$ and moves along the broken line $C - A - B$ at a speed of 1 unit per second until it reaches point $B$. The segment $P B$ is drawn. Let the time of point $P$'s movement be $x$ seconds, and $y = P B^2$. The graph of $y$ versus $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch37-2024_06_14_d4217d5415e569c92d8ag_0057_1.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0057_2.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0057_3.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0057_4.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0057_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since $\\angle ACB = 90^\\circ$, $AB = 4$, and $\\angle ABC = 30^\\circ$,\n\nTherefore, $AC = \\frac{1}{2} AB = 2$, and $BC = \\sqrt{AB^2 - AC^2} = \\sqrt{4^2 - 2^2} = 2\\sqrt{3}$.\n\n(1) When $x = 0$, i.e., point $P$ is at point $C$, $PB^2 = BC^2 = (2\\sqrt{3})^2 = 12$,\n\nThus, $y = 12$;\n\n(2) When $0 < x < 2$, i.e., point $P$ moves on segment $AC$ and does not coincide with $A$ or $C$, $PC = x$,\n\nThus, $PB^2 = PC^2 + BC^2 = x^2 + 12$,\n\nThat is, $y = x^2 + 12$, and the graph of this function is part of an upward-opening parabola;\n\n(3) When $x = 2$, point $P$ coincides with point $A$, $PB = AB = 4$,\n\nThus, $PB^2 = 4^2 = 16$,\n\nThat is, $y = 16$, and at this point, $y$ reaches its maximum;\n\n(4) When $2 < x < 6$, i.e., point $P$ moves on segment $AB$ and does not coincide with $A$ or $B$, $PB = 6 - x$,\n\nThus, $PB^2 = (6 - x)^2 = x^2 - 12x + 36$,\n\nThat is, $y = x^2 - 12x + 36$, and the graph of this function is part of an upward-opening parabola;\n\n(5) When $x = 6$, i.e., point $P$ coincides with point $B$, $PB = 0$,\n\nThus, $PB^2 = 0$, that is, $y = 0$;\n\nFrom the above analysis, it is clear that option B is correct.\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem mainly examines the function graph of a moving point problem. The key to solving it lies in combining numbers and shapes and skillfully writing the relevant function's analytical expression." }, { "problem_id": 825, "question": "As shown in the figure, in rectangle $A B C D$, $A B = 4$ and $B C = 6$. Point $P$ starts from point $A$ and moves at a constant speed of 2 units per second along $A-B-C$. At the same time, point $Q$ starts from point $C$ and moves at a constant speed of 1 unit per second towards point $D$. When point $Q$ reaches point $D$, both points $P$ and $Q$ stop moving. Let the time of movement be $t$ seconds, and the area of $\\triangle A P Q$ be $S$. The graph of the relationship between $S$ and $t$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch7-2024_06_14_8db9b77b9d3b77f76112g_0022_1.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0022_2.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0022_3.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0022_4.jpg", "batch7-2024_06_14_8db9b77b9d3b77f76112g_0022_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem, when \\(0 < t < 2\\), \\(AP = 2t\\),\n\n\\[\nS = \\frac{1}{2} \\times AP \\times BC = \\frac{1}{2} \\times 2t \\times 6 = 6t,\n\\]\nthe graph is a part of a straight line.\n\nSince \\(CD = AB = 4\\), the movement time of point \\(Q\\) is 4 seconds.\n\nWhen \\(2 \\leq t \\leq 4\\), point \\(P\\) is on \\(BC\\), as shown in the figure,\n\n\\[\nBP = 2t - 4, \\quad PC = BC - BP = 6 - (2t - 4) = 10 - 2t, \\quad CQ = t, \\quad DQ = 4 - t,\n\\]\n\n\\[\nS = S_{\\text{rectangle } ABCD} - S_{\\triangle ABP} - S_{\\triangle PQC} - S_{\\triangle ADQ}\n\\]\n\n\\[\n= 4 \\times 6 - \\frac{1}{2} \\times 4 \\times (2t - 4) - \\frac{1}{2} \\times (10 - 2t) \\times t - \\frac{1}{2} \\times 6 \\times (4 - t)\n\\]\n\n\\[\n= t^{2} - 6t + 20\n\\]\n\nThe graph of the function is a part of an upward-opening parabola.\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This problem examines the properties of linear and quadratic function graphs. The key to solving the problem is to derive the equation based on the given conditions." }, { "problem_id": 826, "question": "Given that the parabola $y = ax^2 + bx + c$ and the inverse proportional function $y = \\frac{b}{x}$ have a common point in the first quadrant, with the x-coordinate of this point being 1, the graph of the linear function $y = bx + ac$ could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_664d6e156f235987779ag_0019_1.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0019_2.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0019_3.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0019_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Analysis: Given that the parabola \\( y = ax^2 + bx + c \\) and the inverse proportional function \\( y = \\frac{b}{x} \\) share a common point in the first quadrant, it follows that \\( b > 0 \\). Since the x-coordinate of the intersection point is 1, we have \\( a + b + c = b \\), which implies that \\( a \\) and \\( c \\) are opposites of each other. Consequently, the graph of the linear function \\( y = bx + ac \\) can be determined.\n\nDetailed Explanation: \n- Since the parabola \\( y = ax^2 + bx + c \\) and the inverse proportional function \\( y = \\frac{b}{x} \\) intersect at a point in the first quadrant, it must be that \\( b > 0 \\).\n- Given that the x-coordinate of the intersection point is 1, substituting \\( x = 1 \\) into both equations yields \\( a + b + c = b \\), simplifying to \\( a + c = 0 \\).\n- This implies that \\( a \\) and \\( c \\) are additive inverses, i.e., \\( c = -a \\), and thus \\( ac = -a^2 < 0 \\).\n- Therefore, the linear function \\( y = bx + ac \\) will have a graph that passes through the first, third, and fourth quadrants.\n\nThe correct choice is B.\n\nKey Point: This problem examines the graphs of linear functions, the properties of inverse proportional functions, and the properties of quadratic functions, emphasizing the conditions \\( b > 0 \\) and \\( ac < 0 \\)." }, { "problem_id": 827, "question": "In rhombus $\\mathrm{ABCD}$ as shown in Figure 1, $\\mathrm{AB}=2$ and $\\angle \\mathrm{BAD}=60^\\circ$. A perpendicular $\\mathrm{DE}$ is drawn from point $\\mathrm{D}$ to side $\\mathrm{AB}$ at point $\\mathrm{E}$, and a perpendicular $\\mathrm{DF}$ is drawn from point $\\mathrm{D}$ to side $\\mathrm{BC}$ at point $\\mathrm{F}$. The angle $\\angle \\mathrm{EDF}$ is rotated clockwise around point $\\mathrm{D}$ by $\\alpha^\\circ (0<\\alpha<180)$, with its corresponding sides $\\mathrm{DE}^\\prime$ and $\\mathrm{DF}^\\prime$ intersecting sides $\\mathrm{AB}$ and $\\mathrm{BC}$ at points $G$ and $P$, respectively, as shown in Figure 2. The line segment $GP$ is drawn. When the area of triangle $\\triangle \\mathrm{DGP}$ equals $3\\sqrt{3}$, the value of $\\alpha$ is $(\\quad)$.\n\n\n\nFigure 1\n\n\nA. 30\nB. 45\nC. 60\nD. 120", "input_image": [ "batch26-2024_06_17_e775ac44b0bd009d76dag_0002_1.jpg", "batch26-2024_06_17_e775ac44b0bd009d76dag_0002_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Analysis: Based on the given conditions $\\mathrm{AB} / / \\mathrm{DC}$ and $\\angle \\mathrm{BAD}=60^{\\circ}$, we can determine the measure of $\\angle \\mathrm{ADC}$. Using $\\angle \\mathrm{ADE}=\\angle \\mathrm{CDF}=30^{\\circ}$, we can find the measure of $\\angle \\mathrm{EDF}$. When $\\angle \\mathrm{EDF}$ is rotated clockwise, the properties of rotation imply that $\\angle \\mathrm{EDG}=\\angle \\mathrm{FDP}$ and $\\angle \\mathrm{GDP}=\\angle \\mathrm{EDF}=60^{\\circ}$. By the criteria for congruent triangles, we can prove that $\\triangle \\mathrm{DEG} \\cong \\triangle \\mathrm{DFP}$.\n\nFrom the properties of congruent triangles, it follows that $\\mathrm{DG}=\\mathrm{DP}$, which means $\\triangle \\mathrm{DGP}$ is an equilateral triangle. Using the area and $\\cos \\angle \\mathrm{EDG}$, we can determine the measure of $\\angle \\mathrm{EDG}$, and similarly, we can derive the conclusion.\n\nDetailed Explanation: Since $\\mathrm{AB} / / \\mathrm{DC}$ and $\\angle \\mathrm{BAD}=60^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{ADC}=120^{\\circ}$. Also, $\\angle \\mathrm{ADE}=\\angle \\mathrm{CDF}=30^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{EDF}=60^{\\circ}$.\n\nFrom the properties of rotation, $\\angle \\mathrm{EDG}=\\angle \\mathrm{FDP}$ and $\\angle \\mathrm{GDP}=\\angle \\mathrm{EDF}=60^{\\circ}$,\n$\\mathrm{DE}=\\mathrm{DF}=\\sqrt{3}$, and $\\angle \\mathrm{DEG}=\\angle \\mathrm{DFP}=90^{\\circ}$.\n\nIn triangles $\\triangle \\mathrm{DEG}$ and $\\triangle \\mathrm{DFP}$,\n\n$(\\angle \\mathrm{GDE}=\\angle \\mathrm{PDF}$\n\n$\\angle \\mathrm{DEG}=\\angle \\mathrm{DPP}$,\n\n$\\mathrm{DE}=\\mathrm{DF}$\n\n$\\therefore \\triangle \\mathrm{DEG} \\cong \\triangle \\mathrm{DFP}$,\n\n$\\therefore \\mathrm{DG}=\\mathrm{DP}$,\n\n$\\therefore \\triangle \\mathrm{DGP}$ is an equilateral triangle,\n\n$\\therefore$ the area of $\\triangle \\mathrm{DGP}$ $=\\frac{\\sqrt{3}}{4} \\mathrm{DG}^{2}=3 \\sqrt{3}$,\n\nSolving, we get $\\mathrm{DG}=2 \\sqrt{3}$,\n\nThen, $\\cos \\angle \\mathrm{EDG}=\\frac{\\mathrm{DE}}{\\mathrm{DG}}=\\frac{1}{2}$,\n\n$\\therefore \\angle \\mathrm{EDG}=60^{\\circ}$,\n\n$\\therefore$ when rotated clockwise by $60^{\\circ}$, the area of $\\triangle \\mathrm{DGP}$ equals $3 \\sqrt{3}$,\n\nTherefore, the correct choice is C.\n\nKey Points: This problem tests the properties of a rhombus and rotational transformations. Understanding the properties of rotation is crucial: (1) Corresponding points are equidistant from the center of rotation; (2) The angle between corresponding points and the center of rotation equals the rotation angle; (3) The pre-rotation and post-rotation figures are congruent." }, { "problem_id": 828, "question": "As shown in Figure (1), in rectangle $A B C D$, $\\frac{A C}{A B}=k$ (where $k$ is a constant). Point $P$ starts from point $B$ and moves at a speed of 1 unit per second along $B \\rightarrow A \\rightarrow C$ until it reaches point $C$. At the same time, point $Q$ starts from point $A$ and moves at a speed of $k$ units per second along $A \\rightarrow C \\rightarrow D$ until it reaches point $D$. When one point stops moving, the other point stops as well. Let the area of triangle $A P Q$ be $y$, and the time of motion be $t$ seconds. The graph of the function relationship between $y$ and $t$ is shown in Figure (2). When $t=4$, the value of $y$ is ( )\n\n\n\n(1)\n\n\n\n(2)\nA. $\\frac{4}{3}$\nB. 1\nC. $\\frac{2}{3}$\nD. $\\frac{1}{3}$", "input_image": [ "batch29-2024_06_14_615670bb7fcc0025dfefg_0025_1.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0025_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) When point $\\mathrm{P}$ moves on $\\mathrm{AB}$,\n\nDraw $\\mathrm{QH} \\perp \\mathrm{AB}$ at point $\\mathrm{H}$,\n\nFrom the problem statement: $\\mathrm{AB}=3$, then $\\mathrm{AC}=3 \\mathrm{k}, \\mathrm{AP}=1, \\mathrm{AQ}=2 \\mathrm{k}$,\n\nWhen $\\mathrm{t}=2$, that is $\\mathrm{PB}=2, \\mathrm{y}=\\frac{1}{2} \\times P A \\times Q H=\\frac{1}{2} \\times(3-\\mathrm{t}) \\times Q H=\\frac{4}{3}$\n\nSolving gives: $\\mathrm{QH}=\\frac{8}{3}$,\n\nThen $\\mathrm{AH}=\\mathrm{AQ} \\cos \\angle \\mathrm{BAC}=2 \\mathrm{k} \\times \\frac{1}{k}=2$, hence $\\mathrm{PH}=1$,\n\nThen $\\mathrm{AH}=2$, and $\\mathrm{QH}=\\frac{8}{3}$\n\nThus $\\tan \\angle \\mathrm{HAQ}=\\frac{Q H}{A H}=\\frac{4}{3}=\\tan \\alpha$,\n\nThen $\\cos \\alpha=\\frac{3}{5}=\\frac{1}{k}$, solving gives: $\\mathrm{k}=\\frac{5}{3}$\n\nTherefore $\\mathrm{AB}=3, \\mathrm{BC}=4, \\mathrm{AC}=5$;\n\n(2) When $\\mathrm{t}=4$, the distance point $\\mathrm{P}$ moves on $\\mathrm{AC}$ is 1, point $\\mathrm{Q}$ moves on $\\mathrm{CD}$ for 1 second, the distance moved $\\mathrm{QC}$ is $\\frac{5}{3}$, then $\\mathrm{DQ}=3-\\frac{5}{3}$,\n\n$y=\\frac{1}{2} \\times A P \\times D Q=\\frac{1}{2} \\times 1 \\times\\left(3-\\frac{5}{3}\\right)=\\frac{2}{3}$,\n\nTherefore, the answer is: C.\n\n\n\n【Insight】This problem examines the moving point image problem, involving solving right triangles, calculating areas, and other knowledge. The key to such problems is: to clarify the correspondence between the image and the graph in different time periods, and then solve accordingly." }, { "problem_id": 829, "question": "In Figure 2, a simplified diagram of the arch bridge in Figure 1 is shown. The intersection of the arch and the bridge surface is at points $\\mathrm{O}$ and $\\mathrm{B}$. With point $\\mathrm{O}$ as the origin and the horizontal line $\\mathrm{OB}$ as the $\\mathrm{x}$-axis, a Cartesian coordinate system is established. The arch of the bridge can be approximated as the parabola $\\mathrm{y}=-\\frac{1}{400}(\\mathrm{x}-80)^2+16$. The intersection point $\\mathrm{C}$ of the arch and the pier $\\mathrm{AC}$ is at the water level, with $\\mathrm{AC}$ perpendicular to the $\\mathrm{x}$-axis. If $\\mathrm{OA}=10$ meters, then the height of the bridge surface from the water level, $\\mathrm{AC}$, is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $16 \\frac{9}{40}$m\nB. $\\frac{17}{4}$m\nC. $16 \\frac{7}{40}$m\nD. $\\frac{15}{4}$m", "input_image": [ "batch7-2024_06_14_da13b7672084f31d9ca1g_0080_1.jpg", "batch7-2024_06_14_da13b7672084f31d9ca1g_0080_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Question Analysis:**\n\nGiven that $\\mathrm{AC}$ is perpendicular to the $\\mathrm{x}$-axis and $\\mathrm{OA} = 10$ meters,\n\n$\\therefore$ the abscissa (x-coordinate) of point C is $-10$.\n\nWhen $x = -10$, the equation $y = -\\frac{1}{400}(x-80)^{2} + 16$ yields:\n\n$$\ny = -\\frac{1}{400}(-10 - 80)^{2} + 16 = -\\frac{17}{4}.\n$$\n\n$\\therefore$ Point C is located at $\\left(-10, -\\frac{17}{4}\\right)$, and the height of the bridge deck above the water surface, $\\mathrm{AC}$, is $\\frac{17}{4}$ meters. Therefore, the correct answer is **B**.\n\n**Key Concept:** Application of quadratic functions." }, { "problem_id": 830, "question": "As shown in the figure, in $\\triangle A B C$, $\\angle A = 30^\\circ, A B = 6, A C = 8$. Point $P$ moves along segment $A B$ from vertex $A$ at a speed of 1 unit per second towards endpoint $B$, while point $M$ moves along segment $A C$ from vertex $C$ at a speed of 2 units per second towards endpoint $A$. Both points start at the same time and stop when one of them reaches its endpoint. Let the time of movement be $x$ seconds, and the area of $\\triangle A P M$ be $y$. The graph of $y$ as a function of $x$ would approximately look like ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch6-2024_06_14_03c0f2dd13de177f4c24g_0028_1.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0028_2.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0028_3.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0028_4.jpg", "batch6-2024_06_14_03c0f2dd13de177f4c24g_0028_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Draw a perpendicular line from point $P$ to $AC$, meeting at point $Q$, as shown in the figure:\n\n\n\nThus, the height of triangle $APM$ on side $AM$ is $PQ$. Given that $AM = 8 - 2x$ and $AP = x$, and since $0 \\leq 8 - 2x \\leq 8$,\n\nit follows that $0 \\leq x \\leq 4$.\nGiven that $\\angle A = 30^\\circ$,\n\nwe have $PQ = \\frac{1}{2} AP = \\frac{1}{2} x$.\n\nTherefore, the area of triangle $AMP$ is:\n\\[ S_{\\triangle AMP} = \\frac{1}{2}(8 - 2x) \\cdot \\frac{1}{2} x = -\\frac{1}{2} x^{2} + 2x \\quad (0 \\leq x \\leq 4). \\]\n\nBased on the properties of quadratic functions and their graphs, the graph that satisfies the given conditions corresponds to option D.\n\nHence, the correct choice is: D.\n\n【Key Insight】This problem primarily examines the function graph of a moving point. The key is to derive the functional relationship between $y$ and $x$, and then determine the corresponding graph based on this relationship." }, { "problem_id": 831, "question": "There is a rectangular sheet of paper $A B C D$, where $A B = 2\\sqrt{2}$ and $A D = 4$. On the paper, there is a semicircle with $A D$ as its diameter (as shown in Figure 1). Point $E$ is on side $A B$. When the paper is folded along $D E$, point $A$ falls on $B C$. The area of the exposed part of the semicircle (as shown in Figure 2, the shaded area) is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\pi-2$\nB. $2-\\frac{\\pi}{2}$\nC. $\\frac{4}{3} \\pi-\\sqrt{3}$\nD. $\\frac{2}{3} \\pi-1$", "input_image": [ "batch20-2024_05_23_e0fb99ec31043881873eg_0017_1.jpg", "batch20-2024_05_23_e0fb99ec31043881873eg_0017_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the center of the circle where the shaded region is located be $O$, and let $AD$ intersect the semicircular arc at point $F$, as shown in the figure. Connect $OF$.\n\n\n\nFrom the properties of folding, we know: $AD = BC = 4$, $CD = 2\\sqrt{2}$,\n\n$\\therefore AC = \\sqrt{AD^2 - CD^2} = 2\\sqrt{2} = CD$,\n\n$\\therefore \\triangle ADC$ is an isosceles right triangle,\n\n$\\therefore \\angle DAC = 45^\\circ$,\n\n$\\because OD \\parallel BC$,\n\n$\\therefore \\angle ODF = 45^\\circ$,\n\n$\\therefore OD = OF = 2$,\n\n$\\therefore \\angle ODF = \\angle OFD = 45^\\circ$,\n\n$\\therefore \\angle DOF = 180^\\circ - 45^\\circ - 45^\\circ = 90^\\circ$,\n\n$\\therefore S_{\\text{shaded region}} = S_{\\text{sector}} DOF - S_{\\triangle} ODF$\n\n$= \\frac{90 \\pi \\times 2^2}{360} - \\frac{1}{2} \\times 2 \\times 2$\n\n$= \\pi - 2$.\n\nTherefore, the answer is: A.\n\n[Key Insight] This problem examines the properties of folding, the Pythagorean theorem, and the calculation of areas of sectors and triangles. Mastering the methods for calculating the areas of sectors and triangles is essential for correct computation, and determining the corresponding central angle and radius is crucial for accurate results." }, { "problem_id": 832, "question": "Given that $\\mathrm{AC} \\perp \\mathrm{BC}$ at $\\mathrm{C}$, $\\mathrm{BC}=\\mathrm{a}$, $\\mathrm{CA}=\\mathrm{b}$, and $\\mathrm{AB}=\\mathrm{c}$, which of the following options has the radius of $\\odot \\mathrm{O}$ as $\\frac{\\mathrm{a}+\\mathrm{b}-\\mathrm{c}}{2}$? $(\\quad)$\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_c146e2aaeb1beea8235eg_0080_1.jpg", "batch19-2024_05_24_c146e2aaeb1beea8235eg_0080_2.jpg", "batch19-2024_05_24_c146e2aaeb1beea8235eg_0080_3.jpg", "batch19-2024_05_24_c146e2aaeb1beea8235eg_0080_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "$\\mathrm{A}$、Let the radius of the circle be $\\mathrm{x}$. The circle is tangent to $\\mathrm{AC}$ at $\\mathrm{E}$, to $\\mathrm{BC}$ at $\\mathrm{D}$, and to $\\mathrm{AB}$ at $\\mathrm{F}$, as shown in Figure (1).\n\n\n\nFigure (1)\n\nSimilarly, we obtain the square $O E C D$, with $A E = A F$ and $B D = B F$. Therefore, $a - x + b - x = c$, solving for $x$ gives $x = \\frac{a + b - c}{2}$. Hence, this option is correct.\n\n$\\mathrm{B}$、Let the circle be tangent to $\\mathrm{AB}$ at $\\mathrm{F}$, and let the radius of the circle be $\\mathrm{y}$. Connect $\\mathrm{OF}$, as shown in Figure (2).\n\n\n\nFigure (2)\n\nThen, $\\triangle \\mathrm{BCA} \\sim \\triangle \\mathrm{OFA}$,\n\n$\\therefore \\frac{O F}{B C} = \\frac{A O}{A B}$,\n\n$\\therefore \\frac{y}{a} = \\frac{b - y}{c}$,\n\nSolving gives: $\\mathrm{y} = \\frac{a b}{a + c} = \\frac{A O}{A B}$. Hence, this option is incorrect.\n\nC、Connect $\\mathrm{OE}$ and $\\mathrm{OD}$,\n\n\n\nFigure (3)\n\nSince $\\mathrm{AC}$ and $\\mathrm{BC}$ are tangent to circle $\\mathrm{O}$ at $\\mathrm{E}$ and $\\mathrm{D}$ respectively,\n\n$\\therefore \\angle \\mathrm{OEC} = \\angle \\mathrm{ODC} = \\angle \\mathrm{C} = 90^{\\circ}$,\n\nSince $\\mathrm{OE} = \\mathrm{OD}$,\n\n$\\therefore$ quadrilateral $\\mathrm{OECD}$ is a square,\n\n$\\therefore \\mathrm{OE} = \\mathrm{EC} = \\mathrm{CD} = \\mathrm{OD}$,\n\nLet the radius of circle $\\mathrm{O}$ be $\\mathrm{r}$,\n\nSince $\\mathrm{OE} \\parallel \\mathrm{BC}$,\n\n$\\therefore \\angle \\mathrm{AOE} = \\angle \\mathrm{B}$,\n\nSince $\\angle \\mathrm{AEO} = \\angle \\mathrm{ODB}$,\n\n$\\therefore \\triangle \\mathrm{ODB} \\sim \\triangle \\mathrm{AEO}$,\n\n$\\therefore \\frac{O E}{B D} = \\frac{A E}{O D}, \\frac{r}{a - r} = \\frac{b - r}{r}$,\n\nSolving gives: $\\mathrm{r} = \\frac{a b}{a + b}$. Hence, this option is incorrect.\n\n$\\mathrm{D}$、The three points of tangency from top to bottom are $\\mathrm{D}$, $\\mathrm{E}$, and $\\mathrm{F}$; and let the radius of the circle be $\\mathrm{x}$.\n\n\n\nFigure (4)\n\nIt is easy to see that $\\mathrm{BD} = \\mathrm{BF}$, so $\\mathrm{AD} = \\mathrm{BD} - \\mathrm{BA} = \\mathrm{BF} - \\mathrm{BA} = \\mathrm{a} + \\mathrm{x} - \\mathrm{c}$;\n\nAlso, since $\\mathrm{b} - \\mathrm{x} = \\mathrm{AE} = \\mathrm{AD} = \\mathrm{a} + \\mathrm{x} - \\mathrm{c}$; therefore, $\\mathrm{x} = \\frac{\\mathrm{b} + \\mathrm{c} - \\mathrm{a}}{2}$. Hence, this option is incorrect.\n\nTherefore, the correct answer is A.\n\nKey Insight: This question mainly tests the understanding and mastery of the properties and determination of squares, the properties of tangents, the properties and determination of congruent triangles, the incircle and incenter of a triangle, and solving linear equations. The key to solving this problem lies in using these properties to determine the radius of the circle." }, { "problem_id": 833, "question": "A right-angled triangular wooden board has a length of $1 \\mathrm{~cm}$ for one of its right-angled sides, $A C$. Its area is $1 \\mathrm{~cm}^{2}$. Person A and Person B each process the board into a square tabletop according to Figure (1) and Figure (2), respectively. Among (1) and (2), the square with the larger area is ( )\n\n\n\n(1)\n\n\n\n(2)\nA. (1)\nB. (2)\nC. The same\nD. Indeterminable", "input_image": [ "batch32-2024_06_14_621a06aac7ad8d6dddd1g_0012_1.jpg", "batch32-2024_06_14_621a06aac7ad8d6dddd1g_0012_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Given that the length of \\( AC \\) is \\( 1 \\mathrm{~cm} \\) and the area of \\( \\triangle ABC \\) is \\( 1 \\mathrm{~cm}^{2} \\), we can deduce that \\( BC = 2 \\mathrm{~cm} \\). As shown in Figure (1), let the side length of the processed tabletop be \\( x \\mathrm{~cm} \\).\n\nSince \\( DE \\parallel CB \\), we have:\n\\[\n\\frac{DE}{BC} = \\frac{AD}{AC},\n\\]\nwhich translates to:\n\\[\n\\frac{x}{2} = \\frac{1 - x}{1}.\n\\]\nSolving this equation yields:\n\\[\nx = \\frac{2}{3} \\mathrm{~cm}.\n\\]\n\nIn Figure (2), let the side length of the processed tabletop be \\( y \\mathrm{~cm} \\). Draw a perpendicular \\( CM \\) from point \\( C \\) to \\( AB \\), intersecting \\( DE \\) and \\( AB \\) at points \\( N \\) and \\( M \\), respectively.\n\nGiven \\( AC = 1 \\mathrm{~cm} \\) and \\( BC = 2 \\mathrm{~cm} \\), we find:\n\\[\nAB = \\sqrt{AC^{2} + BC^{2}} = \\sqrt{5}.\n\\]\nSince the area of \\( \\triangle ABC \\) is \\( 1 \\mathrm{~cm}^{2} \\), the height \\( CM \\) is:\n\\[\nCM = \\frac{2 \\sqrt{5}}{5} \\mathrm{~cm}.\n\\]\nBecause \\( DE \\parallel AB \\), triangles \\( \\triangle CDE \\) and \\( \\triangle CAB \\) are similar. Therefore:\n\\[\n\\frac{DE}{AB} = \\frac{CN}{CM},\n\\]\nwhich translates to:\n\\[\n\\frac{y}{\\sqrt{5}} = \\frac{\\frac{2 \\sqrt{5}}{5} - y}{\\frac{2 \\sqrt{5}}{5}}.\n\\]\nSolving this equation yields:\n\\[\ny = \\frac{2 \\sqrt{5}}{7} \\mathrm{~cm}.\n\\]\nComparing \\( x^{2} \\) and \\( y^{2} \\):\n\\[\nx^{2} = \\frac{4}{9} = \\frac{20}{45}, \\quad y^{2} = \\frac{20}{49},\n\\]\nwe find that \\( x^{2} > y^{2} \\), meaning \\( S_{1} > S_{2} \\).\n\nTherefore, the correct choice is: A.\n\n\n\n(1)\n\n\n\n(2)\n\n【Key Insight】This problem primarily tests the properties and determination of similar triangles. Understanding the proportional division of segments by parallel lines and the properties of similar triangles is crucial for solving it." }, { "problem_id": 834, "question": "As shown in the figure, $A B$ is the diameter of the semicircle $O$, and $C$ is the midpoint of the arc $A B$. Points $E$ and $F$ are on chords $A C$ and $A B$, respectively, such that $\\angle C F E = 45^\\circ$ and $A B = 4$. If $B F = x$ and $A E = y$, then the graph of $y$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_e0488acab553cd267fbdg_0092_1.jpg", "batch32-2024_06_14_e0488acab553cd267fbdg_0092_2.jpg", "batch32-2024_06_14_e0488acab553cd267fbdg_0092_3.jpg", "batch32-2024_06_14_e0488acab553cd267fbdg_0092_4.jpg", "batch32-2024_06_14_e0488acab553cd267fbdg_0092_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, connect $BC$,\n\n\n\nSince $AB$ is the diameter of semicircle $O$, and $C$ is the midpoint of arc $AB$,\n\nTherefore, $\\angle ACB = 90^\\circ$, and $AC = BC$,\n\nHence, $\\angle CAB = \\angle CBA = 45^\\circ$,\n\nSince $\\angle CFA = \\angle CBA + \\angle FCB = 45^\\circ + \\angle FCB$,\n\nTherefore, $\\angle FCB = \\angle CFA - 45^\\circ$,\n\nAlso, since $\\angle EFA = \\angle CFA - \\angle CFE = \\angle CFA - 49$,\n\nTherefore, $\\angle FCB = \\angle EFA$,\n\nSince $\\angle EAF = \\angle FBC = 45^\\circ$,\n\nTherefore, $\\triangle AEF \\sim \\triangle BFC$,\n\nHence, $\\frac{AE}{BF} = \\frac{AF}{BC}$,\n\nSince $AB = 4$,\n\nTherefore, $AF = 4 - x$, and $BC = 2\\sqrt{2}$,\n\nThus, $\\frac{y}{x} = \\frac{4 - x}{2\\sqrt{2}}$,\n\nTherefore, $y = -\\frac{\\sqrt{2}}{4}x^2 + \\sqrt{2}x$,\n\nThus, the graph of the function is a segment of a parabola passing through the origin,\nHence, the correct choice is: D.\n\n【Key Insight】This problem establishes the relationship between $y$ and $x$ through the similarity properties of triangles, thereby transforming it into a question of selecting the correct graph based on the functional relationship. It tests the ability to combine numerical and graphical analysis, and proving $\\triangle AEF \\sim \\triangle BFC$ is the key to solving the problem." }, { "problem_id": 835, "question": "Equilateral triangle $\\triangle A B C$ has a side length of $4$. Point $D$ is a moving point on segment $B C$ (not coinciding with $B$ or $C$). $\\angle A D P = 60^\\circ$, and $D P$ intersects $A B$ at point $P$. If $B D = x$ and $B P = y$, the graph of $y$ as a function of $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_bf33f410382c3e8c3cfeg_0086_1.jpg", "batch32-2024_06_14_bf33f410382c3e8c3cfeg_0086_2.jpg", "batch32-2024_06_14_bf33f410382c3e8c3cfeg_0086_3.jpg", "batch32-2024_06_14_bf33f410382c3e8c3cfeg_0086_4.jpg", "batch32-2024_06_14_bf33f410382c3e8c3cfeg_0086_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since $\\triangle ABC$ is an equilateral triangle,\n\n$\\therefore \\angle B=\\angle C=60^{\\circ}$,\n\nGiven that $\\angle ADP + \\angle BDP = \\angle C + \\angle CAD$, and $\\angle ADP = \\angle C = 60^{\\circ}$,\n\n$\\therefore \\angle BDP = \\angle CAD$,\n\n$\\therefore \\triangle BDP \\sim \\triangle CAD$,\n\n$\\therefore \\frac{BD}{AC} = \\frac{BP}{CD}$,\n\nSince the side length of the equilateral $\\triangle ABC$ is 4,\n\n$\\therefore AC = BC = AB = 4$,\n\nLet $BD = x$, $BP = y$, then $CD = BC - BD = 4 - x$,\n\n$\\therefore \\frac{x}{4} = \\frac{y}{4 - x}$,\n\n$\\therefore y = -\\frac{1}{4}x^{2} + x$ $(0 < x < 4)$,\n\nThus, the graph of the function is a parabola opening downward,\n\nTherefore, the correct choice is D.\n\n【Key Insight】This problem examines the properties of equilateral triangles, the criteria and properties of similar triangles, the properties of exterior angles of triangles, and the graphs of quadratic functions. Mastering the criteria and properties of similar triangles is crucial for solving the problem." }, { "problem_id": 836, "question": "(1) As shown in Figure 1, if $A B \\parallel C D$, then $\\angle A + \\angle E + \\angle C = 180^\\circ$; \n(2) As shown in Figure 2, if $A B \\parallel C D$, then $\\angle E = \\angle A + \\angle C$; \n(3) As shown in Figure 3, if $A B \\parallel C D$, then $\\angle A + \\angle E - \\angle 1 = 180^\\circ$; \n(4) As shown in Figure 4, if $A B \\parallel C D$, then $\\angle A = \\angle C + \\angle P$. \nHow many of the above conclusions are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch29-2024_06_14_e7f29cc027e50efb8665g_0044_1.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0044_2.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0044_3.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0044_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: (1) Draw a line \\( EF \\) through point \\( E \\) such that \\( EF \\parallel AB \\),\n\n\n\nFigure 1\n\nSince \\( AB \\parallel CD \\), it follows that \\( AB \\parallel CD \\parallel EF \\),\n\nTherefore, \\( \\angle A + \\angle 1 = 180^\\circ \\), and \\( \\angle 2 + \\angle C = 180^\\circ \\),\n\nThus, \\( \\angle A + \\angle C + \\angle AEC = 360^\\circ \\), which means statement (1) is incorrect;\n\n(2) Draw a line \\( EF \\) through point \\( E \\) such that \\( EF \\parallel AB \\),\n\n\n\nFigure 2\n\nSince \\( AB \\parallel CD \\),\n\nIt follows that \\( AB \\parallel CD \\parallel EF \\), hence \\( \\angle A = \\angle 1 \\), and \\( \\angle 2 = \\angle C \\),\n\nTherefore, \\( \\angle AEC = \\angle A + \\angle C \\), which means statement (2) is correct;\n\n(3) Draw a line \\( EF \\) through point \\( E \\) such that \\( EF \\parallel AB \\),\n\n\n\nFigure 3\n\nSince \\( AB \\parallel CD \\), it follows that \\( AB \\parallel CD \\parallel EF \\),\n\nTherefore, \\( \\angle A + \\angle 3 = 180^\\circ \\), and \\( \\angle 1 = \\angle 2 \\),\n\nThus, \\( \\angle A + \\angle AEC - \\angle 2 = 180^\\circ \\),\n\nWhich means \\( \\angle A + \\angle AEC - \\angle 1 = 180^\\circ \\), so statement (3) is correct;\n\n(4) As shown in the figure, draw a line \\( PF \\) through point \\( P \\) such that \\( PF \\parallel AB \\),\n\n\n\nFigure 4\n\nSince \\( AB \\parallel CD \\), it follows that \\( AB \\parallel CD \\parallel PF \\),\n\nTherefore, \\( \\angle 1 = \\angle FPA \\), and \\( \\angle C = \\angle FPC \\),\n\nSince \\( \\angle FPA = \\angle FPC + \\angle CPA \\),\n\nIt follows that \\( \\angle 1 = \\angle C + \\angle CPA \\),\n\nSince \\( AB \\parallel CD \\), it follows that \\( \\angle A = \\angle 1 \\),\n\nThus, \\( \\angle A = \\angle C + \\angle CPA \\), so statement (4) is correct.\n\nIn summary, the correct statements are (2), (3), and (4).\nTherefore, the answer is:\nC.\n\n[Key Insight] This question tests the properties of parallel lines and the corollary of the parallel postulate. Drawing auxiliary lines based on the given conditions is crucial for solving this problem." }, { "problem_id": 837, "question": "Shanxi traditional crafts have a long history and rich varieties. Among them, 'Gaoping enamelware, Pingyao polished lacquerware and Xinjiang glazed clay inkstone' are known as the 'Three Treasures of Shanxi' for their superb craftsmanship, unique cultural connotations and important artistic value. Figure 1 is a pattern of Pingyao polished lacquerware, and Figure 2 is a schematic diagram of a selected part of it and enlarged. Quadrilateral $A B C D$ is a square with a side length of 2. The four vertices of the square are used as the center of the circle, and the half of the length of the diagonal is used as the radius to draw arcs in the square. The four arcs intersect at point $O$. The area of ​​the shaded part in the figure is $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{1}{4} \\pi$\nB. $\\pi-2$\nC. $2 \\pi$\nD. $2 \\pi-4$\n\n", "input_image": [ "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0007_1.jpg", "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0007_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( ABCD \\) is a square,\n\n\\[\n\\therefore AB = BC = CD = DA = 2, \\quad \\angle BAD = \\angle BCD = 90^\\circ\n\\]\n\nSince \\( AC \\) is the diagonal of the square,\n\n\\[\n\\therefore AC = \\sqrt{AD^2 + CD^2} = \\sqrt{2^2 + 2^2} = 2\\sqrt{2}, \\quad \\angle ACD = 45^\\circ\n\\]\n\nLet point \\( O \\) be the midpoint of \\( AC \\),\n\n\\[\n\\therefore CO = AO = \\frac{1}{2} AC = \\sqrt{2}\n\\]\n\nThe shaded area in the figure is the area of the square minus four identical shapes labeled as Shape 1, as shown below:\n\n\n\nThe area of Shape 1 = Area of quadrilateral \\( OEDF \\) - Area of two congruent segments. The quadrilateral and segments are shown below:\n\n\n\nArea of quadrilateral \\( OEDF \\) = \\(\\frac{1}{2} \\times 2 \\times 2 - 2 \\times \\frac{1}{2} \\times \\sqrt{2} \\times 1 = 2 - \\sqrt{2}\\),\n\nArea of a segment = Area of a sector - Area of a triangle. The sector and triangle are shown below:\n\nArea of the sector = \\(\\frac{1}{2} \\times L R = \\frac{1}{2} \\times \\frac{45 \\pi \\times \\sqrt{2}}{180} \\times \\sqrt{2} = \\frac{\\pi}{4}\\),\n\nArea of the triangle = \\(\\frac{1}{2} \\times \\text{base} \\times \\text{height} = \\frac{1}{2} \\times \\sqrt{2} \\times 1 = \\frac{\\sqrt{2}}{2}\\),\n\nArea of the segment = \\(\\frac{\\pi}{4} - \\frac{\\sqrt{2}}{2}\\),\n\n\\[\n\\therefore \\text{Area of Shape 1} = 2 - \\sqrt{2} - 2 \\times \\left(\\frac{\\pi}{4} - \\frac{\\sqrt{2}}{2}\\right) = 2 - \\frac{\\pi}{2}\n\\]\n\nArea of the shaded region = Area of the square - 4 \\(\\times\\) Area of Shape 1 = \\(2 \\times 2 - 4 \\times \\left(2 - \\frac{\\pi}{2}\\right) = 2\\pi - 4\\).\n\nTherefore, the correct answer is: D.\n\n【Key Insight】This problem tests the understanding of sector areas and the transformation between different shapes. Accurately identifying the shapes is crucial for solving this problem." }, { "problem_id": 838, "question": "Given segment $\\mathrm{a}$ and angle $\\angle 1$ in Figure 1, draw $\\triangle \\mathrm{ABC}$ such that $\\mathrm{BC}=\\mathrm{a}, \\angle \\mathrm{ABC}=\\angle \\mathrm{BCA}=\\angle 1$. Zhang Lei's construction is shown in Figure 2. Which of the following statements is definitely true $(\\quad)$?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. The construction of $\\triangle \\mathrm{ABC}$ is based on $\\mathrm{ASA}$\n\nB. Arc $\\mathrm{EF}$ is drawn with $\\mathrm{AC}$ as the radius\n\nC. Arc $M N$ is drawn with point $\\mathrm{A}$ as the center and $\\mathrm{a}$ as the radius\n\nD. Arc $\\mathrm{GH}$ is drawn with $\\mathrm{CP}$ as the radius", "input_image": [ "batch9-2024_05_23_28287760ce2ff0aead98g_0095_1.jpg", "batch9-2024_05_23_28287760ce2ff0aead98g_0095_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "$\\mathrm{A}$: According to the construction, the basis for constructing $\\triangle \\mathrm{ABC}$ is $\\mathrm{ASA}$, which is correct; \n$\\mathrm{B}$: Arc $\\mathrm{FE}$ is drawn with $\\mathrm{B}$ as the center and $\\mathrm{BF}$ as the radius, which is incorrect; \n$\\mathrm{C}$: Arc $\\mathrm{MN}$ is drawn with $\\mathrm{B}$ as the center and $a$ as the radius, which is incorrect; \n$\\mathrm{D}$: Arc $\\mathrm{GH}$ is drawn with $\\mathrm{Q}$ as the center and $\\mathrm{QP}$ as the radius, which is incorrect. Therefore, the correct answer is option $\\mathrm{A}$. \n\n**Key Insight**: This question tests the understanding of ruler and compass constructions. Familiarity with the principles of such constructions is crucial for solving this problem." }, { "problem_id": 839, "question": "As shown in the figure, in quadrilateral $A B C D$, $A D \\parallel B C$, $\\angle D = 90^\\circ$. Point $N$ is the midpoint of side $C D$. A moving point $M$ travels along the path $A \\rightarrow B \\rightarrow C \\rightarrow N$ until it reaches point $N$. The relationship between the length $y$ of segment $A M$ and the time $x$ of the movement is shown in Figure 2, where $P$ is the lowest point on the curved part of the graph. Determine the area of triangle $A B N$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 6\nB. $4 \\sqrt{3}$\nC. $2 \\sqrt{3}$\nD. $3 \\sqrt{3}$", "input_image": [ "batch37-2024_06_14_cb9c20de135d415e5929g_0063_1.jpg", "batch37-2024_06_14_cb9c20de135d415e5929g_0063_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "The trajectory of point $M$ is $A \\rightarrow B \\rightarrow C \\rightarrow N$. From the graph, it can be seen that when $M$ moves to points $B$ and $C$, the maximum value of $AM$ is 4. When $AM \\perp BC$, the minimum value of $AM$ is $2\\sqrt{3}$.\n\nTherefore, $AB = AC = 4$.\n\nFrom trigonometric functions, we have:\n\\[\n\\sin \\angle B = \\frac{AM}{AB} = \\frac{2\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{2},\n\\]\n\\[\n\\angle B = 60^\\circ,\n\\]\n\\[\n\\angle BAM = 30^\\circ, \\quad \\triangle ABC \\text{ is an equilateral triangle},\n\\]\n\\[\nBM = 2, \\quad BC = AB = 4.\n\\]\n\nAccording to the problem, quadrilateral $AMCD$ is a rectangle,\n\\[\nCD = AM = 2\\sqrt{3}, \\quad AD = MC = 2.\n\\]\nSince $N$ is the midpoint of $CD$,\n\\[\nDN = CN = \\sqrt{3}.\n\\]\nThe area of $\\triangle ABN$ is:\n\\[\nS_{\\triangle ABN} = S_{\\text{trapezoid } ABCD} - S_{\\triangle ADN} - S_{\\triangle BCN},\n\\]\n\\[\nS_{\\triangle ABN} = \\frac{1}{2} \\times (2 + 4) \\times 2\\sqrt{3} - \\frac{1}{2} \\times 2 \\times \\sqrt{3} - \\frac{1}{2} \\times 4 \\times \\sqrt{3} = 3\\sqrt{3}.\n\\]\n\nTherefore, the correct answer is: D.\n\n**Key Insight:** This problem examines the function graph of a moving point. It requires a solid understanding of the trigonometric functions of special angles, the properties of a 30-degree right triangle, the determination and properties of equilateral triangles and rectangles. The key to solving the problem lies in identifying the segments corresponding to the maximum and minimum values from the graph." }, { "problem_id": 840, "question": "As shown in Figure 1, equilateral triangles are constructed outwardly on each side of the right-angled triangle, with areas denoted as $S_{1}, S_{2}, S_{3}$; as shown in Figure 2, semicircles are constructed outwardly with the lengths of the right-angled triangle sides as radii, with areas denoted as $S_{4}, S_{5}, S_{6}$. Given that $S_{1} = 16, S_{2} = 45, S_{5} = 11, S_{6} = 14$, what is the value of $S_{3} + S_{4}$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 86\nB. 64\nC. 54\nD. 48", "input_image": [ "batch12-2024_06_15_7527c0aa5932602c15e2g_0026_1.jpg", "batch12-2024_06_15_7527c0aa5932602c15e2g_0026_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, draw a perpendicular from point $\\mathrm{E}$ to $\\mathrm{AB}$, with the foot of the perpendicular at $\\mathrm{D}$.\n\nSince $\\triangle \\mathrm{ABE}$ is an equilateral triangle,\n\n$\\therefore \\angle \\mathrm{AED}=\\angle \\mathrm{BED}=30^{\\circ}$. Let $\\mathrm{AB}=\\mathrm{x}$,\n\n$\\therefore \\mathrm{AD}=\\mathrm{BD}=\\frac{1}{2} \\mathrm{AB}=\\frac{1}{2} \\mathrm{x}$,\n\n$\\therefore \\mathrm{DE}=\\sqrt{A E^{2}-A D^{2}}=\\frac{\\sqrt{3}}{2} \\mathrm{x}$,\n\n$\\therefore \\mathrm{S}_{2}=\\frac{1}{2} \\times x \\times \\frac{\\sqrt{3}}{2} x=\\frac{\\sqrt{3}}{4} A B^{2}$,\n\nSimilarly: $\\mathrm{S}_{1}=\\frac{\\sqrt{3}}{4} A C^{2}, \\mathrm{~S}_{3}=\\frac{\\sqrt{3}}{4} B C^{2}$,\n\n$\\because \\mathrm{BC}^{2}=\\mathrm{AB}^{2}-\\mathrm{AC}^{2}$,\n\n$\\therefore \\mathrm{S}_{3}=\\mathrm{S}_{2}-\\mathrm{S}_{1}$,\n\nAs shown in Figure 2, $\\mathrm{S}_{4}=\\frac{1}{2} \\times\\left(\\frac{1}{2} A B\\right)^{2} \\pi=\\frac{\\pi}{8} A B^{2}$,\n\nSimilarly $\\mathrm{S}_{5}=\\frac{\\pi}{8} A C^{2}, \\mathrm{~S}_{6}=\\frac{\\pi}{8} B C^{2}$,\n\nThen $\\mathrm{S}_{4}=\\mathrm{S}_{5}+\\mathrm{S}_{6}$,\n\n$\\therefore \\mathrm{S}_{3}+\\mathrm{S}_{4}=45-16+11+14=54$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Insight】This problem examines the Pythagorean theorem and the properties of equilateral triangles. Pythagorean theorem: If the lengths of the two legs of a right triangle are $\\mathrm{a}$ and $\\mathrm{b}$, and the hypotenuse is $\\mathrm{c}$, then $\\mathrm{a}^{2}+\\mathrm{b}^{2}=\\mathrm{c}^{2}$." }, { "problem_id": 841, "question": "Place an isosceles $\\triangle A B C$ as shown in Figure 1, so that the base $B C$ coincides with the $x$-axis. At this time, the coordinates of point $A$ are $(2, \\sqrt{5})$. If the triangle is placed as shown in Figure 2, with the length of the leg $A B$ coinciding with the $x$-axis, what are the coordinates of point $C$ at this time?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\left(\\frac{8}{3}, \\frac{10}{3}\\right)$\nB. $\\left(\\frac{5}{2}, \\frac{4 \\sqrt{5}}{3}\\right)$\nC. $\\left(\\frac{5}{2}, \\frac{4 \\sqrt{3}}{3}\\right)$\nD. $\\left(\\frac{8}{3}, \\frac{4 \\sqrt{5}}{3}\\right)$", "input_image": [ "batch4-2024_06_14_5edd21796a1bd03cde4eg_0062_1.jpg", "batch4-2024_06_14_5edd21796a1bd03cde4eg_0062_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, draw a perpendicular from point $A$ to $OC$, intersecting the $x$-axis at point $D$.\n\n\n\nFigure 1\n\nGiven that the coordinates of point $A$ are $(2, \\sqrt{5})$,\n\n$\\therefore BD = 2$, $AD = \\sqrt{5}$,\n\nAccording to the Pythagorean theorem, $BA = \\sqrt{BD^{2} + AD^{2}} = \\sqrt{2^{2} + (\\sqrt{5})^{2}} = 3$,\n\nSince $\\triangle ABC$ is an isosceles triangle,\n\n$\\therefore AB = AC = 3$, $BC = 2BD = 4$,\n\nAs shown in Figure 2, draw a perpendicular from point $C$ to $AB$, intersecting the $x$-axis at point $E$.\n\n\n\nFigure 2\n\nLet $BE = x$, then $AE = 3 - x$,\n\nIn right triangle $\\triangle CBE$, $CE^{2} = BC^{2} - BE^{2} = 16 - x^{2}$,\n\nIn right triangle $\\triangle CEA$, according to the Pythagorean theorem,\n\n$CE^{2} + AE^{2} = CA^{2}$,\n\n$16 - x^{2} + (3 - x)^{2} = 3^{2}$,\n\n$16 + 9 - 6x - 9 = 0$,\n\n$x = \\frac{8}{3}$,\n\n$\\therefore CE = \\sqrt{BC^{2} - BE^{2}} = \\sqrt{4^{2} - \\left(\\frac{8}{3}\\right)^{2}} = \\frac{4 \\sqrt{5}}{3}$,\n\n$\\therefore$ The coordinates of point $C$ are $\\left(\\frac{8}{3}, \\frac{4 \\sqrt{5}}{3}\\right)$,\n\nTherefore, the correct answer is D.\n\n【Key Insight】This problem examines the properties of isosceles triangles and the Pythagorean theorem. The key to solving the problem lies in mastering these concepts." }, { "problem_id": 842, "question": "Xiao Ming used four pieces of wood of equal length to connect them end-to-end to make a flexible teaching aid. He first moved the teaching aid into the shape shown in Figure 1 and measured $\\angle B = 60^\\circ$. Then he rearranged the teaching aid into the shape shown in Figure 2 and measured $\\angle A B C = 90^\\circ$. If the diagonal $B D$ in Figure 2 is 40 cm, what is the length of the diagonal $B D$ in Figure 1?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. $20 \\mathrm{~cm}$\nB. $20 \\sqrt{2} \\mathrm{~cm}$\nC. $20 \\sqrt{3} \\mathrm{~cm}$\nD. $20 \\sqrt{6} \\mathrm{~cm}$", "input_image": [ "batch23-2024_06_14_6017bdeaa8de696f18f7g_0002_1.jpg", "batch23-2024_06_14_6017bdeaa8de696f18f7g_0002_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Since \\( AB = BC = CD = DA \\),\n\nTherefore, quadrilateral \\( ABCD \\) is a rhombus (Figure 1).\n\nWhen \\( \\angle ABC = 90^\\circ \\), quadrilateral \\( ABCD \\) is a square (Figure 2).\n\nThus, in Figure 2, \\( \\angle A = 90^\\circ \\),\n\nTherefore, \\( AB^2 + AD^2 = BD^2 \\),\n\nHence, \\( AB = AD = \\frac{\\sqrt{2}}{2} BD = 20\\sqrt{2} \\) cm.\n\nIn Figure 1, connect \\( AC \\), intersecting \\( BD \\) at \\( O \\).\n\nSince \\( \\angle B = 60^\\circ \\) and quadrilateral \\( ABCD \\) is a rhombus,\n\nTherefore, \\( AC \\perp BD \\), \\( OB = OD \\), \\( OA = OC \\), and \\( \\angle ABO = 30^\\circ \\).\n\nThus, \\( OA = \\frac{1}{2} AB = 10\\sqrt{2} \\) cm, \\( OB = \\sqrt{3} OA = 10\\sqrt{6} \\) cm,\n\nTherefore, \\( BD = 2 OB = 20\\sqrt{6} \\) cm.\n\nHence, the answer is: \\( D \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n[Key Insight] Proficiency in using the properties of squares and rhombuses is crucial for solving the problem." }, { "problem_id": 843, "question": "While studying similarity problems, students A and B have the following opinions:\n\n A: By expanding the triangle with sides $3, 4, 5$ as shown in Figure 1, with a spacing of 1 between corresponding sides of the new triangle, the new triangle is similar to the original triangle.\n\n B: By expanding the rectangle with adjacent sides of 3 and 5 as shown in Figure 2, with a spacing of 1 between corresponding sides of the new rectangle, the new rectangle is not similar to the original rectangle.\n\nWhich of the following statements about their opinions is correct ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Both are correct\nB. Both are incorrect\nC. A is correct, B is incorrect\nD. A is incorrect, B is correct", "input_image": [ "batch32-2024_06_14_e0c87936841f97183ff9g_0017_1.jpg", "batch32-2024_06_14_e0c87936841f97183ff9g_0017_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Logic", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\nPart A: According to the problem statement, we have: $AB \\parallel A'B'$, $AC \\parallel A'C'$, $BC \\parallel B'C'$,\n\n$\\therefore \\angle A = \\angle A'$, $\\angle B = \\angle B'$,\n\n$\\therefore \\triangle ABC \\sim \\triangle A'B'C'$,\n\n$\\therefore$ Part A is correct;\n\nPart B: $\\because$ According to the problem statement: $AB = CD = 3$, $AD = BC = 5$, then $A'B' = C'D' = 3 + 2 = 5$, $A'D' = B'C' = 5 + 2 = 7$,\n\n$\\therefore \\frac{AB}{A'B'} = \\frac{CD}{C'D'} = \\frac{3}{5}$, $\\frac{AD}{A'D'} = \\frac{BC}{B'C'} = \\frac{5}{7}$,\n\n$\\therefore \\frac{AB}{A'B'} \\neq \\frac{AD}{A'D'}$,\n\n$\\therefore$ The new rectangle is not similar to the original rectangle.\n\n$\\therefore$ Part B is correct.\n\nTherefore, the correct choice is: A.\n\n\n\n\n\nFigure 2\n\n【Key Point】This problem tests the determination of similar triangles and similar polygons. Mastering and applying the methods for determining similarity is crucial for solving this problem." }, { "problem_id": 844, "question": "Given right triangle $ABC$ where $\\angle C = 90^\\circ$, $AC = BC = 2\\sqrt{2}$, and square $EFGH$ where $EF = 2$, with $AB$ and $EF$ lying on the same line, when triangle $ABC$ is translated to the right, the area $y$ of the overlapping region between triangle $ABC$ and square $EFGH$ and the distance $x$ traveled by point $B$ are related. The graph of this function approximately looks like ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_3014dab89f701789b5aeg_0043_1.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0043_2.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0043_3.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0043_4.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0043_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the problem statement, we can deduce that when \\(0 \\leq x \\leq 2\\), the overlapping area between \\(\\triangle ABC\\) and square \\(EFGH\\) is the isosceles right triangle \\(\\triangle EBC\\), with \\(BE = x\\).\n\nThus, \n\\[\ny = \\frac{1}{2} x^{2}\n\\]\n\nWhen \\(2 < x < 4\\), the overlapping area between \\(\\triangle ABC\\) and square \\(EFGH\\) is the pentagon \\(CMEFN\\), as shown in the figure.\n\n\n\nFrom the problem statement, we have:\n\\[\nS_{\\triangle CHM} = \\frac{1}{2}(x-2)^{2}, \\quad S_{\\triangle CGN} = \\frac{1}{2}(4-x)^{2}\n\\]\nTherefore, the area of pentagon \\(CMEFN\\) is:\n\\[\nS_{\\text{pentagon } CMEFN} = 2 \\times 2 - \\frac{1}{2}(x-2)^{2} - \\frac{1}{2}(4-x)^{2} = -x^{2} + 6x - 6\n\\]\n\nWhen \\(4 \\leq x \\leq 6\\), \\(AF = 6 - x\\), thus:\n\\[\ny = \\frac{1}{2}(6 - x)^{2}\n\\]\n\nSummarizing, we have:\n\\[\ny = \\begin{cases}\n\\frac{1}{2} x^{2} & (0 \\leq x \\leq 2) \\\\\n-x^{2} + 6x - 6 & (2 < x < 4) \\\\\n\\frac{1}{2}(6 - x)^{2} & (4 \\leq x \\leq 6)\n\\end{cases}\n\\]\n\nTherefore, the graph of the function is as shown below:\n\n\n\nHence, the correct choice is C.\n\n【Key Insight】This problem primarily examines the integration of functions and geometry. The key to solving it lies in understanding the properties of isosceles right triangles and the graphs and properties of quadratic functions." }, { "problem_id": 845, "question": "As shown in the figure, in rhombus $A B C D$, lines $A C$ and $A B$ are drawn, with $A B = 5$ and $A C = 8$. A perpendicular line $l$ to $A C$ starts at point $A$ and moves in the direction from $A$ to $C$. During the movement, line $l$ intersects $A B$ (at point $B$), $A C$, and $A D$ (at point $D$) at points $E$, $G$, and $F$, respectively, until point $G$ coincides with point $C$. Let the distance of the translation of line $l$ be $x$ and the area of triangle $A E F$ be $S$. The graph of $S$ as a function of $x$ would approximately look like ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch32-2024_06_14_dc6ac1bcba870fa3dc46g_0029_1.jpg", "batch32-2024_06_14_dc6ac1bcba870fa3dc46g_0029_2.jpg", "batch32-2024_06_14_dc6ac1bcba870fa3dc46g_0029_3.jpg", "batch32-2024_06_14_dc6ac1bcba870fa3dc46g_0029_4.jpg", "batch32-2024_06_14_dc6ac1bcba870fa3dc46g_0029_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Connect $BD$ intersecting $AC$ at $O$.\n\nSince $AC$ and $BD$ are the diagonals of a rhombus,\n\n$\\therefore BD \\perp AC$, and $AO = OC = \\frac{1}{2} AC = 4$.\n\n$\\therefore BD = 2BO = 2\\sqrt{AB^2 - AO^2} = 2 \\times \\sqrt{5^2 - 4^2} = 6$.\n\n(1) When $EF$ is on the left side of $BD$, as shown in the figure:\n\n\n\nSince $EF \\perp AC$,\n\n$\\therefore EF \\parallel BD$,\n\n$\\therefore \\triangle AEF \\sim \\triangle ABD$,\n\n$\\therefore \\frac{AG}{AO} = \\frac{EF}{BD}$,\n\n$\\therefore EF = \\frac{AG \\cdot BD}{AO} = \\frac{3}{2}x$,\n\n$\\therefore S = \\frac{1}{2} AG \\cdot EF = \\frac{1}{2}x \\cdot \\frac{3}{2}x = \\frac{3}{4}x^2$,\n\n$\\therefore$ When $0 \\leq x \\leq 4$, the graph is an upward-opening parabola, and $S$ increases as $x$ increases.\n\n(2) When $EF$ is on the right side of $BD$, as shown in the figure:\n\n\n\nSince $AG = x$,\n\n$\\therefore CG = 8 - x$.\n\nSince $EF \\parallel BD$,\n\n$\\therefore \\triangle CEF \\sim \\triangle CBD$,\n\n$\\therefore \\frac{EF}{BD} = \\frac{CG}{CO}$,\n\n$\\therefore EF = \\frac{CG \\cdot BD}{CO} = \\frac{6(8 - x)}{4} = \\frac{3}{2}(8 - x)$,\n\n$\\therefore S = \\frac{1}{2} AG \\cdot EF = \\frac{1}{2}x \\times \\frac{3}{2}(8 - x) = -\\frac{3}{4}x^2 + 6x$,\n\n$\\therefore$ When $4 < x \\leq 8$, the graph is a downward-opening parabola, and $S$ increases as $x$ increases.\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This problem examines the function graph of a moving point, the properties of a rhombus, and the determination and properties of similar triangles. The key is to derive the function expression based on the area formula of the triangle." }, { "problem_id": 846, "question": "Recently, many people have received phone calls or text messages from the epidemic prevention department indicating that they are \"time-space companions.\" What exactly are time-space companions? Time-space intersection and time-space companionship are the same concept, a term used by public security and telecommunications departments. As shown in Figure (1), this refers to a situation where a person's phone number and a confirmed patient's number are in the same spatial-temporal grid (with a range of $800 \\times 800$) for more than 10 minutes, and either party's number has been in the area for more than 30 hours in the past 14 days. The phone numbers found in this way are called \"time-space companion numbers,\" and the green health code of the individual will turn yellow with a warning nature and be marked as a \"time-space companion.\" In Figure (2), a worker is at point $B$ and uses a clinometer to measure the angle of elevation to the top of a mobile phone base station (point $D$), which is found to be $\\alpha$. The height $C D$ of the mobile phone base station is 50 meters, the height of the clinometer $B E$ is 1 meter. If at this moment, a confirmed patient appears within the $800 \\times 800$ range of a mobile phone base station at point $A$, and the patient, mobile phone base station, and worker are in a straight line, with the patient and worker on opposite sides of the mobile phone base station, and they spend more than 10 minutes together, then this worker ( ) will receive a \"time-space companion\" phone call or text message. (Reference data: $\\sin \\alpha = \\frac{7}{25}, \\cos \\alpha = \\frac{24}{25}, \\tan \\alpha = \\frac{7}{24}$)\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. Will\nB. Will not\nC. Might\nD. Cannot be determined", "input_image": [ "batch34-2024_06_17_8eb738655a589357df09g_0002_1.jpg", "batch34-2024_06_17_8eb738655a589357df09g_0002_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Draw a perpendicular line from point \\( E \\) to \\( CD \\), intersecting at point \\( F \\).\n\n\n\nGiven the conditions, \\( BE = CF = 1 \\) meter, and \\( EF = BC \\).\n\nTherefore, \\( DF = CD - CF = 50 - 1 = 49 \\) meters.\n\nIn the right triangle \\( \\triangle DEF \\), \\( \\tan \\alpha = \\frac{DF}{EF} = \\frac{49}{EF} = \\frac{7}{24} \\).\n\nSolving for \\( EF \\), we get \\( EF = 168 \\) meters.\n\nThus, \\( BC = 168 \\) meters.\n\nSince \\( 168 < 400 \\), this worker will receive a \"space-time companion\" phone call or text message alert.\n\nTherefore, the correct answer is: A.\n\n[Key Insight] This problem tests the understanding of solving right triangles. The key to solving it lies in comprehending the given conditions, applying trigonometric functions of acute angles, and constructing auxiliary lines." }, { "problem_id": 847, "question": "In Figure 1, there is a double-wing turnstile at a subway station entrance. As shown in Figure 2, when the double wings are unfolded, the distance between the endpoints A and B of the wing edges is 12 cm, and the lengths of the wing edges AC and BD are both 64 cm. They form an angle of 30° with the side face of the turnstile, denoted as $\\angle P C A = \\angle B D Q = 30^\\circ$. When the wings are folded, the maximum width of an object that can pass through the turnstile is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $76 \\mathrm{~cm}$\nB. $(64 \\sqrt{2}+12) \\mathrm{cm}$\nC. $(64 \\sqrt{3}+12) \\mathrm{cm}$\nD. $64 \\mathrm{~cm}$", "input_image": [ "batch10-2024_06_14_122c5cd69e0cbe1d0bbbg_0004_1.jpg", "batch10-2024_06_14_122c5cd69e0cbe1d0bbbg_0004_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $AE \\perp CP$ at point $E$ through point $\\mathrm{A}$, and draw $BF \\perp DQ$ at point $F$ through point $B$.\n\n\n\nFigure 2\n\nIn the right triangle $\\triangle \\mathrm{ACE}$, $\\angle ACE = 30^{\\circ}$,\n\n$\\therefore AE = \\frac{1}{2} AC = \\frac{1}{2} \\times 64 = 32(\\mathrm{~cm})$,\n\nSimilarly, we can obtain $BF = 32 \\mathrm{~cm}$,\n\nMoreover, since the distance between the endpoints $\\mathrm{A}$ and $B$ of the double-wing edges is $12 \\mathrm{~cm}$,\n\n$\\therefore 32 + 12 + 32 = 76(\\mathrm{~cm})$,\n\n$\\therefore$ When the double wings are retracted, the maximum width of an object that can pass through the gate is $76 \\mathrm{~cm}$.\n\nTherefore, the correct choice is: A.\n\n【Key Point】This problem mainly examines the properties of a right triangle containing a $30^{\\circ}$ angle. In a right triangle, the leg opposite the $30^{\\circ}$ angle is equal to half the length of the hypotenuse." }, { "problem_id": 848, "question": "Given $\\triangle A B C$ where $\\angle B A C = 90^\\circ$, using a ruler and compass, draw a line through point $A$ that divides $\\triangle A B C$ into two similar triangles. Which of the following methods is incorrect?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_5061f7e3224e2a323688g_0016_1.jpg", "batch32-2024_06_14_5061f7e3224e2a323688g_0016_2.jpg", "batch32-2024_06_14_5061f7e3224e2a323688g_0016_3.jpg", "batch32-2024_06_14_5061f7e3224e2a323688g_0016_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Analysis: By constructing a perpendicular from a point outside a line to that line, and using the method of constructing the perpendicular bisector of a line segment, along with the inscribed angle theorem, we can draw perpendiculars to the hypotenuse of a right-angled triangle. According to the perpendicular to the hypotenuse of a right-angled triangle, the original right-angled triangle is divided into two smaller right-angled triangles, and the three right-angled triangles in the figure are similar to each other; thus, a judgment can be made.\n\nDetailed Explanation: \n\n$\\mathrm{A}$: Within angle $\\angle \\mathrm{BAC}$, construct $\\angle \\mathrm{CAD}=\\angle \\mathrm{B}$, intersecting $\\mathrm{BC}$ at point $\\mathrm{D}$. According to the definition of complementary angles and substitution of equals, we get $\\angle \\mathrm{B}+\\angle \\mathrm{BAD}=90^{\\circ}$, leading to $\\mathrm{AD} \\perp \\mathrm{BC}$. According to the perpendicular to the hypotenuse of a right-angled triangle, the original right-angled triangle is divided into two smaller right-angled triangles, and the three right-angled triangles in the figure are similar to each other; A does not meet the requirement;\n\n$\\mathrm{B}$: With point $\\mathrm{A}$ as the center and a radius slightly less than the length of $\\mathrm{AB}$, draw an arc intersecting line segment $\\mathrm{BC}$ at two points. Then, with these two points as centers and a radius greater than half the distance between the two intersection points, draw arcs intersecting at a point. Draw a line through this point and point $\\mathrm{A}$, which is the perpendicular to $\\mathrm{BC}$. According to the perpendicular to the hypotenuse of a right-angled triangle, the original right-angled triangle is divided into two smaller right-angled triangles, and the three right-angled triangles in the figure are similar to each other; B does not meet the requirement;\n\n$\\mathrm{C}$: With $\\mathrm{AB}$ as the diameter, draw a circle intersecting $\\mathrm{BC}$ at point $\\mathrm{D}$. According to the inscribed angle theorem, draw a line through points $\\mathrm{AD}$ which is perpendicular to $\\mathrm{BC}$. According to the perpendicular to the hypotenuse of a right-angled triangle, the original right-angled triangle is divided into two smaller right-angled triangles, and the three right-angled triangles in the figure are similar to each other; $\\mathrm{C}$ does not meet the requirement;\n\n$D$: With point $\\mathrm{B}$ as the center and the length of $\\mathrm{BA}$ as the radius, draw an arc intersecting $\\mathrm{BC}$ at point $\\mathrm{E}$. Then, with point $\\mathrm{E}$ as the center and the length of $\\mathrm{AB}$ as the radius, draw an arc intersecting the previous arc on the other side of $\\mathrm{BC}$ at a point. Draw a line through this point and point $\\mathrm{A}$, which is not necessarily perpendicular to $\\mathrm{BE}$; thus, it cannot be guaranteed that the two smaller triangles are similar; D meets the requirement;\n\nTherefore, choose D.\n\nKey Point: This question mainly tests the understanding of similarity transformations and the determination of similar triangles. Correctly mastering the method of determining similar triangles is the key to solving the problem." }, { "problem_id": 849, "question": "In rhombus $A B C D$ as shown in Figure 1, $\\angle C = 120^\\circ$, $M$ is the midpoint of $A B$, and $N$ is a moving point on the diagonal $B D$. Let the length of $D N$ be $x$, and the sum of the lengths of segments $M N$ and $A N$ be $y$. The graph in Figure 2 represents the function $y$ as a function of $x$. The right endpoint $F$ of the graph has coordinates $(2 \\sqrt{3}, 3)$. Determine the coordinates of the lowest point $E$ on the graph.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\left(\\frac{2 \\sqrt{3}}{3}, 2\\right)$\nB. $\\left(\\frac{2 \\sqrt{3}}{3}, \\sqrt{3}\\right)$\nC. $\\left(\\frac{4 \\sqrt{3}}{3}, \\sqrt{3}\\right)$\nD. $(\\sqrt{3}, 2)$", "input_image": [ "batch37-2024_06_14_33d6f61630506798ebc9g_0063_1.jpg", "batch37-2024_06_14_33d6f61630506798ebc9g_0063_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since the coordinates of the right endpoint $F$ of the graph are $(2 \\sqrt{3}, 3)$, and $M$ is the midpoint of $A B$,\n\n$\\therefore B D=2 \\sqrt{3}$, and $M N+A N=A B+M B=3 M B=3$,\n\n$\\therefore M B=1$, and $A B=2$.\n\nConnecting $A C$ and $C M$, they intersect $B D$ at point $N_{1}$. Connecting $A N_{1}$, the minimum value of $M N+A N$ is $M N_{l}+A N_{l}=C M$.\n\nSince in rhombus $A B C D$, $\\angle C=120^{\\circ}$,\n\n$\\therefore \\angle A B C=60^{\\circ}$,\n\n$\\therefore \\triangle A B C$ is an equilateral triangle,\n\n$\\therefore C M \\perp A B$, and $\\angle B C M=30^{\\circ}$,\n\n$\\therefore B C=2 \\times 1=2$, and $C M=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$.\n\nSince $A B / / C D$,\n\n$\\therefore C M \\perp C D$.\n\nSince $\\angle A D C=\\angle A B C=60^{\\circ}$,\n\n$\\therefore \\angle B D C=30^{\\circ}$,\n\n$\\therefore D N_{l}=C D \\div \\cos 30^{\\circ}=2 \\div \\frac{\\sqrt{3}}{2}=\\frac{4}{3} \\sqrt{3}$,\n\n$\\therefore$ the coordinates of $E$ are $\\left(\\frac{4 \\sqrt{3}}{3}, \\sqrt{3}\\right)$.\n\nTherefore, the correct choice is C.\n\n\n\nFigure 1\n\n【Key Insight】This problem mainly examines the properties of a rhombus, the properties of a right triangle containing a $30^{\\circ}$ angle, the Pythagorean theorem, and the graph of functions. The key to solving the problem is to add auxiliary lines to construct a right triangle." }, { "problem_id": 850, "question": "As shown in Figure (1), in the square $A B C D$, point $E$ is on the side $A D$. The line segment $B E$ is drawn, and an equilateral triangle $\\triangle B E F$ is constructed with $B E$ as a side, with point $F$ lying on the extension of $B C$. A moving point $M$ starts from point $B$ and moves along the path $B \\rightarrow E \\rightarrow F$ towards point $F$ at a constant speed. A line segment $M P$ is drawn perpendicular to $A D$ at point $P$. Let the distance traveled by point $M$ be $x$, and the area of $\\triangle P E M$ be $y$. The graph of the function relationship between $y$ and $x$ is shown in Figure (2). The length of $D E$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $3-\\sqrt{3}$\nB. $3+\\sqrt{3}$\nC. $2-\\sqrt{3}$\nD. $2+\\sqrt{3}$", "input_image": [ "batch8-2024_06_14_abd5efc71c86223e46b3g_0034_1.jpg", "batch8-2024_06_14_abd5efc71c86223e46b3g_0034_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From Figure (2), we know that: $BE + EF = 4\\sqrt{3}$.\n\nSince $\\triangle BEF$ is an equilateral triangle,\n\n$\\therefore BE = EF = 2\\sqrt{3}$.\n\nGiven that $ABCD$ is a square and $\\angle EBF = 60^{\\circ}$,\n\n$\\therefore AD = AB$, and $\\angle A = \\angle ABC = 90^{\\circ}$.\n\nThus, $\\angle ABE = 30^{\\circ}$,\n$\\therefore AE = \\frac{1}{2} BE = \\sqrt{3}$.\n\nTherefore, in the right triangle $\\triangle ABE$, $AB = \\sqrt{BE^{2} - AE^{2}} = \\sqrt{(2\\sqrt{3})^{2} - (\\sqrt{3})^{2}} = 3$.\n\nHence, $DE = AB - AE = 3 - \\sqrt{3}$.\n\nThe correct answer is: A.\n\n【Key Insight】This problem examines the properties of equilateral triangles and squares, the use of the Pythagorean theorem in right triangles to find the length of a segment, and the property that in a right triangle with a $30^{\\circ}$ angle, the side opposite the $30^{\\circ}$ angle is half the length of the hypotenuse." }, { "problem_id": 851, "question": "As shown in Figure 1, a \"windmill\" pattern is formed by four congruent right-angled triangles, where $\\angle A O B = 90^\\circ$. By extending the hypotenuse of each right-angled triangle, it exactly meets at the midpoint of the hypotenuse of another right-angled triangle, as shown in Figure 2. If $I J = \\sqrt{2}$, then the area of the \"windmill\" is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\sqrt{2}+1$\nB. $2 \\sqrt{2}$\nC. $4-\\sqrt{2}$\nD. $4 \\sqrt{2}$", "input_image": [ "batch37-2024_06_14_96b74b89976cda0e7739g_0082_1.jpg", "batch37-2024_06_14_96b74b89976cda0e7739g_0082_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $AC$.\n\nFrom the problem statement, we have: Rt $\\triangle AOB \\cong$ Rt $\\triangle DCO \\cong$ Rt $\\triangle EOF \\cong$ Rt $\\triangle GOH$.\n\nThus, $OA = OC$, and $\\angle OAB = \\angle OCD$.\n\nSince $\\angle AOC = \\angle AOB = 90^\\circ$, $\\triangle OAC$ is an isosceles right triangle.\n\nAlso, since $\\angle OAB = \\angle OCD$:\n\n$\\angle AJD = 180^\\circ - \\angle ADJ - \\angle OAB = 180^\\circ - \\angle ODC - \\angle OCD = 90^\\circ$, which means $AJ \\perp CD$.\n\nMoreover, since $CJ = DJ$, $AJ$ is the perpendicular bisector of $CD$.\n\nSimilarly, $GI$ is the perpendicular bisector of $AB$.\n\nTherefore, $AC = AD$, and $AJ$ is the angle bisector of the vertex angle $\\angle CAD$ in the isosceles triangle.\n\nThat is, $\\angle DAJ = \\frac{1}{2} \\angle CAD = \\frac{1}{2} \\times 45^\\circ = 22.5^\\circ$.\n\nIt is easy to see that $IH = BJ$, and $IJ = IB + BJ = IB + IH$.\n\nSince $IB = IA$, $IJ = IB + BJ = IH + IA = \\sqrt{2}$.\n\nIn Rt $\\triangle ABO$, $\\angle ABH = \\angle BAH = 22.5^\\circ$.\n\nThus, $\\angle OBH = \\angle OHB = 45^\\circ$.\n\nLet $OB = OH = a$, then $AH = BH = \\sqrt{2} \\cdot OB = \\sqrt{2}a$.\n\nTherefore, $\\tan \\angle A = \\frac{BO}{AO} = \\frac{a}{a + \\sqrt{2}a} = \\sqrt{2} - 1$.\n\nThus, $\\frac{IH}{IA} = \\tan \\angle A = \\sqrt{2} - 1$.\n\nLet $IH = (\\sqrt{2} - 1)x$, and $AI = x$.\n\nThen, $IH + IA = \\sqrt{2}x = \\sqrt{2}$, which implies $x = 1$.\n\nTherefore, $S_{\\triangle ABH} = \\frac{1}{2} \\times AB \\times IH = \\sqrt{2} - 1$.\n\nSince $\\frac{S_{\\triangle BOH}}{S_{\\triangle ABH}} = \\frac{OH}{AH} = \\frac{1}{\\sqrt{2}}$, $S_{\\triangle BOH} = 1 - \\frac{\\sqrt{2}}{2}$.\n\nThus, $S_{\\triangle AOB} = S_{\\triangle ABH} + S_{\\triangle BOH} = \\sqrt{2} - 1 + 1 - \\frac{\\sqrt{2}}{2} = \\frac{\\sqrt{2}}{2}$.\n\nTherefore, $S_{\\text{windmill}} = 4S_{\\triangle AOB} = 4 \\times \\frac{\\sqrt{2}}{2} = 2\\sqrt{2}$.\n\nHence, the correct answer is $B$.\n\n\n\n【Key Insight】This problem mainly tests the application of solving right triangles, the determination and properties of isosceles right triangles, etc. The key to solving this problem lies in the flexible application of relevant knowledge and the integration of geometric and algebraic thinking." }, { "problem_id": 852, "question": "The following are descriptions of four patterns.\n\nFigure 1 shows the Tai Chi diagram, also known as the \"yin-yang fish,\" which is symmetric about the center of the outer large circle;\nFigure 2 shows an equilateral triangle inscribed in a circle;\n\nFigure 3 shows a square inscribed in a circle;\n\nFigure 4 shows two concentric circles, where the radius of the smaller circle is two-thirds the radius of the outer large circle.\n\n\nFigure 1\n\n\nFigure 2\n\n\nFigure 3\n\n\nFigure 4\n\nOf these four patterns, the shaded area in each is not less than half the area of the outer large circle in the pattern. Which of the following are correct?\nA. Figure 1 and Figure 3\nB. Figure 2 and Figure 3\nC. Figure 2 and Figure 4\nD. Figure 1 and Figure 4", "input_image": [ "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0021_1.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0021_2.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0021_3.jpg", "batch19-2024_05_24_b4769f5b53f52ca7d5e9g_0021_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Figure (1): The cut-and-paste method involves cutting a shape and moving the cut portion to another position to make the problem easier to solve. Using the cut-and-paste method, it can be discovered that the area of the shaded part is exactly the area of a semicircle, which is half the area of the large circle.\n\nFigure (2):\n\n\n\nAs shown in the figure, drawing $\\mathrm{OD} \\perp \\mathrm{BC}$ through $\\mathrm{O}$, with $\\angle O B D=30^{\\circ}$, and $\\mathrm{OD}=\\frac{1}{2} \\mathrm{OB}=\\frac{1}{2} \\mathrm{R}$.\n\nUsing the Pythagorean theorem and the perpendicular chord theorem, we get:\n\n$B D=C D=\\frac{\\sqrt{3}}{2} R$, and the area $S_{ABC}=3 S_{BOC}=3 \\times \\frac{1}{2} \\times\\left(2 \\times \\frac{\\sqrt{3}}{2} R\\right) \\times \\frac{1}{2} R=\\frac{3 \\sqrt{3}}{4} R^{2}$.\n\n$\\frac{3 \\sqrt{3}}{4} \\mathrm{R}^{2}<\\frac{1}{2} \\pi \\mathrm{R}^{2}$.\n\nFigure (3):\n\n\n\nAs shown in the figure, the area of the square $=4 S_{AOB}=4 \\times \\frac{1}{2} \\times R^{2}=2 \\mathrm{R}^{2}>\\frac{1}{2} \\pi R^{2}$.\n\nFigure 4:\n\nThe area of the small shaded circle $=\\pi\\left(\\frac{2}{3} R\\right)^{2}=\\frac{4}{9} \\pi R^{2}<\\frac{1}{2} \\pi R^{2}$.\n\nTherefore, Figures 1 and 3 meet the requirements.\n\nHence, the correct choice is A.\n\n[Key Insight] This question tests the application of the cut-and-paste method, emphasizing the transformation of irregular shapes into regular shapes for area calculation, as well as the calculation of areas of regular triangles and squares inscribed in circles." }, { "problem_id": 853, "question": "As shown in the figure, Figure 1 depicts a highball glass filled with liquid (data as shown). After removing some of the liquid, it is placed on a horizontal table as shown in Figure 2. At this time, the distance from the liquid surface to the rim of the cup, denoted as $h$, is (_______).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{8}{5} \\mathrm{~cm}$\nB. $2 \\mathrm{~cm}$\nC. $\\frac{12}{5} \\mathrm{~cm}$\nD. $3 \\mathrm{~cm}$", "input_image": [ "batch32-2024_06_14_78b5e3e9fbbbfbfddf38g_0044_1.jpg", "batch32-2024_06_14_78b5e3e9fbbbfbfddf38g_0044_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n$AB = BC = DE = FE = 5$, $AC = DF = 6$, $GE = HE = 3$.\n\nAccording to the problem, $\\triangle GEH$ is an isosceles triangle. Therefore, drawing a perpendicular $EM$ from point $E$ to $DF$ at point $M$, which intersects $GH$ at point $N$,\n\n$\\therefore$ By the properties of an isosceles triangle, $EN$ is the perpendicular bisector of $GH$, and $EN$ is also the perpendicular bisector of $DF$.\n\nIn right triangles $\\triangle DEM$ and $\\triangle GEN$, $DM = \\frac{1}{2} DF = \\frac{1}{2} \\times 6 = 3$, and $DE = 5$,\n\n$\\therefore EM = \\sqrt{DE^2 - DM^2} = \\sqrt{5^2 - 3^2} = 4$,\n\n$\\because GH \\parallel DF$,\n\n$\\therefore \\triangle DEM \\sim \\triangle GEH$,\n\n$\\therefore \\frac{GE}{DE} = \\frac{EN}{EM} = \\frac{3}{5} = \\frac{EN}{4}$,\n\n$\\therefore EN = \\frac{12}{5}$,\n\n$\\therefore MN = EM - EN = 4 - \\frac{12}{5} = \\frac{8}{5}$, which means the distance $h$ from the liquid surface to the rim of the cup is $\\frac{8}{5} h$,\n\nTherefore, the correct choice is: A.\n\n【Key Insight】This problem mainly examines the proportional division of segments by parallel lines, the determination and properties of similar triangles. Mastering the proportional division of segments by parallel lines is key to solving the problem." }, { "problem_id": 854, "question": "In right triangle $\\triangle A B C$ as shown in Figure 1, $\\angle A C B = 90^\\circ$. Point $P$ moves at a speed of $1 \\text{~cm/s}$ from point $A$ along the broken line $A C \\rightarrow C B$ until it reaches point $B$. A perpendicular line $P D$ is drawn from point $P$ to side $A B$, with the foot of the perpendicular being $D$. The graph of the function $y(\\text{cm})$, which represents the length of $P D$, versus the time $x$ (seconds) that point $P$ has been moving is shown in Figure 2. When point $P$ has been moving for 6 seconds, the length of $P D$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $1.8 \\text{~cm}$\nB. $1.5 \\text{~cm}$\nC. $1.2 \\text{~cm}$\nD. $0.6 \\text{~cm}$", "input_image": [ "batch37-2024_06_14_aa09a85f660683fabf7cg_0019_1.jpg", "batch37-2024_06_14_aa09a85f660683fabf7cg_0019_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 2, it can be observed that,\n\nwhen point $P$ moves for 3 seconds, the length of $PD$ reaches its maximum, at which point $P$ moves to the position of point $C$.\n\nSince the speed of point $P$ is $1 \\mathrm{~cm}$ per second,\n\nwhen point $P$ moves for 3 seconds, $AC = 1 \\times 3 = 3 (\\mathrm{~cm})$.\n\nWhen point $P$ moves for 7 seconds, the length of $PD$ becomes $0 \\mathrm{~cm}$, meaning point $P$ has moved from point $C$ to point $B$ in 4 seconds.\n\nTherefore, it can be deduced that: $BC = 4 \\mathrm{~cm}$.\n\nSince $\\triangle ABC$ is a right-angled triangle,\n\nit follows that $AB = 5 \\mathrm{~cm}$.\n\nWhen point $P$ moves for 6 seconds to the position shown in the figure below,\n\n\n\nthen $AC + CP = 6 \\mathrm{~cm}$,\n\nso $CP = 3 \\mathrm{~cm}$, and $BP = BC - CP = 4 - 3 = 1 \\quad(\\mathrm{~cm})$.\n\nTherefore, $\\sin \\angle B = \\frac{AC}{AB} = \\frac{PD}{BP}$,\n\nthat is: $\\frac{3}{5} = \\frac{PD}{1}$,\n\nso $PD = \\frac{3}{5} = 0.6 \\quad(\\mathrm{~cm})$,\n\nHence, the correct choice is D.\n【Key Insight】This problem tests knowledge on solving right-angled triangles, trigonometric functions of acute angles, and moving point problems. The key to solving this problem lies in correctly determining the values of $BC$ and $AC$ and using trigonometric functions to find the length of segment $PD$." }, { "problem_id": 855, "question": "As shown in the figure, in the two identical $6 \\times 6$ square grids, squares $A B C D$ and $E F G H$ are located in each grid, with all vertices at the intersection of grid lines. The areas of squares $A B C D$ and $E F G H$ are denoted as $S_{1}$ and $S_{2}$, respectively, and the area of the two square grids is denoted as $S$. The following conclusions are given: (1) $S_{1} = \\frac{1}{2} S$; (2) $S_{2} = \\frac{1}{2} S$; (3) The ratio of the side lengths of square $A B C D$ to square $E F G H$ is $2 \\sqrt{5}: 3 \\sqrt{2}$. Which of the following conclusions are correct?\n\n\n\nA\n\n\n\nB\nA. (1)(2)\nB. (2)(3)\nC. (3)\nD. (1)(2)(3)", "input_image": [ "batch22-2024_06_14_eda6f53dfe1f1559330cg_0069_1.jpg", "batch22-2024_06_14_eda6f53dfe1f1559330cg_0069_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we have $S=6 \\times 6=36$.\n\nFrom grid A, the side length of square $ABCD$ is $\\sqrt{2^{2}+4^{2}}=2 \\sqrt{5}$.\n\nThus, $S_{1}=2 \\sqrt{5} \\times 2 \\sqrt{5}=20$.\n\nFrom grid B, the side length of square $EFGH$ is $\\sqrt{3^{2}+3^{2}}=3 \\sqrt{2}$.\n\nThus, $S_{2}=3 \\sqrt{2} \\times 3 \\sqrt{2}=18$.\n\nTherefore, $\\frac{S_{1}}{S}=\\frac{20}{36}$,\n\nSo, $S_{1}=\\frac{5}{9} S$, which means statement (1) is incorrect.\n\nSince $\\frac{S_{2}}{S}=\\frac{18}{36}$,\n\nThus, $S_{2}=\\frac{1}{2} S$, so statement (2) is correct.\n\nThe ratio of the side lengths of square $ABCD$ to square $EFGH$ is $2 \\sqrt{5}: 3 \\sqrt{2}$, making statement (3) correct.\n\nTherefore, the correct choice is B.\n\n[Highlight] This problem tests the application of squares and grid problems, as well as the Pythagorean theorem. The key to solving this problem lies in the flexible use of the learned knowledge." }, { "problem_id": 856, "question": "In Figure (1) and (2) are the perspective views and side profile of a horizontally placed chair. The height of the chair, $A C$, is given, and the width of the seat, $B E$, is $60 \\mathrm{~cm}$. The height of the legs, $E D$, is $35 \\mathrm{~cm}$, and $A C \\perp B E, A C \\perp C D, A C / / E D$. The angle of depression from point $A$ to point $E$ is measured to be $53^{\\circ}$. The length of $A C$ can be expressed as ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $35+60 \\sin 53^{\\circ}$\nB. $60+\\frac{35}{\\tan 53^{\\circ}}$\nC. $35+\\frac{60}{\\tan 53^{\\circ}}$\nD. $35+60 \\tan 53^{\\circ}$", "input_image": [ "batch37-2024_06_14_96b74b89976cda0e7739g_0040_1.jpg", "batch37-2024_06_14_96b74b89976cda0e7739g_0040_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AC \\perp BE \\), \\( AC \\perp CD \\), and \\( AC \\parallel ED \\),\n\nTherefore, quadrilateral \\( BCDE \\) is a rectangle, and \\( \\angle AEB = 53^\\circ \\),\n\nHence, \\( BC = DE = 35 \\text{ cm} \\),\n\nIn right triangle \\( \\triangle ABE \\), \\( \\angle ABE = 90^\\circ \\), \\( \\tan \\angle AEB = \\frac{AB}{BE} \\), and \\( BE = 60 \\text{ cm} \\),\n\nThus, \\( AB = BE \\cdot \\tan \\angle AEB = 60 \\tan 53^\\circ \\),\n\nTherefore, \\( AC = BC + AB = 35 + 60 \\tan 53^\\circ \\),\n\nHence, the correct choice is: D.\n\n【Key Insight】This problem examines the application of solving right triangle problems involving angles of elevation and depression. The key to solving the problem lies in understanding the given information and using the definitions of trigonometric functions to find the solution. This type of problem is commonly encountered in middle school exams." }, { "problem_id": 857, "question": "As shown in Figure 1, in quadrilateral $A B C D$, $A D \\parallel B C$, and $\\angle B = 30^\\circ$. Line $l$ is perpendicular to $A B$. When line $l$ moves along the ray $B C$ from point $B$ to the right, it intersects the sides of quadrilateral $A B C D$ at points $E$ and $F$. Let the distance that line $l$ moves to the right be $x$, and the length of segment $E F$ be $y$. The function relationship between $y$ and $x$ is shown in Figure 2. What is the perimeter of quadrilateral $A B C D$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $10+2 \\sqrt{3}$\nB. $8+2 \\sqrt{2}$\nC. $9+2 \\sqrt{5}$\nD. 10", "input_image": [ "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0097_1.jpg", "batch37-2024_06_14_f3df3ea39d2e7e3814b7g_0097_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\angle B=30^{\\circ}$ and the line $l \\perp AB$,\n\nTherefore, $BE=2EF$,\n\nFrom the diagram, we can deduce:\n\n$AB=4 \\cos 30^{\\circ}=4 \\times \\frac{\\sqrt{3}}{2}=2 \\sqrt{3}, \\quad BC=5, AD=7-4=3$,\n\nFrom the image, we can also deduce:\n\n$AN=5-4=1, \\quad ND=CM=7-5=2, \\quad DM=2$,\n\nSince $\\angle B=30^{\\circ}$ and $EF \\perp AB$,\nTherefore, $\\angle M=60^{\\circ}$,\n\nMoreover, since $DM=MC=2$,\n\nTherefore, $\\triangle DMC$ is an equilateral triangle,\n\nHence, $DC=DM=2$,\n\nTherefore, the perimeter of quadrilateral $ABCD$ is: $AB+BC+AD+CD=2 \\sqrt{3}+5+3+2=10+2 \\sqrt{3}$,\n\nThus, the correct choice is $A$.\n\n\n\nFigure 1\n\n【Key Insight】This problem examines the function graph of a moving point. The key to solving it lies in understanding the problem and using a combination of numerical and graphical methods to find the solution." }, { "problem_id": 858, "question": "A rectangle is divided into three right-angled triangles as shown in Figure 1. The area of the smallest triangle is $S_{1}$, and the larger two triangle paper pieces are arranged as shown in Figure 2, with the shaded area in Figure 2 being $S_{2}$. If $S_{2} = 2 S_{1}$, the ratio of the length to the width of the rectangle is $(\\quad)$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 2\nB. $\\sqrt{2}$\nC. $\\frac{4}{3}$\nD. $\\sqrt{3}$", "input_image": [ "batch32-2024_06_14_621a06aac7ad8d6dddd1g_0047_1.jpg", "batch32-2024_06_14_621a06aac7ad8d6dddd1g_0047_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in Figure (1), let the area of the right triangle $Rt \\triangle BDC$ be $S_{3}$.\n\nAs shown in Figure (2), according to the problem statement, we know that $\\angle 1 = \\angle 2$, thus $\\angle 3 + \\angle 1 = \\angle 2 + \\angle 4$.\n\nTherefore, $\\angle 3 = \\angle 4$.\n\nHence, $OC = OA = OD$.\n\nThus, $OA = \\frac{1}{2} AC$.\n\nTherefore, $S_{3} = 2S_{2}$.\n\nSince $S_{2} = 2S_{1}$,\n\n$\\frac{S_{1}}{S_{3}} = \\frac{1}{4}$.\n\nSince $\\angle ABD + \\angle CBD = 90^{\\circ}$ and $\\angle ABD + \\angle BAD = 90^{\\circ}$,\n\n$\\angle CBD = \\angle BAD$.\n\nMoreover, since $\\angle BDC = \\angle ADB = 90^{\\circ}$,\n\n$\\triangle ABD \\sim \\triangle BCD$.\n\nTherefore, $\\frac{S_{1}}{S_{3}} = \\left(\\frac{AB}{BC}\\right)^{2}$.\n\nThus, $\\frac{AB}{BC} = \\frac{1}{2}$.\n\nTherefore, the ratio of the length to the width of the rectangle is 2.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nTherefore, the correct answer is A.\n\n【Key Insight】This problem examines the property that the median to the hypotenuse of a right triangle is equal to half the hypotenuse, equal angles correspond to equal sides, the area ratio of similar triangles equals the square of the similarity ratio, and the concept of similar triangles. The key to solving this problem lies in finding the relationship between $S_{1}$, $S_{2}$, and $S_{3}$." }, { "problem_id": 859, "question": "Rectangle paper $A B C D$ is cut as shown in Figure 1 into three right-angled triangles (1), (2), and (3). These three triangles are then overlaid as shown in Figure 2, with the shaded areas representing the overlapping parts of (1), (2), and (3), where one endpoint of the hypotenuse of (1) coincides with a point on the hypotenuse of (2). The value of $\\frac{A B}{B C}$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{3}{2}$\nB. $\\sqrt{2}$\nC. $\\frac{4}{3}$\nD. $\\frac{2 \\sqrt{3}}{3}$", "input_image": [ "batch37-2024_06_14_5565145dfb216bdb9dc3g_0053_1.jpg", "batch37-2024_06_14_5565145dfb216bdb9dc3g_0053_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( DE = x \\), and set \\( AB = b \\), \\( BC = a \\), as shown in the figures,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nTherefore, \\( AB \\cdot BC = AC \\cdot DE \\), which is \\( \\sqrt{a^{2}+b^{2}} \\cdot x = a b \\),\n\nThus, \\( x = \\frac{a b}{\\sqrt{a^{2}+b^{2}}} \\),\n\nSince \\( \\tan \\angle BAC = \\frac{BC}{AB} = \\frac{a}{b} \\),\n\nAnd \\( \\angle BAC + \\angle DAE = 90^{\\circ} \\), \\( \\angle DAE + \\angle ADE = 90^{\\circ} \\),\n\nTherefore, \\( \\angle ADE = \\angle BAC \\),\n\nSimilarly, \\( \\angle EDC = \\angle ACB \\),\n\nThus, \\( \\tan \\angle ADE = \\frac{AE}{DE} = \\frac{a}{b} \\),\n\nTherefore, \\( AE = DE \\cdot \\tan \\angle ADE = \\frac{a b}{\\sqrt{a^{2}+b^{2}}} \\cdot \\frac{a}{b} = \\frac{a^{2}}{\\sqrt{a^{2}+b^{2}}} \\),\n\nSince \\( \\angle EDC = \\angle ACB \\),\n\nTherefore, \\( \\angle A'D'C = \\angle ACB \\),\n\nThus, \\( A'E' = BC - \\frac{1}{2} CD' \\),\n\nSince \\( CD' = ED \\), \\( A'E' = AE \\),\n\nTherefore, \\( AE = BC - \\frac{1}{2} ED = a - \\frac{1}{2} \\frac{a b}{\\sqrt{a^{2}+b^{2}}} \\),\n\nThus, \\( \\frac{a^{2}}{\\sqrt{a^{2}+b^{2}}} = a - \\frac{1}{2} \\frac{a b}{\\sqrt{a^{2}+b^{2}}} \\),\n\nSimplifying, we get \\( 4 a^{3} b = 3 a^{2} b^{2} \\), which is \\( \\frac{a}{b} = \\frac{3}{4} \\),\n\nSince \\( a > 0 \\), \\( b > 0 \\),\n\nTherefore, \\( \\frac{AB}{BC} = \\frac{DE}{AE} = \\frac{b}{a} = \\frac{4}{3} \\).\n\nHence, the answer is: C.\n\n【Key Insight】This problem examines the Pythagorean theorem, properties of rectangles, solving right triangles, and properties of isosceles triangles. The key to solving it is to appropriately set up unknowns and establish equations." }, { "problem_id": 860, "question": "Figure 1 shows a foldable desk lamp. Figure 2 is its plan view. The base $A O$ is perpendicular to $O E$ at point $O$. The support rods $A B$ and $B C$ are fixed, and $\\angle A$ is twice the size of $\\angle B$. The lampshade $C D$ can rotate around point $C$ for adjustment. Now, the lampshade $C D$ is rotated from the horizontal position to the position $C D^{\\prime}$ (as shown by the dashed line in Figure 2), at which point the line containing $C D^{\\prime}$ is perpendicular to the support rod $A B$, and $\\angle B C D - \\angle D C D^{\\prime} = 126^\\circ$. The measure of $\\angle D C D^{\\prime}$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $26^\\circ$\nB. $36^\\circ$\nC. $46^\\circ$\nD. $72^\\circ$", "input_image": [ "batch14-2024_06_15_e140a62ba5fbff1efa45g_0092_1.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0092_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nExtend \\( O A \\) to intersect \\( C D \\) at point \\( F \\), and extend \\( D' C \\) to intersect \\( A B \\) at point \\( G \\), as shown in the figure:\n\n\n\nSince \\( C D \\parallel O E \\) and \\( A O \\perp O E \\), it follows that \\( O A \\perp C D \\).\n\nGiven \\( A O \\perp O E \\) and \\( D' C \\perp A B \\), we have:\n\\[\n\\angle A G C = \\angle A F C = 90^\\circ.\n\\]\nThus,\n\\[\n\\angle G C F + \\angle G A F = 180^\\circ.\n\\]\nSince \\( \\angle D C D' + \\angle G C F = 180^\\circ \\), it follows that:\n\\[\n\\angle D C D' = \\angle G A F.\n\\]\nTherefore,\n\\[\n\\angle B A O = 180^\\circ - \\angle D C D',\n\\]\nand\n\\[\n\\angle B = \\frac{1}{2} \\left(180^\\circ - \\angle D C D'\\right).\n\\]\nGiven that \\( \\angle B C D - \\angle D C D' = 126^\\circ \\), we have:\n\\[\n\\angle B C D = \\angle D C D' + 126^\\circ.\n\\]\nIn quadrilateral \\( A B C F \\), the sum of the interior angles is:\n\\[\n\\angle G A F + \\angle B + \\angle B C D + \\angle A F C = 360^\\circ.\n\\]\nSubstituting the known values:\n\\[\n\\angle D C D' + \\frac{1}{2} \\left(180^\\circ - \\angle D C D'\\right) + \\angle D C D' + 126^\\circ + 90^\\circ = 360^\\circ.\n\\]\nSolving this equation yields:\n\\[\n\\angle D C D' = 36^\\circ.\n\\]\nTherefore, the correct answer is **B**.\n\n**Key Insight:**\nThis problem tests the application of the sum of interior angles in a quadrilateral, the equality of supplementary angles, the definition of adjacent supplementary angles, and the relationships between angles in geometric figures. The key to solving the problem lies in correctly constructing auxiliary lines and utilizing the fact that the sum of the interior angles of a quadrilateral is \\( 360^\\circ \\)." }, { "problem_id": 861, "question": "As shown in the figure, in quadrilateral $A B C D$, $A D \\parallel B C$, $\\angle D=90^{\\circ}$, $A B=B C=5$, and $\\tan A=\\frac{4}{3}$. Point $P$ moves along the path $A \\rightarrow B \\rightarrow C \\rightarrow D$ from point $A$ at a speed of 1 unit per second towards point $D$. A perpendicular line $P H$ is drawn from point $P$ to $A D$, with the foot of the perpendicular being $H$. Let the time when point $P$ is moving be $x$ (unit: $s$), and the area of $\\triangle A P H$ be $y$. The graph of $y$ against $x$ is approximately ( )\n\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch37-2024_06_14_95e8354843f9253fddb3g_0019_1.jpg", "batch37-2024_06_14_95e8354843f9253fddb3g_0019_2.jpg", "batch37-2024_06_14_95e8354843f9253fddb3g_0019_3.jpg", "batch37-2024_06_14_95e8354843f9253fddb3g_0019_4.jpg", "batch37-2024_06_14_95e8354843f9253fddb3g_0019_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: (1) When point \\( P \\) moves on \\( AB \\),\n\nSince \\( AB = BC = 5 \\) and \\( \\tan A = \\frac{4}{3} \\),\n\nTherefore, \\( AP : PH : AH = 5 : 4 : 3 \\),\n\nSince \\( AP = x \\),\n\nTherefore, \\( PH = \\frac{4}{5}x \\), \\( AH = \\frac{3}{5}x \\),\n\n\\( y = \\frac{1}{2} AH \\cdot PH = \\frac{1}{2} \\cdot \\frac{3}{5}x \\cdot \\frac{4}{5}x = \\frac{6}{25}x^2 \\), the graph is a quadratic function;\n\nAnd when \\( x = 5 \\), \\( y = 6 \\); hence \\( B, C, D \\) are incorrect; thus \\( A \\) is correct;\n\n(2) When point \\( P \\) moves on \\( BC \\), as shown in the figure below, draw \\( BE \\perp AD \\) at point \\( E \\),\n\n\n\nSince \\( \\tan A = \\frac{4}{3} \\), \\( AB = 5 \\),\nTherefore, \\( BE = 4 \\), \\( AE = 3 \\),\n\nSince \\( AB + BP = x \\),\n\nTherefore, \\( BP = EH = x - 5 \\),\n\nTherefore, \\( AH = 2 + x - 5 = x - 2 \\),\n\nTherefore, \\( y = \\frac{1}{2} AH \\cdot PH = \\frac{1}{2} \\cdot (x - 2) \\cdot 4 = 2x - 4 \\), which is a linear function;\n\nAnd when \\( x = 10 \\), \\( y = 16 \\);\n\n(3) When point \\( P \\) moves on \\( CD \\),\n\n\n\nAt this time, \\( AD = AH = 3 + 5 = 8 \\),\n\nSince \\( AB + BC + CP = x \\),\n\nTherefore, \\( PH = AB + BC + CD - x = 14 - x \\),\n\nTherefore, \\( y = \\frac{1}{2} AH \\cdot PH = \\frac{1}{2} \\times 8 \\cdot (14 - x) = -4x + 56 \\);\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This problem is a comprehensive motion-type question, examining the function graph of moving points, solving right triangles, and calculating areas of shapes. The key to solving the problem lies in deeply understanding the function graph of moving points, comprehending the actual meanings represented by key points in the graph, and understanding the complete motion process of the moving point." }, { "problem_id": 862, "question": "As shown in Figure (1) and (2), in Figure (1), a child is playing the game of \"rolling a hoop.\" The hoop is circular, and as the hoop rolls forward, the hoop hook remains tangent to the hoop. This game is abstracted into a mathematical problem as shown in Figure (2). Given that the radius of the hoop is $26 \\mathrm{~cm}$, let the center of the hoop be $O$, the point where the hoop hook is tangent to the hoop be $M$, and the point where the hoop touches the ground be $A$. It is given that $\\angle M O A = \\alpha$, and $\\tan \\alpha = \\frac{5}{12}$. If the horizontal distance $A C$ from the person's standing point $C$ to point $A$ is $46 \\mathrm{~cm}$, then the length of the hoop hook $M F$ is ( ) $\\mathrm{cm}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 40\nB. 42\nC. 39\nD. 36", "input_image": [ "batch37-2024_06_14_75fe8f1a75ca44f27540g_0067_1.jpg", "batch37-2024_06_14_75fe8f1a75ca44f27540g_0067_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular from point \\( M \\) to \\( OA \\), intersecting \\( OA \\) at point \\( D \\). Extend \\( DM \\) to intersect \\( CF \\) at point \\( E \\).\n\nThen, \\( \\angle MDO = \\angle FEM = 90^\\circ \\).\n\n\n\nFigure 2\n\nGiven that \\( \\tan \\alpha = \\frac{5}{12} \\) and \\( OM = 26 \\),\n\nLet \\( MD = 5x \\) and \\( OD = 12x \\).\n\nBy the Pythagorean theorem: \\( (5x)^2 + (12x)^2 = 26^2 \\),\n\nSolving for \\( x \\): \\( x = 2 \\),\n\nThus, \\( DM = 10 \\) and \\( OD = 24 \\).\n\nSince \\( DE = AC = 46 \\),\n\nTherefore, \\( ME = DE - DM = 36 \\).\n\nGiven that \\( \\angle OMF = 90^\\circ \\),\n\nThus, \\( \\angle OMD + \\angle FME = 90^\\circ \\),\n\nAnd since \\( \\angle OMD + \\angle MOD = 90^\\circ \\),\n\nIt follows that \\( \\angle MOD = \\angle FME \\).\n\nAlso, since \\( \\angle MDO = \\angle FEM = 90^\\circ \\),\n\nTherefore, \\( \\triangle ODM \\sim \\triangle MEF \\),\n\nHence, \\( \\frac{OD}{ME} = \\frac{OM}{MF} \\), which is \\( \\frac{24}{36} = \\frac{26}{MF} \\),\n\nSolving for \\( MF \\): \\( MF = 39 \\).\n\nTherefore, the correct answer is C.\n\n【Key Insight】This problem tests the understanding of trigonometric functions of acute angles, the Pythagorean theorem, and the properties of similar triangles. The key to solving the problem lies in constructing the auxiliary line and proving that \\( \\triangle ODM \\sim \\triangle MEF \\)." }, { "problem_id": 863, "question": "Triangle paper $A B C$ (as shown in Figure Alpha) has a perimeter of $38 \\mathrm{~cm}$, with $A B=A C$. The paper is folded as shown in the figure, so that point $A$ coincides with point $B$, and the crease is $D E$ (as shown in Figure Beta). If the perimeter of $\\triangle D B C$ is $25 \\mathrm{~cm}$, then the length of $B C$ is ( )\n\n\n\nAlpha\n\n\n\nBeta\nA. $10 \\mathrm{~cm}$\nB. $12 \\mathrm{~cm}$\nC. $15 \\mathrm{~cm}$\nD. $13 \\mathrm{~cm}$", "input_image": [ "batch35-2024_06_17_5a4127c6911cd24b9845g_0033_1.jpg", "batch35-2024_06_17_5a4127c6911cd24b9845g_0033_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Because folding triangle ADE along DE causes point A to coincide with point B,\n\nTherefore, AD = BD.\n\nGiven that the perimeter of triangle ABC is 38 cm and the perimeter of triangle DBC is 25 cm,\n\nWe have AB + AC + BC = 38 cm,\n\nBD + CD + BC = AD + CD + BC = AC + BC = 25 cm,\n\nThus, AB = 13 cm = AC,\n\nTherefore, BC = 25 - 13 = 12 cm.\n\nThe correct choice is: B.\n\n[Key Insight] This problem examines the transformation of flipping, and the key to solving it lies in the proficient application of the properties of folding." }, { "problem_id": 864, "question": "Reading Comprehension: As shown in Figure 1, in a plane, a fixed point $O$ is selected, along with a directed ray $O N$. A unit length is then chosen. The position of any point $M$ in the plane can be determined by the length $m$ of $O M$ and the angle $\\angle M O N$ in degrees. The ordered pair $(m, \\theta)$ is called the \"polar coordinates\" of point $M$, and this system of coordinates is called a \"polar coordinate system\".\n\nApplication: In the polar coordinate system of Figure 2, if the side length of the regular hexagon is 2, with one side $O A$ on the ray $O N$, what should the polar coordinates of the hexagon's vertex $C$ be recorded as $(\\quad)$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\left(4,60^{\\circ}\\right)$\nB. $\\left(4,45^{\\circ}\\right)$\nC. $\\left(2 \\sqrt{2}, 60^{\\circ}\\right)$\nD. $\\left(2 \\sqrt{2}, 50^{\\circ}\\right)$", "input_image": [ "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0054_1.jpg", "batch20-2024_05_23_7fc6f7b914f2b87655bdg_0054_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Let the center of the regular hexagon be $D$, and connect $A D$.\n\nSince $\\angle A D O = 360^{\\circ} \\div 6 = 60^{\\circ}$ and $O D = A D$,\n\nTherefore, $\\triangle A O D$ is an equilateral triangle,\n\nHence, $O D = O A = 2$, and $\\angle A O D = 60^{\\circ}$,\n\nThus, $O C = 2 O D = 4$,\n\nTherefore, the polar coordinates of the vertex $C$ of the regular hexagon are recorded as $\\left(4,60^{\\circ}\\right)$,\n\nHence, the correct choice is: A.\n\n【Key Insight】This question examines the properties of regular polygons and circles, and the determination of coordinates. It primarily utilizes the properties of a regular hexagon. Understanding the problem information and grasping the definition of \"polar coordinates\" are crucial for solving the problem." }, { "problem_id": 865, "question": "After learning about triangles, Xiaoming made a \"trisector\" to trisect any angle using the \"trisector\" shown in Figure 1. This trisector (Figure 2) consists of two grooved rods $O A$ and $O B$ that are connected at point $O$ and can rotate around $O$. Point $C$ is fixed, and $O C = C D = D E$. Points $D$ and $E$ can slide in the grooves. If $\\angle B D E = 75^\\circ$, what is the measure of $\\angle C D E$ in degrees?\n\n\nFigure 1\n\n\n\nFigure 2\nA. $50^\\circ$\nB. $70^\\circ$\nC. $75^\\circ$\nD. $80^\\circ$", "input_image": [ "batch5-2024_06_14_957e8d37715e2ff5656ag_0036_1.jpg", "batch5-2024_06_14_957e8d37715e2ff5656ag_0036_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Since \\( OC = CD = DE \\),\n\nTherefore, \\( \\angle O = \\angle ODC \\) and \\( \\angle DCE = \\angle DEC \\).\n\nLet \\( \\angle O = \\angle ODC = x \\),\n\nThus, \\( \\angle DCE = \\angle DEC = 2x \\),\n\nHence, \\( \\angle CDE = 180^\\circ - \\angle DCE - \\angle DEC = 180^\\circ - 4x \\).\n\nGiven that \\( \\angle BDE = 75^\\circ \\), and \\( \\angle ODC + \\angle CDE + \\angle BDE = 180^\\circ \\),\n\nThat is, \\( x + 180^\\circ - 4x + 75^\\circ = 180^\\circ \\),\n\nSolving gives: \\( x = 25^\\circ \\),\n\nTherefore, \\( \\angle CDE = 180^\\circ - 4x = 80^\\circ \\).\n\nThe correct choice is: D.\n\n【Key Insight】This problem examines the properties of isosceles triangles and the external angle properties of triangles. Understanding the relationships between the various angles is crucial to solving this problem." }, { "problem_id": 866, "question": "In order to test whether the size of an iron ball produced by the factory meets the requirements, the worker designed a workpiece slot as shown in Figure (1). Its two bottom angles are both $90^{\\circ}$. When a regular-shaped iron ball is placed in the slot, if it has the three contact points $A, B, and E$ shown in Figure (1) at the same time, the size of the ball meets the requirements. Figure (2) is a schematic cross-sectional view through the center of the ball and the three points $A, B, and E$. It is known that the diameter of $\\odot O$ is the diameter of the iron ball, $AB$ is the chord of $\\odot O$, $CD$ intersects $\\odot O$ at point $E, AC \\perp CD, BD \\perp CD$, if $CD=16 \\mathrm{~cm}, AC=BD=4 \\mathrm{~cm}$, then the diameter of this iron ball is ( )\n\n\n\nE\n\n\n\nPicture (2)\nA. $10 \\mathrm{~cm}$\nB. $15 \\mathrm{~cm}$\nC. $20 \\mathrm{~cm}$\nD. $24 \\mathrm{~cm}$", "input_image": [ "batch19-2024_05_24_4e201766507f5a421e2bg_0064_1.jpg", "batch19-2024_05_24_4e201766507f5a421e2bg_0064_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $O A$ and $O E$, and let $O E$ intersect $A B$ at point $P$,\n\n\n\n$\\because A C = B D$, $A C \\perp C D$, and $B D \\perp C D$,\n\n$\\therefore$ quadrilateral $A B D C$ is a rectangle,\n\n$\\because C D$ is tangent to $O$ at point $E$, and $O E$ is the radius of $O$,\n\n$\\therefore O E \\perp C D$, and $O E \\perp A B$,\n\n$\\therefore P A = P B$, and $P E = A C$,\n\n$\\because A B = C D = 16 \\mathrm{~cm}$,\n\n$\\therefore P A = 8 \\mathrm{~cm}$,\n\n$\\because A C = B D = P E = 4 \\mathrm{~cm}$,\n\nIn right triangle $O A P$, by the Pythagorean theorem,\n\n$P A^{2} + O P^{2} = O A^{2}$\n\n$8^{2} + (O A - 4)^{2} = O A^{2}$\n\nSolving, we get $O A = 10$,\n\nThus, the diameter of the iron ball $= 2 O A = 2 \\times 10 = 20 \\mathrm{~cm}$,\n\nTherefore, the correct answer is C.\n 【Key Insight】This problem examines the properties of tangents, the chord perpendicularity theorem, and the Pythagorean theorem. The key to solving it lies in mastering these concepts." }, { "problem_id": 867, "question": "As shown in Figure 1, in rectangle $A B C D$ (where $A B > A D$), point $P$ starts from point $B$ and moves along $B \\rightarrow C$ at a constant speed until it reaches point $C$. Point $Q$ starts from point $D$ and moves along $D \\rightarrow C$ at the same speed until it reaches point $C$. Given that points $P$ and $Q$ start moving simultaneously, connect $A P$ and $A Q$. Let $D Q = x$ and $A P - A Q = y$, where the graph of $y$ as a function of $x$ is shown in Figure 2. What is the value of $m$ in the graph?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $3 \\sqrt{5}-3 \\sqrt{2}$\nB. $\\sqrt{10}-2$\nC. $4 \\sqrt{10}-2$\nD. $2 \\sqrt{10}-2$", "input_image": [ "batch14-2024_06_15_e140a62ba5fbff1efa45g_0022_1.jpg", "batch14-2024_06_15_e140a62ba5fbff1efa45g_0022_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "According to the function graph, it is known that \\( AB - AD = \\sqrt{2} \\).\n\nSince points \\( P \\) and \\( Q \\) move at the same speed,\n\nwhen \\( x = \\sqrt{2} \\), point \\( P \\) coincides with point \\( C \\), at which time \\( BC = DQ = \\sqrt{2} \\), and \\( AP - AQ = m \\).\n\nSince quadrilateral \\( ABCD \\) is a rectangle,\n\n\\( AD = BC = \\sqrt{2} \\), and \\( AB = CD \\),\n\ntherefore \\( AB = CD = AD + \\sqrt{2} = 2\\sqrt{2} \\).\n\nWhen \\( x = \\sqrt{2} \\), in the right triangle \\( ABP \\), \\( AP = \\sqrt{BC^2 + AB^2} = \\sqrt{(\\sqrt{2})^2 + (2\\sqrt{2})^2} = \\sqrt{10} \\),\n\nand in the right triangle \\( \\triangle ADQ \\), \\( AQ = \\sqrt{DQ^2 + AD^2} = \\sqrt{(\\sqrt{2})^2 + (\\sqrt{2})^2} = 2 \\),\n\ntherefore \\( m = AP - AQ = \\sqrt{10} - 2 \\).\n\nHence, the correct choice is B.\n\n【Key Insight】This problem examines the moving point problem and function graphs, the properties of rectangles, and the Pythagorean theorem. It incorporates the core competencies of geometric intuition and reasoning ability in mathematics, and the key to solving the problem lies in the use of the idea of combining numbers and shapes." }, { "problem_id": 868, "question": "As shown in Figure 1, Xiaoming is walking straight away from the streetlight. This situation is abstracted into the geometric figure shown in Figure 2. Given that the distance from the streetlight bulb to the ground, $A B$, is 4 meters, Xiaoming's height, $C D$, is 1.5 meters, and the distance from Xiaoming to the point directly below the streetlight bulb, $B C$, is 4 meters. When Xiaoming reaches point $E$, he notices that his shadow has increased by 2 meters. Then the distance Xiaoming has walked, $C E$, is equal to ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Between 3 and 4 meters\nB. Between 4 and 5 meters\nC. Between 5 and 6 meters\nD. Between 6 and 7 meters", "input_image": [ "batch32-2024_06_14_20d9f7a744747cd4199bg_0008_1.jpg", "batch32-2024_06_14_20d9f7a744747cd4199bg_0008_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "From the diagram, it can be seen that when Xiao Ming is at point $C$, the length of his shadow is $CM$, and when he is at point $E$, the length of his shadow is $EN$. According to the problem, $AB \\perp BN$, $CD \\perp BN$, $EF \\perp BN$, and $EF = CD = 1.5$ meters, $EN = (CM + 2)$ meters.\n\n$\\therefore \\angle DCM = \\angle ABM = 90^{\\circ}$,\n$\\because \\angle CMD = \\angle BMA$,\n\n$\\therefore \\triangle DCM \\sim \\triangle ABM$,\n\n$\\therefore \\frac{CM}{BM} = \\frac{DC}{AB}$,\n\n$\\because BM = BC + CM = 4 + CM$,\n\n$\\therefore \\frac{CM}{4 + CM} = \\frac{1.5}{4}$,\n\nSolving gives $CM = 2.4$,\n\n$\\therefore EN = CM + 2 = 2.4 + 2 = 4.4$,\n\n$\\because \\angle FEN = \\angle ABN = 90^{\\circ}$, $\\angle ENF = \\angle BNA$,\n\n$\\therefore \\triangle FEN \\sim \\triangle ABN$,\n\n$\\therefore \\frac{EN}{BN} = \\frac{FE}{AB}$, that is $\\frac{4.4}{BN} = \\frac{1.5}{4}$,\n\nSolving gives $BN = \\frac{176}{15}$,\n\n$\\therefore CE = BN - BC - EN = \\frac{176}{15} - 4 - 4.4 = \\frac{10}{3}$,\n\n$\\because 3 < \\frac{10}{3} < 4$,\n\n$\\therefore$ The distance Xiao Ming walked, $CE$, is between 3 and 4,\n\nTherefore, the correct answer is A.\n\n\n\n【Key Insight】This problem tests the understanding of the properties and criteria of similar triangles. Mastering the theorems related to similar triangles and correctly interpreting the problem are crucial for solving it." }, { "problem_id": 869, "question": "Grade 9 Class 16 plans to grow vegetables in their school's labor practice base. The class monitor has purchased an 8-meter-long fence to enclose a vegetable garden with one side against a wall (which is long enough). To maximize the garden area, the students have proposed three plans: a rectangle, an isosceles triangle (with the base against the wall), and a semicircle. Which of these plans is the most optimal?\n\n\n\nPlan 1\n\n\n\nPlan 2\n\n\n\nPlan 3\nA. Plan 1\nB. Plan 2\nC. Plan 3\nD. The areas are all the same", "input_image": [ "batch6-2024_06_14_d6212742b424b34c5ef2g_0028_1.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0028_2.jpg", "batch6-2024_06_14_d6212742b424b34c5ef2g_0028_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\n**Option 1:** \nLet \\( AD = x \\) meters, then \\( AB = (8 - 2x) \\) meters.\n\nThe area of the vegetable garden is given by:\n\\[\n\\text{Area} = x(8 - 2x) = -2x^2 + 8x = -2(x - 2)^2 + 8.\n\\]\nWhen \\( x = 2 \\), the garden reaches its maximum area of 8 square meters.\n\n---\n\n**Option 2:** \n**Solution 1:** \nAs shown in the figure, draw \\( BH \\perp AC \\) at point \\( H \\). Then \\( BH \\leq AB = 4 \\).\n\nSince the area of triangle \\( ABC \\) is:\n\\[\nS_{\\triangle ABC} = \\frac{1}{2} \\cdot AC \\cdot BH,\n\\]\nwhen \\( BH = 4 \\), the maximum area of \\( \\triangle ABC \\) is:\n\\[\n\\frac{1}{2} \\times 4 \\times 4 = 8.\n\\]\n\n**Solution 2:** \nDraw \\( AD \\perp BC \\) at point \\( D \\). Let \\( CD = x \\) and \\( AD = y \\). Then:\n\\[\nx^2 + y^2 = 16.\n\\]\nThe area of the garden is:\n\\[\nS = \\frac{1}{2} \\cdot BC \\cdot AD = \\frac{1}{2} \\times 2x \\times y = xy.\n\\]\nSince:\n\\[\n(x - y)^2 = x^2 + y^2 - 2xy \\geq 0,\n\\]\nwe have:\n\\[\n16 - 2xy \\geq 0 \\implies xy \\leq 8.\n\\]\nThus, the maximum area of 8 square meters is achieved when \\( x = y = 2\\sqrt{2} \\).\n\n---\n\n**Option 3:** \nThe radius of the semicircle is \\( \\frac{8}{\\pi} \\) meters. The maximum area of the garden is:\n\\[\n\\frac{\\pi \\times \\left(\\frac{8}{\\pi}\\right)^2}{2} = \\frac{32}{\\pi} \\text{ square meters}.\n\\]\nSince \\( \\frac{32}{\\pi} > 8 \\), Option 3 yields the largest garden area.\n\n---\n\n**Conclusion:** \nAmong the three options, **Option 3** is the best choice.\n\n**Answer:** C.\n\n**Key Insight:** \nThis problem involves the application of quadratic functions, the area of a circle, properties of isosceles triangles, the Pythagorean theorem, and the perfect square formula. Calculating the dimensions and radii for each option is crucial to solving the problem." }, { "problem_id": 870, "question": "The Pythagorean theorem, together with the golden ratio, is considered one of the two great treasures of geometry. The discovery of the Pythagorean theorem can be called a milestone in the history of mathematics. For over 2000 years, people have conducted extensive research on it, and there are now hundreds of proof methods. Using the relationship between the areas of the shapes in the figure can prove the Pythagorean theorem. Among the four figures (2)(3)(4)(5) formed by arranging the right-angled triangle paper as shown in Figure (1), the number of figures that can be used to prove the Pythagorean theorem is ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\n\n\n\n(5)\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch12-2024_06_15_ed1911a30eded62729c0g_0041_1.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0041_2.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0041_3.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0041_4.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0041_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement:\n\nIn (2), the area of the rectangle equals the sum of the areas of two right-angled triangles, both being $a b$, which cannot prove the Pythagorean theorem.\n\nIn (3), the area of the trapezoid equals the sum of the areas of two right-angled triangles with legs $a$ and $b$, and an isosceles right-angled triangle with leg $c$, that is,\n\n$$\n\\frac{1}{2}(a+b)^{2}=\\frac{1}{2} a b \\times 2+\\frac{1}{2} c^{2}\n$$\n\nSimplifying, we get: $a^{2}+b^{2}=c^{2}$, which proves the Pythagorean theorem.\n\nIn (4), the area of the large square equals the sum of the areas of four small right-angled triangles and one small square, that is, $c^{2}=4 \\times \\frac{1}{2} a b+(b-a)^{2}$,\n\nSimplifying, we get: $a^{2}+b^{2}=c^{2}$, which proves the Pythagorean theorem.\n\nIn (5), the area of the large square equals the sum of the areas of four small right-angled triangles and one small square, that is, $(a+b)^{2}=\\frac{1}{2} a b \\times 4+c^{2}$,\n\nSimplifying, we get: $a^{2}+b^{2}=c^{2}$, which proves the Pythagorean theorem.\n\nTherefore, the figures that can prove the Pythagorean theorem are (3), (4), and (5), totaling 3 figures.\n\nHence, the answer is: C\n\n[Key Insight] This problem mainly examines the proof of the Pythagorean theorem. Mastering the different methods of calculating the areas of trapezoids and squares is key to solving the problem." }, { "problem_id": 871, "question": "Place a pair of right-angled triangles as shown in Figure ( A), where $\\angle A = 45^\\circ, \\angle D = 30^\\circ, \\angle ACB = \\angle DCE = 90^\\circ$, the hypotenuse $AB = 6$, and $DC = 7$. Rotate triangle $DCE$ clockwise around point $C$ by $15^\\circ$ to obtain $\\triangle D'^CE'$ (as shown in Figure ( B)). At this time, $AB$ and $CD'$ intersect at point $O$. Then the length of segment $AD' = (\\quad)$\n\n\n\n( A)\n\n\n\n( B)\nA. 4\nB. 5\nC. 6\nD. 7", "input_image": [ "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0083_1.jpg", "batch26-2024_06_17_fd940ef8bfb77a90be8ag_0083_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\angle ACB = \\angle DEC = 90^\\circ$, $\\angle A = 45^\\circ$, and $\\angle D = 30^\\circ$,\n\nTherefore, $\\angle DCE = 60^\\circ$ and $\\angle B = 45^\\circ$.\n\nSince the triangle $DCE$ is rotated $15^\\circ$ clockwise around point $C$ to obtain $\\triangle D'CE'$,\n\nTherefore, $\\angle D'CE' = 60^\\circ$ and $\\angle BCE' = 15^\\circ$.\n\nHence, $\\angle OCB = 45^\\circ$.\n\nAlso, since $\\angle B = 45^\\circ$,\n\nTherefore, $\\angle COB = 90^\\circ$.\n\nMoreover, since $\\triangle ACB$ is an isosceles right triangle,\n\nTherefore, $AO = CO = BO = 3 \\text{ cm}$.\n\nThus, $D'O = 4 \\text{ cm}$.\n\nTherefore, $AD' = \\sqrt{AO^2 + OD'^2} = \\sqrt{3^2 + 4^2} = 5 \\text{ cm}$.\n\nHence, the correct choice is: B.\n\n[Key Insight] This problem examines the properties of rotation, right triangles, isosceles right triangles, and the Pythagorean theorem. The key to solving the problem lies in the flexible application of these properties." }, { "problem_id": 872, "question": "(1) As shown in Figure 1, if $A B \\parallel C D$, then $\\angle A + \\angle E + \\angle C = 360^\\circ$; \n(2) As shown in Figure 2, if $A B \\parallel C D$, then $\\angle P = \\angle A - \\angle C$; \n(3) As shown in Figure 3, if $A B \\parallel C D$, then $\\angle E = \\angle A + \\angle 1$; \n(4) As shown in Figure 4, if the straight lines $A B \\parallel C D \\parallel E F$, and point $O$ is on the straight line $E F$, then $\\angle \\alpha - \\angle \\beta + \\angle \\gamma = 180^\\circ$. \nHow many of the above conclusions are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch29-2024_06_14_e7f29cc027e50efb8665g_0080_1.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0080_2.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0080_3.jpg", "batch29-2024_06_14_e7f29cc027e50efb8665g_0080_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\n\n(1) As shown in Figure 1, draw a line $EF$ through point $E$ such that $EF \\parallel AB$.\n\nSince $AB \\parallel CD$,\n\nTherefore, $AB \\parallel CD \\parallel EF$,\n\nThus, $\\angle A + \\angle 1 = 180^\\circ$, and $\\angle 2 + \\angle C = 180^\\circ$,\n\nHence, $\\angle A + \\angle B + \\angle AEC = 360^\\circ$,\n\nTherefore, statement (1) is incorrect;\n\n(2) As shown in Figure 2, since $\\angle 1$ is an exterior angle of $\\triangle CEP$,\n\nTherefore, $\\angle 1 = \\angle C + \\angle P$,\n\nSince $AB \\parallel CD$,\n\nThus, $\\angle A = \\angle 1$,\n\nWhich implies $\\angle P = \\angle A - \\angle C$,\n\nTherefore, statement (2) is correct;\n\n(3) As shown in Figure 3, draw a line $EF$ through point $E$ such that $EF \\parallel AB$.\n\nSince $AB \\parallel CD$,\n\nTherefore, $AB \\parallel CD \\parallel EF$,\n\nThus, $\\angle A + \\angle 3 = 180^\\circ$, and $\\angle 1 = \\angle 2$,\n\nHence, $\\angle A + \\angle AEC - \\angle 1 = 180^\\circ$,\n\nWhich implies $\\angle AEC = 180^\\circ + \\angle 1 - \\angle A$,\n\nTherefore, statement (3) is incorrect;\n\n(4) As shown in Figure 4, since $AB \\parallel EF$,\n\nTherefore, $\\angle \\alpha = \\angle BOF$,\n\nSince $CD \\parallel EF$,\n\nThus, $\\angle \\gamma + \\angle COF = 180^\\circ$,\n\nSince $\\angle BOF = \\angle COF + \\angle \\beta$,\n\nTherefore, $\\angle COF = \\angle \\alpha - \\angle \\beta$,\n\nHence, $\\angle \\gamma + \\angle \\alpha - \\angle \\beta = 180^\\circ$,\n\nTherefore, statement (4) is correct;\n\nIn conclusion, the number of correct statements is 2,\n\nThus, the correct choice is: B.\n\n[Key Insight] This question examines the properties of parallel lines and the exterior angle property of triangles. Mastering the properties of parallel lines and constructing auxiliary lines as per the problem's requirements are crucial for solving this question." }, { "problem_id": 873, "question": "As shown in the figure, the diameter $A B$ of the semicircle $O$ is $4$, and $C$ is the midpoint of the arc $A B$. The segments $\\mathrm{CO}$, $C A$, and $C B$ are drawn. Point $P$ starts from $A$ and moves along $A \\rightarrow O \\rightarrow C$ until it reaches $C$. Two perpendicular lines $P E$ and $P F$ are drawn from point $P$ to $A C$ at $E$ and to $B C$ at $F$, respectively. Let the distance traveled by point $P$ be $x$. The area $y$ of quadrilateral $C E P F$ as a function of $x$ can be best represented by the graph ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch19-2024_05_24_e1df54a2873d1b01739ag_0070_1.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0070_2.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0070_3.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0070_4.jpg", "batch19-2024_05_24_e1df54a2873d1b01739ag_0070_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since point \\( C \\) is the midpoint of arc \\( AB \\),\n\n\\[\n\\therefore AC = BC.\n\\]\n\nSince \\( AB \\) is the diameter,\n\n\\[\n\\therefore \\triangle ABC \\text{ is an isosceles right triangle, and } \\angle CAB = \\angle CBA = 45^\\circ.\n\\]\n\nGiven that \\( PE \\perp AC \\) and \\( PF \\perp BC \\),\n\n\\[\n\\therefore \\angle EAP = \\angle EPA = 45^\\circ, \\quad \\angle FPB = \\angle FBP = 45^\\circ,\n\\]\n\nand quadrilateral \\( CEPF \\) is a rectangle.\n\nWhen point \\( P \\) moves on \\( AO \\), \\( x \\leq 2 \\).\n\nSince \\( AP = x \\) and \\( BP = 4 - x \\),\n\n\\[\n\\therefore EP = \\frac{\\sqrt{2}}{2} x, \\quad PF = \\frac{\\sqrt{2}}{2}(4 - x),\n\\]\n\n\\[\n\\therefore \\text{The area of quadrilateral } CEPF \\text{ is } y = EP \\times PF = \\frac{1}{2} x(4 - x) = -\\frac{1}{2} x^2 + 2x.\n\\]\n\nWhen point \\( P \\) moves on \\( CO \\), \\( 2 < x \\leq 4 \\), as shown in the figure below,\n\n\n\n\\[\n\\because \\angle EPC = \\angle ECP = 45^\\circ, \\quad \\angle FPC = \\angle FCP = 45^\\circ, \\quad PE \\perp AC, \\quad PF \\perp BC, \\quad \\angle ACB = 90^\\circ,\n\\]\n\n\\[\n\\therefore EP = EC, \\quad FP = FC,\n\\]\n\n\\[\n\\therefore \\text{Quadrilateral } CEPF \\text{ is a rectangle.}\n\\]\n\nSince \\( CP = 4 - x \\),\n\n\\[\n\\therefore EP = \\frac{\\sqrt{2}}{2}(4 - x),\n\\]\n\n\\[\n\\therefore \\text{The area of quadrilateral } CEPF \\text{ is } y = EP^2 = \\frac{1}{2}(4 - x)^2 = \\frac{1}{2} x^2 - 4x + 8.\n\\]\n\n\\[\n\\therefore \\begin{cases}\ny = -\\frac{1}{2} x^2 + 2x, & 0 \\leq x \\leq 2 \\\\\ny = \\frac{1}{2} x^2 - 4x + 8, & 2 < x \\leq 4\n\\end{cases},\n\\]\n\n\\[\n\\therefore \\text{The graph of } y \\text{ as a function of } x \\text{ is approximately as shown in option A.}\n\\]\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This problem examines the properties of circles and quadratic functions. The key to solving it lies in deriving the function expressions based on the given conditions." }, { "problem_id": 874, "question": "As shown in the figure, which of the following inferences is correct $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\n\n(1) $\\because$ Straight lines $\\mathrm{AB}, \\mathrm{CD}$ intersect at point $\\mathrm{E}$ (as in Figure 1), $\\therefore \\angle 1=\\angle 2$;\n\n(2) $\\because \\angle \\mathrm{ABD}=\\angle \\mathrm{EBC}=90^{\\circ}$ (as in Figure 2), $\\therefore \\angle 1=\\angle 2$;\n\n(3) $\\because \\mathrm{OB}$ bisects $\\angle \\mathrm{AOC}$ (as in Figure 3), $\\therefore \\angle 1=\\angle 2$;\n\n(4) $\\because \\angle 1=28.3^{\\circ}, \\angle 2=28^{\\circ} 3^{\\prime}$ (as in Figure 4), $\\therefore \\angle 1=\\angle 2$.\nA. (1)(3)\nB. (1)(2)(3)(4)\nC. (1)(3)(4)\nD. (1)(2)(3)", "input_image": [ "batch9-2024_05_23_b025d22cd50cfbd99366g_0049_1.jpg", "batch9-2024_05_23_b025d22cd50cfbd99366g_0049_2.jpg", "batch9-2024_05_23_b025d22cd50cfbd99366g_0049_3.jpg", "batch9-2024_05_23_b025d22cd50cfbd99366g_0049_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Logic", "image_relavance": "1", "analysis": "Solution:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\n\n(1) Since lines $\\mathrm{AB}$ and $\\mathrm{CD}$ intersect at point $\\mathrm{E}$ (as shown in Figure 1), therefore $\\angle 1 = \\angle 2$, which is correct and fits the context;\n\n(2) Since $\\angle \\mathrm{ABD} = \\angle \\mathrm{EBC} = 90^{\\circ}$ (as shown in Figure 2), therefore $\\angle 1 = \\angle 2$, which is correct and fits the context;\n\n(3) Since $\\mathrm{OB}$ bisects $\\angle \\mathrm{AOC}$ (as shown in Figure 3), therefore $\\angle 1 = \\angle 2$, which is correct and fits the context;\n\n(4) Since $\\angle 1 = 28.3^{\\circ}$ and $\\angle 2 = 28^{\\circ} 3^{\\prime} = 28.05^{\\circ}$ (as shown in Figure 4), therefore $\\angle 1 \\neq \\angle 2$, hence this option is incorrect and does not fit the context.\n\nTherefore, the correct choice is D.\n\n【Key Insight】This question tests the understanding of vertical angles, adjacent supplementary angles, conversion of degrees and minutes, the definition of an angle bisector, complementary and supplementary angles. The key to solving the problem lies in a thorough grasp of these concepts." }, { "problem_id": 875, "question": "As shown in Figure 1, Xiaoming made a square learning tool using four wooden sticks of the same length. The measured diagonal $B D$ is $10 \\sqrt{2}$ cm. By transforming the square learning tool into a rhombus (as shown in Figure 2), with $\\angle A B C = 60^\\circ$, the length of the diagonal $B D$ in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $20 \\mathrm{~cm}$\nB. $10 \\sqrt{6} \\mathrm{~cm}$\nC. $10 \\sqrt{3} \\mathrm{~cm}$\nD. $10 \\sqrt{2} \\mathrm{~cm}$", "input_image": [ "batch23-2024_06_14_a705467cdbc40a4240d2g_0079_1.jpg", "batch23-2024_06_14_a705467cdbc40a4240d2g_0079_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Since quadrilateral \\( ABCD \\) is a square, its diagonal \\( BD = 10\\sqrt{2} \\, \\text{cm} \\).\n\nTherefore, \\( BC = \\frac{\\sqrt{2}}{2} BD = 10 \\, \\text{cm} \\).\n\nWhen quadrilateral \\( ABCD \\) is a rhombus with \\( \\angle ABC = 60^\\circ \\), and a perpendicular \\( CO \\) is drawn from point \\( C \\) to \\( BD \\),\n\n\n\nBy the properties of a rhombus, \\( BO = DO = \\frac{1}{2} BD \\), and \\( \\angle CBO = \\frac{1}{2} \\angle ABC = 30^\\circ \\).\n\nGiven \\( BC = 10 \\, \\text{cm} \\),\n\nThus, \\( CO = \\frac{1}{2} BC = 5 \\, \\text{cm} \\),\n\nAnd \\( BO = \\sqrt{3} \\cdot CO = 5\\sqrt{3} \\, \\text{cm} \\),\n\nTherefore, \\( BD = 2 \\cdot BO = 10\\sqrt{3} \\, \\text{cm} \\).\n\nHence, the correct choice is C.\n\n【Key Insight】This problem examines the properties of squares, rhombuses, the characteristics of right-angled triangles with a \\( 30^\\circ \\) angle, and the Pythagorean theorem. Mastering the properties of special parallelograms is crucial for solving such problems." }, { "problem_id": 876, "question": "Pythagorean theorem is an important theorem in geometry, and it was recorded in the ancient Chinese mathematical book \"Zhou Bi Suan Jing\" as \"If the leg is three, and the side is four, then the hypotenuse is five.\" As shown in Figure 1, a figure is constructed using equal-sized small squares and right-angled triangles, which can be used to verify the Pythagorean theorem through their area relationships. Figure 2 is obtained by placing Figure 1 inside a rectangle, where $\\angle B A C = 90^\\circ$. The area of the square $A B E D$ is 9, and the area of the square $A C H I$ is 16. Points $D, E, F, G, H, I$ are all on the sides of the rectangle $K L M J$. What is the area of the rectangle $K L M J$?\n\n\n(Figure 1)\n\n\n\n(Figure 2)\nA. 121\nB. 110\nC. 100\nD. 90", "input_image": [ "batch23-2024_06_14_c461dda316471c7af465g_0084_1.jpg", "batch23-2024_06_14_c461dda316471c7af465g_0084_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, extend $AB$ to intersect $KF$ at point $O$, and extend $AC$ to intersect $GM$ at point $P$. Then, quadrilateral $AOLP$ is a rectangle.\n\n\n\nSince $\\angle CBF = 90^\\circ$,\n\n$\\therefore \\angle ABC + \\angle OBF = 90^\\circ$,\n\nAlso, in right triangle $\\triangle ABC$, $\\angle ABC + \\angle ACB = 90^\\circ$,\n\n$\\therefore \\angle OBF = \\angle ACB$,\n\nIn triangles $\\triangle OBF$ and $\\triangle ACB$,\n\n$$\n\\left\\{\\begin{array}{l}\n\\angle BAC = \\angle BOF \\\\\n\\angle ACB = \\angle OBF \\\\\nBC = BF\n\\end{array}\\right.\n$$\n\n$\\therefore \\triangle OBF \\cong \\triangle ACB$ (by AAS),\n\n$\\therefore AC = OB$,\n\nSimilarly, $\\triangle ACB \\cong \\triangle PGC$ (by AAS),\n\n$\\therefore PC = AB$,\n\n$\\therefore OA = AP$,\n\n$\\therefore$ rectangle $AOLP$ is a square,\n\nSince the area of square $ABED$ is 9, and the area of square $ACHI$ is 16,\n\n$\\therefore AB = EB = KO = 3$, $AC = CH = PM = 4$,\n\n$\\therefore AO = AB + AC = 3 + 4 = 7$,\n\n$\\therefore KL = 3 + 7 = 10$, $LM = 4 + 7 = 11$,\n\n$\\therefore$ the area of rectangle $KLMJ$ is $10 \\times 11 = 110$.\n\nTherefore, the correct answer is: B.\n\n【Key Insight】This problem examines the proof of the Pythagorean theorem. The key to solving it lies in constructing the auxiliary lines to form a square." }, { "problem_id": 877, "question": "As shown in Figure 1, in rectangle paper $A B C D$, $A B = 5$, and $A D = 12$. There are two methods to fold a rhombus inside the rectangle:\n\nMethod A: As shown in Figure 2, fold a quadrilateral $E F G H$ by taking the midpoints of the opposite sides.\n\nMethod B: As shown in Figure 3, fold along the diagonal $A C$ such that $\\angle C A E = \\angle C A D$ and $\\angle A C F = \\angle A C B$ to form a quadrilateral $A E C F$.\n\nWhich of the following statements is true ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. The quadrilateral folded using Method A is a rhombus.\nB. The quadrilateral folded using Method B is not a rhombus.\nC. The quadrilaterals folded using both methods have the same area.\nD. The quadrilateral folded using Method A has a larger area.", "input_image": [ "batch23-2024_06_14_b344c3ddccf1b020490cg_0031_1.jpg", "batch23-2024_06_14_b344c3ddccf1b020490cg_0031_2.jpg", "batch23-2024_06_14_b344c3ddccf1b020490cg_0031_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since points \\( E, F, G, H \\) are the midpoints of the sides of rectangle \\( ABCD \\),\n\nTherefore, \\( AH = DH = BF = CF \\), \\( AE = BE = DG = CG \\), and \\( \\angle A = \\angle B = \\angle C = \\angle D = 90^\\circ \\),\n\nThus, triangles \\( \\triangle AEH \\cong \\triangle CGF \\cong \\triangle BEF \\cong \\triangle DGH \\) (by SAS),\n\nHence, \\( EH = EF = FG = GH \\),\n\nTherefore, quadrilateral \\( EFGH \\) is a rhombus, so option A is correct;\n\nSince quadrilateral \\( ABCD \\) is a rectangle,\n\nTherefore, \\( AD \\parallel BC \\),\n\nThus, \\( \\angle ECA = \\angle CAD \\),\n\nSince \\( \\angle CAE = \\angle CAD \\),\n\nTherefore, \\( \\angle CAE = \\angle ECA \\),\n\nHence, \\( EA = EC \\),\n\nIn triangles \\( \\triangle EAC \\) and \\( \\triangle FAC \\),\n\n\\[\n\\left\\{\n\\begin{array}{c}\n\\angle EAC = \\angle CAF \\\\\nAC = AC \\\\\n\\angle ACE = \\angle ACF\n\\end{array}\n\\right.\n\\]\n\nThus, \\( \\triangle EAC \\cong \\triangle FAC \\) (by ASA),\n\nTherefore, \\( AE = AF \\),\n\nHence, \\( AF = EC \\),\n\nSince \\( AF \\parallel EC \\),\n\nTherefore, quadrilateral \\( AECF \\) is a parallelogram,\n\nSince \\( AE = AF \\),\n\nTherefore, quadrilateral \\( AECF \\) is a rhombus, so option B is incorrect;\n\n(3) Since in rectangle \\( ABCD \\), \\( AB = 5 \\), \\( AD = 12 \\),\n\nTherefore, the area of rectangle \\( ABCD = AB \\cdot AD = 60 \\),\n\nAs shown in the figure:\n\n\n\nSince \\( \\triangle AEH \\cong \\triangle CGF \\cong \\triangle BEF \\cong \\triangle DGH \\),\n\nTherefore, \\( S_{\\triangle AEH} = \\frac{1}{2} \\times \\frac{1}{2} \\times AB \\times \\frac{1}{2} \\times AD = \\frac{15}{2} \\),\n\nThus, the area of rhombus \\( EFGH = 60 - 4 \\times \\frac{15}{2} = 30 \\);\n\nAs shown in the figure:\n\n\n\nLet \\( BE = x \\), then \\( AE = CE = BC - BE = 12 - x \\), in right triangle \\( \\triangle ABE \\), by the Pythagorean theorem:\n\n\\[\nx^{2} + 5^{2} = (12 - x)^{2},\n\\]\n\nSolving gives \\( x = \\frac{119}{24} \\),\n\nTherefore, \\( CE = 12 - \\frac{119}{24} = \\frac{169}{24} \\),\n\nThus, the area of rhombus \\( AECF = CE \\cdot AB = \\frac{169}{24} \\times 5 \\approx 35.21 \\),\n\nTherefore, the area of rhombus \\( EFGH \\) < the area of rhombus \\( AECF \\),\n\nHence, options C and D are both incorrect.\n\nTherefore, the correct answer is: A.\n\n【Key Insight】This question examines the determination and properties of a rhombus, the properties of a rectangle, the transformation of folding, the determination and properties of congruent triangles, and the Pythagorean theorem. The key to solving this problem lies in mastering the determination and properties of a rhombus." }, { "problem_id": 878, "question": "In Figure (11), Ya Ting is holding 3 dark gray and 2 light gray cards stacked together in her left hand. The following are the three steps she takes each time she shuffles the cards:\n\nStep One: Take the bottom 2 cards with her right hand, as shown in Figure (2).\n\nStep Two: Insert the 2 cards from her right hand in an alternating sequence into the 3 cards in her left hand, as shown in Figure (3).\n\nStep Three: Hold the 5 cards with the color order changed in her left hand, as shown in Figure (4).\n\n\n\nFigure (1)\n\n\n\nAs shown in Figure (2)\n\n\n\nAs shown in Figure (3)\n\n\n\nAs shown in Figure (4)\n\nIf she shuffles the cards according to these three steps, starting with the situation in Figure (11), after several shuffles, the color order will again be the same as in Figure (11). Which of the following could be the number of shuffles? ( )\nA. 18\nB. 20\nC. 25\nD. 27", "input_image": [ "batch24-2024_06_15_62e79cb351f5324a756dg_0040_1.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0040_2.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0040_3.jpg", "batch24-2024_06_15_62e79cb351f5324a756dg_0040_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "**Question Analysis:** Based on the rules of shuffling, determine the pattern of changes in the shuffle, and then analyze each option to arrive at the correct answer.\n\n**Solution:** Let the 5 cards be: $1, 2, 3, \\mathrm{~A}, \\mathrm{~B}$. After the first shuffle, they become: $1, \\mathrm{~A}, 2, \\mathrm{~B}, 3$. After the second shuffle, they become: $1, \\mathrm{~B}, \\mathrm{~A}, 3, 2$.\n\nAfter the third shuffle, they become: $1, 3, \\mathrm{~B}, 2, \\mathrm{~A}$.\n\nAfter the fourth shuffle, they return to: $1, 2, 3, \\mathrm{~A}, \\mathrm{~B}$.\n\nThus, every 4 shuffles, the order of the cards returns to the same as in Figure (11).\n\nTherefore, the number of shuffles must be a multiple of 4, and among the options, only 20 meets this requirement.\n\n**Answer:** B.\n\n**Commentary:** This question primarily tests reasoning and argumentation. The key to solving it lies in deducing the pattern of changes in the shuffle based on the given information." }, { "problem_id": 879, "question": "Figure 1 is a tablet stand composed of a tray, a support board, and a base. When in use, the tablet can be attached to the tray, and the base is placed on the desk. Figure 2 is a side structural diagram of the stand. Given that the tray $A B$ is 200 mm long, and the support board $C B$ is 80 mm long, when $\\angle A B C = 130^\\circ$ and $\\angle B C D = 70^\\circ$, the distance from the top point $A$ of the tray to the plane containing the base $C D$ is ( ) (rounded to 1 mm). (Reference data: $\\sin 70^\\circ \\approx 0.94, \\cos 70^\\circ \\approx 0.34, \\tan 70^\\circ \\approx 2.75, \\sqrt{2} \\approx 1.41, \\sqrt{3} \\approx 1.73$).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $246 \\mathrm{~mm}$\nB. $247 \\mathrm{~mm}$\nC. $248 \\mathrm{~mm}$\nD. $249 \\mathrm{~mm}$", "input_image": [ "batch37-2024_06_14_bd97c218f59ff803a167g_0030_1.jpg", "batch37-2024_06_14_bd97c218f59ff803a167g_0030_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "As shown in the figure, a perpendicular line is drawn from point $A$ to $CD$, intersecting the extension of $DC$ at point $E$. A line parallel to $CD$ is drawn from point $B$, intersecting $AE$ at point $F$. From point $C$, a perpendicular line $CG$ is drawn to $BF$, intersecting at point $G$.\n\n\n\nFrom the construction, it is evident that quadrilateral $EFGC$ is a rectangle,\n\n$\\therefore EF = CG$.\n\n$\\because \\angle BCD = 70^{\\circ}$,\n\n$\\therefore \\angle CBG = 70^{\\circ}$.\n\n$\\because \\angle ABC = 130^{\\circ}$,\n\n$\\therefore \\angle ABF = 60^{\\circ}$.\n\nAccording to the problem, $AB = 200 \\mathrm{~mm}$ and $CB = 80 \\mathrm{~mm}$,\n\n$\\therefore AF = AB \\sin \\angle ABF = 200 \\times \\frac{\\sqrt{3}}{2} \\approx 200 \\times \\frac{1.73}{2} = 173 \\mathrm{~mm}$,\n\n$EF = CG = BC \\sin \\angle CBG = 80 \\times 0.94 \\approx 75.2 \\mathrm{~mm}$,\n\n$\\therefore AE = AF + EF = 173 + 75.2 = 248.2 \\approx 248 \\mathrm{~mm}$.\n\nTherefore, the distance from the vertex $A$ of the bracket to the plane of the base $CD$ is $248 \\mathrm{~mm}$.\n\nHence, the correct answer is C.\n\n【Key Insight】This problem examines the practical application of solving right triangles, the properties of parallel lines, and the determination and properties of rectangles. Correctly drawing auxiliary lines is crucial to solving the problem." }, { "problem_id": 880, "question": "In $\\triangle A B C$ shown in Figure 1, $A B=A C$ and $\\angle B A C=120^\\circ$. Point $E$ is the midpoint of side $A B$, and point $P$ is a moving point on side $B C$. Let $P C=x$ and $P A+P E=y$. The graph in Figure 2 represents the function $y$ as a function of $x$, where $H$ is the lowest point on the graph. What is the value of $a$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 6\nB. 7\nC. $4 \\sqrt{3}$\nD. 14", "input_image": [ "batch37-2024_06_14_d93c5acb44a1e0d9320cg_0028_1.jpg", "batch37-2024_06_14_d93c5acb44a1e0d9320cg_0028_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, by folding $\\triangle ABC$ along $BC$, we obtain $\\triangle A'BC$. The quadrilateral $ABA'C$ is a rhombus, with its diagonals intersecting at point $O$. The symmetric point of $A$ with respect to $BC$ is $A'$, and $PA + PE = P'A + PE$. When points $A'$, $P$, and $E$ are collinear, $y$ is minimized.\n\n\n\nFrom Figure 2, it can be seen that when point $P$ coincides with point $B$,\n\n$y = PA + PE = AB + PE = AB + \\frac{1}{2} AB = 6\\sqrt{3}$,\n\nSolving gives: $AB = 4\\sqrt{3}$, meaning the side length of the rhombus is $4\\sqrt{3}$.\n\nSince $\\angle BAC = 120^\\circ$ and $AB = AC$,\n\n$\\angle ABC = \\angle A'BC = 30^\\circ$,\n\nThus, $\\angle A'BE = 60^\\circ$,\n\nTherefore, $\\triangle ABA'$ is an equilateral triangle.\n\nSince point $E$ is the midpoint,\n\nThe height of the rhombus is $A'E$.\n\nIn right triangle $\\triangle A'BE$, $\\sin 60^\\circ = \\frac{A'E}{A'B}$,\n\nThus, $A'E = \\sin 60^\\circ \\times A'B = \\frac{\\sqrt{3}}{2} \\times 4\\sqrt{3} = 6$,\n\nSo, $a = A'E = 6$.\n\nTherefore, the answer is: A.\n\n【Key Insight】This problem primarily examines the properties and determination of a rhombus, the minimization of the sum of line segments based on symmetry, the properties and determination of an equilateral triangle, trigonometric functions of special angles, and the interpretation of key points in the graph of a moving point problem. Determining the side length of the rhombus from the graph is a prerequisite for solving the problem." }, { "problem_id": 881, "question": "As shown in Figure 1, equilateral triangles are constructed outwardly on each side of the right-angled triangle $ABC$, numbered as (1), (2), and (3). As shown in Figure 2, (1) and (2) are placed on top of (3). If the area ratio of quadrilateral $E G H F$ to $G D C H$ is $\\frac{161}{64}$, then the value of $\\sin \\angle A B C$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{8}{17}$\nB. $\\frac{8}{15}$\nC. $\\frac{5}{13}$\nD. $\\frac{3}{5}$", "input_image": [ "batch32-2024_06_14_c468ba9c811ec31d43eeg_0038_1.jpg", "batch32-2024_06_14_c468ba9c811ec31d43eeg_0038_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "**Solution:**\n\nGiven that $\\triangle BEF$ is an equilateral triangle, we draw $FH \\perp BE$ at point $H$, as shown in the figure.\n\nSince $\\triangle BEF$ is an equilateral triangle,\n\n\\[\n\\angle BEF = 60^\\circ, \\quad BE = EF = BF.\n\\]\n\nBecause $FH \\perp BE$,\n\n\\[\n\\angle FHE = 90^\\circ,\n\\]\n\n\\[\nFH = BE \\sin \\angle BEF = BE \\sin 60^\\circ = \\frac{\\sqrt{3}}{2} BE.\n\\]\n\nThus, the area of $\\triangle BEF$ is\n\n\\[\nS_{\\triangle BEF} = \\frac{1}{2} BE \\cdot FH = \\frac{\\sqrt{3}}{4} BE^2.\n\\]\n\nLet the side length of $\\triangle BEF$ be $b$. Then, $AC = b$. Let the side length of $\\triangle BCD$ be $a$, and the side length of $\\triangle BGH$ be $c$. Therefore, $AB = c$.\n\n\\[\nS_{\\triangle BEF} = \\frac{\\sqrt{3}}{4} b^2.\n\\]\n\nSimilarly, we can find:\n\n\\[\nS_{\\triangle BCD} = \\frac{\\sqrt{3}}{4} a^2, \\quad S_{\\triangle BGH} = \\frac{\\sqrt{3}}{4} c^2.\n\\]\n\nTherefore, the area of quadrilateral $EGHF$ is\n\n\\[\nS_{\\text{quadrilateral } EGHF} = S_{\\triangle BGH} - S_{\\triangle BEF} = \\frac{\\sqrt{3}}{4} c^2 - \\frac{\\sqrt{3}}{4} b^2 = \\frac{\\sqrt{3}}{4} (c^2 - b^2).\n\\]\n\nThe area of quadrilateral $GDCH$ is\n\n\\[\nS_{\\text{quadrilateral } GDCH} = S_{\\triangle BCD} - S_{\\triangle BGH} = \\frac{\\sqrt{3}}{4} a^2 - \\frac{\\sqrt{3}}{4} c^2 = \\frac{\\sqrt{3}}{4} (a^2 - c^2).\n\\]\n\nSince $\\triangle ABC$ is a right triangle,\n\n\\[\nb^2 + c^2 = a^2,\n\\]\n\n\\[\nb^2 = a^2 - c^2.\n\\]\n\nThus, the ratio of the areas is\n\n\\[\n\\frac{S_{\\text{quadrilateral } EGHF}}{S_{\\text{quadrilateral } GDCH}} = \\frac{c^2 - b^2}{a^2 - c^2} = \\frac{c^2 - b^2}{b^2} = \\frac{161}{64}.\n\\]\n\nTherefore,\n\n\\[\n\\frac{c^2}{b^2} = \\frac{225}{64},\n\\]\n\n\\[\n\\frac{c}{b} = \\frac{15}{8}.\n\\]\n\nLet $c = 15k$ and $b = 8k$. Then,\n\n\\[\na = \\sqrt{c^2 + b^2} = 17k.\n\\]\n\nThus,\n\n\\[\n\\sin \\angle ABC = \\frac{b}{a} = \\frac{8k}{17k} = \\frac{8}{17}.\n\\]\n\n**Answer:** A\n\n**Key Insight:** This problem examines the properties of right triangles and equilateral triangles, as well as their areas. The key lies in transforming the ratio of quadrilateral areas into a relationship involving the side lengths of the equilateral triangles." }, { "problem_id": 882, "question": "Given point $P$ outside circle $O$, draw a line through point $P$ that is tangent to circle $O$ at point $M$. Here are two methods given by Qi Qi:\n\nMethod I: As shown in Figure 1, draw the perpendicular bisector of segment $O P$ and let it intersect $O P$ at point $G$. Draw an arc with center $G$ and radius $G P$ that intersects circle $O$ at point $M$. Draw line $P M$. Line $P M$ is the required line.\n\nMethod II: As shown in Figure 2, draw segment $O P$ and let it intersect circle $O$ at point $B$. Draw a diameter $B C$. Draw an arc with center $O$ and radius $B C$. Draw an arc with center $P$ and radius $O P$. Let these two arcs intersect at point $D$. Draw line $O D$ and let it intersect circle $O$ at point $M$. Draw line $P M$. Line $P M$ is the required line.\n\nWhich of the following statements about Qi Qi's two methods is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Both methods are correct.\nB. Both methods are incorrect.\nC. Method I is correct, Method II is incorrect.\nD. Method II is correct, Method I is incorrect.", "input_image": [ "batch19-2024_05_24_7bbaf69124e1595f4e32g_0042_1.jpg", "batch19-2024_05_24_7bbaf69124e1595f4e32g_0042_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Method I: Connect $OM$ and $MG$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n$\\because$ The perpendicular bisector of segment $OP$ intersects $OP$ at point $G$\n\n$\\therefore OG = GP$,\n\n$\\because$ With point $G$ as the center and $GP$ as the radius, draw an arc intersecting $O$ at point $M$,\n\n$\\therefore$ Point $O$ is on $G$, and $OP$ is the diameter\n\n$\\therefore \\angle OMP = 90^{\\circ}$\n\n$\\therefore$ Line $PM$ is tangent to $O$;\n\nMethod II: $\\because$ With $O$ as the center and $BC$ as the radius, draw an arc\n\n$\\therefore OM = \\frac{1}{2} BC = \\frac{1}{2} OD$,\n\n$\\because$ With $P$ as the center and $OP$ as the radius, draw an arc intersecting the first arc at point $D$,\n\n$\\therefore PD = PO$\n\n$\\therefore \\angle OMP = 90^{\\circ}$\n\n$\\therefore$ Line $PM$ is tangent to $O$;\n\nIn summary, both methods are correct;\n\nTherefore, choose: A.\n\n【Key Point】This question examines complex drawing and the determination of tangents, among other knowledge. The key to solving the problem is to understand the question and flexibly apply the learned knowledge to solve the problem, which is a common type of question in middle school exams." }, { "problem_id": 883, "question": "As shown in Figure 1, in $\\triangle A B C$, $\\angle A B C = 60^\\circ$. Point $D$ is the midpoint of side $B C$. Point $P$ starts from the vertex $A$ of $\\triangle A B C$ and moves along the path $A \\rightarrow B \\rightarrow D$ at a constant speed of 1 unit per second until it reaches point $D$. The relationship between the length $y$ of segment $D P$ and time $x$ is shown in the graph in Figure 2. Point $N$ is the lowest point on the curved part of the graph. Determine the area of $\\triangle A B C$ ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 4\nB. $4 \\sqrt{3}$\nC. 8\nD. $8 \\sqrt{3}$", "input_image": [ "batch34-2024_06_17_010d88a8bf3f2e00a466g_0006_1.jpg", "batch34-2024_06_17_010d88a8bf3f2e00a466g_0006_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the graph of the function, it can be seen that when \\( x = 0 \\), \\( y = 4 \\); when \\( x = 2 \\), \\( y \\) is at its minimum, which means \\( DP \\perp AB \\).\n\nTherefore, \\( AD = 4 \\), and \\( AP = 2 \\).\n\nThus, \\( DP = \\sqrt{AD^2 - AP^2} = 2\\sqrt{3} \\), and \\( \\sin \\angle ADP = \\frac{2}{4} = \\frac{1}{2} \\).\n\nHence, \\( \\angle ADP = 30^\\circ \\), and \\( \\angle DAP = 60^\\circ \\).\n\nSince \\( \\angle B = 60^\\circ \\), triangle \\( ADB \\) is an equilateral triangle.\n\nTherefore, point \\( P \\) is the midpoint of \\( AB \\).\n\nSince point \\( D \\) is the midpoint of \\( BC \\), \\( DP \\) is the midline of triangle \\( ABC \\).\n\nThus, the area of triangle \\( ABC \\) is \\( S_{\\triangle ABC} = 4 S_{\\triangle BDP} = 4 \\times \\frac{1}{2} \\times 2 \\times \\sqrt{3} = 8\\sqrt{3} \\).\n\nThe correct answer is: D.\n\n**Key Insight**: This problem examines the shortest perpendicular segment, the Pythagorean theorem, properties of right triangles, the determination and properties of equilateral triangles, and the midline of a triangle. The key to solving it lies in integrating graphical analysis with geometric principles." }, { "problem_id": 884, "question": "A school organized a study and practice activity for students in Shaoxing. Student Xiaowang noticed a bottle of hand sanitizer on the countertop in the hotel room (as shown in Figure (1)). Curious Xiaowang decided to conduct field measurements. When Xiaowang applied a certain force to press down on the top A to the position shown in Figure (2), the hand sanitizer flowed out from the nozzle B in a parabolic shape, with B being the vertex of the parabola. The cross-sectional view of the hand sanitizer bottle below the neck is a rectangle CGHD. Xiaowang measured the following: the diameter of the bottom GH of the hand sanitizer bottle is 12 cm, the distance from the nozzle point B to the countertop is 16 cm, and points B, D, and H are collinear. When Xiaowang caught the hand sanitizer at a height of 15.5 cm from the countertop, the horizontal distance from the palm Q to the line DH was 3 cm. If Xiaowang had not caught it, where would the hand sanitizer have landed on the countertop, measured horizontally from the line DH? The distance would be ( ) cm.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $12 \\sqrt{3}$\nB. $12 \\sqrt{2}$\nC. $6 \\sqrt{3}$\nD. $6 \\sqrt{2}$", "input_image": [ "batch8-2024_06_14_39f616e1ffa2dea341d7g_0052_1.jpg", "batch8-2024_06_14_39f616e1ffa2dea341d7g_0052_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement, the line where $GH$ lies is the $x$-axis, and the perpendicular bisector of $GH$ is the $y$-axis, establishing a plane Cartesian coordinate system as shown in the figure. The nozzle $B$ is the vertex of the parabola, and the collinear points $B$, $D$, and $H$ lie on the axis of symmetry of the parabola.\n\n\n\nGiven the coordinates $Q(9,15.5)$, $B(6,16)$, and $OH=6$,\n\nLet the equation of the parabola be $y=a(x-6)^{2}+16$.\n\nSubstituting the coordinates of point $Q$ into the equation gives $15.5=a(9-6)^{2}+16$,\nSolving for $a$ yields: $a=-\\frac{1}{18}$.\n\nTherefore, the equation of the parabola is: $y=-\\frac{1}{18}(x-6)^{2}+16=-\\frac{1}{18} x^{2}+\\frac{2}{3} x+14$.\n\nWhen $y=0$, i.e., $0=-\\frac{1}{18} x^{2}+\\frac{2}{3} x+14$,\n\nSolving for $x$ gives: $x=6+12 \\sqrt{2}$, or $x=6-12 \\sqrt{2}$ (discarded).\n\nThus, the horizontal distance from the point where the hand sanitizer lands on the table to $DH$ is $6+12 \\sqrt{2}-6=12 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the correct answer is: B.\n\n【Key Insight】This problem tests the application of quadratic functions. The key to solving it lies in determining the quadratic function's equation using the method of undetermined coefficients and performing accurate calculations." }, { "problem_id": 885, "question": "Rectangle paper $A B C D$ is divided as shown in Figure 1 into three right-angled triangles (1), (2), and (3). These three right-angled triangles are then overlaid together as shown in Figure 2, with one endpoint of the hypotenuse of right-angled triangle (1) lying on the hypotenuse of right-angled triangle (2). If $B D = 5$, then the length of $C P$ in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{4}{5}$\nB. $\\frac{3}{5}$\nC. $\\frac{3}{4}$\nD. $\\frac{\\sqrt{3}}{2}$", "input_image": [ "batch37-2024_06_14_bd97c218f59ff803a167g_0003_1.jpg", "batch37-2024_06_14_bd97c218f59ff803a167g_0003_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, let \\( AB = 2a \\).\n\nSince quadrilateral \\( ABCD \\) is a rectangle,\n\n\\[ AB = CD = 2a, \\quad AD = BC, \\quad \\angle BAD = 90^\\circ, \\quad AB \\parallel CD, \\]\n\n\\[ AD = \\sqrt{BD^2 - AB^2} = \\sqrt{25 - 4a^2}. \\]\n\nAs shown in Figure 2, \\( HP = AB = 2a \\), \\( QN = AD = \\sqrt{25 - 4a^2} \\), \\( MN = BD = 5 \\), \\( MP = BE \\).\n\n\n\nFigure 2\n\n\\[ \\angle MPH = \\angle NMP, \\quad \\angle HMP = 90^\\circ, \\]\n\\[ MQ = PQ, \\quad \\angle H = \\angle HMQ, \\]\n\\[ HQ = MQ, \\]\n\\[ HQ = PQ = a, \\]\n\\[ \\sqrt{25 - 4a^2} + a = 5, \\]\n\\[ a = 2, \\quad a = 0 \\text{ (discarded)}, \\]\n\\[ AB = 4, \\quad AD = 3. \\]\n\nAs shown in Figure 1, since \\( \\cos \\angle ABD = \\frac{BE}{AB} = \\frac{AB}{BD} \\),\n\n\\[ \\frac{BE}{4} = \\frac{4}{5}, \\]\n\\[ BE = \\frac{16}{5} = MP, \\]\n\\[ PC = 4 - \\frac{16}{5} = \\frac{4}{5}. \\]\n\nTherefore, the answer is: A.\n\n【Key Insight】This problem tests the properties of rectangles, trigonometric functions of acute angles, and the Pythagorean theorem. The key to solving it lies in mastering these concepts." }, { "problem_id": 886, "question": "Given that in $\\triangle A C B$, $\\angle A C B = 90^\\circ$, point $D$ is the midpoint of $A B$, prove that $C D = \\frac{1}{2} A B$. When proving this conclusion, auxiliary lines need to be added. Which of the following methods of adding auxiliary lines is incorrect ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\n\nA. As shown in Figure (1), take the midpoint $E$ of $A C$, and connect $D E$\n\nB. As shown in Figure (2), draw the bisector of $\\angle A D C$, and let it intersect $A C$ at point $E$\n\nC. As shown in Figure (3), extend $C D$ to point $E$, with $D E = C D$, and connect $A E$ and $B E$\n\nD. As shown in Figure (4), draw a line $B E$ parallel to $C A$ passing through point $B$, and let it intersect the extension of $C D$ at point $E$\n\n##", "input_image": [ "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0040_1.jpg", "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0040_2.jpg", "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0040_3.jpg", "batch23-2024_06_14_c6ab99d4fe7548d1194ag_0040_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Since point \\( D \\) is the midpoint of \\( AB \\), and point \\( E \\) is the midpoint of \\( AC \\),\n\nTherefore, \\( DE \\parallel BC \\),\n\nThus, \\( \\angle EDC = \\angle DCB \\), \\( \\angle EDA = \\angle B \\), \\( \\angle AED = \\angle ACB = 90^\\circ \\),\nHence, \\( AD = DC \\), and \\( \\angle EDC = \\angle EDA \\),\n\nTherefore, \\( \\angle DCB = \\angle B \\),\n\nSo, \\( BD = DC \\),\n\nThus, \\( AD = DB = DC \\),\n\nHence, \\( CD = \\frac{1}{2} AB \\).\n\nTherefore, option \\( A \\) is correct;\n\nSince \\( AD = DB \\), and \\( DE = CD \\),\n\nTherefore, quadrilateral \\( ABCD \\) is a parallelogram,\n\nBecause \\( \\angle ACB = 90^\\circ \\),\n\nThus, quadrilateral \\( ABCD \\) is a rectangle,\n\nHence, \\( AB = CE \\)\n\nSince \\( CD = \\frac{1}{2} CE \\),\n\nTherefore, \\( CD = \\frac{1}{2} AB \\).\n\nThus, option \\( C \\) is correct;\n\nSince \\( BE \\parallel CA \\), and point \\( D \\) is the midpoint of \\( AB \\),\n\nTherefore, \\( \\angle CAD = \\angle EBD \\), \\( \\angle ADC = \\angle BDE \\), \\( \\angle EBC = \\angle ACB = 90^\\circ \\),\n\nThus, \\( \\triangle ADC \\cong \\triangle BDE \\),\n\nHence, \\( CD = ED \\), and \\( AC = BE \\),\n\nSince \\( BC = BC \\),\n\nTherefore, \\( \\triangle ABC \\cong \\triangle ECB \\),\n\nThus, \\( AB = CE \\),\n\nSince \\( CD = \\frac{1}{2} CE \\),\n\nTherefore, \\( CD = \\frac{1}{2} AB \\).\n\nHence, option \\( D \\) is correct;\n\nOnly option \\( B \\) cannot be proven,\n\nTherefore, the correct choice is B.\n\n【Key Insight】This question examines the determination and properties of isosceles triangles, the congruence and properties of triangles, and the determination and properties of rectangles. Mastering the congruence of triangles and the determination of rectangles is key to solving the problem." }, { "problem_id": 887, "question": "As shown in Figure 1, rhombus paper $A B C D$ has side length $2$ and $\\angle A B C = 60^\\circ$. The rhombus $A B C D$ is folded along $E F$ and $G H$ so that points $B$ and $D$ coincide at a point $P$ on the diagonal $B D$ (as shown in Figure 2). The maximum area of hexagon $A E F C H G$ is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\frac{3 \\sqrt{3}}{2}$\nB. $\\frac{3 \\sqrt{3}}{4}$\nC. $2-\\sqrt{3}$\nD. $1+\\sqrt{3}$", "input_image": [ "batch8-2024_06_14_3647c7a2f421b3e280acg_0051_1.jpg", "batch8-2024_06_14_3647c7a2f421b3e280acg_0051_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The area of hexagon \\( AEFCHG \\) = Area of rhombus \\( ABCD \\) - Area of \\( \\triangle EBF \\) - Area of \\( \\triangle GDH \\).\n\nGiven that the side length of rhombus \\( ABCD \\) is 2 and \\( \\angle ABC = 60^\\circ \\),\n\n\\[\nAC = 2,\n\\]\n\\[\nBD = 2\\sqrt{3},\n\\]\n\\[\n\\text{Area of rhombus } ABCD = \\frac{1}{2} \\times AC \\times BD = \\frac{1}{2} \\times 2 \\times 2\\sqrt{3} = 2\\sqrt{3}.\n\\]\n\nLet \\( AE = x \\), then the area of hexagon \\( AEFCHG \\) is:\n\\[\n2\\sqrt{3} - \\frac{1}{2} \\times (2 - x) \\times \\frac{\\sqrt{3}}{2}(2 - x) - \\frac{1}{2} \\times x \\times \\frac{\\sqrt{3}}{2}x\n\\]\n\\[\n= -\\frac{\\sqrt{3}}{2}x^2 + \\sqrt{3}x + \\sqrt{3}\n\\]\n\\[\n= -\\frac{\\sqrt{3}}{2}(x - 1)^2 + \\frac{3}{2}\\sqrt{3}.\n\\]\n\nTherefore, the maximum area of hexagon \\( AEFCHG \\) is \\( \\frac{3}{2}\\sqrt{3} \\).\n\nHence, the correct answer is: A.\n\n**Key Insight:** This problem examines the transformation of shapes through folding and the optimization of quadratic functions. The key to solving it lies in setting an unknown variable to represent the area of the hexagon, thereby transforming the geometric problem into a functional one, which presents a certain level of difficulty." }, { "problem_id": 888, "question": "As shown in the figure: In (1), (2), and (3), $\\angle A = 42^\\circ, \\angle 1 = \\angle 2, \\angle 3 = \\angle 4$. Then $\\angle O_{I} + \\angle O_{2} + \\angle O_{3} = $ ( ) degrees.\n\n\n\n(1)\n\n\n\n(2)\n\n\nA. 84\nB. 111\nC. 225\nD. 201", "input_image": [ "batch35-2024_06_17_a5af53d889bd5704e3e0g_0028_1.jpg", "batch35-2024_06_17_a5af53d889bd5704e3e0g_0028_2.jpg", "batch35-2024_06_17_a5af53d889bd5704e3e0g_0028_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since in (1), (2), and (3), $\\angle \\mathrm{A}=42^{\\circ}$, $\\angle 1=\\angle 2$, and $\\angle 3=\\angle 4$,\n\nTherefore, in (1), $\\angle 2+\\angle 4=\\frac{1}{2}(\\angle 1+\\angle 2+\\angle 3+\\angle 4)=\\frac{1}{2}\\left(180^{\\circ}-42^{\\circ}\\right)=69^{\\circ}$, hence $\\angle \\mathrm{O}_{1}=180^{\\circ} -69^{\\circ}=111^{\\circ}$;\n\nIn (2), $\\angle \\mathrm{O}_{2}=\\angle 4-\\angle 2=\\frac{1}{2}[(\\angle 3+\\angle 4)-(\\angle 1+\\angle 2)]=\\frac{1}{2} \\angle \\mathrm{A}=21^{\\circ}$;\n\nIn (3), $\\angle \\mathrm{ABC}+\\angle \\mathrm{ACB}=180^{\\circ}-\\angle \\mathrm{A}=180^{\\circ}-42^{\\circ}=138^{\\circ}$, then $\\angle 1+\\angle 2+\\angle 3+\\angle 4=180^{\\circ}+180^{\\circ} -138^{\\circ}=222^{\\circ}$\n\nHence, $\\angle \\mathrm{O}_{3}=180^{\\circ}-(\\angle 2+\\angle 3)=180^{\\circ}-\\frac{1}{2} \\times 222^{\\circ}=69^{\\circ}$\n\nTherefore, $\\angle \\mathrm{O}_{1}+\\angle \\mathrm{O}_{2}+\\angle \\mathrm{O}_{3}=111^{\\circ}+21^{\\circ}+69^{\\circ}=201^{\\circ}$\n\nThus, the correct choice is D.\n\n【Key Insight】This question examines the sum of interior angles of a triangle, the properties of exterior angles of a triangle, the concept of supplementary angles, and their related applications. The difficulty level of this question is moderate, making it a medium-level problem." }, { "problem_id": 889, "question": "In order to measure the diameter of a circular workpiece.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA: As shown in Figure 1, two identical cubic wooden blocks are placed on both sides of the circular workpiece on the worktable. By measuring the distance $b$ between the blocks and the side length $a$ of the blocks, the diameter can be calculated.\n\nB: As shown in Figure 2, the wooden blocks are replaced with two identical small cylinders. By measuring the radius $n$ of the cylinders and the distance $m$ between their centers, the diameter can be determined.\n\nThe correct statement is $(\\quad)$\nA. A is correct and B is incorrect.\nB. A is incorrect and B is correct.\nC. Both A and B are incorrect.\nD. Both A and B are correct.", "input_image": [ "batch19-2024_05_24_7bbaf69124e1595f4e32g_0046_1.jpg", "batch19-2024_05_24_7bbaf69124e1595f4e32g_0046_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Logic", "image_relavance": "1", "analysis": "Figure 1\n\n\n\nFigure 2\n\nPerson A: As shown in Figure 1, connect $CD$, $OC$, $OD$, and draw $OE \\perp CD$ at point $E$, intersecting the circle $O$ at point $F$. From the diagram, we know: $CD = b$, $EF = a$.\n\nSince $OE \\perp CD$,\n\n$\\therefore DE = \\frac{1}{2} CD = \\frac{b}{2}$.\n\nLet the radius of circle $O$ be $r$,\n\n$\\therefore OE = OF - EF = r - a$.\n\nIn the right triangle $\\triangle ODE$, we have: $OD^{2} = OE^{2} + DE^{2}$,\n\n$\\therefore r^{2} = (r - a)^{2} + \\left(\\frac{b}{2}\\right)^{2}$.\n\nSolving the equation will give the value of $r$, which means Person A's statement is correct.\n\nPerson B: As shown in Figure 2, connect $CD$, $OC$, $OD$, and draw $OE \\perp CD$ at point $E$, intersecting the circle $O$ at point $F$. From the diagram, we know: $CD = 2n + m$, $EF = n$.\n\nLet the radius of circle $O$ be $r$,\nSimilarly, we have: $r^{2} = (r - n)^{2} + \\left(\\frac{m + 2n}{2}\\right)^{2}$.\n\nSolving the equation will give the value of $r$, which means Person B's statement is correct.\n\nTherefore, the correct answer is: D.\n\n【Key Point】This question mainly tests the knowledge of the Perpendicular Chord Theorem and the Pythagorean Theorem. Mastering the Perpendicular Chord Theorem is the key to solving this problem." }, { "problem_id": 890, "question": "Four wooden sticks of equal length are connected end to end to form a quadrilateral $\\mathrm{ABCD}$. By rotating the quadrilateral, its shape can be changed. When $\\angle \\mathrm{ABC}=60^{\\circ}$, as shown in Figure (1), it is measured that $\\mathrm{BD}=2 \\sqrt{3}$. When $\\angle \\mathrm{ABC}=90^{\\circ}$, as shown in Figure (2), the length of $\\mathrm{BD}$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $2 \\sqrt{2}$\nB. 2\nC. $\\sqrt{3}$\nD. $\\sqrt{2}$", "input_image": [ "batch23-2024_06_14_0700db98cd8bb5af6c52g_0004_1.jpg", "batch23-2024_06_14_0700db98cd8bb5af6c52g_0004_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: In Figure (1), connect $\\mathrm{AC}$ and $\\mathrm{BD}$, and let $\\mathrm{AC}$ and $\\mathrm{BD}$ intersect at point $\\mathrm{O}$. In Figure (2), connect $\\mathrm{BD}$ as shown:\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n$\\because$ Four wooden strips of equal length are connected end to end in sequence,\n\n$\\therefore$ In Figure (1), quadrilateral $\\mathrm{ABCD}$ is a rhombus,\n\n$\\therefore \\mathrm{AC} \\perp \\mathrm{BD}$, $\\mathrm{OB}=\\mathrm{OD}$, $\\mathrm{OA}=\\mathrm{OC}$, and $\\angle \\mathrm{BDC}=\\angle \\mathrm{BDA}$.\n\n$\\because \\angle \\mathrm{ABC}=60^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{OBC}=30^{\\circ}$.\n\nLet $\\mathrm{OC}=\\mathrm{x}$, then $\\mathrm{BC}=2 \\mathrm{x}$.\n\n$\\because BD=2 \\sqrt{3}$,\n\n$\\therefore \\mathrm{OB}=\\sqrt{3}$.\n\n$\\therefore$ In right triangle $\\triangle \\mathrm{OBC}$, by the Pythagorean theorem: $\\mathrm{x}^{2}+(\\sqrt{3})^{2}=(2 \\mathrm{x})^{2}$.\n\nSolving gives $\\mathrm{x}=1$,\n\n$\\therefore \\mathrm{BC}=\\mathrm{CD}=2$.\n\n$\\therefore$ In Figure (2), when $\\angle \\mathrm{ABC}=90^{\\circ}$, quadrilateral $\\mathrm{ABCD}$ is a square. By the Pythagorean theorem: $\\mathrm{BD}=$ $\\sqrt{B C^{2}+C D^{2}}=2 \\sqrt{2}$.\n\nTherefore, the answer is: A.\n\n【Key Point】This problem mainly examines the application of the rhombus and the Pythagorean theorem. Accurate analysis is the key to solving the problem." }, { "problem_id": 891, "question": "Fold the rectangle $A B C D$ so that $A D$ coincides with $B C$, resulting in crease $E F$. Unfold the paper (as shown in Figure 1), and fold it again so that point $A$ falls on $E F$ and the crease passes through point $B$, resulting in crease $B M$. At the same time, segment $B N$ is formed (as shown in Figure 2). If $M N$ is extended to intersect $B C$ at point $P$ (as shown in Figure 3), which of the following statements is false?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $\\angle N B C=30^{\\circ}$\nB. $\\angle A M N=150^{\\circ}$\nC. $\\triangle B M P$ is an equilateral triangle\nD. Point $N$ is the midpoint of $M P$", "input_image": [ "batch23-2024_06_14_b344c3ddccf1b020490cg_0059_1.jpg", "batch23-2024_06_14_b344c3ddccf1b020490cg_0059_2.jpg", "batch23-2024_06_14_b344c3ddccf1b020490cg_0059_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Draw $N Q \\perp B C$ intersecting at point $Q$,\n\n\n\nSince the rectangle is folded along $E F$,\n\n$\\therefore N Q=\\frac{1}{2} A B$\n\nSince $\\triangle A B M \\cong \\triangle N B M$\n$\\therefore A B=N B, \\angle A B M=\\angle N B M, \\angle A M B=\\angle N M B$,\n\n$\\therefore N Q=\\frac{1}{2} N B$,\n\n$\\therefore \\angle N B Q=30^{\\circ}$, which means $\\angle N B C=30^{\\circ}$, so A is correct, not matching the question;\n\nSince $A B C D$ is a rectangle,\n\n$\\therefore \\angle A B N=60^{\\circ}$,\n\n$\\therefore \\angle A B M=\\angle N B M=30^{\\circ}$\n\n$\\therefore \\angle A M B=60^{\\circ}$,\n\nSince $\\angle A M B=\\angle N M B$,\n\n$\\therefore \\angle A M B=\\angle N M B=60^{\\circ}$,\n\n$\\therefore \\angle A M N=\\angle A M B+\\angle N M B=120^{\\circ}$, so B is incorrect, matching the question;\n\nSince $\\angle M B P=\\angle N B C+\\angle N B M=60^{\\circ}, \\angle N M B=60^{\\circ}$,\n\n$\\therefore \\triangle B M P$ is an equilateral triangle, so $\\mathrm{C}$ is correct, not matching the question;\n\nSince $\\triangle B M P$ is an equilateral triangle, $\\angle N B M=\\angle N B C=30^{\\circ}$\n\n$\\therefore B N \\perp M P$, which means point $N$ is the midpoint of $M P$, so D is correct, not matching the question;\n\nTherefore, the answer is: B\n\n【Key Insight】This question tests the properties of rectangles, the properties of folding, the determination of congruent triangles, and the property that the side opposite a $30^{\\circ}$ angle in a right triangle is half the hypotenuse. The key to solving the problem is to be proficient in the properties of rectangles, the properties of folding, the determination of congruent triangles, and the property that the side opposite a $30^{\\circ}$ angle in a right triangle is half the hypotenuse." }, { "problem_id": 892, "question": "As shown in the figure, Figure 1 is a diagram of a highball glass filled with liquid (data as shown). After using some of the liquid, it appears as in Figure 2. At this time, the diameter of the liquid surface $A B=$( )\n\n\nFigure 1\n\n\nFigure 2\nA. $(6-4 \\tan \\frac{\\alpha}{2}) \\mathrm{cm}$\nB. $(6-8 \\tan \\frac{\\alpha}{2}) \\mathrm{cm}$\nC. $(6-4 \\tan \\alpha) \\mathrm{cm}$\nD. $(6-8 \\tan \\alpha) \\mathrm{cm}$", "input_image": [ "batch37-2024_06_14_bd97c218f59ff803a167g_0044_1.jpg", "batch37-2024_06_14_bd97c218f59ff803a167g_0044_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure: Draw $O M \\perp C D$ through point $O$, with the foot of the perpendicular at $M$, and draw $O^{\\prime} N \\perp A B$ through point $O^{\\prime}$, with the foot of the perpendicular at $N$.\n\n\nFigure 1\n\n\nFigure 2\n\n$\\because C D / / A B$\n\n$\\therefore \\triangle C D O \\sim A B O^{\\prime}$, meaning the similarity ratio is $\\frac{C D}{A B}$,\n\n$\\therefore \\frac{C D}{A B}=\\frac{O M}{O^{\\prime} N}$,\n\n$\\because \\tan \\frac{\\alpha}{2}=\\frac{C M}{O M}$,\n$\\therefore O M=\\frac{\\frac{1}{2} C D}{\\tan \\frac{\\alpha}{2}}=\\frac{3}{\\tan \\frac{\\alpha}{2}}$,\n\n$\\because O^{\\prime} N=O M-(15-11)=\\frac{3-4 \\tan \\frac{\\alpha}{2}}{\\tan \\frac{\\alpha}{2}} \\mathrm{~cm}$,\n\n$\\therefore \\frac{6}{A B}=\\frac{\\frac{3}{\\tan \\frac{\\alpha}{2}}}{\\frac{3-4 \\tan \\frac{\\alpha}{2}}{\\tan \\frac{\\alpha}{2}}}=\\frac{3}{3-4 \\tan \\frac{\\alpha}{2}}$,\n\n$\\therefore A B=6-8 \\tan \\frac{\\alpha}{2}$.\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem examines the application of similar triangles and the solution of right triangles. Mastering the determination and properties of similar triangles is crucial for solving the problem." }, { "problem_id": 893, "question": "A is a circular cleaning robot with a diameter of $30 \\mathrm{~cm}$, B is a Reuleaux triangle cleaning robot (an enclosed figure formed by drawing arcs with the vertices of an equilateral triangle $\\triangle A B C$ as centers and the side lengths as radii), and C is a rectangular room with a length of $4 \\mathrm{~m}$ and a width of $3 \\mathrm{~m}$. Now, either A or B is used alone to clean C. The area of the \"blind spots\" that cannot be cleaned is ( )\n\n\nA\n\n\nB\n\n\nC\nA. Larger for A\nB. Larger for B\nC. The same for A and B\nD. Indeterminable", "input_image": [ "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0001_1.jpg", "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0001_2.jpg", "batch20-2024_05_23_cfa64aa3bc59a683d5f2g_0001_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "The text describes a scenario involving a Reuleaux triangle-shaped cleaning robot ( B) with a circumference of \\(30\\pi\\) cm. The Reuleaux triangle is formed by drawing arcs centered at each vertex of an equilateral triangle \\(ABC\\) with side lengths equal to the radius of the arcs. The key steps and conclusions are as follows:\n\n1. **Side Length Calculation**:\n - Since the circumference of the Reuleaux triangle is \\(30\\pi\\) cm, and it consists of three equal arcs, each arc corresponds to a side of the equilateral triangle.\n - Therefore, each side of the equilateral triangle \\(ABC\\) is:\n \\[\n AB = AC = BC = \\frac{1}{3} \\times 30\\pi = 10\\pi \\text{ cm}\n \\]\n - However, this is later corrected to \\(AB = AC = BC = 30\\) cm based on the arc length calculation.\n\n2. **Angle Calculation**:\n - The angles of the equilateral triangle are all \\(60^\\circ\\):\n \\[\n \\angle BAC = \\angle ACB = \\angle ABC = 60^\\circ\n \\]\n\n3. **Arc Length Calculation**:\n - The length of arc \\(BC\\) is given by:\n \\[\n BC = \\frac{60 \\times AC \\times \\pi}{180}\n \\]\n - Substituting the known values:\n \\[\n 10\\pi = \\frac{60 \\times AC \\times \\pi}{180}\n \\]\n - Solving for \\(AC\\):\n \\[\n AC = 30 \\text{ cm}\n \\]\n - Thus, the side lengths of the equilateral triangle are:\n \\[\n AB = AC = BC = 30 \\text{ cm}\n \\]\n\n4. **Circumradius Calculation**:\n - Let \\(O\\) be the circumcenter of the equilateral triangle \\(ABC\\). Connecting \\(AO\\) and \\(CO\\), and extending \\(CO\\) to intersect \\(AB\\) at point \\(D\\), we have:\n \\[\n CD \\perp AB, \\quad AD = BD = \\frac{1}{2} AB = 15 \\text{ cm}\n \\]\n - The angle \\(\\angle OAD\\) is half of \\(\\angle BAC\\):\n \\[\n \\angle OAD = \\angle CAO = \\frac{1}{2} \\angle BAC = 30^\\circ\n \\]\n - In the right triangle \\(OAD\\), using the cosine of \\(\\angle OAD\\):\n \\[\n \\cos 30^\\circ = \\frac{AD}{OA}\n \\]\n - Solving for \\(OA\\):\n \\[\n OA = \\frac{AD}{\\cos 30^\\circ} = \\frac{15}{\\frac{\\sqrt{3}}{2}} = 10\\sqrt{3} \\text{ cm}\n \\]\n - Therefore, the diameter of the circle formed by the rotation of the robot is:\n \\[\n 2OA = 2 \\times 10\\sqrt{3} = 20\\sqrt{3} \\text{ cm}\n \\]\n\n5. **Comparison of Areas**:\n - Comparing \\(20\\sqrt{3}\\) cm and \\(30\\) cm:\n \\[\n 20\\sqrt{3} = \\sqrt{1200}, \\quad 30 = \\sqrt{900}\n \\]\n - Since \\(\\sqrt{1200} > \\sqrt{900}\\), it follows that:\n \\[\n 20\\sqrt{3} > 30\n \\]\n - This implies that the area of the \"dead zone\" (the area the robot cannot clean) is larger for robot B.\n\n6. **Conclusion**:\n - The correct answer is **B**.\n\n**Key Insight**: This problem involves understanding the properties of an equilateral triangle and its circumradius, as well as the geometry of the Reuleaux triangle. The key is to correctly calculate the circumradius and use it to determine the diameter of the robot's rotation, which in turn affects the area it can clean." }, { "problem_id": 894, "question": "As shown in Figure 1, one side of a rectangle has length $x$, and half of its perimeter is $y$. Define $(x, y)$ as the coordinates of this rectangle. As shown in Figure 2, in the Cartesian coordinate system, the lines $x = 1$ and $y = 3$ divide the first quadrant into four regions. Given that the point $A$ corresponding to the coordinates of Rectangle 1 lies on the hyperbola shown, and the point corresponding to the coordinates of Rectangle 2 lies in Region (4).\n\n\nFigure 1\n\n\nFigure 2\n\nThen which of the following statements is correct?\n\nA. The x-coordinate of point $A$ could be greater than 3.\n\nB. When Rectangle 1 is a square, point $A$ is located in Region (2).\n\nC. As point $A$ moves up along the hyperbola, the area of Rectangle 1 decreases.\n\nD. When point $A$ is in Region (1), Rectangle 1 could be congruent to Rectangle 2.", "input_image": [ "batch13-2024_06_15_5d227d10b18c15d49d57g_0002_1.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0002_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Let point \\( A(x, y) \\) (where \\( x, y \\) are both positive numbers).\n\nA. Let the inverse proportional function be: \\( y = \\frac{k}{x} \\) (\\( k \\neq 0 \\)).\n\nFrom the graph, when \\( x = 1 \\), \\( y < 3 \\).\n\nThus, \\( k = xy < 3 \\).\n\nSince \\( x < y \\), \\( x^2 < xy < 3 \\).\n\nTherefore, \\( x < \\sqrt{3} < 3 \\).\n\nThis means the x-coordinate of point \\( A \\) cannot be greater than 3.\n\nHence, option A is incorrect.\n\nB. When rectangle 1 is a square, the side lengths are \\( x \\) and \\( y = 2x \\).\n\nThen, point \\( A \\) is the intersection of the line \\( y = 2x \\) and the hyperbola, as shown in Figure 2. The intersection \\( A \\) is in region (3).\n\nThus, option B is incorrect.\n\nC. When one side is \\( x \\), the other side is \\( y - x \\). The area \\( S = x(y - x) = xy - x^2 = k - x^2 \\).\n\nSince as point \\( A \\) moves upward along the hyperbola, the value of \\( x \\) decreases, the area of rectangle 1 increases.\n\nTherefore, option C is incorrect.\n\nD. When point \\( A \\) is in region (1):\n\nSince point \\( A(x, y) \\), \\( x < 1 \\) and \\( y > 3 \\), the other side is \\( y - x > 2 \\).\n\nRectangle 2 falls in region (4), where \\( x > 1 \\) and \\( y > 3 \\), so the other side \\( y - x > 0 \\).\n\nThus, when point \\( A \\) is in region (1), rectangle 1 may be congruent to rectangle 2.\n\nFor example, if the adjacent sides of the rectangle are \\( 0.9 \\) and \\( 2.9 \\), both rectangles satisfy the conditions and are congruent.\n\nTherefore, option D is correct.\n\nThe correct answer is: D.\n\n[Key Insight] This question tests the understanding of inverse proportional function graphs and new definitions. Understanding the meanings of \\( x \\) and \\( y \\) is crucial, and solving the problem using a combination of numerical and graphical methods is essential." }, { "problem_id": 895, "question": "In a school's science and technology club, Xiaoming followed these steps to create a practical device:\n\n\n\n\nFigure 2\n\n\nFigure 3\n\n(1) Xiaoming took a circular thin iron ring provided by the teacher, found the center $O$, and then chose an arbitrary diameter, which he marked as $A B$ (as shown in Figure 1). He measured $A B$ to be 8 decimeters;\n\n(2) He folded the ring so that point $B$ landed on the center $O$. The folded and unfolded parts of the ring intersected at points $C$ and $D$ (as shown in Figure 2).\n\n(3) He connected points $C$ and $D$ with a thin rubber rod (as shown in Figure 3);\n\n(4) He calculated the length of the rubber rod $C D$.\n\nXiaoming calculated the length of rubber rod $C D$ to be $(\\quad)$\nA. $4 \\sqrt{3}$ decimeters\nB. $2 \\sqrt{3}$ decimeters\nC. $2 \\sqrt{6}$ decimeters\nD. $3 \\sqrt{2}$ decimeters", "input_image": [ "batch19-2024_05_24_5280277849147224ad6cg_0062_1.jpg", "batch19-2024_05_24_5280277849147224ad6cg_0062_2.jpg", "batch19-2024_05_24_5280277849147224ad6cg_0062_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:** Draw \\( O E \\perp C D \\) at point \\( E \\), intersecting \\( \\odot O \\) at point \\( F \\).\n\n\n\nFigure 3\n\nSince \\( C D \\) is the perpendicular bisector of \\( O F \\),\n\n\\[\n\\therefore C O = C F\n\\]\n\n\\[\n\\therefore C O = C F = O F\n\\]\n\n\\[\n\\therefore \\triangle O C F \\text{ is an equilateral triangle}\n\\]\n\n\\[\n\\because O C = 4\n\\]\n\n\\[\n\\therefore C E = O C \\cdot \\cos 30^{\\circ} = 2 \\sqrt{3}\n\\]\n\n\\[\n\\because O E \\perp C D\n\\]\n\n\\[\n\\therefore C E = E D\n\\]\n\n\\[\n\\therefore C D = 2 C E = 4 \\sqrt{3}\n\\]\n\nTherefore, the correct answer is **A**.\n\n**Key Insight:** This problem tests the knowledge of the **Perpendicular Chord Theorem**, **Inscribed Angle Theorem**, and **Transformation by Folding**. Mastery of these fundamental concepts is crucial, as this type of question is commonly encountered in middle school exams." }, { "problem_id": 896, "question": "Enlarge a triangle and a rectangle as shown in the figure, so that the distance between their corresponding sides is 1, resulting in new triangles and rectangles. Which of the following statements is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nA. The new triangle is similar to the original triangle.\n\nB. The new rectangle is similar to the original rectangle.\n\nC. Both the new triangle and the new rectangle are similar to their respective original figures.\n\nD. None of them are similar.", "input_image": [ "batch32-2024_06_14_e37e324738a2af6a062bg_0049_1.jpg", "batch32-2024_06_14_e37e324738a2af6a062bg_0049_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement, we have:\n$AB // A^{\\prime} B^{\\prime}, AC // A^{\\prime} C^{\\prime}, BC // B^{\\prime} C^{\\prime}$,\n$\\therefore \\angle A=\\angle A^{\\prime}, \\quad \\angle B=\\angle B^{\\prime}$,\n\n$\\therefore \\triangle ABC \\sim \\triangle A^{\\prime} B^{\\prime} C^{\\prime}$\n\nGiven that $\\mathrm{AB}=\\mathrm{CD}=\\mathrm{x}, \\quad \\mathrm{AD}=\\mathrm{BC}=\\mathrm{y}$, then $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime}=\\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}=\\mathrm{x}+2, \\mathrm{~A}^{\\prime} \\mathrm{D}^{\\prime}=\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}=\\mathrm{y}+2$,\n\n$\\therefore \\frac{AD}{A^{\\prime} D^{\\prime}}=\\frac{y}{y+2}, \\frac{AB}{A^{\\prime} B^{\\prime}}=\\frac{x}{x+2}$,\n\nSince $x \\neq y$\n\n$\\therefore \\frac{AD}{A^{\\prime} D^{\\prime}} \\neq \\frac{AB}{A^{\\prime} B^{\\prime}}$,\n\n$\\therefore$ The ratio of the corresponding sides of the new rectangle to the original rectangle is not equal,\n\n$\\therefore$ The new rectangle is not similar to the original rectangle.\n\n\n\nTherefore, the correct choice is $A$.\n\n【Key Insight】The problem examines the judgment of similar figures. The key to solving it is understanding that polygons with equal corresponding angles and proportional corresponding sides are called similar polygons." }, { "problem_id": 897, "question": "Pythagorean theorem is an ancient theorem that was recorded in ancient Chinese mathematical book \"Zhou Bi Suan Jing\". Mathematicians have suggested using Figure 1 as a signal to communicate with \"aliens\". As shown in Figure 1, squares are constructed outwardly on each side of the right-angled triangle $ABC$ (with $AB > AC$). Then, the largest square paper is folded upward as shown in Figure 2. If the area of the shaded region in the figure is known, then it is definitely possible to find ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. The area of square $B C M N$\nB. The area of quadrilateral $N P A B$\nC. The area of square $A C D E$\nD. The area of right-angled triangle $A B C$", "input_image": [ "batch5-2024_06_14_7ad1f93a1f17b69ef356g_0045_1.jpg", "batch5-2024_06_14_7ad1f93a1f17b69ef356g_0045_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "According to the problem, in triangles $\\triangle CMP$ and $\\triangle BCK$, $\\angle BCK = \\angle M = 90^\\circ$.\n\nIn the right triangle $\\triangle ABC$, $\\angle ABC + \\angle ACB = 90^\\circ$.\n\nAlso, $\\angle ACB + \\angle ACM = 90^\\circ$.\n\nTherefore, $\\angle ABC = \\angle ACM$.\n\nIn triangles $\\triangle BCK$ and $\\triangle CMP$,\n\n\\[\n\\begin{cases}\n\\angle ABC = \\angle ACM \\\\\nBC = CM \\\\\n\\angle BCK = \\angle M\n\\end{cases},\n\\]\n\nThus, $\\triangle BCK \\cong \\triangle CMP$.\n\nTherefore, the area of $\\triangle BCK$ equals the area of $\\triangle CMP$,\n\n\\[\nS_{\\triangle BCK} = S_{\\triangle CMP},\n\\]\n\n\\[\nS_{\\triangle BCK} - S_{\\triangle ACK} = S_{\\triangle CMP} - S_{\\triangle ACK},\n\\]\n\n\\[\nS_{\\triangle ABC} = S_{\\text{shaded area}},\n\\]\n\nHence, the correct answer is: D.\n\n【Key Insight】This problem examines the proof and properties of congruent triangles. Mastering the methods of proving triangle congruence is crucial for solving such problems." }, { "problem_id": 898, "question": "Fold the rectangular paper $\\mathrm{ABCD}$ along a straight line passing through point $\\mathrm{B}$ so that point $\\mathrm{A}$ falls on point $\\mathrm{F}$ on side $\\mathrm{BC}$, with the crease being $\\mathrm{BE}$ (as shown in Figure 1); then fold again along a straight line passing through point $\\mathrm{E}$ so that point $\\mathrm{D}$ falls on point $\\mathrm{D}^{\\prime}$ on $\\mathrm{BE}$, with the crease being $\\mathrm{EG}$ (as shown in Figure 2); finally, flatten the paper (as shown in Figure 3). The measure of angle $\\alpha$ in Figure 3 is ( \n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $30^{\\circ}$\nB. $25.5^{\\circ}$\nC. $20^{\\circ}$\nD. $22.5^{\\circ}$", "input_image": [ "batch23-2024_06_14_28843a4185f44c6dd7b6g_0029_1.jpg", "batch23-2024_06_14_28843a4185f44c6dd7b6g_0029_2.jpg", "batch23-2024_06_14_28843a4185f44c6dd7b6g_0029_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, as shown in Figure (3), quadrilateral $\\mathrm{ABFE}$ is a square.\n\n$\\therefore \\angle \\mathrm{AEB}=\\angle \\mathrm{FEB}=45^{\\circ}$.\n\nSince $\\mathrm{EG}$ is the fold line, $\\therefore \\angle \\mathrm{BEG}=\\angle \\mathrm{DEG}$.\n\nGiven that $\\angle \\mathrm{AEB}=45^{\\circ}$ and $\\angle \\mathrm{AEB}+\\angle \\mathrm{BEG}+\\angle \\mathrm{DEG}=180^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{DEG}=67.5^{\\circ}$.\n\nThus, $\\angle \\alpha=90^{\\circ}-\\angle \\mathrm{DEG}=22.5^{\\circ}$.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This problem tests the properties of folding, the properties of a square, and the relationships between angles. The difficulty of the problem is not high, and the key to solving it lies in the application of the concept of combining numbers with shapes." }, { "problem_id": 899, "question": "As shown in the figure, the straight line $y = ax + b$ intersects the graph of the function $y = \\frac{k}{x} (x > 0)$ at points $A(1, m)$ and $B(n, 1)$, and it intersects the $x$-axis at point $C$. Given that $\\frac{BC}{AC} = \\frac{1}{3}$, the solution set of the inequality $ax + b > \\frac{k}{x}$ on the number line is represented correctly by ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_8ea622ffe3d86972ab0eg_0019_1.jpg", "batch13-2024_06_15_8ea622ffe3d86972ab0eg_0019_2.jpg", "batch13-2024_06_15_8ea622ffe3d86972ab0eg_0019_3.jpg", "batch13-2024_06_15_8ea622ffe3d86972ab0eg_0019_4.jpg", "batch13-2024_06_15_8ea622ffe3d86972ab0eg_0019_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, draw a perpendicular from point $A$ to $OC$ at point $D$, and from point $B$ to $OC$ at point $E$.\n\n\n\n$\\therefore AD \\parallel BE$,\n\n$\\therefore \\triangle BCE \\sim \\triangle ACD$,\n\n$\\therefore \\frac{BE}{AD} = \\frac{BC}{AC} = \\frac{1}{3}$,\n\n$\\because A(1, m)$ and $B(n, 1)$,\n\n$\\therefore BE = 1$, $AD = m$,\n\n$\\therefore \\frac{1}{m} = \\frac{1}{3}$, which implies $m = 3$,\n$\\therefore$ point $A(1,3)$,\n\n$\\therefore$ the inverse proportional function is $y = \\frac{3}{x}$,\n\nSubstituting point $B$ $(n, 1)$ into the equation gives: $n = 3$,\n\nObserving the graph: when $1 < x < 3$, the graph of the linear function $y = ax + b$ is above the graph of the inverse proportional function $y = \\frac{k}{x}$,\n\n$\\therefore$ the solution set of the inequality $ax + b > \\frac{k}{x}$ is $1 < x < 3$,\n\nRepresenting the solution set on the number axis, as shown below:\n\n\n\nTherefore, the correct choice is: D\n\n【Key Point】This problem mainly examines the intersection of linear and inverse proportional functions, the determination and properties of similar triangles. Mastering the determination and properties of similar triangles and using graphical interpretation is crucial for solving the problem." }, { "problem_id": 900, "question": "As shown in the figure, $\\triangle A B C$ is an equilateral triangle, and $D$ is the midpoint of $B C$. After rotating $\\triangle A B C$ clockwise around center $D$ by $60^{\\circ}$, the resulting figure is ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch26-2024_06_17_4ff1ec270366d069efb8g_0093_1.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0093_2.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0093_3.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0093_4.jpg", "batch26-2024_06_17_4ff1ec270366d069efb8g_0093_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\triangle \\mathrm{ABC}$ is an equilateral triangle,\n\n$\\therefore \\angle \\mathrm{B}=\\angle \\mathrm{ACB}=60^{\\circ}$,\n\nBecause the triangle $\\triangle ABC$ is rotated clockwise by $60^{\\circ}$ around point $\\mathrm{D}$ to obtain $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$, and $D$ is the midpoint of $BC$,\n\n$\\therefore \\angle \\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}=60^{\\circ}, \\angle \\mathrm{BDB}^{\\prime}=60^{\\circ}, \\mathrm{DB}=\\mathrm{DB}^{\\prime}$,\n\n$\\therefore$ point $\\mathrm{B}^{\\prime}$ lies on $\\mathrm{AB}$,\n\nSimilarly, point $\\mathrm{C}$ lies on $\\mathrm{A}^{\\prime} \\mathrm{C}^{\\prime}$, as shown in the figure.\n\nTherefore, the correct choice is D.\n\n\n\n【Insight】This question examines the properties of rotation: the distance from corresponding points to the center of rotation is equal; the angle between the lines connecting corresponding points and the center of rotation equals the rotation angle; the figures before and after rotation are congruent." }, { "problem_id": 901, "question": "As shown in the figure, $A B C D$ is a square piece of paper, $E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. By folding along the crease passing through point $D$, point $A$ is placed on $E F$ (as shown in Figure (2) as point $A^{\\prime}$). The crease intersects $A E$ at point $G$. What is the measure of $\\angle A D G$?\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $15^{\\circ}$\nB. $20^{\\circ}$\nC. $25^{\\circ}$\nD. $30^{\\circ}$", "input_image": [ "batch23-2024_06_14_a705467cdbc40a4240d2g_0053_1.jpg", "batch23-2024_06_14_a705467cdbc40a4240d2g_0053_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( A B C D \\) is a square,\n\n\\[\n\\therefore \\angle C = \\angle A = 90^{\\circ}, \\quad AD = BC = CD = AB.\n\\]\n\nSince \\( E \\) and \\( F \\) are the midpoints of \\( AB \\) and \\( CD \\) respectively,\n\n\\[\n\\therefore EF \\parallel BC,\n\\]\n\n\\[\n\\therefore \\text{quadrilateral } ADFE \\text{ is a rectangle},\n\\]\n\n\\[\n\\therefore \\angle EFD = 90^{\\circ}, \\quad FD = \\frac{1}{2} CD = \\frac{1}{2} AD.\n\\]\n\nAccording to the properties of folding: \\( A'D = AD \\).\n\nIn right triangle \\( \\triangle FAD \\), \\( DF = \\frac{1}{2} A'D \\),\n\n\\[\n\\therefore \\angle FA'D = 30^{\\circ},\n\\]\n\n\\[\n\\therefore \\angle A'DA = \\angle FA'D = 30^{\\circ},\n\\]\n\n\\[\n\\therefore \\angle ADG = \\angle A'DG = \\frac{1}{2} \\angle ADA' = \\frac{1}{2} \\times 30^{\\circ} = 15^{\\circ}.\n\\]\n\nTherefore, the correct answer is: A.\n\n**Key Insight:** This problem examines the properties of squares, the properties of folding, and the properties of right triangles containing a \\( 30^{\\circ} \\) angle. The key to solving this problem lies in deducing that in right triangle \\( \\triangle FAD \\), \\( DF = \\frac{1}{2} A'D \\), which leads to \\( \\angle FA'D = 30^{\\circ} \\)." }, { "problem_id": 902, "question": "Definition: If the angle bisector of a triangle is equal to the side opposite the angle it bisects, then the triangle is called a \"beautiful triangle,\" and this angle bisector is called the \"beautiful line\" of the triangle. Among the following four triangles, where $B D$ bisects $\\angle A B C$, which one has $B D$ as a \"beautiful line\"?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch36-2024_06_17_4ee3413e94120a083933g_0002_1.jpg", "batch36-2024_06_17_4ee3413e94120a083933g_0002_2.jpg", "batch36-2024_06_17_4ee3413e94120a083933g_0002_3.jpg", "batch36-2024_06_17_4ee3413e94120a083933g_0002_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: \n\n**A.** In the right-angled triangle \\( ABC \\), \\( \\angle C = 90^\\circ \\), \\( \\angle A = 30^\\circ \\).\n\nThus, \\( \\angle ABC = 60^\\circ \\).\n\nSince \\( BD \\) bisects \\( \\angle ABC \\),\n\n\\( \\angle DBC = \\angle ABD = \\frac{1}{2} \\angle ABC = 30^\\circ \\).\n\nTherefore, \\( \\angle ABD = \\angle BAD = 30^\\circ \\).\n\nHence, \\( \\angle ADB = 150^\\circ \\).\n\nThus, \\( BD < AB \\).\n\nAlso, since \\( \\angle C = 90^\\circ \\),\n\n\\( \\angle BDC = 60^\\circ \\).\n\nTherefore, \\( BD > BC \\), meaning \\( BD \\) is not a \"beautiful line\".\n\nThus, option **A** does not meet the requirement.\n\n---\n\n**B.** In triangle \\( ABC \\), \\( \\angle A = \\angle C = 60^\\circ \\).\n\nThus, \\( \\triangle ABC \\) is an equilateral triangle,\n\nso \\( \\angle ABC = 60^\\circ \\), and \\( BA = BC \\).\n\nSince \\( BD \\) bisects \\( \\angle ABC \\),\n\n\\( \\angle ABD = \\angle CBD = 30^\\circ \\).\n\nTherefore, \\( \\angle ADB = \\angle CDB = 90^\\circ \\).\n\nThus, \\( BD < AB \\), and \\( BD < BC \\), meaning \\( BD \\) is not a \"beautiful line\".\n\nThus, option **B** does not meet the requirement.\n\n---\n\n**C.** In triangle \\( ABC \\), \\( \\angle A = 90^\\circ \\), \\( \\angle C = 45^\\circ \\).\n\nThus, \\( \\angle ABC = 45^\\circ \\).\n\nSince \\( BD \\) bisects \\( \\angle ABC \\),\n\n\\( \\angle ABD = \\angle CBD = \\frac{1}{2} \\angle ABC = 22.5^\\circ \\).\n\nTherefore, \\( \\angle ADB = 67.5^\\circ \\), and \\( \\angle BDC = 112.5^\\circ \\).\n\nThus, \\( BD > AB \\), and \\( BD < BC \\), meaning \\( BD \\) is not a \"beautiful line\".\n\nThus, option **C** does not meet the requirement.\n\n---\n\n**D.** In triangle \\( ABC \\), \\( \\angle A = 36^\\circ \\), \\( \\angle C = 72^\\circ \\).\n\nThus, \\( \\angle ABC = 180^\\circ - 36^\\circ - 72^\\circ = 72^\\circ \\).\n\nSince \\( BD \\) bisects \\( \\angle ABC \\),\n\n\\( \\angle ABD = \\angle CBD = \\frac{1}{2} \\angle ABC = 36^\\circ \\).\n\nTherefore, \\( \\angle BDC = 180^\\circ - 36^\\circ - 72^\\circ = 72^\\circ \\).\n\nThus, \\( BD = BC \\), meaning \\( BD \\) is a \"beautiful line\".\n\nThus, option **D** meets the requirement.\n\n---\n\n**Answer:** \\( D \\)\n\n**Key Insight:** This question primarily tests the understanding of the definition of a \"beautiful line\". Mastering the definition of a \"beautiful line\" is crucial for solving this problem." }, { "problem_id": 903, "question": "As shown in the figure, Figure 1, Figure 2, and Figure 3 represent the routes taken by individuals A, B, and C from point A to point B (arrows indicate the direction of travel). Point E is the midpoint of AB, and AH > HB. Determine the order of the lengths of their routes as ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. A < BC\nB. B < C < A\nC. C < B < A\nD. A = B = C", "input_image": [ "batch23-2024_06_14_3aea59197e45be81d8deg_0049_1.jpg", "batch23-2024_06_14_3aea59197e45be81d8deg_0049_2.jpg", "batch23-2024_06_14_3aea59197e45be81d8deg_0049_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In Figure 1, the length of the path taken by A is the sum of the lengths of $\\mathrm{AC}$ and $\\mathrm{BC}$.\n\nIn Figure 2, as shown, extend $\\mathrm{ED}$ and $\\mathrm{BF}$ to intersect at $\\mathrm{C}$.\n\n\nSince $\\angle \\mathrm{DEA} = \\angle \\mathrm{B} = 60^{\\circ}$, it follows that $\\mathrm{DE} \\parallel \\mathrm{CF}$.\n\nSimilarly, $\\mathrm{EF} \\parallel \\mathrm{CD}$.\n\nTherefore, quadrilateral $\\mathrm{CDEF}$ is a parallelogram,\n\nHence, $\\mathrm{EF} = \\mathrm{CD}$ and $\\mathrm{DE} = \\mathrm{CF}$.\n\nThis means the length of the path taken by B is $\\mathrm{AD} + \\mathrm{DE} + \\mathrm{EF} + \\mathrm{FB} = \\mathrm{AD} + \\mathrm{CD} + \\mathrm{CF} + \\mathrm{BC} = \\mathrm{AC} + \\mathrm{BC}$.\n\nIn Figure 3, as shown, extend $\\mathrm{AG}$ and $\\mathrm{BK}$ to intersect at $\\mathrm{C}$.\n\n\n\nFollowing a similar proof process as above, $\\mathrm{GH} = \\mathrm{CK}$ and $\\mathrm{CG} = \\mathrm{HK}$. Thus, the length of the path taken by C is $\\mathrm{AG} + \\mathrm{GH} + \\mathrm{HK} + \\mathrm{KB} = \\mathrm{AG} + \\mathrm{CG} + \\mathrm{CK} + \\mathrm{BK} = \\mathrm{AC} + \\mathrm{BC}$.\n\nTherefore, A = B = C.\n\nHence, the correct choice is D." }, { "problem_id": 904, "question": "As shown in Figure 1, four small rectangles with a perimeter of 12 each are placed inside the rectangle $A B C D$. Now, rectangle $E F G H$ is placed inside the large rectangle $A B C D$ such that it overlaps with the four small rectangles, as shown in Figure 2. Given that $A B = 10$ and $B C = 8$, the sum of the perimeters of the overlapping parts is 28. What is the perimeter of rectangle $E F G H$? $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 20\nB. 24\nC. 26\nD. 28", "input_image": [ "batch33-2024_06_14_63e9a6949d940e018a72g_0020_1.jpg", "batch33-2024_06_14_63e9a6949d940e018a72g_0020_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\nGiven that $AB = 10$ and $BC = 8$,\n\nTherefore, $AB + BC + CD + DA = 2(AB + BC) = 36$,\n\nSince four small rectangles, each with a perimeter of 12, are placed inside rectangle $ABCD$,\n\nThus, $AN + AO = BM + BL = CK + CJ = DI + PD = \\frac{1}{2} \\times 12 = 6$,\n\nTherefore, $(AB + BC + CD + DA) - (AN + AO) - (BM + BL) - (CK + CJ) - (DI + PD) = 36 - 6 - 6 - 6 - 6 = 12$, which means $MN + LK + IJ + OP = 12$,\n\nHence, $XW + UV + ST + QR = 12$,\n\nSince the sum of the perimeters of the four overlapping parts is 28,\n\nThus, $EX + EQ + RH + HS + TG + GU + FV + WF = \\frac{1}{2} \\times 28 = 14$,\n\nTherefore, $(EX + EQ + RH + HS + TG + GU + FV + WF) + (XW + UV + ST + QR) = 14 + 12 = 26$,\n\nHence, $EF + FG + HG + EH = 26$, which means the perimeter of rectangle $EFGH$ is 26,\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem tests the understanding of the perimeter of a rectangle. The key to solving it is to grasp that the perimeter of a rectangle is equal to twice the sum of its length and width." }, { "problem_id": 905, "question": "In ancient China, during the Western Zhou Dynasty, mathematician Shang Gao summarized a method of measuring the height of an object using a \"square\" (as shown in Figure 1): Place the square's sides as shown in Figure 2, with one end A (the observer's eye) looking at point E, making the line of sight pass through the other end point C. Label the observer's standing position as point B, and measure the length $B G$. The height of the object $E G$ can then be calculated. Given $a = 30 \\text{ cm}, b = 60 \\text{ cm}, A B = 1.6 \\text{ m}, B G = 2.4 \\text{ m}$, the height of $E G$ is $(\\quad)$\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $1.2 \\text{ m}$\nB. $2.8 \\text{ m}$\nC. $4.8 \\text{ m}$\nD. $6.4 \\text{ m}$", "input_image": [ "batch32-2024_06_14_8e19a707a2b31f303531g_0049_1.jpg", "batch32-2024_06_14_8e19a707a2b31f303531g_0049_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 2, we can deduce that $AF = BG = 2.4 \\mathrm{~m}$, $EF = EG - FG$, and $FG = AB = 1.6 \\mathrm{~m}$.\n\nTherefore, $EF = (EG - 1.6) \\mathrm{m}$.\n\nSince $CD \\perp AF$ and $EF \\perp AF$,\n\nIt follows that $CD // EF$,\n\nThus, $\\triangle ADC \\sim \\triangle AFE$,\n\nHence, $\\frac{CD}{EF} = \\frac{AD}{AF}$,\n\nWhich means $\\frac{30}{(EG - 160)} = \\frac{60}{240}$,\n\nSolving this, we find $EG = 280 \\mathrm{~cm} = 2.8 \\mathrm{~m}$.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This problem tests the application of linear functions and the properties of similar triangles. The key to solving it lies in understanding the given information and using a combination of graphical and algebraic methods to find the solution." }, { "problem_id": 906, "question": "A type of paper clip is shown in Figure 1. Figure 2 is a diagram of the closed position, and Figure 3 is a diagram of the open position (data as shown, units: $\\mathrm{mm}$). The distance between points $B$ and $D$ decreases by ( ) when the clip goes from the closed to the open position.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. $25 \\mathrm{~mm}$\nB. $20 \\mathrm{~mm}$\nC. $15 \\mathrm{~mm}$\nD. $8 \\mathrm{~mm}$", "input_image": [ "batch32-2024_06_14_8e19a707a2b31f303531g_0030_1.jpg", "batch32-2024_06_14_8e19a707a2b31f303531g_0030_2.jpg", "batch32-2024_06_14_8e19a707a2b31f303531g_0030_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, connect $BD$,\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n$\\because AE = CF = 28, \\quad BE = DF = 35$,\n\n$\\therefore \\frac{AE}{AB} = \\frac{AF}{AD} = \\frac{28}{63} = \\frac{4}{9}$, and $\\angle EAF = \\angle BAD$,\n\n$\\therefore \\triangle AEF \\sim \\triangle ABD$,\n\n$\\therefore \\frac{BD}{EF} = \\frac{AB}{AE}$, and $EF = 20$,\n\n$\\therefore \\frac{BD}{20} = \\frac{9}{4}$, solving gives: $BD = 45$,\n\nAs shown in Figure 3, connect $BD$,\n\n$\\because BE \\parallel DF, BE = DF$,\n\n$\\therefore$ quadrilateral $EFDB$ is a parallelogram,\n\n$\\because \\angle BEF = 90^{\\circ}$,\n\n$\\therefore$ quadrilateral $EFDB$ is a rectangle, hence $BD = EF = 20$,\n\n$\\therefore$ the distance between $B$ and $D$ decreases from closed to open by $45 - 20 = 25$ (mm),\n\nTherefore, the answer is: A.\n\n【Key Insight】This problem examines the application of similar triangles, the determination of parallelograms, and the properties of rectangles. Understanding the problem and utilizing the properties of similar triangles to solve practical problems is key to answering this question." }, { "problem_id": 907, "question": "The students in an exploratory learning group were assigned the task of measuring the diameter of the same type of cylindrical workpiece. They used a right-angled triangle with a $60^\\circ$ angle and a ruler. Xiaoming's measurement method is shown in Figure Alpha. He measured $\\mathrm{PC} = 12 \\mathrm{~cm}$. Xiaoliang's measurement method is shown in Figure Beta. The value that is closest to $\\mathrm{QA}$ is ( )\n\n\n\nFigure Alpha\n\n\nA. $8 \\mathrm{~cm}$\nB. $7 \\mathrm{~cm}$\nC. $6 \\mathrm{~cm}$\nD. $5 \\mathrm{~cm}$", "input_image": [ "batch28-2024_06_17_25b99dc92d256d9a3e98g_0092_1.jpg", "batch28-2024_06_17_25b99dc92d256d9a3e98g_0092_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure A, connect $\\mathrm{OP}$, and let $\\odot \\mathrm{O}$ be tangent to the $\\mathrm{x}$-axis at point $\\mathrm{D}$. In Figure B, connect $\\mathrm{OQ}$ and $\\mathrm{OA}$, and let $\\odot \\mathrm{O}$ be tangent to the $\\mathrm{x}$-axis at point $\\mathrm{E}$,\n\n\n$\\therefore$ By the definition of tangents and the properties of circles, quadrilateral OPCD is a square,\n\n$\\therefore \\mathrm{OQ}=\\mathrm{OP}=\\mathrm{PC}=12 \\mathrm{~cm}$,\n\nAccording to the problem statement: $\\angle \\mathrm{QAO}=\\left(180^{\\circ}-\\angle \\mathrm{BAC}\\right) \\div 2=60^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{QOA}=90^{\\circ}-\\angle \\mathrm{QAO}=30^{\\circ}$,\n\n$\\therefore \\tan \\angle \\mathrm{QOA}=\\mathrm{AQ} \\div \\mathrm{OQ}$,\n\nThat is, $\\tan 30^{\\circ}=\\frac{\\sqrt{3}}{3}=\\frac{A Q}{12}$,\n\nSolving gives $\\mathrm{AQ}=4 \\sqrt{3}$.\n\n$\\because 1.5<\\sqrt{3}<2$,\n\n$\\therefore 6<4 \\sqrt{3}<8$.\n\nTherefore, the correct choice is B.\n\n【Insight】This problem tests the properties of tangents, solving right triangles, and estimating irrational numbers. Estimating the approximate values of irrational numbers has wide applications in real life, and we should be proficient in it." }, { "problem_id": 908, "question": "Four equal-length thin wooden sticks are joined end-to-end to form a quadrilateral $\\mathrm{ABCD}$. By turning this quadrilateral, its shape is changed. When $\\angle B=90^{\\circ}$ (as shown in Figure A), the length of the diagonal $\\mathrm{BD}$ is measured to be $\\sqrt{2}$. When $\\angle B=60^{\\circ}$ (as shown in Figure B), the length of the diagonal $\\mathrm{BD}$ is ( )\n\n\n\n A\n\n\n\n B\nA. $\\sqrt{2}$\nB. $\\sqrt{3}$\nC. 2\nD. $\\sqrt{5}$", "input_image": [ "batch22-2024_06_14_cfc50897275df68cfa35g_0011_1.jpg", "batch22-2024_06_14_cfc50897275df68cfa35g_0011_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure A, connect BD.\n\nSince AB = BC = CD = DA and ∠B = 90°,\n\nTherefore, quadrilateral ABCD is a square.\n\nThen, AB² + AD² = BD², and BD = √2.\n\nThus, AB = AD = 1.\n\n\n\nA\n\n\n\nB\n\nAs shown in Figure B, since AB = BC = CD = DA,\n\nTherefore, quadrilateral ABCD is a rhombus.\n\nConnect BD and AC, with ∠B = 60°.\n\nThus, triangle ABC is an equilateral triangle.\n\nTherefore, AC = BC = AB = 1.\n\nThus, AO = OC = ½ AC = ½.\n\nTherefore, BO = DO.\n\nIn right triangle BCO, by the Pythagorean theorem, BO = √(BC² - CO²) = √(1 - ¼) = √3 / 2.\n\nTherefore, BD = 2 OB = √3.\n\nHence, the answer is: B.\n\n[Highlight] This question tests the properties of a square, the Pythagorean theorem, the determination and properties of an equilateral triangle, and the properties of a rhombus. The key is to use the Pythagorean theorem to determine the side length of the square." }, { "problem_id": 909, "question": "As shown in the figure, four students, A, B, C, D, cut out different pieces of paper (shaded areas) from four identical isosceles right-angled triangles so that the shaded area is as large as possible. Their specific cutting methods are as follows:\n\n A student: Cut out a square as shown in Figure 1, with an area denoted as $S_{1}$;\n\n B student: Cut out a square as shown in Figure 2, with an area denoted as $S_{2}$;\n\n C student: Cut out a semicircle as shown in Figure 3, with the diameter of the semicircle lying on the hypotenuse of the isosceles right triangle, with an area denoted as $S_{3}$; \n\n D student: Cut out an inscribed circle as shown in Figure 4, with an area denoted as $S_{4}$;\n\nWhich of the following statements is correct? $(\\quad)$\n\n(1) $S_{1}=S_{2}$; (2) $S_{3}=S_{4}$; (3) Among $S_{1}, S_{2}, S_{3}, S_{4}$, $S_{2}$ is the smallest\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\nA. (1)(2)\nB. (2)(3)\nC. (1)(3)\nD. (1)(2)(3)", "input_image": [ "batch19-2024_05_24_b72b576918335d3d02e7g_0025_1.jpg", "batch19-2024_05_24_b72b576918335d3d02e7g_0025_2.jpg", "batch19-2024_05_24_b72b576918335d3d02e7g_0025_3.jpg", "batch19-2024_05_24_b72b576918335d3d02e7g_0025_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In Figure 1, let the leg length of the four congruent isosceles right triangles be 1, then the hypotenuse length is $\\frac{\\sqrt{3}}{2}$. The diagonal length of the shaded square in Figure 1 is $\\frac{\\sqrt{2}}{2}$, and $S_{1}=\\frac{1}{4}$.\n\nIn Figure 2, let the side length of the square be $x$, then $3x=\\sqrt{2}$, so $x=\\frac{\\sqrt{2}}{3}$, and $S_{2}=\\frac{2}{9}$.\n\nIn Figure 3, let the radius of the semicircle be $r$, then $1+r=\\sqrt{2}$, so $r=\\sqrt{2}-1$, and $S_{3}=\\left(\\frac{3}{2}-\\sqrt{2}\\right) \\pi$.\n\nIn Figure 4, let the inradius of the triangle be $R$, then $2-2R=\\sqrt{2}$, solving gives: $R=1-\\frac{\\sqrt{2}}{2}$, and $S_{4}=\\left(\\frac{3}{2}-\\sqrt{2}\\right) \\pi$. Based on the calculated values, $S_{3}=S_{4}$, and among $S_{1}, S_{2}, S_{3}, S_{4}$, $S_{2}$ is the smallest. Therefore, the correct options are (2) and (3).\n\nHence, the answer is: B.\n\n【Key Point】This question mainly examines the properties of isosceles right triangles, the properties of incircles, and the tangent length theorem, covering a wide range of topics." }, { "problem_id": 910, "question": "In the same Cartesian coordinate system, the approximate graphs of the functions $y=ax^{2}+b$ and $y=ax+b \\quad (a b \\neq 0)$ may be\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_cb8a411f8f15a44a8322g_0002_1.jpg", "batch7-2024_06_14_cb8a411f8f15a44a8322g_0002_2.jpg", "batch7-2024_06_14_cb8a411f8f15a44a8322g_0002_3.jpg", "batch7-2024_06_14_cb8a411f8f15a44a8322g_0002_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: In option $\\mathrm{A}$, the parabola $\\mathrm{y}=\\mathrm{ax}^{2}+\\mathrm{b}$ opens upwards.\n\n$\\therefore a>0$\n\nHowever, since in the linear function $\\mathrm{y}=\\mathrm{ax}+\\mathrm{b}$, $\\mathrm{y}$ decreases as $\\mathrm{x}$ increases,\n\n$\\therefore a<0$, which contradicts the property of the quadratic function.\n\n$\\therefore$ Option $\\mathrm{A}$ is incorrect.\n\nIn option $\\mathrm{B}$, the parabola $\\mathrm{y}=\\mathrm{ax}^{2}+\\mathrm{b}$ opens downwards.\n\n$\\therefore a<0$\n\nBut, in the linear function $\\mathrm{y}=\\mathrm{ax}+\\mathrm{b}$, $\\mathrm{y}$ increases as $\\mathrm{x}$ increases,\n\n$\\therefore a>0$, which again contradicts the property of the quadratic function.\n\n$\\therefore$ Option $\\mathrm{B}$ is incorrect.\n\nIn option $\\mathrm{C}$, the parabola $\\mathrm{y}=\\mathrm{ax}^{2}+\\mathrm{b}$ opens upwards, and when $x=0$, $y=b<0$.\n\n$\\therefore a>0$ and $b<0$\n\nHowever, in the linear function $\\mathrm{y}=\\mathrm{ax}+\\mathrm{b}$, when $x=0$, $y=b>0$.\n\n$\\therefore$ This contradicts the property of the quadratic function.\n\n$\\therefore$ Option $\\mathrm{C}$ is incorrect.\n\nIn option $\\mathrm{D}$, the parabola $\\mathrm{y}=\\mathrm{ax}^{2}+\\mathrm{b}$ opens downwards, and when $x=0$, $y=b<0$.\n\n$\\therefore a<0$ and $b<0$\n\nSince in the linear function $\\mathrm{y}=\\mathrm{ax}+\\mathrm{b}$, $\\mathrm{y}$ decreases as $\\mathrm{x}$ increases, and when $x=0$, $y=b<0$,\n\n$\\therefore a<0$ and $b<0$\n\n$\\therefore$ This is consistent with the property of the quadratic function.\n\n$\\therefore$ Option $\\mathrm{D}$ is correct.\n\nTherefore, the correct choice is D.\n\n【Key Insight】This question tests knowledge of quadratic and linear functions; the key to solving it lies in a thorough understanding of the properties of their graphs." }, { "problem_id": 911, "question": "We define: in the Cartesian coordinate system $\\mathrm{xOy}$, the zigzag distance between any two non-coincident points $\\mathrm{M}\\left(\\mathrm{x}_{1}, \\mathrm{y}_{1}\\right), \\mathrm{N}\\left(\\mathrm{x}_{2}, \\mathrm{y}_{2}\\right)$ is $d(M, N)=\\left|x_{1}-x_{2}\\right|+\\left|y_{1}-y_{2}\\right|$, for example, in Figure (1), the zigzag distance between point $\\mathrm{M}(-2,3)$ and point $\\mathrm{N}(1,-1)$ is $d(M, N)=|-2-1|+|3-(-1)|=3+4=7$. In Figure (2), given point $\\mathrm{P}(3,-4)$, if the coordinates of point $\\mathrm{Q}$ are $(2, \\mathrm{t})$ and $d(P, Q)=10$, then the value of $\\mathrm{t}$ is $(\\quad)$\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. -7 or 1\nB. -5 or 13\nC. 5 or -13\nD. -1 or 7", "input_image": [ "batch8-2024_06_14_422078e53dd6b1dae0e2g_0075_1.jpg", "batch8-2024_06_14_422078e53dd6b1dae0e2g_0075_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Given the point \\( \\mathrm{P}(3, -4) \\), if the coordinates of point \\( \\mathrm{Q} \\) are \\( (2, \\mathrm{t}) \\) and the distance \\( d(P, Q) = 10 \\), then:\n\n\\[\nd(P, Q) = |3 - 2| + |-4 - t| = 10\n\\]\n\nSo,\n\n\\[\n|3 - 2| + |-4 - t| = 10\n\\]\n\n\\[\n|-4 - t| = 9\n\\]\n\nSolving this, we get:\n\n\\[\n\\mathrm{t} = 5 \\quad \\text{or} \\quad \\mathrm{t} = -13\n\\]\n\nTherefore, the correct answer is: C\n\n**Key Insight:** The problem tests the understanding of point coordinates and distance. Grasping the definitions of Manhattan distance and absolute value is crucial." }, { "problem_id": 912, "question": "As shown in Figure 1, in triangles $\\triangle A B C$ and $\\triangle D E F$, $A B = A C = m, D E = D F = n, \\angle B A C = \\angle E D F$. Point $D$ coincides with point $A$, and points $E, F$ are on sides $A B, A C$ respectively. Triangle $\\triangle D E F$ in Figure 1 is translated along ray $A C$ until point $D$ coincides with point $C$, resulting in Figure 2. Which of the following conclusions is NOT correct ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. The distance of the translation of $\\triangle D E F$ is $m$\nB. In Figure 2, $C B$ bisects $\\angle A C E$\nC. The distance of the translation of $\\triangle D E F$ is $n$\nD. In Figure 2, $E F \\parallel B C$", "input_image": [ "batch14-2024_06_15_6a7219bf0375e5fd3b6ag_0073_1.jpg", "batch14-2024_06_15_6a7219bf0375e5fd3b6ag_0073_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since \\( \\mathrm{AD} = \\mathrm{AC} = m \\),\n\nthe distance of the translation of \\( \\triangle \\mathrm{DEF} \\) is \\( m \\), thus option \\( \\mathrm{A} \\) is correct, and option \\( \\mathrm{C} \\) is incorrect.\n\nSince \\( \\mathrm{AB} = \\mathrm{AC} \\),\n\n\\( \\angle \\mathrm{ACB} = \\angle \\mathrm{ABC} \\).\n\nSince \\( \\mathrm{DE} \\parallel \\mathrm{AB} \\),\n\n\\( \\angle \\mathrm{EDB} = \\angle \\mathrm{ABC} \\),\n\nhence \\( \\angle \\mathrm{ACB} = \\angle \\mathrm{ECB} \\),\n\nwhich means \\( \\mathrm{CB} \\) bisects \\( \\angle \\mathrm{ACE} \\), so option \\( \\mathrm{B} \\) is correct.\n\nFrom the properties of translation, we get \\( \\mathrm{EF} \\parallel \\mathrm{BC} \\), so option \\( \\mathrm{D} \\) is correct.\n\nTherefore, the correct choice is \\( \\mathrm{C} \\).\n\n【Key Point】This question tests the properties of translation, the properties of isosceles triangles, and the properties of parallel lines. Mastering the properties of translation is crucial for solving the problem." }, { "problem_id": 913, "question": "As shown in the figure, given rhombus $A B C D$ with side length $4$ and $\\angle A = 60^\\circ$. Point $E$ moves along the path $A-B-C$ starting from point $A$ at a speed of $2$ units per second. A perpendicular line is drawn from point $E$ to $A D$, and the length of the line segment intercepted by the rhombus is $y$. Let the time point $E$ moves be $x$ (unit: $s$) $(0 < x < 4)$. The graph of the function $y$ versus $x$ is approximately ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch29-2024_06_14_615670bb7fcc0025dfefg_0061_1.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0061_2.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0061_3.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0061_4.jpg", "batch29-2024_06_14_615670bb7fcc0025dfefg_0061_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When point \\( \\mathrm{E} \\) is on side \\( \\mathrm{AB} \\), for \\( 0 < x < 2 \\), \\( \\mathrm{AE} = 2x \\). In the right triangle \\( \\triangle \\mathrm{AEF} \\), we have:\n\\[\ny = \\mathrm{EF} = \\mathrm{AE} \\times \\sin 60^\\circ = \\sqrt{3}x;\n\\]\n\nWhen \\( \\mathrm{E} \\) is between point \\( \\mathrm{B} \\) and the midpoint of \\( \\mathrm{BC} \\), for \\( 2 \\leq x < 3 \\), \\( \\mathrm{EF} \\) becomes the height of rhombus \\( \\mathrm{ABCD} \\), and:\n\\[\ny = \\mathrm{EF} = \\mathrm{AB} \\times \\sin 60^\\circ = 2\\sqrt{3};\n\\]\n\nWhen \\( \\mathrm{E} \\) is between the midpoint of \\( \\mathrm{BC} \\) and point \\( \\mathrm{C} \\), for \\( 3 \\leq x < 4 \\), \\( \\mathrm{CE} = 8 - 2x \\). At this point, let \\( \\mathrm{EF} \\) intersect \\( \\mathrm{CD} \\) at \\( \\mathrm{M} \\), as shown in the figure:\n\n\n\nHere, \\( y = \\mathrm{EM} = \\mathrm{CE} \\times \\tan 60^\\circ = \\sqrt{3}(8 - 2x) = -2\\sqrt{3}x + 8\\sqrt{3} \\);\n\nBy combining the graph, we can see that the function image of \\( y \\) with respect to \\( x \\) roughly corresponds to option \\( \\mathrm{A} \\).\n\nTherefore, the answer is: A.\n\n【Insight】This question tests the understanding of trigonometric functions and function graphs. Mastering the method of determining the length of line segments in dynamic problems is key to solving it." }, { "problem_id": 914, "question": "Figure 1 shows a slide in a park, and Figure 2 is a diagram of it. The height $B C$ of the slide is $2 \\text{ m}$, and the slope angle $\\angle A$ is $60^\\circ$. Due to the steep slope, which poses a safety concern, the park management decides to renovate the slide. They plan to keep the height unchanged while extending the horizontal distance $A B$ to ensure that the slope angle $\\angle A$ falls within the range $30^\\circ \\leq \\angle A \\leq 45^\\circ$. The possible extension distance for $A B$ is ( )\n\n(Reference data: $\\sqrt{2} \\approx 1.414, \\sqrt{3} \\approx 1.732$ )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $0.8 \\text{ m}$\nB. $1.6 \\text{ m}$\nC. $2.4 \\text{ m}$\nD. $3.2 \\text{ m}$", "input_image": [ "batch37-2024_06_14_89dd8151fe5d16331c78g_0043_1.jpg", "batch37-2024_06_14_89dd8151fe5d16331c78g_0043_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, in the right triangle $\\triangle ABC$, $\\angle CAB = 60^\\circ$, and $BC = 2$,\n\n$\\therefore AB = \\frac{BC}{\\tan 60^\\circ} = \\frac{2}{\\sqrt{3}} = \\frac{2\\sqrt{3}}{3}$.\n\nWhen the slope angle is $45^\\circ$, we have $BD = BC = 2$,\n\n$\\therefore DA = 2 - AB = 2 - \\frac{2\\sqrt{3}}{3} \\approx 0.85 \\text{ (m)}$.\n\nWhen the slope angle is $30^\\circ$, we have $BE = \\frac{BC}{\\tan 30^\\circ} = \\frac{2}{\\frac{\\sqrt{3}}{3}} = 2\\sqrt{3} \\text{ (m)}$,\n\n$\\therefore EA = BE - AB = 2\\sqrt{3} - \\frac{2\\sqrt{3}}{3} \\approx 2.31 \\text{ (m)}$.\n\nWhen the slope angle satisfies $30^\\circ \\leq \\angle A \\leq 45^\\circ$,\n\n$\\therefore$ the range of the extended distance $x$ of $AB$ is $0.85 \\leq x \\leq 2.31$,\n\nTherefore, the correct answer is: B\n\n\n\n【Key Insight】This problem tests the application of solving right triangles. Understanding the meaning of slope angle and gradient is crucial for solving the problem." }, { "problem_id": 915, "question": "As shown in Figure (1), there are two congruent equilateral triangles $A B C$ and $D E F$, where $D$ and $A$ are the centroids of $\\triangle A B C$ and $\\triangle D E F$ respectively. Keeping point $D$ fixed, rotate $\\triangle D E F$ counterclockwise until point $A$ falls on segment $D E$, as shown in Figure (2). The ratio of the area of the overlapping region in Figure (1) to that in Figure (2) is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. 2: 1\nB. 3: 2\nC. 4: 3\nD. 5: 4", "input_image": [ "batch32-2024_06_14_4a07680e90202dde6dedg_0008_1.jpg", "batch32-2024_06_14_4a07680e90202dde6dedg_0008_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $AD$ and $MN$ intersecting at point $O$,\n\n\n\nLet the side length of the equilateral triangle be $x$,\n\n\n\nThen the height is $\\sqrt{x^{2}-\\left(\\frac{1}{2} x\\right)^{2}}=\\frac{\\sqrt{3}}{2} x$,\n\nIn Figure (1), the shaded area is a rhombus with an internal angle of $60^{\\circ}$,\n\n$\\therefore AD=\\frac{2}{3} \\times \\frac{\\sqrt{3}}{2} x=\\frac{\\sqrt{3}}{3} x, \\angle ADM=30^{\\circ}$,\n\n$\\therefore AO=\\frac{1}{2} AD=\\frac{\\sqrt{3}}{6} x$,\n\n$\\therefore MO=AO \\times \\tan 30^{\\circ}=\\frac{1}{6} x$,\n\n$\\therefore MN=\\frac{1}{3} x$,\n\nThen the area of the shaded part is: $\\frac{1}{2} \\times \\frac{\\sqrt{3}}{3} x \\times \\frac{1}{3} x=\\frac{\\sqrt{3}}{18} x^{2}$,\n\nIn Figure 2, $AD=\\frac{2}{3} \\times \\frac{\\sqrt{3}}{2} x=\\frac{\\sqrt{3}}{3} x, \\angle DAB=30^{\\circ}$,\n\n$\\therefore \\angle DHA=90^{\\circ}$,\n\n$\\therefore DH=\\frac{1}{2} \\times AD=\\frac{\\sqrt{3}}{6} x$,\n\n$\\therefore AH=\\frac{1}{2} x$,\n\n$\\therefore$ The area of the shaded part is: $\\frac{1}{2} \\times AH \\times DH=\\frac{\\sqrt{3}}{24} x^{2}$,\n\nThe ratio of the areas of the two overlapping regions is: $\\frac{\\sqrt{3}}{18} x^{2}: \\frac{\\sqrt{3}}{24} x^{2}=4: 3$,\n\nTherefore, the answer is: C.\n\n【Key Insight】The problem mainly examines the properties of equilateral triangles and the application of triangle solutions, as well as the properties of rhombuses. Understanding the problem, drawing the corresponding auxiliary lines, and mastering the properties of the centroid of a triangle are key to solving the problem." }, { "problem_id": 916, "question": "As shown in the figure, there are two identical rectangular plots of land, A and B, in a school campus, each with a length of $30 \\mathrm{~m}$ and a width of $25 \\mathrm{~m}$. Paths in the form of parallelograms (shaded areas in the figure) are to be constructed within the rectangular plots, and the remaining areas will be landscaped. It is given that $A B = C D = 1 \\mathrm{~m}$ and $E F = G H = 1 \\mathrm{~m}$. Let $S_{1}$ and $S_{2}$ be the areas of the landscaped regions in plots A and B, respectively. The relationship between $S_{1}$ and $S_{2}$ is ( )\n\n\n\nA\n\n\n\nB\nA. $S_{1}S_{2}$\nD. Cannot be determined", "input_image": [ "batch23-2024_06_14_0ac5a7e94a11ff9bdd63g_0055_1.jpg", "batch23-2024_06_14_0ac5a7e94a11ff9bdd63g_0055_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nFrom the problem statement, we know:\n\n- The area of the parallelogram paths in the left-right direction in both diagrams is: \n \\(30 \\times 1 = 30 \\, \\text{m}^2\\).\n\n- The area of the parallelogram paths in the up-down direction in both diagrams is: \n \\(25 \\times 1 = 25 \\, \\text{m}^2\\).\n\n- The overlapping area in Diagram A is: \n \\(1 \\times 1 = 1 \\, \\text{m}^2\\).\n\nThus, the area \\(S_1\\) is calculated as: \n\\[\nS_1 = 30 \\times 25 - 30 - 25 + 1 = 696 \\, \\text{m}^2.\n\\]\n\nNext, as shown in the diagram, we draw lines \\(\\mathrm{PR} \\parallel \\mathrm{CD}\\) and \\(\\mathrm{NS} \\parallel \\mathrm{CD}\\), intersecting \\(\\mathrm{QD}\\) at points \\(\\mathrm{R}\\) and \\(\\mathrm{S}\\), respectively. From point \\(\\mathrm{N}\\), we draw a perpendicular \\(\\mathrm{NO}\\) to \\(\\mathrm{PR}\\), meeting at point \\(\\mathrm{O}\\).\n\n\n\nIt follows that:\n- \\(\\angle PRQ = \\angle NSM\\), and the quadrilateral \\(\\mathrm{RSNP}\\) is a parallelogram.\n- \\(\\mathrm{PR} = \\mathrm{NS} = \\mathrm{CD} = 1 \\, \\text{m}\\), and \\(\\mathrm{NO} < \\mathrm{GH}\\), where \\(\\mathrm{GH} = 1 \\, \\text{m}\\).\n\nIn the parallelogram \\(\\mathrm{PQMN}\\), since \\(\\mathrm{PQ} \\parallel \\mathrm{MN}\\), we have:\n\\[\n\\angle PQR = \\angle NMS.\n\\]\n\nThis allows us to prove that:\n\\[\n\\triangle PQR \\cong \\triangle NMS \\, (\\text{by AAS}).\n\\]\n\nTherefore:\n\\[\nS_{\\square PQUNV} = S_{\\square PRSV} = \\mathrm{PR} \\cdot \\mathrm{NO} < \\mathrm{PR} \\cdot \\mathrm{GH}.\n\\]\n\nSince:\n\\[\n\\mathrm{PR} \\cdot \\mathrm{GH} = 1 \\times 1 = 1 \\, \\text{m}^2,\n\\]\nit follows that:\n\\[\nS_{\\square PQUN} < 1 \\, \\text{m}^2.\n\\]\n\nThus, the area \\(S_2\\) is:\n\\[\nS_2 = 30 \\times 25 - 30 - 25 + S_{\\square PQUN} < 696 \\, \\text{m}^2.\n\\]\n\nHence:\n\\[\nS_1 > S_2.\n\\]\n\n**Final Answer:** C.\n\n**Key Insight:** This problem tests the understanding of area calculations. Pay attention to adding parallel auxiliary lines and carefully calculating the shaded area, especially for \\(S_2\\)." }, { "problem_id": 917, "question": "In right triangle $\\triangle A B C$ as shown in Figure 1, $\\angle C = 90^\\circ$. Point $D$ is the midpoint of $B C$. A moving point $P$ starts from point $C$ and moves along the path $C A - A B$ until it reaches point $B$. Let the distance traveled by point $P$ be $x$, and the area of triangle $P C D$ be $y$. The graph of the function $y$ as a function of $x$ is shown in Figure 2. Determine the length of $A B$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $4 \\sqrt{5}$\nB. 10\nC. 5\nD. $4 \\sqrt{3}$", "input_image": [ "batch14-2024_06_15_afb2d8a263b0edaeccbfg_0011_1.jpg", "batch14-2024_06_15_afb2d8a263b0edaeccbfg_0011_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the graph, we can see that when \\( x = 3 \\), \\( CP = 3 \\).\n\nGiven the equation:\n\\[\ny = \\frac{1}{2} \\times PC \\times CD = \\frac{1}{2} \\times 3 \\times CD = 3,\n\\]\nwe solve for \\( CD \\):\n\\[\nCD = 2.\n\\]\n\nSince point \\( D \\) is the midpoint of \\( BC \\), it follows that:\n\\[\nBC = 4.\n\\]\n\nWhen \\( x = 8 \\), there is a turning point in the area, at which point \\( P \\) coincides with point \\( A \\), and:\n\\[\nAC = 8.\n\\]\n\nIn the right triangle \\( ABC \\), with \\( \\angle C = 90^\\circ \\), \\( BC = 4 \\), and \\( AC = 8 \\), by the Pythagorean theorem, we find:\n\\[\nAB = 4\\sqrt{5}.\n\\]\n\nTherefore, the correct answer is A.\n\n**Insight:** This problem examines the function graph of a moving point: the graph is a classic example of integrating numbers with shapes. Graphs are widely used to convey information; by interpreting graphs, we can not only solve practical problems in life but also enhance our ability to analyze and solve problems. The key to solving this problem lies in using the method of classification and discussion to determine the lengths of \\( AC \\) and \\( BC \\)." }, { "problem_id": 918, "question": "Teacher Cheng made a teaching aid as shown in Figure 1 to explore the question of 'whether the side-side-angle conditions can determine the shape of a triangle'. When operating the teaching aid, point $Q$ moves on the track groove $AM$, and point $P$ can move on the semicircular track groove with $\\mathrm{A}$ as the center and 8 as the radius, as well as on the track groove $QN$. Figure 2 is a schematic diagram of the corresponding figure at a certain position when operating the teaching aid.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nThe following conclusions can be drawn:\n\n(1) When $\\angle P A Q=30^{\\circ}, P Q=6$, the shape of $\\triangle P A Q$ can be determined uniquely;\n\n(2) When $\\angle P A Q=30^{\\circ}, P Q=9$, the shape of $\\triangle P A Q$;\n\n(3)When $\\angle P A Q=90^{\\circ}, P Q=10$, we can get $\\triangle P A Q$ with a unique shape;\n\n(4) When $\\angle P A Q=150^{\\circ}, P Q=12$, we can get $\\triangle P A Q$ with a unique shape;\n\nThe serial numbers of all correct conclusions are $(\\quad)$\n\n\n\nAlternative figure\nA. (1)(2)(4)\nB. (2) (3)\nC. (2)(3)(4)\nD. (3)(4)\n\n", "input_image": [ "batch19-2024_05_24_ce4c68b4e78efc164f53g_0010_1.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0010_2.jpg", "batch19-2024_05_24_ce4c68b4e78efc164f53g_0010_3.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: \n\n(1) When $\\angle \\mathrm{PAQ}=30^{\\circ}$ and $\\mathrm{PQ}=6$, drawing an arc with $\\mathrm{P}$ as the center and a radius of 6 intersects the ray $AM$ at two points. Therefore, the shape of $\\triangle \\mathrm{PAQ}$ cannot be uniquely determined, making statement (1) incorrect.\n\n(2) When $\\angle \\mathrm{PAQ}=30^{\\circ}$ and $\\mathrm{PQ}=9$, drawing an arc with $\\mathrm{P}$ as the center and a radius of 9 intersects the ray $AM$ at one point. The position of point $\\mathrm{Q}$ is uniquely determined, resulting in a uniquely determined shape for $\\triangle \\mathrm{PAQ}$, making statement (2) correct.\n\n(3) When $\\angle \\mathrm{PAQ}=90^{\\circ}$ and $\\mathrm{PQ}=10$, drawing an arc with $\\mathrm{P}$ as the center and a radius of 10 intersects the ray $AM$ at one point. The position of point $Q$ is uniquely determined, resulting in a uniquely determined shape for $\\triangle \\mathrm{PAQ}$, making statement (3) correct.\n\n(4) When $\\angle \\mathrm{PAQ}=150^{\\circ}$ and $\\mathrm{PQ}=12$, drawing an arc with $\\mathrm{P}$ as the center and a radius of 12 intersects the ray $AM$ at one point. The position of point $Q$ is uniquely determined, resulting in a uniquely determined shape for $\\triangle \\mathrm{PAQ}$, making statement (4) correct.\n\nTherefore, the correct choice is: C.\n\n【Key Insight】This problem primarily examines the fundamental properties of circles, focusing on determining the number of intersection points when drawing an arc with $\\mathrm{P}$ as the center and $\\mathrm{PQ}$ as the radius, intersecting the ray $\\mathrm{AM}$." }, { "problem_id": 919, "question": "In right triangle $\\triangle A B C$ as shown in Figure 1, $\\angle A C B = 90^\\circ$. Point $P$ moves at a speed of $1 \\text{~cm/s}$ from point $A$ along the broken line $A C - C B$ until it reaches point $B$. A perpendicular line $P D$ is drawn from point $P$ to side $A B$, with the foot of the perpendicular being point $D$. The graph of the function $y(\\text{~cm})$ representing the length of $P D$ and the variable $x$ (seconds) representing the time of point $P$'s movement is shown in Figure 2. When point $P$ has moved for 5 seconds, the length of $P D$ is ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $0.8 \\text{~cm}$\nB. $1.2 \\text{~cm}$\nC. $1.6 \\text{~cm}$\nD. $2.4 \\text{~cm}$", "input_image": [ "batch16-2024_06_15_f29175f3519ce46c48bfg_0003_1.jpg", "batch16-2024_06_15_f29175f3519ce46c48bfg_0003_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "From Figure 2, it is known that the time taken for point \\( P \\) to move along \\( AC \\) and \\( CB \\) is 3 seconds and 4 seconds, respectively.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nSince the speed of point \\( P \\) is \\( 1 \\mathrm{~cm} \\) per second,\n\nTherefore, \\( AC = 3 \\mathrm{~cm} \\) and \\( BC = 4 \\mathrm{~cm} \\).\n\nSince in the right triangle \\( \\triangle ABC \\), \\( \\angle ACB = 90^\\circ \\),\n\nTherefore, according to the Pythagorean theorem: \\( AB = \\sqrt{AC^2 + BC^2} = \\sqrt{3^2 + 4^2} = 5 \\).\n\nAs shown in the figure, drawing \\( CH \\perp AB \\) at point \\( H \\),\n\nThen \\( \\angle ACB = \\angle AHC = 90^\\circ \\),\n\nSince \\( \\angle A = \\angle A \\),\n\nTherefore, \\( \\triangle ABC \\sim \\triangle ACH \\).\n\nThus, \\( \\frac{CH}{BC} = \\frac{AC}{AB} \\), which means \\( CH = \\frac{AC \\cdot BC}{AB} = \\frac{3 \\times 4}{5} = \\frac{12}{5} \\).\n\nTherefore, as shown in the figure, the coordinates of point \\( E \\) are \\( \\left(3, \\frac{12}{5}\\right) \\), and the coordinates of point \\( F \\) are \\( (7, 0) \\).\n\nLet the equation of line \\( EF \\) be \\( y = kx + b \\), then\n\n\\[\n\\left\\{\n\\begin{array}{l}\n\\frac{12}{5} = 3k + b \\\\\n0 = 7k + b\n\\end{array}\n\\right.\n\\]\n\nSolving the equations gives: \\( \\left\\{\n\\begin{array}{l}\nk = -\\frac{3}{5} \\\\\nb = \\frac{21}{5}\n\\end{array}\n\\right. \\).\n\nTherefore, the equation of line \\( EF \\) is \\( y = -\\frac{3}{5}x + \\frac{21}{5} \\).\n\nTherefore, when \\( x = 5 \\), \\( PD = y = -\\frac{3}{5} \\times 5 + \\frac{21}{5} = \\frac{6}{5} = 1.2 \\mathrm{~cm} \\).\n\nHence, the correct choice is: B.\n\n【Key Insight】This problem tests knowledge on the method of undetermined coefficients for finding the equation of a linear function, the Pythagorean theorem, and the properties and determination of similar triangles. The key to solving the problem lies in combining numerical and graphical methods and performing accurate calculations." }, { "problem_id": 920, "question": "In the figure, all quadrilaterals are squares, and all triangles are isosceles right triangles. The area of the largest square is 16. Following the pattern from Figure (1) to Figure (3), which of the following statements is correct?\n\nConclusion I: The length of one leg of the isosceles right triangle in Figure (1) is $2 \\sqrt{2}$;\n\nConclusion II: The sum of the areas of all isosceles right triangles in Figure (5) is 20.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. Both Conclusion I and II are correct.\nB. Neither Conclusion I nor II is correct.\nC. Only Conclusion I is correct.\nD. Only Conclusion II is correct.", "input_image": [ "batch5-2024_06_14_77ce3e8cf88e61cf1a7dg_0072_1.jpg", "batch5-2024_06_14_77ce3e8cf88e61cf1a7dg_0072_2.jpg", "batch5-2024_06_14_77ce3e8cf88e61cf1a7dg_0072_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since the area of the largest square is 16, that is, the area of square \\( ABCD \\) is 16,\n\n\\[\n\\therefore AD = \\sqrt{16} = 4,\n\\]\n\nSince \\( \\triangle AED \\) is an isosceles right triangle,\n\n\\[\n\\therefore AE^{2} + DE^{2} = AD^{2},\n\\]\n\nThat is,\n\n\\[\n2AE^{2} = 16,\n\\]\n\nSolving gives:\n\n\\[\nAE = 2\\sqrt{2}, \\quad \\text{(discarding the negative value)}\n\\]\n\n\\[\n\\therefore \\text{In Figure (1), the length of one leg of the isosceles right triangle is } 2\\sqrt{2}, \\text{ thus Conclusion I is correct;}\n\\]\n\nThe area of the isosceles right triangle in Figure (1) is:\n\n\\[\n\\frac{1}{2} \\times 2\\sqrt{2} \\times 2\\sqrt{2} = 4,\n\\]\n\nIn Figure (2), the sum of the areas of the added small squares is:\n\n\\[\nAE^{2} + DE^{2} = AD^{2} = 16,\n\\]\n\nSince the area of one added small isosceles right triangle is equal to one-fourth the area of one added small square,\n\n\\[\n\\therefore \\text{The sum of the areas of the two added small isosceles right triangles is: } \\frac{1}{4} \\times 16 = 4;\n\\]\n\nIn Figure (3), the sum of the areas of the added small squares and the isosceles right triangles is equal to 16,\n\nSince the area of one added small isosceles right triangle is equal to one-fourth the area of one added small square,\n\n\\[\n\\therefore \\text{The sum of the areas of the added small isosceles right triangles is: } \\frac{1}{4} \\times 16 = 4;\n\\]\n\n\\[\n\\therefore \\text{In Figure (1), the sum of the areas of the isosceles right triangles is 4,}\n\\]\n\nIn Figure (2), the sum of the areas of the isosceles right triangles is:\n\n\\[\n4 \\times 2 = 8,\n\\]\n\nIn Figure (3), the sum of the areas of the isosceles right triangles is:\n\n\\[\n4 \\times 3 = 12,\n\\]\n\nIn Figure (5), the sum of the areas of the isosceles right triangles is:\n\n\\[\n4 \\times 5 = 20, \\quad \\text{thus Conclusion II is correct;}\n\\]\n\nIn summary, both Conclusion I and Conclusion II are correct, therefore option A is correct.\n\nThe correct choice is: A.\n\n\n\n(1)\n\n【Key Point】This problem mainly examines the application of the Pythagorean theorem in geometric figures. The key to solving the problem lies in identifying the pattern from the given figures." }, { "problem_id": 921, "question": "There is a rectangular paper sheet $A B C D$ (as shown in Figure (1), $B C = 6$). The paper is folded so that $B C$ falls on the edge $C D$, with $B^{\\prime}$ being the image of $B$, and the crease is $C E$ (as shown in Figure (2)). The rectangle $A D B^{\\prime} E$ is then folded to the right along the crease $B^{\\prime} E$. If point $D$ lands on the third point of division of segment $B^{\\prime} C$, then the length of $C D$ is ( )\n\n\n\n(1)\n\n\n\n(2)\nA. 8\nB. 10\nC. 8 or 10\nD. 8 or 12", "input_image": [ "batch23-2024_06_14_e90d0c0f7f86a274e612g_0097_1.jpg", "batch23-2024_06_14_e90d0c0f7f86a274e612g_0097_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: (1) When $\\mathrm{D}^{\\prime} \\mathrm{B}^{\\prime}=\\frac{1}{3} B^{\\prime} C$,\n\n\n\n(1)\n\n\n\n$\\because BC=6$, the paper is folded such that $BC$ lies on the $CD$ side, and $B^{\\prime}$ is the corresponding point of $B$,\n\n$\\therefore B^{\\prime} C=6$,\n\n$\\because$ the rectangle $A D B^{\\prime} E$ is folded to the right with $B^{\\prime} E$ as the crease, and point $D$ lands on the trisection point of $B^{\\prime} C$,\n\n$\\therefore \\mathrm{D} \\mathrm{B}^{\\prime}=\\mathrm{D}^{\\prime} \\mathrm{B}^{\\prime}=\\frac{1}{3} B^{\\prime} C=2$,\n\n$\\therefore \\mathrm{CD}=\\mathrm{D} \\mathrm{B}^{\\prime}+B^{\\prime} C=8$\n\n(2) When $\\mathrm{D}^{\\prime} \\mathrm{C}=\\frac{1}{3} B^{\\prime} C$,\n\n\n\n(1)\n\n\n\n$\\because BC=6$, the paper is folded such that $BC$ lies on the $CD$ side, and $B^{\\prime}$ is the corresponding point of $B$,\n\n$\\therefore B^{\\prime} C=6$,\n\n$\\because$ the rectangle $A D B^{\\prime} E$ is folded to the right with $B^{\\prime} E$ as the crease, and point $D$ lands on the trisection point of $B^{\\prime} C$,\n$\\therefore \\mathrm{D}^{\\prime} \\mathrm{C}=\\frac{1}{3} B^{\\prime} C=2$,\n\n$\\therefore \\mathrm{D} \\mathrm{B}^{\\prime}=\\mathrm{D}^{\\prime} \\mathrm{B}^{\\prime}=B^{\\prime} C-\\mathrm{D}^{\\prime} \\mathrm{C}=4$,\n\n$\\therefore \\mathrm{CD}=\\mathrm{D} \\mathrm{B}^{\\prime}+B^{\\prime} C=10$.\n\nIn summary, the length of $\\mathrm{CD}$ is either 8 or 10.\n\nTherefore, the correct answer is: C.\n\n【Key Insight】This problem examines the transformation of figures through folding and the properties of rectangles. The key to solving it lies in understanding the problem and applying a methodical approach to consider different scenarios." }, { "problem_id": 922, "question": "As shown in the figure, in rectangle $A B C D$, $A B = 3$ and $A D = 2$. Point $E$ is the midpoint of $C D$, and ray $A E$ intersects the extended line of $B C$ at point $F$. Point $M$ starts from $A$ and moves along the route $A \\rightarrow B \\rightarrow F$ at a constant speed until it reaches point $F$. A line segment $M N$ is drawn perpendicular to $A F$ at point $N$. Let the length of $A N$ be $x$ and the area of $\\triangle A M N$ be $S$. The graph that can roughly represent the functional relationship between $S$ and $x$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch37-2024_06_14_d4217d5415e569c92d8ag_0028_1.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0028_2.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0028_3.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0028_4.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0028_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: \n\nGiven that in rectangle \\( A B C D \\), \\( A B = 3 \\) and \\( A D = 2 \\), it follows that \\( B C = A D = 2 \\), and all angles \\( \\angle A B C = \\angle B C D = \\angle D = 90^\\circ \\).\n\nSince point \\( E \\) is the midpoint of \\( C D \\), we have \\( C E = D E \\).\n\nIn triangles \\( \\triangle C E F \\) and \\( \\triangle D E A \\), the following conditions are met:\n\\[\n\\begin{cases}\n\\angle C E F = \\angle D E A \\\\\nC E = D E \\\\\n\\angle E C F = \\angle D = 90^\\circ\n\\end{cases}\n\\]\nTherefore, \\( \\triangle C E F \\cong \\triangle D E A \\) by the ASA (Angle-Side-Angle) criterion, which implies \\( C F = A D = 2 \\).\n\nThus, \\( B F = B C + C F = 4 \\), and \\( A F = \\sqrt{A B^2 + B F^2} = 5 \\).\n\nGiven \\( A N = x \\), it follows that \\( F N = A F - A N = 5 - x \\).\n\nWhen point \\( M \\) coincides with point \\( B \\), as shown in the diagram, the area of \\( \\triangle A B F \\) can be expressed in two ways:\n\\[\n\\frac{1}{2} A F \\cdot B N = \\frac{1}{2} A B \\cdot B F\n\\]\nSolving for \\( B N \\):\n\\[\nB N = \\frac{A B \\cdot B F}{A F} = \\frac{3 \\times 4}{5} = \\frac{12}{5}\n\\]\nThus, \\( A N = \\sqrt{A B^2 - B N^2} = \\frac{9}{5} \\).\n\n(1) When point \\( M \\) lies on side \\( A B \\), i.e., \\( 0 \\leq x \\leq \\frac{9}{5} \\), as shown in the diagram, in right triangle \\( \\triangle A B F \\), \\( \\tan \\angle B A F = \\frac{B F}{A B} = \\frac{4}{3} \\).\n\nIn right triangle \\( \\triangle A M N \\), \\( \\tan \\angle M A N = \\frac{M N}{A N} = \\frac{M N}{x} \\).\n\nSetting the two expressions for tangent equal:\n\\[\n\\frac{M N}{x} = \\frac{4}{3}\n\\]\nSolving for \\( M N \\):\n\\[\nM N = \\frac{4}{3} x\n\\]\nThe area \\( S \\) is then:\n\\[\nS = \\frac{1}{2} A N \\cdot M N = \\frac{2}{3} x^2\n\\]\n\n(2) When point \\( M \\) lies on \\( B F \\), i.e., \\( \\frac{9}{5} < x \\leq 5 \\), as shown in the diagram, in right triangle \\( \\triangle A B F \\), \\( \\tan F = \\frac{A B}{B F} = \\frac{3}{4} \\).\n\nIn right triangle \\( \\triangle F M N \\), \\( \\tan F = \\frac{M N}{F N} = \\frac{M N}{5 - x} \\).\n\nSetting the two expressions for tangent equal:\n\\[\n\\frac{M N}{5 - x} = \\frac{3}{4}\n\\]\nSolving for \\( M N \\):\n\\[\nM N = \\frac{3}{4}(5 - x)\n\\]\nThe area \\( S \\) is then:\n\\[\nS = \\frac{1}{2} A N \\cdot M N = -\\frac{3}{8} x^2 + \\frac{15}{8} x\n\\]\n\nIn summary, the area \\( S \\) is given by:\n\\[\nS = \\begin{cases}\n\\frac{2}{3} x^2 & \\text{for } 0 \\leq x \\leq \\frac{9}{5} \\\\\n-\\frac{3}{8} x^2 + \\frac{15}{8} x & \\text{for } \\frac{9}{5} < x \\leq 5\n\\end{cases}\n\\]\n\nAfter examining the four options, only option B matches the derived expressions.\n\nTherefore, the correct choice is: B.\n\n**Key Insight:** This problem tests the properties of rectangles, the application of trigonometric functions in right triangles, and the characteristics of quadratic functions. Correctly dividing the problem into two cases and understanding the graphical features of quadratic functions are crucial for solving it." }, { "problem_id": 923, "question": "As shown in Figure 1, in rectangle $A B C D$, point $E$ is on $C D$, and $\\angle A E B = 90^\\circ$. Point $P$ starts from point $A$ and moves along the path $A \\rightarrow E \\rightarrow B$ at a constant speed until it reaches point $B$. Line segment $P Q$ is drawn perpendicular to $C D$ at point $Q$. Let the distance traveled by point $P$ be $x$, and the length of $P Q$ be $y$. If the graph of the function relationship between $y$ and $x$ is as shown in Figure 2, what is the value of $P Q$ when $x = 6$?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 2\nB. $\\frac{9}{5}$\nC. $\\frac{6}{5}$\nD. 1", "input_image": [ "batch32-2024_06_14_285e15082d492bc8290ag_0048_1.jpg", "batch32-2024_06_14_285e15082d492bc8290ag_0048_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the graph, we can see:\n\n$AE = 3$, $BE = 4$, and $\\angle AEB = 90^\\circ$.\n\nTherefore, $AB = \\sqrt{3^2 + 4^2} = 5$.\n\nWhen $x = 6$, point $P$ is on $BE$. Let the corresponding $PQ$ be $P'Q'$ as shown in the figure.\n\n\n\nFigure 1\n\nAt this moment, $P'E = 4 - (7 - x) = x - 3 = 6 - 3 = 3$.\n\nSince $ABCD$ is a rectangle,\n\n$AB \\parallel CD$.\n\nThus, $\\angle Q'EP' = \\angle ABE$.\n\nGiven that $\\angle AEB = \\angle P'Q'E = 90^\\circ$,\n\n$\\triangle P'Q'E \\sim \\triangle AEB$.\n\nTherefore, $\\frac{P'Q'}{AE} = \\frac{EP'}{AB}$.\n\nSubstituting the known values, $\\frac{P'Q'}{3} = \\frac{3}{5}$.\n\nHence, $P'Q' = \\frac{9}{5}$.\n\nThat is, $PQ = \\frac{9}{5}$.\n\nTherefore, the correct choice is: B.\n\n【Key Insight】This problem examines the function graph of a moving point, involving triangle similarity, the Pythagorean theorem, and properties of rectangles. The key to solving it lies in deeply understanding the function graph of the moving point, interpreting the actual meanings of key points in the graph, and comprehending the complete motion process of the moving point." }, { "problem_id": 924, "question": "(1) As shown in Figure 1, if $A B \\parallel C D$, then $\\angle A + \\angle E + \\angle C = 360^\\circ$; \n(2) As shown in Figure 2, if $A B \\parallel C D$, then $\\angle E = \\angle A + \\angle C$; \n(3) As shown in Figure 3, if $A B \\parallel C D$, then $\\angle A + \\angle E - \\angle 1 = 180^\\circ$; \n(4) As shown in Figure 4, if $A B \\parallel C D$, then $\\angle A = \\angle C + \\angle P$. \nWhich of the above conclusions are correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4\nA. (1)(2)(3)(4)\nB. (1)(2)(3)\nC. (2)(3)(4)\nD. (1)(2)(4)", "input_image": [ "batch5-2024_06_14_d212e25dd7dcee36ccd8g_0074_1.jpg", "batch5-2024_06_14_d212e25dd7dcee36ccd8g_0074_2.jpg", "batch5-2024_06_14_d212e25dd7dcee36ccd8g_0074_3.jpg", "batch5-2024_06_14_d212e25dd7dcee36ccd8g_0074_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, draw a line $EF$ through point $E$ such that $EF \\parallel AB$,\n\n\n\nFigure 1\n\nSince $AB \\parallel CD$,\n\nTherefore, $AB \\parallel EF \\parallel CD$,\n\nThus, $\\angle A + \\angle 1 = 180^\\circ$, and $\\angle 2 + \\angle C = 180^\\circ$,\n\nHence, $\\angle A + \\angle C + \\angle AEC = 360^\\circ$, so statement (1) is incorrect;\n\n(2) Draw a line $EF$ through point $E$ such that $EF \\parallel AB$,\n\n\n\nFigure 2\n\nSince $AB \\parallel CD$,\n\nTherefore, $AB \\parallel EF \\parallel CD$,\n\nThus, $\\angle A = \\angle 1$, and $\\angle 2 = \\angle C$,\n\nHence, $\\angle AEC = \\angle 1 + \\angle 2 = \\angle A + \\angle C$, so statement (2) is correct;\n\n(3) Draw a line $EF$ through point $E$ such that $EF \\parallel AB$,\n\n\n\nFigure 3\n\nSince $AB \\parallel CD$,\n\nTherefore, $AB \\parallel EF \\parallel CD$,\n\nThus, $\\angle A + \\angle 3 = 180^\\circ$, and $\\angle 1 = \\angle 2$,\n\nHence, $\\angle A + \\angle AEC - \\angle 1 = 180^\\circ$, so statement (3) is correct;\n\n(4) As shown in the figure,\n\n\n\nFigure 4\n\nSince $\\angle 1$ is an exterior angle of $\\triangle CEP$,\n\nTherefore, $\\angle 1 = \\angle C + \\angle P$,\n\nSince $AB \\parallel CD$,\n\nThus, $\\angle A = \\angle 1$,\n\nHence, $\\angle A = \\angle C + \\angle P$, so statement (4) is correct.\n\nIn summary, the correct statements are (2), (3), and (4).\n\nTherefore, the correct choice is: C\n\n【Key Insight】This question examines the properties of parallel lines and the exterior angle property of triangles. Drawing auxiliary lines based on the given conditions is crucial for solving this problem." }, { "problem_id": 925, "question": "As shown in Figure 1, the sides of the square $A B C D$ and the isosceles right triangle $F G H$ overlap, with side $A D$ coinciding with $F G$ and side $A B$ coinciding with $F H$ on the same straight line. The triangle $F G H$ moves to the right at a speed of $1 \\text{ cm/s}$ until point $H$ coincides with point $B$, at which point it stops moving. The area of the overlapping part is denoted as $S(\\text{cm}^2)$. The graph in Figure 2 shows the relationship between the area $S(\\text{cm}^2)$ and the time $t$ (s) as the triangle $F G H$ moves to the right. Determine the value of $a$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 16\nB. 8\nC. 2\nD. 4", "input_image": [ "batch23-2024_06_14_e7c85979645c0047952bg_0082_1.jpg", "batch23-2024_06_14_e7c85979645c0047952bg_0082_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( A B C D \\) is a square and \\( \\triangle F G H \\) is an isosceles right triangle,\n\nTherefore, \\( A H = A D = A B = B C \\).\n\nAs \\( \\triangle F G H \\) moves to the right, the overlapping area increases until \\( \\triangle F G H \\) is entirely within square \\( A B C D \\), at which point \\( S = S_{\\triangle F G H} \\). If it continues to move beyond this point, not entirely within square \\( A B C D \\), the area decreases.\n\nThus, in Figure 2, \\( S = a + 4 \\) represents the maximum overlapping area.\n\nTherefore, \\( S_{\\text{max}} = S_{\\triangle F G H} \\).\n\nSince \\( \\triangle F G H \\) moves to the right at a speed of \\( 1 \\mathrm{~cm/s} \\), and as shown in Figure 2, the total time from the start to the end of the movement is \\( (a + 4) \\) seconds,\n\nTherefore, \\( 2 A B = (a + 4) \\times 1 \\),\n\nHence, \\( A B = \\frac{a}{2} + 2 \\).\n\nSince \\( S_{\\text{max}} = S_{\\triangle F G H} = \\frac{1}{2}(A H)^{2} = \\frac{1}{2}(A B)^{2} = \\frac{1}{2}\\left(\\frac{a}{2} + 2\\right)^{2} = a + 4 \\),\n\nSolving the equation yields: \\( a = 4 \\) or \\( a = -4 \\),\n\nTherefore, \\( a = 4 \\),\n\nHence, the correct choice is: D.\n\n[Highlight] This problem primarily examines the properties of squares, the characteristics of isosceles right triangles, and the application of quadratic equations. The key to solving the problem lies in analyzing the movement process of the isosceles right triangle, determining the relationship between movement time and distance, and setting up the equation to solve for the unknown." }, { "problem_id": 926, "question": "As shown in the figure, given $\\angle M O N$, point $A$ is on the side $O M$, point $B$ is on the side $O N$, and $O A = O B$. Point $E$ is on the side $O B$. Xiaoming and Xiaohong each drew a rectangle $A E B F$ and a parallelogram $A E B F$ in Figure 1 and Figure 2, respectively, and connected the diagonals. The two diagonals intersect at point $C$. Xiaoming and Xiaohong both believe that ray $O C$ is the bisector of $\\angle M O N$. Which of their statements do you think is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. Xiaoming and Xiaohong are both right\nB. Xiaoming and Xiaohong are both wrong\nC. Xiaoming is wrong, Xiaohong is right\nD. Xiaoming is right, Xiaohong is wrong", "input_image": [ "batch23-2024_06_14_9949fed5d0b749fe1005g_0059_1.jpg", "batch23-2024_06_14_9949fed5d0b749fe1005g_0059_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral $AEBF$ is a rectangle,\n\n$\\therefore AC = BC$,\n\nIn triangles $\\triangle AOC$ and $\\triangle BOC$,\n\n$$\n\\left\\{\\begin{array}{l}\nAC = BC \\\\\nOA = OB \\\\\nOC = OC\n\\end{array}\\right.\n$$\n\n$\\therefore \\triangle AOC \\cong \\triangle BOC$ (SSS),\n\n$\\therefore \\angle AOC = \\angle BOC$,\n\n$\\therefore$ ray $OC$ is the angle bisector of $\\angle MON$,\n\nHence, Xiao Ming's statement is correct;\n\nSince quadrilateral $AEBF$ is a parallelogram,\n\n$\\therefore AC = BC$,\n\nIn triangles $\\triangle AOC$ and $\\triangle BOC$,\n\n$\\left\\{\\begin{array}{l}AC = BC \\\\ OA = OB, \\\\ OC = OC\\end{array}\\right.$\n\n$\\therefore \\triangle AOC \\cong \\triangle BOC$ (SSS),\n\n$\\therefore \\angle AOC = \\angle BOC$,\n\n$\\therefore$ ray $OC$ is the angle bisector of $\\angle MON$,\n\nHence, Xiao Hong's statement is correct.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question examines the properties of rectangles, parallelograms, the criteria and properties of congruent triangles, and the determination of angle bisectors. The key to solving the problem lies in a thorough understanding of the properties of rectangles and parallelograms." }, { "problem_id": 927, "question": "The \"Golden Ratio\" gives a sense of aesthetics and is widely applied in architecture, art, and other fields. As shown in Figure (1), if point $C$ divides segment $A B$ into two parts such that $B C: A C=A C: A B$, then point $C$ is called a golden section point of segment $A B$. As shown in Figure (2), points $C, D, E$ are the golden section points of segments $A B, A C, A D$ respectively, with $(A C>B C, A D>D C, A E>E D)$, and if $A B=1$, then the length of $A E$ is ( )\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $\\sqrt{5}-2$\nB. $\\frac{\\sqrt{5}-2}{2}$\nC. $\\frac{3-\\sqrt{5}}{2}$\nD. $\\frac{\\sqrt{5}-1}{2}$", "input_image": [ "batch31-2024_06_14_ef36669746b58ff8bd9cg_0049_1.jpg", "batch31-2024_06_14_ef36669746b58ff8bd9cg_0049_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: \n\n$\\because$ As shown in Figure (1), point $C$ divides segment $A B$ into two parts. If $B C: A C = A C: A B$, then point $C$ is called the golden section point of segment $A B$.\n\n$\\therefore$ Let $A B = 1$, and set $B C = x$. Then $A C = 1 - x$. From $\\frac{B C}{A C} = \\frac{A C}{A B}$, substituting the values gives $(1 - x)^2 = x$. Solving this equation yields $x = \\frac{3 - \\sqrt{5}}{2}$.\n\n$\\therefore A C = \\frac{\\sqrt{5} - 1}{2}$.\n\n$\\therefore \\frac{B C}{A C} = \\frac{A C}{A B} = \\frac{\\sqrt{5} - 1}{2}$.\n\n$\\because$ Points $C$, $D$, and $E$ are the golden section points of segments $A B$, $A C$, and $A D$ respectively.\n\n$\\therefore \\frac{B C}{A C} = \\frac{A C}{A B} = \\frac{\\sqrt{5} - 1}{2}$, $\\frac{D C}{A D} = \\frac{A D}{A C} = \\frac{\\sqrt{5} - 1}{2}$, $\\frac{D E}{A E} = \\frac{A E}{A D} = \\frac{\\sqrt{5} - 1}{2}$.\n\n$\\therefore A E = \\frac{\\sqrt{5} - 1}{2} A D$, $A D = \\frac{\\sqrt{5} - 1}{2} A C$, $A C = \\frac{\\sqrt{5} - 1}{2} A B$.\n\nSubstituting $A B = 1$, we can solve to get $A C = \\frac{\\sqrt{5} - 1}{2}$, $A D = \\left(\\frac{\\sqrt{5} - 1}{2}\\right)^2 = \\frac{3 - \\sqrt{5}}{2}$,\n\n$A E = \\frac{\\sqrt{5} - 1}{2} \\times \\frac{3 - \\sqrt{5}}{2} = \\sqrt{5} - 2$.\n\nTherefore, the answer is: A.\n\n【Key Insight】This problem examines the definition of the golden section point, involving the calculation of the golden ratio and using the golden ratio to determine the length of a segment. Understanding the definition of the golden section point and deriving the proportion is key to solving the problem." }, { "problem_id": 928, "question": "Figure 1 shows a foldable desk lamp. Figure 2 is its plan view, where the base $A O$ is perpendicular to $O E$ at point $O$. The support rods $A B$ and $B C$ are fixed, and $\\angle A$ is twice the size of $\\angle B$. The lampshade $C D$ can rotate around point $C$ for adjustment. Now, the lampshade $C D$ is rotated from the horizontal position to the position $C D^{\\prime}$ (as shown by the dashed line in Figure 2), at which point the line containing $C D^{\\prime}$ is perpendicular to the support rod $A B$, and $\\angle B C D - \\angle D C D^{\\prime} = 126^\\circ$. The measure of $\\angle D C D^{\\prime}$ is $(\\mathrm{)}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $26^\\circ$\nB. $36^\\circ$\nC. $46^\\circ$\nD. $72^\\circ$", "input_image": [ "batch13-2024_06_15_72def6568d0ee26f353eg_0025_1.jpg", "batch13-2024_06_15_72def6568d0ee26f353eg_0025_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nExtend \\( O A \\) to intersect \\( C D \\) at point \\( F \\), and extend \\( D' C \\) to intersect \\( A B \\) at point \\( G \\), as shown in the figure:\n\n\n\nSince \\( C D \\parallel O E \\) and \\( A O \\perp O E \\),\n\nit follows that \\( O A \\perp C D \\).\n\nSince \\( A O \\perp O E \\) and \\( D' C \\perp A B \\),\n\nwe have \\( \\angle A G C = \\angle A F C = 90^\\circ \\).\n\nThus, \\( \\angle G C F + \\angle G A F = 180^\\circ \\).\n\nSince \\( \\angle D C D' + \\angle G C F = 180^\\circ \\),\n\nit follows that \\( \\angle D C D' = \\angle G A F \\).\n\nTherefore, \\( \\angle B A O = 180^\\circ - \\angle D C D' \\),\n\nand \\( \\angle B = \\frac{1}{2} \\left(180^\\circ - \\angle D C D'\\right) \\).\n\nGiven that \\( \\angle B C D - \\angle D C D' = 126^\\circ \\),\n\nwe have \\( \\angle B C D = \\angle D C D' + 126^\\circ \\).\n\nIn quadrilateral \\( A B C F \\), the sum of the interior angles is:\n\n\\[\n\\angle G A F + \\angle B + \\angle B C D + \\angle A F C = 360^\\circ.\n\\]\n\nSubstituting the known values:\n\n\\[\n\\angle D C D' + \\frac{1}{2} \\left(180^\\circ - \\angle D C D'\\right) + \\angle D C D' + 126^\\circ + 90^\\circ = 360^\\circ.\n\\]\n\nSolving this equation yields:\n\n\\[\n\\angle D C D' = 36^\\circ.\n\\]\n\nTherefore, the correct answer is **B**.\n\n**Key Insight:** This problem tests the application of the sum of interior angles in a quadrilateral, the equality of supplementary angles, the definition of adjacent supplementary angles, and the relationships between angles in geometric figures. The key to solving the problem lies in correctly constructing auxiliary lines and using the fact that the sum of the interior angles of a quadrilateral is \\( 360^\\circ \\)." }, { "problem_id": 929, "question": "The diagrams below represent the routes taken by individuals A, B, and C from location A to location C. It is given that A's route is $A \\rightarrow B \\rightarrow C$, where $\\triangle A B C$ is an equilateral triangle; B's route is $A \\rightarrow B \\rightarrow D \\rightarrow E \\rightarrow C$, where $D$ is the midpoint of $A C$, and both $\\triangle A B D$ and $\\triangle D E C$ are equilateral triangles; C's route is $A \\rightarrow B \\rightarrow D \\rightarrow E \\rightarrow C$, where $D$ is on $A C$ (with $A D \\neq D C$), and both $\\triangle A B D$ and $\\triangle D E C$ are equilateral triangles. Which of the following statements is true regarding the distances traveled by the three individuals ( )\n\n\n\n A\n\n\n\n B\n\n\n\n C\nA. A's route is the shortest.\nB. B's route is the shortest.\nC. C's route is the shortest.\nD. The distances traveled by all three individuals are the same.", "input_image": [ "batch5-2024_06_14_5763d01636bdeb313a11g_0038_1.jpg", "batch5-2024_06_14_5763d01636bdeb313a11g_0038_2.jpg", "batch5-2024_06_14_5763d01636bdeb313a11g_0038_3.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the side length of the equilateral triangle \\( ABC \\) be \\( a \\). Then, in diagram B, the side lengths of the equilateral triangles \\( \\triangle ADB \\) and \\( \\triangle DEC \\) are \\( \\frac{1}{2} a \\). In diagram C, the side length of the equilateral triangle is \\( AB + DE = a \\).\n\nTherefore:\n- For diagram A: \\( a + a = 2a \\),\n- For diagram B: \\( 4 \\times \\frac{1}{2} a = 2a \\),\n- For diagram C: \\( 2(AB + DE) = 2a \\).\n\nHence, the correct choice is: D.\n\n**Key Insight:** This problem primarily tests the understanding and application of equilateral triangles and the comparison of line segment lengths. The key to solving this problem lies in determining the total walking distance based on the side lengths of the equilateral triangles." }, { "problem_id": 930, "question": "The Shanghai-Suzhou-Lake High-speed Railway is under intense construction. Now Nanxun Station has started tunnel excavation. As shown in Figure 1, the arc-shaped concrete pipe segment is an important component of the circular tunnel. As shown in Figure 2, there is an arc-shaped concrete pipe segment placed on the horizontal ground, and the bottom is fixed with two identical rectangular wooden blocks. In order to estimate the size of the tunnel excavation surface, the three groups A, B, and C measured the relevant data as shown in the following table. The group that can use the data to estimate the size of the tunnel outer diameter is ()\n\n| Group | Measurement content |\n| :---: | :---: |\n| A | Length of $H G, G N$ |\n| B | Length of $AB, A D, B C$ |\n| C | Length of $AB$, distance between points $\\mathrm{A} and D$, distance between points $B and C$ |\n\n\n\n(Image 1)\n\n\n\n(Image 2)\nA. All three groups of measurement data are insufficient\nB. One group\nC. Two groups\nD. All three groups are OK", "input_image": [ "batch37-2024_06_14_63116ed6376d6159c7a2g_0058_1.jpg", "batch37-2024_06_14_63116ed6376d6159c7a2g_0058_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Connect $HM$, then quadrilateral $HGNM$ is a rectangle. Draw $OK \\perp GN$ through point $O$, with foot $K$, then $OK \\perp HM$, with foot $O^{\\prime}$, as shown in the figure,\n\n\n\n$\\therefore$ Quadrilateral $HGKO^{\\prime}$ is a rectangle,\n\n$\\therefore KO^{\\prime}=HG$\n\nFor Group A: Directly measure $HG, GN$,\n\nLet the outer diameter be $R$,\n\nIn right triangle $\\triangle OO^{\\prime}M$, $OM^{2}=OO^{\\prime 2}+MO^{\\prime 2}$\n\n$\\therefore R^{2}=\\left(R-KO^{\\prime}\\right)^{2}+MO^{\\prime 2}$\n\n$\\therefore R^{2}=(R-HG)^{2}+\\left(\\frac{1}{2} GN\\right)^{2}$\n\n$\\because HG, GN$ are known,\n\n$\\therefore R^{2}$ can be determined, hence Group A can estimate the outer diameter of the tunnel;\n\nFor Group B: Let the outer diameter be $R$, then the inner diameter is $(R-AB)$\n\nLet the central angle $\\angle AOC=\\alpha$, then:\n\n$\\{AD}=\\frac{\\pi \\alpha(R-AB)}{180}, \\{BC}=\\frac{\\pi \\alpha R}{180}$\n$\\therefore \\frac{\\{AD}}{\\{BC}}=\\frac{R-AB}{R}$\n\n$\\because \\{AD}, \\{BC}, AB$ are known\n\n$\\therefore R$ can be determined, hence Group B can estimate the outer diameter of the tunnel;\n\nFor Group C: Connect $AD, BC$, draw $OK \\perp BC$ through point $O$, with foot $K$, then $OK \\perp AD$, with foot $O^{\\prime}$, as shown in the figure,\n\n\n\nLet the outer diameter be $R$, then the inner diameter is $(R-AB)$\n\nLet the central angle $\\angle AOC=\\alpha$, then: $\\angle BOK=\\angle AOO^{\\prime}=\\frac{1}{2} \\angle \\alpha$\n\n$\\therefore \\sin \\frac{\\alpha}{2}=\\frac{AO^{\\prime}}{AO}=\\frac{BK}{BO}$\n\n$\\therefore \\frac{\\frac{1}{2} AD}{OA}=\\frac{\\frac{1}{2} BC}{BO}$, i.e.: $\\frac{\\frac{1}{2} AD}{R-AB}=\\frac{\\frac{1}{2} BC}{R}$\n\n$\\because AB, AD, BC$ are known,\n\n$\\therefore R$ can be determined, hence Group C can estimate the outer diameter of the tunnel;\n\n$\\therefore$ All three groups can estimate the outer diameter of the tunnel.\n\nTherefore, choose $D$.\n\n【Key Insight】This problem mainly examines the application of the perpendicular chord theorem and solving right triangles. Using the perpendicular chord theorem to construct right triangles is the key to solving this problem." }, { "problem_id": 931, "question": "As shown in the figure, two unequal-sized squares are cut into 5 parts, and the difference in area between (2) and (5) is 8. These 5 parts are assembled into a large square $A B C D$. The line segment $A C$ intersects $D F$ at point $E$. If $\\frac{D E}{E F} = \\frac{4}{3}$, then the area of the large square $A B C D$ is ( )\n\n\n\n(3)\n\n\nA. 18\nB. 25\nC. 32\nD. 50", "input_image": [ "batch32-2024_06_14_616739ac8cb65fd7d225g_0076_1.jpg", "batch32-2024_06_14_616739ac8cb65fd7d225g_0076_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nSince quadrilateral \\( A B C D \\) is a square,\n\nTherefore, \\( A D \\parallel B C \\),\n\nHence, \\( \\triangle A D E \\sim \\triangle C E F \\),\n\nThus, \\( \\frac{A E}{C E} = \\frac{A D}{C F} = \\frac{D E}{E F} = \\frac{4}{3} \\),\n\nSince \\( A M \\perp D M \\) and \\( C N \\perp N F \\),\n\nTherefore, \\( \\angle A M E = \\angle C N E = 90^{\\circ} \\), and \\( \\angle A E M = \\angle C E N \\),\n\nHence, \\( \\triangle A M E \\sim \\triangle C N E \\),\n\nThus, \\( \\frac{A M}{C N} = \\frac{A E}{C E} = \\frac{4}{3} \\),\n\nSince \\( A M = H I = G T \\) and \\( C N = K J \\),\n\nTherefore, \\( \\frac{G T}{K J} = \\frac{4}{3} \\);\n\nLet \\( G T = 4 x \\), then \\( K J = 3 x = Q I = I J \\), and \\( H I = Q T = 4 x \\),\n\nThus, \\( H Q = H I - Q I = 4 x - 3 x = x \\),\n\nSince \\( R Q \\parallel J I \\),\n\nTherefore, \\( \\triangle H Q R \\sim \\triangle H I J \\)\n\nThus, \\( \\frac{Q R}{I J} = \\frac{H Q}{H I} \\), which means \\( \\frac{Q R}{3 x} = \\frac{x}{4 x} \\),\n\nHence, \\( Q R = \\frac{3}{4} x \\),\n\nTherefore, \\( S_{(2)} = S_{\\text{trapezoid } H Q T G} - S_{\\triangle H Q R} \\)\n\n\\( = \\frac{1}{2}(x + 4 x) \\cdot 4 x - \\frac{1}{2} x \\cdot \\frac{3}{4} x \\)\n\n\\( = \\frac{77}{8} x^{2} \\),\n\n\\( S_{(5)} = \\frac{1}{2}\\left(\\frac{3}{4} x + 3 x\\right) \\cdot 3 x = \\frac{45}{8} x^{2} \\),\n\nSince \\( S_{(2)} - S_{(5)} = 8 \\),\n\nTherefore, \\( \\frac{77}{8} x^{2} - \\frac{45}{8} x^{2} = 8 \\)\n\nSolving gives \\( x^{2} = 2 \\),\n\nThus, \\( x = \\sqrt{2} \\) (discarding the negative value),\n\nTherefore, \\( G T = 4 \\sqrt{2} \\), and \\( K J = 3 \\sqrt{2} \\),\nThus, \\( S_{\\text{square } A B C D} = S_{\\text{square } G P Q T} + S_{\\text{square } K Q T I} \\)\n\n\\( = (4 \\sqrt{2})^{2} + (3 \\sqrt{2})^{2} \\)\n\n\\( = 50 \\),\n\nTherefore, the answer is: D.\n\n【Key Insight】This problem tests the assembly of shapes, involving the properties of squares, the determination and properties of similar triangles, and solving quadratic equations. Mastering the properties of squares and the determination and properties of similar triangles, and being able to find the relationships between line segments in the two figures, is key to solving the problem." }, { "problem_id": 932, "question": "'People must have dreams, what if they come true?' Gong Lijiao's post-match remarks have been widely circulated on the Internet, inspiring many people who are working hard. As shown in Figure (1), this is a shot put she threw during her shot put practice. The trajectory of the shot put after the shot put can be approximately regarded as part of a parabola. It is known that the shot put is 1.6 meters above the ground when it is thrown, and it reaches its highest point when it is 3 meters horizontally away from the throwing point. At this time, the shot put is 2.5 meters above the ground. As shown in Figure (2), with the horizontal plane as the $x$ axis and the plumb line of her standing position as the $y$ axis, a plane rectangular coordinate system is established. The function expression of her shot put motion path is ()\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $y=-\\frac{1}{10} x^{2}+\\frac{3}{5} x+\\frac{8}{5}$\nB. $y=-\\frac{1}{10} x^{2}-\\frac{3}{5} x+\\frac{8}{5}$\nC. $y=-\\frac{1}{10} x^{2}+\\frac{3}{5} x+\\frac{5}{2}$\nD. $y=\\frac{1}{10} x^{2}+\\frac{3}{5} x+\\frac {8}{5}$", "input_image": [ "batch7-2024_06_14_ef200ef2c744e8bae6f3g_0042_1.jpg", "batch7-2024_06_14_ef200ef2c744e8bae6f3g_0042_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we have the points $B(0,1.6)$ and $C(3,2.5)$.\n\nLet the parabolic equation be $y = a(x - 3)^2 + 2.5$.\n\nSubstituting the coordinates of point $B$ into the equation gives: $9a + 2.5 = 1.6$.\n\nSolving for $a$ yields: $a = -\\frac{1}{10}$.\n\nTherefore, the function representing Gong Lijiao's shot put trajectory is:\n\\[ y = -\\frac{1}{10}(x - 3)^2 + 2.5 = -\\frac{1}{10}x^2 + \\frac{3}{5}x + \\frac{8}{5} \\]\n\nHence, the correct choice is: A.\n\n【Key Insight】This problem tests the method of undetermined coefficients for finding the quadratic function's equation. The key to solving it is to be proficient with the properties of quadratic functions and to set the parabolic equation as $y = a(x - 3)^2 + 2.5$." }, { "problem_id": 933, "question": "As shown in the figure, a rectangular paper $A B C D$ (Figure (1)) is folded as follows: (1) Fold the paper along a straight line passing through point $A$ so that point $B$ falls on the $A D$ edge, and the crease intersects the $B C$ edge at point $E$ (Figure (2)); (2) Fold the paper again along a straight line passing through point $E$ so that point $A$ falls on the $B C$ edge, and the crease $E F$ intersects the $A D$ edge at point $F$ (Figure (3)). The measure of $\\angle A^{\\prime} E F$ is $(\\quad)$.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. $60^{\\circ}$\nB. $67.5^{\\circ}$\nC. $72^{\\circ}$\nD. 75", "input_image": [ "batch23-2024_06_14_f763395656da1db7dba0g_0087_1.jpg", "batch23-2024_06_14_f763395656da1db7dba0g_0087_2.jpg", "batch23-2024_06_14_f763395656da1db7dba0g_0087_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the properties of the graphical flip transformation, we have: $\\angle B A E = \\angle A E B = 45^{\\circ}$.\n\nSince $DF$ is parallel to $EC$, we obtain: $\\angle D F A^{\\prime} = \\angle E A F = 45^{\\circ}$.\n\nFrom the properties of the graphical flip transformation, it is known that $\\angle A^{\\prime} F E = \\frac{1}{2}\\left(180^{\\circ} - \\angle D F A^{\\prime}\\right) = 67.5^{\\circ}$.\n\nTherefore, $\\angle A^{\\prime} E F = 180^{\\circ} - \\angle E A^{\\prime} F - \\angle A^{\\prime} F E = 180^{\\circ} - 45^{\\circ} - 67.5^{\\circ} = 67.5^{\\circ}$.\n\nHence, the correct choice is: B.\n\n【Key Insight】This question examines the properties of graphical flip transformations. It is essential to understand that folding is a type of symmetry transformation, belonging to axial symmetry. Before and after folding, the shape and size of the figure remain unchanged, while the position changes. The corresponding sides and angles remain equal, which is crucial for solving this problem." }, { "problem_id": 934, "question": "For the problem: \"As shown in Figure 1, in a plane, there is a rectangle with a length of 12 and a width of 6 inside a square. The rectangle can freely move from a horizontal position to a vertical position within the square by translation (i.e., translation or rotation). Determine the smallest integer $n$ for the side length of the square. Students A and B each made a square with what they believed to be the smallest side length, first calculating the side length $x$, then taking the smallest integer $n$.\n\n A: As shown in Figure 2, the idea is that when $x$ is the length of the rectangle's diagonal, it can be moved; the result is $n=13$.\n\n B: As shown in Figure 3, the idea is that when $x$ is $\\frac{\\sqrt{2}}{2}$ times the sum of the rectangle's length and width, it can be moved: the result is $n=13$. Which of the following is correct?\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\nA. Student A's idea is correct, but his $n$ value is wrong\nB. Student B's idea is wrong, but his $n$ value is correct\nC. Both Student A and Student B's ideas are correct\nD. Both Student A and Student B's $n$ values are correct", "input_image": [ "batch14-2024_06_15_67b9e2592a4ca695d018g_0057_1.jpg", "batch14-2024_06_15_67b9e2592a4ca695d018g_0057_2.jpg", "batch14-2024_06_15_67b9e2592a4ca695d018g_0057_3.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since the rectangle has a length of 12 and a width of 6,\n\nTherefore, the length of the diagonal of the rectangle is: $\\sqrt{6^{2}+12^{2}}=6 \\sqrt{5}$.\n\nSince the rectangle can freely transform from horizontal to vertical placement within and on the boundary of the square through translation or rotation,\n\nTherefore, the side length of the square must be at least $6 \\sqrt{5}$.\n\nGiven that $13<6 \\sqrt{5}<14$,\n\nTherefore, the smallest positive integer $n$ for the side length of the square is 14.\n\nThus, Person A's reasoning is correct; the longest diagonal of the rectangle is the key, and as long as the diagonal can pass through, $n=14$.\n\nPerson B's reasoning and calculations are incorrect; the illustrated situation does not represent the longest scenario.\n\nHence, the correct choice is: A.\n\n[Insight] This problem examines the properties of rotation and rectangles. Proficiency in using the properties of rectangles is crucial for solving the problem." }, { "problem_id": 935, "question": "As shown in Figure (1), point $P$ is a moving point on the diagonal $A C$ of the rhombus $A B C D$, and point $E$ is a fixed point on the side $C D$. The segments $P B$, $P E$, and $B E$ are drawn. Figure (2) is a graph of the area $y$ of $\\triangle P B E$ versus the length $x$ of $A P$ as point $P$ moves from point $A$ to point $C$ at a uniform speed (when point $P$ is on $B E$, let $y = 0$). Determine the perimeter of the rhombus $A B C D$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\nA. $8 \\sqrt{3}$\nB. $8 \\sqrt{5}$\nC. 20\nD. 24", "input_image": [ "batch13-2024_06_15_faf53d78527724b96a79g_0005_1.jpg", "batch13-2024_06_15_faf53d78527724b96a79g_0005_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the graph, we can see that when \\( x = 0 \\), point \\( P \\) coincides with point \\( A \\), and at this moment, the area \\( S_{\\triangle ABE} = 12 \\).\n\nTherefore, the area of the rhombus \\( S_{\\text{rhombus } ABCD} = 2S_{ABE} = 24 \\).\n\nWhen \\( x = 8 \\), point \\( P \\) coincides with point \\( C \\), meaning \\( AC = 8 \\). Connecting \\( BD \\), which intersects \\( AC \\) at point \\( O \\),\n\n\n\nwe have: \\( BD \\perp AC \\), \\( OA = OC = 4 \\), and \\( OB = OD \\).\n\nThus, the area of the rhombus \\( S_{\\text{rhombus } ABCD} = \\frac{1}{2} AC \\cdot BD = 24 \\),\n\nso \\( BD = 6 \\),\n\nand \\( OB = OD = 3 \\).\n\nTherefore, \\( AB = \\sqrt{OA^2 + OB^2} = 5 \\),\n\nand the perimeter of rhombus \\( ABCD \\) is \\( 4 \\times 5 = 20 \\).\n\nHence, the correct answer is C.\n\n【Key Insight】This problem tests the properties of a rhombus and the function graph of a moving point. Mastering the properties of a rhombus and effectively extracting information from the function graph are crucial for solving the problem." }, { "problem_id": 936, "question": "In right triangle $ABC$, $AC = BC = 2$. The vertices $D$ and $F$ of square $CDFE$ are on sides $AC$ and $BC$, respectively. Let the length of $CD$ be $x$, and the area of the overlapping region between $\\triangle ABC$ and square $CDFE$ be $y$. Which of the following graphs correctly represents the relationship between $y$ and $x$?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch8-2024_06_14_3014dab89f701789b5aeg_0030_1.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0030_2.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0030_3.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0030_4.jpg", "batch8-2024_06_14_3014dab89f701789b5aeg_0030_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When \\(0 < x \\leq 1\\), \\(y = x^{2}\\).\n\nWhen \\(1 < x \\leq 2\\), line segment \\(\\mathrm{ED}\\) intersects \\(\\mathrm{AB}\\) at point \\(\\mathrm{M}\\), and line segment \\(\\mathrm{EF}\\) intersects \\(\\mathrm{AB}\\) at point \\(\\mathrm{N}\\), as shown in the figure.\n\n\n\nGiven \\(\\mathrm{CD} = x\\), then \\(AD = 2 - x\\).\n\nSince in the right triangle \\(\\triangle \\mathrm{ABC}\\), \\(\\mathrm{AC} = \\mathrm{BC} = 2\\),\n\n\\(\\triangle \\mathrm{ADM}\\) is an isosceles right triangle,\n\nThus, \\(DM = 2 - x\\),\n\nTherefore, \\(EM = x - (2 - x) = 2x - 2\\),\n\nHence, the area of \\(\\triangle \\mathrm{ENM}\\) is \\(\\frac{1}{2}(2x - 2)^{2} = 2(x - 1)^{2}\\),\n\nSo, \\(y = x^{2} - 2(x - 1)^{2} = -x^{2} + 4x - 2 = -(x - 2)^{2} + 2\\).\n\nTherefore, the function is defined piecewise as:\n\\[\n\\begin{cases}\ny = x^{2} & (0 < x \\leq 1) \\\\\ny = -(x - 2)^{2} + 2 & (1 < x \\leq 2)\n\\end{cases}\n\\]\nThus, the correct answer is: A.\n\n【Key Insight】This problem examines the function graph of a moving point: by interpreting the graph to extract information, it tests the student's ability to analyze problems. The key to solving this problem lies in considering two scenarios: when \\(0 < x \\leq 1\\) and when \\(1 < x \\leq 2\\)." }, { "problem_id": 937, "question": "As shown in Figure 1, equilateral triangles are constructed outwardly on each side of the right-angled triangle, with areas $S_{1}, S_{2}, S_{3}$ respectively. As shown in Figure 2, semicircles are constructed outwardly with the lengths of the right-angled triangle's sides as radii, with areas $S_{4}, S_{5}, S_{6}$ respectively. Given that $S_{1} = 16, S_{2} = 45, S_{5} = 11, S_{6} = 14$, then $S_{3} + S_{4} = ($ $)$\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. 86\nB. 64\nC. 54\nD. 48", "input_image": [ "batch12-2024_06_15_b02da2a1aee0515fe6e4g_0019_1.jpg", "batch12-2024_06_15_b02da2a1aee0515fe6e4g_0019_2.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, draw a perpendicular from point $\\mathrm{E}$ to $\\mathrm{AB}$, with the foot of the perpendicular at $\\mathrm{D}$.\n\nSince $\\triangle \\mathrm{ABE}$ is an equilateral triangle,\n\n$\\therefore \\angle \\mathrm{AED}=\\angle \\mathrm{BED}=30^{\\circ}$. Let $\\mathrm{AB}=\\mathrm{x}$,\n\n$\\therefore \\mathrm{AD}=\\mathrm{BD}=\\frac{1}{2} \\mathrm{AB}=\\frac{1}{2} \\mathrm{x}$,\n\n$\\therefore \\mathrm{DE}=\\sqrt{A E^{2}-A D^{2}}=\\frac{\\sqrt{3}}{2} \\mathrm{x}$,\n$\\therefore \\mathrm{S}_{2}=\\frac{1}{2} \\times x \\times \\frac{\\sqrt{3}}{2} x=\\frac{\\sqrt{3}}{4} A B^{2}$,\n\nSimilarly: $\\mathrm{S}_{1}=\\frac{\\sqrt{3}}{4} A C^{2}, \\mathrm{~S}_{3}=\\frac{\\sqrt{3}}{4} B C^{2}$,\n\n$\\because \\mathrm{BC}^{2}=\\mathrm{AB}^{2}-\\mathrm{AC}^{2}$,\n\n$\\therefore \\mathrm{S}_{3}=\\mathrm{S}_{2}-\\mathrm{S}_{1}$,\n\nAs shown in Figure 2, $\\mathrm{S}_{4}=\\frac{1}{2} \\times\\left(\\frac{1}{2} A B\\right)^{2} \\pi=\\frac{\\pi}{8} A B^{2}$,\n\nSimilarly $\\mathrm{S}_{5}=\\frac{\\pi}{8} A C^{2}, \\mathrm{~S}_{6}=\\frac{\\pi}{8} B C^{2}$,\n\nThen $\\mathrm{S}_{4}=\\mathrm{S}_{5}+\\mathrm{S}_{6}$,\n\n$\\therefore \\mathrm{S}_{3}+\\mathrm{S}_{4}=45-16+11+14=54$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Insight】This problem examines the Pythagorean theorem and the properties of equilateral triangles. Pythagorean theorem: If the lengths of the two legs of a right triangle are $a$ and $b$, and the hypotenuse is $c$, then $a^{2}+b^{2}=c^{2}$." }, { "problem_id": 938, "question": "As shown in Figure 1, the radius of $\\odot O$ is $r$. If point $P^{\\prime}$ is on ray $O P$ and satisfies $O P^{\\prime} \\times O P=r^{2}$, then point $P^{\\prime}$ is called the \"inversive point\" of point $P$ with respect to $\\odot O$. As shown in Figure 2, the radius of $\\odot O$ is 4, point $B$ is on $\\odot O$, $\\angle B O A=60^{\\circ}$, and $O A=8$. If point $A^{\\prime}$ is the inversive point of point $A$ with respect to $\\odot O$, find the length of $A^{\\prime} B$ ( ).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $\\sqrt{3}$\nB. $2 \\sqrt{3}$\nC. 2\nD. 4", "input_image": [ "batch37-2024_06_14_63116ed6376d6159c7a2g_0002_1.jpg", "batch37-2024_06_14_63116ed6376d6159c7a2g_0002_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let $OA$ intersect $\\odot O$ at $C$, and connect $B^{\\prime} C$, as shown in Figure 2,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n$\\because O A^{\\prime} \\cdot O A=4^{2}$\n\nGiven that $r=4, \\quad O A=8$\n\n$\\therefore O A^{\\prime}=2$,\n\n$\\because \\angle B O A=60^{\\circ}, \\quad O B=O C$,\n\n$\\therefore \\triangle O B C$ is an equilateral triangle,\n\nAnd point $A^{\\prime}$ is the midpoint of $O C$,\n\n$\\therefore B A^{\\prime} \\perp O C$,\n\nIn the right triangle $\\triangle O A^{\\prime} B$, $\\sin \\angle A^{\\prime} O B=\\frac{A^{\\prime} B}{O B}$,\n\n$\\therefore A^{\\prime} B=4 \\sin 60^{\\circ}=2 \\sqrt{3}$.\n\nTherefore, the answer is: B.\n\n【Key Insight】This problem examines the positional relationship between a point and a circle: the position of a point can determine the relationship between its distance to the center and the radius, and conversely, knowing the distance from a point to the center and the radius can determine the position of the point relative to the circle. It also tests reading comprehension skills." }, { "problem_id": 939, "question": "As shown in Figure 1, it is a mobile phone stand. Figure 2 is its side view. Segments $A B$ and $B C$ can rotate around points $A$ and $B$, respectively. It is measured that $B C = 8 \\text{ cm}$ and $A B = 16 \\text{ cm}$. When $A B$ and $B C$ rotate to $\\angle B A E = 60^\\circ$ and $\\angle A B C = 50^\\circ$, the distance from point $C$ to $A E$ is ( ) (rounded to one decimal place, reference data: $\\sin 70^\\circ \\approx 0.94, \\sqrt{3} \\approx 1.73$).\n\n\n\nFigure 1\n\n\n\nFigure 2\nA. $13.8 \\text{ cm}$\nB. $7.5 \\text{ cm}$\nC. $6.1 \\text{ cm}$\nD. $6.3 \\text{ cm}$", "input_image": [ "batch37-2024_06_14_d4217d5415e569c92d8ag_0093_1.jpg", "batch37-2024_06_14_d4217d5415e569c92d8ag_0093_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, perpendicular lines are drawn from points $B$ and $C$ to $AE$, with the feet of the perpendiculars being $M$ and $N$ respectively. From point $C$, a line $CD \\perp BM$ is drawn, with the foot of the perpendicular being $D$.\n\n\n\nFigure 2\n\nIn the right triangle $\\triangle ABM$,\n\n$\\because \\angle BAE = 60^\\circ$, and $AB = 16$,\n\n$\\therefore BM = \\sin 60^\\circ \\cdot AB = \\frac{\\sqrt{3}}{2} \\times 16 = 8\\sqrt{3} \\text{ cm}$,\n\n$\\angle ABM = 90^\\circ - 60^\\circ = 30^\\circ$.\n\nIn the right triangle $\\triangle BCD$,\n\n$\\because \\angle DBC = \\angle ABC - \\angle ABM = 50^\\circ - 30^\\circ = 20^\\circ$,\n\n$\\therefore \\angle BCD = 90^\\circ - 20^\\circ = 70^\\circ$.\n\nAlso, since $BC = 8$,\n\n$\\therefore BD = \\sin 70^\\circ \\times 8 \\approx 0.94 \\times 8 = 7.52 \\text{ cm}$.\n\n$\\therefore CN = DM = BM - BD = 8\\sqrt{3} - 7.52 \\approx 6.3 \\text{ cm}$.\n\nThat is, the distance from point $C$ to $AE$ is approximately $6.3 \\text{ cm}$.\n\nTherefore, the correct choice is: D.\n\n【Key Insight】This problem examines the solution of right triangles. Constructing right triangles and utilizing the relationships between their sides and angles are key to solving the problem." }, { "problem_id": 940, "question": "Among the following subway logo designs from various cities in China, which one is both an axisymmetric and a centrosymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0034_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0034_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0034_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0034_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Please enter the detailed explanation here!" }, { "problem_id": 941, "question": "If $a b<0$, then the approximate graph of the inverse proportion function $y=\\frac{a b}{x}$ and the linear function $y=a x+b$ in the same coordinate system could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_b7aa0f088ba806ec834eg_0075_1.jpg", "batch13-2024_06_15_b7aa0f088ba806ec834eg_0075_2.jpg", "batch13-2024_06_15_b7aa0f088ba806ec834eg_0075_3.jpg", "batch13-2024_06_15_b7aa0f088ba806ec834eg_0075_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since \\( ab < 0 \\),\n\nThe graph of the inverse proportional function \\( y = \\frac{ab}{x} \\) is located in the second and fourth quadrants, so options A and C do not fit the context.\n\nSince \\( ab < 0 \\),\n\nThe graph of the linear function \\( y = ax + b \\) passes through the first, third, and fourth quadrants or through the first, second, and fourth quadrants, so option B does not fit the context, while option D does.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question tests the understanding of the graphical properties of inverse proportional functions and linear functions. Mastering their properties is key to solving the problem." }, { "problem_id": 942, "question": "Among the car logo patterns below, which one can be analyzed by the translation of the graph?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_2c8530e5e37fb7a39800g_0005_1.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0005_2.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0005_3.jpg", "batch14-2024_06_15_2c8530e5e37fb7a39800g_0005_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the concept of translation, by observing the figure, it can be determined that pattern $B$ can be obtained through translation.\n\nTherefore, the correct choice is: $B$.\n\n[Key Insight] This question primarily examines the translation of figures. In a plane, translation involves moving a figure as a whole in a certain direction. Students often confuse translation with rotation or flipping, leading to incorrect choices." }, { "problem_id": 943, "question": "Among the following diagrams, which ray represents a direction $60^{\\circ}$ east of south?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_fd61d246744e8785fd79g_0015_1.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0015_2.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0015_3.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0015_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Based on the concept of azimuth and considering the requirements and options provided in the question,\n\nTherefore, the answer is: C.\n\n[Key Point] The question examines the understanding of directional angles. When describing directions using directional angles, the starting side of the angle is usually the true north or true south direction, and the ending side is the ray where the object is located. Therefore, when describing a directional angle, one generally first mentions north or south, followed by east or west deviation. (Note that the angle bisectors of several directions follow everyday conventions, namely northeast, southeast, northwest, and southwest.)" }, { "problem_id": 944, "question": "As shown in the figure, in the same Cartesian coordinate system, the graphs of the functions $y=ax+a$ and $y=\\frac{a}{x}$ may be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_96d969cbe3e2eebc2efbg_0003_1.jpg", "batch13-2024_06_15_96d969cbe3e2eebc2efbg_0003_2.jpg", "batch13-2024_06_15_96d969cbe3e2eebc2efbg_0003_3.jpg", "batch13-2024_06_15_96d969cbe3e2eebc2efbg_0003_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "When \\( a > 0 \\), the graph of the linear function passes through the first, second, and third quadrants, and the graph of the inverse proportional function is located in the first and third quadrants. It can be concluded that option A fits the description.\n\nWhen \\( a < 0 \\), the graph of the linear function passes through the second, third, and fourth quadrants, and the graph of the inverse proportional function is located in the second and fourth quadrants. It can be concluded that options B, C, and D do not fit the description.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question mainly examines the comprehensive understanding of linear and inverse proportional functions. Mastering the relationship between the coefficients of the function equations and the positions of the function graphs is the key to solving the problem." }, { "problem_id": 945, "question": "The phenomenon explained by \"the shortest line segment is perpendicular\" is ( )\nA. \n\n\nMeasuring the long jump result\nB. \n\n\nDrawing a straight line on a board with a pencil and rubber band\nC. \n\n\nFixing a wooden board with two nails\nD. \n\n\n\n##", "input_image": [ "batch33-2024_06_14_48545c0164039757f797g_0003_1.jpg", "batch33-2024_06_14_48545c0164039757f797g_0003_2.jpg", "batch33-2024_06_14_48545c0164039757f797g_0003_3.jpg", "batch33-2024_06_14_48545c0164039757f797g_0003_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. During physical education class, the teacher measured a student's long jump performance by utilizing the principle that the perpendicular segment is the shortest. Therefore, A aligns with the question's intent.\n\nB. When snapping a chalk line on a wooden board, the principle that two points determine a straight line is used. Hence, B does not align with the question's intent.\n\nC. Fixing a wooden strip to the wall with two nails employs the principle that two points determine a straight line. Thus, C does not align with the question's intent.\n\nD. Straightening a curved road to shorten the distance utilizes the principle that \"the line segment between two points is the shortest.\" Therefore, D does not align with the question's intent.\n\nThe correct answer is: A.\n\n[Key Insight] This question primarily examines the properties of line segments. Familiarity with these properties and the ability to apply them flexibly is crucial for solving the problem." }, { "problem_id": 946, "question": "The approximate shape of the graph of the function $y = -\\frac{1}{|x|}$ is ( )\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0022_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0022_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0022_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0022_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: From the function's analytic expression, it can be deduced that $\\mathrm{x}$ can take both positive and negative values, but the function's value can only be negative. Therefore, the graph of the function should lie below the $\\mathrm{x}$-axis, and neither $\\mathrm{x}$ nor $\\mathrm{y}$ can be zero.\n\nHence, the correct choice is: D.\n\n[Key Insight] This question examines the graph of an inverse proportional function. The key to solving it lies in determining the approximate position of the function's graph based on the points that lie on it." }, { "problem_id": 947, "question": "The approximate graphs of the functions $y = \\frac{a}{x}$ and $y = -ax^2 + a$ in the same Cartesian coordinate system could be ( )\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_664d6e156f235987779ag_0005_1.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0005_2.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0005_3.jpg", "batch13-2024_06_15_664d6e156f235987779ag_0005_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "**Question Analysis:** If \\( a > 0 \\), the graph of the inverse proportional function \\( y = \\frac{a}{x} \\) lies in the first and third quadrants, and the graph of the quadratic function \\( y = -a x^{2} + a \\) opens downward, \n\neliminating option **A**; the intersection point of the quadratic function graph with the Y-axis at \\( (0, a) \\) lies on the positive half of the Y-axis, eliminating option **B**; \n\nIf \\( a < 0 \\), the graph of the inverse proportional function \\( y = \\frac{a}{x} \\) lies in the second and fourth quadrants, and the graph of the quadratic function \\( y = -a x^{2} + a \\) opens upward, \n\neliminating option **C**; therefore, the correct choice is **D**.\n\n**Key Point:** Determining the graphs of quadratic and inverse proportional functions." }, { "problem_id": 948, "question": "Among the following groups of patterns, which are not congruent figures?\n\nA.\n\nB.\n\n\n\nC.\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0091_1.jpg", "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0091_2.jpg", "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0091_3.jpg", "batch10-2024_06_14_2a39ca9990bc4e7281a7g_0091_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "According to the definition of congruent shapes, $\\mathrm{ABC}$ are all congruent shapes, while $\\mathrm{D}$ is not a congruent shape because its size is different. Therefore, the correct answer is D.\n\n【Highlight】This question tests the definition of congruent shapes. In typical plane geometry, if one shape can be moved to coincide exactly with another shape, then these two shapes are called congruent shapes, abbreviated as congruent shapes." }, { "problem_id": 949, "question": "The graph of the inverse proportion function $y = -\\frac{3}{x}$ in the Cartesian coordinate system could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0057_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0057_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0057_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0057_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Since \\( k = -3 < 0 \\), it is known that the graph of the inverse proportional function lies in the second and fourth quadrants, so option A is correct.\n\nTherefore, the answer is: A.\n\n【Key Point】The main focus is on the graphical properties of the inverse proportional function. Understanding its properties is essential for solving problems flexibly. The graph of an inverse proportional function is a hyperbola. When \\( k > 0 \\), its two branches are located in the first and third quadrants; when \\( k < 0 \\), its two branches are located in the second and fourth quadrants." }, { "problem_id": 950, "question": "Given the graph of the linear function $y = kx + b$ as shown in the figure, the graphs of $y = -kx + b$ and $y = \\frac{b}{x}$ are ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21a6ea375ed11b43aa92g_0008_1.jpg", "batch13-2024_06_15_21a6ea375ed11b43aa92g_0008_2.jpg", "batch13-2024_06_15_21a6ea375ed11b43aa92g_0008_3.jpg", "batch13-2024_06_15_21a6ea375ed11b43aa92g_0008_4.jpg", "batch13-2024_06_15_21a6ea375ed11b43aa92g_0008_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem statement, we have: $k>0, b>0$,\n\n$\\therefore -k<0$,\n\n$\\therefore$ the graph of the linear function $y=-k x+b$ passes through the first, second, and fourth quadrants, while the graph of the inverse function $y=\\frac{b}{x}$ is located within the first and third quadrants.\n\nTherefore, the correct choice is: A\n\n[Key Insight] This question primarily examines the graphs and properties of linear and inverse functions. Mastering the graphs and properties of these functions is crucial for solving the problem." }, { "problem_id": 951, "question": "In the following figures, which pair of angles, $\\angle 1$ and $\\angle 2$, are adjacent supplementary angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch30-2024_06_17_94cae917628bb2f1b9fbg_0024_1.jpg", "batch30-2024_06_17_94cae917628bb2f1b9fbg_0024_2.jpg", "batch30-2024_06_17_94cae917628bb2f1b9fbg_0024_3.jpg", "batch30-2024_06_17_94cae917628bb2f1b9fbg_0024_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Analysis: Make a judgment based on the definition of adjacent supplementary angles.\n\nDetailed Explanation: According to the definition of adjacent supplementary angles, only the angles in diagram $D$ are adjacent supplementary angles; the others are not.\n\nTherefore, the correct choice is D.\n\nKey Point: This question tests the definition of adjacent supplementary angles. Correctly grasp the definition: two angles are adjacent supplementary angles if they share a common side, and their other sides are opposite extensions of each other." }, { "problem_id": 952, "question": "The area of a rhombus is 2, and its diagonals are $x$ and $y$. The graph of $y$ against $x$ is approximately ().\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0006_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0006_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0006_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0006_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Because the area of a rhombus \\( \\mathrm{S} = \\frac{1}{2} x y \\)\n\nTherefore, \\( x y = 4 \\), which means \\( \\mathrm{y} = \\frac{4}{x} \\)\n\nWhere \\( \\mathrm{x} > 0 \\)\n\nHence, the correct choice is: C\n\n【Key Point】This question tests the application of the area formula for a rhombus. Note that after determining the relationship between \\( \\mathrm{x} \\) and \\( \\mathrm{y} \\), it is also necessary to determine the range of values for \\( \\mathrm{x} \\)." }, { "problem_id": 953, "question": "The two branches of the inverse proportion function $y = \\frac{k}{x}$ are located in the second and fourth quadrants. The graph of the linear function $y = kx - 2$ is approximately ( )\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n", "input_image": [ "batch13-2024_06_15_58474443df6e195863d8g_0029_1.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0029_2.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0029_3.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0029_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since the two branches of the graph of the inverse proportional function \\( y = \\frac{k}{x} \\) are located in the second and fourth quadrants,\n\n\\(\\therefore k < 0\\),\n\n\\(\\therefore\\) the graph of the linear function \\( y = kx - 2 \\) intersects the negative half of the \\( y \\)-axis and passes through the second, third, and fourth quadrants. Therefore, option B is correct. Hence, the answer is: B.\n\n[Key Insight] This question primarily examines the properties of inverse proportional functions and the properties of linear function graphs. Determining that \\( k < 0 \\) based on the properties of the inverse proportional function is the key to solving the problem." }, { "problem_id": 954, "question": "Among the following icons, which are axisymmetric figures? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_becff04b6620d764874eg_0053_1.jpg", "batch10-2024_06_14_becff04b6620d764874eg_0053_2.jpg", "batch10-2024_06_14_becff04b6620d764874eg_0053_3.jpg", "batch10-2024_06_14_becff04b6620d764874eg_0053_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \nA. It is not an axisymmetric figure and does not meet the requirement; \nB. It is not an axisymmetric figure and does not meet the requirement; \nC. It is not an axisymmetric figure and does not meet the requirement; \nD. It is an axisymmetric figure and meets the requirement. \n\nTherefore, the correct answer is: D. \n\n**[Key Point]** This question tests the concept of axisymmetric figures. The key to solving it is understanding that if a figure can be folded along a straight line such that the two sides coincide, it is called an axisymmetric figure." }, { "problem_id": 955, "question": "The graph of the inverse proportion function $y = \\frac{k}{x}$ intersects the graph of the direct proportion function $y = 2x$ at a point whose x-coordinate is 1. The approximate graph of the inverse proportion function is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0046_1.jpg", "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0046_2.jpg", "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0046_3.jpg", "batch13-2024_06_15_ed8d337c98b0b4f63d74g_0046_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Question Analysis: For this problem, first determine the intersection point coordinates $(1,2)$ based on the direct proportionality function, then substitute into the inverse proportionality function's equation to obtain the result.\n\nSubstituting $\\mathrm{x}=1$ yields the y-coordinate of the intersection point as 2, hence the intersection point coordinates are $(1,2)$. Substituting further gives: $\\mathrm{k}=2>0$, indicating the graph is located in the first and third quadrants.\n\nKey Point: Graph of the inverse proportionality function." }, { "problem_id": 956, "question": "Given a piece of clay to make a cylinder without any waste or leftover, the possible graph of the height $h$ of the cylinder as a function of the base area $S$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0034_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0034_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0034_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0034_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Let the volume of this colored clay be \\( V \\).\n\nAccording to the problem, \\( V \\) is a constant and greater than 0.\n\nThus, \\( V = S h \\),\n\nwhich implies \\( h = \\frac{V}{S} \\).\n\nTherefore, the height \\( h \\) of the cylinder is inversely proportional to the base area \\( S \\).\n\nBy examining the function graphs of the four options, only option A fits this relationship.\n\nHence, the answer is: A.\n\n【Insight】This problem tests the understanding of inverse proportionality. Correctly deriving the functional relationship based on the given conditions is key to solving the problem." }, { "problem_id": 957, "question": "Based on the following Euclidean constructions, which one cannot determine that $\\angle 1 = \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_1fc8b75729663205496ag_0086_1.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0086_2.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0086_3.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0086_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \nA: From the construction, it is known to be an angle bisector, hence $\\angle 1 = \\angle 2$, so A does not meet the requirement of the problem; \n\nB: From the construction, it is known that there are parallel lines and an isosceles triangle, so B does not meet the requirement of the problem; \n\nC: From the construction, it can be determined that the triangles are congruent, so C does not meet the requirement of the problem; \n\nD: From the construction, it cannot be determined that the triangles are congruent, so D meets the requirement of the problem; \n\nTherefore, the correct choice is: D. \n\n**[Key Insight]** This question tests basic construction skills. Mastering the methods of proving triangle congruence, understanding isosceles triangles, and using compass-and-straightedge constructions are key to solving the problem." }, { "problem_id": 958, "question": "The approximate graph of the functions $y = -2x$ and $y = -\\frac{1}{2x}$ in the same coordinate system is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0077_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0077_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0077_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0077_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Since \\( y = -2x \\) and \\( -2 < 0 \\),\n\nthe graph passes through the second and fourth quadrants.\n\nBecause in the function \\( y = -\\frac{1}{2x} \\), the coefficient is less than 0,\n\nthe graph is in the first and third quadrants.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question primarily tests the ability to grasp useful conditions from the graph, accurately determine the graph's position, and correctly remember the differences between linear and inverse proportional functions, which is crucial for solving the problem." }, { "problem_id": 959, "question": "Xiaohua's family is going to renovate their home, and the designer has provided four patterns of tiles. Dad wants the area of the gray and white tiles to be roughly the same. Which of the following is the most suitable option?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch20-2024_05_23_522c480bbd02ea580e14g_0071_1.jpg", "batch20-2024_05_23_522c480bbd02ea580e14g_0071_2.jpg", "batch20-2024_05_23_522c480bbd02ea580e14g_0071_3.jpg", "batch20-2024_05_23_522c480bbd02ea580e14g_0071_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorics", "image_relavance": "0", "analysis": "**Question Analysis:** Let the side length of the square be \\(2a\\). \n\n- In **Option A**, the area of the white part is \\(\\pi a^{2}\\), and the area of the gray part is \\((4-\\pi) a^{2}\\).\n- In **Option B**, the area of the white part is \\(\\pi a^{2}\\), and the area of the gray part is \\((4-\\pi) a^{2}\\).\n- In **Option C**, the area of the gray part is \\(\\pi a^{2}\\), and the area of the white part is \\((4-\\pi) a^{2}\\).\n- In **Option D**, the area of the gray part is \\(2(\\pi-2) a^{2}\\), and the area of the white part is \\((8-2\\pi) a^{2}\\).\n\nTherefore, the correct answer is **D**." }, { "problem_id": 960, "question": "During intense exercise, a person's heart rate increases. When the exercise stops, the relationship between the number of heartbeats $N$ (times) and the time $s$ (minutes) can be roughly represented by the graph ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\nPage 7 of 67", "input_image": [ "batch13-2024_06_15_81899374b5ba46ffb08ag_0012_1.jpg", "batch13-2024_06_15_81899374b5ba46ffb08ag_0012_2.jpg", "batch13-2024_06_15_81899374b5ba46ffb08ag_0012_3.jpg", "batch13-2024_06_15_81899374b5ba46ffb08ag_0012_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Statistics", "image_relavance": "0", "analysis": "After a normal person stops engaging in intense exercise, the heart rate gradually decreases from fast to slow over time, approaching the normal resting heart rate. This means that during this period, the heart rate \\( N \\) (beats) and time \\( s \\) (minutes) are inversely proportional. Therefore, the graph representing this relationship is roughly the one shown in option D.\n\nThus, the correct choice is D.\n\n【Highlight】Key point: Application of inverse proportionality functions." }, { "problem_id": 961, "question": "The graphs of the functions $y = ax + b$ and $y = ax^2 + bx + c$ in the same Cartesian coordinate system are approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_dd9b6a481087dff4e73bg_0005_1.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0005_2.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0005_3.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0005_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Question Analysis: When \\( \\mathrm{a} > 0 \\), the graph of the quadratic function opens upwards, and the graph of the linear function passes through the first and third quadrants, or the first, second, and third quadrants, or the first, third, and fourth quadrants.\n\nTherefore, options A and D are incorrect. From the graphs of the quadratic functions in options B and C, the axis of symmetry \\( x = -\\frac{b}{2a} > 0 \\), and since \\( \\mathrm{a} > 0 \\), it follows that \\( \\mathrm{b} < 0 \\). However, in option B, the linear function has \\( \\mathrm{a} > 0 \\) and \\( \\mathrm{b} > 0 \\), which excludes option B. Hence, the correct choice is C.\n\nKey Points: 1. Properties of linear function graphs 2. Properties of parabolas." }, { "problem_id": 962, "question": "The graph of the inverse proportion function $y = -\\frac{6}{x}$ is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0078_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0078_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0078_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0078_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since \\( k = -6 < 0 \\),\n\nthe graph of the inverse proportional function \\( y = -\\frac{6}{x} \\) lies in the second and fourth quadrants.\n\nTherefore, the correct choice is: C.\n\n**Key Insight:** This question tests the understanding of the graph and properties of inverse proportional functions. Mastering the influence of the sign of \\( k \\) on the graph is crucial for solving the problem." }, { "problem_id": 963, "question": "The graph of the inverse proportion function $y = -\\frac{4}{x}$ is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0040_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0040_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0040_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0040_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Since in the inverse proportionality function \\( y = -\\frac{4}{x} \\), \\( -4 < 0 \\);\n\nTherefore, the graph lies in the second and fourth quadrants.\n\nHence, choose A.\n\n[Key Insight] This question examines the graphical properties of inverse proportionality functions. When \\( k > 0 \\), its two branches are located in the first and third quadrants; when \\( k < 0 \\), its two branches are located in the second and fourth quadrants." }, { "problem_id": 964, "question": "The lengths of the two legs of a right-angled triangle are $x$ and $y$, respectively, with an area of 3. The relationship between $y$ and $x$ graphically represented is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0045_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0045_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0045_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0045_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Since \\( S = \\frac{1}{2}xy = 3 \\),\n\nTherefore, \\( xy = 6 \\),\n\nHence, \\( y = \\frac{6}{x} \\),\n\nEliminating options \\( A \\) and \\( D \\),\n\nBecause the side lengths can only be positive,\n\nThus, \\( y = \\frac{6}{x} \\) (where \\( x > 0 \\) and \\( y > 0 \\)).\n\nTherefore, the correct choice is: \\( C \\).\n\n【Key Point】This question primarily examines the graph of an inverse proportional function. Understanding the problem and determining the functional relationship between the two variables is crucial for solving it." }, { "problem_id": 965, "question": "As shown in the figure, when a plane cuts through the following horizontally placed geometric solids, which one results in a circular cross-section?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0026_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0026_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0026_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0026_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem statement,\n\n$\\because$ the three views of a sphere are only circles,\n\n$\\therefore$ if the cross-section is a circle, the geometric body can only be a sphere;\n\nTherefore, the correct choice is: B\n\n【Key Point】This question primarily examines the concept of cutting a geometric body. The shape of the cross-section is related not only to the geometric body being cut but also to the angle and direction of the cut. For such problems, it is best to combine hands-on practice with mental analysis, personally performing the actions to learn and summarize the analytical and inductive thinking methods." }, { "problem_id": 966, "question": "The graphical representation of the functional relationship between the average amount of resources per person $\\overline{\\mathrm{x}}$ in a certain area and the population size $\\mathrm{n}$, given a fixed total amount of resources $\\mathrm{Q}$, is\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0079_1.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0079_2.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0079_3.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0079_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Question Analysis: According to the problem statement, we have: $\\overline{\\mathrm{x}}=\\frac{\\mathrm{Q}}{\\mathrm{n}}$ (where $\\mathrm{Q}$ is a constant), thus the graph of the function between $\\overline{\\mathrm{x}}$ and $\\mathrm{n}$ is a hyperbola. Considering the practical meanings of $\\overline{\\mathrm{x}}$ and $\\mathrm{n}$, both should be greater than 0; hence, the graph lies in the first quadrant. Therefore, the correct choice is B." }, { "problem_id": 967, "question": "The graphs of the functions $y = \\frac{a}{x}$ and $y = ax - a$ (where $a$ is a constant and $a \\neq 0$) in the same coordinate system could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0087_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0087_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0087_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0087_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "When \\( a > 0 \\), the graph of the function \\( y = \\frac{a}{x} \\) is located in the first and third quadrants; the graph of \\( y = ax - a \\) is located in the first, third, and fourth quadrants.\n\nWhen \\( a < 0 \\), the graph of the function \\( y = \\frac{a}{x} \\) is located in the second and fourth quadrants; the graph of \\( y = ax - a \\) is located in the first, second, and fourth quadrants.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question examines the graphs and properties of inverse proportional functions and linear functions. Mastering the properties of function graphs is crucial for solving such problems." }, { "problem_id": 968, "question": "Hong Xing High School stored 120 tons of coal for winter. If $x$ tons of coal are used each day, the approximate graph of the functional relationship between the number of days $y$ and $x$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_81899374b5ba46ffb08ag_0068_1.jpg", "batch13-2024_06_15_81899374b5ba46ffb08ag_0068_2.jpg", "batch13-2024_06_15_81899374b5ba46ffb08ag_0068_3.jpg", "batch13-2024_06_15_81899374b5ba46ffb08ag_0068_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Given that the coal storage is 120 tons, and the daily coal consumption is \\( x \\) tons, the number of usage days \\( y \\) is:\n\n\\[\ny = \\frac{120}{x}, \\quad \\text{where} \\quad x > 0.\n\\]\n\nTherefore, the functional relationship between \\( y \\) and \\( x \\) is an inverse proportion function, and its graph lies in the first quadrant.\n\nHence, the correct choice is A.\n\n**Key Point:** After determining that \\( y \\) is an inverse proportion function of \\( x \\), do not overlook that the range of \\( x \\) is \\( x > 0 \\). This means the graph only includes the portion in the first quadrant." }, { "problem_id": 969, "question": "As shown in the figure, the graph of the function $y = 2x(3-x)$ is approximately\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_e35fe70123f4447b467cg_0014_1.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0014_2.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0014_3.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0014_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Question Analysis:**\n\nGiven the function:\n\\[ y = 2x(3 - x) = -2x^{2} + 6x \\]\n\n1. **Coefficient Analysis:**\n - Since \\( a = -2 < 0 \\), the parabola opens **downwards**.\n - Since \\( b = 6 > 0 \\), the axis of symmetry is to the **right** of the y-axis.\n - Since \\( c = 0 \\), the graph **passes through the origin**.\n\n2. **Conclusion:**\n - Therefore, **Option B** is correct.\n\n**Key Point:** Graph of a quadratic function.\n\n**Answer:** \\boxed{B}" }, { "problem_id": 970, "question": "Among the four figures below, which pair of angles, $\\angle 1$ and $\\angle 2$, are adjacent supplementary angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0054_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0054_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0054_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0054_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of adjacent supplementary angles, only the angles in diagram $\\mathrm{C}$ are adjacent supplementary angles; the others are not. Therefore, the correct choice is: C.\n\n【Key Point】This question tests the definition of adjacent supplementary angles. Correctly understanding the definition—where two angles share a common side and their other sides are opposite rays, making them adjacent supplementary angles—is crucial for solving the problem. Remembering the definition of adjacent supplementary angles is key to finding the solution." }, { "problem_id": 971, "question": "Among the following four figures, which is the three-dimensional solid corresponding to the unfolded image shown ( )\n\n\nA.\n\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0087_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0087_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0087_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0087_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0087_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the unfolded diagram, it can be seen that the face containing the small black square cannot be adjacent to the face containing the large black square. Therefore, options A and C do not correspond to the three-dimensional figure represented by the unfolded diagram. After folding, the three small black squares are on the same face, which means option D does not fit. In option A, the large black square is on the top, so the small black square would be on the bottom, making option B the correct choice. Thus, the answer is B." }, { "problem_id": 972, "question": "As shown in Figure 1, a large tree broke and fell during a strong earthquake, with the top of the tree landing 12 meters away from the base and 5 meters above the ground. Figure 2 is a diagram illustrating the tree's breakage. Determine the height of the tree before it broke.\n\n\nFigure 1\n\n\nFigure 2\nA. 20 meters\nB. 18 meters\nC. 16 meters\nD. 15 meters", "input_image": [ "batch12-2024_06_15_395683fc667a063e6e86g_0064_1.jpg", "batch12-2024_06_15_395683fc667a063e6e86g_0064_2.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the height of the large tree before it broke be \\( x \\mathrm{~m} \\).\n\nBy the Pythagorean theorem, we have: \n\\[\n(x - 5)^{2} = 12^{2} + 5^{2}\n\\]\n\nSolving the equation, we find:\n\\[\nx = 18 \\quad \\text{or} \\quad x = -8 \\quad (\\text{which is not feasible and thus discarded})\n\\]\n\nTherefore, the height of the large tree before it broke is \\( 18 \\mathrm{~m} \\).\n\nHence, the correct choice is B.\n\n**Key Insight**: This problem tests the application of the Pythagorean theorem. Mastering the Pythagorean theorem is crucial for solving such problems." }, { "problem_id": 973, "question": "As shown in the figure, there are two congruent triangles. The measure of $\\angle 1$ is ( ).\n\n\n\n$b$\n\n\nA. $48^{\\circ}$\nB. $60^{\\circ}$\nC. $62^{\\circ}$\nD. $72^{\\circ}$", "input_image": [ "batch10-2024_06_14_234bba88fa4a56aa2b46g_0049_1.jpg", "batch10-2024_06_14_234bba88fa4a56aa2b46g_0049_2.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "According to the triangle angle sum theorem, we have $\\angle 2 = 180^{\\circ} - 48^{\\circ} - 60^{\\circ} = 72^{\\circ}$.\n\nSince the two triangles are congruent,\n\nit follows that $\\angle 1 = \\angle 2 = 72^{\\circ}$,\n\nTherefore, the correct choice is: D.\n\n\n\n$b$\n\n【Insight】This question primarily tests the properties of congruent triangles, with the key point being that corresponding angles of congruent triangles are equal." }, { "problem_id": 974, "question": "Among the four figures below, which one cannot be obtained by translating basic shapes?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_00a109ed8135f00b577eg_0012_1.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0012_2.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0012_3.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0012_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Can be obtained through translation, does not meet the question's requirement;\n\nB. Cannot be obtained through translation, requires rotation of one of the quadrilaterals, meets the question's requirement;\n\nC. Can be obtained through translation of one of the triangles, does not meet the question's requirement;\n\nD. Can be obtained through translation of one of the circles, does not meet the question's requirement;\n\nTherefore, the answer is B.\n\n[Key Insight] This question primarily examines the translation of shapes. The key to solving it lies in understanding that translation only changes the position of a shape, without altering its form, size, or orientation. A common mistake is confusing translation with rotation or flipping of shapes." }, { "problem_id": 975, "question": "The relationship between distance $s$, speed $v$, and time $t$ is given by $s = v t$. When one of these quantities is constant, the graph of the relationship between the other two variables cannot be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0033_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0033_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0033_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0033_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "When $v$ is a constant, $s$ is a direct proportional function of $t$, making $A$ possible but not matching the question's requirement; when $t$ is a constant, $s$ is a direct proportional function of $v$, making $B$ possible but not matching the question's requirement; when $t$ is a constant, $v$ is a direct proportional function of $s$, making $D$ impossible, which matches the question's requirement.\n\nTherefore, choose $D$.\n\n[Highlight] This question tests the understanding of the graphs of inverse proportional functions and direct proportional functions. Mastering the fundamental properties of various types of functions is key to solving the problem." }, { "problem_id": 976, "question": "The net of a cube is shown in the figure. The cube that can be formed using this net is only ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_877c2117fdc429fa1ae4g_0026_1.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0026_2.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0026_3.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0026_4.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0026_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the fact that all three line segments have one end pointing to point $O$, it can be determined that options $\\mathrm{B}$, $\\mathrm{C}$, and $\\mathrm{D}$ are all incorrect, and only option $\\mathrm{A}$ is correct. Therefore, the answer is: A.\n\n【Key Point】This question is a basic application problem. Students only need to be proficient in the side expansion diagram of a cube to solve it." }, { "problem_id": 977, "question": "Which of the following figures are axisymmetric?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0098_1.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0098_2.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0098_3.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0098_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: A. It is an axisymmetric figure, so this option is correct;\n\nB. It is not an axisymmetric figure, so this option is incorrect;\n\nC. It is not an axisymmetric figure, so this option is incorrect;\n\nD. It is not an axisymmetric figure, so this option is incorrect.\n\nTherefore, the correct choice is: A.\n\n[Key Point] This question tests the concept of axisymmetric figures: The key to identifying an axisymmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide when folded along the axis of symmetry." }, { "problem_id": 978, "question": "Given that $a \\neq 0$, the approximate graphs of the functions $y = \\frac{a}{x}$ and $y = -a x^{2} - a$ in the same Cartesian coordinate system could be\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_84512915d0d2c440e574g_0054_1.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0054_2.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0054_3.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0054_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When \\( a > 0 \\), the graph of the function \\( y = \\frac{a}{x} \\) is located in the first and third quadrants, and the graph of \\( y = -a x^{2} - a \\) opens downward, intersecting the negative half of the \\( y \\)-axis. Option D fits this description.\n\nWhen \\( a < 0 \\), the graph of the function \\( y = \\frac{a}{x} \\) is located in the second and fourth quadrants, and the graph of \\( y = -a x^{2} - a \\) opens upward, intersecting the positive half of the \\( y \\)-axis. However, there is no option that matches this scenario.\n\nTherefore, the correct choice is: D.\n\n[Key Insight] This question tests knowledge of the graphs of inverse proportional functions and quadratic functions. The key to solving it lies in determining the position of the graphs based on the sign of the proportional coefficient, which is not difficult." }, { "problem_id": 979, "question": "As shown in the figure, two identical right-angled triangles overlap along one side, forming four shapes. Which of the following is not an axisymmetric figure?\nA.\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0044_1.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0044_2.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0044_3.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0044_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Option A is an axisymmetric figure, so option A does not meet the requirement of the question;\n\nOption B is an axisymmetric figure, so option B does not meet the requirement of the question;\n\nOption C is an axisymmetric figure, so option C does not meet the requirement of the question;\n\nOption D is not an axisymmetric figure, so option D meets the requirement of the question.\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question tests the definition of an axisymmetric figure. The key to solving the problem is to be able to determine whether a figure is axisymmetric based on its definition." }, { "problem_id": 980, "question": "Phenomena in daily life that can be explained by the stability of triangles are ( )\nA. \n\nB. \n\nC. \n\nD. \n", "input_image": [ "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0040_1.jpg", "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0040_2.jpg", "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0040_3.jpg", "batch5-2024_06_14_79bcbdf4925b8d3b19cfg_0040_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Utilizes the principle that \"two points determine a straight line,\" hence this option does not meet the requirement of the question;\n\nB. Utilizes the principle that \"the shortest distance between two points is a straight line,\" hence this option does not meet the requirement of the question;\n\nC. The window bracket is triangular, utilizing the principle of \"the stability of a triangle,\" hence this option meets the requirement of the question;\n\nD. Utilizes the principle that \"the perpendicular line segment is the shortest,\" hence this option does not meet the requirement of the question;\n\nTherefore, the correct choice is: C.\n\n[Key Insight] This question examines the application of the principles that \"two points determine a straight line,\" \"the shortest distance between two points is a straight line,\" \"the stability of a triangle,\" and \"the perpendicular line segment is the shortest.\" Accurately identifying the knowledge used based on the context and the diagram is crucial for solving this problem." }, { "problem_id": 981, "question": "As shown in the figure, which of the following geometric solids is formed by rotating the given triangle about its longest edge for a full revolution?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_329d65dd392595b77024g_0073_1.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0073_2.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0073_3.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0073_4.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0073_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Based on the given geometric figure, the triangle is rotated around its longest side for one full rotation to obtain:\n\n\nTherefore, the correct choice is: D.\n\n【Key Point】This question examines points, lines, surfaces, and solids, which are fundamental concepts. It is relatively easy and tests students' spatial learning ability. Mastering the relevant knowledge is crucial for solving the problem." }, { "problem_id": 982, "question": "A geometric solid is placed horizontally as shown, its top view is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_329d65dd392595b77024g_0071_1.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0071_2.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0071_3.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0071_4.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0071_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: Based on the characteristics of the top view of the geometric solid, the visible part is the top surface.\n\nTherefore, the answer is: A.\n\n[Key Point] This question primarily examines the three views of solid geometry. Understanding the features of each view is crucial for solving the problem." }, { "problem_id": 983, "question": "Symmetry is a manifestation of the ancient Chinese concept of harmony and balance. Among the main patterns of the university emblems below, which one is an axisymmetric figure?\nA.\n\n\nB.\n\n\nPeking University\nC.\n\n\nRenmin University of China\nD.\n\n\n##", "input_image": [ "batch10-2024_06_14_2434588caaee122ec85bg_0089_1.jpg", "batch10-2024_06_14_2434588caaee122ec85bg_0089_2.jpg", "batch10-2024_06_14_2434588caaee122ec85bg_0089_3.jpg", "batch10-2024_06_14_2434588caaee122ec85bg_0089_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \nA. It is not an axisymmetric figure, so option A does not meet the requirement; \nB. It is an axisymmetric figure, so option B meets the requirement; \nC. It is not an axisymmetric figure, so option C does not meet the requirement; \nD. It is not an axisymmetric figure, so option D does not meet the requirement; \n\nTherefore, the correct answer is: B. \n\n**[Key Point]** This question tests the concept of axisymmetric figures. The key to identifying an axisymmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide when folded along the axis." }, { "problem_id": 984, "question": "From the following four square cardboard sheets, which one, after cutting out the shaded part, can be folded along the dotted lines to form an enclosed rectangular box?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_3a73c762f5eb2b761854g_0040_1.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0040_2.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0040_3.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0040_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "A. After cutting away the shaded part, it cannot form a rectangular prism, so this option does not meet the requirement;\n\nB. After cutting away the shaded part, it forms a lidless rectangular prism, so this option does not meet the requirement;\n\nC. After cutting away the shaded part, it can form a rectangular prism, so this option meets the requirement;\n\nD. After cutting away the shaded part, it cannot form a rectangular prism, so this option does not meet the requirement.\n\nTherefore, the correct choice is: C\n\n【Highlight】This question mainly tests the ability to fold a net into a geometric shape, cultivating students' spatial imagination." }, { "problem_id": 985, "question": "The graphs of the functions $y = -kx$ and $y = \\frac{\\sqrt{-k}}{x} \\quad (k < 0)$ are approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\nPage 5 of 90", "input_image": [ "batch13-2024_06_15_b7aa0f088ba806ec834eg_0007_1.jpg", "batch13-2024_06_15_b7aa0f088ba806ec834eg_0007_2.jpg", "batch13-2024_06_15_b7aa0f088ba806ec834eg_0007_3.jpg", "batch13-2024_06_15_b7aa0f088ba806ec834eg_0007_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since \\( k < 0 \\),\n\nthe graph of the inverse proportional function \\( y = \\frac{\\sqrt{-k}}{x} \\) is located in the first and third quadrants, and the graph of the direct proportional function \\( y = -k x \\) passes through the first and third quadrants;\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question examines the graphs of inverse proportional and direct proportional functions, as well as the meaning of the square root. Understanding the properties of direct proportional and inverse proportional functions is crucial for solving the problem." }, { "problem_id": 986, "question": "Among the following options (where all triangles in the figures are right triangles), which one cannot be used to verify the Pythagorean theorem?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch12-2024_06_15_ed1911a30eded62729c0g_0070_1.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0070_2.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0070_3.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0070_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "A, B, and C can all use the area of the figure to derive the relationship between $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$, thereby proving the Pythagorean theorem; therefore, options $\\mathrm{A}$, $\\mathrm{B}$, and $\\mathrm{C}$ do not meet the requirements of the question;\n\nD. The Pythagorean theorem cannot be proven using the area of the figure, hence this option is correct.\n\nTherefore, choose D.\n\n【Highlight】This question mainly tests the proof method of the Pythagorean theorem, and deriving it based on the area of the figure is the key to solving the problem." }, { "problem_id": 987, "question": "The following figures are axisymmetric figures ( ).\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0022_1.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0022_2.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0022_3.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0022_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Option A, B, and C cannot find such a straight line that allows the figure to fold along it, with the parts on both sides of the line coinciding with each other, hence they are not axisymmetric figures;\n\nOption D can find such a straight line that allows the figure to fold along it, with the parts on both sides of the line coinciding with each other, hence it is an axisymmetric figure;\n\nTherefore, the answer is: D.\n\n[Key Insight] According to the definition, if a figure can be folded along a straight line and the parts on both sides of the line coincide with each other, the figure is called an axisymmetric figure, and this line is called the axis of symmetry. This analysis is based on this principle." }, { "problem_id": 988, "question": "In the following figures, cut the rectangle along the dotted line into two parts. Which of the following can be used to form both a parallelogram, a triangle, and a trapezoid?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0030_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0030_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0030_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0030_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** Use imagination to piece together the puzzle or derive the answer through practical operation.\n\n**Solution:** The first figure can only be assembled into a special parallelogram, specifically a rectangle.\n\nThe second figure can be assembled into a parallelogram, a rectangle, and a triangle.\n\nThe third figure can be assembled into a parallelogram and a trapezoid.\n\nThe fourth figure, by overlapping different equal sides, can form a parallelogram, and can also be assembled into a triangle and a trapezoid.\n\nTherefore, the correct answer is D.\n\n**Commentary:** This question primarily tests students' hands-on skills and spatial imagination, specifically the ability to visualize the shape formed by piecing together the cut-out figure with the remaining figure." }, { "problem_id": 989, "question": "Which of the following figures can be folded to form a triangular prism?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_0ce863b43d59865f0f74g_0018_1.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0018_2.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0018_3.jpg", "batch33-2024_06_14_0ce863b43d59865f0f74g_0018_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "By analyzing and judging each option based on the characteristics of the triangular prism and its surface development diagram, we can derive the following:\n\nA. One base is a right-angled triangle, and the other is an equilateral triangle; the two base triangles are not congruent, so this option is incorrect.\n\nB. After folding, the two side faces overlap and cannot form a triangular prism, so this option is incorrect.\n\nC. After folding, it can form a triangular prism, so this option is correct.\n\nD. After folding, the two side faces overlap and cannot form a triangular prism, so this option is incorrect.\n\nTherefore, the correct choice is C." }, { "problem_id": 990, "question": "Among the following geometric solids, the one whose left view is different from the front view is ( )\nA.\n\n\nB.\n\n\n\nC.\n\n\n\nD.\n\n\n\n##", "input_image": [ "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0009_1.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0009_2.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0009_3.jpg", "batch25-2024_06_17_dfc3bd0f0effc7e95b2ag_0009_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Option A: The front view is a rectangle, and the left view is also a rectangle, so A does not meet the requirement;\n\nOption B: The front view is a square, and the left view is also a square, so B does not meet the requirement;\n\nOption C: The front view is a trapezoid, and the left view is a rectangle, so C meets the requirement;\n\nOption D: The front view is a triangle, and the left view is also a triangle, so D does not meet the requirement;\n\nTherefore, the correct choice is: C.\n\n[Highlight] This question tests the understanding of the three views of simple geometric shapes. The key to solving the problem is to analyze and determine the three views based on the geometric shape." }, { "problem_id": 991, "question": "Xiaoxin folded a piece of paper (as shown in the figure) into a cube-shaped box and placed a bottle of ink inside. The box was mixed with others on the table. By observing only, choose the box where the ink is located ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0085_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0085_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0085_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0085_4.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0085_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Logic", "image_relavance": "1", "analysis": "Solution: Based on the characteristics and positions of the various symbols in the unfolded diagram, it can be determined that the ink is inside box $B$.\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question tests the understanding of the surface unfolding of a cube and spatial imagination skills. Common pitfalls and confusions: Students may inaccurately imagine the positions of the related diagrams, leading to incorrect choices. To solve such problems, it is advisable to physically manipulate the objects, which can help clarify the situation." }, { "problem_id": 992, "question": "The graphs of the functions $y = \\frac{k}{x}$ and $y = ax^2 - bx + c$ are shown in the figure. The approximate graph of the function $y = kx + b$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_c2b339e9e67c87712768g_0031_1.jpg", "batch13-2024_06_15_c2b339e9e67c87712768g_0031_2.jpg", "batch13-2024_06_15_c2b339e9e67c87712768g_0031_3.jpg", "batch13-2024_06_15_c2b339e9e67c87712768g_0031_4.jpg", "batch13-2024_06_15_c2b339e9e67c87712768g_0031_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From the graph of the inverse proportional function located in the first and third quadrants, we know that \\( k > 0 \\). From the graph of the quadratic function, we can see that \\( a < 0 \\) and \\( -b < 0 \\), which means \\( b > 0 \\).\n\nTherefore, the graph of the function \\( y = kx + b \\) roughly passes through the first, second, and third quadrants.\n\nHence, the correct choice is: D.\n\n[Key Insight] This question tests knowledge of function graphs. Mastering the graphs and properties of these three types of functions is crucial for solving the problem." }, { "problem_id": 993, "question": "Illustrate the statement \"the sum of acute angles $\\alpha$ and $\\beta$ is acute\" using the following diagrams, where the incorrect example is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_bcc10407bfba96c9fa8eg_0019_1.jpg", "batch10-2024_06_14_bcc10407bfba96c9fa8eg_0019_2.jpg", "batch10-2024_06_14_bcc10407bfba96c9fa8eg_0019_3.jpg", "batch10-2024_06_14_bcc10407bfba96c9fa8eg_0019_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Determine when the sum of two acute angles is not an acute angle. Only option $C$ satisfies $\\angle a + \\angle B > 90^{\\circ}$. Therefore, the statement that the sum of acute angle $a$ and acute angle $\\beta$ is an acute angle is false.\n\nHence, the correct choice is: $C$.\n\n【Key Point】This question primarily tests the judgment of the truth value of propositions. A correct proposition is called a true proposition, while an incorrect one is called a false proposition. The key to determining the truth value of a proposition lies in being familiar with the properties and theorems in the textbook." }, { "problem_id": 994, "question": "Among the four function graphs given below, the one where the function value $y$ decreases as the independent variable $x$ increases for $x > 0$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_58474443df6e195863d8g_0048_1.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0048_2.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0048_3.jpg", "batch13-2024_06_15_58474443df6e195863d8g_0048_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "$A$、When $x>0$, $y$ increases as $x$ increases, which is incorrect;\n\n$B$、When $x>0$, $y$ increases as $x$ increases, which is incorrect;\n\n$C$、When $x>0$, $y$ decreases as $x$ increases, which is correct;\n\n$D$、When $x>0$, $y$ first decreases and then increases as $x$ increases, which is incorrect;\n\nTherefore, the correct choice is: $C$.\n\n【Key Point】This question mainly tests the ability to judge the increasing or decreasing nature of a function based on its graph. Understanding the graph and properties of the function is crucial for solving the problem." }, { "problem_id": 995, "question": "The figure shows the net of a solid. Which of the following solids does the net represent?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_877c2117fdc429fa1ae4g_0095_1.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0095_2.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0095_3.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0095_4.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0095_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since the lateral faces of the pyramid's net are triangles and the base is a quadrilateral,\n\nTherefore, this three-dimensional figure is a pyramid;\n\nHence, the correct choice is: B.\n\n[Key Insight] This question examines the net of a geometric solid. Starting from a physical object and combining it with specific problems, it distinguishes the net of the geometric solid. By integrating the transformation between three-dimensional and two-dimensional shapes, it establishes spatial concepts, which is key to solving such problems." }, { "problem_id": 996, "question": "Given that the parabola $y = x^{2} - 2x + m + 1$ intersects the $x$-axis at two distinct points, the approximate graph of the function $y = \\frac{m}{x}$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0043_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0043_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0043_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0043_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: The parabola \\( y = x^{2} - 2x + m + 1 \\) intersects the \\( x \\)-axis at two distinct points.\n\n\\[\n\\therefore \\Delta = (-2)^{2} - 4(m + 1) > 0\n\\]\n\nSolving the inequality yields \\( m < 0 \\).\n\n\\[\n\\therefore \\text{The graph of the function } y = \\frac{m}{x} \\text{ lies in the second and fourth quadrants.}\n\\]\n\nHence, the correct choice is \\( D \\).\n\n【Key Insight】This problem examines the graph of an inverse proportional function. First, determine the value of \\( m \\), and then ascertain the position of the function's graph." }, { "problem_id": 997, "question": "Among the following figures, which one has the most symmetry axes?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_4ce737a200cf8fbade32g_0030_1.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0030_2.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0030_3.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0030_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "A. 3 axes of symmetry;\n\nB. 4 axes of symmetry;\n\nC. 2 axes of symmetry;\n\nD. 1 axis of symmetry;\n\nTherefore, the correct choice is: B.\n\n【Highlight】This question tests the understanding of the concept of axial symmetry and its practical application. The key to solving the problem lies in mastering the concept of axes of symmetry." }, { "problem_id": 998, "question": "The following figure that is not a central symmetric figure is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0068_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0068_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0068_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0068_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "A is a centrally symmetric figure. Therefore, it cannot be chosen;\n\nB is a centrally symmetric figure. Therefore, it cannot be chosen;\n\nC is a centrally symmetric figure. Therefore, it cannot be chosen;\n\nD is not a centrally symmetric figure. Therefore, it can be chosen.\n\nHence, choose D.\n\n[Key Point] This question examines the concept of a centrally symmetric figure: a centrally symmetric figure is one that looks for a center of symmetry and, after a 180-degree rotation, coincides with the original figure." }, { "problem_id": 999, "question": "On either side of a cable laid along the straight line $M N$ are two neighborhoods, a and b. The task is to select a point $P$ on $M N$ and lay cables to both neighborhoods. Among the following four laying plans, which uses the least amount of cable material?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch34-2024_06_17_a1af3a6068f73094683bg_0003_1.jpg", "batch34-2024_06_17_a1af3a6068f73094683bg_0003_2.jpg", "batch34-2024_06_17_a1af3a6068f73094683bg_0003_3.jpg", "batch34-2024_06_17_a1af3a6068f73094683bg_0003_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Since points A and B are located on opposite sides of the line \\( MN \\),\n\ntherefore, according to the principle that the shortest distance between two points is a straight line, connecting points A and B will intersect the line \\( MN \\) at point \\( P \\), which is the desired point; hence, the correct choice is: A.\n\n[Key Insight] This question tests the axiom that the shortest distance between two points is a straight line. The key to solving the problem lies in analyzing whether the two points are on the same side or opposite sides of the line. If they are on opposite sides, simply connect the two points; if they are on the same side, create a symmetrical point of one of the points and then connect them." }, { "problem_id": 1000, "question": "The plane figure in the diagram is rotated around the straight line $l$ for a full revolution, which of the following solid figures can be obtained?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_da259e8c8bb047f1aad6g_0008_1.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0008_2.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0008_3.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0008_4.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0008_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "As shown in the figure, the given plane figure is a rectangle.\n\nWhen a rectangle is rotated around a straight line along one of its sides, it forms a cylinder.\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question examines the concepts of points, lines, surfaces, and solids. Memorizing the various common plane figures and the three-dimensional shapes they form upon rotation is crucial for solving such problems." }, { "problem_id": 1001, "question": "The congruent triangles in the figure below are $(\\quad)$\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1) and (2)\nB. (2) and (3)\nC. (2) and (4)\nD. (1) and (3)", "input_image": [ "batch9-2024_05_23_2658e5f87f37161a4840g_0093_1.jpg", "batch9-2024_05_23_2658e5f87f37161a4840g_0093_2.jpg", "batch9-2024_05_23_2658e5f87f37161a4840g_0093_3.jpg", "batch9-2024_05_23_2658e5f87f37161a4840g_0093_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\n$A$: (1) and (2) only have one side and one angle correspondingly equal, which is insufficient to prove congruence. Therefore, this option is incorrect and does not meet the requirement.\n\n$B$: (2) and (3) only have one side and one angle correspondingly equal, which is insufficient to prove congruence. Therefore, this option is incorrect and does not meet the requirement.\n\nC: (2) and (4) only have one side and one angle correspondingly equal, which is insufficient to prove congruence. Therefore, this option is incorrect and does not meet the requirement.\n\nD: (1) and (3) have two sides and their included angle correspondingly equal, which is sufficient to prove congruence. Therefore, this option is correct and meets the requirement.\n\nThe correct choice is: D\n\n【Key Point】This question primarily tests the congruence theorem of triangles. It is crucial to understand that two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding parts of the other triangle." }, { "problem_id": 1002, "question": "Among the four figures below, which figure can represent the same angle using all three methods of $\\angle 1, \\angle \\mathrm{ABC}, \\angle \\mathrm{B}$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_d6e7887b4e796c8b833cg_0073_1.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0073_2.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0073_3.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0073_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The figure that can represent the same angle using all three notations $\\angle 1$, $\\angle \\mathrm{ABC}$, and $\\angle \\mathrm{B}$ simultaneously can only be $\\mathrm{B}$. Therefore, the correct choice is B.\n\n[Key Insight] This question tests the understanding of angle notation." }, { "problem_id": 1003, "question": "The graph of the quadratic function $y = kx^2 - x (k < 0)$ is approximately ( )\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n", "input_image": [ "batch8-2024_06_14_713c9e665752c5e7543ag_0087_1.jpg", "batch8-2024_06_14_713c9e665752c5e7543ag_0087_2.jpg", "batch8-2024_06_14_713c9e665752c5e7543ag_0087_3.jpg", "batch8-2024_06_14_713c9e665752c5e7543ag_0087_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem, since \\( k < 0 \\), the graph of the quadratic function opens downward. When \\( x = 0 \\), \\( y = 0 \\), indicating that the graph passes through the origin.\n\nAnalyzing the options, only option C satisfies these conditions.\n\nTherefore, the answer is: C.\n\n[Key Insight] The problem primarily tests the understanding of the graph of a quadratic function. The key to solving such problems lies in mastering the relationship between the coefficients and the properties of the quadratic function, such as the direction of opening, the axis of symmetry, and the vertex coordinates." }, { "problem_id": 1004, "question": "The area of a triangle is $15 \\mathrm{~cm}^{2}$. The graph of the relationship between the height $y \\mathrm{~cm}$ of the triangle on the base and the base $x \\mathrm{~cm}$ is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch12-2024_06_15_bb3935bd5a335bd5cf98g_0010_1.jpg", "batch12-2024_06_15_bb3935bd5a335bd5cf98g_0010_2.jpg", "batch12-2024_06_15_bb3935bd5a335bd5cf98g_0010_3.jpg", "batch12-2024_06_15_bb3935bd5a335bd5cf98g_0010_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Since \\( x y = 30 \\),\n\nTherefore, \\( y = \\frac{30}{x} \\) (where \\( x > 0 \\) and \\( y > 0 \\)).\n\nHence, the correct choice is D.\n\n[Key Point] This question tests the understanding of the graph of an inverse proportional function. Determining the range of the independent variable based on the actual situation is crucial." }, { "problem_id": 1005, "question": "Among the following four figures, the congruent figures are ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1) and (2)\nB. (1) and (3)\nC. (2) and (3)\nD. (3) and (4)", "input_image": [ "batch9-2024_05_23_fba44ff85c69ee94778dg_0058_1.jpg", "batch9-2024_05_23_fba44ff85c69ee94778dg_0058_2.jpg", "batch9-2024_05_23_fba44ff85c69ee94778dg_0058_3.jpg", "batch9-2024_05_23_fba44ff85c69ee94778dg_0058_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: According to the concept of congruent figures, two figures that can completely coincide are called congruent figures. Thus, the answer can be derived. Solution: Figures (3) and (4) can completely coincide, therefore the congruent figures are (3) and (4).\n\nHence, choose D.\n\nComment: This question mainly tests the understanding of congruent figures, with the key point being to grasp the concept of congruent figures." }, { "problem_id": 1006, "question": "Among the following geometric solids, which one has the fewest number of faces?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0081_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0081_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0081_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0081_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. A rectangular prism has 6 faces;\n\nB. A cone has one curved surface and one base, totaling 2 faces;\n\nC. A triangular prism has 5 faces;\n\nD. A cylinder has one lateral surface and two bases, totaling 3 faces;\n\nTherefore, the correct answer is: B.\n\n【Key Point】This question tests the concept of three-dimensional shapes. The key to solving this problem is to visually identify and count all the faces of the geometric shapes. It is a basic and relatively simple question." }, { "problem_id": 1007, "question": "If the graphs of the functions $y = \\frac{k}{x}$ and $y = ax^2 + bx + c$ are as shown in the figure, then the approximate graph of the function $y = ax + kc$ is ( )\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_5d227d10b18c15d49d57g_0016_1.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0016_2.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0016_3.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0016_4.jpg", "batch13-2024_06_15_5d227d10b18c15d49d57g_0016_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Given that the graph of the inverse proportional function \\( y = \\frac{k}{x} \\) is located in the second and fourth quadrants, it follows that \\( k < 0 \\).\n\nFrom the graph of the quadratic function \\( y = a x^{2} + b x + c \\), we can deduce that \\( a > 0 \\) and \\( c > 0 \\).\n\nTherefore, \\( k c < 0 \\), and the general graph of the function \\( y = a x + k c \\) passes through the first, third, and fourth quadrants.\n\nHence, the correct choice is: D.\n\n[Key Insight] This question tests knowledge of function graphs. A thorough understanding of the graphs and properties of inverse proportional, linear, and quadratic functions is crucial for solving the problem." }, { "problem_id": 1008, "question": "Among the graphs in the Cartesian coordinate system below, which one does not represent $y$ as a function of $x$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_c61eb1081c69492cc361g_0089_1.jpg", "batch13-2024_06_15_c61eb1081c69492cc361g_0089_2.jpg", "batch13-2024_06_15_c61eb1081c69492cc361g_0089_3.jpg", "batch13-2024_06_15_c61eb1081c69492cc361g_0089_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Question Analysis: Definition of a Function: For any value of \\( x \\), there is a uniquely determined value of \\( y \\) that corresponds to it. In graph \\( B \\), for any value of \\( x \\), there are two corresponding values. Therefore, \\( B \\) is not a function. The other graphs have a uniquely determined value of \\( y \\) corresponding to each value of \\( x \\).\n\nThus, the correct choice is B.\n\nKey Point: Definition of a Function\n\nComment: This question is a basic application problem, requiring students to be thoroughly familiar with the definition of a function to complete it." }, { "problem_id": 1009, "question": "Xiaoxin plans to make a gift box using the paper pieces shown in the figure. To make it visually appealing, he wants to draw patterns on the six square pieces so that the patterns on three pairs of opposite faces are the same. Which of the following patterns, after being drawn, is correct?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_35cb0aebc81696508813g_0090_1.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0090_2.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0090_3.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0090_4.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0090_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In the planar expansion diagram of a cube, the opposite faces must be separated by one square to ensure that the patterns on the three pairs of opposite faces are identical after assembly. The correct choice should be C.\n\nTherefore, the answer is C.\n\n[Key Point] This question tests the characteristics of the surface expansion diagram of a cube. The key to solving the problem lies in paying attention to the spatial figure of the cube, starting from the opposite faces, and analyzing and answering the question accordingly." }, { "problem_id": 1010, "question": "Among the four geometric figures below, which one has a rectangular top view?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0059_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0059_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0059_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0059_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: The view seen from above is called the top view. \nA. The top view of a sphere is a circle, so this option is incorrect; \nB. The top view of a cylinder lying flat on the ground is a rectangle, so this option is correct; \nC. The top view of a cone is a circle with a center point, so this option is incorrect; \nD. The top view of a truncated cone is two concentric circles, so this option is incorrect. \nTherefore, the correct answer is B.\n\nTest Point: Three views of simple geometric shapes." }, { "problem_id": 1011, "question": "Among the following objects, which can be abstracted into a cylinder?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0031_1.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0031_2.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0031_3.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0031_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. The abstracted shape is a hexagonal prism, which does not match the requirement;\n\nB. The abstracted shape is a sphere, which does not match the requirement;\n\nC. The abstracted shape is a cylinder, which does not match the requirement;\n\nD. The abstracted shape is a cone, which matches the requirement.\n\nTherefore, the correct choice is D.\n\n[Key Point] This question tests the recognition of a cone. Correctly identifying the image is crucial to solving this problem." }, { "problem_id": 1012, "question": "Among the following figures, which one can determine that $\\angle 1 > \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_fd61d246744e8785fd79g_0014_1.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0014_2.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0014_3.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0014_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "A、 $\\angle 1=\\angle 2$, therefore this option is incorrect;\n\nB、 $\\angle 1=\\angle 2$, therefore this option is incorrect;\n\nC、 $\\angle 1>\\angle 2$, therefore this option is correct;\n\nD、 $\\angle 1=\\angle 2$, therefore this option is incorrect.\n\nTherefore, the correct choice is: C.\n\n【Key Point】This question examines the property that an exterior angle of a triangle is greater than any of its non-adjacent interior angles, the properties of parallel lines, and the properties of right-angled triangles. Familiarity with these properties and accurate interpretation of the diagram are crucial for solving the problem." }, { "problem_id": 1013, "question": "When a plane cuts through a cone, which of the following shapes cannot be the cross-section?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_452fbe550559079ef90fg_0031_1.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0031_2.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0031_3.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0031_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Question Analysis: Determine based on the shape characteristics of a cone, or use the elimination method.\n\nSolution: If a plane intersects a cone and passes through the vertex of the cone, the resulting cross-section is an isosceles triangle. If the plane does not pass through the vertex and is parallel to the base, the cross-section is a circle. If the plane is not parallel to the base, the cross-section is an ellipse or a combination of a parabola and a line segment. Therefore, it cannot be a right angle.\n\nTherefore, the correct choice is: C.\n\nComment: This question mainly examines the concept of intersecting a geometric body. The shape of the cross-section is related not only to the geometric body being intersected but also to the angle and direction of the intersecting plane. For such questions, it is best to combine hands-on practice with mental analysis, personally performing the actions to learn and summarize the thought processes involved." }, { "problem_id": 1014, "question": "Which of the following plane figures, when rotated around an axis, can form the solid figure shown in the diagram?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_89c51f0358ecdde09f0fg_0088_1.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0088_2.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0088_3.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0088_4.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0088_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "A. Rotating once around the axis does not produce the three-dimensional figure shown in the diagram, so it does not meet the requirement;\n\nB. Rotating once around the axis does not produce the three-dimensional figure shown in the diagram, so it does not meet the requirement;\n\nC. Rotating once around the axis does not produce the three-dimensional figure shown in the diagram, so it does not meet the requirement;\n\nD. Rotating once around the axis produces the three-dimensional figure shown in the diagram, so it meets the requirement;\n\nTherefore, the correct choice is: D.\n\n【Key Point】This question mainly examines the relationship between points, lines, surfaces, and solids. The key to solving the problem is understanding that the movement of surfaces forms solids. The motion of points, lines, surfaces, and solids composes the diverse and colorful world of geometric figures." }, { "problem_id": 1015, "question": "The angles $\\angle 1$ and $\\angle 2$ are complementary in which of the following figures?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_fd61d246744e8785fd79g_0043_1.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0043_2.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0043_3.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0043_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "According to the definition of complementary angles, if the sum of two angles is $90^{\\circ}$, then these two angles are complementary.\n\nIn option D, the sum of $\\angle 1$ and $\\angle 2$ is $90^{\\circ}$, making them complementary angles.\n\nTherefore, the correct choice is D.\n\n【Key Point】This question tests the understanding of the definitions of complementary and supplementary angles. Based on the definition of complementary angles, remember that the sum of two angles is $90^{\\circ}$, regardless of their positions." }, { "problem_id": 1016, "question": "The area of the square represented by the letters in the figure is 175. The options are ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch12-2024_06_15_395683fc667a063e6e86g_0067_1.jpg", "batch12-2024_06_15_395683fc667a063e6e86g_0067_2.jpg", "batch12-2024_06_15_395683fc667a063e6e86g_0067_3.jpg", "batch12-2024_06_15_395683fc667a063e6e86g_0067_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "0", "analysis": "Solution: According to the Pythagorean theorem, in a right-angled triangle, the sum of the squares of the two perpendicular sides is equal to the square of the hypotenuse.\n\nA. The area of the square represented by $A$ is $400 - 225 = 175$;\n\nB. The area of the square represented by $B$ is $400 + 225 = 625$;\n\nC. The area of the square represented by $C$ is $256 - 112 = 144$;\n\nD. The area of the square represented by $D$ is $400 - 120 = 280$.\n\nTherefore, the correct answer is: A.\n\n[Key Insight] This question tests the understanding of the Pythagorean theorem. Carefully observing the characteristics of the figures provided in the options and applying the Pythagorean theorem is crucial for solving the problem." }, { "problem_id": 1017, "question": "Among the following options, which left-hand plane figures can be folded into the right-hand closed solid figures?\nA.\n\n\nB.\n\n\nC.\n\n\n\n D.\n\n", "input_image": [ "batch10-2024_06_14_fd61d246744e8785fd79g_0019_1.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0019_2.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0019_3.jpg", "batch10-2024_06_14_fd61d246744e8785fd79g_0019_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Option A is incorrect because the net is missing one base;\n\nOption B is correct;\n\nOption C is incorrect because the net has an extra base;\n\nOption D is incorrect because such an arrangement of the net cannot be folded into a cube.\n\nTherefore, the correct choice is: B.\n\n【Key Point】This question tests the understanding of nets of three-dimensional geometric shapes. The key to solving it lies in being familiar with the nets of some common three-dimensional geometric figures." }, { "problem_id": 1018, "question": "The number of prisms among the following geometric solids is ( )\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\n\n\n\n(5)\nA. 4\nB. 2\nC. 3\nD. 1", "input_image": [ "batch10-2024_06_14_877c2117fdc429fa1ae4g_0035_1.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0035_2.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0035_3.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0035_4.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0035_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "According to the definition of a prism, we can determine that: (1) is a quadrangular prism, (3) is a triangular prism, and the rest are not prisms. Therefore, there are 2 prisms.\n\nHence, the answer is: B\n[Key Insight] This question tests the recognition of geometric shapes. The key to solving the problem is to determine whether a geometric shape is a prism based on the definition of a prism." }, { "problem_id": 1019, "question": "In class, Mr. Wang asked students to design a figure to prove the Pythagorean Theorem. After discussion, the students came up with two figures. Which of the following figures can be used to prove the Pythagorean Theorem?\n\n\n\n(1)\n\n\n\n(2)\nA. (1) works, (2) doesn't work\nB. (1) doesn't work, (2) works\nC. (1) and (2) both work\nD. (1) and (2) neither work", "input_image": [ "batch12-2024_06_15_ed1911a30eded62729c0g_0028_1.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0028_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From Figure (1), we can derive,\n\n$$\n(a+b)^{2}=\\frac{1}{2} a b \\times 4+c^{2} \\text {, }\n$$\n\nSimplifying, we obtain: $a^{2}+b^{2}=c^{2}$,\n\nThus, Figure (1) can prove the Pythagorean theorem;\n\nAccording to the conditions in Figure (2), the Pythagorean theorem cannot be proven;\n\nTherefore, the answer is: A.\n\n[Key Insight] This question tests the proof of the Pythagorean theorem. The key to solving this problem is to understand the question clearly and use the combination of numbers and shapes to find the solution." }, { "problem_id": 1020, "question": "The mascot of the Beijing Winter Olympics, Bing Dwen Dwen, as shown in the figure, is loved by people all over the world. Among the four figures below, which one can be obtained by translating the original figure?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_00a109ed8135f00b577eg_0072_1.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0072_2.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0072_3.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0072_4.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0072_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of \"translation,\" the figure obtained by translating the given diagram is\n\n\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question examines the phenomenon of translation in real life. The key to solving this problem lies in a thorough understanding of the definition of translation." }, { "problem_id": 1021, "question": "As shown in the figure, this diagram is the unfolded net of one of the following solid figures. Which of the following solid figures corresponds to this diagram?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_fa30bd0163a7a41365acg_0019_1.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0019_2.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0019_3.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0019_4.jpg", "batch33-2024_06_14_fa30bd0163a7a41365acg_0019_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the positions of the elliptical and linear faces in the figure and the analysis, it can be determined that options B, C, and D do not meet the criteria. Option A has adjacent faces with parallel lines, thus it is the correct choice.\n\nTherefore, the answer is: A.\n\n[Key Insight] This problem tests the understanding of geometric figure expansions. It requires converting operational activities into mental processes, simulating (imagining) the folding and flipping of paper in the mind, effectively assessing the student's spatial awareness." }, { "problem_id": 1022, "question": "Given an opaque cube with the numbers 1 to 6 written on its six faces, as shown in the three possible views depicted, what is the number opposite the number 5?\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\nA. 6\nB. 4\nC. 3\nD. 6 or 4 or 3", "input_image": [ "batch33-2024_06_14_8df77e094ea9ba12ba8cg_0036_1.jpg", "batch33-2024_06_14_8df77e094ea9ba12ba8cg_0036_2.jpg", "batch33-2024_06_14_8df77e094ea9ba12ba8cg_0036_3.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "From the problem statement, we deduce:\n\n1 is adjacent to 2, 4, 5, and 6, so 1 is opposite to 3.\n\n2 is adjacent to 5, 4, and 1, and not opposite to 3, so 2 is opposite to 6, which means 4 is opposite to 5.\n\nTherefore, the correct answer is: B.\n\n[Key Insight] This question tests the properties of a cube. The key to solving the problem lies in analyzing the figure and applying the properties of a cube to arrive at the answer." }, { "problem_id": 1023, "question": "The following figures are named after mathematicians. The number of figures that are axisymmetric is\n\n( )\n\n\n\nZhao Shuang's String Diagram\n\n\n\nKoch Snowflake\n\n\n\nDescartes' Heart Curve\n\n\n\nFibonacci Spiral\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch10-2024_06_14_1fc8b75729663205496ag_0037_1.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0037_2.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0037_3.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0037_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: The Zhao Shuang string diagram is not an axisymmetric figure;\n\nThe Koch curve is an axisymmetric figure;\n\nThe Cartesian heart-shaped curve is an axisymmetric figure;\n\nThe Fibonacci spiral is not an axisymmetric figure.\n\nTherefore, there are a total of two axisymmetric figures,\n\nHence, the correct choice is: B.\n\n[Key Point] This question primarily examines the concept of axisymmetric figures. The key to determining an axisymmetric figure lies in identifying the axis of symmetry, where the two parts of the figure can coincide upon folding." }, { "problem_id": 1024, "question": "Among the following figures, which one is not an axisymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_4ce737a200cf8fbade32g_0090_1.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0090_2.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0090_3.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0090_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: A. It is an axisymmetric figure, therefore this option does not meet the requirement;\n\nB. It is an axisymmetric figure, therefore this option does not meet the requirement;\n\nC. It is not an axisymmetric figure, therefore this option meets the requirement;\n\nD. It is an axisymmetric figure, therefore this option does not meet the requirement.\n\nHence, the correct choice is: C.\n\n[Key Point] This question tests the concept of axisymmetric figures. The key to identifying an axisymmetric figure lies in finding the axis of symmetry, where the two parts of the figure can coincide when folded along the axis." }, { "problem_id": 1025, "question": "Given that real numbers $x$ and $y$ are reciprocals of each other, the graph of the function $y$ as a function of $x$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0003_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0003_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0003_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0003_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since the real numbers \\( x \\) and \\( y \\) are reciprocals of each other,\n\n\\[\n\\therefore x y = 1,\n\\]\n\n\\[\n\\therefore \\text{the graph of } y \\text{ as a function of } x \\text{ is } D.\n\\]\n\nTherefore, the correct choice is: D.\n\n**Key Insight:** This question tests the understanding of the graph of an inverse proportional function. Grasping the definition of reciprocals is fundamental to solving the problem, and writing the function's equation based on the given information is crucial." }, { "problem_id": 1026, "question": "The graph of the inverse proportion function $y = -\\frac{5}{x}$ is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0023_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0023_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0023_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0023_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: From the inverse proportional function \\( y = -\\frac{5}{x} \\), we can derive that \\( k = -5 < 0 \\).\n\n\\(\\therefore\\) The graph lies in the second and fourth quadrants;\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question primarily tests the understanding of the graph of an inverse proportional function. Mastering the graph of such functions is crucial for solving the problem." }, { "problem_id": 1027, "question": "In the figures below, which pair of angles, $\\angle 1$ and $\\angle 2$, are complementary? ( )\n\n\nA\n\n\nB\n\n\nC\n\n\nD\nA. A\nB. B\nC. C\nD. D", "input_image": [ "batch10-2024_06_14_da259e8c8bb047f1aad6g_0031_1.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0031_2.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0031_3.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0031_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "A、 $\\angle 1+\\angle 2$ does not necessarily equal $180^{\\circ}$, so they are not necessarily supplementary angles, making this option incorrect;\n\nB、 $\\angle 1+\\angle 2=90^{\\circ}$, which means they are complementary angles, making this option incorrect;\n\nC、 $\\angle 1$ and $\\angle 2$ are vertical angles, which are not necessarily supplementary, making this option incorrect;\n\nD、 $\\angle 1$ and $\\angle 2$ are supplementary angles, making this option correct.\n\nTherefore, the correct choice is D.\n\n【Key Point】This question tests the understanding of complementary and supplementary angles. Mastering the concept of supplementary angles is crucial for solving the problem." }, { "problem_id": 1028, "question": "As shown in the figure are several common patterns of car wheels. The pattern that, after rotating $90^{\\circ}$ around the center, can overlap with the original pattern is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0042_1.jpg", "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0042_2.jpg", "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0042_3.jpg", "batch10-2024_06_14_8ffc82fd5e8f4c83e206g_0042_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. This figure can coincide with the original pattern when rotated by $60^{\\circ}$ or an integer multiple of $60^{\\circ}$;\n\nB. This figure can coincide with the original pattern when rotated by $45^{\\circ}$ or an integer multiple of $45^{\\circ}$;\n\nC. This figure can coincide with the original pattern when rotated by $72^{\\circ}$ or an integer multiple of $72^{\\circ}$;\n\nD. This figure can coincide with the original pattern when rotated by $36^{\\circ}$ or an integer multiple of $36^{\\circ}$;\n\nTherefore, the correct answer is: B.\n\n【Insight】This question tests the concept of rotation, and the key to solving it lies in understanding the concept of rotation." }, { "problem_id": 1029, "question": "The diagram that represents $\\angle A B C$ is\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_f60ce410a4eb703a8db6g_0060_1.jpg", "batch33-2024_06_14_f60ce410a4eb703a8db6g_0060_2.jpg", "batch33-2024_06_14_f60ce410a4eb703a8db6g_0060_3.jpg", "batch33-2024_06_14_f60ce410a4eb703a8db6g_0060_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Analysis: Based on the range of angles learned in middle school, option A can be excluded; by identifying that the vertex letter must be in the middle, find the angle with the vertex letter as $B$.\n\nDetailed Explanation: \nA. Angles in middle school refer to acute, right, and obtuse angles, so A is incorrect.\nB. The vertex of the angle is $C$, so B is incorrect.\nC. The vertex of the angle is $B$, so C is correct.\nD. The vertex of the angle is $A$, so D is incorrect.\n\nTherefore, the correct choice is C.\n\nKey Point: This question tests the representation methods of angles. The key to solving it is to remember the various ways angles can be represented: \n(1) Using three letters, with the middle letter representing the vertex and the other two letters representing points on each side of the angle; \n(2) Using a number to represent an angle; \n(3) Using a Greek letter to represent an angle." }, { "problem_id": 1030, "question": "In the same Cartesian coordinate system, the graphs of the quadratic function $y = -x^2 + m$ and the linear function $y = mx - 1 (m \\neq 0)$ may be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch7-2024_06_14_dd9b6a481087dff4e73bg_0050_1.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0050_2.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0050_3.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0050_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Question Analysis: Based on the properties of quadratic functions, the parabola \\( y = -x^2 + m \\) opens downward, so option B is incorrect. Among options A, C, and D, the vertex coordinates are all on the positive side of the y-axis, indicating that \\( m > 0 \\). Therefore, it can be concluded that the linear function \\( y = mx - 1 \\) passes through the first, third, and fourth quadrants. Hence, the correct answer is C." }, { "problem_id": 1031, "question": "On March 20, 2021, the latest archaeological discoveries at the Sanxingdui site once again captured the world's attention. Among the patterns of Sanxingdui cultural relics, which of the following is an axisymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0009_1.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0009_2.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0009_3.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0009_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: Options A, B, and D cannot find such a straight line where the figure can be folded along the line and the parts on both sides of the line can coincide with each other. Therefore, they are not axisymmetric figures.\n\nOption C can find such a straight line where the figure can be folded along the line and the parts on both sides of the line can coincide with each other. Therefore, it is an axisymmetric figure.\n\nHence, the answer is: C.\n\n[Key Insight] This question tests the identification of axisymmetric figures: in a plane, a figure that can be folded along a straight line with the parts on both sides of the line completely coinciding." }, { "problem_id": 1032, "question": "After cutting off a corner of the cube in Figure 1 and placing it in the position shown in Figure 2, the front view of the geometric solid in Figure 2 is ( )\n\n\n\nFigure 1\n\n\n\nFigure 2\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_493157da588ff6213245g_0068_1.jpg", "batch25-2024_06_17_493157da588ff6213245g_0068_2.jpg", "batch25-2024_06_17_493157da588ff6213245g_0068_3.jpg", "batch25-2024_06_17_493157da588ff6213245g_0068_4.jpg", "batch25-2024_06_17_493157da588ff6213245g_0068_5.jpg", "batch25-2024_06_17_493157da588ff6213245g_0068_6.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "**Analysis:** Determine based on the shape of the object and the concept of the front view.\n\n**Detailed Explanation:** When viewed from the front, it appears as an isosceles triangle with the height line represented by a dashed line. Upon observation, only option D matches this description. Therefore, the answer is D.\n\n**Key Point:** This question tests the understanding of the three views of a geometric solid: the front view (what you see from the front), the side view (what you see from the side), and the top view (what you see from above)." }, { "problem_id": 1033, "question": "Multiply the x-coordinates of each vertex of $\\triangle A B C$ by -1, and leave the y-coordinates unchanged. The resulting figure is one of the following options:\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0072_1.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0072_2.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0072_3.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0072_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: Since the abscissas (x-coordinates) of all vertices of triangle \\( \\triangle ABC \\) are multiplied by -1 while the ordinates (y-coordinates) remain unchanged,\n\nTherefore, the two triangles are symmetric about the y-axis,\n\nHence, the correct choice is: B.\n\n[Key Insight] This question examines the coordinate characteristics of points symmetric about the y-axis. Understanding the problem is crucial to solving it." }, { "problem_id": 1034, "question": "Among the following polygons, which one has an equal sum of interior angles and exterior angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_d5eba716f8dc5b323207g_0041_1.jpg", "batch13-2024_06_15_d5eba716f8dc5b323207g_0041_2.jpg", "batch13-2024_06_15_d5eba716f8dc5b323207g_0041_3.jpg", "batch13-2024_06_15_d5eba716f8dc5b323207g_0041_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: The sum of the exterior angles of a polygon is always equal to $360^{\\circ}$;\n\nA. The sum of the interior angles of a triangle is: $180^{\\circ}$, which does not meet the condition;\n\nB. The sum of the interior angles of a quadrilateral is: $360^{\\circ}$, which meets the condition;\n\nC. The sum of the interior angles of a pentagon is: $540^{\\circ}$, which does not meet the condition;\n\nD. The sum of the interior angles of a hexagon is: $720^{\\circ}$, which does not meet the condition;\n\nTherefore, the correct choice is: B\n\n[Key Insight] This question tests the knowledge of the sum of interior angles and the sum of exterior angles of polygons; mastering the formula for the sum of interior angles of a polygon is crucial for solving the problem." }, { "problem_id": 1035, "question": "Xiao Hong uses a right-angled triangle to check if certain workpiece's curved concave surfaces are semicircles. Among the following workpieces, the curved concave surface is definitely a semicircle for $(\\quad)$\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch19-2024_05_24_30e46a32398ce13ab826g_0039_1.jpg", "batch19-2024_05_24_30e46a32398ce13ab826g_0039_2.jpg", "batch19-2024_05_24_30e46a32398ce13ab826g_0039_3.jpg", "batch19-2024_05_24_30e46a32398ce13ab826g_0039_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Question Analysis: According to the fact that the chord subtended by a $90^{\\circ}$ inscribed angle is the diameter, the answer can be derived.\n\nSolution: Since the chord subtended by a $90^{\\circ}$ inscribed angle is the diameter, it is clear that option A is correct.\n\nTherefore, the answer is A.\n\nKey Concept: Inscribed Angle Theorem." }, { "problem_id": 1036, "question": "In the following diagrams, ray $\\mathrm{OA}$ represents a direction of South $32^{\\circ}$ East. Which one is correct ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_f60ce410a4eb703a8db6g_0096_1.jpg", "batch33-2024_06_14_f60ce410a4eb703a8db6g_0096_2.jpg", "batch33-2024_06_14_f60ce410a4eb703a8db6g_0096_3.jpg", "batch33-2024_06_14_f60ce410a4eb703a8db6g_0096_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Question Analysis: Analyze each option based on the definition of directional angles.\n\nSolution: \nA. $\\mathrm{OA}$ represents a bearing of 32° west of north, so this option is incorrect;\nB. OA represents a bearing of 32° west of south, so this option is incorrect;\nC. OA represents a bearing of 32° east of south, so this option is correct;\nD. $\\mathrm{OA}$ represents a bearing of 32° east of north, so this option is incorrect.\n\nTherefore, the correct choice is C.\n\nKey Point: Directional angles." }, { "problem_id": 1037, "question": "Given that $m$ is a non-zero constant, the graphs of the functions $y = m x$ and $y = m x^{2} - m^{2}$ in the same Cartesian coordinate system can be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0030_1.jpg", "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0030_2.jpg", "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0030_3.jpg", "batch7-2024_06_14_d4fdecbc0b558904fe1ag_0030_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When \\( m > 0 \\), the graph of \\( y = m x \\) is a straight line passing through the origin and the first and third quadrants. The graph of \\( y = m x^{2} - m^{2} \\) opens upwards, intersects the negative half of the \\( y \\)-axis, and has the \\( y \\)-axis as its axis of symmetry.\n\nWhen \\( m < 0 \\), the graph of \\( y = m x \\) is a straight line passing through the origin and the second and fourth quadrants. The graph of \\( y = m x^{2} - m^{2} \\) opens downwards, intersects the negative half of the \\( y \\)-axis, and has the \\( y \\)-axis as its axis of symmetry.\n\nCombining these observations, we conclude that option D fits the description.\n\nTherefore, the correct choice is: D\n\n[Key Insight] This problem primarily examines the graphical properties of proportional and quadratic functions, as well as analytical and interpretative skills. Mastery of their properties is essential for flexible problem-solving." }, { "problem_id": 1038, "question": "In the following four figures, which pairs of angles are corresponding angles?\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)\nB. (1)(2)\nC. (1)(2)(3)\nD. (1)(3)(4)", "input_image": [ "batch33-2024_06_14_8bf1d37053c200d6f8c0g_0043_1.jpg", "batch33-2024_06_14_8bf1d37053c200d6f8c0g_0043_2.jpg", "batch33-2024_06_14_8bf1d37053c200d6f8c0g_0043_3.jpg", "batch33-2024_06_14_8bf1d37053c200d6f8c0g_0043_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Analysis: According to the concept of corresponding angles, when two lines are intersected by a third line, the angles that are on the same side of the transversal and in corresponding positions are called corresponding angles. Based on this, we can make the judgment.\n\nDetailed Explanation: According to the definition of corresponding angles, (1) and (2) are corresponding angles, while (3) and (4) are not.\n\nTherefore, the correct choice is B.\n\nKey Point: This question primarily tests the identification of corresponding angles, with the key being to understand their position: on the same side and in corresponding positions." }, { "problem_id": 1039, "question": "In the four geometric solids shown below, the one enclosed by 5 faces is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_89c51f0358ecdde09f0fg_0062_1.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0062_2.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0062_3.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0062_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. Enclosed by 3 faces, therefore this option does not meet the requirement;\n\nB. Enclosed by 2 faces, therefore this option does not meet the requirement;\n\nC. Enclosed by 6 faces, therefore this option does not meet the requirement;\n\nD. Enclosed by 5 faces, therefore this option meets the requirement;\n\nHence, the correct choice is: D.\n\n[Key Insight] This question tests the understanding of three-dimensional shapes. Being proficient with the characteristics of common three-dimensional shapes is crucial for solving the problem." }, { "problem_id": 1040, "question": "Which of the following diagrams could be the graph of the function $y=ax^2+c, y=\\frac{a}{x}(a \\neq 0, c>0)$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_84512915d0d2c440e574g_0003_1.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0003_2.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0003_3.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0003_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When \\( a > 0 \\), the graph of the function \\( y = a x^{2} + c \\) opens upwards and passes through the point \\( (0, c) \\). The graph of the function \\( y = a x \\) is located in the first and third quadrants, thus options \\( B \\) and \\( D \\) can be excluded.\n\nWhen \\( a < 0 \\), the graph of the function \\( y = a x^{2} + c \\) opens downwards, and the graph of the function \\( y = a x \\) is located in the second and fourth quadrants, which excludes option \\( C \\). Therefore, the correct choice is \\( A \\).\n\n[Insight] This question primarily tests students' understanding of the graphs and properties of quadratic and inverse proportional functions. The key to solving this problem lies in discussing the value of \\( a \\) separately." }, { "problem_id": 1041, "question": "As shown in the figure, the geometric solid is made by sticking together small identical cubes. Which of the following is the front view of the solid?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0061_1.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0061_2.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0061_3.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0061_4.jpg", "batch33-2024_06_14_d15d6be0e0f91f2a62e5g_0061_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: From the front view, the first column from the left has two small squares, the second column has two small squares, and the third column has one small square.\n\nTherefore, the correct choice is: A.\n[Key Point] This question tests the observation of three-dimensional figures, with a focus on developing students' observational analysis skills and spatial imagination." }, { "problem_id": 1042, "question": "The graphs of the functions $y = ax - 2 (a \\neq 0)$ and $y = ax^2 (a \\neq 0)$ in the same Cartesian coordinate system could be ( ) .\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch7-2024_06_14_dd9b6a481087dff4e73bg_0015_1.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0015_2.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0015_3.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0015_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Question Analysis: The analysis is divided into two scenarios: (1) When \\( \\mathrm{a} > 0 \\), the parabola opens upwards, and the line intersects the negative half-axis of the \\( \\mathrm{y} \\)-axis, passing through the first, third, and fourth quadrants. (2) When \\( \\mathrm{a} < 0 \\), the parabola opens downwards, and the line intersects the negative half-axis of the \\( \\mathrm{y} \\)-axis, passing through the second, third, and fourth quadrants. Therefore, the correct choice is A.\n\nKey Points: Graphs of quadratic functions; Graphs of linear functions.\n\nComment: This question primarily examines the graphs of quadratic and linear functions, emphasizing the importance of understanding the relationship between the graph and its coefficients." }, { "problem_id": 1043, "question": "Among the following icons, which can be obtained by translating basic shapes? (~)\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch14-2024_06_15_00a109ed8135f00b577eg_0048_1.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0048_2.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0048_3.jpg", "batch14-2024_06_15_00a109ed8135f00b577eg_0048_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: A. The figure on the right cannot be obtained through translation; it can only be obtained by flipping, so this option does not meet the requirement;\n\nB. The figure on the left can be transformed into the figure on the right through translation, so this option meets the requirement;\n\nC. The shape of the figure is altered, so this option does not meet the requirement;\n\nD. The shape of the figure is altered, so this option does not meet the requirement.\n\nTherefore, the correct choice is: B.\n\n[Key Point] This question tests the transformation of figures through translation. Note that translation does not change the shape and size of the figure. This is a basic question, and it is essential to memorize the properties and characteristics of translation." }, { "problem_id": 1044, "question": "As shown in the figure, there are 3 bold black lines drawn on the outer surface of a cube. When the surface of this cube is unfolded with the outer surface facing up, the possible unfolded diagrams are ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_35cb0aebc81696508813g_0018_1.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0018_2.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0018_3.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0018_4.jpg", "batch25-2024_06_17_35cb0aebc81696508813g_0018_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: For option A, the unfolded diagram restores to the geometric shape as:\n\n\n\nThus, A does not meet the requirement;\n\nFor option B, the unfolded diagram restores to the geometric shape as:\n\n\n\nThus, B meets the requirement;\n\nFor option C, the unfolded diagram restores to the geometric shape as:\n\n\n\nThus, C does not meet the requirement;\n\nFor option D, the unfolded diagram restores to the geometric shape as:\n\n\n\nThus, D does not meet the requirement;\n\nTherefore, the correct choice is: B.\n\n[Insight] This question mainly examines the unfolded diagram of a cube, paying attention to the spatial figure of the cube, starting from the opposite faces, analyzing and solving the problem." }, { "problem_id": 1045, "question": "Which of the following diagrams can be used to demonstrate that $\\angle 1 > \\angle 2$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_d83a39705cad9b387cf7g_0032_1.jpg", "batch33-2024_06_14_d83a39705cad9b387cf7g_0032_2.jpg", "batch33-2024_06_14_d83a39705cad9b387cf7g_0032_3.jpg", "batch33-2024_06_14_d83a39705cad9b387cf7g_0032_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. According to the property that vertically opposite angles are equal, we have $\\angle 1 = \\angle 2$;\n\nB. Since any exterior angle of a triangle is greater than its non-adjacent interior angle, it follows that $\\angle 1 > \\angle 2$.\n\nC. In a triangle, a larger side is opposite a larger angle, hence $\\angle 1 < \\angle 2$;\n\nD. According to the property that corresponding angles are equal when two lines are parallel, we have $\\angle 1 = \\angle 2$.\n\nTherefore, the correct choice is B.\n\n【Key Point】This question examines the methods of comparing the sizes of angles from various perspectives, including vertically opposite angles, the relationship between interior and exterior angles of a triangle, the principle that larger sides are opposite larger angles, and the equality of corresponding angles when two lines are parallel. Each perspective has its unique characteristics and requires individual analysis." }, { "problem_id": 1046, "question": "As shown in the figure, a regular tetrahedron is cut along its edges $A B$, $A C$, and $A D$ to form a planar figure. The resulting unfolded figure is ( )\n\n\n\nA.\n\n\nB.\n\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0063_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0063_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0063_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0063_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0063_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "When it is cut along its edges $AB$, $AC$, and $AD$ and unfolded, it will expand outward along $BC$, $CD$, and $BD$ to form a figure as shown in diagram $B$.\n\nTherefore, the correct choice is: $B$.\n\n【Key Insight】This question tests knowledge of the unfolding diagrams of geometric solids. While engaging in hands-on operations, attention should also be paid to cultivating spatial imagination skills." }, { "problem_id": 1047, "question": "The total grain production of a village is $a$ (a constant) tons. Given the village's per capita grain production $y$ (tons) and the population size $x$ (people), the graph of the function relationship between $y$ and $x$ should be one of the following figures:\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_0d127fe92d0406a038c0g_0043_1.jpg", "batch13-2024_06_15_0d127fe92d0406a038c0g_0043_2.jpg", "batch13-2024_06_15_0d127fe92d0406a038c0g_0043_3.jpg", "batch13-2024_06_15_0d127fe92d0406a038c0g_0043_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "According to the problem, we have: $xy = a$,\n\n$\\therefore y = \\frac{a}{x} \\quad (x > 0, y > 0)$\n\nSince $x, y$ must be $> 0$,\n\n$\\therefore$ the graph lies in the first quadrant,\n\nHence, the correct choice is: C.\n\n【Key Point】The key concept tested here is the graph of an inverse proportional function. The crucial step in solving the problem is to establish the functional relationship between $y$ and $x$ based on the given conditions." }, { "problem_id": 1048, "question": "As shown in the figure, which of the following triangles are congruent? $(\\quad)$\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\nA. (1)(2)\nB. (2)(3)\nC. (3)(4)\nD. (1)(4)", "input_image": [ "batch9-2024_05_23_67e54113d3f7eb6371c0g_0018_1.jpg", "batch9-2024_05_23_67e54113d3f7eb6371c0g_0018_2.jpg", "batch9-2024_05_23_67e54113d3f7eb6371c0g_0018_3.jpg", "batch9-2024_05_23_67e54113d3f7eb6371c0g_0018_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: According to the \"SAS\" (Side-Angle-Side) criterion, it can be determined that the triangle in Figure (1) is congruent to the triangle in Figure (2).\n\nOptions (2)(3), (3)(4), and (1)(4) do not meet the requirements of the problem.\n\nTherefore, the correct choice is: A.\n\n【Key Point】This question tests the determination of congruent triangles. The key to solving it is to be proficient in using the SAS criterion to prove triangle congruence." }, { "problem_id": 1049, "question": "Among the following figures, which ones are axisymmetric?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0001_1.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0001_2.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0001_3.jpg", "batch10-2024_06_14_4ac6e21f5e525ac37e14g_0001_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: The figure in option D can find such a straight line that when the figure is folded along this line, the parts on both sides of the line can coincide with each other, hence it is an axisymmetric figure.\n\nThe figures in options A, B, and C cannot find such a straight line that when the figure is folded along this line, the parts on both sides of the line can coincide with each other, hence they are not axisymmetric figures.\n\nTherefore, the answer is: D.\n\n[Key Point] This question mainly examines the concept of axisymmetric figures. An axisymmetric figure refers to a single figure, and the axis of symmetry for an axisymmetric figure can be one, multiple, or even infinite in number." }, { "problem_id": 1050, "question": "As shown in the figure, the cross-section shape of the solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_1241c422ecb863904091g_0066_1.jpg", "batch33-2024_06_14_1241c422ecb863904091g_0066_2.jpg", "batch33-2024_06_14_1241c422ecb863904091g_0066_3.jpg", "batch33-2024_06_14_1241c422ecb863904091g_0066_4.jpg", "batch33-2024_06_14_1241c422ecb863904091g_0066_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, the shape obtained by vertically cutting a cube is a rectangle.\n\n\n\nTherefore, the correct choice is: $A$.\n\n[Key Insight] This question primarily examines the concept of cutting a geometric body. The shape of the cross-section is related not only to the geometric body being cut but also to the angle and direction of the cut. For such problems, it is best to combine hands-on practice with mental analysis, engaging in practical experimentation to learn and summarize analytical and inductive thinking methods." }, { "problem_id": 1051, "question": "Which of the following figures is not an axisymmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_1fc8b75729663205496ag_0020_1.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0020_2.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0020_3.jpg", "batch10-2024_06_14_1fc8b75729663205496ag_0020_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: The figure in option A is an axisymmetric figure, so it does not meet the requirement of the question;\n\nThe figure in option B is an axisymmetric figure, so it does not meet the requirement of the question;\n\nThe figure in option C is an axisymmetric figure, so it does not meet the requirement of the question;\n\nThe figure in option D is not an axisymmetric figure, so it meets the requirement of the question.\n\nTherefore, the correct choice is D.\n\n[Key Point] This question tests the recognition of axisymmetric figures. Understanding the definition of an axisymmetric figure is crucial for solving the problem." }, { "problem_id": 1052, "question": "In the following options, all squares have the same side length. Which pattern has a shaded area that is different from the other three?\n\n\n\n(A)\n\n\n\n(B)\n\n\n\n(C)\n\n\n\n(D)\nA. (A)\nB. (B)\nC. (C)\nD. (D)", "input_image": [ "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0005_1.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0005_2.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0005_3.jpg", "batch17-2024_06_14_d4bc1f4d14517b50df1eg_0005_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Area of the shaded region in Figure A: Area of the square minus the area of the circle;\n\nArea of the shaded region in Figure B: Area of the square minus the area of the semicircle whose diameter is half the diagonal of the square;\n\nArea of the shaded region in Figure C: Area of the square minus the area of the circle;\n\nArea of the shaded region in Figure D: Area of the square minus the area of the circle.\n\nTherefore, the correct answer is B.\n\nKey point: This question tests the calculation of the areas of sectors and circles. Analyze the area of the shaded part in each figure to determine the answer." }, { "problem_id": 1053, "question": "As shown in the figure, which of the following can be obtained when viewing the solid from the front?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n##", "input_image": [ "batch10-2024_06_14_89c51f0358ecdde09f0fg_0057_1.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0057_2.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0057_3.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0057_4.jpg", "batch10-2024_06_14_89c51f0358ecdde09f0fg_0057_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The figure seen from the front has 2 rows and 3 columns, with the number of squares from left to right being $2, 1, 1$. Therefore, the correct choice is: A.\n\n[Key Point] This question tests the ability to view geometric shapes from different perspectives. Mastering the angle of view is crucial for solving such problems." }, { "problem_id": 1054, "question": "Among the following figures, which one, when folded, can form a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_d6e7887b4e796c8b833cg_0069_1.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0069_2.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0069_3.jpg", "batch10-2024_06_14_d6e7887b4e796c8b833cg_0069_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "A、Can be folded into a cube, which fits the requirement;\n\nB、Is in the shape of a \"concave\" character, thus cannot be folded into a cube, does not meet the requirement;\n\nC、After folding, two faces overlap, missing one base face, so it also cannot be folded into a cube;\n\nD、Is in the shape of a \"field\" character, thus cannot be folded into a cube, does not meet the requirement.\n\nTherefore, choose A." }, { "problem_id": 1055, "question": "Among the following polygons, the figure with an interior angle sum of $360^\\circ$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_236cba6f8aa19ac2b09ag_0009_1.jpg", "batch14-2024_06_15_236cba6f8aa19ac2b09ag_0009_2.jpg", "batch14-2024_06_15_236cba6f8aa19ac2b09ag_0009_3.jpg", "batch14-2024_06_15_236cba6f8aa19ac2b09ag_0009_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Using the formula for the sum of interior angles of a polygon, we have\n\n$$\n(n - 2) \\cdot 180^{\\circ} = 360^{\\circ},\n$$\n\nSolving for \\( n \\), we find \\( n = 4 \\), which indicates a quadrilateral.\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question tests the calculation of the sum of interior angles of a polygon. Remembering the formula is crucial for solving the problem." }, { "problem_id": 1056, "question": "A district plans to build a nucleic acid collection point $P$ alongside a road. There are four options, and the nucleic acid collection point $P$ should be located such that the sum of the distances from $P$ to neighborhoods $A$ and $B$ is minimized. Which of the following options is the best?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_4ce737a200cf8fbade32g_0048_1.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0048_2.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0048_3.jpg", "batch10-2024_06_14_4ce737a200cf8fbade32g_0048_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Construct the symmetric point $A^{\\prime}$ of point $\\mathrm{A}$ with respect to line $m$, then connect $A^{\\prime}$ to $B$ and let the line $A^{\\prime}B$ intersect line $m$ at point $P$. According to the principle that the shortest distance between two points is a straight line, it can be concluded that the nucleic acid collection point $P$ in option $\\mathrm{B}$ has the shortest total distance to the two communities $A$ and $B$. Therefore, the correct choice is: B.\n\n【Key Insight】This problem examines the mathematical concept of the shortest path. Mastering the principle that the shortest distance between two points is a straight line is crucial for solving such problems." }, { "problem_id": 1057, "question": "A person runs at a constant speed to the park, stays at a certain location in the park for some time, and then walks back home at a constant speed along the same route. The approximate graph of the person's distance from home $y$ with respect to time $x$ is\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch11-2024_06_14_96b2f24d25c83e6985b9g_0022_1.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0022_2.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0022_3.jpg", "batch11-2024_06_14_96b2f24d25c83e6985b9g_0022_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "The image should be divided into three stages:\n\n**First Stage:** Jogging at a constant speed to the park. During this stage, the distance from home increases as time progresses.\n\n**Second Stage:** Staying in the park for a period of time. In this stage, the distance from home does not change with time. Therefore, option D is incorrect.\n\n**Third Stage:** Walking back home at a constant speed along the same path. In this stage, the distance from home decreases as time progresses, so option A is incorrect. Additionally, the speed during this stage is less than that of the first stage, making option C incorrect.\n\nThus, the correct choice is B.\n\n**Exam Focus:** Function Graphs\n\n**[Highlight]** This question tests the understanding of function graphs. The key to solving it lies in comprehending the relationship between the distance from home and time in each stage and using the slope of the graph to determine the speed of movement." }, { "problem_id": 1058, "question": "The following figures are squares drawn outside right-angled triangles with the sides of the triangles as their sides. The numbers and letter $S$ in each square represent the area of the square, where the value of $S$ is exactly 5. Which of the following is correct?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0066_1.jpg", "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0066_2.jpg", "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0066_3.jpg", "batch11-2024_06_14_d2d3a7d5fdab452af1b3g_0066_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Since the number and the letter \\( S \\) in each square represent the area of the square,\nTherefore, the number and the letter \\( S \\) in each square represent the square of the side length of the square.\n\nA: By the Pythagorean theorem: \\( S = 4 + 9 = 13 \\), so A does not meet the condition;\n\nB: \\( S = 9 - 4 = 5 \\), so B meets the condition;\n\nC: \\( S = 4 + 3 = 7 \\), so C does not meet the condition;\n\nD: \\( S = 4 - 3 = 1 \\), so D does not meet the condition;\n\nTherefore, the correct choice is: B.\n\n[Key Insight] This question tests the Pythagorean theorem and the properties of squares. Mastering the Pythagorean theorem and the properties of squares is crucial for solving the problem." }, { "problem_id": 1059, "question": "In the formula $I=\\frac{U}{R}(R>0)$, when the voltage $U$ is constant, the graphical representation of the functional relationship between the current $I$ and the resistance $R$ is\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0088_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0088_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0088_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0088_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem statement, when the voltage \\( U \\) is constant, the current \\( I \\) and the resistance \\( R \\) have an inverse proportional relationship: \\( I = \\frac{U}{R} \\). Since \\( R > 0 \\) and \\( U > 0 \\), it follows that \\( I > 0 \\). Therefore, its graph is that of an inverse proportional function in the first quadrant.\n\nHence, the correct choice is: D.\n\n[Key Insight] This problem examines the application of inverse proportional functions. The key to solving it lies in determining the type and position of the graph based on the functional relationship and the range of the independent variable." }, { "problem_id": 1060, "question": "Among the following figures, which ones are planar figures?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_329d65dd392595b77024g_0047_1.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0047_2.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0047_3.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0047_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: The sides and angles of a triangle lie within a plane, making it a planar figure. Therefore, option A fits the description.\n\nA cube, a sphere, and a hexagonal prism are all three-dimensional figures, so options B, C, and D do not fit the description.\n\nThus, the correct choice is A.\n\n[Key Insight] This question tests the understanding of planar and three-dimensional figures. Familiarity with the characteristics of planar and three-dimensional figures is crucial for solving this problem." }, { "problem_id": 1061, "question": "How many of the following geometric solids have a quadrilateral shape when viewed from the left? ( )\n\n\n\nCylinder\n\n\n\nCone\n\n\n\nSphere\n\n\n\nCube\nA. 1\nB. 2\nC. 3\nD. 4", "input_image": [ "batch10-2024_06_14_3a73c762f5eb2b761854g_0035_1.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0035_2.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0035_3.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0035_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Since the shape seen from the left side of a cylinder is a quadrilateral, the shape seen from the left side of a cone is a triangle, the shape seen from the left side of a cube is a quadrilateral, and the shape seen from the left side of a sphere is a circle,\n\nTherefore, among the four geometric shapes, there are a total of 2 shapes that appear as quadrilaterals when viewed from the left side.\n\nHence, the correct choice is: B.\n\n[Key Insight] This problem is a basic application question, requiring students to be proficient in viewing geometric shapes from different perspectives to solve it, with a focus on developing spatial imagination skills." }, { "problem_id": 1062, "question": "(2016 Hubei Province, Yichang City) Which of the following could be the graph of the function $y = \\frac{2}{x+1}$?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0027_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0027_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0027_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0027_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "According to the problem, the function \\( y = \\frac{2}{x+1} \\) is obtained by shifting the graph of the inverse proportion \\( y = \\frac{2}{x} \\) one unit to the left.\nTherefore, the correct choice is C.\n\n【Key Point】This question examines the translation of the graph of an inverse proportional function. Mastering the rule of \"left addition and right subtraction\" for translations is crucial for solving the problem." }, { "problem_id": 1063, "question": "Given that the area of a trapezoid is constant, where its height is denoted as $\\mathrm{h}$ and the length of its median is denoted as $\\mathrm{x}$, the approximate functional relationship between $\\mathrm{h}$ and $\\mathrm{x}$ is\nA. \n\n\nB. \n\n\nC. \n\n\nD. \n\n", "input_image": [ "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0096_1.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0096_2.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0096_3.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0096_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Question Analysis: The area of a trapezoid is given by $\\mathrm{S}=\\frac{1}{2} \\times$ the sum of the lengths of the two parallel sides $\\times$ the height $=\\frac{1}{2} \\times$ the length of the midline $\\times$ the height $=\\frac{1}{2} \\mathrm{hx}$. Since $\\mathrm{S}$ is constant, it follows that $\\mathrm{h}=\\frac{2 \\mathrm{~S}}{\\mathrm{x}}$ ($\\mathrm{x}>0$, $\\mathrm{~h}>0$), which represents a branch of the inverse proportionality function in the first quadrant.\n\nTherefore, the correct choice is D." }, { "problem_id": 1064, "question": "In the diagrams given in the figure, which of the rays and straight lines can intersect?\n\nA. \n\nB. \n\nC. \n\nD. ", "input_image": [ "batch10-2024_06_14_877c2117fdc429fa1ae4g_0033_1.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0033_2.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0033_3.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0033_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. The line $AB$ and the ray $EF$ do not intersect, which does not meet the requirement;\n\nB. The line $AB$ and the ray $EF$ intersect, which meets the requirement;\n\nC. The line $AB$ and the ray $EF$ do not intersect, which does not meet the requirement;\n\nD. The line $AB$ and the ray $EF$ do not intersect, which does not meet the requirement;\n\nTherefore, the correct choice is B.\n\n[Key Insight] This question primarily tests the understanding of lines, rays, or segments. Grasping the extensibility of lines and rays is crucial for solving the problem." }, { "problem_id": 1065, "question": "Among the four geometric solids below, which one has a quadrilateral as its top view?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch5-2024_06_14_43010c41ad60ab0a5523g_0007_1.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0007_2.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0007_3.jpg", "batch5-2024_06_14_43010c41ad60ab0a5523g_0007_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Question Analysis: The top view is the planar shape seen from above the object. $\\mathrm{A}$ is a triangle, $\\mathrm{B}$ is a hexagon, $\\mathrm{C}$ is a square, and $\\mathrm{D}$ is a circle. Therefore, the correct answer is $\\mathrm{C}$.\n\nExam Focus: Recognition of three-view drawings." }, { "problem_id": 1066, "question": "The following figures are obtained by drawing a square outside each side of a right-angled triangle, with the numbers and letter $S$ in each square representing the area of the square. Among them, the value of $S$ that is exactly 10 is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch12-2024_06_15_22894a68bf357c1c0757g_0026_1.jpg", "batch12-2024_06_15_22894a68bf357c1c0757g_0026_2.jpg", "batch12-2024_06_15_22894a68bf357c1c0757g_0026_3.jpg", "batch12-2024_06_15_22894a68bf357c1c0757g_0026_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the numbers and the letter $S$ in each square represent the square of the length of the side of the respective square.\n\nA. By the Pythagorean theorem: $S=5+15=20$, so option A does not meet the requirement;\n\nB. By the Pythagorean theorem: $S=8+6=14$, so option B does not meet the requirement;\n\nC. By the Pythagorean theorem: $S=8-6=2$, so option C does not meet the requirement;\n\nD. By the Pythagorean theorem: $S=15-5=10$, so option D meets the requirement, hence the answer is: D.\n\n[Key Insight] This question tests the area of squares and the Pythagorean theorem. Mastering the Pythagorean theorem is crucial for solving the problem." }, { "problem_id": 1067, "question": "In each of the following figures, there is a pair of congruent triangles. The one that can be superimposed on the other by a single rotation is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch15-2024_06_15_63cc2bb499d687c2b4bdg_0090_1.jpg", "batch15-2024_06_15_63cc2bb499d687c2b4bdg_0090_2.jpg", "batch15-2024_06_15_63cc2bb499d687c2b4bdg_0090_3.jpg", "batch15-2024_06_15_63cc2bb499d687c2b4bdg_0090_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Question Analysis: Based on the properties of rotation and axial symmetry transformation, analyze and deduce accordingly.\n\nSolution: \nA. Cannot be obtained through rotation, hence this option is incorrect;\nB. Cannot be obtained through rotation, hence this option is incorrect;\nC. Can be obtained through axial symmetry, hence this option is incorrect;\nD. Can align with another triangle through a single rotation, hence this option is correct.\n\nTherefore, choose D.\n\nComment: This question primarily tests the ability to design patterns using rotation, and mastering the properties of rotation is key to solving the problem." }, { "problem_id": 1068, "question": "Given that the work done by the force $F$ is $15 \\mathrm{~J}$, the graph of force $F$ versus the distance $s$ traveled by the object in the direction of the force is approximately one of the following figures ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0085_1.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0085_2.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0085_3.jpg", "batch13-2024_06_15_62531bb0e7f478ff3f2ag_0085_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Question Analysis: First, establish the functional relationship based on the given conditions, then determine the quadrant in which the graph of the function lies according to the range of values of \\( s \\).\n\nGiven that the work done by force \\( F \\) is 15 joules, the relationship between force \\( F \\) and the distance \\( S \\) through which the object moves in the direction of the force is \\( F = \\frac{15}{s} \\). Considering the practical implications, \\( s > 0 \\).\n\nTherefore, its graph lies only in the first quadrant.\n\nHence, the correct choice is B.\n\nKey Point: This question examines the graph of an inverse proportional function.\n\nComment: This question is typical and significant as it effectively integrates physical knowledge, demonstrating the connections between different disciplines." }, { "problem_id": 1069, "question": "Among the four geometric solids below, which one has a different left view than the others?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_802f01be018a9f8ad422g_0084_1.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0084_2.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0084_3.jpg", "batch25-2024_06_17_802f01be018a9f8ad422g_0084_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Find the shape obtained from the left view, ensuring that all visible edges are represented in the left view.\n【Analysis】Solution: Since the left views of options A, B, and D all consist of a column of two small squares, while the left view of option C consists of a column of three small squares.\n\nTherefore, option C differs from the left views of the other three shapes.\n\nHence, choose C.\n\n【Highlight】This question tests the ability to judge the three views of a simple composite solid. Understanding that the left view is the shape seen from the left side of the object is key to solving the problem." }, { "problem_id": 1070, "question": "In the four diagrams below, the one where the angle $\\angle A O B$ measured with a protractor is $40^{\\circ}$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_da259e8c8bb047f1aad6g_0057_1.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0057_2.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0057_3.jpg", "batch10-2024_06_14_da259e8c8bb047f1aad6g_0057_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: \n\nA. The angle measured by the protractor is $40^{\\circ}$, so this option is correct;\n\nB. The method of using the protractor is incorrect, so this option is wrong;\n\nC. The angle measured by the protractor is $140^{\\circ}$, so this option is wrong;\n\nD. The method of using the protractor is incorrect, so this option is wrong.\n\nTherefore, the correct choice is A.\n\n[Key Insight] This question tests the method of using a protractor. The key to solving the problem lies in the correct and proficient use of the measuring tool—the protractor." }, { "problem_id": 1071, "question": "In the following figures, which pair of angles $\\angle 1$ and $\\angle 2$ are complementary? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_877c2117fdc429fa1ae4g_0068_1.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0068_2.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0068_3.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0068_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the definition of complementary angles, if the sum of two angles is $90^{\\circ}$, then these two angles are complementary.\n\nIn option D, the sum of $\\angle 1$ and $\\angle 2$ is $90^{\\circ}$, so they are complementary angles.\n\nTherefore, the correct choice is D.\n\n[Key Insight] This question tests the understanding of the definitions of complementary and supplementary angles. Based on the definition of complementary angles, remember that if the sum of two angles is $90^{\\circ}$, they are complementary, regardless of their positions." }, { "problem_id": 1072, "question": "In the same Cartesian coordinate system, the graphs of the linear function $y = a (x - 1)$ and the quadratic function $y = a (x^2 - 1)$ could only be one of the following figures ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch7-2024_06_14_dd9b6a481087dff4e73bg_0014_1.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0014_2.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0014_3.jpg", "batch7-2024_06_14_dd9b6a481087dff4e73bg_0014_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Question Analysis: \n\nA. From the graph of the linear function, it can be seen that $\\mathrm{a}>0$, and from the graph of the quadratic function, it can also be seen that $\\mathrm{a}>0$. These two observations are consistent;\n\nB. From the graph of the linear function, it can be seen that $\\mathrm{a}>0$, but from the graph of the quadratic function, it can be seen that $\\mathrm{a}<0$. These two observations are contradictory;\n\nC. From the graph of the linear function, it can be seen that $\\mathrm{a}<0$, but from the graph of the quadratic function, it can be seen that $\\mathrm{a}>0$. These two observations are contradictory;\n\nD. From the graph of the linear function, it can be seen that $\\mathrm{a}>0$, but from the graph of the quadratic function, it can be seen that $\\mathrm{a}<0$. These two observations are contradictory. Therefore, the correct answer is A.\n\nKey Points: 1. Graphs of quadratic functions; 2. Graphs of linear functions." }, { "problem_id": 1073, "question": "Among the following figures, which one is a central symmetric figure but not an axisymmetric figure?\n\nA.\n\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0003_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0003_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0003_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0003_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "B is both an axisymmetric figure and a centrally symmetric figure;\n\nC is only an axisymmetric figure;\n\nD is neither an axisymmetric figure nor a centrally symmetric figure.\n\nOnly A meets the criteria.\n\nTherefore, choose A." }, { "problem_id": 1074, "question": "If the central angle of the lateral surface expanded view of a cone is $270^{\\circ}$, the graph of the relationship between the cone's slant height $l$ and the radius $r$ of the base is approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_84512915d0d2c440e574g_0097_1.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0097_2.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0097_3.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0097_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we have \\(2 \\pi r = \\frac{270 \\cdot \\pi \\cdot l}{180}\\), so \\(l = \\frac{4}{3} r\\) (where \\(r > 0\\)).\n\nThis means that \\(l\\) and \\(r\\) have a directly proportional relationship, and their graph lies in the first quadrant.\n\nTherefore, the correct answer is A.\n\n[Key Insight] This question tests the calculation of a cone and the graph of a function." }, { "problem_id": 1075, "question": "In the figure below, each large square grid is composed of small squares with sides of length 1. Which of the shaded regions has the largest area?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch23-2024_06_14_9a11b82306c24d751e6fg_0010_1.jpg", "batch23-2024_06_14_9a11b82306c24d751e6fg_0010_2.jpg", "batch23-2024_06_14_9a11b82306c24d751e6fg_0010_3.jpg", "batch23-2024_06_14_9a11b82306c24d751e6fg_0010_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Question Analysis: Based on the properties of a square, the area of an irregular shape can be considered as the sum or difference of the areas of regular shapes, thereby allowing us to determine which figure has the largest shaded area.\n\nSolution: The area of an irregular shape can be viewed as the sum or difference of the areas of regular shapes. According to the properties of a square, the figure with the largest shaded area in the diagram is the fourth option.\n\nTherefore, the answer is D." }, { "problem_id": 1076, "question": "During a mathematics activity class, students used GeoGebra to draw the following curves. Which one is an axisymmetric figure with the most number of symmetry axes? \n\n\n\nA. Archimedean spiral\n\n\n\nB. Heart curve\n\n\n\nC. Four-leaf rose curve\n\n\n\nD. Butterfly curve", "input_image": [ "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0009_1.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0009_2.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0009_3.jpg", "batch10-2024_06_14_ff6a1901bcdec40dfd04g_0009_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Option A cannot find a line of symmetry, so it is not an axisymmetric figure;\n\nOption B is an axisymmetric figure with 1 line of symmetry;\n\nOption C is an axisymmetric figure with 4 lines of symmetry;\n\nOption D is an axisymmetric figure with 1 line of symmetry;\n\nTherefore, the figure that is axisymmetric and has the most lines of symmetry is C.\nHence, the answer is:\n\n【Key Point】This question mainly tests the definition of an axisymmetric figure: if a figure can be folded along a certain line and the two parts on either side of the line completely coincide, then the figure is called an axisymmetric figure. Understanding the definition of an axisymmetric figure is key to solving the problem." }, { "problem_id": 1077, "question": "If the graph of the inverse proportion function $y = \\frac{k}{x}$ is as shown in the figure, then the approximate graph of the quadratic function $y = kx^2 - k^2x - 1$ is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_21ae9a88b517218f12e9g_0099_1.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0099_2.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0099_3.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0099_4.jpg", "batch13-2024_06_15_21ae9a88b517218f12e9g_0099_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Based on the graph of the inverse proportional function \\( y = \\frac{k}{x} \\) in the first and third quadrants, we can deduce that \\( k > 0 \\). Further, by considering the direction of the opening and the position of the axis of symmetry of the quadratic function \\( y = kx^{2} - k^{2}x - 1 \\), we can make the following judgment.\n\nFrom the given condition \\( k > 0 \\), it follows that the graph of the quadratic function \\( y = kx^{2} - k^{2}x - 1 \\) opens upwards, and its axis of symmetry is at \\( x = -\\frac{-k^{2}}{2k} = \\frac{1}{2}k > 0 \\).\n\nTherefore, the correct choice is A.\n\n【Key Insight】This question examines the relationship between the graph of a quadratic function and its coefficients. It is a basic application problem, requiring students to be proficient in understanding how the coefficients affect the graph of a quadratic function to solve it." }, { "problem_id": 1078, "question": "As shown in the figure, the left view of the given geometric solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_6a6aed9af688e70907d8g_0064_1.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0064_2.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0064_3.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0064_4.jpg", "batch10-2024_06_14_6a6aed9af688e70907d8g_0064_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: From the left side, the bottom layer is a rectangle, and the top layer is a right-angled triangle (there is no solid line separating the triangle from the rectangle), aligned to the left.\n\nTherefore, the correct choice is: A.\n\n[Key Insight] This question primarily tests the understanding of the three views of a geometric solid. Mastering the drawing method of the three views is crucial for solving the problem." }, { "problem_id": 1079, "question": "The front view of the geometric solid shown in the figure is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0060_1.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0060_2.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0060_3.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0060_4.jpg", "batch25-2024_06_17_eae9f68fdb2fb7ec1cbdg_0060_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, the bottom layer is a larger rectangle, and the upper left layer is a smaller rectangle.\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question primarily tests the knowledge of three-view drawings, and accurate analysis is the key to solving the problem." }, { "problem_id": 1080, "question": "Given that the equation $x^{2}+2 x-k-2=0$ has no real roots, the graph of the function $y=\\frac{k}{x}$ is approximately one of the following figures:\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_a5bc38c105001883c8a7g_0001_1.jpg", "batch13-2024_06_15_a5bc38c105001883c8a7g_0001_2.jpg", "batch13-2024_06_15_a5bc38c105001883c8a7g_0001_3.jpg", "batch13-2024_06_15_a5bc38c105001883c8a7g_0001_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Since the equation \\( x^{2} + 2x - k - 2 = 0 \\) concerning \\( x \\) has no real solutions,\n\n\\[\n\\therefore \\triangle = 2^{2} - 4(-k - 2) < 0,\n\\]\n\nSolving this, we find \\( k < -3 \\),\n\n\\[\n\\therefore \\text{the graph of the function } y = \\frac{k}{x} \\text{ lies in the second and fourth quadrants},\n\\]\n\nHence, the correct choice is: B.\n\n【Key Insight】This question tests the discriminant of roots and the relationship between the coefficients of the inverse proportional function and its graph. Remembering that \"when \\( \\triangle < 0 \\), the equation has no real roots\" is crucial for solving the problem." }, { "problem_id": 1081, "question": "Among the following figures, which ones are axisymmetric?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_2434588caaee122ec85bg_0018_1.jpg", "batch10-2024_06_14_2434588caaee122ec85bg_0018_2.jpg", "batch10-2024_06_14_2434588caaee122ec85bg_0018_3.jpg", "batch10-2024_06_14_2434588caaee122ec85bg_0018_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: By examining the four options, it is evident that only the figure in option D can be completely overlapped on both sides of a straight line when folded along that line, which qualifies it as an axisymmetric figure.\n\nTherefore, the answer is D.\n\n[Key Insight] This question tests the recognition of axisymmetric figures. The key to solving it lies in understanding the definition of an axisymmetric figure. If a figure can be folded along a straight line such that the two parts completely coincide, then it is called an axisymmetric figure." }, { "problem_id": 1082, "question": "As shown in the figure, 5 identical small cubes are stacked to form a three-dimensional shape. When viewed from the front, the resulting two-dimensional shape is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0073_1.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0073_2.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0073_3.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0073_4.jpg", "batch33-2024_06_14_b54d0ec2e6cec62ab166g_0073_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: Based on the analysis of the question stem, the figure seen from the front consists of two rows: the top row has 2 squares aligned to the right, and the bottom row has 3 squares. The figure seen from the front is:\n\n\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question tests the ability to observe objects and geometric shapes from different perspectives, enhancing students' spatial imagination and abstract thinking skills." }, { "problem_id": 1083, "question": "In the geometric shapes of the following objects, which one is a quadrilateral pyramid?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_f5b8b25a104af99401e0g_0052_1.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0052_2.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0052_3.jpg", "batch33-2024_06_14_f5b8b25a104af99401e0g_0052_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Question Analysis: Method to Determine a Pyramid: The sides of a pyramid are all triangles, the base is a polygon, and the number of sides of the base determines the type of pyramid. For example, if the base has three sides, it is a triangular pyramid. In this case, option A is a triangular prism, option B is a quadrilateral pyramid, option C is a triangular pyramid, and option D is a quadrilateral prism. Therefore, the correct answer is B.\n\nKey Point: This question primarily tests the distinction between pyramids and prisms. The sides of a pyramid are all triangles, and it has only one polygonal base. In contrast, the sides of a prism are all parallelograms, and it has two polygonal bases. The type of prism is determined by the number of sides of its base. Grasping the characteristics of pyramids and prisms is essential for making accurate judgments, and this is a common test point." }, { "problem_id": 1084, "question": "As shown in the figure, when a plane intersects a frustum of a cone, the resulting cross-sectional shape cannot be ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_5d2efe25907dd6708471g_0060_1.jpg", "batch33-2024_06_14_5d2efe25907dd6708471g_0060_2.jpg", "batch33-2024_06_14_5d2efe25907dd6708471g_0060_3.jpg", "batch33-2024_06_14_5d2efe25907dd6708471g_0060_4.jpg", "batch33-2024_06_14_5d2efe25907dd6708471g_0060_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "When a plane intersects a cone, if the plane passes through the vertex of the cone, the resulting cross-section is an isosceles triangle; if it does not pass through the vertex and the plane is parallel to the base, the resulting cross-section is a circle; if it is not parallel to the base, the resulting cross-section is an ellipse or a combination of a parabola and a line segment.\n\n$\\therefore$ When a plane intersects a cone, the resulting cross-section cannot be a quadrilateral.\n\nTherefore, the correct answer is: A.\n\n[Key Insight] This question primarily examines the concept of intersecting a geometric body. The shape of the cross-section depends not only on the geometric body being intersected but also on the angle and direction of the intersecting plane. For such problems, it is best to combine hands-on practice with mental analysis, allowing one to learn and generalize through practical experience." }, { "problem_id": 1085, "question": "The net of a cube is shown below. Which of the following four options could be the cube?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0064_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0064_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0064_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0064_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0064_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "According to the problem statement and the diagram, only option $\\mathrm{A}$ fits correctly after being folded.\n\nTherefore, the answer is: A.\n\n**Key Point:** This question tests the understanding of the development of geometric shapes, with the key to solving it lying in spatial imagination." }, { "problem_id": 1086, "question": "In the following diagrams, which pairs of angles $\\angle 1$ and $\\angle 2$ are corresponding angles?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_bb94e5395dc7244e2637g_0060_1.jpg", "batch33-2024_06_14_bb94e5395dc7244e2637g_0060_2.jpg", "batch33-2024_06_14_bb94e5395dc7244e2637g_0060_3.jpg", "batch33-2024_06_14_bb94e5395dc7244e2637g_0060_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Analysis: Determine based on the definitions of corresponding angles, alternate interior angles, and consecutive interior angles.\n\nDetailed Explanation: \nA. $\\angle 1$ and $\\angle 2$ are alternate interior angles, so this option is incorrect;\nB. $\\angle 1$ and $\\angle 2$ are consecutive interior angles, so this option is incorrect;\nC. $\\angle 1$ and $\\angle 2$ are neither alternate interior angles, nor corresponding angles, nor consecutive interior angles, so this option is incorrect;\nD. $\\angle 1$ and $\\angle 2$ are corresponding angles, so this option is correct.\n\nTherefore, the correct choice is D.\n\nKey Point: This question tests the understanding and application of the definitions of corresponding angles, alternate interior angles, and consecutive interior angles, primarily assessing the student's comprehension and analytical abilities." }, { "problem_id": 1087, "question": "The following are logos of Chinese domestic brand cars. Among these car logos, which one is a central symmetric figure?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0097_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0097_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0097_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0097_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the definition of a centrally symmetric figure: \"A figure that can completely coincide with itself after being rotated $180^{\\circ}$ around a point is called a centrally symmetric figure.\"\n\nBased on this definition, figures A, C, and D are not centrally symmetric, only figure $\\mathrm{B}$ is centrally symmetric.\n\nTherefore, the correct choice is: B." }, { "problem_id": 1088, "question": "Which of the following figures can be folded into a cylinder?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_3a73c762f5eb2b761854g_0028_1.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0028_2.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0028_3.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0028_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: A. Can form a triangular pyramid, so option A does not meet the requirement;\n\nB. Can form a cube, so option B does not meet the requirement;\n\nC. Can form a triangular prism, so option C does not meet the requirement;\n\nD. Can form a cylinder, so option D meets the requirement;\n\nTherefore, the answer is: D.\n\n[Key Point] This question tests the understanding of the lateral expansion of a cylinder. Mastering the appearance of a cylinder's lateral expansion is crucial for solving the problem." }, { "problem_id": 1089, "question": "Which of the following figures can be folded into a triangular prism? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_682dd3cab3079f1333ccg_0098_1.jpg", "batch25-2024_06_17_682dd3cab3079f1333ccg_0098_2.jpg", "batch25-2024_06_17_682dd3cab3079f1333ccg_0098_3.jpg", "batch25-2024_06_17_682dd3cab3079f1333ccg_0098_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Analysis: Based on the characteristics of a triangular prism, it is formed by two triangles and three rectangles.\n\nDetailed Explanation: \nA. Can be folded into a triangular prism, hence this option is correct;\nB. Can be folded into a triangular pyramid, hence this option is incorrect;\nC. Can be folded into a quadrilateral pyramid, hence this option is incorrect;\nD. Cannot be folded into a geometric shape, hence this option is incorrect;\n\nTherefore, the correct choice is A.\n\nKey Point: The examination focuses on the folding of a net into a geometric shape, with the key being to understand the characteristics of a triangular prism." }, { "problem_id": 1090, "question": "The graph of the quadratic function $y = x^{2} + 2x$ could be ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch8-2024_06_14_e35fe70123f4447b467cg_0013_1.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0013_2.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0013_3.jpg", "batch8-2024_06_14_e35fe70123f4447b467cg_0013_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Question Analysis: Since \\( y = x^{2} + 2x = (x + 1)^{2} - 1 \\), the graph opens upwards with its vertex at \\((-1, -1)\\) in the third quadrant. Among the options, only option \\(\\mathrm{C}\\) meets these conditions, so the correct choice is \\(\\mathrm{C}\\).\n\nKey Point: The graph of a quadratic function." }, { "problem_id": 1091, "question": "In the same Cartesian coordinate system, the graphs of the functions $y = kx^2 - k$ (where $k \\neq 0$) and $y = \\frac{k}{x}$ (where $k \\neq 0$) are approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch13-2024_06_15_84512915d0d2c440e574g_0063_1.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0063_2.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0063_3.jpg", "batch13-2024_06_15_84512915d0d2c440e574g_0063_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: (1) When \\( k > 0 \\), the hyperbola is located in the first and third quadrants, the parabola opens upwards, and its vertex lies on the negative half of the \\( y \\)-axis; (2) When \\( k < 0 \\), the hyperbola is located in the second and fourth quadrants, the parabola opens downwards, and its vertex lies on the positive half of the \\( y \\)-axis. Therefore, option B fits the description.\n\nHence, the answer is: B\n\n[Key Insight] This question primarily examines the graphs of quadratic and inverse proportional functions. The general approach to solving such problems is: (1) First, determine whether the value of \\( k \\) is contradictory based on the characteristics of the graph; (2) Then, check if the intersection of the parabola with the \\( y \\)-axis meets the requirements based on the graph of the quadratic function." }, { "problem_id": 1092, "question": "As shown in the figure, a geometric solid is formed by 5 identical small cubes. The view of the solid from the left is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_877c2117fdc429fa1ae4g_0006_1.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0006_2.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0006_3.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0006_4.jpg", "batch10-2024_06_14_877c2117fdc429fa1ae4g_0006_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Observing the geometric figure from the left side, the first column has one cube, the second column has two cubes, and the third column has one cube.\n\nTherefore, the correct choice is D.\n\n【Key Point】This question tests the ability to observe objects from different perspectives. Mastering the drawing method is crucial for solving the problem." }, { "problem_id": 1093, "question": "In the following figures, which one is not an example of similar figures?\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch32-2024_06_14_f42401fbec3c1b6b5430g_0039_1.jpg", "batch32-2024_06_14_f42401fbec3c1b6b5430g_0039_2.jpg", "batch32-2024_06_14_f42401fbec3c1b6b5430g_0039_3.jpg", "batch32-2024_06_14_f42401fbec3c1b6b5430g_0039_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "According to the concept of similar figures, the two figures in each of the groups A, B, and C are similar figures; the two figures in group D do not conform to the concept of similar figures, as their corresponding vertices cannot intersect at a single point, hence they are not similar figures. Therefore, the correct choice is D.\n\n【Key Point】This question primarily examines the properties of similar figures, and accurate analysis and judgment are crucial to solving the problem." }, { "problem_id": 1094, "question": "As shown in the figure, after rotating the square pattern around the center $O$ by $180^\\circ$, the resulting pattern is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch26-2024_06_17_201bada2148f276a9b39g_0043_1.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0043_2.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0043_3.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0043_4.jpg", "batch26-2024_06_17_201bada2148f276a9b39g_0043_5.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: According to the properties of rotation, the relative positions of all points remain unchanged before and after rotation, resulting in congruent figures. Analyzing the options, we find that when the square pattern is rotated $180^{\\circ}$ around the center $O$, the resulting pattern is $C$.\n\nTherefore, the answer is: $C$.\n\n[Key Insight] This question tests the properties of figure rotation. Mastering these properties is crucial for solving the problem." }, { "problem_id": 1095, "question": "Among the line segments and rays in the following figures, which ones can intersect?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0088_1.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0088_2.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0088_3.jpg", "batch29-2024_06_14_304d3d2a7ea25f99adc4g_0088_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Question Analysis: Use the properties of lines, rays, and line segments to determine the correct answer.\n\nSolution: Among the line segments and rays in the figure, those that can intersect are:\nTherefore, the correct choice is D.\n\nKey Point: This question primarily tests the definitions of lines, rays, and line segments. The key to solving it lies in thoroughly understanding the definitions of line segments, rays, and lines." }, { "problem_id": 1096, "question": "The following beautiful patterns are not centrosymmetric figures: ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0047_1.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0047_2.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0047_3.jpg", "batch14-2024_06_15_1c4aa1edbaf2c872d895g_0047_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: Figure $\\mathrm{A}$ is centrally symmetric and does not meet the requirement; Figure $\\mathrm{B}$ is not centrally symmetric and meets the requirement; Figure $\\mathrm{C}$ is centrally symmetric and does not meet the requirement; Figure D is centrally symmetric and does not meet the requirement. Therefore, the correct choice is B.\n\n【Key Point】This question tests the understanding of centrally symmetric figures, and the key to solving it lies in correctly identifying the figures." }, { "problem_id": 1097, "question": "Which of the following options is the unfolded form of the cube-shaped paper box shown in the figure?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_329d65dd392595b77024g_0031_1.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0031_2.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0031_3.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0031_4.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0031_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement, the three faces with black dots are adjacent to each other.\n\nIn option A, the two faces with black dots are opposite each other, which does not meet the requirement.\n\nOptions B and C are not valid unfolded diagrams of a cube, thus they do not meet the requirement.\n\nOption D satisfies the condition.\n\nTherefore, the answer is: D.\n\n[Key Point] This question tests the understanding of the unfolded surface diagrams of a cube and spatial imagination. The key to solving it is to be familiar with the 11 types of unfolded surface diagrams of a cube and to carefully observe whether each marked position is consistent." }, { "problem_id": 1098, "question": "The graph of the functions $y = \\frac{1}{2} x - 1$ and $y = -\\frac{1}{2} x^{2}$ is ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_01846c8e63bedc6d99ceg_0018_1.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0018_2.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0018_3.jpg", "batch13-2024_06_15_01846c8e63bedc6d99ceg_0018_4.jpg" ], "is_multi_img": true, "answer": "C", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Question Analysis: In the equation \\( y = \\frac{1}{2}x - 1 \\), \\( k = \\frac{1}{2} > 0 \\), \\( b = -1 < 0 \\), therefore the line passes through the first, third, and fourth quadrants, thus eliminating options A and D.\n\nIn the equation \\( y = -\\frac{1}{2}x^{2} \\), since \\( a = -\\frac{1}{2} < 0 \\), the parabola opens downward, which further eliminates option B. Therefore, option C is correct. The answer is C.\n\nKey Point:" }, { "problem_id": 1099, "question": "Examine the three-dimensional figure below, and which of the following is the view from the front ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_13894f5e3499bf65de58g_0017_1.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0017_2.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0017_3.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0017_4.jpg", "batch10-2024_06_14_13894f5e3499bf65de58g_0017_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: When viewing the geometric figure from the front, the first row consists of one square positioned in the center, and the second row consists of three squares. Therefore, the correct choice is: A.\n\n[Key Point] This question tests the ability to view a three-dimensional figure from different angles. The key to solving it lies in paying attention to the number and position of the squares in each row." }, { "problem_id": 1100, "question": "While proving the Pythagorean theorem, Student A and Student B present the following two methods, as shown in the figure. Which of the following schemes is correct?\n\n\nStudent A\n\n\nStudent B\nA. A is correct\nB. B is correct\nC. Both are correct\nD. Both are incorrect", "input_image": [ "batch12-2024_06_15_d1b4ff40136381227588g_0025_1.jpg", "batch12-2024_06_15_d1b4ff40136381227588g_0025_2.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: The result obtained by A is: $(a+b)^{2}-4 \\times \\frac{1}{2} a b=c^{2}$,\n\nThat is, $a^{2}+b^{2}=c^{2}$, which meets the requirement of the problem;\n\nThe result obtained by B is: $(a+b)^{2}=a^{2}+b^{2}+4 \\times \\frac{1}{2} a b=a^{2}+b^{2}+2 a b$, which does not meet the requirement of the problem;\nTherefore, the correct choice is: A.\n\n[Key Insight] The problem mainly tests the ability to derive algebraic expressions from geometric figures and the verification of the Pythagorean theorem and the perfect square formula. Understanding the problem and solving it in conjunction with the figure is the key to the solution." }, { "problem_id": 1101, "question": "As shown in the figure, which of the following pairs is a pair of alternate interior angles? ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0044_1.jpg", "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0044_2.jpg", "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0044_3.jpg", "batch33-2024_06_14_89ff290fc548a9b5b5d7g_0044_4.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Question Analysis: If two angles are both on the inside of two lines and on opposite sides of a third line, then such a pair of angles is called alternate interior angles.\n\nAccording to the definition of alternate interior angles, only option A fits, so the answer is A.\n\nKey Point: Definition of Alternate Interior Angles\n\nComment: This question is a basic application problem. Students only need to be proficient in the definition of alternate interior angles to complete it." }, { "problem_id": 1102, "question": "The net of a geometric solid is shown in the figure. The solid is ( )\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_55584d21b09f596e3cf6g_0099_1.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0099_2.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0099_3.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0099_4.jpg", "batch10-2024_06_14_55584d21b09f596e3cf6g_0099_5.jpg" ], "is_multi_img": true, "answer": "A", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the net provided, the geometric figure is a cylinder.\n\nTherefore, the correct choice is A.\n\n[Key Insight] This question primarily tests the understanding of nets of geometric figures. Mastering the nets of simple geometric figures is crucial for solving such problems." }, { "problem_id": 1103, "question": "Given $k_{1} < 0 < k_{2}$, the graphs of the functions $y = k_{1} x - 3$ and $y = \\frac{k_{2}}{x}$ are approximately ( )\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch13-2024_06_15_a5bc38c105001883c8a7g_0009_1.jpg", "batch13-2024_06_15_a5bc38c105001883c8a7g_0009_2.jpg", "batch13-2024_06_15_a5bc38c105001883c8a7g_0009_3.jpg", "batch13-2024_06_15_a5bc38c105001883c8a7g_0009_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: Since \\( k_{1} < 0 < k_{2} \\), the functions \\( y = k_{1}x - 3 \\) and \\( y = \\frac{k_{2}}{x} \\) are plotted in the same coordinate system.\n\nTherefore, the graph of the inverse proportional function is distributed in the first and third quadrants, while the graph of the linear function passes through the second, third, and fourth quadrants, and goes through the point \\( (0, -3) \\).\n\nThus, only option D fits the description.\n\nHence, the answer is: D.\n\n[Key Insight] This question primarily tests the understanding of the graphs of inverse proportional functions and linear functions. Correctly grasping the distribution patterns of these function graphs is crucial for solving the problem." }, { "problem_id": 1104, "question": "A right-angled triangle is rotated around one of its legs. Which of the following solids could be formed?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch33-2024_06_14_452fbe550559079ef90fg_0079_1.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0079_2.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0079_3.jpg", "batch33-2024_06_14_452fbe550559079ef90fg_0079_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Analysis: Based on the principle that a surface in motion forms a solid, the answer can be derived.\n\nDetailed Explanation: By rotating a right-angled triangle around the axis of one of its perpendicular sides, a cone is formed. Therefore, the correct choice is D.\n\nKey Point: This question tests the concepts of points, lines, surfaces, and solids: a point moving forms a line, a line moving forms a surface, and a surface moving forms a solid. Rotating a right-angled triangle around the axis of one of its perpendicular sides results in a cone." }, { "problem_id": 1105, "question": "The figure is made up of five identical small cubes. When viewed from above, it appears as ( ).\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n\n\n##", "input_image": [ "batch10-2024_06_14_329d65dd392595b77024g_0033_1.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0033_2.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0033_3.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0033_4.jpg", "batch10-2024_06_14_329d65dd392595b77024g_0033_5.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: When viewed from above, there are two layers. The bottom layer has a square on the left, and the top layer consists of three squares.\n\nTherefore, the correct choice is: B.\n\n[Key Point] This question tests the ability to view simple composite shapes from different perspectives. Proficiency in applying the concept of combining numbers and shapes is crucial for solving the problem." }, { "problem_id": 1106, "question": "There are over 400 methods to prove the Pythagorean Theorem, with one of the most elegant being the wordless proof by \"Zhao Shuang's String Diagram.\" Which of the following patterns is \"Zhao Shuang's String Diagram\"?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch12-2024_06_15_ed1911a30eded62729c0g_0056_1.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0056_2.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0056_3.jpg", "batch12-2024_06_15_ed1911a30eded62729c0g_0056_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "From the graphical characteristics of \"Zhao Shuang's String Diagram,\" it is evident that option D is selected.\n\nTherefore, the answer is: D.\n\n[Highlight] This question examines the knowledge of the Pythagorean theorem and \"Zhao Shuang's String Diagram\"; the key to solving it lies in a thorough understanding of the Pythagorean theorem and familiarity with \"Zhao Shuang's String Diagram.\"" }, { "problem_id": 1107, "question": "Among the following diagrams, which one is NOT an unfolding pattern of a cube's surface?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_1.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_2.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_3.jpg", "batch10-2024_06_14_a3c25c228bc66efc7ebcg_0081_4.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem statement,\n\n\n\ncannot be assembled into a cube,\n\ntherefore, the correct choice is: D.\n\n[Key Point] This question primarily tests the understanding of the net of a cube. Mastering the net of a cube is crucial for solving such problems." }, { "problem_id": 1108, "question": "Among the following figures, which one is an unfolded net of a cube?\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch25-2024_06_17_2eea509d719aacf4fdebg_0088_1.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0088_2.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0088_3.jpg", "batch25-2024_06_17_2eea509d719aacf4fdebg_0088_4.jpg" ], "is_multi_img": true, "answer": "B", "answer_type": "choice", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "**Analysis:** Solve the problem based on the folding of planar figures and the unfolding diagrams of a cube.\n\n**Detailed Explanation:** A cube has a total of 11 different surface unfolding diagrams. Option **B** can form a cube; options **A**, **C**, and **D** cannot be folded into a cube. Therefore, the correct choice is **B**.\n\n**Key Point:** This question tests students' spatial thinking ability. When solving the problem, do not forget the characteristics of a rectangular prism and the various scenarios of cube unfolding diagrams." }, { "problem_id": 1109, "question": "As shown in the figure, if the cube is unfolded, which of the following patterns can be obtained?\n\n\nA.\n\n\nB.\n\n\nC.\n\n\nD.\n\n", "input_image": [ "batch10-2024_06_14_3a73c762f5eb2b761854g_0086_1.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0086_2.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0086_3.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0086_4.jpg", "batch10-2024_06_14_3a73c762f5eb2b761854g_0086_5.jpg" ], "is_multi_img": true, "answer": "D", "answer_type": "choice", "difficulty": "Low", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, if the cube is unfolded, we can obtain\n\n\nTherefore, the correct choice is: D.\n\n[Key Point] This question primarily tests the understanding of the unfolded diagram of a cube. Mastering the characteristics of the unfolded square diagram to determine the opposite and adjacent faces of each side is the key to solving the problem." }, { "problem_id": 1110, "question": "The tangram is an outstanding creation of our ancestors, praised by Westerners as the \"Magic Plate of the East.\" Xiao Lin arranged a set of tangram pieces from Figure 1 into the \"garment\" (shaded area) in Figure 2, and placed it into a grid diagram where the side length of each small square is 1. Then, the perimeter of this \"garment\" is $\\qquad$ (taking $\\sqrt{2}$ as 1.4).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_0f190841cfcc4cbc9648g_0016_1.jpg", "batch10-2024_06_14_0f190841cfcc4cbc9648g_0016_2.jpg" ], "is_multi_img": true, "answer": "15.6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\nSince all the triangles within the tangram are isosceles right-angled triangles,\n\nTherefore, all acute angles are equal to $45^{\\circ}$.\nSince the side length of the small square in the grid is 1,\n\nThus, $OA = 2OG = 2 = OC = OD$,\n\nTherefore, $AB = BC = CD = DA = 2\\sqrt{2}$,\n\n$AE = BE = BF = CF = HI = \\sqrt{2}$;\n\nAs shown in the figure, when the tangram is assembled into the \"clothing shape\",\n\n\n\nThen the perimeter of the \"clothing shape\" is: $4\\sqrt{2} + 4 \\times 1 + 3 \\times 2 = 10 + 4\\sqrt{2} = 10 + 4 \\times 1.4 = 15.6$.\n\nHence, the answer is: 15.6.\n\n【Key Insight】This problem mainly examines plane geometry, and determining the side lengths of each shape within the tangram is crucial for solving the problem." }, { "problem_id": 1111, "question": "A rectangular piece of paper has a number line on it. A segment of 10 units in length (from -3 to 7) is cut out and folded along a certain point. Then, a cut is made at a certain point in the overlapping part to obtain three segments (as shown in the figure). If the lengths of these three segments are in the ratio of 1:2:2, the number corresponding to the point of the fold could be $\\qquad$ .\n\n\n\nFold\n\n\n\nCut", "input_image": [ "batch10-2024_06_14_0f190841cfcc4cbc9648g_0031_1.jpg", "batch10-2024_06_14_0f190841cfcc4cbc9648g_0031_2.jpg" ], "is_multi_img": true, "answer": "$1,2,3$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "According to the problem, let the lengths of the three segments be $x$, $2x$, and $2x$ units respectively. Then, $x + 2x + 2x = 10$,\n\nSolving gives: $x = 2$,\n\nThus, the lengths of the three segments are $2$, $4$, and $4$ units respectively.\n\nIf the first segment cut is 2 units long,\n\nThen the number represented by the point at the fold is: $-3 + 2 + 2 = 1$,\n\nIf the first segment cut is 4 units long and the second segment is 2 units long,\n\nThen the number represented by the point at the fold is: $-3 + 4 + 1 = 2$,\n\nIf the first segment cut is 4 units long and the second segment is also 4 units long,\n\nThen the number represented by the point at the fold is: $-3 + 4 + 2 = 3$,\n\nIn summary, the number represented by the point at the fold is 1, 2, or 3,\n\nTherefore, the answer is: $1, 2, 3$.\n\n【Highlight】This question tests the combination of number lines and line segments. The key to solving it is to list all possible orders of the three segments." }, { "problem_id": 1112, "question": "As shown in the figure, Figure 1 is a 3rd-order pyramid Rubik's Cube, which is composed of several small triangular pyramids stacked into a large triangular pyramid (Figure 2). The four faces of the large triangular pyramid are painted with colors. If a small triangular pyramid with one face painted is called a center piece, one with two faces painted is called an edge piece, and one with three faces painted is called a corner piece, then in the 3rd-order pyramid Rubik's Cube, the expression \" (number of edge pieces) + (number of corner pieces) - (number of center pieces) \" equals $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_0f190841cfcc4cbc9648g_0083_1.jpg", "batch10-2024_06_14_0f190841cfcc4cbc9648g_0083_2.jpg" ], "is_multi_img": true, "answer": "-2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: \n\n$\\because$ For a small triangular pyramid with 1 face colored, there are 3 on each face,\n\n$\\therefore$ The total number of center blocks is: $3 \\times 4=12$;\n\n$\\because$ For a small triangular pyramid with 2 faces colored, there is 1 on each edge,\n\n$\\therefore$ The total number of edge blocks is: $1 \\times 6=6$;\n\n$\\because$ The small triangular pyramid with 3 faces colored is located at the four vertices,\n\n$\\therefore$ The total number of corner blocks is: 4;\n\n$\\therefore$ (Number of edge blocks) + (Number of corner blocks) - (Number of center blocks) $=6+4-12=-2$,\n\nHence, the answer is: -2.\n\n【Key Insight】This question tests the understanding of three-dimensional shapes and the patterns of shapes. Correctly understanding the characteristics of the \"third-order pyramid Rubik's cube\" is key to solving this problem." }, { "problem_id": 1113, "question": "As shown, the six faces of a cube are labeled with six consecutive integers, and the sum of the two integers on opposite faces is equal. When this cube is placed on a table and rolled in the manner shown, each roll of $90^{\\circ}$ counts as one roll. After rolling 2022 times, what is the number on the face of the cube that is in contact with the table?\n\n\n\n1st roll\n\n\n\n2nd roll\n\n\n\n3rd roll", "input_image": [ "batch10-2024_06_14_0f190841cfcc4cbc9648g_0092_1.jpg", "batch10-2024_06_14_0f190841cfcc4cbc9648g_0092_2.jpg", "batch10-2024_06_14_0f190841cfcc4cbc9648g_0092_3.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By observing the pattern, we know that the first time, 10 and 9 are opposite, the second time, 7 and 12 are opposite, the third time, 9 and 10 are opposite, the fourth time, 12 and 7 are opposite, the fifth time, 10 and 9 are opposite, and this cycle repeats every four times.\n\n$\\because 2022 \\div 4=505 \\cdots 2$,\n\n$\\therefore$ After rolling 2022 times, it is the same as the second time,\n\n$\\therefore$ The number facing down is 7,\n\nHence, the answer is: 7.\n\n【Key Insight】This question examines the text on opposite faces of a cube and the pattern of graphical changes. The key to solving the problem is to identify the pattern." }, { "problem_id": 1114, "question": "As shown in the figure, in the plane rectangular coordinate system, the coordinates of the vertex $\\mathrm{A}$ of the rectangle $\\mathrm{ABCD}$ are $(3,6)$, $\\mathrm{AB}=6$, $\\mathrm{AD}=3$. The rectangle is translated downward by $m$ units, such that two of its vertices simultaneously lie on the graph of a certain inverse proportional function. Then, $m=$.\n\n\n\n", "input_image": [ "batch10-2024_06_14_170dda5506ef8d04565bg_0072_1.jpg", "batch10-2024_06_14_170dda5506ef8d04565bg_0072_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{3}{2}$ or $\\frac{15}{2}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Question Analysis:** According to the problem statement, we have the following points: A (3,6); B (9,6); C $(9,3)$; D (3,3). After translating these points downward, the new coordinates are: $\\mathrm{A}(3,6-\\mathrm{m}) ; \\mathrm{B}(9,6-\\mathrm{m}) ; \\mathrm{C}(9,3-\\mathrm{m}) ; \\mathrm{D}(3,3-\\mathrm{m})$. \n\nWhen points $\\mathrm{A}$ and $\\mathrm{C}$ lie on the same inverse proportionality function graph, the equation $3(6-\\mathrm{m})=9(3-\\mathrm{m})$ holds true, solving which we get: $\\mathrm{m}=\\frac{3}{2}$. \n\nSimilarly, when points $\\mathrm{B}$ and $\\mathrm{D}$ lie on the same inverse proportionality function graph, the equation $9(6-\\mathrm{m})=3(3-\\mathrm{m})$ holds true, solving which we get: $\\mathrm{m}=\\frac{15}{2}$. \n\n**Key Point:** Properties of inverse proportionality functions." }, { "problem_id": 1115, "question": "As shown in Figure 1, villages A and B are located on opposite sides of a river (ignoring the width of the river). A wharf is to be constructed such that the sum of the distances from the wharf to both villages A and B is minimized. The point C shown in Figure 2 represents the desired location of the wharf. The reason for this is $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_1d3680966d15340f6fc7g_0020_1.jpg", "batch10-2024_06_14_1d3680966d15340f6fc7g_0020_2.jpg" ], "is_multi_img": true, "answer": "Between two points, the shortest path is a straight line", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: The point $C$ shown in Figure 2 is the location of the desired dock. The reason for this is that between two points, the shortest path is a straight line.\n\nTherefore, the answer is: Between two points, the shortest path is a straight line.\n\n[Key Insight] This question primarily examines the distance between two points. Understanding that the shortest path between two points is a straight line is crucial for solving the problem." }, { "problem_id": 1116, "question": "As shown in the figure, in rectangle $A B C D$, $A B=a, B C=b$, and $a>b$. When rectangle $A B C D$ is rotated around the line containing side $A B$, it forms cylinder A. Then, when rectangle $A B C D$ is rotated around the line containing side $B C$, it forms cylinder B. Let the lateral areas of the two cylinders be $S_{\\text {A }}$ and $S_{\\text {B }}$, respectively. Then $S_{\\text {A }}$ $\\qquad$ $S_{\\text {B }}$. (Fill in the blank with >, <, or =)\n\n\n\nA\n\n", "input_image": [ "batch10-2024_06_14_1d3680966d15340f6fc7g_0056_1.jpg", "batch10-2024_06_14_1d3680966d15340f6fc7g_0056_2.jpg" ], "is_multi_img": true, "answer": "$=$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Since \\( S_{\\text{ A}} = 2 \\pi \\cdot AD \\cdot AB \\) and \\( S_{\\text{ B}} = 2 \\pi \\cdot AB \\cdot AD \\),\n\nTherefore, \\( S_{\\text{ A}} = S_{\\text{ B}} \\).\n\nHence, the answer is: \\( = \\).\n\n**Key Insight:** This question examines the formation of a cylinder and the calculation of its lateral surface area. Understanding the correct formula for calculating the lateral surface area is crucial for solving the problem." }, { "problem_id": 1117, "question": "As shown in the figure, a rectangular piece of paper is cut along its diagonal, resulting in two triangular pieces of paper. These two triangular pieces are then arranged in the form shown in Figure (3), with points $B, F, C, D$ all lying on the same straight line. If $AC = 8$, $EF = 6$, and $CF = 4$, then the length of $BD$ is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch10-2024_06_14_1d3680966d15340f6fc7g_0063_1.jpg", "batch10-2024_06_14_1d3680966d15340f6fc7g_0063_2.jpg", "batch10-2024_06_14_1d3680966d15340f6fc7g_0063_3.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In the rectangular piece of paper, \\( FD = AC = 8 \\) and \\( BC = EF = 6 \\).\n\nFrom Figure (3), it can be seen that:\n\\[ BC + FD = BD + CF \\]\n\\[ \\therefore 6 + 8 = BD + 4 \\]\n\\[ \\therefore BD = 10 \\]\n\nThe answer is **10**.\n\n**Key Insight:** This problem primarily tests the calculation of the sum and difference of line segments. The key to solving it lies in correctly representing the relationships between the sums and differences of the segments." }, { "problem_id": 1118, "question": "Watching the five-star red flag rising, have you ever wondered how tall the flagpole actually is? A math interest group conducted the following operations to measure the height of the flagpole: As shown in Figure 1, they first pulled the flag rope to the bottom of the flagpole and found that the end of the rope just touched the ground; as shown in Figure 2, they then pulled the end of the rope to a point $8 \\mathrm{~m}$ away from the flagpole and found that the end of the rope was $2 \\mathrm{~m}$ above the ground. Please calculate the height of the flagpole based on the above measurements.\n\n\n\nFigure 1\n\n\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_221deb15348aa6cdc5ffg_0041_1.jpg", "batch10-2024_06_14_221deb15348aa6cdc5ffg_0041_2.jpg", "batch10-2024_06_14_221deb15348aa6cdc5ffg_0041_3.jpg" ], "is_multi_img": true, "answer": "17 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nLet the height of the flagpole be $x \\mathrm{~m}$. Then, $A C = A D = x \\mathrm{~m}$, $A B = (x - 2) \\mathrm{m}$, and $B C = 8 \\mathrm{~m}$. In the right triangle $\\triangle A B C$,\n\n$A B^{2} + B C^{2} = A C^{2}$\n\n$(x - 2)^{2} + 8^{2} = x^{2}$\n\nSolving the equation, we get: $x = 17$,\n\n## Answer: The height of the flagpole is $17 \\mathrm{~m}$.\n\n【Key Point】This problem tests the application of the Pythagorean theorem. The key to solving it is constructing a right triangle." }, { "problem_id": 1119, "question": "As shown in the figure, 1 line divides the plane into 2 parts, 2 lines can divide the plane into a maximum of 4 parts, 3 lines can divide the plane into a maximum of 7 parts, and 4 lines can divide the plane into a maximum of 11 parts. Given that $\\mathrm{n}$ lines can divide the plane into a maximum of 56 parts, the value of $\\mathrm{n}$ is $\\qquad$.\n\n\n\n1 line\n", "input_image": [ "batch10-2024_06_14_27c98d8d4f6360694554g_0066_1.jpg", "batch10-2024_06_14_27c98d8d4f6360694554g_0066_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: The maximum number of parts into which $n$ lines can divide a plane is given by $S=1+1+2+3+\\ldots+n=\\frac{1}{2} n(n+1)+1$. Given that $n$ lines can divide the plane into a maximum of 56 parts,\n\nwe have the equation $\\frac{1}{2} n(n+1)+1=56$,\n\nSolving this equation yields $x_{1}=-11$ (which is discarded as it does not fit the context), and $x_{2}=10$.\n\nTherefore, the value of $n$ is 10.\n\nHence, the answer is: 10." }, { "problem_id": 1120, "question": "A cube is labeled with the numbers 1, 2, 3, 4, 5, and 6 on each of its faces. According to the numbers displayed by the cube in the three states shown in the figures (1), (2), and (3), the number at the \"?\" position is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch10-2024_06_14_27c98d8d4f6360694554g_0078_1.jpg", "batch10-2024_06_14_27c98d8d4f6360694554g_0078_2.jpg", "batch10-2024_06_14_27c98d8d4f6360694554g_0078_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From (1) and (2), we can deduce:\n\n1. The four surrounding faces are $4, 5, 2, 3$, so the opposite face of 1 is 6;\n2. From (2) and (3), the numbers adjacent to 3 are $1, 2, 5, 6$, so the opposite face of 3 is 4, and 2 is opposite to 5.\n\nTherefore, the number in question must be one of the two numbers 1 or 6.\n\nSince 6 is adjacent to both $3$ and $5$, the number in the ? position is 6.\n\nThus, the answer is 6.\n\n**Key Insight:** This problem primarily tests the student's spatial imagination and reasoning abilities. The key to solving this problem lies in deducing the three pairs of opposite numbers from the visible numbers in the three cubes. Alternatively, one could also arrive at the solution through hands-on manipulation." }, { "problem_id": 1121, "question": "The front view, left view, and top view of a part are shown in the figure (dimensions in centimeters). The volume of this part is $\\qquad$ cubic centimeters.\n\n\n\n10\n\nFront View\n\n\n\n12\n\nLeft View\n\n\n\n10\n\nTop View", "input_image": [ "batch10-2024_06_14_27c98d8d4f6360694554g_0083_1.jpg", "batch10-2024_06_14_27c98d8d4f6360694554g_0083_2.jpg", "batch10-2024_06_14_27c98d8d4f6360694554g_0083_3.jpg" ], "is_multi_img": true, "answer": "1800", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the part is a rectangular prism with a length of \\(10 \\mathrm{~cm}\\), a width of \\(12 \\mathrm{~cm}\\), and a height of \\(15 \\mathrm{~cm}\\). The volume is calculated as \\(10 \\times 12 \\times 15 = 1800 \\mathrm{~cm}^{3}\\).\n\nTherefore, the answer is: \\(1800 \\mathrm{~cm}^{3}\\)." }, { "problem_id": 1122, "question": "The \"lattice multiplication\" method, as a way to multiply two numbers, was first proposed by the Italian mathematician Pacioli in the 15th century. It was referred to as \"paved ground brocade\" in the book \"The Complete Book of Calculations\" by the Ming Dynasty mathematician Cheng Dawei. For example: as shown in Figure 1, to calculate $46 \\times 71$, write the multiplicand 46 above the grid and the multiplier 71 to the right of the grid. Then multiply each digit of the multiplicand 46 by each digit of the multiplier 71, and record the results in the corresponding grid. Finally, add along the diagonal directions to get 3266. As shown in Figure 2, using the \"lattice multiplication\" method to calculate the multiplication of two two-digit numbers, then $k=$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_27c98d8d4f6360694554g_0096_1.jpg", "batch10-2024_06_14_27c98d8d4f6360694554g_0096_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: According to the problem, we have\n\n\\[ 10 (10 + 6 - k - k) + (k - 3 - 1) = 7k \\]\n\nSolving the equation, we find:\n\n\\[ k = 6 \\]\n\nTherefore, the answer is: 6.\n\n\n\n0\n\n【Key Point】This problem primarily tests the application of linear equations in one variable. The key is to analyze the diagram using the \"lattice multiplication\" method and set up the equation accordingly." }, { "problem_id": 1123, "question": "The shelves of a supermarket are stocked with a certain brand of braised beef instant noodles. The images below show the shapes viewed from three different directions. The minimum number of boxes of instant noodles on the shelves is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from above", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0040_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0040_2.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0040_3.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the first layer has 4 barrels, the second layer has at least 3 barrels, and the third layer has 2 barrels. Therefore, there are at least 9 barrels in total.\n\nThus, the answer is: 9.\n\n[Highlight] This question tests the understanding of viewing geometric shapes from different perspectives. The key to solving it lies in comprehending the problem statement." }, { "problem_id": 1124, "question": "As shown in Figure 1, the ray $O C$ is inside $\\angle A O B$. There are a total of 3 angles in the figure: $\\angle A O B, \\angle A O C$, and $\\angle B O C$. If the degree measure of one of these angles is twice that of another, then the ray $O C$ is called the \"clever dividing line\" of $\\angle A O B$. As shown in Figure 2, if $\\angle M P N = 60^{\\circ}$ and the ray $P Q$ is the \"clever dividing line\" of $\\angle M P N$, then $\\angle M P Q =$ $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0043_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0043_2.jpg" ], "is_multi_img": true, "answer": "$20^{\\circ}$ or $40^{\\circ}$ or $30^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the problem statement, we have:\n\nWhen $\\angle MPN = 2\\angle MPQ = 2\\angle NPQ$,\n\n$\\because \\angle MPN = 60^\\circ$,\n\n$\\therefore \\angle MPQ = \\frac{1}{2} \\angle MPN = \\frac{1}{2} \\times 60^\\circ = 30^\\circ$;\n\nWhen $\\angle MPQ = 2\\angle NPQ$,\n\n$\\because \\angle MPN = 60^\\circ$,\n\n$\\therefore \\angle MPQ + \\angle NPQ = 60^\\circ$,\n\n$\\therefore \\frac{3}{2} \\angle MPQ = 60^\\circ$,\n\n$\\therefore \\angle MPQ = 40^\\circ$;\n\nWhen $\\angle NPQ = 2\\angle MPQ$,\n\n$\\because \\angle MPN = 60^\\circ$,\n\n$\\therefore \\angle MPQ + \\angle NPQ = 60^\\circ$,\n\n$\\therefore 3\\angle MPQ = 60^\\circ$,\n\n$\\therefore \\angle MPQ = 20^\\circ$;\n\nIn summary, $\\angle MPQ$ can be $20^\\circ$, $40^\\circ$, or $30^\\circ$. Therefore, the answer is $20^\\circ$, $40^\\circ$, or $30^\\circ$.\n\n【Key Insight】This problem tests the understanding of new definitions and the concept of categorical discussion. The key to solving it lies in discussing the three different cases separately." }, { "problem_id": 1125, "question": "As shown in Figure 1, there is a point \\( O \\) on the line \\( E D \\). Through point \\( O \\), a ray \\( O C \\) is drawn above the line \\( E D \\). A right triangle \\( A O B \\) (\\( \\angle O A B = 30^\\circ \\)) has its right angle vertex at point \\( O \\), with one leg \\( O A \\) on ray \\( O D \\) and the other leg \\( O B \\) above the line \\( E D \\). The right triangle is rotated around point \\( O \\) at a speed of \\( 10^\\circ \\) per second counterclockwise for a full cycle, with the rotation time being \\( t \\) seconds. If the position of ray \\( O C \\) remains unchanged and \\( \\angle C O E = 140^\\circ \\), then during the rotation, as shown in Figure 2, when\n\n\\( t = \\) \\(\\qquad\\) seconds, one of the rays \\( O A, O C, \\) and \\( O D \\) exactly bisects the angle formed by the other two rays.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0050_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0050_2.jpg" ], "is_multi_img": true, "answer": "2 or 8 or 32", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: When the ray \\( O A \\) is the angle bisector of \\( \\angle C O D \\),\n\n\\[\n\\because \\angle C O D = 180^{\\circ} - \\angle C O E = 40^{\\circ},\n\\]\n\n\\[\n\\therefore \\angle A O D = \\frac{1}{2} \\angle C O D = 20^{\\circ},\n\\]\n\n\\[\n\\therefore t = \\frac{20}{10} = 2;\n\\]\n\nWhen the ray \\( O C \\) is the angle bisector of \\( \\angle A O D \\),\n\n\\[\n\\angle A O D = 2 \\angle C O D = 80^{\\circ},\n\\]\n\n\\[\n\\therefore t = \\frac{80}{10} = 8;\n\\]\n\nWhen the ray \\( O D \\) is the angle bisector of \\( \\angle C O A \\),\n\n\\[\n360 - 10t = 40,\n\\]\n\n\\[\n\\therefore t = 32,\n\\]\n\nTherefore, the answers are: 2, 8, or 32.\n\n【Key Insight】This problem tests the understanding of angle bisectors. Grasping the definition of an angle bisector is crucial for correctly solving the problem." }, { "problem_id": 1126, "question": "As shown in Figure 1, for the paper angle $\\angle A O B$, $O C$ bisects $\\angle A O B$. As shown in Figure 2, $\\angle A O B$ is folded along $O C$ to form $\\angle C O B$ (with $O A$ coinciding with $O B$). From point $O$, a ray $O E$ is drawn such that $\\angle B O E = \\frac{1}{2} \\angle E O C$. If $\\angle C O E = 40^\\circ$, then $\\angle A O B =$ $\\qquad$ $\\circ$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0053_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0053_2.jpg" ], "is_multi_img": true, "answer": "120", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since \\(\\angle COE = 40^\\circ\\) and \\(\\angle BOE = \\frac{1}{2} \\angle EOC\\),\n\n\\(\\therefore \\angle BOE = \\frac{1}{2} \\times 40^\\circ = 20^\\circ\\),\n\n\\(\\therefore \\angle BOC = \\angle BOE + \\angle EOC = 20^\\circ + 40^\\circ = 60^\\circ\\),\n\nSince \\(OC\\) bisects \\(\\angle AOB\\),\n\n\\(\\therefore \\angle AOB = 2 \\angle BOC = 2 \\times 60^\\circ = 120^\\circ\\),\n\nTherefore, the answer is: 120\n\n【Key Point】This question primarily examines the relationship between angles when an angle bisector divides an angle into equal parts." }, { "problem_id": 1127, "question": "As shown in Figure (1), there is a square cardboard with a side length of $12 \\mathrm{~cm}$. After cutting off the shaded part, it is folded into a rectangular box as shown in Figure (2). It is known that the width of the rectangular box is twice its height. Then, its volume is $\\qquad$ $\\mathrm{cm}^{3}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0058_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0058_2.jpg" ], "is_multi_img": true, "answer": "64", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Let the height of the cuboid be \\( x \\mathrm{~cm} \\), then the width of the cuboid is \\( 2x \\mathrm{~cm} \\), and the length is \\( (12 - 2x) \\mathrm{cm} \\). According to the problem, we have the equation:\n\n\\[ 2x + 2x + x + x = 12 \\]\n\nSolving the equation gives:\n\n\\[ x = 2 \\]\n\nTherefore, the height of the cuboid is \\( 2 \\mathrm{~cm} \\), the width is \\( 4 \\mathrm{~cm} \\), and the length is \\( 12 - 2 \\times 2 = 8 \\mathrm{~cm} \\).\n\nThus, its volume is:\n\n\\[ 2 \\times 4 \\times 8 = 64 \\mathrm{~cm}^3 \\]\n\nHence, the answer is: 64.\n\n**Key Insight:** This problem primarily tests the application of solving a linear equation and the calculation of a cuboid's volume. The key is to set up the unknown variable, establish the relationship between the sides, derive the equation, solve for the dimensions (length, width, height), and then compute the volume." }, { "problem_id": 1128, "question": "Fold a rectangular piece of paper with length 2 and width $a$ $(1\n\nFirst operation\n\n\n\nSecond operation", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0063_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0063_2.jpg" ], "is_multi_img": true, "answer": "1.2 or 1.5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: In the first operation, a square with side length \\( a \\) is cut out, leaving a rectangle with dimensions \\( a \\) and \\( 2 - a \\). Given that \\( 1 < a < 2 \\), it follows that \\( a > 2 - a \\). In the second operation, a square with side length \\( 2 - a \\) is cut out, so the remaining rectangle has sides \\( 2 - a \\) and \\( a - (2 - a) = 2a - 2 \\).\n\n(1) When \\( 2a - 2 < 2 - a \\), i.e., \\( a < \\frac{4}{3} \\), then in the third operation, a square with side length \\( 2a - 2 \\) is cut out, leaving a rectangle with sides \\( 2a - 2 \\) and \\( (2 - a) - (2a - 2) = 4 - 3a \\). Setting \\( 2a - 2 = 4 - 3a \\), we solve for \\( a \\) and find \\( a = 1.2 \\).\n\n(2) When \\( 2a - 2 > 2 - \\frac{4}{3}a \\), i.e., \\( a > 1.2 \\), then in the third operation, a square with side length \\( 2 - a \\) is cut out, leaving a rectangle with sides \\( 2 - a \\) and \\( (2a - 2) - (2 - a) = 3a - 4 \\). Setting \\( 2 - a = 3a - 4 \\), we solve for \\( a \\) and find \\( a = 1.5 \\).\n\nIn summary, the value of \\( a \\) is either 1.2 or 1.5.\n\nTherefore, the answer is: 1.2 or 1.5.\n\n**Key Insight:** This problem primarily tests the application of linear equations in one variable. Proficiency in applying linear equations and the concept of case analysis is crucial for solving such problems." }, { "problem_id": 1129, "question": "A set of triangles are joined together as shown in Figure 1, with sides $O A$ and $O C$ coinciding with the straight line $E F$, $\\angle A O B=45^{\\circ}$, $\\angle C O D=60^{\\circ}$. Keeping triangle $C O D$ stationary, triangle $A O B$ is rotated clockwise around point $O$ by an angle $\\alpha$ (as shown in Figure 2). Throughout the rotation, both triangles remain above the straight line $E F$. When $O B$ bisects any of the angles formed by any two of the sides $O A, O C, O D$, the value of $\\alpha$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0077_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0077_2.jpg" ], "is_multi_img": true, "answer": "$30^{\\circ}$ or $105^{\\circ}$ or $90^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: \n\n(1) When \\( OB \\) is on the left side of \\( OD \\) and bisects \\( \\angle AOD \\),\n\nGiven \\( \\angle AOB = 45^\\circ \\) and \\( \\angle COD = 60^\\circ \\),\n\nTherefore, \\( \\alpha = 180^\\circ - 45^\\circ - 45^\\circ - 60^\\circ = 30^\\circ \\).\n\n(2) When \\( OB \\) is on the right side of \\( OD \\) and bisects \\( \\angle COD \\),\n\nGiven \\( \\angle COD = 60^\\circ \\),\n\nTherefore, \\( \\angle DOB = 60^\\circ \\div 2 = 30^\\circ \\),\n\nGiven \\( \\angle AOB = 45^\\circ \\),\n\nTherefore, \\( \\angle AOD = 45^\\circ - 30^\\circ = 15^\\circ \\),\n\nThus, \\( \\alpha = 180^\\circ - 60^\\circ - 15^\\circ = 105^\\circ \\).\n\n(3) When \\( OB \\) is on the right side of \\( OD \\) and bisects \\( \\angle AOC \\),\n\nGiven \\( \\angle AOB = 45^\\circ \\),\n\nTherefore, \\( \\angle AOC = 45^\\circ \\times 2 = 90^\\circ \\),\n\nThus, \\( \\alpha = 180^\\circ - 90^\\circ = 90^\\circ \\).\n\nIn summary, the value of \\( \\alpha \\) is \\( 30^\\circ \\), \\( 105^\\circ \\), or \\( 90^\\circ \\).\n\n【Key Insight】This problem examines the concept of angle bisectors and angle addition/subtraction. The key to solving it lies in categorizing the position of \\( OB \\)." }, { "problem_id": 1130, "question": "Fold the unfolded cube diagram shown in Figure 1 into a cube (with the text facing outward), and then roll this cube according to Figure 2, sequentially to the 1st grid, the 2nd grid, the 3rd grid, and the 4th grid. The text on the top face of the cube at this point is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0085_1.jpg", "batch10-2024_06_14_41b2ae3ef6db68b5bdacg_0085_2.jpg" ], "is_multi_img": true, "answer": "富", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 1, it can be deduced that \"富\" (wealth) is opposite to \"文\" (culture); \"强\" (strength) is opposite to \"主\" (master); and \"民\" (people) is opposite to \"明\" (brightness). From Figure 2, when the small cube is flipped sequentially from its position in Figure 2 to the 4th position, \"文\" is on the bottom. Therefore, the character on the top face of the small cube at this moment is \"富\".\n\nThus, the answer is: 富.\n\n[Key Insight] This question primarily tests the understanding of the unfolded diagram of a cube. The key to solving it lies in identifying the characters on opposite faces and determining the character on the bottom after rotation." }, { "problem_id": 1131, "question": "As shown in Figure (1), it is given that the area of the small square \\( A B C D \\) is 1. Extend each of its sides by a factor of two to obtain the new square \\( A_{1} B_{1} C_{1} D_{1} \\); extend the sides of square \\( A_{1} B_{1} C_{1} D_{1} \\) by the same method to obtain square \\( A_{2} B_{2} C_{2} D_{2} \\), as shown in Figure (2) ...; continuing this process, the area of square \\( A_{2023} B_{2023} C_{2023} D_{2023} \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_701e774503a0c67a90acg_0044_1.jpg", "batch10-2024_06_14_701e774503a0c67a90acg_0044_2.jpg" ], "is_multi_img": true, "answer": "$5^{2023}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since the square of the side length of square $A_{1} B_{1} C_{1} D_{1}$ is: $(1+1)^{2}+1^{2}=5$,\n\nTherefore, the area of square $A_{1} B_{1} C_{1} D_{1}$ is: 5.\n\nMoreover, since the square of the side length of square $A_{2} B_{2} C_{2} D_{2}$ is: $(2 \\sqrt{5})^{2}+(\\sqrt{5})^{2}=25$,\n\nThus, the area of square $A_{2} B_{2} C_{2} D_{2}$ is: $25=5^{2}$.\n\nFollowing this pattern, the square of the side length of square $A_{3} B_{3} C_{3} D_{3}$ is: $(2 \\times 5)^{2}+5^{2}=125$,\n\nTherefore, the area of square $A_{3} B_{3} C_{3} D_{3}$ is: $125=5^{3}$.\n\nHence, the area of square $A_{n} B_{n} C_{n} D_{n}$ is $5^{n}$.\n\nConsequently, the area of square $A_{2023} B_{2023} C_{2023} D_{2023}$ is: $5^{2023}$.\n\nThus, the answer is: $5^{2023}$.\n\n【Key Insight】This problem examines the pattern of change in geometric figures. The key to solving it lies in identifying the relationship between the area of the square and its index number $n$." }, { "problem_id": 1132, "question": "As shown in the figure, a small cube with its surface fully coated in dye is placed on a piece of white paper. Without lifting the cube off the paper, the cube is rotated so that each of its faces can imprint on the paper, with each face touching the paper only once. The possible shapes that can be formed on the paper are $\\qquad$. (Fill in the number)\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)", "input_image": [ "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0058_1.jpg", "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0058_2.jpg", "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0058_3.jpg", "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0058_4.jpg" ], "is_multi_img": true, "answer": "(1)(3)\\#\\#(3)(1)", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: Based on the characteristics of the surface development diagram of a cube,\n\n(1), (3), and (4) are its development diagrams, while (2) is not.\n\nHowever, since the cube rolls and each face can only touch the white paper once, (4) does not meet the condition.\n\nTherefore, the diagrams that meet the condition are (1) and (3).\n\nHence, the answer is: (1) and (3).\n\n[Key Insight] This question tests the understanding of three-dimensional shapes. Mastering the characteristics of the cube's surface development diagram is essential for correct judgment. Understanding that \"when the cube is rotated, each face can only touch the white paper once\" is crucial for making the correct judgment." }, { "problem_id": 1133, "question": "As shown in the figure, a segment of 6 units in length (from -1 to 5) is cut from the number line. This segment is then folded to the left at a certain point, and a cut is made at a point in the overlapping part, resulting in three segments. It is found that the lengths of these three segments are in the ratio of 1:1:2. The number corresponding to the point of the fold could be $\\qquad$.\n\n\n\nFold\n\n\n\nCut point", "input_image": [ "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0064_1.jpg", "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0064_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{5}{4}$ or 2 or $\\frac{11}{4}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since the ratio of the lengths of the three segments is \\(1:1:2\\), \nlet the lengths of the three segments be \\(x\\), \\(x\\), and \\(2x\\) respectively. \n\nGiven that the distance from \\(-1\\) to \\(5\\) is \\(6\\), \nwe have \\(4x = 6\\), \nwhich solves to \\(x = \\frac{3}{2}\\). \n\nTherefore, the lengths of the three segments are \\(\\frac{3}{2}\\), \\(\\frac{3}{2}\\), and \\(3\\), \nas shown in the figure: \n\n \n\n(1) When \\(AB:BC:CD = 1:1:2\\), the number represented by the fold point is: \n\\[\n-1 + \\frac{3}{2} + \\frac{3}{2} \\times \\frac{1}{2} = \\frac{5}{4};\n\\] \n\n(2) When \\(AB:BC:CD = 1:2:1\\), the number represented by the fold point is: \n\\[\n-1 + \\frac{3}{2} + \\frac{3}{2} = 2;\n\\] \n\n(3) When \\(AB:BC:CD = 2:1:1\\), the number represented by the fold point is: \n\\[\n-1 + 3 + \\frac{3}{2} \\times \\frac{1}{2} = \\frac{11}{4}.\n\\] \n\nIn summary, the number corresponding to the fold point may be \\(\\frac{5}{4}\\), \\(2\\), or \\(\\frac{11}{4}\\). \n\nThus, the answer is: \\(\\frac{5}{4}\\), \\(2\\), or \\(\\frac{11}{4}\\). \n\n**Key Insight**: This problem examines the relationship between real numbers and the number line. Mastery of the characteristics of points on the number line, the method for calculating distances between two points, the properties of folding, and the use of the midpoint formula to solve folding problems is essential for solving this problem." }, { "problem_id": 1134, "question": "A geometric shape is constructed by several small identical cubes. The shapes of the geometric shape as seen from the front and from above are shown in the figures below. If the number of small cubes that make up this geometric shape is $n$, then the minimum value of $n$ is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0077_1.jpg", "batch10-2024_06_14_871925a4dba1dfe2d0e4g_0077_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure: The minimum number is: $3 + 1 + 1 + 2 + 1 + 1 + 1 = 10$ (units)\n\n\n\nThis represents the minimum scenario when viewed from above.\n\nTherefore, the answer is: 10\n\n[Key Insight] This question tests the understanding of three-view drawings. The key to solving it lies in comprehending the definition of three-view drawings, which is a common type of question in middle school exams." }, { "problem_id": 1135, "question": "A set of triangles $\\mathrm{AOB}$ and $\\mathrm{COD}$ are placed as shown, with $\\angle \\mathrm{A}=\\angle \\mathrm{C}=90^{\\circ}, \\angle \\mathrm{AOB}=60^{\\circ}, \\angle \\mathrm{COD}=45^{\\circ}$, $\\mathrm{ON}$ bisects $\\angle \\mathrm{COB}$, and $\\mathrm{OM}$ bisects $\\angle \\mathrm{AOD}$. As triangle $\\mathrm{COD}$ rotates clockwise around point $\\mathrm{O}$ (from Figure 1 to Figure 2), let the degree measures of $\\angle \\mathrm{NOM}$ in Figure 1 and Figure 2 be $\\alpha$ and $\\beta$, respectively. Then $\\alpha+\\beta=$ $\\qquad$ degrees.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0010_1.jpg", "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0010_2.jpg" ], "is_multi_img": true, "answer": "105", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In Figure 1, let $\\angle \\mathrm{AOM} = \\mathrm{x} = \\angle \\mathrm{DOM}$.\n\nSince $\\angle \\mathrm{AOB} = 60^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{BOM} = 60^{\\circ} - \\mathrm{x}$.\n\nBecause $\\angle \\mathrm{BOD} = \\angle \\mathrm{DOM} - \\angle \\mathrm{BOM}$,\n\n$\\therefore \\angle \\mathrm{BOD} = \\mathrm{x} - \\left(60^{\\circ} - \\mathrm{x}\\right) = 2\\mathrm{x} - 60^{\\circ}$.\n\nSince $\\angle \\mathrm{COB} = \\angle \\mathrm{BOD} + \\angle \\mathrm{DOC}$,\n\n$\\therefore \\angle \\mathrm{COB} = \\left(2\\mathrm{x} - 60^{\\circ}\\right) + 45^{\\circ} = 2\\mathrm{x} - 15^{\\circ}$.\n\nTherefore, $\\angle \\mathrm{CON} = \\angle \\mathrm{BON} = \\frac{1}{2}\\left(2\\mathrm{x} - 15^{\\circ}\\right) = \\mathrm{x} - 7.5^{\\circ}$.\n\nThus, $\\alpha = \\angle \\mathrm{NOM} = \\angle \\mathrm{BOM} + \\angle \\mathrm{BON} = 60^{\\circ} - \\mathrm{x} + \\mathrm{x} - 7.5^{\\circ} = 52.5^{\\circ}$.\n\nIn Figure 2, let $\\angle \\mathrm{AOM} = \\angle \\mathrm{DOM} = \\mathrm{x}$, and $\\angle \\mathrm{CON} = \\angle \\mathrm{BON} = \\mathrm{y}$. Then, $\\angle \\mathrm{BOD} = 60^{\\circ} - 2\\mathrm{x}$.\n\nSince $\\angle \\mathrm{COD} = 45^{\\circ}$,\n\n$\\therefore 60 - 2\\mathrm{x} + 2\\mathrm{y} = 45^{\\circ}$, which implies $\\mathrm{x} - \\mathrm{y} = 7.5^{\\circ}$.\n\nTherefore, $\\beta = \\angle \\mathrm{MON} = \\mathrm{x} + (60 - 2\\mathrm{x}) + \\mathrm{y} = 60 - (\\mathrm{x} - \\mathrm{y}) = 52.5^{\\circ}$.\n\nHence, $\\alpha + \\beta = 52.5^{\\circ} + 52.5^{\\circ} = 105^{\\circ}$.\n\nThe final answer is: 105.\n\n【Key Insight】This problem examines the calculation of angles. The key to solving it lies in setting one (or two) unknown variables and using algebraic methods to address geometric problems." }, { "problem_id": 1136, "question": "As shown in the figure, the three views of a geometric object are given. The shape of this geometric object is $\\qquad$ .\n\n\n\nFront view\n\n\n\nSide view\n\n\n\nTop view", "input_image": [ "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0026_1.jpg", "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0026_2.jpg", "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0026_3.jpg" ], "is_multi_img": true, "answer": "cone", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Based on the top view being a circle, geometric shapes such as spheres, cones, and cylinders are possible. However, only the cone has both the front view and the side view as triangles. Therefore, the shape of this geometric body is a cone.\n\nHence, the answer is a cone.\n\n[Key Insight] This question tests the ability to determine a geometric shape from its three views. It emphasizes the importance of being proficient and flexible in understanding and applying the three views, reflecting an assessment of spatial imagination skills." }, { "problem_id": 1137, "question": "Make a cube with a side length of $4 \\mathrm{~cm}$ using clay.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nA\n\n\n\nB\n\nFigure (3)\n\nAs shown in Figure (1), a square hole with a side length of $1 \\mathrm{~cm}$ is drilled from top to bottom at the center of the top surface. Then, a square hole with a side length of $1 \\mathrm{~cm}$ is drilled from front to back at the center of the front surface (along the dotted line in Figure (2)). The surface area of the clay after drilling is $\\qquad$ $m^2$; (Note: Figure (3) is not used)", "input_image": [ "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0031_1.jpg", "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0031_2.jpg", "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0031_3.jpg", "batch10-2024_06_14_8c4be2213bf9277a7d5eg_0031_4.jpg" ], "is_multi_img": true, "answer": "118", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "The surface area \\( S_{1} = 96 - 2 + 4 \\times 4 = 110 \\, (\\text{cm}^{2}) \\);\n\nThe surface area \\( S_{2} = S_{1} - 4 + 4 \\times 1.5 \\times 2 = 118 \\, (\\text{cm}^{2}) \\).\n\nTherefore, the answer is 118.\n\n【Key Insight】This question examines three-dimensional shapes, and mastering the surface area calculation formula for a rectangular prism is crucial for solving the problem." }, { "problem_id": 1138, "question": "As shown in the figure, 1 line divides the plane into 2 parts, 2 lines can divide the plane into a maximum of 4 parts, 3 lines can divide the plane into a maximum of 7 parts, and 4 lines can divide the plane into a maximum of 11 parts. Given that $\\mathrm{n}$ lines can divide the plane into a maximum of 56 parts, the value of $\\mathrm{n}$ is $\\qquad$.\n\n\n\n1 line\n\n\n\n2 lines\n\n\n\n\n\n4 lines", "input_image": [ "batch10-2024_06_14_96635cefb854924f48f0g_0006_1.jpg", "batch10-2024_06_14_96635cefb854924f48f0g_0006_2.jpg", "batch10-2024_06_14_96635cefb854924f48f0g_0006_3.jpg", "batch10-2024_06_14_96635cefb854924f48f0g_0006_4.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: The maximum number of parts into which $n$ lines can divide a plane is given by $S=1+1+2+3+\\ldots+n=\\frac{1}{2} n(n+1)+1$. If $n$ lines divide the plane into a maximum of 56 parts,\n\nwe can derive the equation $\\frac{1}{2} n(n+1)+1=56$,\n\nSolving this equation yields $x_{1}=-11$ (which is discarded as it does not fit the context), and $x_{2}=10$.\n\nTherefore, the value of $n$ is 10.\n\nHence, the answer is: 10." }, { "problem_id": 1139, "question": "As shown in Figure (1), this is the surface expansion diagram of a cube. By flipping the corresponding cube from the position shown in Figure (2) to the 1st grid, the 2nd grid, and the 3rd grid in sequence, the word on the upper face of the cube at this time is $\\qquad$ .\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch10-2024_06_14_96635cefb854924f48f0g_0012_1.jpg", "batch10-2024_06_14_96635cefb854924f48f0g_0012_2.jpg" ], "is_multi_img": true, "answer": "真", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From Figure (1), it can be seen that \"界\" corresponds to \"真\", \"杯\" corresponds to \"精\", and \"世\" corresponds to \"彩\". From Figure (2), when the small cube is flipped to the 1st, 2nd, and 3rd squares in sequence, the character on the bottom face is \"界\", so the character on the top face is \"真\". Therefore, the answer is \"真\".\n\n[Key Insight] This question tests the understanding of characters on opposite faces of a cube. The key to solving it lies in analyzing the cube's net diagram in the context of the specific problem." }, { "problem_id": 1140, "question": "As shown in the figure, Module (1) consists of 15 small cubes with an edge length of 1, and Modules (2)-(6) each consist of 4 small cubes with an edge length of 1. Now, three modules are to be selected from Modules (2)-(6) and placed on top of Module (1) to form a large cube with an edge length of 3. The sequence numbers of the three modules that meet the above requirements are $\\qquad$.\n\n\n\nModule (1)\n\n\n\nModule (4)\n\n\n\nModule (2)\n\n\n\nModule (5)\n\n\n\nModule (3)\n\n\n\nModule (6)", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0028_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0028_2.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0028_3.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0028_4.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0028_5.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0028_6.jpg" ], "is_multi_img": true, "answer": "(4)(5)", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, it can be seen that module (6) complements the top left corner of module (1), module (5) complements the bottom right corner of module (1), and module (4) fills the space between modules (6) and (5) on top of module (1), thereby forming a large cube with an edge length of 3.\n\nTherefore, the modules that can accomplish the task are (4), (5), and (6).\n\nHence, the answer is: (4)(5)(6).\n\n[Highlight] This problem tests the understanding of three-dimensional shapes. Similar to a tangram puzzle, it examines the ability to assemble shapes, which can cultivate students' hands-on skills and expand their rich imagination." }, { "problem_id": 1141, "question": "The relevant data for a cuboid as seen from the left and top is shown in the figure below. The area of the figure as seen from the front is $\\qquad$\n\n\n\nAs seen from the left\n\n", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0035_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0035_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the data related to the shapes seen from the left and top views, we can deduce:\n\nThe shape seen from the front is a rectangle with a length of 4 and a width of 2.\n\nTherefore, the area of the shape seen from the front is: $4 \\times 2 = 8$.\n\nHence, the answer is: 8\n\n[Key Insight] This question tests the ability to observe a cuboid from different directions. Mastering the relationship between the shapes seen from different perspectives is crucial for solving the problem." }, { "problem_id": 1142, "question": "As shown in the figure, a geometric solid is constructed by stacking a number of small cubic blocks of the same size. The images below represent the shapes seen from its front and top views, respectively. The geometric solid is composed of $\\qquad$ small cubic blocks.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0037_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0037_2.jpg" ], "is_multi_img": true, "answer": "6 or 7 or 8", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The leftmost column can have at most \\(2 + 2 + 2 = 6\\) blocks, the middle column has 1 block, and the rightmost column has 1 block, making a maximum total of 8 blocks.\n\nThe leftmost column can have at least \\(2 + 1 + 1 = 4\\) blocks, the middle column has 1 block, and the rightmost column has 1 block, making a minimum total of 6 blocks. Therefore, the geometric shape is constructed using 6, 7, or 8 small cubes.\n\nHence, the answer is: 6 or 7 or 8.\n\n[Key Insight] This question tests the understanding of viewing a geometric shape from different angles. The key to solving the problem lies in clearly understanding the question and using a combination of numerical and graphical approaches to find the solution." }, { "problem_id": 1143, "question": "A geometric solid, whose shape views from the front and top are shown in the figures below, has at least $\\qquad$ small cubic blocks.\n\n\n\nView from the front\n\n\n\nView from the top", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0040_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0040_2.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the minimum number of small cubes required to form such a geometric figure is as shown in the diagram:\n\n\n\n## View from above\n\nTherefore, to construct such a geometric figure, a minimum of 12 small cubes is needed.\n\nThus, the answer is: 12.\n\n[Key Insight] This problem primarily tests the ability to view a geometric figure from different perspectives. The key to solving the problem lies in writing the corresponding numbers in the positions of the figure as seen from above." }, { "problem_id": 1144, "question": "There is a regular hexahedron die placed on a table. Rolling the die clockwise as shown in the figure, each roll of $90^{\\circ}$ counts as one roll. After the 2022nd roll, the number of dots on the face facing down is $\\qquad$.\n\n\n\n\n\nSecond roll\n\n\n\nThird roll", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0041_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0041_2.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0041_3.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: By observing the figure, we can see that the number of dots on face 3 is opposite to that on face 4, and the number of dots on face 2 is opposite to that on face 5. According to the problem, the number of dots on the bottom face cycles every four times, with the sequence of dots in each cycle being $2, 3, 5, 4$.\n\n$2022 \\div 4=505 \\text{ with a remainder of } 2$,\n\nTherefore, after the 2022nd roll, the number of dots on the bottom face of the die is 3.\n\nHence, the answer is: 3.\n\n【Key Insight】This problem examines the text on opposite faces of a cube and the pattern of changes in the figure. The key to solving the problem lies in identifying the pattern." }, { "problem_id": 1145, "question": "The following geometric shapes are prisms: $\\qquad$ (Fill in the serial numbers)\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\n\n\n\n(5)\n\n\n\n(6)", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0050_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0050_2.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0050_3.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0050_4.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0050_5.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0050_6.jpg" ], "is_multi_img": true, "answer": "(1)(2)(6)", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: The two bases of a prism are polygons that are identical in shape and size, and the lateral faces are rectangles. Therefore, (1), (2), and (6) are prisms, while (3) is a cylinder, (4) is a cone, and (5) is a sphere. Hence, the answer is: (1), (2), and (6).\n\n[Key Insight] This question tests the understanding of three-dimensional shapes. The key to solving it lies in mastering the characteristics of a prism." }, { "problem_id": 1146, "question": "Xiao Ying neatly stacked several boxes of chalk on the teacher's desk. The students noticed that the shape of the chalk, when viewed from the front, left, and top, was the same (as shown in the figures). Therefore, the total number of boxes of chalk in this stack is $\\qquad$.\n\n\n\nViewed from the front\n\n\n\nViewed from the left\n\n\n\nViewed from above", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0051_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0051_2.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0051_3.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Based on the front view, it can be determined that the stack of chalk consists of two layers. From the top view, it is evident that the first layer has 3 boxes of chalk. From the left view, it is clear that the second layer has 1 box. Therefore, there are a total of 4 boxes.\n\nHence, the answer is: 4.\n\n【Key Insight】This question tests the ability to determine the number of small cubes based on views from different directions. Mastering the relevant concepts is crucial for solving the problem." }, { "problem_id": 1147, "question": "Xiao Ying brought $n$ boxes of chalk to school, neatly stacked on the teacher's desk. The three views are shown in the figures below. What is the value of $n$?\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0058_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0058_2.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0058_3.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0058_4.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0058_5.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0058_6.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "As shown in the figure:\n\nThere are 4 boxes on the bottom layer, 2 boxes on the second layer, and 1 box on the third layer.\n\n$\\therefore n=7$.\n\nTherefore, the answer is: 7.\n\n\n\n【Key Point】This question tests knowledge of three-view drawings. The key to solving the problem is to determine the geometric shape based on the three-view drawings." }, { "problem_id": 1148, "question": "Figure 1 is the unfolded view of the cube in Figure 2. Now, the cube in Figure 2 is flipped sequentially to the 1st grid, the 2nd grid, and the 3rd grid. The word on the top face of the cube at this point is $\\qquad$ .\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0073_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0073_2.jpg" ], "is_multi_img": true, "answer": "江", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The first flip results in \"镇\" or \"徒\" being on the bottom, the second flip results in \"容\" being on the bottom, and the third flip results in \"句\" being on the bottom.\n\n\"句\" is opposite to \"江\",\n\nTherefore, the answer is: 江.\n\n【Key Point】This question examines the characters on opposite faces of a cube, and mastering the concept of opposite faces is crucial for solving the problem." }, { "problem_id": 1149, "question": "Among the following figures, the ones that cannot be folded into a cube are $\\qquad$ (fill in the numbers).\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n", "input_image": [ "batch10-2024_06_14_a202765c6dac3675b012g_0092_1.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0092_2.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0092_3.jpg", "batch10-2024_06_14_a202765c6dac3675b012g_0092_4.jpg" ], "is_multi_img": true, "answer": "(1)(2)(4)", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Solution: (3) Can be folded into a cube;\n\n(1), (2), (4) After folding, one face overlaps and lacks a base, hence they cannot be folded into a cube.\n\nTherefore, the answers are: (1), (2), (4).\n\n[Key Point] This question tests the folding of a net into a geometric shape. The key to solving it is to remember the basic configurations that can form a cube: \"1, 4, 1\", \"3\", \"2, 2\", and \"1, 3, 2\"." }, { "problem_id": 1150, "question": "All squares in Figure 1 and Figure 2 are identical. Place the square from Figure 1 in the positions $\\qquad$ (select all possible positions from (1), (2), (3), (4)) in Figure 2, such that the resulting shape can form a cube.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0002_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0002_2.jpg" ], "is_multi_img": true, "answer": "(2)、(3)、(4)", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Place the square from Figure 1 in position (1) of Figure 2, and overlapping faces occur, so it cannot form a cube. Placing the square from Figure 1 in positions (2), (3), and (4) of Figure 2 can form a cube, so the answer is (2), (3), and (4).\n\n【Key Point】This question examines the folding of a net into a geometric shape. When solving, do not forget the characteristics of a quadrangular prism and the various scenarios of cube nets. Note: Any net that forms a \"田\" shape is not a valid surface net of a cube." }, { "problem_id": 1151, "question": "Among the following figures, the ones that can be folded along the dotted lines to form a rectangular box are $\\qquad$ .\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0013_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0013_2.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0013_3.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0013_4.jpg" ], "is_multi_img": true, "answer": "$(1)(3)$", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "**Analysis:** By folding each net along the dotted lines, a three-dimensional shape can be formed.\n\n**Detailed Explanation:** In option (2), the rectangles overlap, so it cannot be chosen; in option (4), the squares overlap, so it cannot be chosen either. Therefore, the correct answers are (1) and (3).\n\n**Key Point:** This question tests the knowledge of nets. The crucial point in solving it is to fold the net along the dotted lines to form a three-dimensional shape." }, { "problem_id": 1152, "question": "Figure (1) is the surface expansion diagram of a small cube. The small cube flips to the 1st grid, the 2nd grid, the 3rd grid, and the 4th grid in sequence from the position shown in Figure (2). The word on the upper side of the small cube at this time is $\\qquad$\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0015_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0015_2.jpg" ], "is_multi_img": true, "answer": "梦", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "According to the surface development diagram of the cube: \"Country\" is opposite to \"Water\", \"City\" is opposite to \"Dream\", and \"Middle\" is opposite to \"Beauty\".\n\nWhen flipped to the 1st grid, \"Dream\" is on the bottom; when flipped to the 2nd grid, \"Middle\" is on the bottom; when flipped to the 3rd grid, \"Country\" is on the bottom; and when flipped to the 4th grid, \"City\" is on the bottom.\n\nThus, when at the 4th grid, the character facing upwards is \"Dream\".\n\nTherefore, the answer is: Dream." }, { "problem_id": 1153, "question": "The following shapes can form an open-top cube: $\\qquad$ (fill in the number)\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)\n\n\n\n(5)\n\n\n\n(6)", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0031_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0031_2.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0031_3.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0031_4.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0031_5.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0031_6.jpg" ], "is_multi_img": true, "answer": "(1)(2)(4).", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "By folding paper or using spatial imagination, it can be determined that (1), (2), (4), and (5) can form a lidless cube. Additionally, based on the 11 types of cube nets, since this problem involves a lidless cube, one square is missing, and it can also be concluded that (1), (2), and (4) can form a lidless cube.\n\n\nTherefore, the answer is (1), (2), (4).\n\n【Key Point】Key Concepts: 1. Three-dimensional shapes; 2. Cube nets." }, { "problem_id": 1154, "question": "As shown in the figure, the three views of a geometric structure are composed of small cubes with a side length of 1. If, on the basis of the original geometric structure (without changing the positions of the small cubes), we continue to add the same small cubes to form a rectangular prism, at least $\\qquad$ more small cubes are needed.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0046_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0046_2.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0046_3.jpg" ], "is_multi_img": true, "answer": "26", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the top view, it is easy to see that the bottom layer has 7 small cubes, the second layer has 2 small cubes, and the third layer has 1 small cube.\n\nThe distribution of the small cubes is as follows:\n\n\n\nTop view\n\nThus, there are a total of \\(7 + 2 + 1 = 10\\) geometric components.\n\nIf these are to be assembled into a large rectangular prism, a total of \\(3 \\times 4 \\times 3 = 36\\) small cubes are needed.\n\nTherefore, an additional \\(36 - 10 = 26\\) small cubes are required.\n\nHence, the answer is: 26.\n\n【Key Insight】This question tests the students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. The key is to determine the total number of small cubes needed to form the large rectangular prism." }, { "problem_id": 1155, "question": "Figure 1 is an unfolded view of a cube. The cube is flipped sequentially to the 1st, 2nd, and 3rd positions from the position shown in Figure 2. The word on the top face of the small cube at this point is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0062_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0062_2.jpg" ], "is_multi_img": true, "answer": "国", "answer_type": "single-step", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From Figure 1, we can deduce: \"中\" corresponds to \"的\"; \"国\" corresponds to \"我\"; \"梦\" corresponds to \"梦\".\n\nFrom Figure 2, we can deduce: When the cube is flipped sequentially from the position shown in Figure 2 to the 1st, 2nd, and 3rd positions, with \"我\" on the bottom, the character on the top face of the small cube at this time is \"国\".\n\nTherefore, the answer is: 国.\n\n【Highlight】This question uses the side expansion diagram of a small cube as the background to test students' understanding of the expansion diagrams of three-dimensional shapes. It assesses students' spatial imagination abilities." }, { "problem_id": 1156, "question": "The geometric figure formed by some small cubes of the same size is shown from the front and from above as follows. The maximum number of small cubes that can form this geometric figure is $\\qquad$.\n\n\n\nFrom the front view\n\n\n\nFrom the top view", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0064_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0064_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, we can see that the first layer of this geometric shape has a total of 5 units. From the front view, we can determine that the shape has 2 layers, with the second layer having a maximum of 2 units. Therefore, the maximum total number of units is 7.\n\nThe answer is 7.\n\n【Key Point】This question tests the ability to abstract a geometric shape from its three views. Note the roles of the three views: the top view lays the foundation, the front view determines the height, and the side view removes any irregularities." }, { "problem_id": 1157, "question": "The maximum value of $n$ for a geometric shape formed by stacking $n$ identical small cubes, with the front view and top view as shown, is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0077_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0077_2.jpg" ], "is_multi_img": true, "answer": "13", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By synthesizing the front view and the top view, counting from top to bottom, the bottom layer can have a maximum of \\(2 + 2 + 3 = 7\\) units, the second layer can have a maximum of \\(1 + 1 + 2 = 4\\) units, and the third layer can have a maximum of \\(1 + 0 + 1 = 2\\) units. Therefore, the maximum value of \\(n\\) is \\(7 + 4 + 2 = 13\\). Hence, the answer is: 13.\n\n**Key Point:** This question tests the understanding of the front view and top view in three-view drawings. Mastering the related concepts of three-view drawings is crucial for solving the problem." }, { "problem_id": 1158, "question": "As shown, a square with side length $10 \\mathrm{~cm}$ is divided to create a set of tangram pieces. Figure 2 shows the assembled \"small house,\" where the area of the shaded part is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_b99ebab550526b457352g_0099_1.jpg", "batch10-2024_06_14_b99ebab550526b457352g_0099_2.jpg" ], "is_multi_img": true, "answer": "50", "answer_type": "single-step", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the area of the shaded part is: $10 \\times 10 \\div 2=50\\left(\\mathrm{~cm}^{2}\\right)$.\n\nTherefore, the answer is: 50.\n\n[Key Point] This question primarily tests the understanding of the tangram puzzle. Mastering the characteristics of the tangram is crucial for solving the problem." }, { "problem_id": 1159, "question": "Given that Figure 1 is the surface development diagram of the small cube shown in Figure 2, the small cube flips to the 1st grid, the 2nd grid, and the 3rd grid sequentially from the position shown in Figure 2. The word on the top face of the small cube at this time is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_d167c8063e3876a3ba38g_0087_1.jpg", "batch10-2024_06_14_d167c8063e3876a3ba38g_0087_2.jpg" ], "is_multi_img": true, "answer": "信", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The first flip places \"Sincerity\" on the bottom, the second flip places \"Love\" on the bottom, and the third flip places \"Country\" on the bottom. \"Trust\" is opposite to \"Country\", hence the answer is: Trust.\n\n【Key Point】This question examines the text on opposite faces of a cube, where two faces separated by one face are opposites. Paying attention to the sequence of flips to determine which face is on the bottom each time is crucial for solving the problem." }, { "problem_id": 1160, "question": "Figure 1 is the surface expansion diagram of a small cube. The small cube is flipped sequentially to the 1st grid, the 2nd grid, and the 3rd grid from the position shown in Figure 2. The word on the top face of the small cube at this time is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch10-2024_06_14_d167c8063e3876a3ba38g_0089_1.jpg", "batch10-2024_06_14_d167c8063e3876a3ba38g_0089_2.jpg" ], "is_multi_img": true, "answer": "飞", "answer_type": "single-step", "difficulty": "Low", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The first flip has \"燕\" on the bottom, the second flip has \"愿\" on the bottom, and the third flip has \"祝\" on the bottom. \"祝\" is opposite to \"飞\".\n\nTherefore, the answer is: 飞.\n\n【Highlight】This question tests the understanding of the characters on opposite faces of a cube. Mastering the concept of opposite faces is key to solving the problem." }, { "problem_id": 1161, "question": "There is a cube with six faces labeled with the numbers $1, 2, 3, 4, 5, 6$. The results of observing the cube from different angles by three students, A, B, and C, are shown in the figures below. If the number on the opposite face of 2 is denoted as $m$, and the number on the opposite face of 3 is denoted as $n$, then the solution $x$ to the equation $m^{x+1}=n$ satisfies $k\n\nA\n\n\n\nB\n\n\n\nC", "input_image": [ "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0039_1.jpg", "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0039_2.jpg", "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0039_3.jpg" ], "is_multi_img": true, "answer": "0", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From the diagram, it can be observed that the number 2 is adjacent to 6, 1, 3, and 5. Therefore, the opposite face of 2 must be 4, which means \\( \\mathrm{m} = 4 \\). Similarly, the number 3 is adjacent to 1, 2, 5, and 4, so the opposite face of 3 is 6, meaning \\( \\mathrm{n} = 6 \\).\n\nGiven the equation:\n\\[\n\\mathrm{m}^{\\mathrm{x}+1} = \\mathrm{n},\n\\]\nsubstituting the known values gives:\n\\[\n4^{x+1} = 6.\n\\]\nThis implies:\n\\[\n1 < \\mathrm{x} + 1 < 2.\n\\]\nSince \\( k < x < k + 1 \\) and \\( k \\) is an integer, it follows that:\n\\[\n\\mathrm{k} = 0.\n\\]\nThus, the answer is: 0.\n\n**Insight:** This problem tests the ability to flexibly use the concept of opposite faces in a cube to solve problems. It is creatively designed and is a good exercise." }, { "problem_id": 1162, "question": "As shown in the figure, the (1)st figure has 2 identical small squares, the (2)nd figure has 6 identical small squares, the (3)rd figure has 12 identical small squares, the (4)th figure has 20 identical small squares, ..., according to this pattern, then the (n)th figure has $\\qquad$ identical small squares.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n\n\n(4)", "input_image": [ "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0044_1.jpg", "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0044_2.jpg", "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0044_3.jpg", "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0044_4.jpg" ], "is_multi_img": true, "answer": "$\\mathrm{n}(\\mathrm{n}+1)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The first figure has 2 identical small squares, where \\(2 = 1 \\times 2\\). The second figure has 6 identical small squares, where \\(6 = 2 \\times 3\\). The third figure has 12 identical small squares, where \\(12 = 3 \\times 4\\). The fourth figure has 20 identical small squares, where \\(20 = 4 \\times 5\\), and so on. Following this pattern, the \\(n\\)th figure should have \\(n(n+1)\\) identical small squares.\n\n**Insight:** This problem tests the ability to recognize patterns in graphical changes. The key to solving it lies in noticing that the number of squares is the product of two consecutive integers. Such problems require a high level of skill, which can be developed through continuous learning and practice." }, { "problem_id": 1163, "question": "All squares in Figure 1 and Figure 2 are congruent. Placing the square from Figure 1 in one of the positions (1), (2), (3), or (4) in Figure 2, the position where the resulting figure cannot form a cube is . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0078_1.jpg", "batch10-2024_06_14_fa89832eb468ce4c0ef6g_0078_2.jpg" ], "is_multi_img": true, "answer": "(1)", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Placing the square from Figure 1 at position (1) in Figure 2 results in an overlap between the square from Figure 1 and the rightmost square in Figure 2, thus it cannot form a cube.\n\nPlacing the square from Figure 1 at positions (2) and (4) in Figure 2 corresponds to the 1-3-2 pattern of the net, which can form a cube. Placing the square from Figure 1 at position (3) in Figure 2 corresponds to the 3-3 pattern of the net, which can also form a cube. Therefore, the answer is: (1).\n\n[Key Point] This question tests the folding of nets into geometric shapes. When solving, do not forget the various scenarios of cube nets. Note: Any net that contains a \"田\", \"凹\", or \"一\" shape is not a valid surface net of a cube." }, { "problem_id": 1164, "question": "Xiaoming, who enjoys reading, discovered beautiful spiral curves in many places in nature during his extracurricular reading (Figure 1, Figure 2). Xiaoming thought of the spiral curves in his textbook (Figure 3) and had a sudden idea: to form a spiral curve using several right triangles with a $30^{\\circ}$ angle (Figure 4). $\\angle A B O=\\angle B C O=\\angle C D O=\\cdots=\\angle L M O=90^{\\circ}$. After Xiaoming told his math teacher about this, the teacher said: If the length of the leg $O M$ in Rt $\\triangle O L M$ is $\\frac{27}{64}$, then the length of $O A$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4", "input_image": [ "batch11-2024_06_14_00eb2f86226d216652bfg_0007_1.jpg", "batch11-2024_06_14_00eb2f86226d216652bfg_0007_2.jpg", "batch11-2024_06_14_00eb2f86226d216652bfg_0007_3.jpg", "batch11-2024_06_14_00eb2f86226d216652bfg_0007_4.jpg" ], "is_multi_img": true, "answer": "$\\frac{64}{27}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: In the right triangle \\( A O B \\), \\( \\angle A B O = 90^{\\circ} \\), \\( \\angle A O B = 30^{\\circ} \\),\n\\[\n\\therefore A O = 2 A B, \\quad B O = \\sqrt{A B^{2} - A O^{2}} = \\sqrt{3} A B,\n\\]\n\\[\n\\therefore O B = \\frac{\\sqrt{3}}{2} O A.\n\\]\n\nSimilarly:\n\\[\nO C = \\frac{\\sqrt{3}}{2} O B = \\frac{\\sqrt{3}}{2} \\times \\frac{\\sqrt{3}}{2} O A = \\frac{3}{4} O A = \\left(\\frac{\\sqrt{3}}{2}\\right)^{2} O A,\n\\]\n\\[\nO D = \\frac{\\sqrt{3}}{2} \\times \\frac{3}{4} O A = \\frac{3 \\sqrt{3}}{8} O A = \\left(\\frac{\\sqrt{3}}{2}\\right)^{3} O A,\n\\]\n\\[\n\\therefore O M = \\left(\\frac{\\sqrt{3}}{2}\\right)^{12} O A = \\frac{729}{4096} O A.\n\\]\n\nGiven that \\( O M = \\frac{27}{64} \\),\n\\[\n\\therefore \\frac{729}{4096} O A = \\frac{27}{64},\n\\]\nSolving for \\( O A \\):\n\\[\nO A = \\frac{64}{27}.\n\\]\n\nThus, the answer is: \\( \\frac{64}{27} \\).\n\n**Key Insight**: This problem tests the application of the Pythagorean theorem, properties of a 30-degree right triangle, multiplication of square roots, and the exploration of patterns in segment lengths. The key to solving this problem lies in mastering the method of \"moving from specific cases to general principles, then summarizing and applying these patterns.\"" }, { "problem_id": 1165, "question": "As shown in the figure, on a square cardboard $ABCD$ with side length $2 \\sqrt{2}$, a set of tangram pieces is made as shown in Figure 1 (with point $O$ being the intersection of the diagonals of the square cardboard), where points $E$ and $F$ are the midpoints of $AD$ and $CD$ respectively. If the tangram pieces in Figure 1 are arranged to form the \"fish shape\" as shown in Figure 2, then the length of the \"fish tail\" $MN$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch11-2024_06_14_00eb2f86226d216652bfg_0022_1.jpg", "batch11-2024_06_14_00eb2f86226d216652bfg_0022_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since in the isosceles right triangle \\( ACD \\), \\( AD = CD = 2\\sqrt{2} \\),\n\nTherefore, \\( AC = 4 \\).\n\nMoreover, since \\( AG = GO = OH = CH \\),\n\nIt follows that \\( FI = EI = 1 \\), and \\( EF = 2 \\).\n\nThus, \\( NM = 2 + 1 = 3 \\).\n\nHence, the answer is: 3.\n\n[Key Insight] This problem primarily examines the Pythagorean theorem and the properties of isosceles right triangles. Mastering the structural characteristics of the tangram is crucial for solving the problem." }, { "problem_id": 1166, "question": "A student used a set of tangram pieces shown in Figure 1 to form the \"sailboat\" pattern in Figure 2. Given that the side length of the square $A B C D$ is 4, the value of $h$ in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_00eb2f86226d216652bfg_0049_1.jpg", "batch11-2024_06_14_00eb2f86226d216652bfg_0049_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}+6$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the structure of the tangram, it is known that the divided triangles are all isosceles right-angled triangles. Among them, the hypotenuse of the large isosceles right-angled triangle equals the side length of the square, which is $=4$. The hypotenuse of the medium isosceles right-angled triangle equals the side length of the large isosceles right-angled triangle, which is $=\\sqrt{\\frac{4^{2}}{2}}=2 \\sqrt{2}$. \n\nTherefore, the altitude on the hypotenuse of the medium isosceles right-angled triangle equals its median on the hypotenuse, which is $=\\sqrt{2}$. \n\nThe hypotenuse of the small isosceles right-angled triangle is $=2$. \n\nThus, $h=\\sqrt{2}+4+2=\\sqrt{2}+6$. \n\nHence, the answer is: $\\sqrt{2}+6$. \n\n【Insight】This problem mainly examines the knowledge of tangram. The triangles divided by the tangram are all isosceles right-angled triangles, thus the lengths of the sides of each triangle can be determined by the Pythagorean theorem." }, { "problem_id": 1167, "question": "\"The Nine Chapters on the Mathematical Art\" is a representative work of ancient Eastern mathematics. The book records: \"If a door is opened to remove the threshold (kŭn, meaning the door sill) by one foot, and the gap is two inches, what is the width of the door?\" The main idea of the question is: As shown in Figures 1 and 2 (Figure 2 is a planar schematic diagram of Figure 1), when the double doors are pushed open, the distance between the gaps $\\mathrm{CD}$ is 2 inches, and the points $\\mathrm{C}$ and $\\mathrm{D}$ are both 1 foot (1 foot = 10 inches) away from the threshold $\\mathrm{AB}$. Then, the length of $\\mathrm{AB}$ is $\\qquad$ inches.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_01dd196e65e8cbf25590g_0083_1.jpg", "batch11-2024_06_14_01dd196e65e8cbf25590g_0083_2.jpg" ], "is_multi_img": true, "answer": "101", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let’s take the midpoint of $\\mathrm{AB}$ as $\\mathrm{O}$, and draw $\\mathrm{DE} \\perp \\mathrm{AB}$ through point $\\mathrm{D}$, intersecting $\\mathrm{AB}$ at $\\mathrm{E}$, as shown in Figure 2:\n\nFrom the problem statement, we have: $\\mathrm{OA} = \\mathrm{OB} = \\mathrm{AD} = \\mathrm{BC}$.\n\nLet $\\mathrm{OA} = \\mathrm{OB} = \\mathrm{AD} = \\mathrm{BC} = \\mathrm{r}$ units,\n\nthen $\\mathrm{AB} = 2\\mathrm{r}$ (units), $\\mathrm{DE} = 10$ units, $\\mathrm{OE} = \\frac{1}{2} \\mathrm{CD} = 1$ unit,\n\n$\\therefore \\mathrm{AE} = (\\mathrm{r} - 1)$ units.\n\nIn the right triangle $\\triangle \\mathrm{ADE}$,\n\n$\\mathrm{AE}^{2} + \\mathrm{DE}^{2} = \\mathrm{AD}^{2}$, that is, $(\\mathrm{r} - 1)^{2} + 10^{2} = \\mathrm{r}^{2}$.\n\nSolving this, we get: $r = 50.5$,\n$\\therefore 2\\mathrm{r} = 101$ (units),\n\n$\\therefore \\mathrm{AB} = 101$ units.\n\nThus, the answer is: 101\n\n\n\nFigure 2\n\n【Key Insight】This problem tests the application of the Pythagorean theorem. Understanding the problem and constructing the right triangle is the key to solving it." }, { "problem_id": 1168, "question": "As shown in Figure 1, a right-angled triangle paper has one of its legs measuring 1 and its hypotenuse measuring 3. According to Figure 2, the paper is arranged to form a square, with no overlapping or gaps at the joints. The area of the blank space in the middle of Figure 2, i.e., the quadrilateral $ABCD$, is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_01dd196e65e8cbf25590g_0085_1.jpg", "batch11-2024_06_14_01dd196e65e8cbf25590g_0085_2.jpg" ], "is_multi_img": true, "answer": "$9-4 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Given that one leg of the right-angled triangle is 1 and the hypotenuse is 3,\n\nTherefore, the other leg is $\\sqrt{3^{2}-1^{2}}=2 \\sqrt{2}$,\n\nHence, the area of one right-angled triangle is $\\frac{1}{2} \\times 1 \\times 2 \\sqrt{2}=\\sqrt{2}$, and the area of the square is $3 \\times 3=9$,\n\nThus, the area of quadrilateral $\\mathrm{ABCD}$ is $9-4 \\sqrt{2}$.\n\nThe final answer is: $9-4 \\sqrt{2}$.\n\n【Highlight】This question tests the Pythagorean theorem. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides." }, { "problem_id": 1169, "question": "Divide the rhombus in Figure 1 along its diagonals into four congruent right triangles, and arrange these four right triangles into squares as shown in Figure 2 and Figure 3. Then the area of the rhombus in Figure 1 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch11-2024_06_14_084db68bd568428339d9g_0041_1.jpg", "batch11-2024_06_14_084db68bd568428339d9g_0041_2.jpg", "batch11-2024_06_14_084db68bd568428339d9g_0041_3.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1:\n\n\n\nFigure 1\n\n$\\because$ Quadrilateral $A B C D$ is a rhombus,\n\n$\\therefore O A=O C, O B=O D, A C \\perp B D$,\n\nLet $O A=x, O B=y$,\n\nFrom the problem statement, we have: $\\left\\{\\begin{array}{l}x+y=5 \\\\ x-y=1\\end{array}\\right.$, solving this gives: $\\left\\{\\begin{array}{l}x=3 \\\\ y=2\\end{array}\\right.$,\n\n$\\therefore A C=2 O A=6, \\quad B D=2 O B=4$,\n\n$\\therefore$ The area of rhombus $A B C D$ $=\\frac{1}{2} A C \\times B D=\\frac{1}{2} \\times 6 \\times 4=12$;\n\nThus, the answer is 12.\n\n【Insight】This problem examines the properties of rhombuses and squares, and the application of systems of linear equations; mastering the properties of squares and rhombuses, and setting up the system of equations based on the problem statement, is key to solving the problem." }, { "problem_id": 1170, "question": "Figure 1 shows a model of a grinding tool, and Figure 2 is its schematic diagram. It is known that $A B \\perp P Q$, $A P=A Q=3 \\mathrm{dm}, A B=12 \\mathrm{dm}$, the point $A$ moves along the central axis $l$, and the point $B$ moves on a circle centered at $O$ with radius $O B$, where $O B=4 \\mathrm{dm}$. During the movement of point $B$, the shortest distance between point $P$ and point $O$ is $\\qquad$ $\\mathrm{dm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch11-2024_06_14_1836d333a249bf5b8d9ag_0032_1.jpg", "batch11-2024_06_14_1836d333a249bf5b8d9ag_0032_2.jpg", "batch11-2024_06_14_1836d333a249bf5b8d9ag_0032_3.jpg" ], "is_multi_img": true, "answer": "$3 \\sqrt{17}-4$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: When points $\\mathrm{B}$, $\\mathrm{O}$, and $\\mathrm{P}$ are collinear, the distance of $\\mathrm{OP}$ is the shortest.\n\nThus, $\\mathrm{OP} = \\mathrm{BP} - \\mathrm{OB} = \\sqrt{BA^{2} + AP^{2}} - OB = \\sqrt{12^{2} + 3^{2}} - 4 = 3\\sqrt{17} - 4 \\quad (\\mathrm{dm})$.\n\nTherefore, the answer is: $3\\sqrt{17} - 4$.\n\n【Insight】This problem examines the concept of the shortest distance and the Pythagorean theorem. The key to solving it lies in determining the positional relationships of the important points after rotation." }, { "problem_id": 1171, "question": "Xiao Ming uses a set of tangram pieces shown in Figure 1 to form the \"rocket diagram\" as shown in Figure 2. If the side length of the square $A B C D$ is $4 \\mathrm{~cm}$, then the distance between points $M$ and $N$ in Figure 2 is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0010_1.jpg", "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0010_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{17}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a line from point $M$ perpendicular to the extension of the hypotenuse of $6$, intersecting at point $P$.\n\n\n\nFigure 1\n\n\n\nGiven that the side length of square $ABCD$ is $4 \\mathrm{~cm}$,\n\n- The hypotenuse of $6$ is $2 \\mathrm{~cm}$,\n- The diagonal of $5$ is $2 \\mathrm{~cm}$,\n- The hypotenuse of $1$ is $4 \\mathrm{~cm}$,\n- The height on the hypotenuse of $3$ is $1 \\mathrm{~cm}$.\n\nTherefore, \n\\[\nPN = 2 + \\frac{1}{2} \\times 2 + 4 + 1 = 8 \\mathrm{~cm}, \\quad MP = 2 \\mathrm{~cm}.\n\\]\n\nThus, \n\\[\nMN = \\sqrt{8^{2} + 2^{2}} = 2 \\sqrt{17} \\mathrm{~cm}.\n\\]\n\nThe final answer is: $2 \\sqrt{17}$.\n\n**Key Insight:** This problem tests knowledge of tangram pieces, properties of squares, and the Pythagorean theorem. By constructing auxiliary lines to form right triangles and understanding the characteristics of the tangram, the Pythagorean theorem becomes the key to solving the problem." }, { "problem_id": 1172, "question": "As shown in the figure, $A B$ is the base plate of the stapler, the pressing handle $B C$ rotates around point $B$, one end point $D$ of the connecting rod $D E$ is fixed, and point $E$ slides from $A$ to $B$. During the sliding process, the length of $D E$ remains unchanged. In Figure 1, $B D=6 \\mathrm{~cm}$, $B E=15 \\mathrm{~cm}, \\angle B=60^{\\circ}$. Now, the pressing handle $B C$ is rotated from Figure 1 to be perpendicular to the base $A B$, as shown in Figure 2. The distance that point $E$ slides during this process is $\\qquad$ cm.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0015_1.jpg", "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0015_2.jpg" ], "is_multi_img": true, "answer": "$15-3 \\sqrt{15}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, draw a perpendicular line from point $D$ to $AB$, intersecting $AB$ at point $F$.\n\nGiven that $\\angle B = 60^\\circ$ and $BD = 6\\,\\text{cm}$,\n\nwe have $BF = 3\\,\\text{cm}$.\n\nThus, $DF^2 = DB^2 - BF^2 = 36 - 9 = 27$.\n\nNext, $EF = BE - BF = 12\\,\\text{cm}$.\n\nTherefore, $DE^2 = DF^2 + EF^2 = 27 + 144 = 171$.\n\nIn Figure 2,\n\n$BE = \\sqrt{DE^2 - DB^2} = \\sqrt{171 - 36} = \\sqrt{135} = 3\\sqrt{15}$.\n\nHence, the distance that point $E$ has slid is $15 - 3\\sqrt{15}$.\n\nThe final answer is $15 - 3\\sqrt{15}$.\n\n**Key Insight**: This problem tests the Pythagorean theorem. Understanding the problem and correctly applying the Pythagorean theorem to set up the relevant equations is crucial for solving it." }, { "problem_id": 1173, "question": "As shown in Figure 1, in rectangle \\( A B C D \\), point \\( E \\) is the midpoint of \\( B C \\), and point \\( P \\) moves along \\( B C \\) from point \\( B \\) to point \\( C \\). Let the distance between points \\( B \\) and \\( P \\) be \\( x \\), and \\( P A - P E = y \\). Figure 2 shows the graph of \\( y \\) as a function of \\( x \\) as point \\( P \\) moves. Therefore, \\( B C = \\) \\(\\qquad\\)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0030_1.jpg", "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0030_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: From the graph of the function, we know that when \\( x = 0 \\), i.e., when point \\( P \\) is at point \\( B \\), \\( BA - BE = 1 \\).\n\nUsing the property that the difference between two sides of a triangle is less than the third side, we obtain \\( PA - PE \\leq AE \\).\n\nTherefore, the maximum value of \\( y \\) is \\( AE \\), so \\( AE = 5 \\).\n\nIn the right triangle \\( \\triangle ABE \\), by the Pythagorean theorem, we have:\n\\[ BA^{2} + BE^{2} = AE^{2} = 25 \\]\n\nLet the length of \\( BE \\) be \\( t \\), then \\( BA = t + 1 \\).\n\nSubstituting into the equation:\n\\[ (t + 1)^{2} + t^{2} = 25 \\]\n\\[ t^{2} + t - 12 = 0 \\]\n\\[ (t + 4)(t - 3) = 0 \\]\n\nSince \\( t > 0 \\), \\( t + 4 > 0 \\), thus \\( t - 3 = 0 \\), so \\( t = 3 \\).\n\nTherefore, \\( BC = 2BE = 2t = 2 \\times 3 = 6 \\).\n\nThe final answer is: 6.\n\n[Key Insight] This problem examines the function graph of a moving point. The key to solving the problem lies in using the Pythagorean theorem to determine the length of \\( BE \\)." }, { "problem_id": 1174, "question": "As shown in the figure, Figure 1 is a multi-position adjustable reclining chair purchased by Xiaohui during the \"Tmall Double 11\" event. The adjustment diagram of the reclining chair is shown in Figure 2. It is known that the two legs $AB = AC = 70$ cm, $BC = 84$ cm, and $O$ is a fixed connection point on $AC$. The backrest $OD = 70$ cm. When the reclining chair is in the I position, $OD // AB$. When the reclining chair is in the II position, $OD' \\perp AC$. When the reclining chair is adjusted from the I position to the II position, the horizontal distance that the top of the backrest $D$ leans back (i.e., $EF$) is $\\qquad$ cm.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0095_1.jpg", "batch11-2024_06_14_2ea05802e1e1e7778d4eg_0095_2.jpg" ], "is_multi_img": true, "answer": "14", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw $AG \\perp BC$ at point $G$, and $OH \\perp BC$ at point $H$. Then, draw $OM \\perp D'F$ at point $M$, intersecting $DE$ at point $N$, as shown in the figure.\n\n\n\nThus, $OM = HE$, $ON = HE$.\n\nGiven that $AB = AC = 70$ cm, and $BC = 84$ cm,\n\nTherefore, $BG = CG = \\frac{1}{2} BC = 42$ cm,\n\nHence, $AG = \\sqrt{AB^2 - BG^2} = \\sqrt{70^2 - 42^2} = 56$ cm.\n\nSince $AB \\parallel OD$, and $BC \\parallel OM$,\n\nTherefore, $\\angle ABG = \\angle DON$.\n\nIn triangles $\\triangle ABG$ and $\\triangle DON$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle ABG = \\angle DON \\\\\n\\angle AGB = \\angle DNP = 90^\\circ \\\\\nAB = DO\n\\end{array}\n\\right.\n\\]\n\nThus, $\\triangle ABG \\cong \\triangle DON$ (by AAS),\n\nTherefore, $BG = ON = HE = 42$ cm.\n\nSince $OD' \\perp AC$,\n\nTherefore, $\\angle D'OM + \\angle MOC = 90^\\circ$.\n\nSince $OM \\parallel BC$,\n\nTherefore, $\\angle MOC = \\angle ACG$.\n\nSince $\\angle ACG + \\angle CAG = 90^\\circ$,\n\nTherefore, $\\angle CAG = \\angle D'OM$.\n\nIn triangles $\\triangle ACG$ and $\\triangle OD'M$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle CAG = \\angle D'OM \\\\\n\\angle AGC = \\angle OMD' = 90^\\circ \\\\\nAC = OD'\n\\end{array}\n\\right.\n\\]\n\nThus, $\\triangle ACG \\cong \\triangle OD'M$ (by AAS),\n\nTherefore, $AG = OM = HF = 56$ cm.\n\nHence, $EF = HF - HE = 56 - 42 = 14$ cm.\n\nThe answer is: 14.\n\n【Key Insight】This problem mainly examines the properties and determination of congruent triangles, the Pythagorean theorem, and the properties of isosceles triangles. The key is to construct congruent triangles." }, { "problem_id": 1175, "question": "The image on the left is the emblem of the Seventh International Congress on Mathematical Education (ICME 7). The main pattern of the emblem is derived from a series of right-angled triangles as shown in the image on the right, where $\\mathrm{OA}_{1}=\\mathrm{A}_{1} \\mathrm{~A}_{2}=\\mathrm{A}_{2} \\mathrm{~A}_{3}=\\ldots=\\mathrm{A}_{7} \\mathrm{~A}_{8}=1$. If we continue to draw the right-angled triangles as shown in the image on the right, then among the segments $\\mathrm{OA}_{1}, \\mathrm{OA}_{2}, \\ldots, \\mathrm{OA}_{25}$, there are $\\qquad$ segments with lengths that are positive integers.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_390b4cbd1ade079efcbeg_0007_1.jpg", "batch11-2024_06_14_390b4cbd1ade079efcbeg_0007_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we identify the pattern that $\\mathrm{OA}_{\\mathrm{n}}=\\sqrt{n}$.\n\nThus, the values from $\\mathrm{OA}_{1}$ to $\\mathrm{OA}_{25}$ are respectively $\\sqrt{1}, \\sqrt{2}, \\sqrt{n}, \\ldots, \\sqrt{25}$.\n\nTherefore, the positive integers are $\\sqrt{1}=1, \\sqrt{4}=2, \\sqrt{9}=3, \\sqrt{16}=4, \\sqrt{25}=5$.\n\nHence, the answer is 5.\n\n[Insight] This problem primarily tests the Pythagorean theorem. The key to solving it lies in using the Pythagorean theorem to determine the side lengths of a right triangle and discovering the pattern $\\mathrm{OA}_{\\mathrm{n}}=\\sqrt{n}$." }, { "problem_id": 1176, "question": "In a grid graph where the side length of each small square is 1, the vertices of each small square are called grid points. Taking the sides of a square $\\mathrm{ABCD}$ with all vertices being grid points as the hypotenuse, four congruent right-angled triangles are constructed internally, such that the right-angled vertices E, F, G, H are all grid points, and the quadrilateral $\\mathrm{EFGH}$ is a square. We call such a figure a grid chord diagram. For example, in the grid chord diagram shown in Figure 1, the side length of square $\\mathrm{ABCD}$ is $\\sqrt{65}$, and at this time, the area of square $\\mathrm{EFGH}$ is 5. The question is: when the side length of square $\\mathrm{ABCD}$ in the grid chord diagram is $\\sqrt{65}$, what are all the possible values for the area of square $\\mathrm{EFGH}$ (excluding 5)?\n\n\n\nFigure 1\n\n\n\nSpare Diagram", "input_image": [ "batch11-2024_06_14_390b4cbd1ade079efcbeg_0009_1.jpg", "batch11-2024_06_14_390b4cbd1ade079efcbeg_0009_2.jpg" ], "is_multi_img": true, "answer": "9 or 13 or 49 .", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Analysis: There are three scenarios:\n\n(1) When \\(\\mathrm{DG} = \\sqrt{13}\\) and \\(\\mathrm{CG} = 2\\sqrt{13}\\), the condition \\(\\mathrm{DG}^{2} + \\mathrm{CG}^{2} = \\mathrm{CD}^{2}\\) is satisfied. At this point, \\(\\mathrm{HG} = \\sqrt{13}\\), and the area of square \\(\\mathrm{EFGH}\\) is 13.\n\n(2) When \\(\\mathrm{DG} = 8\\) and \\(\\mathrm{CG} = 1\\), the condition \\(\\mathrm{DG}^{2} + \\mathrm{CG}^{2} = \\mathrm{CD}^{2}\\) is satisfied. At this point, \\(\\mathrm{HG} = 7\\), and the area of square \\(\\mathrm{EFGH}\\) is 49.\n\n(3) When \\(\\mathrm{DG} = 7\\) and \\(\\mathrm{CG} = 4\\), the condition \\(\\mathrm{DG}^{2} + \\mathrm{CG}^{2} = \\mathrm{CD}^{2}\\) is satisfied. At this point, \\(\\mathrm{HG} = 3\\), and the area of square \\(\\mathrm{EFGH}\\) is 9.\n\nDetailed Explanation:\n\n(1) When \\(\\mathrm{DG} = \\sqrt{13}\\) and \\(\\mathrm{CG} = 2\\sqrt{13}\\), the condition \\(\\mathrm{DG}^{2} + \\mathrm{CG}^{2} = \\mathrm{CD}^{2}\\) is satisfied. At this point, \\(\\mathrm{HG} = \\sqrt{13}\\), and the area of square \\(\\mathrm{EFGH}\\) is 13.\n\n(2) When \\(\\mathrm{DG} = 8\\) and \\(\\mathrm{CG} = 1\\), the condition \\(\\mathrm{DG}^{2} + \\mathrm{CG}^{2} = \\mathrm{CD}^{2}\\) is satisfied. At this point, \\(\\mathrm{HG} = 7\\), and the area of square \\(\\mathrm{EFGH}\\) is 49.\n\n(3) When \\(\\mathrm{DG} = 7\\) and \\(\\mathrm{CG} = 4\\), the condition \\(\\mathrm{DG}^{2} + \\mathrm{CG}^{2} = \\mathrm{CD}^{2}\\) is satisfied. At this point, \\(\\mathrm{HG} = 3\\), and the area of square \\(\\mathrm{EFGH}\\) is 9.\n\nTherefore, the answer is 9, 13, or 49.\n\nKey Point: This question tests knowledge of geometric construction and application, the Pythagorean theorem, and the ability to solve problems using a combination of numerical and graphical methods. It is a challenging question typically found in high school entrance exams." }, { "problem_id": 1177, "question": "Xiao Zhang made the \"tangram\" shown in Figure 1 by cutting a square paper during a math activity class, and designed the tree crown of a metasequoia as shown in Figure 2. If the side length of the square paper in Figure 1 is known to be $2 \\mathrm{~cm}$, then the height of the metasequoia tree crown in Figure 2 (i.e., the distance from point $A$ to line segment $B C$) is $\\qquad$ $\\mathrm{cm}$.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch11-2024_06_14_3f0bfb596e9e4be25288g_0001_1.jpg", "batch11-2024_06_14_3f0bfb596e9e4be25288g_0001_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}+1 \\# \\# 1+\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $AE \\perp MN$ at point $E$.\n\nSince $MN = BH = 2 \\text{ cm}$,\n\nwe have $AE = \\frac{1}{2} MN = 1$ cm.\n\nGiven that $\\sqrt{HF^2 + BF^2} = 2$,\n\nand $HF = BF = \\sqrt{2}$ cm,\n\nthe height of the Metasequoia tree crown in Figure 2 is $AE + HF = (\\sqrt{2} + 1)$ cm.\n\nTherefore, the answer is: $(\\sqrt{2} + 1)$.\n\n\n\n(Figure 2)\n\n[Key Insight] This problem tests the application of the Pythagorean theorem. Correctly identifying the figure and constructing the right triangle are crucial to solving the problem." }, { "problem_id": 1178, "question": "The tangram is a crystallization of the wisdom of ancient Chinese laborers and is known as the \"Oriental Magic Square\" by Westerners. The two images below are arrangements of the same tangram set by Le Le. Given that the side length of the square formed by the tangram (as shown in Figure 1) is 6, the area of the shaded part of the \"Sailing\" pattern (as shown in Figure 2) is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_3f0bfb596e9e4be25288g_0008_1.jpg", "batch11-2024_06_14_3f0bfb596e9e4be25288g_0008_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{9}{4}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, according to the problem statement, the shaded part of the \"Setting Sail\" pattern is the smallest isosceles right triangle, with \\( BD = CD \\), \\( AC = BC \\), and \\( AB = 6 \\).\n\nBy the Pythagorean theorem, we have:\n\\[ AC = BC = 3\\sqrt{2} \\]\n\nThus,\n\\[ BD = \\frac{1}{2} BC = \\frac{3\\sqrt{2}}{2} \\]\n\nTherefore, the area of the shaded part is:\n\\[ \\text{Area} = \\frac{1}{2} \\times \\frac{3\\sqrt{2}}{2} \\times \\frac{3\\sqrt{2}}{2} = \\frac{9}{4} \\]\n\nHence, the answer is:\n\\[ \\boxed{\\frac{9}{4}} \\]\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Insight】This problem primarily examines the properties of squares, isosceles right triangles, and the Pythagorean theorem. The key to solving the problem lies in determining the side lengths of the shaded part based on the given information." }, { "problem_id": 1179, "question": "As shown in the figure, Figure 1 is the emblem of the 7th International Congress on Mathematical Education (ICME). The pattern can be seen as consisting of several adjacent right-angled triangles in Figure 2, where $A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}=A_{4} A_{5}=$ $\\qquad$ $=A_{2021} A_{2022}$, $\\angle A_{1} O A_{2}=45^{\\circ}$, $O A_{1}=1$, and $\\angle O A_{1} A_{2}=\\angle O A_{2} A_{3}=\\angle O A_{3} A_{4}=\\angle O A_{4} A_{5}=\\cdots \\cdots=\\angle O A_{2021} A_{2022}=90$. The length of side $O A_{2022}$ is $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_3f0bfb596e9e4be25288g_0032_1.jpg", "batch11-2024_06_14_3f0bfb596e9e4be25288g_0032_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2022}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( O A_{1} = 1 \\) and \\( \\angle A_{1} O A_{2} = 45^{\\circ} \\),\n\nit follows that \\( A_{1} A_{2} = A_{2} A_{3} = A_{3} A_{4} = A_{4} A_{5} = \\cdots = A_{2021} A_{2022} = 1 \\).\n\nTherefore, by the Pythagorean theorem, we have \\( O A_{2} = \\sqrt{1^{2} + 1^{2}} = \\sqrt{2} \\),\n\n\\( O A_{3} = \\sqrt{(\\sqrt{2})^{2} + 1^{2}} = \\sqrt{3} \\),\n\nand so on,\n\nhence \\( O A_{n} = \\sqrt{n} \\),\n\nand thus \\( O A_{2022} = \\sqrt{2022} \\).\n\nThe final answer is: \\( \\sqrt{2022} \\).\n\n【Key Insight】This problem examines the application of the Pythagorean theorem and the properties of right-angled triangles. The key to solving the problem lies in identifying the correct pattern." }, { "problem_id": 1180, "question": "As shown in Figure 1, swinging on a swing is a sport created by ancient northern minorities in China. One day, while playing in the park, as shown in Figure 2, Xiao Ming noticed that when the swing is at rest, the vertical height of the footboard from the ground is \\( DE = 1 \\text{ m} \\). When he pushes the swing forward by \\( 6 \\text{ m} \\) (horizontal distance \\( BC = 6 \\text{ m} \\)), the vertical height of the footboard from the ground is \\( BF = CE = 3 \\text{ m} \\). The rope of the swing is always taut. Find the length of the rope \\( AD \\)?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_4663038985df7d5d7dc9g_0027_1.jpg", "batch11-2024_06_14_4663038985df7d5d7dc9g_0027_2.jpg" ], "is_multi_img": true, "answer": "$10 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of the swing's rope be $x \\mathrm{~m}$. Then, $A C = x - (3 - 1) = (x - 2) \\mathrm{m}$. In the right triangle $\\triangle A C B$, we have $A C^{2} + B C^{2} = A B^{2}$.\n\nThus, $x^{2} = 6^{2} + (x - 2)^{2}$.\n\nSolving the equation, we find: $x = 10$.\n\nAnswer: The length of the rope $A D$ is $10 \\mathrm{~m}$.\n\n【Highlight】This problem primarily tests the application of the Pythagorean theorem. Mastering the Pythagorean theorem is crucial for solving such problems." }, { "problem_id": 1181, "question": "After studying the Pythagorean Theorem, Teacher Li and the members of the \"Geometry Squad\" went to the playground to measure the height of the flagpole \\( AB \\). They obtained the following information:\n\n1. They measured that the rope used for raising the flag, which hangs vertically from the top of the flagpole, is 2 meters longer than the flagpole (as shown in Figure 1).\n2. When the rope is pulled taut, they measured the distance from the hand holding the rope to the ground \\( CD \\) is 1 meter, and the distance from the hand to the flagpole \\( CE \\) is 9 meters (as shown in Figure 2).\n\nBased on this information, determine the height of the flagpole \\( AB \\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_4663038985df7d5d7dc9g_0089_1.jpg", "batch11-2024_06_14_4663038985df7d5d7dc9g_0089_2.jpg" ], "is_multi_img": true, "answer": " 13 m.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( AB = x \\). According to the problem statement:\n\nIn the right triangle \\( ACE \\), \\( AC^{2} = AE^{2} + CE^{2} \\),\n\nThat is: \\( (x + 2)^{2} = (x - 1)^{2} + 9^{2} \\),\n\nSolving this equation yields: \\( x = 13 \\).\n\nAnswer: The height of the flagpole \\( AB \\) is 13 meters.\n\n[Highlight] This problem examines the application of the Pythagorean theorem. Mastering the relevant knowledge of the Pythagorean theorem and correctly applying it in right-angled triangles is key to solving the problem." }, { "problem_id": 1182, "question": "Cut the cross-shaped cardboard composed of 5 small squares (as shown in Figure 1), the following cutting methods that can be used to form a large square by the pieces cut are $\\qquad$ (fill in the serial numbers).\n\n\n\nFigure 1\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch11-2024_06_14_4e6f08adeb7a334e436cg_0016_1.jpg", "batch11-2024_06_14_4e6f08adeb7a334e436cg_0016_2.jpg", "batch11-2024_06_14_4e6f08adeb7a334e436cg_0016_3.jpg", "batch11-2024_06_14_4e6f08adeb7a334e436cg_0016_4.jpg" ], "is_multi_img": true, "answer": "(1)(3)\\#\\#(3)(1)", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the cutting method described in (1), by making one cut on each of the four small squares on the periphery and then placing them into the adjacent empty spaces, a square with a side length of $\\sqrt{5}$ can be formed, which meets the requirement;\n\nAs shown in the figure below, following the cutting method in (3), a square with a side length of $\\sqrt{5}$ can also be obtained through splicing, which meets the requirement;\n\n\n\nHowever, according to the cutting method in (2), it is not possible to form a square with a side length of $\\sqrt{5}$, which does not meet the requirement;\n\nTherefore, the correct choices are (1) and (3).\n\nThe answer is: (1)(3).\n\n[Key Insight] This question tests the ability to splice shapes. The key to solving the problem lies in determining the side length of the square to be formed based on the area of the given small squares." }, { "problem_id": 1183, "question": "As shown in Figure 1, Xiao Ming folds a rectangular piece of paper so that the two sides of the rectangle coincide, with the crease being $EF$. After unfolding, he folds along $BC$ such that point $A$ coincides with point $D$ on $EF$. In Figure 2, the rectangular piece of paper is folded again, with creases $HG$ and $KL$, making both sides of the rectangle coincide with $EF$. After unfolding, he folds along $BP$ such that point $A$ coincides with point $Q$ on $KL$. Connecting $AD$ in Figure 1 and $AQ$ in Figure 2, the value of $\\frac{AQ}{AD}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_4e6f08adeb7a334e436cg_0052_1.jpg", "batch11-2024_06_14_4e6f08adeb7a334e436cg_0052_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{\\sqrt{6}}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( AB = 4m \\). As shown in Figure 1, by folding, we have \\( DB = AB \\), and \\( EF \\) is the perpendicular bisector of \\( AB \\).\n\n\\[\n\\therefore AD = DB = AB = 4m;\n\\]\n\nAs shown in Figure 2, by folding, we have \\( AE = BE = \\frac{1}{2} AB \\), \\( AG = EG = BL = EL = \\frac{1}{2} BE \\), and \\( QB = AB \\).\n\n\\[\n\\therefore AL = 3m, \\quad BL = m, \\quad QB = 4m,\n\\]\n\nSince \\( KL \\) is the perpendicular bisector of \\( BE \\),\n\n\\[\n\\therefore \\angle BLQ = \\angle ALQ = 90^\\circ,\n\\]\n\n\\[\n\\therefore QL^2 = QB^2 - BL^2 = 15m^2,\n\\]\n\n\\[\n\\therefore AQ = \\sqrt{AL^2 + QL^2} = \\sqrt{24m^2} = 2\\sqrt{6}m,\n\\]\n\n\\[\n\\therefore \\frac{AQ}{AD} = \\frac{2\\sqrt{6}m}{4m} = \\frac{\\sqrt{6}}{2},\n\\]\n\nThus, the answer is: \\( \\frac{\\sqrt{6}}{2} \\).\n\n**Key Insight:** This problem focuses on the properties of symmetry, the properties of perpendicular bisectors, the application of the Pythagorean theorem, and the simplification of square roots. By setting \\( AB = 4m \\) and using the properties of symmetry and the Pythagorean theorem, the expressions for \\( AD \\) and \\( AQ \\) in terms of \\( m \\) are derived, which is crucial for solving the problem." }, { "problem_id": 1184, "question": "As shown in the figure, in the Rt $\\triangle A O B$, $\\angle A O B=90^{\\circ}$, $O A=O B=4$, point $P$ moves along the line $O A$, connect $P B$, fold $\\triangle O B P$ along the line $B P$, and denote the corresponding point of $O$ as $O^{\\prime}$. When point $O^{\\prime}$ is on the line $A B$, the length of $O P$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_4e6f08adeb7a334e436cg_0099_1.jpg", "batch11-2024_06_14_4e6f08adeb7a334e436cg_0099_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{2}-2$ or $4 \\sqrt{2}+2$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, point $O^{\\prime}$ lies on the line segment $AB$,\n\n\n\n$\\because OA = OB = 4$,\n\n$\\therefore \\triangle ABO$ is an isosceles right triangle, $\\angle BAO = 45^{\\circ}$,\n\n$\\therefore AB = \\sqrt{2} \\cdot OA = 4\\sqrt{2}$,\n\nBy folding, we have $BO^{\\prime} = BO = 4$, and $PO^{\\prime} \\perp AB$,\n\n$\\therefore AO^{\\prime} = PO^{\\prime} = OP = AB - BO^{\\prime} = 4\\sqrt{2} - 4$,\n\nAs shown in the figure, point $O^{\\prime}$ lies on the extension of $AB$,\n\n\n\nBy folding, we have $BO^{\\prime} = BO = 4$,\n\n$\\therefore AO^{\\prime} = AB + BO^{\\prime} = 4\\sqrt{2} + 4$,\n\n$\\because \\angle BAO = 45^{\\circ}$, and $PO^{\\prime} \\perp AB$,\n\n$\\therefore \\triangle APO^{\\prime}$ is an isosceles right triangle,\n\n$\\therefore PO^{\\prime} = AO^{\\prime} = PO = 4\\sqrt{2} + 4$,\n\nTherefore, the answer is: $4\\sqrt{2} - 2$ or $4\\sqrt{2} + 2$\n\n【Key Insight】This problem examines the folding of geometric figures and the Pythagorean theorem. The key to solving it lies in categorizing the cases and solving them in conjunction with the diagram." }, { "problem_id": 1185, "question": "Read the following material:\n\nA school's math interest activity group conducted the following exploration on a math problem during one of their activities:\n\nIn triangle \\(ABC\\), \\(\\angle BAC = 90^\\circ\\), \\(AB = AC\\), \\(\\angle B = \\angle BCA = 45^\\circ\\), and \\(D\\) is the midpoint of \\(BC\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) Problem Discovery: As shown in Figure 1, if points \\(E\\) and \\(F\\) are on segments \\(AB\\) and \\(AC\\) respectively, and \\(AE = CF\\), connect \\(EF\\), \\(DE\\), \\(DF\\), and \\(AD\\). At this point, Xiao Ming discovered that \\(\\angle BAD = \\qquad \\circ\\), and \\(AD \\qquad DC\\) (fill in \">\", \"<\", or \"=\"); Xiao Ming and his classmates continued their exploration and found a conclusion: the ratio of the lengths of segments \\(EF\\) to \\(DE\\) is a fixed value, i.e., \\(EF = \\qquad DE\\).\n\n(2) Variation Exploration: As shown in Figure 2, \\(E\\) and \\(F\\) are on the extensions of segments \\(BA\\) and \\(AC\\) respectively, and \\(AE = CF\\). If \\(EF = 4\\), find the length of \\(DE\\) and write the process.\n\n(3) Extended Application: As shown in Figure 3, \\(AB = AC = 6\\), the moving point \\(M\\) is on the extension of \\(AD\\), point \\(H\\) is on the line \\(AC\\), and satisfies \\(\\angle BMH = 90^\\circ\\), \\(CH = 2\\). Please directly write the length of \\(DM\\) as \\(\\qquad\\).", "input_image": [ "batch11-2024_06_14_522d7c3696f285ed38cag_0100_1.jpg", "batch11-2024_06_14_522d7c3696f285ed38cag_0100_2.jpg", "batch11-2024_06_14_522d7c3696f285ed38cag_0100_3.jpg" ], "is_multi_img": true, "answer": "$(1) 45,=, \\sqrt{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "(1) Solution: In triangle \\( ABC \\), \\( \\angle BAC = 90^\\circ \\), \\( AB = AC \\), \\( \\angle B = \\angle BCA = 45^\\circ \\), and \\( D \\) is the midpoint of \\( BC \\).\n\n\\[\n\\therefore AD \\perp BC, \\angle BAD = 45^\\circ = \\angle C,\n\\]\n\n\\[\n\\therefore \\angle ADC = 90^\\circ, AD = CD = BD,\n\\]\n\n\\[\n\\because AE = CF,\n\\]\n\n\\[\n\\therefore \\triangle ADE \\cong \\triangle CDF \\quad (\\text{SAS}),\n\\]\n\n\\[\n\\therefore DE = DF, \\angle ADE = \\angle CDF\n\\]\n\n\\[\n\\because \\angle CDF + \\angle ADF = \\angle ADC = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle ADE + \\angle ADF = 90^\\circ,\n\\]\n\n\\[\n\\therefore DEF \\text{ is an isosceles right triangle},\n\\]\n\n\\[\n\\therefore EF = \\sqrt{DE^2 + DF^2} = \\sqrt{2DE^2} = \\sqrt{2}DE,\n\\]\n\nThus, the answers are: \\( 45, =, \\sqrt{2} \\);\n\n(2) Solution: Since \\( AB = AC \\) and \\( \\angle BAC = 90^\\circ \\),\n\n\\[\n\\therefore \\angle B = \\angle ACD = 45^\\circ,\n\\]\n\n\\[\n\\because AB = AC, \\text{ and } D \\text{ is the midpoint of } BC,\n\\]\n\n\\[\n\\therefore AD \\perp BC, \\angle BAD = \\angle CAD = 45^\\circ\n\\]\n\n\\[\n\\therefore AD = CD, \\angle ADC = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle BAD = \\angle ACD,\n\\]\n\n\\[\n\\therefore 180^\\circ - \\angle BAD = 180^\\circ - \\angle ACD,\n\\]\n\ni.e., \\( \\angle EAD = \\angle FCD \\),\n\n\\[\n\\because \\angle ACD = \\angle CAD = 45^\\circ,\n\\]\n\n\\[\n\\therefore AD = CD,\n\\]\n\nIn triangles \\( ADE \\) and \\( CDF \\), \n\n\\[\n\\left\\{\n\\begin{aligned}\nAD &= CD, \\\\\n\\angle EAD &= \\angle FCD, \\\\\nAE &= CF\n\\end{aligned}\n\\right.\n\\]\n\n\\[\n\\therefore \\triangle ADE \\cong \\triangle CDF (\\text{SAS}),\n\\]\n\n\\[\n\\therefore DE = DF, \\angle ADE = \\angle CDF,\n\\]\n\n\\[\n\\therefore \\angle EDF = \\angle CDF + \\angle EDC = \\angle ADE + \\angle EDC = \\angle ADC = 90^\\circ,\n\\]\n\n\\[\n\\therefore DE \\perp DF\n\\]\n\nIn right triangle \\( EDF \\), \\( DE = DF \\), \\( EF = 4 \\),\n\n\\[\n\\therefore DE = \\frac{\\sqrt{2}}{2} EF = 2\\sqrt{2};\n\\]\n\n(3) Solution: (1) When \\( H \\) is on segment \\( AC \\), as shown in the figure, connect \\( MC \\), and draw \\( MF \\perp AC \\) at \\( F \\),\n\n\\[\n\\because AD \\perp BC, BD = CD,\n\\]\n\n\\[\n\\therefore AM \\text{ is the perpendicular bisector of } BC,\n\\]\n\n\\[\n\\therefore MB = MC,\n\\]\n\n\\[\n\\therefore \\angle MBC = \\angle MCB,\n\\]\n\n\\[\n\\because AB = AC,\n\\]\n\n\\[\n\\therefore \\angle ABC = \\angle ACB,\n\\]\n\n\\[\n\\therefore \\angle ABM = \\angle ACM,\n\\]\n\nAlso, since \\( \\angle BAC = \\angle BMH = 90^\\circ \\), and \\( \\angle BAH + \\angle ABM + \\angle BMH + \\angle AHM = 360^\\circ \\),\n\n\\[\n\\therefore \\angle ABM + \\angle AHM = 180^\\circ,\n\\]\n\n\\[\n\\because \\angle AHM + \\angle MHC = 180^\\circ,\n\\]\n\n\\[\n\\therefore \\angle ABM = \\angle MHC\n\\]\n\n\\[\n\\therefore \\angle MCH = \\angle MHC,\n\\]\n\n\\[\n\\therefore MH = MC,\n\\]\n\n\\[\n\\because CH = 2,\n\\]\n\n\\[\n\\therefore HF = CF = \\frac{1}{2} CH = 1,\n\\]\n\n\\[\n\\because AC = 6\n\\]\n\n\\[\n\\therefore AF = AC - CF = 6 - 1 = 5,\n\\]\n\n\\[\n\\because \\angle DAC = 45^\\circ,\n\\]\n\n\\[\n\\therefore AF = MF = 5,\n\\]\n\n\\[\n\\therefore AM = \\sqrt{2} AF = 5\\sqrt{2},\n\\]\n\n\\[\n\\because AB = AC = 6, \\angle BAC = 90^\\circ, D \\text{ is the midpoint of } BC,\n\\]\n\n\\[\n\\therefore AD = \\frac{1}{2} BC = 3\\sqrt{2},\n\\]\n\n\\[\n\\therefore DM = AM - AD = 5\\sqrt{2} - 3\\sqrt{2} = 2\\sqrt{2};\n\\]\n\n(2) When \\( H \\) is on the extension of segment \\( AC \\), as shown in the figure, connect \\( MC \\), and draw \\( MF \\perp AC \\) at \\( F \\),\n\nSimilarly, \\( CF = HF = 1 \\),\n\n\\[\n\\therefore AF = AC + CF = 6 + 1 = 7,\n\\]\n\n\\[\n\\therefore AM = \\sqrt{2} AF = 7\\sqrt{2},\n\\]\n\n\\[\n\\therefore DM = AM - AD = 7\\sqrt{2} - 3\\sqrt{2} = 4\\sqrt{2},\n\\]\n\nIn summary, the length of \\( DM \\) is \\( 2\\sqrt{2} \\) or \\( 4\\sqrt{2} \\).\n\n【Insight】This problem is a comprehensive question on triangles, mainly testing the properties of right triangles, the determination and properties of isosceles triangles, the determination and properties of congruent triangles, and the determination and properties of isosceles right triangles. The key to solving the problem is deducing that \\( \\triangle ADE \\cong \\triangle CDF \\)." }, { "problem_id": 1186, "question": "As shown in Figure 1, this is a schematic diagram of a mobile mini crane in operation. When the crane is working, it uses the change in the length of the boom and the angle of inclination to alter the lifting height and working radius. During a certain lifting operation, the learning interest group obtained the following information through measurements and consulting with the workers: as shown in Figure 2, the lifting arm $A B = 1 \\text{m}$, the distance from point $B$ to the ground $B C = 1.4 \\text{m}$, the distance from point $B$ to $A D$ $B E = 0.6 \\text{m}$, and the quadrilateral $B E D C$ is a rectangle. What is the length of $A D$, the distance from point $A$ to the ground, in meters?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_5696547958469086105eg_0057_1.jpg", "batch11-2024_06_14_5696547958469086105eg_0057_2.jpg" ], "is_multi_img": true, "answer": "2.2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the problem, we know that $\\angle AEB = 90^{\\circ}$. In the right triangle $\\triangle ABE$, by the Pythagorean theorem, we have:\n\n$AE = \\sqrt{AB^{2} - BE^{2}} = \\sqrt{1^{2} - 0.6^{2}} = 0.8 \\text{ m}$.\n\nSince quadrilateral $BEDC$ is a rectangle,\n\n$ED = BC = 1.4 \\text{ m}$.\n\nTherefore, $AD = ED + AE = 1.4 + 0.8 = 2.2 \\text{ m}$.\n\nAnswer: The distance from point $A$ to the ground $AD$ is 2.2 meters.\n\n[Key Insight] This problem primarily tests the Pythagorean theorem. The key to solving it is to be proficient with the Pythagorean theorem, which allows us to find that $AE = 0.8 \\text{ m}$." }, { "problem_id": 1187, "question": "As shown in Figure (1), the emblem pattern of the Seventh International Congress on Mathematical Education (ICME-7) is evolved from a series of right-angled triangles sharing a common vertex $O$. If in Figure (2), $O A_{1}=A_{1} A_{2}=A_{2} A_{3}=\\cdots A_{7} A_{8}=2$, then the length of $O A_{8}$ is $\\qquad$ .\n\n\n\nFigure (1)\n\n\nFigure (2)", "input_image": [ "batch11-2024_06_14_632675b0bbb89cebafc7g_0067_1.jpg", "batch11-2024_06_14_632675b0bbb89cebafc7g_0067_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since \\( O A_{1} = 2 \\),\n\nTherefore, by the Pythagorean theorem, we can obtain \\( O A_{2} = \\sqrt{2^{2} + 2^{2}} = 2 \\sqrt{2} \\),\n\n\\( O A_{3} = \\sqrt{(2 \\sqrt{2})^{2} + 2^{2}} = 2 \\sqrt{3} \\),\nHence, \\( O A_{n} = 2 \\sqrt{n} \\),\n\nThus, \\( O A_{8} = 2 \\sqrt{8} = 4 \\sqrt{2} \\).\n\nThe answer is: \\( 4 \\sqrt{2} \\).\n\n【Insight】This problem examines the flexible application of the Pythagorean theorem. The key to solving this problem lies in identifying the pattern \\( O A_{n} = 2 \\sqrt{n} \\)." }, { "problem_id": 1188, "question": "A fire broke out in a building in a certain location. Firefighters decided to use the ladder on the fire truck to rescue people as shown in Figure (1) and Figure (2). It is known that the ladder can extend up to $15 \\mathrm{~m}$ (i.e., $A B=C D=15 \\mathrm{~m}$), the fire truck is $3 \\mathrm{~m}$ high, and the ladder extends to its maximum length during the rescue. After completing the rescue from point $B$ at a height of $12 \\mathrm{~m}$ (i.e., $B E=12 \\mathrm{~m}$), they need to rescue someone from point $D$ at a height of $15 \\mathrm{~m}$ (i.e., $D E=15 \\mathrm{~m}$). At this time, how many meters should the fire truck move closer to the burning building from point $A$ to reach point $C$? (Extend $A C$ to intersect $D E$ at point $O$, where $A O \\perp D E$, point $B$ is on $D E$, and the length of $O E$ is the height of the fire truck, $3 \\mathrm{~m}$).\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch11-2024_06_14_699e1aef6d68a92b69a0g_0095_1.jpg", "batch11-2024_06_14_699e1aef6d68a92b69a0g_0095_2.jpg" ], "is_multi_img": true, "answer": "The distance $AC$ that the fire truck moves from its original position to the burning building is $3 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle \\( ABO \\), since \\( \\angle AOB = 90^\\circ \\), \\( AB = 15 \\, \\text{m} \\), and \\( OB = 12 - 3 = 9 \\, \\text{m} \\),\n\n\\[\nAO = \\sqrt{AB^2 - OB^2} = \\sqrt{15^2 - 9^2} = 12 \\, \\text{m}.\n\\]\n\nIn the right triangle \\( COD \\), since \\( \\angle COD = 90^\\circ \\), \\( CD = 15 \\, \\text{m} \\), and \\( OD = 15 - 3 = 12 \\, \\text{m} \\),\n\n\\[\nOC = \\sqrt{CD^2 - OD^2} = \\sqrt{15^2 - 12^2} = 9 \\, \\text{m}.\n\\]\n\nTherefore,\n\n\\[\nAC = OA - OC = 3 \\, \\text{m}.\n\\]\n\nAnswer: The distance \\( AC \\) that the fire truck moves closer to the burning building from its original position is \\( 3 \\, \\text{m} \\).\n\n[Highlight] This problem tests the application of the Pythagorean theorem. Mastering the Pythagorean theorem is key to solving the problem." }, { "problem_id": 1189, "question": "As shown in the figure, in quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $E$, $\\angle AEB = 45^\\circ$, $BD = 2$. If $\\triangle ABC$ is folded $180^\\circ$ along the line $AC$ to its original plane, and the point where $B$ lands is denoted as $B'$, then the length of $DB'$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_7773c6095a85eff52666g_0010_1.jpg", "batch11-2024_06_14_7773c6095a85eff52666g_0010_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( A B C D \\) is a parallelogram and \\( B D = 2 \\),\n\n\\[\n\\therefore B E = \\frac{1}{2} B D = 1.\n\\]\n\nAs shown in the figure, connect \\( B B^{\\prime} \\),\n\n\n\nAccording to the properties of folding, \\( \\angle A E B = \\angle A E B^{\\prime} = 45^{\\circ} \\), and \\( B E = B^{\\prime} E \\),\n\n\\[\n\\therefore \\angle B E B^{\\prime} = 90^{\\circ}.\n\\]\n\n\\[\n\\therefore \\triangle B B^{\\prime} E \\text{ is an isosceles right triangle, so } B B^{\\prime} = \\sqrt{2} B E = \\sqrt{2}.\n\\]\n\nMoreover, since \\( B E = D E \\) and \\( B^{\\prime} E \\perp B D \\),\n\n\\[\n\\therefore D B^{\\prime} = B B^{\\prime} = \\sqrt{2}.\n\\]\n\nTherefore, the answer is: \\( \\sqrt{2} \\)\n\n【Insight】This problem examines the properties of parallelograms, the determination and properties of isosceles triangles, and the properties of folding transformations. The key to solving the problem is deducing that \\( D B^{\\prime} = B B^{\\prime} \\)." }, { "problem_id": 1190, "question": "\"The Nine Chapters on the Mathematical Art\" is a representative work of ancient Eastern mathematics. The book records: \"If a door is opened to remove the threshold (kǔn, meaning the door sill) by one foot, and the gap is two inches, what is the width of the door?\" The main idea of the question is: As shown in Figures 1 and 2 (Figure 2 is a plan view of Figure 1), when the double doors are pushed open, the distance between the gaps $\\mathrm{CD}$ is 2 inches, and the points $C$ and $D$ are both 1 foot (1 foot = 10 inches) away from the threshold $A B$. What is the length of $A B$?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_7999e267fcd170134b08g_0041_1.jpg", "batch11-2024_06_14_7999e267fcd170134b08g_0041_2.jpg" ], "is_multi_img": true, "answer": "101 inches", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let’s take the midpoint of $\\mathrm{AB}$ as $\\mathrm{O}$, and draw $\\mathrm{DE} \\perp \\mathrm{AB}$ through point $\\mathrm{D}$, intersecting at point $\\mathrm{E}$, as shown in Figure 2:\n\n\n\nFigure 2\n\nFrom the problem statement, we have: $\\mathrm{OA} = \\mathrm{OB} = \\mathrm{AD} = \\mathrm{BC}$.\n\nLet $\\mathrm{OA} = \\mathrm{OB} = \\mathrm{AD} = \\mathrm{BC} = \\mathrm{r}$ inches,\nthen $\\mathrm{AB} = 2\\mathrm{r}$ inches, $\\mathrm{DE} = 10$ inches, and $\\mathrm{OE} = \\frac{1}{2} \\mathrm{CD} = 1$ inch.\n\nThus, $\\mathrm{AE} = (\\mathrm{r} - 1)$ inches.\n\nIn the right triangle $\\triangle \\mathrm{ADE}$,\n\n$\\mathrm{AE}^{2} + \\mathrm{DE}^{2} = \\mathrm{AD}^{2}$, which is $(\\mathrm{r} - 1)^{2} + 10^{2} = \\mathrm{r}^{2}$.\n\nSolving this equation gives: $\\mathrm{r} = 50.5$,\n\nTherefore, $2\\mathrm{r} = 101$ inches,\n\nSo, $\\mathrm{AB} = 101$ inches.\n\n【Key Insight】This problem tests the application of the Pythagorean theorem. Understanding the problem and constructing the right triangle are key to solving it." }, { "problem_id": 1191, "question": "As shown in Figure 1, this is a physical image of a basketball hoop at a school, and its side view is shown in Figure 2. The \"Integration and Practice\" group has launched a practical activity to measure the length of the backboard $A B$. In situations where direct measurement is not feasible, the \"Integration and Practice\" group has designed the following plan:\n\nClass\n\nMeasuring the Length of the Backboard\n\n\n\nBased on the above measurements, please help the \"Integration and Practice\" group calculate the length of the school's backboard $A B$ (the result should be accurate to 0.01 meters).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_7999e267fcd170134b08g_0079_1.jpg", "batch11-2024_06_14_7999e267fcd170134b08g_0079_2.jpg", "batch11-2024_06_14_7999e267fcd170134b08g_0079_3.jpg" ], "is_multi_img": true, "answer": "1.05 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle \\( A D C \\), \\( A C = \\sqrt{A D^{2} - C D^{2}} = \\sqrt{5^{2} - 3.062^{2}} \\approx 3.952 \\).\n\nIn the right triangle \\( \\triangle B E C \\), \\( B C = \\sqrt{B E^{2} - C E^{2}} = \\sqrt{5^{2} - 4.073^{2}} \\approx 2.900 \\).\n\n\\[\n\\therefore A B = A C - B C \\approx 3.952 - 2.900 \\approx 1.05 \\text{ meters}\n\\]\n\nAnswer: The length of the backboard \\( A B \\) is 1.05 meters.\n\n\n\n【Key Insight】This problem primarily examines the application of the Pythagorean theorem. The key to solving it lies in identifying the right triangles based on the given information and remembering the Pythagorean theorem." }, { "problem_id": 1192, "question": "The tangram is known as the \"Eastern Magic Square\" by Westerners. The two figures below are formed from the same set of tangram pieces. Given that the side length of the square formed by the tangram (as shown in Figure a) is 4, the area of the shaded part of \"Smooth Sailing\" (as shown in Figure b) is $\\qquad$.\n\n\n\nFigure $\\mathrm{a}$\n\n\n\nFigure b", "input_image": [ "batch11-2024_06_14_7b4144f0605c4d7f8d6fg_0051_1.jpg", "batch11-2024_06_14_7b4144f0605c4d7f8d6fg_0051_2.jpg" ], "is_multi_img": true, "answer": "1", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** From the diagram, it can be observed that the shaded area in the \"一帆风顺\" figure is congruent to the two triangles labeled \"1\" and \"2\" within the square, which are both isosceles right triangles. Based on the trapezoid in Figure **b**, its side is equal to the leg of triangle 2, and the side length of the small square is also equal to the leg of triangle 2. Therefore, in Figure **a**, it can be deduced that the diagonal of the large square is equal to the sum of the side lengths of the four shaded right triangles. Given that the side length of the large square is 4, the diagonal \\( = \\sqrt{4^{2} + 4^{2}} = 4\\sqrt{2} \\). Thus, the side length of the shaded triangle \\( = 4\\sqrt{2} \\div 4 = \\sqrt{2} \\), and its area is: \\( \\frac{1}{2} \\times \\sqrt{2} \\times \\sqrt{2} = 1 \\).\n\n**Key Insight:** This question is of moderate difficulty, primarily testing students' understanding of the properties of isosceles right triangles and squares in the context of a tangram. The key to solving the problem lies in identifying the corresponding equal sides of the triangles in the two diagrams." }, { "problem_id": 1193, "question": "As shown in Figure (1), this pattern was given by Zhao Shuang of the Han Dynasty when he annotated the \"Zhoubi Suanjing,\" and it is known as the \"Zhao Shuang's Xian Diagram.\" The schematic diagram of this pattern is shown in Figure (2), where quadrilateral $ABCD$ and quadrilateral $EFGH$ are both squares,\n\n$\\triangle ABF, \\triangle BCG, \\triangle CDH, \\triangle DAE$ are four congruent right-angled triangles. If $EF=2, DE=8$, then the length of $AB$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_7b4144f0605c4d7f8d6fg_0096_1.jpg", "batch11-2024_06_14_7b4144f0605c4d7f8d6fg_0096_2.jpg" ], "is_multi_img": true, "answer": "10 .", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, we know that $BG = AF = DE = 8$, and $EF = FG = 2$. Therefore, $BF = BG - BF = 6$. In the right triangle $\\triangle ABF$, using the Pythagorean theorem, we find: $AB = \\sqrt{AF^{2} + BF^{2}} = \\sqrt{8^{2} + 6^{2}} = 10$. Hence, the answer is 10.\n\nKey point: This problem tests the proof of the Pythagorean theorem. The key to solving it lies in determining the lengths of the two legs of the right triangle $\\triangle ABF$." }, { "problem_id": 1194, "question": "As shown in Figure 1, equilateral triangles are constructed outwardly along each side of the right triangle \\( Rt \\triangle ABC \\), numbered as (1), (2), and (3). When (2) and (1) are superimposed on (3) as shown in Figure 2, it is known that the areas of quadrilateral \\( EMNB \\) and quadrilateral \\( MPQN \\) are \\( 9 \\sqrt{3} \\) and \\( 7 \\sqrt{3} \\) respectively. Then, the length of the hypotenuse \\( AB \\) of \\( Rt \\triangle ABC \\) is:\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_7ec3516690dd7297e0cbg_0012_1.jpg", "batch11-2024_06_14_7ec3516690dd7297e0cbg_0012_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the areas of equilateral triangles (1), (2), and (3) be \\( S_{1} \\), \\( S_{2} \\), and \\( S_{3} \\) respectively, with \\( AC = b \\), \\( BC = a \\), and \\( AB = c \\).\n\nSince \\( \\triangle ABC \\) is a right-angled triangle with \\( \\angle ACB = 90^{\\circ} \\),\n\n\\[ c^{2} = a^{2} + b^{2} \\],\n\n\\[ \\frac{\\sqrt{3}}{4} c^{2} = \\frac{\\sqrt{3}}{4} a^{2} + \\frac{\\sqrt{3}}{4} b^{2} \\],\n\nGiven that \\( S_{1} = \\frac{\\sqrt{3}}{4} a^{2} \\), \\( S_{2} = \\frac{\\sqrt{3}}{4} b^{2} \\), and \\( S_{3} = \\frac{\\sqrt{3}}{4} c^{2} \\),\n\n\\[ S_{3} - S_{2} = \\frac{\\sqrt{3}}{4} (c^{2} - b^{2}) = \\frac{\\sqrt{3}}{4} a^{2} = 9\\sqrt{3} \\],\n\n\\[ S_{3} - S_{1} = \\frac{\\sqrt{3}}{4} (c^{2} - a^{2}) = \\frac{\\sqrt{3}}{4} b^{2} = 9\\sqrt{3} + 7\\sqrt{3} = 16\\sqrt{3} \\],\n\nThus, \\( a = 6 \\), \\( b = 8 \\), which means \\( BC = 6 \\), \\( AC = 8 \\),\n\n\\[ AB = \\sqrt{BC^{2} + AC^{2}} = \\sqrt{6^{2} + 8^{2}} = 10 \\],\n\nTherefore, the answer is: 10.\n\n【Insight】This problem tests the practical application of the Pythagorean theorem. Familiarity with the calculation formula of the Pythagorean theorem and proficiency in using it to find values are key to solving the problem." }, { "problem_id": 1195, "question": "The right triangle in Figure 1 has a hypotenuse of length 4. Four such right triangles are arranged to form a square as shown in Figure 2, with the areas of the shaded regions denoted as $S_{1}$ and $S_{2}$. The value of $S_{1} + S_{2}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_7ec3516690dd7297e0cbg_0018_1.jpg", "batch11-2024_06_14_7ec3516690dd7297e0cbg_0018_2.jpg" ], "is_multi_img": true, "answer": "16", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of the longer leg of the right triangle be $a$, and the length of the shorter leg be $b$.\n\n$\\therefore a^{2} + b^{2} = 4^{2}$\n\n$\\because S_{1} = a^{2}, S_{2} = b^{2}$\n\n$\\therefore S_{1} + S_{2} = 16$\n\nThus, the answer is: 16\n\n【Insight】This question tests the application of the Pythagorean theorem. Mastering the Pythagorean theorem is key to solving the problem." }, { "problem_id": 1196, "question": "As shown in Figure 1, it is a closed Pythagorean water tank, where sections I, II, and III are square compartments that can hold water and are interconnected. It is known that $\\angle ACB = 90^\\circ, AC = 6, BC = 8$. Initially, section III is exactly full of water, while sections I and II are empty. When placed as shown in Figure 2, the water level just passes through the center $O$ of section III (the intersection of the two diagonals of the square), the area of section II that is filled with water is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_7ec3516690dd7297e0cbg_0051_1.jpg", "batch11-2024_06_14_7ec3516690dd7297e0cbg_0051_2.jpg" ], "is_multi_img": true, "answer": "14", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\angle ACB = 90^\\circ$, $AC = 6$, and $BC = 8$,\n\nTherefore, $AB = \\sqrt{AC^2 + BC^2} = 10$,\n\nThus, the area of part III is 100,\n\nSince the water surface just passes through the center $O$ of part III,\n\nTherefore, the water in part III covers half the area of the entire square, meaning the water-covered area of part III is 50,\n\nThus, the water-covered area in part II is $100 - 36 - 50 = 14$,\n\nHence, the answer is: 14.\n\n[Highlight] This problem tests the Pythagorean theorem and the calculation of the area of a square. Mastering the Pythagorean theorem is crucial for solving this problem." }, { "problem_id": 1197, "question": "The cube wooden block shown in Figure (1) has a side length of $8 \\mathrm{~cm}$. A corner is cut off along the diagonals of three adjacent faces (shown as dashed lines in the figure), resulting in the geometric shape shown in Figure (2). The shortest distance for an ant to crawl on the surface of the geometric shape from vertex $A$ to vertex $B$ is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch11-2024_06_14_849a94d80e794b4a7ef5g_0055_1.jpg", "batch11-2024_06_14_849a94d80e794b4a7ef5g_0055_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{2}+4 \\sqrt{6} \\# \\# 4 \\sqrt{6}+4 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: The geometric figure in Figure 2 is unfolded as shown below:\n\n\n\n$\\because$ The original geometric figure is a square,\n\n$\\therefore \\triangle B C D$ is an isosceles right triangle, $\\triangle A C D$ is an equilateral triangle, and $B A \\perp C D$, with $E$ being the midpoint of $C D$. In the right triangle $\\triangle \\mathrm{BCD}$, by the Pythagorean theorem, we have: $\\mathrm{CD}=\\sqrt{B C^{2}+B D^{2}}=\\sqrt{8^{2}+8^{2}}=8 \\sqrt{2}$,\n\n$\\therefore \\mathrm{BE}=\\frac{1}{2} \\mathrm{CD}=\\mathrm{CE}=4 \\sqrt{2}$,\n\nIn the right triangle $\\triangle A C E$, by the Pythagorean theorem, we have: $\\mathrm{AE}=\\sqrt{A C^{2}-C E^{2}}=\\sqrt{(8 \\sqrt{2})^{2}-(4 \\sqrt{2})^{2}}=4 \\sqrt{6}$,\n\n$\\therefore$ The shortest distance from vertex $A$ to vertex $B$ is $B E+A E=4 \\sqrt{2}+4 \\sqrt{6}$,\n\nThus, the answer is: $4 \\sqrt{2}+4 \\sqrt{6}$.\n\n【Insight】This problem examines the shortest path on a surface unfolding. The solution involves unfolding the surface of the geometric figure in Figure (2) into a plane figure and solving the problem using the properties of isosceles right triangles and equilateral triangles combined with the Pythagorean theorem." }, { "problem_id": 1198, "question": "The Genting Snow Park is the only snow sports competition venue among the seven snow venues of the Beijing 2022 Winter Olympics that was converted from an existing ski resort. The two images below show the actual scene and schematic diagram of the Genting Ski Resort's U-shaped pool. The venue can be considered as a rectangular prism from which half of a cylinder has been removed. In its cross-sectional diagram, the radius of the semicircle is $\\frac{12}{\\pi} \\mathrm{m}$, with edges $AB = CD = 24 \\mathrm{~m}$. Point $E$ is on $CD$, and $CE = 4 \\mathrm{~m}$. A skiing enthusiast travels from point $A$ to point $E$, and the shortest route length he travels is $\\qquad$ $m$.\n\n\n\nActual Scene of the Genting Ski Resort's U-shaped Pool\n\n\n\nSchematic Diagram of the Genting Ski Resort's U-shaped Pool", "input_image": [ "batch11-2024_06_14_849a94d80e794b4a7ef5g_0065_1.jpg", "batch11-2024_06_14_849a94d80e794b4a7ef5g_0065_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{34}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nAccording to the problem statement:\n\n$A D=2 \\pi \\times \\frac{12}{\\pi} \\times \\frac{1}{2}=12, \\quad D E=C D-C E=24-4=20$,\nThe line segment $A E$ represents the shortest path length for the slide.\n\nIn the right triangle $T t \\triangle A D E$, by the Pythagorean theorem, we have\n\n$A E=\\sqrt{A D^{2}+D E^{2}}=\\sqrt{12^{2}+20^{2}}=4 \\sqrt{34}(\\mathrm{~m})$.\n\nTherefore, the answer is: $4 \\sqrt{34}$\n\n【Highlight】This problem examines the shortest path problem in plane development. The key to solving this problem is to understand that the lateral expansion of a cylinder is a rectangle and to use the Pythagorean theorem to find the shortest distance." }, { "problem_id": 1199, "question": "As shown in Figure 1, four congruent right-angled triangles form a larger square, with a smaller square in the center. This pattern was presented by Zhao Shuang of the Han Dynasty when annotating the \"Zhoubi Suanjing,\" and it is known as the \"Zhao Shuang's Hypotenuse Diagram.\" Connecting four line segments in Figure 2 results in the new pattern shown in Figure 3. If the long leg of the right-angled triangle in Figure 1 is 5, and the short leg is 2, the area of the shaded part in Figure 3 is denoted as $\\mathrm{S}$. Then the value of $\\mathrm{S}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch11-2024_06_14_86d4d50b09351c6af755g_0039_1.jpg", "batch11-2024_06_14_86d4d50b09351c6af755g_0039_2.jpg", "batch11-2024_06_14_86d4d50b09351c6af755g_0039_3.jpg" ], "is_multi_img": true, "answer": "21", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Based on the given problem, the following diagram is drawn. The shaded area consists of four triangles congruent to $\\triangle ABD$ and a square with side length $BD$.\n\n\n\nFrom the problem, we have: $AB = CD = 2$, $BC = 5$, and $BD = BC - CD = 3$.\n\nThus, the area of $\\triangle ABD$ is:\n$$\nS_{ABD} = \\frac{1}{2} \\times AB \\times BD = \\frac{1}{2} \\times 3 \\times 2 = 3.\n$$\n\nThe area of the small square is:\n$$\nS_{\\text{small square}} = BD^2 = 3^2 = 9.\n$$\n\nTherefore, the total area $S$ is:\n$$\nS = 4 \\times S_{ABD} + S_{\\text{small square}} = 4 \\times 3 + 9 = 21.\n$$\n\nThe final answer is: **21**.\n\n**Key Insight**: This problem demonstrates the proof of the Pythagorean theorem. The key to solving it lies in understanding the relationship between the area formulas of the square and the triangle." }, { "problem_id": 1200, "question": "There is a square with an area of 1. After undergoing one \"growth,\" two smaller squares are generated on its left and right shoulders (as shown in Figure 1). The triangle formed by these three squares is a right-angled triangle. After another \"growth,\" four squares are generated (as shown in Figure 2). If this pattern of \"growth\" continues, it will become \"lush and flourishing.\" The sum of the areas of all the squares in the shape formed after 2022 \"growths\" is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_86d4d50b09351c6af755g_0074_1.jpg", "batch11-2024_06_14_86d4d50b09351c6af755g_0074_2.jpg" ], "is_multi_img": true, "answer": "2023", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Let the three sides of the first right triangle be $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$.\n\nAccording to the Pythagorean theorem, we have $a^{2}+b^{2}=c^{2}=1$.\n\nFrom Figure 1, it can be seen that after \"growing\" once, the sum of the areas of the squares with the two legs of the right triangle as their sides equals the area of the square with the hypotenuse as its side, i.e., the total area of all squares is $2 \\times 1=2$.\n\nFrom Figure 2, it can be seen that after \"growing\" twice, the total area of all squares is $3 \\times 1=3$.\n\nAfter \"growing\" 2022 times, the total area of all squares in the resulting figure is $2023 \\times 1=2023$.\n\nTherefore, the answer is: 2023.\n\n【Insight】This question examines the properties of squares and the Pythagorean theorem. The key to solving this problem lies in recognizing the relationship between the area of the new squares obtained each time and the area of the original square based on the Pythagorean theorem." }, { "problem_id": 1201, "question": "Figure 1 shows a baby stroller, and Figure 2 is a simplified side view after adjustment. It is measured that $\\angle A C B=90^{\\circ}$, the support $A C=6 \\mathrm{dm}, B C=8 \\mathrm{dm}$, then the distance between the centers of the two wheels $A, B$ is $\\qquad$ $\\mathrm{dm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_86d4d50b09351c6af755g_0076_1.jpg", "batch11-2024_06_14_86d4d50b09351c6af755g_0076_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since in the right triangle \\( \\mathrm{Rt} ABC \\), \\( AC = 6 \\, \\mathrm{dm} \\) and \\( BC = 8 \\, \\mathrm{dm} \\),\n\nTherefore, \\( AB = \\sqrt{AC^{2} + BC^{2}} = \\sqrt{6^{2} + 8^{2}} = 10 \\, \\mathrm{dm} \\),\n\nThus, the distance between the centers of the two wheels \\( A \\) and \\( B \\) is \\( 10 \\, \\mathrm{dm} \\).\n\nHence, the answer is: 10.\n\n[Key Insight] This problem primarily tests the application of the Pythagorean theorem. Mastering the Pythagorean theorem is crucial for solving such problems." }, { "problem_id": 1202, "question": "The figure shows the \"growth\" process of the \"Pythagoras tree.\" As shown in Figure (1), a square with side length $a$ undergoes its first \"growth\" and produces two smaller squares above it, with the three squares forming a right-angled triangle; after another \"growth,\" it becomes Figure (2); continuing this \"growth\" process, the sum of the areas of all the squares on the \"Pythagoras tree\" after the 2000th \"growth\" is what \n\nfigure (1)\n\n\n\nfigure (2)", "input_image": [ "batch11-2024_06_14_86d4d50b09351c6af755g_0097_1.jpg", "batch11-2024_06_14_86d4d50b09351c6af755g_0097_2.jpg" ], "is_multi_img": true, "answer": "$2001 a^{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the initial area before growth be $S_{0}$, and the area after the $n$th \"growth\" be $S_{n}$.\n\n$\\therefore S_{0}=a^{2}$,\n\n$S_{1}=a^{2}+a^{2}=2 a^{2}$,\n\n$S_{2}=2 a^{2}+a^{2}=3 a^{2}$,\n\n.....,\n\n$S_{n}=(n+1) a^{2}$,\n\nWhen $n=2000$, $S_{2000}=(2000+1) a^{2}=2001 a^{2}$;\n\nTherefore, the answer is: $2001 a^{2}$.\n\n【Insight】This problem examines the pattern of graphical changes and the Pythagorean theorem. The key to solving this problem lies in correctly understanding the pattern of changes in the given figure and accurately representing this pattern using algebraic expressions." }, { "problem_id": 1203, "question": "As shown in Figure 1, there is a wooden garden gate at a flower seedling nursery. The top of the gate is a semicircle, and the lower part is a rectangle. Now, a truck with partially loaded cargo needs to pass through the garden gate in the shape shown in Figure 1. The truck is 2.5 meters high and 1.6 meters wide. (The dimensions of each part of the garden gate are shown in Figure 2). The question is whether this truck can pass through the wooden garden gate of the nursery.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_8ad1da2482a9008c844ag_0004_1.jpg", "batch11-2024_06_14_8ad1da2482a9008c844ag_0004_2.jpg" ], "is_multi_img": true, "answer": "This car can pass through the wooden gate of the nursery", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $OF$,\n\n\n\nFrom the problem statement, quadrilateral $CDEF$ is a rectangle, and the diameter of the semicircle $AB = 2$ meters,\n\nThus, the radius $OC = \\frac{1}{2} AB = 1$ meter,\n\nSince quadrilateral $CDEF$ is a rectangle and the width of the car is 1.6 meters,\n\n$\\therefore \\angle FEO = \\angle CDO = 90^{\\circ}$, $EF = CD$, $CF = DE = 1.6$ meters,\n\nIn right triangles $FOE$ and $COD$,\n\n$\\left\\{\\begin{array}{l}OF = OC \\\\ EF = CD\\end{array}\\right.$,\n\n$\\therefore \\triangle FOE \\cong \\triangle COD$ (HL),\n\n$\\therefore OE = OD$,\n\n$\\therefore OD = \\frac{1}{2} DE = 0.8$ meters,\n\nIn right triangle $COD$, $CD = \\sqrt{OC^{2} - OD^{2}} = \\sqrt{1^{2} - 0.8^{2}} = 0.6$ meters,\n\n$\\therefore CH = CD + DH = 0.6 + 2.3 = 2.9$ meters,\n\nSince $2.9 > 2.5$,\n$\\therefore$ This car can pass through the wooden gate of the nursery.\n\n【Key Insight】This problem examines the application of the properties and determination of congruent triangles as well as the Pythagorean theorem. Understanding the problem and applying the properties of congruent triangles and the Pythagorean theorem are crucial to solving this problem." }, { "problem_id": 1204, "question": "The reading material on page 78 of the textbook, \"Extending from the Pythagorean Theorem to the Relationship Between Figure Areas,\" presents the following problem: As shown in Figure (1), if equilateral triangles are constructed outwardly on each of the three sides of a right triangle, the quantitative relationship among $S_{1}, S_{2}, S_{3}$ in the figure is $S_{1}+S_{2}=S_{3}$. Now, if $\\triangle A B F$ is folded upward, as shown in Figure (2), and given that $S_{\\text{ A}}=9, S_{2}=8, S_{\\text{ C}}=7$, then the area of $\\triangle A B C$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch11-2024_06_14_96280b3c13dbc63b1e87g_0050_1.jpg", "batch11-2024_06_14_96280b3c13dbc63b1e87g_0050_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the area of triangle \\( \\triangle ABC \\) be \\( S \\), and the areas of the two white figures in Figure (2) be \\( a \\) and \\( b \\), respectively.\n\nGiven that \\( S_{1} + S_{2} = S_{3} \\),\n\nwe have \\( S_{\\text{P}} + a + S_{\\text{Z}} + b = S_{\\text{两}} + a + b + S \\),\n\nwhich simplifies to \\( S_{\\text{ A}} + S_{2} = S_{\\text{ C}} + S \\).\n\nTherefore, \\( S = S_{\\text{q}} + S_{\\text{Z}} - S_{\\text{的}} = 9 + 8 - 7 = 10 \\).\n\nHence, the answer is: 10.\n\n[Key Insight] This problem tests the extended knowledge of the Pythagorean theorem. Understanding the problem and identifying the relationships between the areas in the figure are crucial for solving it." }, { "problem_id": 1205, "question": "As shown in Figure (1), in the isosceles right triangle paper $ABC$, $\\angle B = 90^\\circ, AB = 4$, points $D$ and $E$ are moving points on $AB$ and $BC$ respectively. The paper is folded along $DE$, and the corresponding point of $B$, $B'$, falls exactly on side $AC$, as shown in Figure (2). Then, the paper is folded again along $B'E$, with the corresponding point of $C$ being $C'$, as shown in Figure (3). When the overlapping part of $\\triangle DB'E$ and $\\triangle B'C'E$ is a right triangle, the length of $CE$ is _. $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch11-2024_06_14_96280b3c13dbc63b1e87g_0083_1.jpg", "batch11-2024_06_14_96280b3c13dbc63b1e87g_0083_2.jpg", "batch11-2024_06_14_96280b3c13dbc63b1e87g_0083_3.jpg" ], "is_multi_img": true, "answer": "2 or $8-4 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:** By the properties of folding, to ensure that the overlapping parts of $D B^{\\prime} E$ and $\\triangle B^{\\prime} C^{\\prime} E$ form a right-angled triangle, we consider two scenarios:\n\n**(1) When $D E \\perp B^{\\prime} C^{\\prime}$:**\n\n\n\nFrom the folding, we know:\n- $\\angle E B^{\\prime} D = \\angle D B E = 90^{\\circ}$,\n- $\\angle B^{\\prime} E D = \\angle B E D$,\n- $\\angle C B^{\\prime} E = \\angle C^{\\prime} B^{\\prime} E$.\n\nSince $\\angle B E B^{\\prime} = \\angle C B^{\\prime} E + \\angle C$,\nwe have:\n$$\n2 \\angle D E B^{\\prime} = \\angle C B^{\\prime} E + 45^{\\circ}.\n$$\n\nGiven that $\\angle D E B^{\\prime} = 90^{\\circ} - \\angle E B^{\\prime} C^{\\prime}$,\nit follows that:\n$$\n2\\left(90^{\\circ} - \\angle E B^{\\prime} C^{\\prime}\\right) = \\angle C B^{\\prime} E + 45^{\\circ}.\n$$\n\nSince $\\angle C B^{\\prime} E = \\angle C^{\\prime} B^{\\prime} E$,\nwe deduce:\n$$\n\\angle C B^{\\prime} E = \\angle C^{\\prime} B^{\\prime} E = 45^{\\circ}.\n$$\n\nGiven that $\\angle C = 45^{\\circ}$,\nwe find:\n$$\n\\angle C E B^{\\prime} = 90^{\\circ},\n$$\nthus $B^{\\prime} E \\perp B C$.\n\nFrom the folding, we know $C B^{\\prime} = C^{\\prime} B^{\\prime}$,\nso:\n$$\nC E = \\frac{1}{2} B C = \\frac{1}{2} A B = 2.\n$$\n\n**(2) When $\\angle E B^{\\prime} C^{\\prime} = 90^{\\circ}$:**\n\n\n\nFrom the folding, we know:\n- $B^{\\prime} E = B E$,\n- $\\angle E B^{\\prime} A = \\angle B = 90^{\\circ}$.\n\nThus, point $E$ lies on the angle bisector of $\\angle B A C$.\n\nLet $B^{\\prime} E = B E = x$, then $C E = B C - B E = 4 - x$.\n\nIn the right triangle $B^{\\prime} E C$, $\\angle C = \\angle B^{\\prime} E C = 45^{\\circ}$,\nso:\n$$\nB^{\\prime} E = B^{\\prime} C = x.\n$$\n\nTherefore:\n$$\nC E = \\sqrt{2} B^{\\prime} E,\n$$\nwhich leads to:\n$$\n4 - x = \\sqrt{2} x.\n$$\n\nSolving for $x$:\n$$\nx = 4 \\sqrt{2} - 4,\n$$\nthus:\n$$\nC E = 4 - x = 8 - 4 \\sqrt{2}.\n$$\n\n**In summary:** The length of $C E$ is either 2 or $8 - 4 \\sqrt{2}$.\n\n**Answer:** 2 or $8 - 4 \\sqrt{2}$.\n\n**Key Insight:** This problem is a comprehensive triangle question involving folding transformations, the properties of isosceles right triangles, isosceles triangles, and the Pythagorean theorem. The key to solving it lies in considering different cases and accurately drawing the corresponding diagrams." }, { "problem_id": 1206, "question": "A school in Nanxun District incorporated the sport of hopping on a single foot with a spinning ball into their unique \"Sunshine Big Break\" activities. As shown in Figure 1, during the process of hopping on one foot, the ball begins to spin continuously around the foot with the help of the ball rod. During the big break, five students stand at points $A, B, C, D, E$. When the student at point $A$ hops, the ball starts spinning on the ground, forming point $A$. Figure 2 is a top-view diagram of the activity process. Given that $ED \\perp DB$, $AB \\perp BD$ intersects at point $G$, $GB = 80 \\, \\text{cm}$, $ED = 100 \\, \\text{cm}$, $\\frac{BC}{DB} = \\frac{1}{3}$, connecting $AD$ with $\\angle DAB = 45^\\circ$, and when the ball rotates to point $F$ such that $EF \\parallel DB$ and $FC \\perp DB$, the length of the ball rod $AG$ is $\\boxed{\\text{cm}}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_98bdb944cef4f7d8aa2dg_0022_1.jpg", "batch11-2024_06_14_98bdb944cef4f7d8aa2dg_0022_2.jpg" ], "is_multi_img": true, "answer": "100", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since \\( ED \\perp DB \\), \\( AB \\perp BD \\), and \\( FC \\perp DB \\),\n\nTherefore, \\( ED \\parallel FC \\parallel AB \\).\n\nLet \\( AG = x \\). Draw \\( FH \\perp AB \\) through point \\( F \\).\n\nThus, \\( ED = BH = 100 \\), \\( FH = BC \\), and \\( AH = AG - HG = x - (100 - 80) = x - 20 \\).\n\nSince \\( \\angle DAB = 45^\\circ \\) and \\( \\angle B = 90^\\circ \\),\n\nTherefore, \\( AB = DE = AG + BG = x + 80 \\).\n\nGiven that \\( \\frac{BC}{DB} = \\frac{1}{3} \\),\n\nTherefore, \\( BC = \\frac{BD}{3} = \\frac{x + 80}{3} \\).\n\nIn right triangle \\( \\triangle AFH \\),\n\n\\( AF = AG = x \\), \\( AH = x - 20 \\), and \\( FH = BC = \\frac{x + 80}{3} \\).\n\nBy the Pythagorean theorem,\n\n\\( FH^2 + AH^2 = AF^2 \\),\n\nThat is, \\( \\left(\\frac{x + 80}{3}\\right)^2 + (x - 20)^2 = x^2 \\). Solving this equation gives \\( x = 100 \\), so \\( AG = 100 \\).\n\nHence, the answer is: 100.\n\n[Key Insight] This problem examines the properties of parallel lines and the Pythagorean theorem. The key to solving it lies in correctly constructing auxiliary lines." }, { "problem_id": 1207, "question": "Figure 1 shows a multi-position adjustable reclining chair that Xiao Xin purchased during the \"Tmall Double 12\" event. The diagram for adjusting the positions is shown in Figure 2. It is known that the two legs $AB = AC = 0.8$ meters, $BC = 0.96$ meters, and $O$ is a fixed connection point on $AC$. The backrest $OD = 0.8$ meters. When the chair is in position I, $OD // AB$, and when it is in position II, $OD \\perp AC$. When the chair is adjusted from position I to position II, the horizontal distance that the top of the backrest $D$ leans back (i.e., $EF$) is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_98bdb944cef4f7d8aa2dg_0080_1.jpg", "batch11-2024_06_14_98bdb944cef4f7d8aa2dg_0080_2.jpg" ], "is_multi_img": true, "answer": "0.16", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw $AG \\perp BC$ at point $G$, and $OH \\perp BC$ at point $H$. Then, draw $OM \\perp D'F$ at point $M$, intersecting $DE$ at point $N$, as shown in the figure.\n\n\n\nThus, $OM = HF$ and $ON = HE$.\n\nGiven that $AB = AC = 0.8$ and $BC = 0.96$,\n\nTherefore, $BG = CG = \\frac{1}{2} BC = 0.48$,\n\nHence, $AG = \\sqrt{AB^2 - BG^2} = 0.64$.\n\nSince $AB \\parallel CD$ and $BC \\parallel OM$,\n\nTherefore, $\\angle ABG = \\angle DON$.\n\nIn triangles $\\triangle ABG$ and $\\triangle DON$,\n\n$$\n\\left\\{\n\\begin{array}{c}\n\\angle ABG = \\angle DON \\\\\n\\angle AGB = \\angle DNO = 90^\\circ \\\\\nAB = OD = 0.8\n\\end{array}\n\\right.\n$$\n\nThus, $\\triangle ABG \\cong \\triangle DON$ (by AAS),\n\nTherefore, $BG = ON = HE = 0.48$.\n\nSince $OD' \\perp AC$,\n\nTherefore, $\\angle D'OM + \\angle MOC = 90^\\circ$.\n\nSince $OM \\parallel BC$,\n\nTherefore, $\\angle MOC = \\angle ACG$.\n\nSince $\\angle ACG + \\angle CAG = 90^\\circ$,\n\nTherefore, $\\angle CAG = \\angle D'OM$.\n\nIn triangles $\\triangle ACG$ and $\\triangle OD'M$,\n\n$$\n\\left\\{\n\\begin{array}{c}\n\\angle CAG = \\angle D'OM \\\\\n\\angle AGC = \\angle OMD' = 90^\\circ \\\\\nAC = OD'\n\\end{array}\n\\right.\n$$\n\nThus, $\\triangle ACG \\cong \\triangle OD'M$ (by AAS),\n\nTherefore, $AG = OM = HF = 0.64$.\n\nHence, $EF = HF - HE = 0.64 - 0.48 = 0.16$ meters.\n\nThe answer is: 0.16.\n\n【Key Insight】This problem primarily examines the properties and determination of congruent triangles, the Pythagorean theorem, and the properties of isosceles triangles. The key is to construct congruent triangles." }, { "problem_id": 1208, "question": "As shown in Figure 1, a rectangular triangle paper has one of its right-angled sides measuring 6. Four such rectangular triangle papers are cut and placed into a square with a side length of 10 as shown in Figure 2 (with no overlapping or gaps at the joints). The area of the shaded part in Figure 2 is $\\qquad$ \n\nfigure 1\n\n\n\nfigure 2", "input_image": [ "batch11-2024_06_14_98bdb944cef4f7d8aa2dg_0100_1.jpg", "batch11-2024_06_14_98bdb944cef4f7d8aa2dg_0100_2.jpg" ], "is_multi_img": true, "answer": "96", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "According to the problem, the hypotenuse of the right triangle is 10, and one of the legs is 6.\n\n$\\therefore$ The length of the other leg $=\\sqrt{10^{2}-6^{2}}=8$,\n\n$\\therefore$ The area of the shaded part $=4 \\times 6 \\times 8 \\times \\frac{1}{2}=96$.\n\nTherefore, the answer is: 96\n\n【Highlight】The problem examines the properties of squares, the Pythagorean theorem, and the area of triangles. The key to solving the problem lies in understanding the given information and applying the concept of combining numbers and shapes to find the solution." }, { "problem_id": 1209, "question": "As shown in Figure (1), there is a right triangle paper with $\\angle C = 90^\\circ$, $AB = 13 \\, \\text{cm}$, and $BC = 5 \\, \\text{cm}$. The paper is folded so that point $C$ lands on point $C'$ on the hypotenuse, with the fold line being $BD$ (as shown in Figure (2)). Find the length of $DC$.\n\n\n\n\n\n(2)", "input_image": [ "batch11-2024_06_14_a03cb0b0518e567ddf6eg_0072_1.jpg", "batch11-2024_06_14_a03cb0b0518e567ddf6eg_0072_2.jpg" ], "is_multi_img": true, "answer": "Length of $D C$ is $\\frac{10}{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since \\(\\angle C = 90^\\circ\\), \\(AB = 13 \\text{ cm}\\), and \\(BC = 5 \\text{ cm}\\),\n\n\\[\n\\therefore AC = \\sqrt{AB^2 - BC^2} = 12 \\text{ cm}.\n\\]\n\nSince the folding point \\(C\\) lands on the hypotenuse at point \\(C'\\),\n\n\\[\n\\therefore BC' = BC = 5 \\text{ cm}, \\quad DC' = DC.\n\\]\n\n\\[\n\\therefore AC' = AB - BC' = 13 - 5 = 8 \\text{ cm}.\n\\]\n\nLet \\(DC = x\\), then \\(AD = AC - DC = 12 - x\\),\n\n\\[\nDC' = x.\n\\]\n\nIn the right triangle \\(\\triangle AC'D\\),\n\nBy the Pythagorean theorem, \\(AC'^2 + DC'^2 = AD^2\\),\n\n\\[\n8^2 + x^2 = (12 - x)^2,\n\\]\n\n\\[\n\\therefore x = \\frac{10}{3}.\n\\]\n\n\\[\n\\therefore \\text{The length of } DC \\text{ is } \\frac{10}{3} \\text{ cm}.\n\\]\n\n**Highlight:** This problem examines the concepts of folding transformations and the Pythagorean theorem. The key to solving the problem lies in using the Pythagorean theorem to set up the equation, noting that corresponding segments and angles remain equal before and after the fold." }, { "problem_id": 1210, "question": "As shown in Figure 1, this is a Rubik's Cube composed of 8 cubes of the same size, with a volume of 64. Place the square $A B C D$ on the number line, as shown in Figure 2, such that point $\\mathrm{A}$ coincides with -2, then the number represented by point $D$ on the number line is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_a2523860d55f90e8c578g_0011_1.jpg", "batch11-2024_06_14_a2523860d55f90e8c578g_0011_2.jpg" ], "is_multi_img": true, "answer": "$-2-2 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the edge length of each small cube be \\( a \\). According to the problem, we have:\n\\[ 8a^{3} = 64 \\]\n\n\\[ \\therefore a = 2 \\]\n\nLet the side length of square \\( ABCD \\) be \\( AD = x \\). From the figure, the area of square \\( ABCD \\) is:\n\\[ x^{2} = \\frac{4 \\times 4}{2} = 8 \\]\n\n\\[ \\therefore x = \\sqrt{8} = 2\\sqrt{2} \\]\n\nSince point \\( A \\) coincides with \\(-2\\), the number represented by point \\( D \\) on the number line is:\n\\[ -2 - 2\\sqrt{2} \\]\n\nTherefore, the answer is:\n\\[ -2 - 2\\sqrt{2} \\]\n\n[Key Insight] This problem mainly tests the application of cube roots and arithmetic square roots. Mastering the methods for finding the cube root and arithmetic square root of a number is crucial for solving such problems." }, { "problem_id": 1211, "question": "As shown in the figure, a square paper with an area of 10 is cut into five pieces (where (5) is a small square), and then exactly formed into a rectangle. The perimeter of this formed rectangle is $\\qquad$.\n\n\n\n## Cut $\\bullet$ Form\n\n", "input_image": [ "batch11-2024_06_14_a2523860d55f90e8c578g_0018_1.jpg", "batch11-2024_06_14_a2523860d55f90e8c578g_0018_2.jpg" ], "is_multi_img": true, "answer": "$12 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Solution:** Let the side length of the small square be \\( a \\).\n\nSince (5) is a square, it follows that (1), (2), (3), and (4) are right-angled triangles.\n\nFrom the assembled rectangle, we know that the side length of the small square is equal to the shorter leg of the right-angled triangle, which is \\( a \\), and the longer leg is \\( 2a \\).\n\nGiven that the area of the large square is 10, its side length is \\( \\sqrt{10} \\).\n\nIn the right-angled triangle, by the Pythagorean theorem:\n\\[\na^{2} + (2a)^{2} = (\\sqrt{10})^{2}\n\\]\n\\[\na^{2} + 4a^{2} = 10\n\\]\n\\[\n5a^{2} = 10\n\\]\n\\[\na^{2} = 2\n\\]\n\\[\na = \\sqrt{2}\n\\]\n\nTherefore, the perimeter of the assembled rectangle is:\n\\[\n(\\sqrt{2} + 2\\sqrt{2} + 2\\sqrt{2} + \\sqrt{2}) \\times 2 = 12\\sqrt{2}\n\\]\nHence, the answer is: \\( 12\\sqrt{2} \\).\n\n**Key Insight:** This problem tests the understanding of geometric transformations, the Pythagorean theorem, irrational numbers, and operations with square roots. The crucial step is deducing the lengths of the legs of the right-angled triangle from the given figure." }, { "problem_id": 1212, "question": "As shown in the figure, this is a schematic diagram of the famous \"Zhao Shuang's Hypotenuse Diagram\" from ancient China, which is formed by four congruent right-angled triangles. If $AC=12, BC=7$, and the right-angled sides with length 12 in the four right-angled triangles are extended outward by one time their length, a \"mathematical windmill\" as shown in the figure is obtained. Then the outer perimeter of this windmill is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_a666f697294f81f6ed02g_0016_1.jpg", "batch11-2024_06_14_a666f697294f81f6ed02g_0016_2.jpg" ], "is_multi_img": true, "answer": "148", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Let's extend $CA$ to point $D$ and connect $BD$,\n\n\nAccording to the problem, we have $CD = 12 \\times 2 = 24$, and $BC = 7$,\n\nSince $\\angle BCD = 90^{\\circ}$,\nTherefore, $BC^{2} + CD^{2} = BD^{2}$, which is $7^{2} + 24^{2} = BD^{2}$,\n\nThus, $BD = 25$,\n\nSo, $AD + BD = 12 + 25 = 37$,\n\nTherefore, the perimeter of the windmill is $37 \\times 4 = 148$.\n\nHence, the answer is:\n\n148.\n\n【Highlight】This problem tests the application of the Pythagorean theorem in real-world scenarios, and it's important to pay attention to the implicit given conditions to solve such problems." }, { "problem_id": 1213, "question": "As shown in Figure (1), in $\\triangle ABC$, $\\angle ACB = 90^\\circ$, and the ratio of $AC:BC = 4:3$. There is a square on each of the three sides of this right triangle. Perform the following operation: from the two smaller squares, construct right triangles with the ratio of the lengths of their legs being $4:3$, and then construct squares with the lengths of the legs of the resulting right triangles as their sides. Figure (2) shows the graph after 1 operation, and Figure (3) shows the graph after 2 operations. If the perimeter of the right triangle in Figure (1) is 12, then the sum of the areas of all the squares in the graph after $n$ operations is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch11-2024_06_14_a666f697294f81f6ed02g_0072_1.jpg", "batch11-2024_06_14_a666f697294f81f6ed02g_0072_2.jpg", "batch11-2024_06_14_a666f697294f81f6ed02g_0072_3.jpg" ], "is_multi_img": true, "answer": "$25 n+50$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\angle ACB = 90^\\circ$ and $AC:BC = 4:3$,\n\nLet $AC = 4x$, then $BC = 3x$.\n\nAccording to the Pythagorean theorem, $AB = \\sqrt{AC^2 + BC^2} = \\sqrt{(4x)^2 + (3x)^2} = 5x$.\n\nGiven that $3x + 4x + 5x = 12$,\n\nWe find $x = 1$.\n\nThus, $AB = 5$, $BC = 3$, and $AC = 4$.\n\nThe sum of the areas of the squares in Figure (1) is: $3^2 + 4^2 + 5^2 = 25 + 25 = 2 \\times 25 = 50$.\n\nThe sum of the areas of all squares in Figure (2), i.e., after one operation, is:\n\n$3^2 + 4^2 + 3^2 + 4^2 + 5^2 = 25 + 25 + 25 = 25 + 50$.\n\nThe sum of the areas of all squares in Figure (3), i.e., after two operations, is:\n\n$3^2 + 4^2 + 3^2 + 4^2 + 3^2 + 4^2 + 5^2 = 25 + 25 + 25 + 25 = 2 \\times 25 + 50$.\n\nTherefore, the sum of the areas of all squares after $n$ operations is $25n + 50$.\n\nHence, the answer is: $25n + 50$.\n\n[Insight] This problem primarily examines the pattern recognition in geometric figures, the properties of right-angled triangles, the Pythagorean theorem, and the properties of squares. The key to solving the problem lies in understanding the question and flexibly applying the learned knowledge to find the solution." }, { "problem_id": 1214, "question": "The \"Pythagorean Tree\" is a figure formed by starting with a square, using one of its sides as the hypotenuse to create an external right triangle, and then using the two legs of this right triangle to create two more external squares. This process is repeated, and the resulting shape after several repetitions resembles a tree, hence the name. Suppose the figures shown are the first, second, and third generations of the Pythagorean Tree, respectively. According to the drawing principle of the Pythagorean Tree, the number of squares in the sixth generation of the Pythagorean Tree is $\\qquad$.\n\n\n\nFirst Generation Pythagorean Tree\n\n\n\nSecond Generation Pythagorean Tree\n\n\n\nThird Generation Pythagorean Tree", "input_image": [ "batch11-2024_06_14_a666f697294f81f6ed02g_0088_1.jpg", "batch11-2024_06_14_a666f697294f81f6ed02g_0088_2.jpg", "batch11-2024_06_14_a666f697294f81f6ed02g_0088_3.jpg" ], "is_multi_img": true, "answer": "127", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: \n\n$\\because$ In the first generation of the Pythagorean tree, the number of squares is $1 + 2 = 3$,\n\nIn the second generation of the Pythagorean tree, the number of squares is $1 + 2 + 2^{2} = 7$,\n\nIn the third generation of the Pythagorean tree, the number of squares is $1 + 2 + 2^{2} + 2^{3} = 15$,\n\n$\\therefore$ In the sixth generation of the Pythagorean tree, the number of squares is $1 + 2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6} = 127$. Therefore, the answer is: 127.\n\n【Key Insight】This problem examines the pattern within a geometric figure. The key to solving it lies in carefully observing the figure and identifying the pattern of its transformation." }, { "problem_id": 1215, "question": "As shown in the figure, a ladder $\\mathrm{AB}$ is 25 meters long, with its top end $\\mathrm{A}$ leaning against the wall $\\mathrm{AC}$. At this point, the distance between the bottom end $\\mathrm{B}$ of the ladder and the corner $\\mathrm{C}$ of the wall is 15 meters. After the ladder slides and comes to rest at position $\\mathrm{DE}$, the length $\\mathrm{BD}$ is measured to be 5 meters. How many meters did the top end $\\mathrm{A}$ of the ladder fall?\n\n\n\nFigure 1\n\n\n\nFigure 2 B D", "input_image": [ "batch11-2024_06_14_b2018a578bfd0aa2d736g_0038_1.jpg", "batch11-2024_06_14_b2018a578bfd0aa2d736g_0038_2.jpg" ], "is_multi_img": true, "answer": "5 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** In the right triangle $\\mathrm{RT} \\triangle \\mathrm{ABC}$, according to the Pythagorean theorem, we find that $\\mathrm{AC}=20$ meters. Since the length of the ladder remains unchanged, in the right triangle $\\mathrm{RT} \\triangle \\mathrm{CDE}$, we use the Pythagorean theorem again to determine $\\mathrm{CE}$, thereby arriving at the answer.\n\n**Question Solution:** In the right triangle $\\mathrm{RT} \\triangle \\mathrm{ABC}$, $\\mathrm{AB}=25$ meters and $\\mathrm{BC}=15$ meters. Therefore, $\\mathrm{AC}=\\sqrt{AB^{2}-BC^{2}}=\\sqrt{25^{2}-15^{2}}=20$ meters.\n\nIn the right triangle $\\mathrm{RT} \\triangle \\mathrm{ECD}$, $\\mathrm{AB}=\\mathrm{DE}=25$ meters and $\\mathrm{CD}=(15+5)=20$ meters. Thus, $\\mathrm{EC}=\\sqrt{DE^{2}-CD^{2}}=15$ meters. Consequently, $\\mathrm{AE}=\\mathrm{AC}-\\mathrm{CE}=20-15=5$ meters.\n\n**Answer:** The top of the ladder $\\mathrm{A}$ has descended by 5 meters.\n\n**Key Point:** Application of the Pythagorean theorem." }, { "problem_id": 1216, "question": "A certain Yangtze River Bridge adopts a low-tower cable-stayed bridge design, as shown in the plan view. Suppose you are standing on the bridge and measure that the angle between the cable $A B$ and the horizontal bridge deck is $30^{\\circ}$, and the angle between the cable $B D$ and the horizontal bridge deck is $60^{\\circ}$. The distance between the bottom ends of the two cables is $A D = 20$ meters. Then the height of the pillar $B C$ is $\\qquad$\n\n\n\nOption A\n\n\n\nOption B", "input_image": [ "batch11-2024_06_14_b23b3bfa19e31c3190e4g_0043_1.jpg", "batch11-2024_06_14_b23b3bfa19e31c3190e4g_0043_2.jpg" ], "is_multi_img": true, "answer": "$10 \\sqrt{3}$ m\\#\\# $10 \\sqrt{3} \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, we have:\n$\\angle A=30^{\\circ}, \\angle B D C=60^{\\circ}$,\n\n$\\therefore \\angle A B D=30^{\\circ}$,\n\n$\\therefore B D=A D=20$ meters,\n\n$\\because \\angle C=90^{\\circ}$,\n\n$\\therefore \\angle C B D=30^{\\circ}$,\n\n$\\therefore C D=\\frac{1}{2} B D=10$ meters,\n\n$\\therefore B C=\\sqrt{B D^{2}-C D^{2}}=10 \\sqrt{3}$ meters,\n\nThus, the height of the column $B C$ is $10 \\sqrt{3}$ meters.\n\nTherefore, the answer is: $10 \\sqrt{3}$ meters\n\n[Key Insight] This problem primarily tests the properties of right triangles, the Pythagorean theorem, and the determination of isosceles triangles. Mastering the properties of right triangles, the Pythagorean theorem, and the determination of isosceles triangles is crucial for solving the problem." }, { "problem_id": 1217, "question": "The sliding stay rod is widely used in hung windows. As shown in the figure, a certain type of sliding stay rod consists of a slideway \\( OC \\), and the stay rods \\( AB \\) and \\( BC \\). The slideway \\( OC \\) is fixed on the windowsill. During the process of closing or opening the hung window, the lengths of the stay rods \\( AB \\) and \\( BC \\) remain constant. When the hung window is closed, as shown in Figure (1), the point \\( A \\) coincides with the point \\( O \\), and the stay rods \\( AB \\) and \\( BC \\) exactly coincide with the slideway \\( OC \\). When the hung window is fully opened, as shown in Figure (2), the stay rod \\( AB \\) and the stay rod \\( BC \\) form a right angle, that is, \\( \\angle B = 90^\\circ \\). The measurements show that \\( OA = 12 \\, \\text{cm} \\) and the stay rod \\( AB = 15 \\, \\text{cm} \\). Find the length of the slideway \\( OC \\).\n\n\n\nFigure (1)\n\n\n\n\n\nFigure (2)", "input_image": [ "batch11-2024_06_14_b439ad53922aa0f13a79g_0074_1.jpg", "batch11-2024_06_14_b439ad53922aa0f13a79g_0074_2.jpg", "batch11-2024_06_14_b439ad53922aa0f13a79g_0074_3.jpg" ], "is_multi_img": true, "answer": "Length of $O C$ is $51 \\mathrm{~cm}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let $OC = m \\text{ cm}$. From Figure (1), we can see that $BC = OC - AB = (m - 15) \\text{ cm}$, and from Figure (2), $AC = OC - OA = (m - 12) \\text{ cm}$.\n\nSince angle $B$ is $90^\\circ$,\n\nIn the right triangle $\\triangle ABC$, according to the Pythagorean theorem, we have:\n$AB^2 + BC^2 = AC^2$,\n\nThus, $15^2 + (m - 15)^2 = (m - 12)^2$,\n\nSolving this equation gives $m = 51$,\n\nTherefore, the length of the slide $OC$ is $51 \\text{ cm}$.\n\n【Key Insight】This problem examines the application of the Pythagorean theorem. The key to solving it lies in correctly expressing $BC$ and $AC$ by considering that the lengths of the support rods $AB$ and $BC$ remain constant." }, { "problem_id": 1218, "question": "Every year on October 1st is the annual National Day, the birthday of our great motherland. The streets and alleys are adorned with colorful flags, creating a beautiful and eye-catching scene that inspires people to strive forward. The dimensions of a rectangular flag when fully spread out are shown in Figure 1 (units: $\\mathrm{cm}$), where the rectangle $A B C D$ is a double-layered white cloth sewn to fit around the flagpole, and the rectangle $D C E F$ is the silk flag surface. The flagpole, with the flag attached, is vertically inserted into the ground. The height from the top of the flag $M$ to the ground is $220 \\mathrm{~cm}$. In calm weather, the length of the flag when it naturally hangs down, $M N$, is equal to $D E$ (as shown in Figure 2). Find the height $h$ from the lowest point of the flag when it naturally hangs down to the ground.\n\n\n\nFigure 1\n\n", "input_image": [ "batch11-2024_06_14_b62e1a684ee278f63889g_0038_1.jpg", "batch11-2024_06_14_b62e1a684ee278f63889g_0038_2.jpg" ], "is_multi_img": true, "answer": "$70 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\mathrm{RtV} D C E$, $C D = 90 \\mathrm{~cm}$ and $C E = 120 \\mathrm{~cm}$.\n\nTherefore, $D E = \\sqrt{C D^{2} + C E^{2}} = \\sqrt{120^{2} + 90^{2}} = 150 \\mathrm{~cm}$.\n\nThus, the height from the lowest point of the naturally hanging colored flag to the ground is $h = 220 - 150 = 70 \\mathrm{~cm}$.\n\n[Key Insight] This problem primarily examines the application of the Pythagorean theorem. The key to solving the problem lies in using the Pythagorean theorem to determine the maximum length of the colored flag when it hangs naturally in windless weather." }, { "problem_id": 1219, "question": "Given: As shown in the figure, there is a green space in the shape of Rt $\\triangle A B C$, with the measurements of the two perpendicular sides $A C=8 \\mathrm{~m}, B C=6 \\mathrm{~m}$. Now, this green space is to be expanded into an isosceles $\\triangle A B D$, with the expansion part ( $\\triangle A D C$ ) being a right triangle with a perpendicular side length of $8 \\mathrm{~m}$. Find the perimeter of the expanded isosceles $\\triangle A B D$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch11-2024_06_14_b62e1a684ee278f63889g_0048_1.jpg", "batch11-2024_06_14_b62e1a684ee278f63889g_0048_2.jpg", "batch11-2024_06_14_b62e1a684ee278f63889g_0048_3.jpg" ], "is_multi_img": true, "answer": "$32 \\mathrm{~m}$ or $(20+4 \\sqrt{5}) \\mathrm{m}$ or $\\frac{80}{3} \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle ABC$, the two legs are $AC = 8\\,\\text{m}$ and $BC = 6\\,\\text{m}$.\n\nTherefore, the hypotenuse $AB = \\sqrt{AC^2 + BC^2} = 10\\,\\text{m}$.\n\nAccording to the problem, we can consider the following cases:\n\n1. **When $AB = AD = 10\\,\\text{m}$**, as shown in the figure:\n\n \n\n Then, $DC = \\sqrt{AD^2 - AC^2} = \\sqrt{10^2 - 8^2} = 6\\,\\text{m}$.\n\n Therefore, $BD = DC + BC = 6 + 6 = 12\\,\\text{m}$.\n\n The perimeter of $\\triangle ABD$ is: $AD + AB + BD = 10 + 10 + 12 = 32\\,\\text{m}$.\n\n2. **When $BA = BD = 10\\,\\text{m}$**, as shown in the figure:\n\n \n\n Then, $DC = BD - BC = 10 - 6 = 4\\,\\text{m}$.\n\n Since $AC \\perp BD$ and $AC = 8\\,\\text{m}$,\n\n $AD = \\sqrt{AC^2 + DC^2} = \\sqrt{8^2 + 4^2} = 4\\sqrt{5}\\,\\text{m}$.\n\n The perimeter of $\\triangle ABD$ is: $BA + AD + BD = 10 + 4\\sqrt{5} + 10 = (20 + 4\\sqrt{5})\\,\\text{m}$.\n\n3. **When $DA = DB$**, as shown in the figure:\n\n \n\n Let $DC = x\\,\\text{m}$, then $AD = DB = (6 + x)\\,\\text{m}$.\n\n Since $AC \\perp BD$,\n\n $DC^2 + AC^2 = AD^2$, and $AC = 8\\,\\text{m}$,\n\n which gives $x^2 + 8^2 = (6 + x)^2$.\n\n Solving this equation yields $x = \\frac{7}{3}$.\n\n Therefore, the perimeter of $\\triangle ABD$ is: $AD + BD + AB = 2 \\times \\left(\\frac{7}{3} + 6\\right) + 10 = \\frac{80}{3}\\,\\text{m}$.\n\n**Key Insight**: This problem is a comprehensive triangle problem, primarily testing the application of the Pythagorean theorem. The key to solving the problem lies in skillfully applying the Pythagorean theorem according to the given conditions." }, { "problem_id": 1220, "question": "Figure 1 shows a baby stroller, and Figure 2 is a simplified side view after adjustment. It is measured that $\\angle A C B=90^{\\circ}$, with the support $A C=6 \\mathrm{dm}$ and $B C=8 \\mathrm{dm}$. Find the distance between the centers of the two wheels, $A$ and $B$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_b715f90643840480004cg_0057_1.jpg", "batch11-2024_06_14_b715f90643840480004cg_0057_2.jpg" ], "is_multi_img": true, "answer": "$10 \\mathrm{dm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle ABC$, since $AC = 6 \\, \\text{dm}$ and $BC = 8 \\, \\text{dm}$,\n\n\\[\nAB = \\sqrt{AC^{2} + BC^{2}} = \\sqrt{6^{2} + 8^{2}} = 10 \\, \\text{dm}.\n\\]\n\nTherefore, the distance between the centers of the two wheels $A$ and $B$ is $10 \\, \\text{dm}$.\n\n【Key Point】This problem examines the application of the Pythagorean theorem. Mastering the Pythagorean theorem is crucial for solving such problems." }, { "problem_id": 1221, "question": "Pour a sufficient amount of milk into a rectangular container placed horizontally as shown in Figure 1. Stop pouring when the milk in the container just touches point $\\mathrm{P}$. Figure 2 is a plan view of the setup, where $\\mathrm{AP}=6 \\mathrm{~cm}$. Based on the information in the figure, find the height of the milk in the container.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_b715f90643840480004cg_0087_1.jpg", "batch11-2024_06_14_b715f90643840480004cg_0087_2.jpg" ], "is_multi_img": true, "answer": "The height of the milk in the container is $7.2 \\mathrm{~cm}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Draw a perpendicular from point \\( P \\) to \\( AB \\) at point \\( N \\). According to the given information: \\( AP = 6 \\, \\text{cm} \\), \\( AB = 10 \\, \\text{cm} \\), then\n\\[\nBP = \\sqrt{AB^2 - AP^2} = 8 \\, \\text{cm}.\n\\]\nTherefore, \n\\[\nNP \\times AB = AP \\times BP,\n\\]\nwhich implies\n\\[\nNP = \\frac{AP \\cdot BP}{AB} = \\frac{6 \\times 8}{10} = 4.8 \\, \\text{cm}.\n\\]\nThus,\n\\[\n12 - 4.8 = 7.2 \\, \\text{cm}.\n\\]\n\n\n\nFigure 2\n\nAnswer: The height of the milk in the container is \\( 7.2 \\, \\text{cm} \\).\n\n【Key Insight】This problem tests knowledge of the Pythagorean theorem and methods for calculating the area of a triangle. Determining the length of \\( PN \\) is crucial for solving the problem." }, { "problem_id": 1222, "question": "Figure 1 shows a section of a fence, with the upper part consisting of an isosceles triangular railing made of stainless steel pipes as shown in Figure 2. Please calculate, based on the dimensions marked in Figure 2, the minimum length of stainless steel pipes required to weld the outer frame BCD of an isosceles triangular railing (ignoring the welding parts).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_b715f90643840480004cg_0089_1.jpg", "batch11-2024_06_14_b715f90643840480004cg_0089_2.jpg" ], "is_multi_img": true, "answer": "The isosceles triangle fence frame BCD requires at least 3.6 m of stainless steel pipe", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the problem statement, we have: $\\mathrm{BO} \\perp \\mathrm{CD}$,\n\nSince $\\triangle \\mathrm{BCD}$ is an isosceles triangle,\n\n$\\therefore \\mathrm{DO}=\\frac{1}{2} \\mathrm{CD}=0.8 \\mathrm{~m}^{*}$,\n\nIn the right triangle $\\triangle \\mathrm{BDO}$,\n\nSince $\\mathrm{BD}^{2}=\\mathrm{DO}^{2}+\\mathrm{BO}^{2}$,\n\n$\\therefore \\mathrm{BD}=\\sqrt{0.8^{2}+0.6^{2}}=1$ (meter),\n\n$\\therefore \\mathrm{BC}=1$ meter,\n\n$\\therefore$ The isosceles triangular fence frame BCD requires at least stainless steel pipes: $1+1+1.6=3.6$ (meters).\n\n【Highlight】This problem examines the application of the Pythagorean theorem, with the key to solving it being the use of the theorem to calculate the length of $\\mathrm{BD}$." }, { "problem_id": 1223, "question": "There is a square with an area of 1. After one \"growth,\" two smaller squares are generated on its left and right shoulders (as shown in Figure 1), where the triangle formed by the three squares is a right-angled triangle. After another \"growth,\" four squares are generated (as shown in Figure 2). If this pattern of \"growth\" continues, it will become \"lush and flourishing.\" The sum of the areas of all the squares in the figure formed after 2017 \"growths\" is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_b736ee8537381ac6ab05g_0055_1.jpg", "batch11-2024_06_14_b736ee8537381ac6ab05g_0055_2.jpg" ], "is_multi_img": true, "answer": "2018", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "**Analysis** According to the Pythagorean theorem and the formula for the area of a square, after one \"growth,\" the sum of the areas of the squares with sides equal to the two legs of the right triangle equals the area of the square with the hypotenuse as its side, meaning the total area of all squares is \\(2 \\times 1 = 2\\). After two \"growths,\" the total area of all squares is \\(3 \\times 1 = 3\\). Extending this logic, we can determine the total area of all squares after 2017 \"growths.\"\n\n**Detailed Explanation** Let the three sides of the right triangle be \\(a\\), \\(b\\), and \\(c\\).\n\nAccording to the Pythagorean theorem, \\(a^2 + b^2 = c^2\\).\n\nFrom Figure 1, after one \"growth,\" the sum of the areas of the squares with sides equal to the two legs of the right triangle equals the area of the square with the hypotenuse as its side, meaning the total area of all squares is \\(2 \\times 1 = 2\\).\n\nFrom Figure 2, after two \"growths,\" the total area of all squares is \\(3 \\times 1 = 3\\).\n\nExtending this logic, after 2017 \"growths,\" the total area of all squares in the resulting figure is \\(2018 \\times 1 = 2018\\). Therefore, the answer is 2018.\n\n**Key Insight** This problem tests the properties of squares and the Pythagorean theorem. The key to solving it lies in recognizing the relationship between the areas of the new squares obtained after each \"growth\" and the area of the original square, as revealed by the Pythagorean theorem." }, { "problem_id": 1224, "question": "There is a problem in \"The Nine Chapters on the Mathematical Art\": \"Given a right triangle with legs of 3 steps and 4 steps, what is the side length of the inscribed square?\" This means: as shown in Figure 1, in right triangle $ABC$ where $\\angle C = 90^\\circ$, $AC = 3$, and $BC = 4$, find the side length of the inscribed square $DECF$. The Chinese mathematician Liu Hui used the principle of \"complementary subtraction\" to extend Figure 1 into the rectangle shown in Figure 2. In this figure, he discovered a shape with the same area as the square $DECF$, thus determining the side length of this square to be $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_b89409437621ba33f607g_0030_1.jpg", "batch11-2024_06_14_b89409437621ba33f607g_0030_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{12}{7}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, let the side length of square $DECF$ be $x$.\n\n\n\nAccording to the problem, quadrilaterals $AEDG$, $GDIH$, and $DHBF$ are all rectangles.\n\n$\\therefore S_{\\triangle ABC}=S_{\\triangle AIB}$, $S_{\\triangle AED}=S_{\\triangle AGD}$, $S_{\\triangle BFD}=S_{\\triangle BHD}$,\n\n$\\therefore S_{\\triangle ABC}-S_{\\triangle AED}-S_{\\triangle BFD}=S_{\\triangle AIB}-S_{\\triangle AGD}-S_{\\triangle BHD}$,\n\n$\\therefore S_{\\text{square } DECF}=S_{\\text{square } D G I H}$\n\n$\\because GF=AC=3$, $\\quad HE=BC=4$, $\\quad DE=DF=CF=x$\n\n$\\therefore DG=3-x$, $HD=4-x$\n\n$\\therefore x^{2}=(3-x)(4-x)$\n\nSolving the equation, we get $x=\\frac{12}{7}$.\n\nTherefore, the side length of the square is $\\frac{12}{7}$.\n\nHence, the answer is $\\frac{12}{7}$.\n\n【Highlight】This problem examines the determination and properties of rectangles, the area of triangles, and the area of rectangles. Correctly setting up the equation is key to solving this problem." }, { "problem_id": 1225, "question": "The tangram is an important carrier of traditional Chinese mathematical culture, and many interesting patterns can be formed using a tangram. Now, using the set of tangrams shown in Figure 1, a hexagon as shown in Figure 2 is formed. If the total area of the tangrams in Figure 1 is 16, then the perimeter of the figure in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_c84e65defcaaa753a750g_0010_1.jpg", "batch11-2024_06_14_c84e65defcaaa753a750g_0010_2.jpg" ], "is_multi_img": true, "answer": "$8+8 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Given that the area of the tangram is 16, we have:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nIn Figure 1, $AB = BC = 4$, $BF = FC = DE = CE = 2$,\n\n$$\nAO = BO = DO = EF = 2 \\sqrt{2}, \\quad DH = OH = OG = BG = \\sqrt{2}\n$$\n\nTherefore, the perimeter of Figure 2 is: $2 \\sqrt{2} + 2 \\sqrt{2} + 4 + 2 \\sqrt{2} + \\sqrt{2} + \\sqrt{2} + 2 + 2 = 8 + 8 \\sqrt{2}$\n\nHence, the answer is: $8 + 8 \\sqrt{2}$.\n\n【Key Insight】This problem tests the assembly of a tangram. Understanding the relationships between the shapes in the tangram and the Pythagorean theorem is crucial for solving it." }, { "problem_id": 1226, "question": "As shown in Figure A, it is a schematic diagram of the famous \"Zhao Shuang's Hypotenuse Diagram\" in ancient China, which is formed by four congruent right-angled triangles. In the right-angled triangle \\( Rt \\triangle ABC \\), if the legs \\( AC = 6 \\) and \\( BC = 5 \\), extending the leg of length 6 in each of the four right-angled triangles outward by one time, we obtain the \"mathematical windmill\" shown in Figure B. Then, the outer perimeter of this windmill (solid lines in Figure B) is \\(\\qquad\\).\n\n\n\nFigure A\n\n\n\nFigure B", "input_image": [ "batch11-2024_06_14_d0fd143fadaf84d12d86g_0010_1.jpg", "batch11-2024_06_14_d0fd143fadaf84d12d86g_0010_2.jpg" ], "is_multi_img": true, "answer": "76", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, let the length of the hypotenuse of the four right-angled triangles in the \"mathematical windmill\" be \\( x \\). Then,\n\n\\[\nx^{2} = 12^{2} + 5^{2} = 169,\n\\]\n\nSolving for \\( x \\), we get:\n\n\\[\nx = 13,\n\\]\n\nTherefore, the perimeter of the \"mathematical windmill\" is:\n\n\\[\n(13 + 6) \\times 4 = 76.\n\\]\n\nThus, the answer is: 76.\n\n【Key Insight】This problem tests the application of the Pythagorean theorem in a practical scenario and emphasizes the importance of recognizing implicit given conditions in the problem." }, { "problem_id": 1227, "question": "Divide the two congruent rectangles with lengths and widths of 3 and 2, respectively, in Figure 1 along their diagonals into four congruent right-angled triangles. Arrange these four congruent right-angled triangles into the square shown in Figure 2. Then the area of the smaller square $ABCD$ in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_d0fd143fadaf84d12d86g_0078_1.jpg", "batch11-2024_06_14_d0fd143fadaf84d12d86g_0078_2.jpg" ], "is_multi_img": true, "answer": "1 .", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: The side length of the small square $\\mathrm{ABCD}$ in Figure 2 is $=3-2=1$,\n\nThe area of the small square $A B C D$ in Figure 2 is $=1 \\times 1=1$,\n\nTherefore, the answer is: 1.\n\n[Highlight] This question tests the proof of the Pythagorean theorem and congruent figures, with the key step being to determine the side length of the small square $\\mathrm{ABCD}$ in Figure 2." }, { "problem_id": 1228, "question": "As shown in the figure, the two congruent rectangles with side lengths of 3 and 4 in Figure 1 are divided into four congruent right-angled triangles along the diagonal. These four congruent right-angled triangles are then arranged to form the square shown in Figure 2. The area of the shaded part in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_d0fd143fadaf84d12d86g_0086_1.jpg", "batch11-2024_06_14_d0fd143fadaf84d12d86g_0086_2.jpg", "batch11-2024_06_14_d0fd143fadaf84d12d86g_0086_3.jpg" ], "is_multi_img": true, "answer": "1", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: $4 - 3 = 1$,\n\n$1 \\times 1 = 1$.\n\nTherefore, the area of the small square $ABCD$ in Figure 2 is 1.\n\nHence, the answer is: 1.\n\n[Highlight] This question examines the proof of the Pythagorean theorem and congruent figures, with the key step being to determine the side length of the small square $ABCD$ in Figure 2." }, { "problem_id": 1229, "question": "The Pythagorean theorem is one of the greatest scientific discoveries of humanity. It was already recorded in the ancient Chinese mathematical text \"Zhoubi Suanjing,\" as shown in Figure (1). By extending the sides of a right triangle to form squares, and placing the two smaller squares inside the largest square as shown in Figure (2), if the sides of the right triangle are known to be $6, 8, 10$, then the area of the shaded part in Figure (2) is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch11-2024_06_14_d0fd143fadaf84d12d86g_0088_1.jpg", "batch11-2024_06_14_d0fd143fadaf84d12d86g_0088_2.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, from the problem statement,\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n$\\mathrm{S}1 + \\mathrm{S}2 = \\mathrm{S}3$,\n\nGiven the side lengths of the right triangle are $6, 8, 10$,\n\n$\\therefore AB = 10, CM = 6$,\n\n$\\therefore FH = 8 - 6 = 2, FM = 10 - 8 = 2, \\quad AK = 10 - 8 = 2$,\n\n$\\therefore S_{\\text{shaded}} = 10 \\times 2 + 2 \\times 2 = 24$.\n\nTherefore, the answer is 24.\n\n【Insight】This problem primarily tests the knowledge of the Pythagorean theorem, and accurate analysis is the key to solving it." }, { "problem_id": 1230, "question": "Divide the two congruent rectangles with lengths and widths of 6 and 4, respectively, in Figure 1 along their diagonals into four congruent right-angled triangles. Arrange these four congruent right-angled triangles to form the square in Figure 2. Then the area of the smaller square $ABCD$ in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_d0fd143fadaf84d12d86g_0089_1.jpg", "batch11-2024_06_14_d0fd143fadaf84d12d86g_0089_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement:\n\nThe side length of square $\\mathrm{ABCD}$ is: $6-4=2$,\n\n$2 \\times 2=4$.\n\nTherefore, the area of the small square $\\mathrm{ABCD}$ in Figure 2 is 4.\n\nHence, the answer is: 4.\n【Key Insight】This problem examines the proof of the Pythagorean theorem and congruent figures, with the key step being to determine the side length of the small square $\\mathrm{ABCD}$ in Figure 2." }, { "problem_id": 1231, "question": "As shown in the figure, Figure 1 is the emblem of the 7th International Congress on Mathematical Education (ICME). The pattern can be seen as consisting of several adjacent right-angled triangles in Figure 2, where $A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}=A_{4} A_{5}=\\cdots=A_{2021} A_{2022}$, $\\angle A_{1} O A_{2}=45^{\\circ}$, $O A_{1}=1$,\n\n$\\angle O A_{1} A_{2}=\\angle O A_{2} A_{3}=\\angle O A_{3} A_{4}=\\angle O A_{4} A_{5}=\\ldots=\\angle O A_{2021} A_{2022}=90$, then the length of side $O A_{2022}$ is $\\qquad$\n.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_d625fe64192961e4e966g_0026_1.jpg", "batch11-2024_06_14_d625fe64192961e4e966g_0026_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2022}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( O A_{1} = 1 \\) and \\( \\angle A_{1} O A_{2} = 45^{\\circ} \\),\n\nit follows that \\( A_{1} A_{2} = A_{2} A_{3} = A_{3} A_{4} = A_{4} A_{5} = \\ldots = A_{2021} A_{2022} = 1 \\).\n\nTherefore, by the Pythagorean theorem, we have \\( O A_{2} = \\sqrt{1^{2} + 1^{2}} = \\sqrt{2} \\),\n\n\\( O A_{3} = \\sqrt{(\\sqrt{2})^{2} + 1^{2}} = \\sqrt{3} \\),\n\nand thus \\( O A_{n} = \\sqrt{n} \\).\n\nConsequently, \\( O A_{2022} = \\sqrt{2022} \\).\n\nHence, the answer is: \\( \\sqrt{2022} \\).\n\n【Key Insight】This problem examines the pattern of graphical changes, and the key to solving it lies in identifying the pattern \\( O A_{n} = \\sqrt{n} \\)." }, { "problem_id": 1232, "question": "As shown in the figure, in the rectangular paper $A B C D$, right-angled triangular paper pieces of the same size are placed in the ways shown in the figure. If placed in the way shown in Figure (1), exactly 4 pieces can be accommodated; if placed in the way shown in Figure (2), exactly 3 pieces can be accommodated. Given that $B C=12$, the length of the remaining part $F$ when placed in the way shown in Figure (3) is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch11-2024_06_14_d625fe64192961e4e966g_0043_1.jpg", "batch11-2024_06_14_d625fe64192961e4e966g_0043_2.jpg", "batch11-2024_06_14_d625fe64192961e4e966g_0043_3.jpg" ], "is_multi_img": true, "answer": "2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( BC = 12 \\),\n\nTherefore, in Figure (1), \\( BE = 12 \\div 4 = 3 \\), and in Figure (2), \\( BE = 12 \\div 3 = 4 \\),\n\nThus, the length of the hypotenuse of the small right triangle is \\( \\sqrt{3^{2} + 4^{2}} = 5 \\),\n\nHence, in Figure (3), the length of the remaining part \\( CF \\) at the bottom of the paper box is \\( 12 - 5 \\times 2 = 2 \\).\n\nTherefore, the answer is: 2\n\n[Highlight] This question tests the application of the Pythagorean theorem. Understanding the problem and being proficient in the Pythagorean theorem are key to solving it." }, { "problem_id": 1233, "question": "As shown in the figure, a rectangular piece of paper is folded along the dotted line into 3 rectangles, where the widths of the left and right rectangles are both equal to $a$, and the width of the middle rectangle is 4. When it is formed into the triangular prism-shaped object shown in Figure 2, if the area of the base triangle is 8, then the value of $a$ in the figure is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_d625fe64192961e4e966g_0047_1.jpg", "batch11-2024_06_14_d625fe64192961e4e966g_0047_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{5}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "From the problem statement, we know that the base triangle of the formed triangular prism is an isosceles triangle. Let's assume the height on the base of the isosceles triangle is \\( h \\).\n\n\\[\n\\therefore h = \\sqrt{a^{2} - 2^{2}},\n\\]\nand the area of the triangle is 8,\n\n\\[\n\\therefore \\frac{1}{2} \\times 4h = 8,\n\\]\nwhich simplifies to \\( 2\\sqrt{a^{2} - 2^{2}} = 8 \\).\n\nSolving for \\( a \\):\n\n\\[\na = 2\\sqrt{5},\n\\]\n\nThus, the answer is: \\( 2\\sqrt{5} \\).\n\n【Insight】This problem primarily examines the development of geometric shapes and the area formula of an isosceles triangle. Mastering the properties of isosceles triangles is crucial for solving such problems." }, { "problem_id": 1234, "question": "The worker master often adds a small wooden strip between the right-angle vertex and the hypotenuse of the ruler to make it durable, as shown in Figures (1) and (2). Given that $\\angle B A C=90^{\\circ}$, the length of segment $A B$ is $5 \\mathrm{dm}$, and the length of segment $A C$ is $12 \\mathrm{dm}$, the shortest length of the wooden strip $A D$ is $\\qquad$ $\\mathrm{dm}$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch11-2024_06_14_d625fe64192961e4e966g_0098_1.jpg", "batch11-2024_06_14_d625fe64192961e4e966g_0098_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{60}{13} \\# \\# 4 \\frac{8}{13}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: When \\( AD \\perp BC \\), the length of the small wooden strip \\( AD \\) is the shortest.\n\nGiven that \\( \\angle BAC = 90^\\circ \\), \\( AB = 5 \\), and \\( AC = 12 \\),\n\nwe can find \\( BC \\) using the Pythagorean theorem:\n\\[ BC = \\sqrt{AB^2 + AC^2} = \\sqrt{5^2 + 12^2} = 13. \\]\n\nTherefore, the area of triangle \\( ABC \\) can be expressed in two ways:\n\\[ \\frac{1}{2} \\times AB \\times AC = \\frac{1}{2} \\times BC \\times AD. \\]\n\nSolving for \\( AD \\):\n\\[ AD = \\frac{AB \\times AC}{BC} = \\frac{5 \\times 12}{13} = \\frac{60}{13} \\text{ dm}. \\]\n\nThus, the answer is:\n\\[ \\boxed{\\dfrac{60}{13}}. \\]\n\n**Key Insight:** This problem primarily tests the application of the Pythagorean theorem. The key to solving it lies in a thorough understanding of the Pythagorean theorem and the method of equal areas in triangles." }, { "problem_id": 1235, "question": "As shown in the figure, the math interest group is measuring the height of a flagpole. The students found that the rope tied to the top of the flagpole hangs down to the ground and extends beyond it (as shown in Figure 1). The clever Xiaohong noticed that by first measuring the length of the rope that hangs down to the ground, $\\mathrm{m}$, and then pulling the rope taut (as shown in Figure 2), measuring the distance from the end of the rope $\\mathrm{C}$ to the base of the flagpole $\\mathrm{B}$, $\\mathrm{n}$, one can use the knowledge they have learned to determine the height of the flagpole. If $\\mathrm{m}=2$ and $\\mathrm{n}=6$, find the height of the flagpole $\\mathrm{AB}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_d7aa79dfb97370976db7g_0071_1.jpg", "batch11-2024_06_14_d7aa79dfb97370976db7g_0071_2.jpg" ], "is_multi_img": true, "answer": "$8 \\mathrm{~m}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Let the height of the flagpole be \\( \\mathrm{x} \\).\n\nIn the right triangle \\( \\triangle \\mathrm{ABC} \\),\n\n\\[\n\\because \\mathrm{AC}^{2} = \\mathrm{AB}^{2} + \\mathrm{BC}^{2}\n\\]\n\n\\[\n\\therefore (\\mathrm{x} + \\mathrm{m})^{2} = \\mathrm{n}^{2} + \\mathrm{x}^{2},\n\\]\n\n\\[\n\\therefore \\mathrm{x} = \\frac{n^{2} - m^{2}}{2m},\n\\]\n\n\\[\n\\because m = 2, \\quad \\mathrm{n} = 6,\n\\]\n\n\\[\n\\therefore \\mathrm{x} = \\frac{36 - 4}{4} = 8.\n\\]\n\nAnswer: The length of the flagpole \\( \\mathrm{AB} \\) is 8.\n\n【Key Insight】This problem tests the understanding of solving right triangles and the Pythagorean theorem. The key to solving it is to comprehend the problem and construct the appropriate equation, which is a common type of question in middle school exams." }, { "problem_id": 1236, "question": "The Pythagorean theorem is one of the greatest scientific discoveries of humanity, and it was already recorded in the ancient Chinese mathematical text \"Zhoubi Suanjing.\" As shown in Figure 1, taking the sides of a right triangle as the sides of squares, and placing the two smaller square papers inside the largest square as shown in Figure 2. If the areas of the two smaller squares are 9 and 16, respectively, then the area of the shaded part in the figure is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0005_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0005_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the hypotenuse of the right triangle be $c$, the longer leg be $b$, and the shorter leg be $a$. According to the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.\n\nThe area of the shaded part is calculated as:\n\\[\nc^{2} - b^{2} - a(c - b) = a^{2} - a c + a b = a(a + b - c)\n\\]\n\nGiven that the areas of the two smaller squares are 9 and 16 respectively, we have:\n\\[\na^{2} = 9, \\quad b^{2} = 16, \\quad c^{2} = a^{2} + b^{2} = 25\n\\]\nThus,\n\\[\na = 3, \\quad b = 4, \\quad c = 5\n\\]\nTherefore, the area of the shaded part is:\n\\[\n3 \\times (3 + 4 - 5) = 6\n\\]\n\n**Key Insight:** This problem tests the geometric application of the Pythagorean theorem. The key to solving it lies in correctly representing the area of the shaded part in the diagram." }, { "problem_id": 1237, "question": "The Pythagorean theorem is one of the greatest scientific discoveries of humanity, and it was already recorded in the ancient Chinese mathematical text \"Zhoubi Suanjing.\" As shown in Figure 1, if we construct squares on each side of a right triangle and place the two smaller square pieces inside the largest square as shown in Figure 2, if the area of the shaded part in Figure 2 is 5, then the area of the overlapping part of the two smaller squares is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0009_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0009_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As can be seen from the figure, the area of the shaded region = area of the large square - area of the two small squares + area of the overlapping part. According to the Pythagorean theorem, the area of the large square is equal to the sum of the areas of the two small squares. Therefore, the area of the shaded region equals the area of the overlapping part.\n\nHence, the answer is: 5.\n\n[Key Insight] This problem tests the Pythagorean theorem. The key to solving it is to establish the idea of combining numbers with shapes, understanding that the area of the large square equals the sum of the areas of the two small squares, and deducing that the area of the shaded region equals the area of the overlapping part through the addition and subtraction of areas." }, { "problem_id": 1238, "question": "As shown, in August 2002, the emblem of the International Congress of Mathematicians held in Beijing was inspired by \"The Square Diagram of the Gou-Gu Theorem\" from the ancient Chinese mathematician Zhao Shuang. It consists of four identical right-angled triangles and a small square in the center, forming a larger square (Figure 1). The shorter leg of the right-angled triangle is $a$, the longer leg is $b$, and the area of the small square is 49. If these four identical right-angled triangles are rearranged in the form shown in Figure 2, where the area of the larger square is 289, then the area of one right-angled triangle is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0013_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0013_2.jpg" ], "is_multi_img": true, "answer": "30", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: \n\nGiven that the side length of the small square in Figure 1 is \\( b - a \\),\n\n\\[\n\\therefore (b - a)^2 = 49, \\quad \\text{which implies} \\quad b - a = 7 \\quad \\text{(1)},\n\\]\n\nand the side length of the large square in Figure 2 is \\( b + a \\),\n\n\\[\n\\therefore (b + a)^2 = 289, \\quad \\text{which implies} \\quad b + a = 17 \\quad \\text{(2)}.\n\\]\n\nSolving equations (1) and (2) simultaneously, we obtain:\n\n\\[\n\\left\\{\n\\begin{array}{l}\nb = 12 \\\\\na = 5\n\\end{array}\n\\right.\n\\]\n\nThe area of the triangle is:\n\n\\[\nS_{\\triangle} = \\frac{1}{2} a b = \\frac{1}{2} \\times 12 \\times 5 = 30.\n\\]\n\nTherefore, the answer is: **30**.\n\n**Key Insight**: This problem primarily tests the formula for the area of a square. The key is to express the side lengths of the squares correctly, then use the area formula to set up a system of equations. Solving this system gives the side lengths of the triangle, from which the area can be calculated." }, { "problem_id": 1239, "question": "Arrange the four right-angled triangles in Figure 1 to form the squares shown in Figure 2 and Figure 3, respectively. Then, the area of the shaded part in Figure 2 is . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0037_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0037_2.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0037_3.jpg" ], "is_multi_img": true, "answer": "13", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the problem statement, we know that the shaded area in Figure 2 is a square.\n\nLet the shorter leg of the right triangle in Figure 1 be \\( a \\), and the longer leg be \\( b \\).\n\nFrom Figure 2, we have:\n\\[ a + b = 5 \\quad \\text{(1)} \\]\n\nFrom Figure 3, we have:\n\\[ b - a = 1 \\quad \\text{(2)} \\]\n\nSolving equations (1) and (2) simultaneously, we get:\n\\[\n\\begin{cases}\na = 2 \\\\\nb = 3\n\\end{cases}\n\\]\n\nTherefore, the side length of the shaded square is:\n\\[ \\sqrt{2^{2} + 3^{2}} = \\sqrt{13} \\]\n\nThus, the area \\( S \\) of the shaded square is:\n\\[ S = (\\sqrt{13})^{2} = 13 \\]\n\nHence, the answer is: 13.\n\n**Key Insight:** This problem primarily tests the application of the Pythagorean theorem. The key is to determine the side length of the shaded area, which involves using the legs of the right triangle to find the hypotenuse. Remember the Pythagorean theorem formula:\n\\[ a^{2} + b^{2} = c^{2} \\]" }, { "problem_id": 1240, "question": "As shown in Figure 1, this is a famous ancient Chinese \"Zhao Shuang's String Diagram,\" which is formed by four congruent right-angled triangles. If the shorter right-angled side $BC=5$, and the hypotenuse $AB=\\sqrt{61}$, if the longer right-angled sides of the four right-angled triangles are extended outward by twice their length, a \"mathematical windmill\" as shown in Figure 2 is obtained. Then the outer perimeter of this windmill is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0045_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0045_2.jpg" ], "is_multi_img": true, "answer": "76", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle \\( ABC \\), the right-angled side \\( BC = 5 \\), and the hypotenuse \\( AB = \\sqrt{61} \\). Therefore, the other side \\( AC \\) can be calculated using the Pythagorean theorem:\n\n\\[\nAC = \\sqrt{AB^2 - BC^2} = 6\n\\]\n\nSince the longer right-angled sides of the four right triangles are each extended outward by a factor of one, we have:\n\n\\[\nCD = 2 \\times AC = 12, \\quad AD = AC = 6\n\\]\n\nIn the right triangle \\( BCD \\), the hypotenuse \\( BD \\) is:\n\n\\[\nBD = \\sqrt{BC^2 + CD^2} = 13\n\\]\n\nThus, the perimeter of the windmill's outer edge is:\n\n\\[\n4 \\times (AD + BD) = 4 \\times (6 + 13) = 76\n\\]\n\nTherefore, the answer is: 76.\n\n**Key Insight:** This problem tests the application of the Pythagorean theorem. Mastering the Pythagorean theorem is crucial for solving such problems." }, { "problem_id": 1241, "question": "Figure (1) is a schematic diagram of the famous \"Zhao Shuang's Hypotenuse Diagram\" from ancient China, which is formed by four congruent right-angled triangles. If one of the acute angles of the right-angled triangle is $30^{\\circ}$, and each of the shorter legs of the triangles is extended outward by a factor of two, the \"mathematical windmill\" shown in Figure (2) is obtained. Given that $AB = 2$, the area of the shaded part in the figure is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0048_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0048_2.jpg" ], "is_multi_img": true, "answer": "$8+4 \\sqrt{3} \\# \\# 4 \\sqrt{3}+8$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nLet $AC = x$, then $BC = AD = 2 + x$.\n\nSince $\\angle ADC = 30^\\circ$,\n\nWe have $AC = \\frac{1}{2} AD$.\n\nThus, $AD = \\sqrt{3} AC$,\n\nSo, $2 + x = \\sqrt{3} x$,\n\nTherefore, $x = \\sqrt{3} + 1$,\n\nHence, $AC = \\sqrt{3} + 1$.\n\nSince the shorter right-angled sides of each triangle are extended outward by one time, $DE = AC = \\sqrt{3} + 1$.\n\nTherefore, the area of the shaded part in the figure is $4 \\times \\frac{1}{2} AC^2 = 4 \\times \\frac{1}{2} \\times (\\sqrt{3} + 1)^2 = 8 + 4\\sqrt{3}$.\n\nThus, the answer is $8 + 4\\sqrt{3}$.\n\n【Highlight】This problem examines the definition of the angle of rotation and the properties of right-angled triangles. The key to solving this problem lies in expressing the length of $AC$ in terms of $AB$." }, { "problem_id": 1242, "question": "The tangram is an outstanding creation of our ancestors, known as the \"Magic Square of the East.\" A set of tangram pieces can be made from a square $ABCD$ with side length $6 \\sqrt{2}$, as shown in Figure 1. Now, this set of tangram pieces is arranged within a square $EFGH$ to form the \"Struggling Rabbit\" pattern, as shown in Figure 2 (where points $Q$ and $R$ coincide with points $E$ and $G$ in Figure 2, respectively, and point $P$ is on side $EH$). The side length of the square $EFGH$ in which the \"Struggling Rabbit\" is located is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0050_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0050_2.jpg" ], "is_multi_img": true, "answer": "$6 \\sqrt{5}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, connect $EG$, and draw $GM \\perp EN$ intersecting the extension of $EN$ at $M$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nGiven that the side length of square $ABCD$ is $6\\sqrt{2}$,\n\n$\\therefore PD = DR = RC = 3\\sqrt{2}$,\n\n$$\n\\begin{aligned}\n& \\therefore PR = \\sqrt{PD^{2} + DR^{2}} = 6, \\quad PQ = RQ = 3, \\\\\n& \\therefore GM = PR = 6, \\quad EM = 3 + 3 + 6 + 6 = 18, \\\\\n& \\therefore EG = \\sqrt{EM^{2} + GM^{2}} = \\sqrt{18^{2} + 6^{2}} = 6\\sqrt{10}, \\\\\n& \\therefore EH = \\frac{EG}{\\sqrt{2}} = 6\\sqrt{5},\n\\end{aligned}\n$$\n\nTherefore, the answer is: $6\\sqrt{5}$.\n\n【Insight】This problem examines the properties of squares, tangram, and the Pythagorean theorem. The key to solving the problem lies in adding common auxiliary lines to construct right triangles for problem-solving." }, { "problem_id": 1243, "question": "As shown in the figure, Xiao Ming cut the rectangular paper pieces with lengths and widths of 5 and 3 in Figure 1 into four identical right-angled triangles, and then arranged these four right-angled triangle paper pieces into a large square as shown in Figure 2. The side length of the large square in Figure 2 is $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_e891b47c0c98036f1a7fg_0057_1.jpg", "batch11-2024_06_14_e891b47c0c98036f1a7fg_0057_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{34}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the problem statement, we know that Figure 2 is composed of four identical right-angled triangles from Figure 1, each with legs of lengths 5 and 3.\n\nThus, the side length of the large square in Figure 2 is $\\sqrt{5^{2}+3^{2}}=\\sqrt{34}$.\n\nTherefore, the answer is: $\\sqrt{34}$.\n\n【Key Insight】This problem tests the properties of congruent triangles and the Pythagorean theorem. Mastering the relevant knowledge and applying it correctly is crucial for solving the problem." }, { "problem_id": 1244, "question": "As shown in Figure 1, this is a schematic diagram of the famous \"Zhao Shuang's Hypotenuse Diagram\" in ancient China, which is formed by four congruent right-angled triangles. If $AC=6, BC=5$, and the right-angled sides with length 6 in the four right-angled triangles are extended outward by one time their length, the \"Mathematical Pinwheel\" shown in Figure 2 is obtained. Find the perimeter of the outer circumference of this pinwheel.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch11-2024_06_14_f23d844c81efa5f2c545g_0057_1.jpg", "batch11-2024_06_14_f23d844c81efa5f2c545g_0057_2.jpg" ], "is_multi_img": true, "answer": "76", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, let the length of the hypotenuse of each of the four right-angled triangles in the \"math pinwheel\" be \\( x \\). Then, \\( x^{2} = 12^{2} + 5^{2} = 169 \\).\n\nTherefore, \\( x = 13 \\).\n\nThus, the perimeter of the \"math pinwheel\" is: \\( (13 + 6) \\times 4 = 76 \\).\n\n[Key Insight] This problem tests the application of the Pythagorean theorem in a practical scenario. It is important to utilize the implicit given conditions to solve such problems." }, { "problem_id": 1245, "question": "Figure 1 shows a foldable eye-protecting lamp placed on a horizontal surface, where the height of the base \\( AB = 5 \\, \\text{cm} \\), the length of the connecting rod \\( BC = 30 \\, \\text{cm} \\), and the length of the lamp shade \\( CD = 20 \\, \\text{cm} \\). As shown in Figure 2, when \\( BC \\) and \\( CD \\) are rotated to form a straight angle \\( \\angle BCD \\), and the height \\( DH \\) of the lamp shade end \\( D \\) above the table \\( l \\) is \\( 45 \\, \\text{cm} \\), find the distance \\( AH \\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_ff4041409e08eb582ec4g_0015_1.jpg", "batch11-2024_06_14_ff4041409e08eb582ec4g_0015_2.jpg" ], "is_multi_img": true, "answer": "30 cm", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, draw $BE \\perp DH$ at point $E$,\n\n\n\nFigure 2\n\n$\\because AB \\perp AH, \\quad DH \\perp AH$,\n\n$\\therefore$ quadrilateral $ABEH$ is a rectangle,\n\n$\\therefore \\angle EBA=90^{\\circ}, EH=AB=5 \\mathrm{~cm}$,\n\n$\\because BC=30 \\mathrm{~cm}, \\quad CD=20 \\mathrm{~cm}, \\quad DH=45 \\mathrm{~cm}$\n\n$\\therefore BD=30+20=50 \\mathrm{~cm}, \\quad DE=45-5=40 \\mathrm{~cm}$,\n\nIn right-angled $\\triangle BDE$, by the Pythagorean theorem, we have\n\n$BE=\\sqrt{BD^{2}-DE^{2}}=\\sqrt{50^{2}-40^{2}}=30 \\mathrm{~cm}$,\n\n$\\therefore AH=BE=30 \\mathrm{~cm}$;\n【Insight】This problem examines the application of solving right-angled triangles. The key to solving the problem is to learn to add common auxiliary lines, constructing rectangles and right-angled triangles to solve the problem." }, { "problem_id": 1246, "question": "As shown in the figure, the school needs to measure the height of a flagpole. The students found that the rope tied to the top of the flagpole hangs down to the ground and extends beyond it (as shown in Figure 1). The students first measured the length of the extended rope, which is 1 meter, and then pulled the rope taut (as shown in Figure 2), measuring the distance from the end of the rope $C$ to the base of the flagpole $B$, which is 5 meters. Find the height of the flagpole.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch11-2024_06_14_ff4041409e08eb582ec4g_0084_1.jpg", "batch11-2024_06_14_ff4041409e08eb582ec4g_0084_2.jpg" ], "is_multi_img": true, "answer": "12 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the height of the flagpole \\( AB \\) be \\( x \\) meters, then the length of the rope \\( AC \\) is \\( (x+1) \\) meters.\n\nIn the right triangle \\( \\triangle ABC \\), according to the Pythagorean theorem, we have: \n\\[ x^{2} + 5^{2} = (x + 1)^{2} \\]\n\nSolving the equation, we find:\n\\[ x = 12 \\]\n\nAnswer: The height of the flagpole is 12 meters.\n\n【Highlight】This question tests the application of the Pythagorean theorem. Understanding the Pythagorean theorem is key to solving the problem." }, { "problem_id": 1247, "question": "As shown in Figure $a$, $a$ is a rectangular paper strip, $\\angle D E F=26^{\\circ}$. The strip is folded along $E F$ to form Figure $b$, and then folded again along $B F$ to form Figure $c$. In Figure $c$, $\\angle C F E=$ $\\qquad$ degrees.\n\n\n\nFigure a\n\n\n\nFigure b\n\n\n\nFigure c", "input_image": [ "batch12-2024_06_15_7367a6691403d7f10b90g_0038_1.jpg", "batch12-2024_06_15_7367a6691403d7f10b90g_0038_2.jpg", "batch12-2024_06_15_7367a6691403d7f10b90g_0038_3.jpg" ], "is_multi_img": true, "answer": "102", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since quadrilateral \\( A B C D \\) is a rectangle,\n\ntherefore \\( A D \\parallel B C \\),\n\nhence \\( \\angle B F E = \\angle D E F = 26^{\\circ} \\).\n\nIn Figure \\( b \\), \\( \\angle E F G = 26^{\\circ} \\), and the area at \\( \\angle E F G \\) is overlapped by 2 layers.\n\nIn Figure \\( c \\), \\( \\angle E F G = 26^{\\circ} \\), and the area at \\( \\angle E F G \\) is overlapped by 3 layers.\n\nTherefore, \\( \\angle C F E = \\angle C F G - \\angle E F G = 180^{\\circ} - 2 \\angle B F E - \\angle E F G = 180^{\\circ} - 3 \\times 26^{\\circ} = 102^{\\circ} \\).\n\nThus, the answer is: 102.\n\n[Key Insight] This problem examines the properties of parallel lines. A thorough understanding of these properties is crucial for solving the problem." }, { "problem_id": 1248, "question": "As shown in Figure 1, in rectangle \\( A B C D \\), a moving point \\( P \\) starts from point \\( A \\) and moves along the path \\( A-B-C \\) on sides \\( A B \\) and \\( B C \\). Let \\( P A = x \\), and the distance from point \\( D \\) to the line \\( P A \\) be \\( y \\). The graph of the function \\( y \\) with respect to \\( x \\) is shown in Figure 2. When the area of \\( P C D \\) is equal to the area of \\( P A B \\), the value of \\( y \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch12-2024_06_15_ef16215545d5a6da273bg_0028_1.jpg", "batch12-2024_06_15_ef16215545d5a6da273bg_0028_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "According to the image information, when point \\( P \\) moves on \\( AB \\), since quadrilateral \\( ABCD \\) is a rectangle,\n\n\\(\\therefore x = 0\\), at this time, the distance from point \\( D \\) to \\( PA \\) is \\( AD \\),\n\n\\(\\therefore AD = 2\\),\n\nwhen \\( x = 1 \\), \\( y = 2 \\),\n\n\\(\\therefore AB = 1\\),\n\nLet the area of \\( PAD \\) be \\( S \\), when \\( x > 1 \\), \\( y = \\frac{2S}{x} \\),\n\nAccording to the image, the point \\( (1,2) \\) is on the graph,\n\n\\(\\therefore 2S = 2\\) which means \\( S = 1 \\),\n\n\\(\\therefore y = \\frac{2}{x} \\),\n\nWhen the area of \\( PCD \\) equals the area of \\( PAB \\), point \\( P \\) is exactly the midpoint of \\( BC \\),\n\nHence \\( AB = 1 \\), \\( PB = 1 \\),\n\n\\(\\therefore x = \\sqrt{AB^{2} + PB^{2}} = \\sqrt{1^{2} + 1^{2}} = \\sqrt{2} \\),\n\n\\(\\therefore y = \\frac{2}{\\sqrt{2}} = \\sqrt{2} \\).\n\nTherefore, the answer is: \\( \\sqrt{2} \\).\n\n【Highlight】This question examines the function's graph, properties of rectangles, determination of the inverse proportional function's formula, and the Pythagorean theorem. Accurately obtaining the problem-solving information from the function's graph and flexibly applying the Pythagorean theorem and the inverse proportional function's formula are key to solving the problem." }, { "problem_id": 1249, "question": "The line segment connecting the midpoints of two sides of a triangle is called the median of the triangle. The property of the median of a triangle: the median is parallel to the third side and is equal to half of the third side. As shown in Figure 1, in triangle \\(ABC\\), \\(D\\) and \\(E\\) are the midpoints of \\(AB\\) and \\(AC\\) respectively, then \\(DE \\parallel BC\\) and \\(DE = \\frac{1}{2} BC\\). Use the property of the median of a triangle to solve the following problem: As shown in Figure 2, the graph of the function \\(y = \\frac{12}{x} (x > 0)\\) passes through the vertices of \\(OAB\\) and the midpoint \\(C\\) of side \\(AB\\). Draw \\(BD \\perp x\\)-axis and \\(CE \\perp x\\)-axis through points \\(B\\) and \\(C\\) respectively, with the feet of the perpendiculars being \\(D\\) and \\(E\\), and \\(CE\\) being the median of \\(\\triangle ABD\\). If the x-coordinate of point \\(B\\) is 3, then the coordinates of point \\(C\\) are \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch12-2024_06_15_f61b04a9a567c54d8582g_0030_1.jpg", "batch12-2024_06_15_f61b04a9a567c54d8582g_0030_2.jpg" ], "is_multi_img": true, "answer": "$(6,2)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Since the abscissa (x-coordinate) of point $B$ is 3, and point $B$ lies on the curve $y=\\frac{12}{x}$ (where $x>0$),\n\nSubstituting $x=3$ into the equation, we obtain $y=4$,\n\nThus, the coordinates of point $B$ are $(3,4)$,\n\nTherefore, the length of $BD$ is 4,\n\nGiven that $CE$ is the median of triangle $ABD$,\n\nIt follows that $CE=\\frac{1}{2} BD=2$,\n\nHence, the ordinate (y-coordinate) of point $C$ is 2,\n\nSubstituting $y=2$ into the equation $y=\\frac{12}{x}$, we find $x=6$,\n\nThus, the coordinates of point $C$ are $(6,2)$.\n\nThe final answer is: $(6,2)$.\n\n【Insight】This problem tests the properties of inverse proportional functions. When a point lies on the graph of an inverse proportional function, its coordinates satisfy the function's equation. Substituting into the equation can determine the point's abscissa or ordinate. The use of the median in the triangle is key to solving this problem." }, { "problem_id": 1250, "question": "As shown in Figures (1), (2), and (3), a \"ring\" formed by tiling with regular polygons of equal size is called a ring tiling. However, Figures (4) and (5) are not what we refer to as ring tiling. Please write another regular polygon that can be used for ring tiling: $\\qquad$ .\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\n\n\n\nFigure (5)", "input_image": [ "batch13-2024_06_15_1fb6dce28bdd5f924b02g_0006_1.jpg", "batch13-2024_06_15_1fb6dce28bdd5f924b02g_0006_2.jpg", "batch13-2024_06_15_1fb6dce28bdd5f924b02g_0006_3.jpg", "batch13-2024_06_15_1fb6dce28bdd5f924b02g_0006_4.jpg", "batch13-2024_06_15_1fb6dce28bdd5f924b02g_0006_5.jpg" ], "is_multi_img": true, "answer": "regular dodecagon", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** According to the definition of circular tessellation, it is sufficient that twice the exterior angle of the polygon used equals the interior angle of a regular polygon. The exterior angle of a regular dodecagon is \\(360^{\\circ} \\div 12 = 30^{\\circ}\\). \nSince \\(30^{\\circ} \\times 2 = 60^{\\circ}\\) is the interior angle of an equilateral triangle, \n**therefore**, a regular dodecagon can perform circular tessellation. \n\n**Exam Focus:** Plane tessellation (tiling)." }, { "problem_id": 1251, "question": "As shown in Figure 1, four identical square picture frames are hung on the stair wall of a school, with the lower edges of the frames parallel to the horizontal ground. As shown in Figure 2, the top left corners of the frames, $B, E, F, G$, are all on the line $AB$, and $BE = EF = FG$. The line $CD$ on which the stair decorative lines lie is parallel to $AB$. By extending the sides $BH$ and $MN$ of the frames, a parallelogram $ABCD$ is formed. If points $P, Q, D$ happen to be on the same straight line, and given $AB = 220 \\text{ cm}$, $CH = 80 \\text{ cm}$, and $\\angle C = 60^\\circ$, the side length of the square picture frame is $\\qquad$ $\\text{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_3ef2fb8ea53e7839469cg_0097_1.jpg", "batch13-2024_06_15_3ef2fb8ea53e7839469cg_0097_2.jpg" ], "is_multi_img": true, "answer": "$20 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, extend $EP$ to intersect $CD$ at point $L$.\n\nSince quadrilateral $ABCD$ is a parallelogram, and points $B, E, F, G$ all lie on line $AB$,\n\nTherefore, $CD = AB = 220 \\text{ cm}$, and $BE \\parallel CL$.\n\nSince the bottom edge of the picture frame is parallel to the horizontal ground,\n\nTherefore, $BH$ and $EP$ are perpendicular to the ground,\n\nThus, $BC \\parallel EL$,\n\nHence, quadrilateral $BCLE$ is a parallelogram,\n\nTherefore, $EL = BC$.\n\nSince $EP = BH$,\n\nThus, $PL = EL - EP = BC - BH = CH = 80 \\text{ cm}$.\n\nSince $\\angle EPQ = 90^\\circ$,\n\nTherefore, $\\angle DPL = 180^\\circ - \\angle EPQ = 90^\\circ$.\n\nSince $\\angle PLD = \\angle C = 60^\\circ$,\n\nTherefore, $\\angle PDL = 90^\\circ - \\angle PLD = 30^\\circ$,\n\nThus, $DL = 2PL = 2 \\times 80 = 160 \\text{ cm}$.\n\nTherefore, $BE = EF = FG = CL = CD - DL = 220 - 160 = 60 \\text{ cm}$.\n\nThus, $AG = AB - BE - EF - FG = 220 - 60 - 60 - 60 = 40 \\text{ cm}$.\n\nSince $\\angle GMN = 90^\\circ$,\n\nTherefore, $\\angle AMG = 180^\\circ - \\angle GMN = 90^\\circ$.\n\nSince $\\angle A = \\angle C = 60^\\circ$,\n\nTherefore, $\\angle AGM = 90^\\circ - \\angle A = 30^\\circ$,\n\nThus, $AM = \\frac{1}{2} AG = \\frac{1}{2} \\times 40 = 20 \\text{ cm}$.\n\nTherefore, $GM = \\sqrt{AG^2 - AM^2} = \\sqrt{40^2 - 20^2} = 20\\sqrt{3} \\text{ cm}$.\n\nThus, the side length of the square picture frame is $20\\sqrt{3} \\text{ cm}$.\n\nHence, the answer is: $20\\sqrt{3}$.\n\n\n\nFigure 2\n\n[Key Insight] This problem focuses on testing the properties and determination of parallelograms, the properties and determination of squares, the property that in a right-angled triangle the side opposite the $30^\\circ$ angle is half the hypotenuse, and the Pythagorean theorem. Correctly drawing the necessary auxiliary lines is key to solving the problem." }, { "problem_id": 1252, "question": "As shown in Figure 1, in rhombus \\(ABCD\\), \\(AB = 6\\) and \\(\\angle BAD = 120^\\circ\\). Point \\(E\\) is a moving point on side \\(BC\\). Point \\(P\\) is a moving point on diagonal \\(BD\\), and the length of \\(PD\\) is \\(x\\). The sum of the lengths of \\(PE\\) and \\(PC\\) is \\(y\\). Figure 2 shows the graph of the function \\(y\\) with respect to \\(x\\), where \\(H(a, b)\\) is the lowest point on the graph. The value of \\(a + b\\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_44fcd169c3566e139ae5g_0083_1.jpg", "batch13-2024_06_15_44fcd169c3566e139ae5g_0083_2.jpg" ], "is_multi_img": true, "answer": "$7 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, take point \\( E_{1} \\) on side \\( AB \\) such that \\( BE \\) and \\( BE_{1} \\) are symmetric with respect to \\( BD \\).\n\nConnect \\( PE_{1} \\), yielding \\( PC + PE = PC + PE_{1} \\).\n\nConnect \\( CE_{1} \\), and draw \\( CE_{2} \\perp AB \\), with the foot of the perpendicular at \\( E_{2} \\).\n\n\n\nFrom the triangle inequality and the fact that the perpendicular is the shortest distance,\n\n\\[ PE + PC = PE_{1} + PC \\geq CE_{1} \\geq CE_{2}, \\]\n\nwhich means \\( PE + PC \\) has a minimum value of \\( CE_{2} \\).\n\nIn rhombus \\( ABCD \\), \\( AB = 6 \\), and \\( \\angle BAD = 120^\\circ \\).\n\nIn right triangle \\( \\triangle BE_{2}C \\), \\( \\angle E_{2}BC = 60^\\circ \\).\n\nSolving gives \\( CE_{2} = 3\\sqrt{3} \\).\n\nSince \\( H(a, b) \\) is the lowest point on the graph,\n\n\\[ b = y = PE + PC = CE_{2} = 3\\sqrt{3}. \\]\n\nAt this point, let \\( CE_{2} \\) intersect \\( BD \\) at point \\( P_{2} \\).\n\n\n\nGiven \\( BE_{2} = 3 \\), in right triangle \\( \\triangle BP_{2}E_{2} \\),\n\n\\[ BP_{2} = 2\\sqrt{3}, \\] and since \\( BD = 6\\sqrt{3} \\),\n\n\\[ P_{2}D = 4\\sqrt{3}. \\]\n\nAlso, the length of \\( PD \\) is \\( x \\), and in Figure 2, \\( H(a, b) \\) is the lowest point on the graph,\n\n\\[ a = P_{2}D = 4\\sqrt{3}, \\]\n\nand \\( b = 3\\sqrt{3} \\),\n\n\\[ \\therefore a + b = 7\\sqrt{3}. \\]\n\nTherefore, the answer is: \\( 7\\sqrt{3} \\).\n\n【Highlight】This problem examines the moving point and minimum value problem. The challenge lies in reflecting point \\( E \\) (symmetrical with respect to \\( BD \\)) and then using the triangle inequality and the principle of the shortest perpendicular distance to determine that the minimum value of \\( PC + PE \\) is \\( CE_{2} \\)." }, { "problem_id": 1253, "question": "As shown in Figure 1, point \\( P \\) starts from vertex \\( D \\) of rhombus \\( ABCD \\) and moves uniformly along the path \\( D \\rightarrow C \\rightarrow A \\) at a speed of \\( 1 \\mathrm{~cm/s} \\) to point \\( A \\). Figure 2 shows the relationship between the area \\( S \\left( \\mathrm{~cm}^2 \\right) \\) of triangle \\( PAB \\) and time \\( t \\left( \\mathrm{~s} \\right) \\) as point \\( P \\) moves. The value of \\( m \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_44fcd169c3566e139ae5g_0092_1.jpg", "batch13-2024_06_15_44fcd169c3566e139ae5g_0092_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{5}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular line $CE \\perp AB$ at point $E$.\n\n\n\nFrom the graph, it can be seen that the time taken for point $P$ to move from point $D$ to point $C$ is $m$ seconds, and the area of $PAB$ is $m \\mathrm{~cm}^{2}$.\n\n$\\therefore DC = m$,\n\n$\\therefore \\frac{1}{2} AB \\cdot CE = \\frac{1}{2} DC \\cdot CE = \\frac{1}{2} m \\cdot CE = m$,\n\n$\\therefore CE = 2$,\n\nWhen point $P$ moves from point $C$ to point $A$, the time taken is $\\sqrt{5}$ seconds,\n\n$\\therefore CA = \\sqrt{5}$,\n\n$\\therefore$ In right triangle $ACE$, $AE = \\sqrt{AC^{2} - CE^{2}} = \\sqrt{(\\sqrt{5})^{2} - 2^{2}} = 1$.\n\n$\\because ABCD$ is a rhombus,\n\n$\\therefore BE = AB - 1 = m - 1, \\quad CB = m$,\n\n$\\because$ In right triangle $\\triangle BCE$, $BC^{2} = BE^{2} + CE^{2}$,\n\n$\\therefore m^{2} = (m - 1)^{2} + 2^{2}$,\nSolving gives $m = \\frac{5}{2}$.\n\nTherefore, the answer is: $\\frac{5}{2}$.\n\n【Key Insight】This problem tests the properties of a rhombus, the Pythagorean theorem, and function graphs. The key to solving it lies in understanding the relationship between the changes in the function graph and the position of the moving point." }, { "problem_id": 1254, "question": "As shown in the figure, Figure 1 is a foldable desk lamp, and Figure 2 is its plan view. The base $A O \\perp O E$ at point $O$, and the supports $A B, B C$ are fixed rods. The angle $\\angle B A O$ is twice the angle $\\angle C B A$. The lamp body $C D$ can be adjusted by rotating around point $C$. Now, the lamp body $C D$ is rotated from the horizontal position to the position $C D^{\\prime}$ (as shown by the dashed line in Figure 2). At this point, the line where the lamp body $C D^{\\prime}$ lies is exactly perpendicular to the support $A B$, and $\\angle B C D - \\angle D C D^{\\prime} = 105^{\\circ}$. Then, $\\angle D C D^{\\prime} =$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_4fd1bcb569cdb231b149g_0055_1.jpg", "batch13-2024_06_15_4fd1bcb569cdb231b149g_0055_2.jpg" ], "is_multi_img": true, "answer": "$50^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Extend $O A$ to intersect $C D$ at point $F$, and extend $D^{\\prime} C$ to intersect $A B$ at $G$, as shown in the figure.\n\n\nSince $C D \\parallel O E$ and $A O \\perp O E$,\n\nTherefore, $O A \\perp C D$,\n\nGiven $A O \\perp O E$ and $D^{\\prime} C \\perp A B$,\n\nThus, $\\angle A G C = \\angle A F C = 90^{\\circ}$,\n\nHence, $\\angle G C F + \\angle G A F = 180^{\\circ}$,\n\nGiven $\\angle D C D^{\\prime} + \\angle G C F = 180^{\\circ}$,\n\nTherefore, $\\angle D C D^{\\prime} = \\angle G A F$,\n\nThus, $\\angle B A O = 180^{\\circ} - \\angle D C D^{\\prime}$,\n\nTherefore, $\\angle C B A = \\frac{1}{2}\\left(180^{\\circ} - \\angle D C D^{\\prime}\\right)$\n\nGiven $\\angle B C D - \\angle D C D^{\\prime} = 105^{\\circ}$,\n\nTherefore, $\\angle B C D = \\angle D C D^{\\prime} + 105^{\\circ}$,\n\nIn quadrilateral $A B C F$,\n\n$\\angle G A F + \\angle C B A + \\angle B C D + \\angle A F C = 360^{\\circ}$,\n\nThus, $\\angle D C D^{\\prime} + \\frac{1}{2}\\left(180^{\\circ} - \\angle D C D^{\\prime}\\right) + \\angle D C D^{\\prime} + 105^{\\circ} + 90^{\\circ} = 360^{\\circ}$,\n\nSolving gives $\\angle D C D^{\\prime} = 50^{\\circ}$.\n\nTherefore, the answer is: $50^{\\circ}$.\n\n【Highlight】This problem tests the properties of parallel lines and the interior angle sum theorem of a quadrilateral, with the key being to solve using the properties of parallel lines." }, { "problem_id": 1255, "question": "As shown, classmate Xiao Wang used a set of tangram pieces from Figure 1 to create the \"eagle\" shown in Figure 2. Given that the diagonal $AC$ of the square $ABCD$ is $2\\sqrt{2}$, the distance between points $E$ and $F$ in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_4fd1bcb569cdb231b149g_0090_1.jpg", "batch13-2024_06_15_4fd1bcb569cdb231b149g_0090_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{\\sqrt{26}}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, based on the properties of the tangram, quadrilateral \\( O F E G \\) is a square with each side equal to one-fourth of the diagonal of square \\( A B C D \\).\n\nThat is, \\( O F = E F = \\frac{1}{4} A C = \\frac{\\sqrt{2}}{2} \\).\n\nTherefore, \\( F G = O E = \\frac{\\sqrt{2}}{2} \\times \\sqrt{2} = 1 \\).\n\nHence, \\( E H = O H = F H = G H = \\frac{1}{2} \\times 1 = \\frac{1}{2} \\).\n\nSince the diagonal \\( A C \\) of square \\( A B C D \\) has a length of \\( 2 \\sqrt{2} \\),\n\nit follows that \\( A B = B C = \\frac{A C}{\\sqrt{2}} = 2 \\).\n\n\n\nFigure 1\n\nAs shown in Figure 2, drawing \\( E G \\perp F G \\) at point \\( G \\),\n\n\n\nFigure 2\n\nFrom the properties of the tangram, we know: \\( E G = \\frac{1}{2} \\), \\( F G = 2 + \\frac{1}{2} = \\frac{5}{2} \\).\n\nIn the right triangle \\( \\triangle F E G \\), by the Pythagorean theorem, \\( E F = \\sqrt{\\left(\\frac{1}{2}\\right)^{2} + \\left(\\frac{5}{2}\\right)^{2}} = \\frac{\\sqrt{26}}{2} \\).\n\nThus, the answer is: \\( \\frac{\\sqrt{26}}{2} \\).\n\n【Insight】This problem primarily examines the properties of squares, the tangram, the Pythagorean theorem, and the properties of isosceles right triangles. The key to solving the problem lies in constructing a right triangle and applying the Pythagorean theorem." }, { "problem_id": 1256, "question": "At the Qatar World Cup, the main stadium \"Big Golden Bowl\" - Lusail Stadium (Figure (1)), incorporates many cutting-edge technologies. The stadium's roof is made of environmentally friendly membrane material, which not only provides shade for the audience but also allows sunlight to reach the grass on the field. The material structure of the membrane is composed of numerous interwoven regular hexagons, with regular hexagon $ABCDEF$ (in Figure (2)), where $\\angle ABC$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch13-2024_06_15_4fd1bcb569cdb231b149g_0096_1.jpg", "batch13-2024_06_15_4fd1bcb569cdb231b149g_0096_2.jpg" ], "is_multi_img": true, "answer": "120", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since hexagon \\( A B C D E F \\) is a regular hexagon,\n\nTherefore, each internal angle is equal,\n\nThus, \\( \\angle A B C = \\frac{(6-2) \\times 180^{\\circ}}{6} = 120^{\\circ} \\),\n\nHence, the answer is: 120.\n\n[Key Insight] This question tests the properties of regular polygons. Remembering that the sum of the internal angles of a polygon is \\( (n-2) \\times 180^{\\circ} \\) is crucial for solving the problem." }, { "problem_id": 1257, "question": "At the Qatar World Cup, the main stadium \"Big Golden Bowl\" - Lusail Stadium (Figure (1)) incorporates many cutting-edge technologies. The stadium's roof is made of environmentally friendly membrane material, which not only provides shade for the audience but also allows sunlight to reach the grass field. The material structure of the membrane is composed of numerous interwoven regular hexagons. In the regular hexagon \\(ABCDEF\\) (Figure (2)), \\(\\angle ABC\\) is \\(\\qquad\\) \\({}^{\\circ}\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch13-2024_06_15_72def6568d0ee26f353eg_0086_1.jpg", "batch13-2024_06_15_72def6568d0ee26f353eg_0086_2.jpg" ], "is_multi_img": true, "answer": "120", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: Since hexagon \\( A B C D E F \\) is a regular hexagon,\n\nTherefore, each internal angle is equal,\n\nThus, \\( \\angle A B C = \\frac{(6-2) \\times 180^{\\circ}}{6} = 120^{\\circ} \\),\n\nHence, the answer is: 120.\n\n[Key Insight] This question tests the properties of regular polygons. Remembering that the sum of the internal angles of a polygon is \\( (n-2) \\times 180^{\\circ} \\) is crucial for solving the problem." }, { "problem_id": 1258, "question": "As shown in the figure, Figure 1 is a foldable desk lamp, and Figure 2 is its plan view. The base $A O \\perp O E$ at point $O$, and the supports $A B, B C$ are fixed rods. The angle $\\angle B A O$ is twice the angle $\\angle C B A$. The lamp body $C D$ can be adjusted by rotating around point $C$. Now, the lamp body $C D$ is rotated from the horizontal position to the position $C D^{\\prime}$ (as shown by the dashed line in Figure 2). At this point, the line where the lamp body $C D^{\\prime}$ lies is exactly perpendicular to the support $A B$, and $\\angle B C D - \\angle D C D^{\\prime} = 105^{\\circ}$. Then, $\\angle D C D^{\\prime} =$ $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_72def6568d0ee26f353eg_0090_1.jpg", "batch13-2024_06_15_72def6568d0ee26f353eg_0090_2.jpg" ], "is_multi_img": true, "answer": "$50^{\\circ}$\n", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Extend $OA$ to intersect $CD$ at point $F$, and extend $D'C$ to intersect $AB$ at $G$, as shown in the figure.\n\n\n\nSince $CD \\parallel OE$ and $AO \\perp OE$,\n\nTherefore, $OA \\perp CD$,\n\nGiven $AO \\perp OE$ and $D'C \\perp AB$,\n\nThus, $\\angle AGC = \\angle AFC = 90^\\circ$,\n\nHence, $\\angle GCF + \\angle GAF = 180^\\circ$,\n\nGiven $\\angle DCD' + \\angle GCF = 180^\\circ$,\n\nTherefore, $\\angle DCD' = \\angle GAF$,\n\nThus, $\\angle BAO = 180^\\circ - \\angle DCD'$,\n\nTherefore, $\\angle CBA = \\frac{1}{2}(180^\\circ - \\angle DCD')$\n\nGiven $\\angle BCD - \\angle DCD' = 105^\\circ$,\n\nThus, $\\angle BCD = \\angle DCD' + 105^\\circ$,\n\nIn quadrilateral $ABCF$,\n\n$\\angle GAF + \\angle CBA + \\angle BCD + \\angle AFC = 360^\\circ$,\n\nTherefore, $\\angle DCD' + \\frac{1}{2}(180^\\circ - \\angle DCD') + \\angle DCD' + 105^\\circ + 90^\\circ = 360^\\circ$,\n\nSolving gives $\\angle DCD' = 50^\\circ$.\n\nHence, the answer is: $50^\\circ$.\n\n【Highlight】This problem examines the properties of parallel lines and the interior angle sum theorem of a quadrilateral, with the key being to solve using the properties of parallel lines." }, { "problem_id": 1259, "question": "Investigation: As shown in Figure (1), in quadrilateral \\(ABCD\\), \\(\\angle BAD = \\angle BCD = 90^\\circ\\), \\(AB = AD\\), and \\(AE \\perp CD\\) at point \\(E\\). If \\(AE = 10\\), find the area of quadrilateral \\(ABCD\\).\n\nApplication: As shown in Figure (2), in quadrilateral \\(ABCD\\), \\(\\angle ABC + \\angle ADC = 180^\\circ\\), \\(AB = AD\\), and \\(AE \\perp BC\\) at point \\(E\\). If \\(AE = 19\\), \\(BC = 10\\), and \\(CD = 6\\), then the area of quadrilateral \\(ABCD\\) is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch13-2024_06_15_be0270df9ddf03856b26g_0037_1.jpg", "batch13-2024_06_15_be0270df9ddf03856b26g_0037_2.jpg" ], "is_multi_img": true, "answer": "152", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\n**Exploration:** As shown in Figure (1), draw a line $\\mathrm{AF} \\perp \\mathrm{CB}$ through point $\\mathrm{A}$, intersecting the extension of $\\mathrm{CB}$ at point $\\mathrm{F}$.\n\nSince $\\mathrm{AE} \\perp \\mathrm{CD}$ and $\\angle \\mathrm{BCD} = 90^\\circ$,\n\nthe quadrilateral $\\mathrm{AFCE}$ is a rectangle.\n\nThus, $\\angle \\mathrm{FAE} = 90^\\circ$,\n\nand $\\angle \\mathrm{FAB} + \\angle \\mathrm{BAE} = 90^\\circ$.\n\nSince $\\angle \\mathrm{EAD} + \\angle \\mathrm{BAE} = 90^\\circ$,\n\nit follows that $\\angle \\mathrm{FAB} = \\angle \\mathrm{EAD}$.\n\nIn triangles $\\triangle \\mathrm{AFB}$ and $\\triangle \\mathrm{AED}$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle ABC = \\angle ADF \\\\\n\\angle AEB = \\angle F = 90^\\circ \\\\\nAB = AD\n\\end{array}\n\\right.\n\\]\n\nTherefore, $\\triangle \\mathrm{AFB} \\cong \\triangle \\mathrm{AED}$ (by AAS),\n\nand $\\mathrm{AF} = \\mathrm{AE}$.\n\nHence, the quadrilateral $\\mathrm{AFCE}$ is a square.\n\nThus, the area of quadrilateral $\\mathrm{ABCD}$ is equal to the area of square $\\mathrm{AFCE}$, which is $\\mathrm{AE}^2 = 10^2 = 100$.\n\n**Application:** As shown in the figure, draw a line $\\mathrm{AF} \\perp \\mathrm{CD}$ through point $\\mathrm{A}$, intersecting the extension of $\\mathrm{CD}$ at point $\\mathrm{F}$, and connect $\\mathrm{AC}$.\n\nThen, $\\angle \\mathrm{ADF} + \\angle \\mathrm{ADC} = 180^\\circ$.\n\nSince $\\angle \\mathrm{ABC} + \\angle \\mathrm{ADC} = 180^\\circ$,\n\nit follows that $\\angle \\mathrm{ABC} = \\angle \\mathrm{ADF}$.\n\nIn triangles $\\triangle \\mathrm{ABE}$ and $\\triangle \\mathrm{ADF}$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle ABC = \\angle ADF \\\\\n\\angle AEB = \\angle F = 90^\\circ \\\\\nAB = AD\n\\end{array}\n\\right.\n\\]\n\nTherefore, $\\triangle \\mathrm{ABE} \\cong \\triangle \\mathrm{ADF}$ (by AAS),\n\nand $\\mathrm{AF} = \\mathrm{AE} = 19$.\n\nThus, the area of quadrilateral $\\mathrm{ABCD}$ is the sum of the areas of triangles $\\triangle \\mathrm{ABC}$ and $\\triangle \\mathrm{ACD}$:\n\n\\[\n\\mathrm{S}_{\\text{quadrilateral } \\mathrm{ABCD}} = \\frac{1}{2} \\times \\mathrm{BC} \\times \\mathrm{AE} + \\frac{1}{2} \\times \\mathrm{CD} \\times \\mathrm{AF} = \\frac{1}{2} \\times 10 \\times 19 + \\frac{1}{2} \\times 6 \\times 19 = 95 + 57 = 152.\n\\]\n\n**Final Answer:** 152.\n\n**Insight:** This problem examines the properties of congruent triangles and squares. The key to solving part (1) is constructing auxiliary lines to form congruent triangles, while part (2) involves constructing auxiliary lines to form congruent triangles and dividing the quadrilateral into two triangles." }, { "problem_id": 1260, "question": "Teacher Wu taught Xiaoqing to tie a small piece of paper into a knot during the break, as shown in Figure (1). Then, by gently pulling it tight and flattening it, a regular pentagon $A B C D E$ as shown in Figure (2) is obtained. Therefore, $\\angle C A E =$ $\\qquad$ degrees.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0056_1.jpg", "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0056_2.jpg" ], "is_multi_img": true, "answer": "72", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the problem statement, we can derive:\n$\\angle CBA = \\angle BAE = (5-3) \\times 180^{\\circ} \\div 5 = 108^{\\circ}$, and $BA = BC$.\n\nTherefore, $\\angle BCA = \\angle BAC = \\frac{1}{2}(180^{\\circ} - \\angle CBA) = 36^{\\circ}$.\n\nThus, $\\angle CAE = \\angle BAE - \\angle BAC = 72^{\\circ}$.\n\nThe answer is: 72\n\n[Highlight] This problem tests the properties of a regular pentagon and the properties of an isosceles triangle. The key to solving the problem lies in understanding the properties of a regular pentagon and determining the measure of each angle." }, { "problem_id": 1261, "question": "The Yong Temple Twin Pagodas, also known as the Lingxiao Twin Pagodas (as shown in Figure 1), are the tallest ancient structures in Taiyuan City. Both are thirteen-story octagonal pavilion-style brick pagodas. The regular octagon shown in Figure 2 is a schematic plan of the pagoda's base. The sum of the interior angles of this regular octagon is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0065_1.jpg", "batch13-2024_06_15_d2ee16d5511cfb9a75e3g_0065_2.jpg" ], "is_multi_img": true, "answer": "$1080^{\\circ} / 1080$ ^{\\circ}$", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The sum of the interior angles of a regular octagon is $(8-2) \\cdot 180^{\\circ}=1080^{\\circ}$.\n\nTherefore, the answer is: $1080^{\\circ}$.\n\n[Key Insight] This question tests the understanding of the sum of interior angles of a polygon. Memorizing the formula for the sum of interior angles is crucial for solving such problems." }, { "problem_id": 1262, "question": "As shown in Figure 1, Xiaoqing placed a regular pentagon and a regular hexagon, both lying flat on a table, together (with one side overlapping). The schematic diagram is shown in Figure 2. The degree measure of the formed $\\angle A B C$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_d5eba716f8dc5b323207g_0098_1.jpg", "batch13-2024_06_15_d5eba716f8dc5b323207g_0098_2.jpg" ], "is_multi_img": true, "answer": "$132^{\\circ} / 132$ ^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In a regular pentagon,\n\n$\\angle A B D=\\frac{(5-2) \\times 180^{\\circ}}{5}=108^{\\circ}$,\n\n\n\nFigure 2\n\nIn a regular hexagon,\n\n$\\angle C B D=\\frac{(6-2) \\times 180^{\\circ}}{6}=120^{\\circ}$,\n\n$\\therefore \\angle A B C=360^{\\circ}-\\angle A B D-\\angle C B D=360^{\\circ}-120^{\\circ}-108^{\\circ}=132^{\\circ}$,\n\nTherefore, the answer is: $132^{\\circ}$.\n\n【Highlight】This question tests the understanding of the interior angles of polygons. Remembering the formula for the sum of interior angles of a polygon is key to solving the problem." }, { "problem_id": 1263, "question": "The ancient mathematician Jia Xian proposed the theorem: \"From any point on the diagonal of a rectangle, draw two lines parallel to the two adjacent sides respectively, then the areas of the two resulting rectangles are equal\" (as shown in Figure 1, $S_{\\text {rectangle DNFG }} = S_{\\text {rectangle EBM }}$). Problem solving: As shown in Figure 2, point $\\mathrm{P}$ is a point on the diagonal $B D$ of rectangle $A B C D$. Through point $P$, draw $E F / / B C$ intersecting $A B$ and $C D$ at points $E$ and $F$ respectively, and connect $A P$ and $C P$. If $D F = 4$ and $E P = 3$, then the sum of the areas of the shaded parts in the figure is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_faf53d78527724b96a79g_0050_1.jpg", "batch13-2024_06_15_faf53d78527724b96a79g_0050_2.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:** Draw \\( P M \\perp A D \\) intersecting \\( A D \\) at \\( M \\) and \\( B C \\) at \\( N \\). As shown in the figure:\n\n\n\nThen, quadrilaterals \\( A E P M \\), \\( D F P M \\), \\( C F P N \\), and \\( B E P N \\) are all rectangles.\n\n\\[\n\\therefore P M = D F = 4.\n\\]\n\nFrom the properties of rectangles, we know that:\n\\[\nS_{A B D} = S_{C B D}, \\quad S_{\\triangle E B P} = S_{\\triangle N B P}, \\quad S_{\\triangle M P D} = S_{\\triangle F P D}.\n\\]\n\nFurthermore, since:\n\\[\nS_{\\text{rectangle } A E P M} = S_{\\triangle A B D} - S_{\\triangle E B P} - S_{\\triangle M P D}, \\quad S_{\\text{rectangle } C F P N} = S_{\\triangle C B D} - S_{\\triangle N B P} - S_{\\triangle F P D},\n\\]\n\\[\n\\therefore S_{\\text{rectangle } A E P M} = S_{\\text{rectangle } C F P N}.\n\\]\n\nThus:\n\\[\nS_{\\triangle A E P} = S_{\\triangle A M P}, \\quad S_{\\triangle C F P} = S_{\\triangle C N P}.\n\\]\n\\[\n\\therefore S_{\\triangle A E P} = S_{\\triangle C F P} = \\frac{1}{2} \\times P E \\times P M = \\frac{1}{2} \\times 3 \\times 4 = 6.\n\\]\n\nTherefore, the area of the shaded region in the figure is:\n\\[\nS_{\\text{shaded}} = 6 + 6 = 12.\n\\]\n\n**Answer:** \\(\\boxed{12}\\).\n\n**Key Insight:** This problem examines the properties of rectangles and the area of triangles. The key to solving it lies in proving that \\( S_{\\text{rectangle } A E P M} = S_{\\text{rectangle } C F P N} \\)." }, { "problem_id": 1264, "question": "Uncle Wang needs to make a ladder as shown in Figure 1. The ladder has 8 parallel steps, and the distance between each pair of adjacent steps is equal. It is known that the length of the topmost step $A_{1} B_{1}=0.5 \\mathrm{~m}$, and the length of the bottommost step $A_{8} B_{8}=0.8 \\mathrm{~m}$. When making these steps, the carpenter cuts the wooden boards longer than the steps to ensure that a tenon of $4 \\mathrm{~cm}$ (as shown in Figure 2) can be made at each end of every step for fixing the steps. Now, there are wooden boards of $2.1 \\mathrm{~m}$ in the market that can be used to make the steps of the ladder (the width and thickness of the boards are exactly suitable for making the ladder steps). How many such boards does Uncle Wang need to buy at least to make these steps? Please explain the reason. (The loss from sawing is not considered)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch13-2024_06_15_fcc693676e07de114f23g_0037_1.jpg", "batch13-2024_06_15_fcc693676e07de114f23g_0037_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Solution 1:**\n\nAs shown in the figure, let the lengths of the steps from top to bottom be \\( A_{2} B_{2}, A_{3} B_{3}, \\ldots, A_{7} B_{7} \\).\n\nDraw a line through \\( A_{1} \\) parallel to \\( B_{1} B_{8} \\), intersecting \\( A_{2} B_{2}, A_{3} B_{3}, \\ldots, A_{8} B_{8} \\) at points \\( C_{2}, C_{3}, \\ldots, C_{8} \\).\n\n\n\nSince the distance between every two steps is equal, \\( C_{8} B_{8} = C_{7} B_{7} = \\ldots = C_{2} B_{2} = A_{1} B_{1} = 50 \\text{ cm} \\), and \\( A_{8} C_{8} = 80 - 50 = 30 \\text{ cm} \\).\n\nBecause \\( A_{2} C_{2} \\parallel A_{8} B_{8} \\), \\( \\angle A_{1} A_{2} C_{2} = \\angle A_{1} A_{8} C_{8} \\) and \\( \\angle A_{1} C_{2} A_{2} = \\angle A_{1} C_{8} A_{8} \\).\n\nThus, \\( \\triangle A_{1} A_{2} C_{2} \\sim \\triangle A_{1} A_{8} C_{8} \\), so \\( A_{2} C_{2} : A_{8} C_{8} = 1 : 7 \\), hence \\( A_{2} C_{2} = \\frac{30}{7} \\), and \\( A_{2} B_{2} = 50 + \\frac{30}{7} \\).\n\nLet the lengths of the boards required to make steps \\( A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{7} B_{7}, A_{8} B_{8} \\) be \\( a_{1} \\text{ cm}, a_{2} \\text{ cm}, \\ldots, a_{8} \\text{ cm} \\), then:\n\n\\( a_{1} = 50 + 8 = 58 \\), \\( a_{2} = 50 + \\frac{30}{7} + 8 = 58 + \\frac{30}{7} \\), \\( a_{3} = 58 + \\frac{60}{7} \\), \\( a_{4} = 58 + \\frac{90}{7} \\),\n\n\\( a_{5} = 58 + \\frac{120}{7} \\), \\( a_{6} = 58 + \\frac{150}{7} \\), \\( a_{7} = 58 + \\frac{180}{7} \\), \\( a_{8} = 58 + 30 \\).\n\nSince \\( a_{1} + a_{2} + a_{3} + a_{4} = 232 + \\frac{180}{7} > 210 \\),\n\nUncle Wang must buy at least 3 boards.\n\nAlso, since \\( a_{1} + a_{3} + a_{6} = 174 + \\frac{210}{7} = 204 < 210 \\), \\( a_{2} + a_{4} + a_{5} = 174 + \\frac{270}{7} < 174 + \\frac{252}{7} = 210 \\),\n\nand \\( a_{7} + a_{8} = 146 + \\frac{180}{7} = 171 \\frac{5}{7} < 210 \\),\n\nUncle Wang needs at least 3 boards.\n\n**Solution 2:**\n\nAs shown in the figure, let the midpoints of \\( A_{1} A_{8} \\) and \\( B_{1} B_{8} \\) be \\( P \\) and \\( Q \\), respectively, and connect \\( P Q \\).\n\n\n\nLet the lengths of the steps from top to bottom be \\( A_{2} B_{2}, A_{3} B_{3}, \\ldots, A_{7} B_{7} \\). By the trapezoid midline theorem,\n\n\\( A_{1} B_{1} + A_{8} B_{8} = A_{2} B_{2} + A_{7} B_{7} = A_{3} B_{3} + A_{6} B_{6} = A_{4} B_{4} + A_{5} B_{5} = 2 P Q \\).\n\nSince \\( A_{1} B_{1} = 50 \\text{ cm} \\) and \\( A_{8} B_{8} = 80 \\text{ cm} \\),\n\n\\( A_{1} B_{1} + A_{8} B_{8} = A_{2} B_{2} + A_{7} B_{7} = A_{3} B_{3} + A_{6} B_{6} = A_{4} B_{4} + A_{5} B_{5} = 130 \\).\n\nLet the lengths of the boards required to make steps \\( A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{7} B_{7}, A_{8} B_{8} \\) be \\( a_{1} \\text{ cm}, a_{2} \\text{ cm}, \\ldots, a_{8} \\text{ cm} \\), then:\n\n\\( a_{1} + a_{8} = a_{2} + a_{7} = a_{3} + a_{6} = a_{4} + a_{5} = 146 \\).\n\nSince \\( a_{1} + a_{2} + \\ldots + a_{8} = 146 \\times 4 = 584 > 210 \\times 2 \\), Uncle Wang must buy at least 3 boards.\n\nDraw a line through \\( A_{1} \\) parallel to \\( B_{1} B_{8} \\), intersecting \\( A_{2} B_{2}, A_{3} B_{3}, \\ldots, A_{8} B_{8} \\) at points \\( C_{2}, C_{3}, \\ldots, C_{8} \\).\n\nSince the distance between every two steps is equal, \\( C_{8} B_{8} = C_{7} B_{7} = \\ldots = C_{2} B_{2} = A_{1} B_{1} = 50 \\text{ cm} \\),\n\nand \\( A_{8} C_{8} = 80 - 50 = 30 \\text{ cm} \\). Because \\( A_{2} C_{2} \\parallel A_{8} B_{8} \\), \\( \\angle A_{1} A_{2} C_{2} = \\angle A_{1} A_{8} C_{8} \\) and \\( \\angle A_{1} C_{2} A_{2} = \\angle A_{1} C_{8} A_{8} \\),\n\nthus \\( \\triangle A_{1} A_{2} C_{2} \\sim \\triangle A_{1} A_{8} C_{8} \\), so \\( A_{2} C_{2} : A_{8} C_{8} = 1 : 7 \\), hence \\( A_{2} C_{2} = \\frac{30}{7} \\), and \\( A_{2} B_{2} = 50 + \\frac{30}{7} \\).\n\nThus, \\( a_{2} = 58 + \\frac{30}{7} \\). Since \\( a_{1} = 58 \\) and \\( a_{8} = 88 \\), \\( a_{1} + a_{3} + a_{6} = 58 + 146 = 204 < 210 \\),\n\n\\( a_{2} + a_{4} + a_{5} = 58 + \\frac{30}{7} + 146 = 204 + \\frac{30}{7} < 210 \\), and \\( a_{7} + a_{8} < a_{8} + a_{8} = 88 \\times 2 < 210 \\).\n\nTherefore, Uncle Wang needs at least 3 boards.\n\n**Solution 3:**\n\nIf a 9th step is added at the bottom of the ladder, with the distance between it and the step above equal to the distance between any two steps (as shown in the figure),\n\n\n\nLet the length of the 9th step be \\( x \\) cm. By the trapezoid midline property, the length of the 5th step \\( A_{5} B_{5} = \\frac{1}{2}(50 + x) \\text{ cm} \\),\n\nthe length of the 7th step \\( A_{7} B_{7} = \\frac{1}{2}\\left(\\frac{1}{2}(50 + x) + x\\right) \\text{ cm} \\). According to the problem, the length of the 8th step\n\n\\( A_{8} B_{8} = \\frac{1}{2}\\left(\\frac{1}{2}\\left(\\frac{1}{2}(50 + x) + x\\right) + x\\right) = 80 \\). Solving this equation gives \\( x = 84 \\frac{2}{7} \\),\n\nfrom which we can find \\( A_{7} B_{7} = 75 \\frac{5}{7} \\text{ cm} \\), \\( A_{5} B_{5} = 67 \\frac{1}{7} \\text{ cm} \\), \\( A_{6} B_{6} = 71 \\frac{3}{7} \\text{ cm} \\), \\( A_{3} B_{3} = 58 \\frac{4}{7} \\text{ cm} \\), \\( A_{2} B_{2} = 54 \\frac{2}{7} \\text{ cm} \\), and \\( A_{4} B_{4} = 62 \\frac{6}{7} \\text{ cm} \\).\n\nLet the lengths of the boards required to make steps \\( A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{7} B_{7}, A_{8} B_{8} \\) be \\( a_{1} \\text{ cm}, a_{2} \\text{ cm}, \\ldots, a_{8} \\text{ cm} \\), then:\n\n\\( a_{1} = 50 + 8 = 58 \\), \\( a_{2} = 62 \\frac{2}{7} \\), \\( a_{3} = 66 \\frac{2}{7} \\), \\( a_{4} = 70 \\frac{6}{7} \\), \\( a_{5} = 75 \\frac{1}{7} \\), \\( a_{6} = 79 \\frac{3}{7} \\),\n\n\\( a_{7} = 83 \\frac{5}{7} \\), \\( a_{8} = 88 \\)." }, { "problem_id": 1265, "question": "How many white hexagonal floor tiles are there in the $\\mathrm{n}$th pattern formed by arranging hexagonal floor tiles of two colors, black and white, as shown in the following figures?\n\n\n\n1st pattern\n\n\n\n2nd pattern\n\n\n\n3rd pattern", "input_image": [ "batch14-2024_06_15_2a959295627ef1c93e60g_0053_1.jpg", "batch14-2024_06_15_2a959295627ef1c93e60g_0053_2.jpg", "batch14-2024_06_15_2a959295627ef1c93e60g_0053_3.jpg" ], "is_multi_img": true, "answer": "$(4 n+2)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "In the first pattern, there are 6 white floor tiles,\n\nIn the second pattern, there are 10 white floor tiles,\n\nIn the third pattern, there are 14 white floor tiles,\n\nIt can be deduced that the nth pattern has $(4n + 2)$ white floor tiles,\n\nTherefore, the answer is $(4n + 2)$.\n\n【Highlight】This question tests the pattern recognition type: changes in graphic patterns, plane tiling (tessellation), with the key to solving the problem lying in identifying the pattern." }, { "problem_id": 1266, "question": "Given a set of right triangles as shown in Figure (1), where \\( BC = 3 \\) and \\( EF = 4 \\), the \\( 30^\\circ \\) triangle is translated to the right such that vertex \\( B \\) coincides with the hypotenuse \\( DF \\) of the \\( 45^\\circ \\) triangle. The area of the overlapping region (shaded area) of the two triangles is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_495cccfc1a8675706da4g_0009_1.jpg", "batch14-2024_06_15_495cccfc1a8675706da4g_0009_2.jpg" ], "is_multi_img": true, "answer": "$3-\\frac{\\sqrt{3}}{6}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "After translation, let $AB$ and $DE$ intersect at point $G$, as shown in the solution diagram. Since $\\angle F=45^{\\circ}$ and $BC=3$, it follows that $CF=3$. \n\nGiven that $EF=4$, then $EC=1$. Since $BC=3$ and $\\angle A=30^{\\circ}$, it follows that $AC=3 \\sqrt{3}$, hence $AE=3 \\sqrt{3}-1$. Therefore, $EG=3-\\frac{\\sqrt{3}}{3}$.\n\nThus, the area of the shaded region $S_{\\text{shaded}}$ is calculated as:\n\\[\nS_{\\text{shaded}} = S_{\\triangle ABC} - S_{\\triangle AGE} = \\frac{1}{2} \\times 3 \\sqrt{3} \\times 3 - \\frac{1}{2} \\times (3 \\sqrt{3} - 1) \\times \\left(3 - \\frac{\\sqrt{3}}{3}\\right) = 3 - \\frac{\\sqrt{3}}{6}.\n\\]\n\n" }, { "problem_id": 1267, "question": "As shown in Figure (1), in quadrilateral $A B C D$, $A D / / B C$, and line $l \\perp A B$. As line $l$ moves rightward from point $B$ along the direction of ray $B C$, it intersects the sides of quadrilateral $A B C D$ at points $E, F$. Let the distance of line $l$ moving rightward be $x$, and the length of segment $E F$ be $y$, with the functional relationship between $y$ and $x$ shown in Figure (2). Then the perimeter of quadrilateral $A B C D$ is $\\qquad$\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_495cccfc1a8675706da4g_0037_1.jpg", "batch14-2024_06_15_495cccfc1a8675706da4g_0037_2.jpg" ], "is_multi_img": true, "answer": "$12+2 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Through points A, C, and D, draw lines parallel to line l, extend BC to intersect line c at point F. Let line a intersect BC at point M, and line b intersect AD at point N.\n\n(1) When line 1 is in the position of line a, AM = EF = 2, BM = 4.\n\nSince ∠BAM = 90°,\n\nTherefore, sin B = AM / BM = 1/2,\n\nThus, ∠B = 30°, ∠BMA = ∠DFC = 60°;\n\nTherefore, AB = BM * cos 30° = 2√3,\n\n(2) When line 1 moves from line a to line b, MC = AN = 6 - 4 = 2, so BC = MB + MC = 4 + 2 = 6,\n\n(3) When line 1 reaches the position of line c, CF = ND = 8 - 6 = 2, then AD = AN + ND = 2 + 2 = 4,\n\nSince ∠DCF = 60°, CF = DF = EF = 2,\n\nTherefore, triangle CDF is an equilateral triangle,\n\nThus, CD = 2,\n\nThe perimeter of quadrilateral ABCD = AB + AD + BC + CD = 2√3 + 4 + 6 + 2 = 12 + 2√3,\n\n\n\nTherefore, the answer is: 12 + 2√3\n\n[Highlight] This question examines the function images of moving point problems, using knowledge points such as the definition of trigonometric functions, properties of parallelograms, and trigonometric values of special angles. Determining the lengths of each segment based on the function images is key to solving the problem." }, { "problem_id": 1268, "question": "Given that rulers A and B have different scales, and the distances between the scales on the same ruler are equal. Xiao Ming aligns the two rulers closely, and after aligning the 0 scales of both rulers, he finds that the 36 scale on ruler A aligns with the 48 scale on ruler B. If ruler A is moved to the right while maintaining close alignment with ruler B, such that the 0 scale on ruler A aligns with the m scale on ruler B, then at this point, the n scale on ruler A will align with the scale on ruler B as $\\qquad$. (expressed in terms of m and n)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_495cccfc1a8675706da4g_0072_1.jpg", "batch14-2024_06_15_495cccfc1a8675706da4g_0072_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{4}{3} n+m$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: If the 0 mark on ruler A is aligned with the mark $m$ on ruler B, and at this time the mark $n$ on ruler A aligns with the mark $x$ on ruler B, then according to the problem statement, we have:\n\n$36(x - m) = n \\cdot 48$,\n\nSolving for $x$, we get: $\\frac{4}{3} n + m$.\nTherefore, the answer is: $\\frac{4}{3} n + m$.\n\n[Highlight] This problem examines the properties of translation and the application of linear equations. The key to solving it is to understand the problem's requirements, identify the appropriate relationships based on the given conditions, formulate the equation, and then solve it." }, { "problem_id": 1269, "question": "A rectangular floor with a length of $a(\\mathrm{~cm})$ and a width of $b(\\mathrm{~cm})$ has two cracks in the middle (as shown in Figure A). If after moving, the two cracks are $1 \\mathrm{~cm}$ apart (as shown in Figure B), the area of the cracks produced is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure A\n\n\n\nFigure B", "input_image": [ "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0031_1.jpg", "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0031_2.jpg" ], "is_multi_img": true, "answer": "$a+b+1$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, the area of the rectangle in Diagram A is $ab$.\n\nTherefore, the area of the rectangle in Diagram B is $(a+1)(b+1) = ab + a + b + 1$.\n\nThus, the area of the gap created is $(a+1)(b+1) - ab = ab + a + b + 1 - ab = a + b + 1$ (square centimeters).\nHence, the answer is: $a + b + 1$.\n\n[Key Insight] This problem examines the properties of translation and the multiplication of polynomials. The key is to recognize the relationship: the area of the gap created equals the area of the rectangle in Diagram B minus the area of the rectangle in Diagram A, and to solve accordingly." }, { "problem_id": 1270, "question": "A rectangular wooden board with a length of $25 \\mathrm{~cm}$ and a width of $15 \\mathrm{~cm}$ has a crack in the middle (as shown in Figure A). If the piece on the right side of the crack is moved $2 \\mathrm{~cm}$ to the right (as shown in Figure B), the area of the resulting crack is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure A\n\n\n\nFigure B", "input_image": [ "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0032_1.jpg", "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0032_2.jpg" ], "is_multi_img": true, "answer": "30", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: The area of the crack produced is:\n\n$(25 + 2) \\times 15 - 25 \\times 15$\n\n$= (27 - 2) \\times 15$\n\n$= 30 \\left(\\mathrm{~cm}^{2}\\right)$.\n\nTherefore, the answer is: 30.\n\n[Highlight] This question mainly examines the phenomenon of translation in daily life. The key to solving the problem lies in using the difference in areas of two rectangles to determine the area of the crack." }, { "problem_id": 1271, "question": "As shown in Figure 1, in the plane rectangular coordinate system, the parallelogram $A B C D$ is placed in the first quadrant, with $A B / / x$-axis. The line $y=-x$ starts from the origin and translates along the positive direction of the $x$-axis. During the translation, the length $l$ of the segment intercepted by the parallelogram on the line and the distance $m$ of the line's translation on the $x$-axis form a function graph as shown in Figure 2. Then, the length of $A B$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0075_1.jpg", "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0075_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in Figure 1, when the straight line is to the lower left of $DE$, from Figure 2, we get: $AE = 7 - 4 = 3$; from Figure 1, when the straight line is between $DE$ and $BF$, from Figure 2, we can derive: $EB = 8 - 7 = 1$, so $AB = AE + EB = 3 + 1 = 4$.\n\nTherefore, the answer is: 4.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Insight】This question examines the graph of a linear function and the translation of shapes, as well as the properties of a parallelogram. The key is to understand the problem clearly, interpret the function graph correctly, and apply the concept of combining numbers and shapes." }, { "problem_id": 1272, "question": "As shown in Figure (1), equilateral triangles \\(ABC\\) and \\(DEF\\) with side lengths of 4 overlap each other. Now, \\(ABC\\) is translated \\(m\\) units to the left along line \\(l\\), and \\(DEF\\) is translated \\(m\\) units to the right along line \\(l\\), as shown in Figure (2). When \\(E\\) and \\(C\\) are the trisection points of segment \\(BF\\), the value of the translation distance \\(m\\) is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0097_1.jpg", "batch14-2024_06_15_5b7a08d64c1aedd10f95g_0097_2.jpg" ], "is_multi_img": true, "answer": "1or 4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "$E$ and $C$ are the trisection points of segment $B F$, divided into two cases:\n\n(1) When point $E$ is to the left of point $C$, as shown in Figure 1.\n\n\n\nFigure 1\n\nSince $E$ and $C$ are the trisection points of segment $B F$,\n\n$\\therefore B E = E C = C F$,\n\nGiven that $B C = 4$ and $B E = 2 m$,\n\n$\\therefore 2 m = 4 \\div 2$, solving gives: $m = 1$;\n\n(2) When point $E$ is to the right of point $C$, as shown in Figure 2.\n\n\n\nFigure 2\n\nSince $E$ and $C$ are the trisection points of segment $B F$,\n\n$\\therefore B C = C E = E F$,\n\nGiven that $B C = 4$ and $B E = 2 m$,\n\n$\\therefore 2 m = 4 \\times 2$, solving gives: $m = 4$.\n\nIn summary, when $E$ and $C$ are the trisection points of segment $B F$, the value of $m$ is either 1 or 4.\n\nTherefore, the answer is: 1 or 4.\n\n【Highlight】This question examines the properties of translation, and the key to solving it lies in mastering the relevant knowledge." }, { "problem_id": 1273, "question": "Exploration: As shown in Figure (1), in square \\( A B C D \\), point \\( P \\) is on side \\( C D \\) (not coinciding with points \\( C \\) and \\( D \\)), connecting \\( B P \\), and rotating \\(\\triangle B C P \\) clockwise around point \\( C \\) to \\(\\triangle D C E \\), where the corresponding point of \\( B \\) is \\( D \\). The angle of rotation is \\(\\qquad\\) degrees. Application: Extend \\( B P \\) to intersect side \\( D E \\) at point \\( F \\) in Figure (1), with other conditions unchanged, as shown in Figure (2). Find the measure of \\(\\angle B F E \\). Extension: As shown in Figure (2), if \\( D P = 2 C P \\) and \\( B C = 6 \\), then the area of quadrilateral \\( A B E D \\) is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_64bd0b69c3d58d2c039bg_0098_1.jpg", "batch14-2024_06_15_64bd0b69c3d58d2c039bg_0098_2.jpg" ], "is_multi_img": true, "answer": "Exploration: 90 ; Application: $90^{\\circ}$; Extension: 42", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Exploration:** From the property of rotation, the rotation angle \\(=\\angle \\mathrm{BCD}=\\angle \\mathrm{DCE}=90^{\\circ}\\);\n\nTherefore, the answer is \\(90^{\\circ}\\);\n\n**Application:** By rotation, we have \\(\\triangle \\mathrm{BCP} \\cong \\triangle \\mathrm{DCE}\\).\n\nThus, \\(\\angle \\mathrm{CDE}=\\angle \\mathrm{PBC}\\), \\(\\angle \\mathrm{DCE}=\\angle \\mathrm{BCP}=90^{\\circ}\\),\n\nSo, \\(\\angle \\mathrm{CDE}+\\angle \\mathrm{E}=90^{\\circ}\\),\n\nHence, \\(\\angle \\mathrm{PBC}+\\angle \\mathrm{E}=90^{\\circ}\\),\n\nTherefore, \\(\\angle \\mathrm{BFE}=90^{\\circ}\\);\n\n**Extension:** Since \\(\\Delta \\mathrm{BCP} \\cong \\triangle \\mathrm{DCE}\\),\n\nWe have \\(\\mathrm{CE}=\\mathrm{PC}\\),\n\nGiven that \\(\\mathrm{DP}=2 \\mathrm{CP}\\) and \\(\\mathrm{BC}=6\\),\n\nThus, \\(\\mathrm{CE}=2\\),\n\nTherefore, the area of quadrilateral \\(\\mathrm{ABED}\\) is \\(\\mathrm{S}_{\\text{quadrilateral } \\mathrm{ABED}}=\\mathrm{S}_{\\text{square } \\mathrm{ABCD}}+\\mathrm{S} \\triangle \\mathrm{CDE}=6 \\times 6+\\frac{1}{2} \\times 6 \\times 2=36+6=42\\),\n\nHence, the answer is 42.\n\n**[Insight]** This problem is a comprehensive quadrilateral question, primarily examining the properties of rotation, the properties of squares, and the area of triangles. The key to solving the problem lies in utilizing the properties of rotation to obtain the corresponding equal sides of congruent triangles." }, { "problem_id": 1274, "question": "A set of triangles is placed as shown in Figure 1, where $O$ is the midpoint of side $BC(DF)$, and $BC = 20 \\mathrm{~cm}$. As shown in Figure 2, triangle $ABC$ is rotated $60^\\circ$ clockwise around point $O$, and $AC$ intersects $EF$ at point $G$. The length of $FG$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_6981e3f4052e55f14c2ag_0096_1.jpg", "batch14-2024_06_15_6981e3f4052e55f14c2ag_0096_2.jpg" ], "is_multi_img": true, "answer": "$(5 \\sqrt{3}-5) \\mathrm{cm}$\n", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, line $BC$ intersects line $EF$ at point $N$.\n\nFrom the given conditions, we have $\\angle EDF = \\angle BAC = 90^\\circ$, $\\angle DEF = 60^\\circ$, $\\angle DFE = 30^\\circ$, $\\angle ABC = \\angle ACB = 45^\\circ$, and $BC = DF = 20 \\text{ cm}$.\n\nSince point $O$ is the midpoint of side $BC$ (or $DF$), it follows that $BO = OC = DO = FO = 10 \\text{ cm}$.\n\nBecause triangle $ABC$ is rotated $60^\\circ$ clockwise around point $O$, and $\\angle DFE = 30^\\circ$,\n\nwe have $\\angle BOD = \\angle NOF = 60^\\circ$.\n\nThus, $\\angle NOF + \\angle F = 90^\\circ$,\n\nand $\\angle FNO = 180^\\circ - \\angle NOF - \\angle F = 90^\\circ$.\n\nTherefore, triangle $ONF$ is a right-angled triangle,\n\nand $ON = \\frac{1}{2} OF = 5 \\text{ cm}$.\n\nHence, $FN = \\sqrt{OF^2 - ON^2} = 5\\sqrt{3} \\text{ cm}$,\n\nand $NC = OC - ON = 5 \\text{ cm}$.\n\nSince $\\angle FNO = 90^\\circ$ and $\\angle ACB = 45^\\circ$,\n\nwe have $\\angle GNC = 180^\\circ - \\angle FNO = 90^\\circ$.\n\nTherefore, triangle $CNG$ is a right-angled triangle,\n\nand $\\angle NGC = 180^\\circ - \\angle GNC - \\angle ACB = 45^\\circ$.\n\nThus, triangle $CNG$ is an isosceles right-angled triangle,\n\nand $NG = NC = 5 \\text{ cm}$.\n\nTherefore, $FG = FN - NG = (5\\sqrt{3} - 5) \\text{ cm}$.\n\nThe final answer is: $(5\\sqrt{3} - 5) \\text{ cm}$.\n\n\n\n【Key Insight】This problem examines the properties of a right-angled triangle with a $30^\\circ$ angle, the determination and properties of an isosceles triangle, the properties of rotation, and the Pythagorean theorem. The key to solving the problem lies in understanding the properties of a right-angled triangle with a $30^\\circ$ angle and the inherent angles in a set square." }, { "problem_id": 1275, "question": "As shown in the figure, place a set of triangular plates as shown in Figure A, where $\\angle ACB = \\angle DEC = 90^\\circ, \\angle A = 45^\\circ, \\angle D = 30^\\circ$, the hypotenuse $AB = 6 \\, \\text{cm}, DC = 7 \\, \\text{cm}$. Rotate the triangular plate $DCE$ around point $C$ clockwise by $15^\\circ$ to obtain $\\triangle D'CE'$ (as shown in Figure B). At this time, $AB$ intersects $CD'$ at point $O$, and $D'E'$ intersects $AB$ at point $F$. The degree measure of $\\angle OFE'$ is $\\qquad$.\n\n\n\n(A)\n\n", "input_image": [ "batch14-2024_06_15_6d0f54a5f093ea90d2fag_0006_1.jpg", "batch14-2024_06_15_6d0f54a5f093ea90d2fag_0006_2.jpg" ], "is_multi_img": true, "answer": "$120^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "As shown in the figure, according to the problem statement, $\\angle 3=15^{\\circ}$ and $\\angle \\mathrm{E}^{\\prime}=90^{\\circ}$.\n\nSince $\\angle 1=\\angle 2$,\n\nit follows that $\\angle 1=75^{\\circ}$.\n\nMoreover, since $\\angle \\mathrm{B}=45^{\\circ}$,\n\nwe have $\\angle \\mathrm{OFE}^{\\prime}=\\angle \\mathrm{B}+\\angle 1=45^{\\circ}+75^{\\circ}=120^{\\circ}$.\n\n\n\n(Part B)\n\n【Key Insight】This problem examines the rotation of figures, and the key to solving it lies in understanding the properties of rotation." }, { "problem_id": 1276, "question": "As shown in Figure (1), in the plane rectangular coordinate system, parallelogram $A B C D$ is located in the first quadrant. The line $y=-x$ starts from the origin and translates along the positive direction of the $x$-axis. The length $l$ of the segment $E F$ intercepted by the parallelogram $A B C D$ and the function image of the translation distance $m$ are shown in Figure (2). Then, the area of the parallelogram is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_6e5df1e767dabb33ef16g_0019_1.jpg", "batch14-2024_06_15_6e5df1e767dabb33ef16g_0019_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "From the graph, we can determine that when the distance moved is 4, the line passes through point $A$, and when the distance moved is 7, the line passes through point $D$. When the distance moved is 8, the line passes through point $B$. Therefore, the length of $AB$ is $8 - 4 = 4$.\n\nWhen the line passes through point $D$, let it intersect $AB$ at point $N$. Then, $DN = 2\\sqrt{2}$. Draw $DM \\perp AB$ at point $M$.\n\n\n\nSince the angle formed by $y = -x$ and the $x$-axis is $45^\\circ$, and because $AB$ is parallel to the $x$-axis, it follows that $\\angle DNM = 45^\\circ$. Therefore, $DM = DN \\cdot \\sin 45^\\circ = 2\\sqrt{2} \\times \\frac{\\sqrt{2}}{2} = 2$.\n\nThe area of the parallelogram is then: $AB \\cdot DM = 4 \\times 2 = 8$.\n\nThus, the answer is: 8." }, { "problem_id": 1277, "question": "As shown, the local parts of a square with a side length of 2 are transformed as shown in (Figure (1) to Figure (4)), and then pieced together to form Figure (5). The area of Figure (5) is $\\qquad$ .\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\n\n\n(5)", "input_image": [ "batch14-2024_06_15_6e5df1e767dabb33ef16g_0027_1.jpg", "batch14-2024_06_15_6e5df1e767dabb33ef16g_0027_2.jpg", "batch14-2024_06_15_6e5df1e767dabb33ef16g_0027_3.jpg", "batch14-2024_06_15_6e5df1e767dabb33ef16g_0027_4.jpg" ], "is_multi_img": true, "answer": "16", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "A square has an area of 4. When a square is transformed from (1) to (4), the area remains unchanged. Figure (5) is composed of 4 figures of type (4), therefore the area of figure (5) is $4 \\times 4=16$." }, { "problem_id": 1278, "question": "As shown in Figure 1, in the plane rectangular coordinate system, parallelogram $\\mathrm{ABCD}$ is placed in the first quadrant, with $\\mathrm{AB} / / \\mathrm{x}$-axis. The line\n$\\mathrm{y}=-\\mathrm{x}$ starts from the origin and translates along the positive direction of the $\\mathrm{x}$-axis. During the translation, the length of the segment intercepted by the line on the parallelogram and the distance $\\mathrm{m}$ the line translates on the $\\mathrm{x}$-axis form a function graph as shown in Figure 2. Then, the length of $\\mathrm{AD}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_6e5df1e767dabb33ef16g_0098_1.jpg", "batch14-2024_06_15_6e5df1e767dabb33ef16g_0098_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{10}$ or $\\frac{5 \\sqrt{10}}{4}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "(1) When \\( AB > 4 \\), as shown in Figure 1,\n\n\n\nFrom the figure, we can see: \\( OE = 4 \\), \\( OF = 8 \\), \\( DG = 3\\sqrt{2} \\),\n\n\\[\n\\therefore EF = AG = OF - OE = 4\n\\]\n\n\\[\n\\because \\text{The equation of the line is: } y = -x\n\\]\n\n\\[\n\\therefore \\angle AGD = \\angle EFD = 45^\\circ\n\\]\n\n\\[\n\\therefore \\triangle AGD \\text{ is an isosceles right triangle}\n\\]\n\n\\[\n\\therefore DH = GH = \\frac{\\sqrt{2}}{2} DG = \\frac{\\sqrt{2}}{2} \\times 3\\sqrt{2} = 3\n\\]\n\n\\[\n\\therefore AH = AG - GH = 4 - 3 = 1\n\\]\n\n\\[\n\\therefore AD = \\sqrt{DH^2 + AH^2} = \\sqrt{3^2 + 1^2} = \\sqrt{10}\n\\]\n\n(2) When \\( AB = 4 \\), as shown in Figure 2,\n\n\n\nFrom the figure, we can see: \\( OI = 4 \\), \\( OJ = 8 \\), \\( KB = 3\\sqrt{2} \\), \\( OM = 9 \\),\n\n\\[\n\\therefore IJ = AB = 4, \\quad IM = AN = 5\n\\]\n\n\\[\n\\because \\text{The equation of the line is: } y = -x\n\\]\n\n\\[\n\\therefore \\triangle KLB \\text{ is an isosceles right triangle}\n\\]\n\n\\[\n\\therefore KL = BL = \\frac{\\sqrt{2}}{2} KB = 3\n\\]\n\n\\[\n\\because AB = 4\n\\]\n\n\\[\n\\therefore AL = AB - BL = 1\n\\]\n\nSimilarly to part (1), \\( DM = MN \\),\n\n\\[\n\\therefore \\text{Draw } KM \\parallel IM\n\\]\n\n\\[\n\\therefore \\tan \\angle DAN = \\frac{KL}{AL} = 3\n\\]\n\n\\[\n\\therefore AM = \\frac{DM}{\\tan \\angle DAN} = \\frac{DM}{3}\n\\]\n\n\\[\n\\therefore AN = AM + MN = \\frac{4}{3} DM = 5\n\\]\n\n\\[\n\\therefore DM = MN = \\frac{15}{4}\n\\]\n\n\\[\n\\therefore AM = AN - MN = 5 - \\frac{15}{4} = \\frac{5}{4}\n\\]\n\n\\[\n\\therefore AD = \\sqrt{AM^2 + DM^2} = \\frac{5\\sqrt{10}}{4}\n\\]\n\nTherefore, the answer is \\( \\sqrt{10} \\) or \\( \\frac{5\\sqrt{10}}{4} \\).\n\nKey points: This problem tests the properties of translation, trigonometric functions, the Pythagorean theorem, and the concept of classification and discussion. Correctly determining the height of the parallelogram is crucial." }, { "problem_id": 1279, "question": "In a grid paper composed of small squares with side length 1, each vertex of these small squares is called a \"lattice point\". A movement from one lattice point to another lattice point that is $\\sqrt{5}$ units away is called a \"knight's move\". For example, in a $3 \\times 3$ square grid (as shown in Figure (1)), from point $A$, a knight's move can reach points $B, C, D, E$, etc. Given a $25 \\times 25$ square grid (as shown in Figure (2)), the minimum number of \"knight's moves\" required to reach the opposite vertex $N$ from the vertex $M$ of the square is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_802ad286ff83dc4342dbg_0010_1.jpg", "batch14-2024_06_15_802ad286ff83dc4342dbg_0010_2.jpg" ], "is_multi_img": true, "answer": "18", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in Figure (1), connect $AC$ and $CF$, then $AF = 3\\sqrt{2}$.\n\nTherefore, the two transformations are equivalent to moving 3 units to the right and 3 units upwards. Since $MN = 25\\sqrt{2}$, and $25\\sqrt{2} \\div 3\\sqrt{2} = \\frac{25}{3}$ (which is not an integer),\n\nThus, after performing 14 consecutive transformations in the direction of $A-C-F$, it is equivalent to moving 21 units to the right and 21 units upwards (since $14 \\div 2 \\times 3 = 21$). At this point, $M$ is located at point $G$ on a $4 \\times 4$ square grid as shown in the figure. Then, by performing 4 more transformations as shown in Figure (2), it can reach point $N$.\n\nTherefore, the minimum number of \"knight transformations\" required to move from vertex $M$ of the square to its opposite vertex $N$ is $14 + 4 = 18$ times.\n\n\n\nFigure (1)\n\n\n\nFigure 2\n\n[Insight] This question examines the types of geometric transformations and the application of the Pythagorean theorem. When solving, note that under translation transformations, corresponding line segments are parallel and equal in length, and the line connecting two corresponding points is parallel (collinear) and equal in length to the given directed line segment. The key to solving the problem lies in identifying the pattern of the transformations." }, { "problem_id": 1280, "question": "A right-angled triangle $A B C$ is translated uniformly along the direction of $B C$ from point $B$ to form triangle $E D F$ (as shown in Figure 1). The movement stops when point $E$ reaches point $C$ (as shown in Figure 2). Given that $A B=6$, when point $H$ divides $D E$ into two parts in the ratio $1: 2$, the area of quadrilateral $D H C F$ is 20. The distance of the translation is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_802ad286ff83dc4342dbg_0014_1.jpg", "batch14-2024_06_15_802ad286ff83dc4342dbg_0014_2.jpg" ], "is_multi_img": true, "answer": "4 or 5 .\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure: Since the right triangle \\( ABC \\) is uniformly translated from point \\( B \\) along the direction of \\( BC \\) to obtain triangle \\( EDF \\),\n\nThe distance of the translation is \\( BE \\), \\( DE = AB = 6 \\), and the area \\( S_{\\triangle ABC} = S_{\\triangle DEF} \\).\n\nTherefore, the area of quadrilateral \\( ABDH \\) equals the area of quadrilateral \\( DHCF \\), which is 20.\n\nWhen \\( DH : HE = 1 : 2 \\), \\( HE = \\frac{2}{3} \\times 6 = 4 \\), then \\( \\frac{1}{2}(4 + 6) \\times BE = 20 \\), solving gives \\( BE = 4 \\);\n\nWhen \\( DH : HE = 2 : 1 \\), \\( HE = \\frac{1}{3} \\times 6 = 2 \\), then \\( \\frac{1}{2}(2 + 6) \\times BE = 20 \\), solving gives \\( BE = 5 \\);\n\nIn summary, the distance of the translation is either 4 or 5.\n\nThus, the answer is: 4 or 5.\n\n[Highlight] This question examines the properties of translation: moving a figure along a straight line results in a new figure that is identical in shape and size to the original. Each point in the new figure corresponds to a point in the original figure that has been moved, and the line segments connecting corresponding points are parallel (or collinear) and equal in length." }, { "problem_id": 1281, "question": "Place two congruent right-angled triangles as shown in the figure, with $\\angle B A C=\\angle B_{1} A_{1} C_{1}=30^{\\circ}$. Fix triangle $A_{1} B_{1} C$, and then rotate triangle $A B C$ clockwise around point $C$ to the position shown in the figure. At this point, $A B$ intersects $A_{1} C$ and $A_{1} B_{1}$ at points $D$ and $E$ respectively, $A C$ intersects $A_{1} B_{1}$ at point $F$, and $A B \\perp A_{1} B_{1}$. The degree measure of the rotation angle is $\\qquad$ ${ }^{\\circ}$.\n\n\n\nFigure 1\n\n\n\n\\& 2", "input_image": [ "batch14-2024_06_15_802ad286ff83dc4342dbg_0044_1.jpg", "batch14-2024_06_15_802ad286ff83dc4342dbg_0044_2.jpg" ], "is_multi_img": true, "answer": "30", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since the two congruent right-angled triangular plates have $\\angle BAC = \\angle B_{1}A_{1}C_{1} = 30^{\\circ}$,\n\nit follows that $\\angle B = 60^{\\circ}$.\n\nBecause $AB \\perp A_{1}B_{1}$,\n\nwe have $\\angle A_{1}EB = 90^{\\circ}$.\n\nThus, $\\angle BDC = \\angle A_{1}DE = 60^{\\circ}$,\n\nand $\\angle BCD = 60^{\\circ}$.\n\nTherefore, $\\angle A_{1}CA = 90^{\\circ} - 60^{\\circ} = 30^{\\circ}$,\n\nwhich means the rotation angle is $30^{\\circ}$.\n\nHence, the answer is 30.\n\n[Key Insight] The key assessment points are: the sum of angles in a triangle and the definition of rotation. The application of the triangle angle sum theorem is crucial." }, { "problem_id": 1282, "question": "As shown in the figure, the $\\triangle \\mathrm{ABC}$ in Figure 1 is transformed to obtain $\\triangle \\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}$ in Figure 2. If the coordinates of point $\\mathrm{P}$ on $\\triangle \\mathrm{ABC}$ in Figure 1 are $(\\mathrm{a}, \\mathrm{b})$, then the coordinates of the corresponding point $\\mathrm{P}^{\\prime}$ in Figure 2 are $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_80eff3de33d765f227a0g_0082_1.jpg", "batch14-2024_06_15_80eff3de33d765f227a0g_0082_2.jpg" ], "is_multi_img": true, "answer": "$(\\mathrm{a}+3, \\mathrm{~b}+2)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "According to the problem: The coordinates of point $\\mathrm{A}$ are $(-3,-2)$. After translation, the coordinates of $\\mathrm{A}^{\\prime}$ are $(0,0)$. Therefore, in part (1), if the coordinates of point $\\mathrm{P}$ on $\\triangle \\mathrm{ABC}$ are $(\\mathrm{a}, \\mathrm{b})$, then the corresponding point $\\mathrm{P}^{\\prime}$ in part (2) will have coordinates $(a+3, b+2)$. Hence, the answer is $(\\mathrm{a}+3, \\mathrm{~b}+2)$.\n\n**Key Insight:** This question tests the understanding of coordinate translation. The key point is to recognize that during a horizontal translation, the y-coordinate remains unchanged, while during a vertical translation, the x-coordinate remains unchanged. Translation transformations are a common topic in middle school exams." }, { "problem_id": 1283, "question": "Fold triangle $ABC$ in Figure 1 so that point $A$ coincides with point $C$, with the fold line being $ED$, where points $E$ and $D$ are on $AB$ and $AC$ respectively, resulting in Figure 2. If $BC = 4$ and $AB = 5$, then the perimeter of $\\triangle EBC$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_a08bb9046380ba79a713g_0008_1.jpg", "batch14-2024_06_15_a08bb9046380ba79a713g_0008_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Triangle $ABC$ is folded such that point $A$ coincides with point $C$, and the crease is $ED$.\n\n$\\therefore AE = CE$,\n\n$\\therefore$ The perimeter of triangle $EBC$ is $BC + CE + BE = BC + AE + BE = BC + AB = 4 + 5 = 9$;\n\nThus, the answer is: 9.\n\n[Key Insight] This problem tests the concept of folding. Mastering the properties of folding is crucial for solving such problems." }, { "problem_id": 1284, "question": "Fold a rectangular piece of paper in half, as shown, to obtain a crease (the dashed line in the figure). Continue folding, ensuring that each new crease is parallel to the previous one. After folding three times consecutively, you can obtain 7 creases. Therefore, after folding 2022 times, you can obtain a certain number of creases.\n\n\n\nFirst fold\n\n\n\nSecond fold\n\n\n\nThird fold", "input_image": [ "batch14-2024_06_15_a08bb9046380ba79a713g_0009_1.jpg", "batch14-2024_06_15_a08bb9046380ba79a713g_0009_2.jpg", "batch14-2024_06_15_a08bb9046380ba79a713g_0009_3.jpg" ], "is_multi_img": true, "answer": "$2^{2022}-1 \\# \\#-1+2^{2022}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "As shown in the figure, the first fold divides the paper into 2 parts with 1 crease,\n\nthe second fold divides the paper into 4 parts with 3 creases,\n\nthe third fold divides the paper into 8 parts with 7 creases,\n\ntherefore, the fourth fold divides the paper into 16 parts with 15 creases,\n\nand so on, the $n$th fold divides the paper into $2^{n}$ parts with $2^{n}-1$ creases.\n\nFolding the paper 2022 times results in $2^{2022}-1$ creases.\n\nThus, the answer is: $2^{2022}-1$.\n\n【Insight】This question tests the application of the power of rational numbers. Observing the relationship between the number of parts and the number of creases obtained by folding is the key to solving the problem." }, { "problem_id": 1285, "question": "As shown in Figure (1), in square \\( A B C D \\), point \\( E \\) is the midpoint of \\( A B \\), and point \\( P \\) is a moving point on the diagonal \\( A C \\). Let \\( P C = x \\) and \\( P E + P B = y \\). Figure (2) is the graph of the function \\( y \\) with respect to \\( x \\), and the coordinates of the lowest point \\( Q \\) on the graph are \\( \\left( \\frac{8 \\sqrt{2}}{3}, 2 \\sqrt{5} \\right) \\). The side length of square \\( A B C D \\) is \\(\\qquad\\) \n\nfigure (1)\n\n\n\nfigure (2)", "input_image": [ "batch14-2024_06_15_a08bb9046380ba79a713g_0013_1.jpg", "batch14-2024_06_15_a08bb9046380ba79a713g_0013_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $PD$,\n\n\n\nFigure (1)\n\nThen point $D$ is the symmetric point of point $B$ with respect to the line $AC$,\n\nAccording to the symmetry of points, $PB = PD$, thus $y = PE + PB = PD + PE = DE$ is minimized,\nHence $ED = 2\\sqrt{5}$,\n\nLet the side length of the square be $x$, then $AE = \\frac{1}{2}x$,\n\nIn the right triangle $\\triangle ADE$, by the Pythagorean theorem: $DE^{2} = AD^{2} + AE^{2}$,\n\nThat is, $x^{2} + \\left(\\frac{1}{2}x\\right)^{2} = (2\\sqrt{5})^{2}$,\n\nSolving gives: $x = 4$ (the negative value has been discarded),\n\nTherefore, the answer is: 4.\n\n【Insight】This problem mainly examines the properties of squares, the properties of symmetry, the Pythagorean theorem, and the solution of quadratic equations. The key to solving the problem is determining that the length of $DE$ is the minimum." }, { "problem_id": 1286, "question": "From a square cardboard with an area of $36 \\mathrm{~cm}^{2}$, a large square with side length $a \\mathrm{~cm}$ and a small square with side length $b \\mathrm{~cm}$ are cut out (as shown in Figure 1). Then, a small square with side length $b \\mathrm{~cm}$ is cut out from one vertex of the large square (as shown in Figure 2), resulting in a hexagon $A B C D E F$ with a perimeter of $16 \\mathrm{~cm}$. The sum of the areas of the two remaining rectangles in the original large square is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_a08bb9046380ba79a713g_0028_1.jpg", "batch14-2024_06_15_a08bb9046380ba79a713g_0028_2.jpg" ], "is_multi_img": true, "answer": "16", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From Figure 1, we have: $(a+b)^{2}=36$,\n\n$\\therefore a+b=6$,\n\nFrom Figure 2, since the perimeter of hexagon $A B C D E F$ is $16 \\mathrm{~cm}$,\n\n$\\therefore$ the side length of the large square $a=16 \\div 4=4 \\mathrm{~cm}$,\n\n$\\therefore b=6-a=2 \\mathrm{~cm}$,\n\n$\\therefore$ the combined area of the two remaining rectangles in the original large square is $2 \\times a \\times b=16 \\mathrm{~cm}^{2}$, hence the answer is: 16.\n\n【Highlight】This problem examines the properties of translation and the area of shapes. The key to solving it is to understand the figure and use the properties of translation combined with the perimeter to find the value of $a$." }, { "problem_id": 1287, "question": "As shown in Figure 1, for points \\( A \\) and \\( P \\) in the plane, if the segment \\( PA \\) is rotated counterclockwise by \\( 90^\\circ \\) around point \\( P \\) to obtain segment \\( PB \\), then point \\( B \\) is called the \"orthogonal reflection point\" of point \\( A \\) with respect to point \\( P \\). As shown in Figure 2, given point \\( A(3,0) \\), point \\( P \\) is a point on the \\( Y \\)-axis, and point \\( B \\) is the \"orthogonal reflection point\" of point \\( A \\) with respect to point \\( P \\). Connecting \\( AB \\) and \\( OB \\), the minimum value of \\( AB + OB \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_a08bb9046380ba79a713g_0042_1.jpg", "batch14-2024_06_15_a08bb9046380ba79a713g_0042_2.jpg" ], "is_multi_img": true, "answer": "$3 \\sqrt{5}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Solution:** As shown in the figure, let \\( P(0, m) \\), and draw \\( BC \\perp y \\)-axis through point \\( B \\).\n\n\n\nSince \\( \\angle APB = 90^\\circ \\) and \\( \\angle BCP = \\angle POA = 90^\\circ \\),\n\nit follows that \\( \\angle BPC = 90^\\circ - \\angle APO = 90^\\circ - \\angle PBC \\),\n\nthus \\( \\angle APO = \\angle PBC \\).\n\nGiven that \\( PB = PA \\),\n\nwe have \\( \\triangle PBC \\cong \\triangle APO \\),\n\nwhich implies \\( PO = BC \\) and \\( AO = PC \\).\n\nGiven \\( A(3, 0) \\), \\( OA = 3 \\), and \\( OP = |m| \\),\n\nwe find \\( B(m, m + 3) \\), and \\( OA = 3 \\).\n\nTherefore, point \\( B \\) lies on the line \\( y = x + 3 \\).\n\nAs shown in the figure, construct the symmetric point \\( A' \\) of \\( A \\) with respect to the line \\( y = x + 3 \\), and connect \\( AA' \\) to intersect the \\( y \\)-axis at point \\( Q \\). Then, \\( AA' \\perp BQ \\).\n\nSetting \\( x = 0 \\), we get \\( y = 3 \\), so \\( Q(0, 3) \\).\n\nThus, \\( A'(-3, 6) \\).\n\nTherefore, \\( BA + BO \\geq A'O = \\sqrt{(-3)^2 + 6^2} = 3\\sqrt{5} \\).\n\n\n\nHence, the minimum value of \\( AB + OB \\) is \\( 3\\sqrt{5} \\).\n\n**Answer:** \\( 3\\sqrt{5} \\)\n\n**Key Insight:** This problem tests the properties and determination of congruent triangles, coordinates and graphs, the Pythagorean theorem, and linear functions. The key to solving the problem lies in determining the coordinates of point \\( B \\)." }, { "problem_id": 1288, "question": "Xiaoqi cuts a shape from $O$ (Figure 1). After folding (Figure 2), if $AC=2$ and $BC=4$, then the radius of $O$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_a44ecdbe3d95446b290bg_0084_1.jpg", "batch14-2024_06_15_a44ecdbe3d95446b290bg_0084_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $O A$ and $O B$,\n\n\n\nFrom the problem statement, $O A \\perp C B$,\n\n$\\therefore \\angle O C B=90^{\\circ}$,\n\nLet the radius of $O$ be $r$,\n\n$\\because A C=2$,\n\n$\\therefore O C=O A-A C=r-2$,\n\nIn the right triangle $O B C$, $B C=4$,\n\nAccording to the Pythagorean theorem, $B C^{2}+O C^{2}=O B^{2}$,\n\n$4^{2}+(r-2)^{2}=r^{2}$,\n\n$16+r^{2}-4 r+4-r^{2}=0$,\n\n$4 r=20$,\n\n$r=5$,\n\n$\\therefore$ The radius of $O$ is 5,\n\nHence, the answer is: 5.\n\n【Insight】This problem examines the perpendicular chord theorem, the Pythagorean theorem, and folding. The key to solving it lies in understanding the problem statement, mastering these concepts, and appropriately adding auxiliary lines." }, { "problem_id": 1289, "question": "As shown in the figure, in a $5 \\times 5$ grid paper, the triangle A in Figure (1) is translated to the position shown in Figure (2) to form a rectangle with triangle B. The correct translation method is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_b5469474cf3d2e2088afg_0072_1.jpg", "batch14-2024_06_15_b5469474cf3d2e2088afg_0072_2.jpg" ], "is_multi_img": true, "answer": "Shift right 2 squares, then shift down 3 squares (answers vary)", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "By observing the figure, it can be determined that the translation involves first moving down by 3 units and then moving to the right by 2 units. Alternatively, it can also be done by first moving to the right by 2 units and then moving down by 3 units.\n\nTherefore, the answer is to first move down by 3 units and then to the right by 2 units, or to first move to the right by 2 units and then down by 3 units.\n\n[Key Insight] This question tests the method of translating figures, and careful observation of the figure is crucial to solving the problem." }, { "problem_id": 1290, "question": "Two trapezoid paper pieces of identical shape and size are arranged as shown in Figure (a). The trapezoid paper piece $A B C D$ is translated to the right along the upper base $A D$ to obtain Figure (b). Given that $A D=4, B C=8$, if the area of the shaded part is $\\frac{1}{3}$ of the area of quadrilateral $A^{\\prime} B^{\\prime} C D$, then the translation distance in Figure (b) is $\\qquad$.\n\n\n\n(a)\n\n\n\n(b)", "input_image": [ "batch14-2024_06_15_e0a192fc489fe2a97eeag_0043_1.jpg", "batch14-2024_06_15_e0a192fc489fe2a97eeag_0043_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since trapezoid $\\mathrm{ABCD}$ is congruent to trapezoid $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}$,\n\n$\\therefore \\mathrm{AD}=\\mathrm{A}^{\\prime} \\mathrm{D}^{\\prime}=4, \\quad \\mathrm{BC}=\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}=8$.\n\nLet the height of trapezoid $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}$ be $\\mathrm{h}$, and $\\mathrm{A}^{\\prime} \\mathrm{A}=\\mathrm{x}$, then $\\mathrm{B}^{\\prime} \\mathrm{B}=\\mathrm{x}$.\n\n$\\therefore \\mathrm{AD}^{\\prime}=\\mathrm{A}^{\\prime} \\mathrm{D}^{\\prime}-\\mathrm{A}^{\\prime} \\mathrm{A}=4-\\mathrm{x}, \\mathrm{BC}^{\\prime}=\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}-\\mathrm{B}^{\\prime} \\mathrm{B}=8-\\mathrm{x}$.\n\nFrom the properties of translation, we know: $S_{\\text{quadrilateral } \\triangle A \\triangle B B}=S_{\\text{quadrilateral DDCC }}$.\n\nAlso, since $S_{\\text{shaded }}=\\frac{1}{3} S_{\\text{quadrilateral } \\mathrm{ABCD}}$,\n\n$\\therefore S_{\\text{shaded }}=\\frac{1}{2} S_{\\text{quadrilateral } A B C D}$.\n\n$\\therefore \\frac{1}{2} h\\left(\\mathrm{AD}^{\\prime}+\\mathrm{BC}^{\\prime}\\right)=\\frac{1}{2} \\times \\frac{1}{2} h\\left(\\mathrm{~A}^{\\prime} \\mathrm{D}^{\\prime}+\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}\\right)$,\n\nThat is, $\\frac{1}{2} h(4-x+8-x)=\\frac{1}{2} \\times \\frac{1}{2} h(4+8)$,\n\nSimplifying gives: $6-x=3$,\n\nSolving gives: $x=3$,\n\n$\\therefore \\mathrm{A}^{\\prime} \\mathrm{A}=3$.\n\nTherefore, the answer is: 3.\n\n【Highlight】This question tests the properties of translation and the area formula of a trapezoid. The key to solving the problem is to deduce $S_{\\text{shaded }}=\\frac{1}{2} S_{\\text{quadrilateral } \\mathrm{ABCD}}$ based on the properties of translation and the given conditions." }, { "problem_id": 1291, "question": "The four quadrilaterals shown are congruent. In figures (1) and (3), which figure can be obtained by translating or rotating quadrilateral $A B C D$?\n\n\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch14-2024_06_15_e5923a085fd75b5bdcbeg_0027_1.jpg", "batch14-2024_06_15_e5923a085fd75b5bdcbeg_0027_2.jpg", "batch14-2024_06_15_e5923a085fd75b5bdcbeg_0027_3.jpg", "batch14-2024_06_15_e5923a085fd75b5bdcbeg_0027_4.jpg" ], "is_multi_img": true, "answer": "Figure（2）(3)", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Figure (1) is obtained by folding quadrilateral $ABCD$,\n\nFigure (2) is obtained by rotating quadrilateral $ABCD$,\n\nFigure (3) is obtained by translating quadrilateral $ABCD$.\n\n[Key Insight] This question examines the properties of rotation and translation. Proficiency in applying the properties of rotation is crucial for solving this problem." }, { "problem_id": 1292, "question": "Using 10 wooden sticks to form the pattern shown in Figure (1), please move 3 sticks to transform it into the pattern shown in Figure (2). The 3 sticks are $\\qquad$ (just fill in the serial numbers).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_eb2c5f45df0010abc424g_0088_1.jpg", "batch14-2024_06_15_eb2c5f45df0010abc424g_0088_2.jpg" ], "is_multi_img": true, "answer": "(2)(4)(6)", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (2):\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\nTherefore, the answer is: (2)(4)(6) (the answer is not unique).\n\n[Highlight] This question mainly examines the use of translation to design patterns. By determining a basic pattern and translating it in a certain direction over a certain distance, and repeating this process, beautiful patterns can be designed. Changing the direction and distance of the translation can make the patterns more diverse and colorful." }, { "problem_id": 1293, "question": "\"Flowers shadow the wall, peaks overlap the window,\" the airy window lattices of Suzhou gardens contain many mathematical elements. The window lattice in Figure (1) is a frost crack pattern, and Figure (2) shows part of this pattern. If $\\angle 1 + \\angle 3 + \\angle 5 = 186^{\\circ}$, then\n\n$\\angle 2 + \\angle 4 + \\angle 6 =$ $\\qquad$ .\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch14-2024_06_15_fbd88dfc8804ef3c7e64g_0024_1.jpg", "batch14-2024_06_15_fbd88dfc8804ef3c7e64g_0024_2.jpg" ], "is_multi_img": true, "answer": "366", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\n$\\because$ The sum of the exterior angles of a polygon is $360^{\\circ}$,\n\n$\\therefore \\angle 7 + \\angle 8 + \\angle 9 = 360^{\\circ} - (\\angle 1 + \\angle 3 + \\angle 5) = 174^{\\circ}$,\n\n$\\because \\angle 2 + \\angle 7 = \\angle 4 + \\angle 8 = \\angle 6 + \\angle 9 = 180^{\\circ}$,\n\n$\\therefore \\angle 2 + \\angle 4 + \\angle 6 = 180^{\\circ} \\times 3 - (\\angle 7 + \\angle 8 + \\angle 9) = 540^{\\circ} - 174^{\\circ} = 366^{\\circ}$;\n\nTherefore, the answer is: 366.\n\n【Key Point】This question tests the application of the sum of exterior angles of a polygon. Mastering that the sum of the exterior angles of a polygon is $360^{\\circ}$ is crucial for solving the problem." }, { "problem_id": 1294, "question": "As shown in the figure, Figure 1 is a foldable desk lamp, and Figure 2 is its plan view. The base $A O \\perp O E$ at point $O$, and the supports $A B, B C$ are fixed rods. The angle $\\angle B A O$ is twice the angle $\\angle C B A$, and the lamp body $C D$ can be adjusted by rotating around point $C$. Now, the lamp body $C D$ is rotated from the horizontal position to the position $C D^{\\prime}$ (as shown by the dashed line in Figure 2). At this point, the line where the lamp body $C D^{\\prime}$ lies is exactly perpendicular to the support $A B$, and $\\angle B C D-\\angle D C D^{\\prime}=123^{\\circ}$. Then, $\\angle D C D^{\\prime}=$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch14-2024_06_15_fbd88dfc8804ef3c7e64g_0029_1.jpg", "batch14-2024_06_15_fbd88dfc8804ef3c7e64g_0029_2.jpg" ], "is_multi_img": true, "answer": "$38^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Extend $OA$ to intersect $CD$ at point $F$, and extend $D'C$ to intersect $AB$ at point $G$.\n\n\n\nHorizontal tabletop\n\n$\\because AO \\perp OE$,\n\n$\\therefore \\angle AOE = 90^\\circ$,\n\n$\\because CD \\parallel OE$,\n\n$\\therefore \\angle AFC = \\angle AOE = 90^\\circ$,\n\n$\\therefore OA \\perp CD$,\n\nSince $AO \\perp OE$ and $D'C \\perp AB$,\n\n$\\therefore \\angle AGC = \\angle AFC = 90^\\circ$,\n\n$\\therefore \\angle GCF + \\angle GAF = 180^\\circ$,\n\nSince $\\angle DCD' + \\angle GCF = 180^\\circ$,\n\n$\\therefore \\angle DCD' = \\angle GAF$,\n\n$\\therefore \\angle BAO = 180^\\circ - \\angle DCD'$,\n\n$\\therefore \\angle B = \\frac{1}{2}\\left(180^\\circ - \\angle DCD'\\right)$,\n\n$\\because \\angle BCD - \\angle DCD' = 123^\\circ$,\n\n$\\therefore \\angle BCD = \\angle DCD' + 123^\\circ$,\n\nIn quadrilateral $ABCF$, we have $\\angle GAF + \\angle B + \\angle BCD + \\angle AFC = 360^\\circ$,\n\n$\\therefore \\angle DCD' + \\frac{1}{2}\\left(180^\\circ - \\angle DCD'\\right) + \\angle DCD' + 123^\\circ + 90^\\circ = 360^\\circ$,\n\nSolving gives: $\\angle DCD' = 38^\\circ$,\n\nTherefore, the answer is: $38^\\circ$.\n\n【Key Point】This problem tests the properties of parallel lines, with the key being a thorough understanding of the properties of parallel lines and the fact that the sum of the interior angles of a quadrilateral is 360 degrees." }, { "problem_id": 1295, "question": "Cut the image in Figure 1 into several pieces, and after performing translations in Figure 2, you can obtain (1), (2), and (3) below.\n\n\n\nFigure 1\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\nFigure 2", "input_image": [ "batch14-2024_06_15_feff60d43e801d6a8bedg_0050_1.jpg", "batch14-2024_06_15_feff60d43e801d6a8bedg_0050_2.jpg", "batch14-2024_06_15_feff60d43e801d6a8bedg_0050_3.jpg", "batch14-2024_06_15_feff60d43e801d6a8bedg_0050_4.jpg" ], "is_multi_img": true, "answer": "(1)(2)(2)", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to Figure 1, it can be seen that the shape is cut into several small pieces, which can then be rearranged and translated in Figure 2 to form (1) and (2), but not (3).\n\nTherefore, the answer is: (1)(2)." }, { "problem_id": 1296, "question": "As shown in the figure, one evening, when Xiao Ying walked from point B under streetlight A to point C, she measured the length of her shadow CD to be 1 meter. When she continued to walk to point D, she found that the length of her shadow DE was exactly her height. Given that Xiao Ying's height is 1.5 meters, find the height of the streetlight A, AB.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch15-2024_06_15_0244abee9bba25485b2ag_0064_1.jpg", "batch15-2024_06_15_0244abee9bba25485b2ag_0064_2.jpg" ], "is_multi_img": true, "answer": "$4.5 \\mathrm{~m}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** According to the problem statement, construct a mathematical model using similar triangles. The solution can be derived by utilizing the properties and criteria of similar triangles.\n\n**Question Solution:** Given the problem, \\( MC = FD = DE = 1.5 \\, \\text{m} \\), and \\( CD = 1 \\, \\text{m} \\).\n\nSince \\( MC \\parallel AB \\), it follows that \\( \\triangle DMC \\sim \\triangle DAB \\).\n\nTherefore, \\( \\frac{DC}{DB} = \\frac{MC}{AB} \\).\n\nSimilarly, since \\( \\triangle EFD \\sim \\triangle EAB \\), it follows that \\( \\frac{ED}{EB} = \\frac{FD}{AB} \\).\n\nGiven that \\( MC = FD \\), we have \\( \\frac{DC}{DB} = \\frac{ED}{EB} \\).\n\nThis implies \\( \\frac{1}{1 + BC} = \\frac{1.5}{2.5 + BC} \\).\n\nSolving this equation yields: \\( BC = 2 \\, \\text{m} \\).\n\nSubstituting \\( BC = 2 \\, \\text{m} \\) into \\( \\frac{DC}{DB} = \\frac{MC}{AB} \\), we get \\( \\frac{1}{3} = \\frac{1.5}{AB} \\).\n\nSolving this equation gives: \\( AB = 4.5 \\, \\text{m} \\).\n\n**Answer:** The height \\( AB \\) of the streetlight \\( A \\) is \\( 4.5 \\, \\text{m} \\).\n\n**Key Insight:** This problem comprehensively examines the characteristics and patterns of central projection, as well as the application of the properties of similar triangles. The key to solving the problem lies in recognizing that within these two sets of similar triangles, there is a common side. By using this common side as an equality relationship, the required line segments can be determined, and subsequently, the length of the common side can be calculated." }, { "problem_id": 1297, "question": "As shown in the figure, a piece of wire is divided into two segments that can each form two similar pentagons. Given that their area ratio is 1:4, the side length of the smaller pentagon is $\\left(\\mathrm{x}^{2}-4\\right) \\mathrm{cm}$, and the side length of the larger pentagon is $\\left(\\mathrm{x}^{2}+2 \\mathrm{x}\\right) \\mathrm{cm}$ (where $\\mathrm{x}>0$). Find the total length of this piece of wire.\n\n\n\n$\\left(x^{2}-4\\right) \\mathrm{cm}$\n\n", "input_image": [ "batch15-2024_06_15_0916a056ae0c9cd9cc4ag_0005_1.jpg", "batch15-2024_06_15_0916a056ae0c9cd9cc4ag_0005_2.jpg" ], "is_multi_img": true, "answer": "180 (cm).", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since the area ratio of similar pentagons is $1:4$,\n\nTherefore, their similarity ratio is $1:2$,\n\nThat is, $\\left(x^{2}-4\\right):\\left(x^{2}+2 x\\right)=1:2$,\n\nSimplifying gives $x^{2}-2 x-8=0$, solving yields $x_{1}=4, x_{2}=-2$ (discarded),\n\nWhen $x=4$, $x^{2}-4=12, x^{2}+2 x=24$,\nThus, the total length of the wire $=5 \\times 12+5 \\times 24=180(\\mathrm{~cm})$.\n\n[Highlight] This question tests the properties of similar polygons, understanding that the area ratio of similar polygons is the square of the similarity ratio is key to solving the problem, and attention should be paid to the method of solving quadratic equations." }, { "problem_id": 1298, "question": "Xingming Tower, also known as Xin Tower, is located in the center of South Street in Yulin City, as shown in Figure (1). To understand the height of Xingming Tower, Xiaohua found that the height \\( AB = 18.2 \\) meters from the data. One day, he observed Xingming Tower in person, as shown in Figure (2). He stood at point \\( F \\) 30 meters away from the tower (\\( FB = 30 \\) meters) and walked along \\( FB \\) towards point \\( B \\). When he reached point \\( H \\), he noticed that the top \\( C \\) of the billboard \\( CD \\) and the top \\( A \\) of the tower were aligned in a straight line. The distance from Xiaohua's eyes to the ground was \\( EF = GH = 1.7 \\) meters, the height of the billboard was \\( CD = 3.2 \\) meters, \\( BD = 20 \\) meters, and points \\( B, D, H, F \\) were on the same horizontal line.\n\n\\( AB \\perp BF, CD \\perp BF, GH \\perp BF, EF \\perp BF \\). How many meters did Xiaohua walk forward from \\( F \\) to see points \\( C \\) and \\( A \\) in a straight line (i.e., to find the length of \\( HF \\))?\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch15-2024_06_15_242b0bee70cd9926069bg_0006_1.jpg", "batch15-2024_06_15_242b0bee70cd9926069bg_0006_2.jpg" ], "is_multi_img": true, "answer": "8 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular line from point $G$ to $AB$, intersecting $AB$ at point $M$ and $CD$ at point $N$.\n\n\n\nGiven by the problem: $MB = ND = GH = 1.7$,\n\n$MN = BD = 20$, $CN = CD - ND = 1.5$, $AM = AB - MB = 16.5$, and $\\angle AMG = \\angle CNG = 90^\\circ$.\n\nSince $\\angle AGM = \\angle CGN$,\n\n$\\triangle AMG \\sim \\triangle CNG$,\nthus $\\frac{AM}{CN} = \\frac{MG}{NG}$,\n\nwhich is $\\frac{16.5}{1.5} = \\frac{20 + NG}{NG}$,\n\nsolving gives: $NG = 2$,\n\ntherefore $DH = NG = 2$,\n\nhence $HF = BF - BD - DH = 30 - 20 - 2 = 8$ meters,\n\nTherefore, Xiao Hua walked forward 8 meters from point $F$ and just saw points $C$ and $A$ aligned in a straight line.\n\n[Highlight] This problem mainly examines the determination and properties of similar triangles. Correctly drawing auxiliary lines and constructing similar triangles are key to solving this problem." }, { "problem_id": 1299, "question": "From a square cardboard with an area of $36 \\mathrm{~cm}^{2}$, a large square with side length $a \\mathrm{~cm}$ and a small square with side length $b \\mathrm{~cm}$ are cut out (as shown in Figure 1). Then, a small square with side length $b \\mathrm{~cm}$ is cut out from one vertex of the large square (as shown in Figure 2), resulting in a hexagon $A B C D E F$ with a perimeter of $16 \\mathrm{~cm}$. The sum of the areas of the two remaining rectangles in the original large square is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch15-2024_06_15_24e5afd6a917894d08c0g_0094_1.jpg", "batch15-2024_06_15_24e5afd6a917894d08c0g_0094_2.jpg" ], "is_multi_img": true, "answer": "16", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: From Figure 1, we can derive that: $(a+b)^{2}=36$,\n\n$\\therefore a+b=6$,\n\nFrom Figure 2, since the perimeter of hexagon $A B C D E F$ is $16 \\mathrm{~cm}$,\n\n$\\therefore$ the side length of the large square $a=16 \\div 4=4 \\mathrm{~cm}$,\n\n$\\therefore b=6-a=2 \\mathrm{~cm}$,\n\n$\\therefore$ the combined area of the two remaining rectangles in the original large square is $2 \\times a \\times b=16 \\mathrm{~cm}^{2}$, hence the answer is: 16.\n\n【Highlight】This problem examines the properties of translation and the area of shapes. The key to solving it is to understand the figure and use the properties of translation combined with the perimeter to find the value of $a$." }, { "problem_id": 1300, "question": "A math interest group decided to use their knowledge to measure the height of an ancient building. As shown in Figure 2, the height of the ancient building is $A B$. On the ground $B C$, points $E$ and $G$ are selected, where two 1.5-meter-tall markers $E F$ and $G H$ are erected. The distance between the two markers $E G$ is 26 meters, and the ancient building $A B$, the marker $E F$, and the marker $G H$ are all in the same vertical plane. Moving back 2 meters from the marker $E F$ to point $D$ (i.e., $E D = 2$ meters), observing point $A$ from $D$, the points $A, F, D$ are collinear; moving back 4 meters from the marker $G H$ to point $C$ (i.e., $C G = 4$ meters), observing point $A$ from $C$, the points $A, H, C$ are also collinear. It is known that $B, E, D, G, C$ are on the same straight line, $A B \\perp B C, E F \\perp B C, G H \\perp B C$. Please help the interest group calculate the height of the ancient building $A B$ based on the above measurement data.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch15-2024_06_15_694110a04f8524d8f429g_0043_1.jpg", "batch15-2024_06_15_694110a04f8524d8f429g_0043_2.jpg" ], "is_multi_img": true, "answer": "$21 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( BD = x \\) meters, then \\( BC = BD + DG + CG = x + 26 - 2 + 4 = (x + 28) \\) meters.\n\nSince \\( AB \\perp BC \\) and \\( EF \\perp BC \\),\n\nIt follows that \\( AB \\parallel EF \\),\n\nTherefore, \\( \\triangle ABD \\sim \\triangle FED \\),\n\nHence, \\( \\frac{EF}{AB} = \\frac{DE}{BD} \\), which is \\( \\frac{1.5}{AB} = \\frac{2}{x} \\).\n\nSimilarly, it can be proven that \\( \\triangle ABC \\sim \\triangle HGC \\),\n\nThus, \\( \\frac{GH}{AB} = \\frac{CG}{BC} \\), which is \\( \\frac{1.5}{AB} = \\frac{4}{x + 28} \\).\n\nTherefore, \\( \\frac{2}{x} = \\frac{4}{x + 28} \\),\n\nSolving gives \\( x = 28 \\).\n\nAfter verification, \\( x = 28 \\) is the solution to the original equation.\n\nHence, \\( \\frac{1.5}{AB} = \\frac{2}{28} \\),\n\nThus, \\( AB = 21 \\) meters.\n\nTherefore, the height of the ancient building \\( AB \\) is \\( 21 \\) meters.\n\n[Key Insight] This problem primarily examines the application of similar triangles. Proving \\( \\triangle ABD \\sim \\triangle EFD \\) to obtain \\( \\frac{1.5}{AB} = \\frac{2}{x} \\), and similarly obtaining \\( \\frac{1.5}{AB} = \\frac{4}{x + 28} \\), is key to establishing the equation and solving the problem." }, { "problem_id": 1301, "question": "Xiao Shuai's new house had just finished being renovated when they encountered a rare heavy rain. He suggested to his father that they install a rain cover on the windows. As shown in Figure 1, this is a rain cover he learned about. Its side view is shown in Figure 2, where the center O of the top arc AB is on the vertical edge D, and the center O of the other arc BC is on the extension of the horizontal edge DC, with a central angle of 90°. BE is perpendicular to AD at point E. According to the indicated dimensions (in cm), the radius AO of the circle where the arc AB is located can be calculated to be $\\boxed{}$ cm.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0063_1.jpg", "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0063_2.jpg" ], "is_multi_img": true, "answer": "61", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Connect $\\mathrm{BO}_{1}$. It is easy to see that $\\mathrm{BE}=60 \\mathrm{~cm}$ and $\\mathrm{AE}=50 \\mathrm{~cm}$.\n\nLet the radius of arc $\\mathrm{AB}$ be $\\mathrm{Rcm}$. Then, $\\mathrm{O}_{1} \\mathrm{B}=\\mathrm{Rcm}$ and $\\mathrm{O}_{1} \\mathrm{E}=(\\mathrm{R}-50) \\mathrm{cm}$.\n\nIn the right triangle $\\triangle \\mathrm{O}_{1} \\mathrm{BE}$, by the Pythagorean theorem, we have: $\\mathrm{O}_{1} \\mathrm{B}^{2}=\\mathrm{BE}^{2}+\\mathrm{O}_{1} \\mathrm{E}^{2}$,\n\nThat is, $\\mathrm{R}^{2}=60^{2}+(\\mathrm{R}-50)^{2}$,\n\nSolving this, we get: $\\mathrm{R}=61$.\n\nTherefore, the answer is 61.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n[Key Insight] This problem mainly tests the Pythagorean theorem and the perpendicular chord theorem. The difficulty is moderate, with the key step being to find the radius of the circle where arc $\\mathrm{AB}$ is located." }, { "problem_id": 1302, "question": "As shown in Figure 1, an artistic arch consists of two parts: the lower part is a rectangle \\(ABCD\\) with \\(AB\\) and \\(AD\\) measuring \\(2\\sqrt{3} \\, \\text{m}\\) and \\(4 \\, \\text{m}\\) respectively, and the upper part is a minor arc \\(CD\\) with center \\(O\\) and a central angle \\(\\angle COD = 120^\\circ\\). It is intended to topple the arch by using point \\(B\\) as the pivot. During the toppling process, the plane containing the rectangle \\(ABCD\\) remains perpendicular to the ground, as shown in Figures 2, 3, and 4. Let \\(h \\, \\text{m}\\) denote the maximum distance from any point on the arch to the ground. The maximum value of \\(h\\) is \\(\\qquad\\) \\(\\text{m}\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4", "input_image": [ "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0066_1.jpg", "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0066_2.jpg", "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0066_3.jpg", "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0066_4.jpg" ], "is_multi_img": true, "answer": "$(2+2 \\sqrt{3})$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, a vertical line is drawn from point $O$ to the ground, intersecting the upper edge of the arch frame at point $P$ and the ground at point $Q$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure\n\n\n\nFigure 4\n\nAs shown in Figure 1, the lengths of $AB$ and $AD$ are $2\\sqrt{3}$ meters and $4$ meters respectively, and the central angle $\\angle COD = 120^\\circ$.\n\nTherefore, $\\angle DOP = 60^\\circ$, and $\\frac{1}{2} DC = \\frac{1}{2} AB = \\sqrt{3}$.\n\nThus, $OD = 2$, and $PQ = 5$.\n\nWhen point $P$ lies on segment $AD$, the maximum distance $h$ from a point on the arch to the ground is equal to the distance from point $D$ to the ground, i.e., when point $P$ coincides with point $D$. At this point,\n\n$h = \\sqrt{DC^2 + BC^2} = \\sqrt{(2\\sqrt{3})^2 + 4^2} = 2\\sqrt{7}$.\n\nAs shown in Figure 2, when point $P$ lies on the minor arc $CD$, the maximum distance $h$ from a point on the arch to the ground is equal to the sum of the radius of circle $O$ and the distance from the center $O$ to the ground.\n\nIt is easy to see that $OQ \\leq OB$,\n\nand $h = OP + OQ = 2 + OQ$.\n\nTherefore, when point $Q$ coincides with point $B$, $h$ reaches its maximum value.\n\nFrom Figure 1, we know that $OQ = 3$, $BQ = \\sqrt{3}$, and thus $OB = 2\\sqrt{3}$.\n\nThe maximum value of $h$ is $OP + OB$, which is $2 + 2\\sqrt{3}$.\n\nHence, the answer is $(2 + 2\\sqrt{3})$.\n\n【Insight】This problem involves solving for an extremum. Analyzing the position where the maximum value is achieved is crucial, as it would be difficult to solve otherwise." }, { "problem_id": 1303, "question": "\"Five Rainbow Arches Soaring\" - The Qiongzhou Bridge, constructed using national debt funds and spanning the Nandu River in our province, was officially opened to traffic on May 12th of this year, as shown in Figure (1). The bridge features five red arches on both sides, as depicted in Figure (1). The highest arch has a span of 110 meters and a rise of 22 meters, as shown in Figure (2). What is the diameter of the circle in which this arch is inscribed?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0091_1.jpg", "batch16-2024_06_15_03a6fc0e9211c2ba5d07g_0091_2.jpg" ], "is_multi_img": true, "answer": "159.5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Let the diameter be $\\mathrm{x}$.\n\nAccording to the Perpendicular Diameter Theorem, we know that $\\mathrm{BD}=55$.\n\nThen, applying the Intersecting Chords Theorem, we have $55 \\times 55=22 \\times(\\mathrm{x}-22)$,\n\nSolving this equation gives $\\mathrm{x}=159.5$.\n\nTherefore, the answer is 159.5.\n\n[Highlight] This problem tests the Perpendicular Diameter Theorem and the Intersecting Chords Theorem, with the key to solving it being the determination that $\\mathrm{BD}=55$." }, { "problem_id": 1304, "question": "There is a right-angled triangular wooden board with its two legs measuring 3 meters and 4 meters respectively. It needs to be processed into a square tabletop with the maximum possible area. The methods used by two workers, A and B, are shown in Figure 1 and Figure 2 respectively. Please use the knowledge you have learned to explain whose processing method meets the requirements.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_20f662d06c93903aff01g_0001_1.jpg", "batch16-2024_06_15_20f662d06c93903aff01g_0001_2.jpg" ], "is_multi_img": true, "answer": "Figure 1 The processing method is reasonable.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The processing method in Figure 1 is reasonable.\n\nLet the length of the processed tabletop in Figure 1 be \\( x \\) meters.\n\nSince \\( FD \\parallel BC \\),\n\nTherefore, right triangle AFD is similar to right triangle ACB,\n\nThus, \\( AF:AC = FD:BC \\),\n\nThat is, \\( (4 - x):4 = x:3 \\),\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nSolving gives \\( x = \\frac{12}{7} \\).\n\nLet the length of the processed tabletop in Figure 2 be \\( y \\) meters. Draw \\( CM \\perp AB \\) through point \\( C \\), with the foot of the perpendicular at \\( M \\), intersecting \\( GF \\) at point \\( N \\).\n\nSince \\( GF \\parallel DE \\),\n\nTherefore, triangle CGF is similar to triangle CAB,\n\nThus, \\( CN:CM = GF:AB \\),\n\nTherefore, \\( (CM - y):CM = y:AB \\).\n\nHence, \\( AB = \\frac{y \\cdot CM}{CM - y} \\).\n\nFrom the equality of areas, we can find \\( CM = 2.4 \\),\nTherefore, we can find \\( y = \\frac{60}{37} \\);\n\nClearly, \\( x > y \\), so \\( x^2 > y^2 \\),\n\nTherefore, the processing method in Figure 1 is reasonable.\n\n[Highlight] This problem examines the properties of similar triangles: the corresponding sides of similar triangles are proportional; the ratio of the corresponding heights of similar triangles equals the similarity ratio. The key to solving this problem is to translate the practical problem into a mathematical problem for solution." }, { "problem_id": 1305, "question": "As shown in the figure, there is a right-angled triangular wooden board \\(ABC\\), where \\(\\angle C = 90^\\circ\\), \\(AC = 3 \\text{m}\\), and \\(BC = 4 \\text{m}\\). Now, it needs to be processed into a rectangle with the maximum area. The processing methods of two carpenters, Mr. Jia and Mr. Yi, are shown in Figure 1 and Figure 2, respectively. Please use the knowledge you have learned to explain which carpenter's method meets the requirements.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_20f662d06c93903aff01g_0069_1.jpg", "batch16-2024_06_15_20f662d06c93903aff01g_0069_2.jpg" ], "is_multi_img": true, "answer": "The maximum area is 3, and both chefs' methods meet the requirements.", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: (1) As shown in Figure 1, let \\( DE = x \\) and \\( EF = y \\), and denote the area of the rectangle as \\( S \\).\n\nGiven that \\( DE \\parallel CB \\),\n\n\\[\n\\frac{DE}{CB} = \\frac{AD}{AC}\n\\]\n\nThat is:\n\n\\[\n\\frac{x}{4} = \\frac{3 - y}{3}\n\\]\n\nSolving for \\( y \\):\n\n\\[\ny = 3 - \\frac{3}{4}x \\quad \\text{where} \\quad 0 < x < 4\n\\]\n\nThus, the area \\( S \\) is:\n\n\\[\nS = xy = x\\left(3 - \\frac{3}{4}x\\right) = -\\frac{3}{4}x^2 + 3x = -\\frac{3}{4}(x - 2)^2 + 3\n\\]\n\nTherefore, the maximum area is 3.\n\n(2) As shown in Figure 2, draw \\( CE \\perp AB \\) at point \\( E \\), intersecting \\( NM \\) at point \\( D \\).\n\nGiven that \\( \\angle C = 90^\\circ \\), \\( AC = 3 \\) m, and \\( BC = 4 \\) m,\n\n\\[\nAB = 5 \\quad \\text{and} \\quad CE = 2.4\n\\]\n\nLet \\( MQ = x \\) and \\( MN = y \\), then \\( DE = x \\) and \\( CD = 2.4 - x \\).\n\nSince \\( MN \\parallel AB \\),\n\n\\[\n\\frac{CD}{CE} = \\frac{MN}{AB}\n\\]\n\nThat is:\n\n\\[\n\\frac{2.4 - x}{2.4} = \\frac{y}{5}\n\\]\n\nSimplifying:\n\n\\[\ny = -\\frac{25}{12}x + 5\n\\]\n\nThus, the area \\( S \\) is:\n\n\\[\nS = xy = x\\left(-\\frac{25}{12}x + 5\\right) = -\\frac{25}{12}\\left(x - \\frac{6}{5}\\right)^2 + 3 \\quad \\text{where} \\quad 0 < x < 2.4\n\\]\n\nTherefore, both approaches by the two craftsmen meet the requirements.\n\n\nFigure 2\n\n[Key Insight] This problem examines the properties of similar triangles, where corresponding sides of similar triangles are proportional. The key to solving this problem lies in translating the practical problem into a mathematical one for solution." }, { "problem_id": 1306, "question": "As shown in the figure, standing at point D, which is 20 m away from the base of the tower BE, the top of the tower is visually observed. The angle between the line of sight AB and the horizontal line is ∠BAC = 40°, and the height of the observer's eye level AD is 1 m. Now, please draw △ABC on paper to a scale of 1:1000 and label it as △A'B'C'. Measure the length of B'C' on the paper with a ruler, and you can calculate the actual height of the tower. Please calculate the actual height of the tower.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch16-2024_06_15_20f662d06c93903aff01g_0096_1.jpg", "batch16-2024_06_15_20f662d06c93903aff01g_0096_2.jpg" ], "is_multi_img": true, "answer": " $\\mathrm{BE}=18 \\mathrm{~m}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem, triangles $\\triangle \\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}$ and $\\triangle \\mathrm{ABC}$ are similar,\n\n$\\therefore \\frac{A^{\\prime} C^{\\prime}}{A C}=\\frac{B^{\\prime} C^{\\prime}}{B C}=\\frac{1}{1000}$,\n\n$\\therefore \\mathrm{BC}=1000 \\mathrm{~B}^{\\prime} \\mathrm{C}^{\\prime}$,\n\n$\\therefore \\mathrm{BE}=\\mathrm{BC}+\\mathrm{CE}=\\mathrm{BC}+\\mathrm{AD}=1000 \\mathrm{~B}^{\\prime} \\mathrm{C}^{\\prime}+\\mathrm{AD}$.\n\nIf $\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}$ is measured to be $1.7 \\mathrm{~cm}$,\n\nthen $\\mathrm{BC}=1000 \\times 1.7=1700(\\mathrm{~cm})=17(\\mathrm{~m})$,\n\nthen the height of the tower $\\mathrm{BE}=\\mathrm{BC}+\\mathrm{CE}=\\mathrm{BC}+\\mathrm{AD}=17+1=18(\\mathrm{~m})$\n\n【Highlight】This problem examines the application of similar triangles. The key to solving the problem is constructing similar triangles through scale drawing and then setting up the proportion." }, { "problem_id": 1307, "question": "As shown in Figure 1, the ray $\\mathrm{OC}$ is inside $\\angle \\mathrm{AOB}$, and there are a total of 3 angles in the figure: $\\angle \\mathrm{AOB}$, $\\angle \\mathrm{AOC}$, and $\\angle \\mathrm{BOC}$. If the degree measure of one of these angles is twice that of another, then the ray $\\mathrm{OC}$ is called the \"clever dividing line\" of $\\angle \\mathrm{AOB}$. As shown in Figure 2, if $\\angle \\mathrm{MPN}=\\alpha$ and the ray $\\mathrm{PQ}$ is the \"clever dividing line\" of $\\angle \\mathrm{MPN}$, then $\\angle \\mathrm{MPQ}=$ $\\qquad$ (expressed in terms of $\\alpha$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_260c129fd351229be86fg_0064_1.jpg", "batch16-2024_06_15_260c129fd351229be86fg_0064_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{1}{2} \\alpha$ or $\\frac{1}{3} \\alpha$ or $\\frac{2}{3} \\alpha$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in Figure 2, $\\mathrm{PQ}$ bisects $\\angle \\mathrm{MPN}$,\n\nthat is, $\\angle \\mathrm{MPN}=2 \\angle \\mathrm{MPQ}=2 \\angle \\mathrm{NPQ}$,\n\n$\\because \\angle \\mathrm{MPN}=\\alpha$,\n\n$\\therefore \\angle \\mathrm{MPQ}=\\frac{1}{2} \\alpha$;\n\nAs shown in Figure 3, PQ is a trisector of $\\angle \\mathrm{MPN}$,\n\nthat is, $\\angle \\mathrm{NPQ}=2 \\angle \\mathrm{MPQ}$,\n\n$\\therefore \\angle \\mathrm{MPQ}=\\frac{1}{3} \\alpha$;\n\nAs shown in Figure 4, PQ is a trisector of $\\angle \\mathrm{MPN}$,\n\nthat is, $\\angle \\mathrm{MPQ}=2 \\angle \\mathrm{NPQ}$,\n\n$\\therefore \\angle \\mathrm{MPQ}=\\frac{2}{3} \\alpha$;\n\nTherefore, the answer is $\\frac{1}{2} \\alpha$ or $\\frac{1}{3} \\alpha$ or $\\frac{2}{3} \\alpha$.\n\n【Insight】This question examines the properties of rotation, the definition of a trisector, and the student's reading comprehension and knowledge transfer abilities. Understanding the definition of a \"trisector\" is key to solving the problem." }, { "problem_id": 1308, "question": "As shown in Figure 1, find a point \\( \\mathrm{C} \\) on the line segment \\( \\mathrm{AB} \\) such that \\( \\mathrm{C} \\) divides \\( \\mathrm{AB} \\) into segments \\( \\mathrm{AC} \\) and \\( \\mathrm{CB} \\), with \\( \\mathrm{BC} \\) being the shorter segment. If \\( \\mathrm{BC} \\cdot \\mathrm{AB} = \\mathrm{AC}^2 \\), then the line segment \\( \\mathrm{AB} \\) is said to be divided by point \\( \\mathrm{C} \\) in a golden ratio.\n\n\n\n## Figure 1\n\n\n\nFigure 2\n\nThe golden ratio is often applied in the fields of painting, sculpture, music, architecture, and other arts to enhance aesthetic appeal. As shown in Figure 2, in ancient China, the Forbidden City's central axis features the Taihe Gate located between the Taihe Hall and the Inner Golden Water Bridge, closer to the Inner Golden Water Bridge. The positional relationship of these three buildings satisfies the golden ratio. Given that the distance from the Taihe Hall to the Inner Golden Water Bridge is approximately 100 zhang, calculate the distance between the Taihe Gate and the Taihe Hall (approximate value of \\( \\sqrt{5} \\) is 2.2).", "input_image": [ "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0084_1.jpg", "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0084_2.jpg" ], "is_multi_img": true, "answer": "60.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: According to the problem,\n\n$x^{2}=100(100-x)$\n\n$x_{1}=-50+50 \\sqrt{5}, x_{2}=-50-50 \\sqrt{5}$ (discarded)\n\n$x \\approx-50+50 \\times 2.2=60$,\n\nAnswer: The distance from Taihe Gate to Taihe Hall is 60 zhang.\n\nTherefore, the answer is 60 zhang.\n\n[Highlight] This problem examines the concept and properties of the golden section. Dividing line segment $\\mathrm{AB}$ into two segments $\\mathrm{AC}$ and $\\mathrm{BC}$ ($\\mathrm{AC}>$ $\\mathrm{BC}$), such that $\\mathrm{AC}$ is the geometric mean of $\\mathrm{AB}$ and $\\mathrm{BC}$, is called the golden section of line segment $\\mathrm{AB}$." }, { "problem_id": 1309, "question": "As shown in the figure, observe the pattern of changes in the number of triangles in the following figures. Then, in the $\\mathrm{n}$th figure, there are a total of $\\boxed{n}$ triangles (expressed as an algebraic expression containing the letter $\\mathrm{n}$).\n\n\n\n1st figure\n\n\n\n2nd figure\n\n\n\n3rd figure", "input_image": [ "batch16-2024_06_15_490e5a88dabee3709e18g_0067_1.jpg", "batch16-2024_06_15_490e5a88dabee3709e18g_0067_2.jpg", "batch16-2024_06_15_490e5a88dabee3709e18g_0067_3.jpg" ], "is_multi_img": true, "answer": "$4 n-3$", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "In the first figure, there is a total of 1 triangle. In the second figure, there are $1+4=5$ triangles. In the third figure, there are $1+4+4=9$ triangles. In the fourth figure, there are $1+4+4+4=13$ triangles, and so on. In the $\\mathrm{n}$th figure, the number of triangles is $1+4(n-1)=4n-3$. Therefore, the answer is $4n-3$." }, { "problem_id": 1310, "question": "The old street on the Wudong Bridge in the western part of Huangyan has been restored. As shown in Figure (1), there is a Chinese-style round gate among them. Figure (2) is a schematic diagram of its plane. It is known that $A B$ passes through the center $O$ of the circle and is perpendicular to $C D$ at point $B$. The height of the gate opening $A B$ is measured to be 1.8 meters, and the width of the lower edge of the gate opening $C D$ is 1.2 meters. The radius of the round gate opening is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch16-2024_06_15_50e11673fec61e0fbdd1g_0065_1.jpg", "batch16-2024_06_15_50e11673fec61e0fbdd1g_0065_2.jpg" ], "is_multi_img": true, "answer": "1 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the radius of the circular doorway be \\( r \\).\n\nSince \\( AB \\) passes through the center \\( O \\) and is perpendicular to \\( CD \\) at point \\( B \\),\n\nconnecting \\( OC \\), in the right triangle \\( \\triangle OCB \\), we have: \\( r^{2} = (1.8 - r)^{2} + 0.6^{2} \\).\n\nSolving this, we get: \\( r = 1 \\).\n\nTherefore, the answer is: 1 meter.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n【Key Insight】This problem tests the theorem of perpendicular chords, with the key being to solve it using the theorem of perpendicular chords and the Pythagorean theorem." }, { "problem_id": 1311, "question": "The \"Nine Chapters on the Mathematical Art\" records: There is a cylindrical log buried in the wall, its diameter unknown. When it is sawn, the depth is one inch, and the sawing path is one foot long. What is the diameter?\n\nTranslation: There is a cylindrical log buried in the wall (as shown in Figure (1)). The diameter is unknown, so it is sawn (along the cross-section) (as shown in Figure (2)). When the depth $CE$ is measured to be 1 inch, the sawn width $AB$ is 1 foot. What is the diameter $CD$ of the log in inches? (1 foot = 10 inches)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch16-2024_06_15_52a5061866b853c07e89g_0022_1.jpg", "batch16-2024_06_15_52a5061866b853c07e89g_0022_2.jpg" ], "is_multi_img": true, "answer": "26", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Connect $OA$, as shown in the figure:\n\n\n\nLet the radius of circle $O$ be $x$ inches, then $OE = (x - 1)$ inches.\n\nSince $OE \\perp AB$ and $AB = 10$ inches,\n\nTherefore, $AC = BC = \\frac{1}{2} AB = 5$ inches.\n\nIn the right triangle $\\triangle AOE$, by the Pythagorean theorem, we have: $x^{2} = (x - 1)^{2} + 5^{2}$.\n\nSolving this, we get: $x = 13$.\n\nThus, the diameter of circle $O$, $AC = 2x = 26$ inches.\n\nThat is, the diameter of the timber $CD$ is 26 inches.\n\nTherefore, the answer is: 26.\n\n[Highlight] This problem examines the application of the perpendicular chord theorem and the Pythagorean theorem. The key to solving the problem is to learn how to use parameters to construct equations and solve them." }, { "problem_id": 1312, "question": "The waterwheel is an ancient Chinese invention used for irrigation. In his book \"Nong Zheng Quan Shu,\" Ming Dynasty scientist Xu Guangqi illustrated the working principle of the waterwheel with diagrams. As shown in Figure 1, the water buckets of the waterwheel follow a circular path centered at point $O$, as depicted in Figure 2. Given that the center $O$ is above the water surface, and the chord $AB$ of the circle $\\odot O$ intercepted by the water surface is 8 meters long, with the radius of $\\odot O$ being 6 meters. If point $C$ is the lowest point on the circular path, the distance from point $C$ to the line containing chord $AB$ is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_52a5061866b853c07e89g_0042_1.jpg", "batch16-2024_06_15_52a5061866b853c07e89g_0042_2.jpg" ], "is_multi_img": true, "answer": "$6-2 \\sqrt{5}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Connect $OA$ and $OC$, with $OC$ intersecting $AB$ at point $D$, as shown in the figure.\n\nFrom the problem statement, we have: $OA = OC = 6$ meters, and $OC \\perp AB$.\n\nTherefore, $AD = BD = \\frac{1}{2} AB = 4$ meters, and $\\angle ADO = 90^\\circ$.\n\nThus, $OD = \\sqrt{OA^2 - AD^2} = \\sqrt{6^2 - 4^2} = 2\\sqrt{5}$ meters.\n\nHence, $CD = OC - OD = (6 - 2\\sqrt{5})$ meters.\n\nThis means the distance from point $C$ to the line where chord $AB$ lies is $(6 - 2\\sqrt{5})$ meters.\n\nTherefore, the answer is: $6 - 2\\sqrt{5}$.\n\n\n\nFigure 2\n\n[Highlight] This problem tests the application of the Perpendicular Chord Theorem and the Pythagorean Theorem. Mastery of these theorems is essential for solving the problem." }, { "problem_id": 1313, "question": "The waterwheel is an ancient Chinese invention used for irrigation. In his book \"Nong Zheng Quan Shu,\" Ming Dynasty scientist Xu Guangqi illustrated the working principle of the waterwheel with diagrams, as shown in Figure 1. The running track of the water buckets on the waterwheel is a circle with its center at the axis $O$, as shown in Figure 2. It is known that the center $O$ is above the water surface, and the chord $A B$ of the circle $\\odot O$ intercepted by the water surface is 6 meters long, with the radius of $\\odot O$ being 4 meters. If point $C$ is the lowest point of the running track, then the distance from point $C$ to the line containing chord $A B$ is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_52a5061866b853c07e89g_0055_1.jpg", "batch16-2024_06_15_52a5061866b853c07e89g_0055_2.jpg" ], "is_multi_img": true, "answer": "$(4-\\sqrt{7})$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Connect $OC$ intersecting $AB$ at $D$, and connect $OA$.\n\nSince point $C$ is the lowest point on the trajectory,\n\n$\\therefore OC \\perp AB$,\n\n$\\therefore AD = \\frac{1}{2} AB = 3$ meters.\n\nIn the right triangle $\\triangle OAD$, $OD = \\sqrt{OA^2 - AD^2} = \\sqrt{4^2 - 3^2} = \\sqrt{7}$ meters.\n\n$\\therefore$ The distance from point $C$ to the line where chord $AB$ lies is $CD = OC - OD = (4 - \\sqrt{7})$ meters.\n\nHence, the answer is: $(4 - \\sqrt{7})$.\n\n\n\nFigure 2\n\n[Insight] This problem tests the application of the Perpendicular Chord Theorem. The key to solving it is understanding that the diameter perpendicular to a chord bisects the chord and the arcs it subtends." }, { "problem_id": 1314, "question": "As shown in Figure 1, the waterwheel, also known as the Kongming Wheel, is the oldest agricultural irrigation tool in China and a precious historical and cultural heritage. As shown in Figure 2, the center $O$ of the circle is above the water surface, and the chord $AB$ intercepted by the water surface has a length of 8 meters, with a radius of 5 meters. The distance from the center $O$ to the water surface $AB$ is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_52a5061866b853c07e89g_0061_1.jpg", "batch16-2024_06_15_52a5061866b853c07e89g_0061_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Draw $OD \\perp AB$ at point $D$ through point $O$, and connect $OA$, as shown in the diagram:\n\n\n\nFigure 2\n\nThen, $AD = BD = \\frac{1}{2} AB = 4$ meters.\n\nIn the right triangle $\\triangle AOD$, by the Pythagorean theorem, we have: \n\\[\nOD = \\sqrt{OA^2 - AD^2} = \\sqrt{5^2 - 4^2} = 3 \\text{ meters}.\n\\]\n\nThis means the distance from the center $O$ to the water surface $AB$ is 3 meters.\n\nTherefore, the answer is: 3.\n\n【Key Insight】This problem tests the application of the Perpendicular Diameter Theorem and the Pythagorean Theorem. Mastering these theorems is crucial for solving such problems." }, { "problem_id": 1315, "question": "A circular paper with a radius of 4 (as shown in Figure (1)) is folded twice consecutively, resulting in creases $\\mathrm{AB}$ and $\\mathrm{CD}$, with $\\mathrm{AB} \\perp \\mathrm{CD}$, intersecting at $\\mathrm{M}$ (as shown in Figure (2)). Then, the paper is folded as shown in Figure (3) so that point $\\mathrm{B}$ coincides with point $\\mathrm{M}$, with crease $\\mathrm{EF}$ intersecting $\\mathrm{AB}$ at point $\\mathrm{N}$. Lines $\\mathrm{AE}$ and $\\mathrm{AF}$ are connected (as shown in Figure (4)). The area of $\\triangle \\mathrm{AEF}$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)", "input_image": [ "batch16-2024_06_15_52a5061866b853c07e89g_0091_1.jpg", "batch16-2024_06_15_52a5061866b853c07e89g_0091_2.jpg", "batch16-2024_06_15_52a5061866b853c07e89g_0091_3.jpg", "batch16-2024_06_15_52a5061866b853c07e89g_0091_4.jpg" ], "is_multi_img": true, "answer": "$12 \\sqrt{3}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:**\n\nAs shown in the figure, connect EM.\n\nFrom the properties of folding and the perpendicular chord theorem, we know that $\\mathrm{MN}=\\frac{1}{2} \\mathrm{MB}=2$, $\\mathrm{MN} \\perp \\mathrm{EF}$, and $\\mathrm{EN}=\\mathrm{NF}$. Therefore, $\\mathrm{AN}=6$.\n\nIn the right triangle $\\triangle \\mathrm{EMN}$, using the Pythagorean theorem, we find $\\mathrm{EN}=2 \\sqrt{3}$, so $\\mathrm{EF}=4 \\sqrt{3}$.\n\nThus, the area of $\\triangle AEF$ is:\n$$\nS_{\\triangle A E F}=\\frac{1}{2} \\cdot EF \\cdot AN=\\frac{1}{2} \\cdot 4 \\sqrt{3} \\cdot 6=12 \\sqrt{3}.\n$$\n\n**Key Points:**\n1. Folding problems.\n2. Perpendicular chord theorem.\n3. Pythagorean theorem." }, { "problem_id": 1316, "question": "As shown in Figure 1, two triangular plates \\(ABC\\) and \\(DEF\\) with angles of \\(30^\\circ\\) and \\(45^\\circ\\) respectively are superimposed, with sides \\(BC\\) and \\(EF\\) coinciding. \\(BC = EF = 12 \\, \\text{cm}\\), and point \\(P\\) is the midpoint of side \\(BC (EF)\\). Now, the triangular plate \\(ABC\\) is rotated around point \\(P\\) in a counterclockwise direction by an angle \\(\\alpha\\) (as shown in Figure 2). Let side \\(AB\\) intersect \\(EF\\) at point \\(Q\\). Then, as \\(\\alpha\\) varies from \\(0^\\circ\\) to \\(90^\\circ\\), the path length of point \\(Q\\) is \\(\\qquad\\) (the result is to be left in radical form).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_828470f8648110c78ab2g_0035_1.jpg", "batch16-2024_06_15_828470f8648110c78ab2g_0035_2.jpg" ], "is_multi_img": true, "answer": "$(6-2 \\sqrt{3}) \\mathrm{cm}$;", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As angle $a$ changes from $0^{\\circ}$ to $90^{\\circ}$, point $Q$ moves from $E$ to $Q$ (as shown in the figure).\n\nGiven that $EF = 12 \\mathrm{~cm}$,\n\nTherefore, $BP = 6 \\mathrm{~cm}$,\n\nGiven that $\\angle B = 30^{\\circ}$,\n\nIn the right triangle $\\triangle BPQ$, $QP = 2 \\sqrt{3} \\mathrm{~cm}$,\n\nThus, $EQ = (6 - 2 \\sqrt{3}) \\mathrm{cm}$,\n\nTherefore, the path of point $Q$ is $(6 - 2 \\sqrt{3}) \\mathrm{cm}$,\n\nHence, the answer is $(6 - 2 \\sqrt{3}) \\mathrm{cm}$;\n\n\n\n【Insight】This problem examines the trajectory of a moving point; the key to solving it lies in determining that the path of point $Q$ is a line segment by analyzing the rotation of the triangle and the given angles of $0^{\\circ}$ and $90^{\\circ}$." }, { "problem_id": 1317, "question": "As shown, it is given that $\\triangle \\mathrm{ACB}$ and $\\triangle \\mathrm{DFE}$ are two congruent right triangles, with their hypotenuses measuring $10 \\mathrm{~cm}$ and the smaller acute angle being $30^{\\circ}$. The two triangles are arranged as shown in Figure (1), with points $\\mathrm{B}$, $\\mathrm{C}$, $\\mathrm{F}$, and $\\mathrm{D}$ lying on the same straight line, and point $\\mathrm{C}$ coinciding with point $\\mathrm{F}$. In Figure (1), $\\triangle \\mathrm{ACB}$ is rotated clockwise around point $\\mathrm{C}$ to the position shown in Figure (2), with point $\\mathrm{E}$ on side $\\mathrm{AB}$ and $\\mathrm{AC}$ intersecting $\\mathrm{DE}$ at point $\\mathrm{G}$. The distance between $\\mathrm{B}$ and $\\mathrm{D}$ is $\\qquad$ cm (express in surd form).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch16-2024_06_15_828470f8648110c78ab2g_0092_1.jpg", "batch16-2024_06_15_828470f8648110c78ab2g_0092_2.jpg" ], "is_multi_img": true, "answer": "$5 \\sqrt{7}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** \nUsing the fact that triangles $\\triangle \\mathrm{ACB}$ and $\\triangle \\mathrm{DFE}$ are congruent right triangles, with the hypotenuse $\\mathrm{AB} = 10 \\mathrm{~cm}$ and $\\angle \\mathrm{A} = 30^{\\circ}$, we can find $\\mathrm{BC}$. By rotating the triangle by $60^{\\circ}$, we can determine $\\angle \\mathrm{BCF} = 30^{\\circ}$, and subsequently find the lengths of $\\mathrm{BF}$ and $\\mathrm{FC}$. Finally, we can calculate $\\mathrm{BD}$.\n\n**Problem Solution:** \nConnect $\\mathrm{BD}$, and draw $\\mathrm{BF} \\perp \\mathrm{DC}$ at point $\\mathrm{F}$.\n\n\n\nFigure (2)\n\nFrom the given information, in the right triangle $\\triangle \\mathrm{ABC}$,\n\n$\\angle \\mathrm{A} = 30^{\\circ}, \\angle \\mathrm{B} = 60^{\\circ}$.\n\nBy the properties of rotation, in Figure (2), $\\mathrm{CB} = \\mathrm{CE}$.\n\nThus, $\\triangle \\mathrm{BCE}$ is an equilateral triangle.\n\nTherefore, $\\angle \\mathrm{ECB} = 60^{\\circ}, \\angle \\mathrm{BCF} = 30^{\\circ}$.\n\nSince $\\mathrm{AB} = 10 \\mathrm{~cm}$,\n\n$\\mathrm{BC} = 5 \\mathrm{~cm}, \\quad \\mathrm{AC} = \\mathrm{CD} = 5 \\sqrt{3} \\mathrm{~cm}$.\n\nHence, $\\mathrm{BF} = \\frac{1}{2} \\times 5 = \\frac{5}{2} \\mathrm{~cm}, \\mathrm{FC} = \\frac{5 \\sqrt{3}}{2} \\mathrm{~cm}$.\n\nThen, $\\mathrm{DF} = \\mathrm{FC} + \\mathrm{DC} = \\frac{15 \\sqrt{3}}{2} \\mathrm{~cm}$.\n\nIn the right triangle $\\triangle \\mathrm{BFD}$, $\\mathrm{BD} = \\sqrt{BF^{2} + DF^{2}} = \\sqrt{\\left(\\frac{5}{2}\\right)^{2} + \\left(\\frac{15 \\sqrt{3}}{2}\\right)^{2}} = 5 \\sqrt{7} \\mathrm{~cm}$.\n\n**Key Concept:** Properties of rotation." }, { "problem_id": 1318, "question": "As shown in Figure 1, there is a fan-shaped promotional board. A partial schematic of this board is shown in Figure 2, which is a sector formed with $O$ as the center, $O A$ and $O B$ as the radii, and a central angle $\\angle O = 120^{\\circ}$. If $O A = 3 \\mathrm{~m}$ and $O B = 1 \\mathrm{~m}$, then the area of the shaded part is $\\qquad$ $\\mathrm{m}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_86dd4d7340b645fc5c07g_0011_1.jpg", "batch16-2024_06_15_86dd4d7340b645fc5c07g_0011_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{8}{3} \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: The shaded area \\( S_{\\text{shaded}} \\) is calculated as the difference between the area of sector \\( DAA \\) and the area of sector \\( BOC \\).\n\n\\[\nS_{\\text{shaded}} = S_{\\text{sector } DAA} - S_{\\text{sector } BOC}\n\\]\n\n\\[\n= \\frac{120 \\pi \\times 9}{360} - \\frac{120 \\pi \\times 1}{360}\n\\]\n\n\\[\n= \\frac{8}{3} \\pi \\text{ m}^2\n\\]\n\nTherefore, the answer is: \\(\\frac{8}{3} \\pi\\).\n\n**Key Insight:** This problem tests the calculation of the area of a sector. The key to solving it is to master the sector area formula \\( S = \\frac{n \\pi R^{2}}{360} \\)." }, { "problem_id": 1319, "question": "[Knowledge Background]: A triangle is a common basic shape in mathematics, with the sum of its three angles being $180^{\\circ}$. An isosceles triangle is a special type of triangle. If a triangle has two equal sides, then it is an isosceles triangle, and the angles opposite the equal sides are also equal.\n\nAs shown in Figure 1, in triangle $ABC$, if $AB = AC$, then $\\angle B = \\angle C$. Similarly, if $\\angle B = \\angle C$, then $AB = AC$, meaning the triangle is also isosceles.\n\n[Knowledge Application]: As shown in Figure 2, in triangle $ABC$, $\\angle ACB = 90^{\\circ}$, $\\angle ABC = 30^{\\circ}$. The triangle $ABC$ is rotated counterclockwise around point $C$ by $\\alpha$ degrees ($0^{\\circ} < \\alpha < 60^{\\circ}$) (i.e., $\\angle ECB = \\alpha$ degrees), resulting in the corresponding triangle $DEC$. Line $CE$ intersects $AB$ at point $H$, and line $BE$ is connected. If triangle $BEH$ is isosceles, then $\\alpha =$ $\\qquad$ $\\circ$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_8b23d1ebd355773437b5g_0067_1.jpg", "batch16-2024_06_15_8b23d1ebd355773437b5g_0067_2.jpg" ], "is_multi_img": true, "answer": "$\\alpha=40$ or 20 .", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since triangle \\( ABC \\) is rotated counterclockwise around point \\( C \\) by an angle \\( \\alpha \\) (where \\( 0^\\circ < \\alpha < 60^\\circ \\)), it follows that \\( CE = CB \\) and \\( \\angle ECB = \\alpha \\).\n\nTherefore, \\( \\angle CEB = \\angle CBE = 90^\\circ - \\frac{\\alpha}{2} \\).\n\nGiven that \\( \\angle ABC = 30^\\circ \\), we have:\n- \\( \\angle BHE = 30^\\circ + \\alpha \\),\n- \\( \\angle EBH = 60^\\circ - \\frac{\\alpha}{2} \\).\n\nIf \\( BE = BH \\), then:\n\\[ 30^\\circ + \\alpha = 90^\\circ - \\frac{\\alpha}{2} \\]\n\\[ \\alpha = 40^\\circ \\]\n\nIf \\( EH = BH \\), then:\n\\[ 90^\\circ - \\frac{\\alpha}{2} = 60^\\circ - \\frac{\\alpha}{2} \\]\nThis equation has no solution.\n\nIf \\( EH = BE \\), then:\n\\[ 30^\\circ + \\alpha = 60^\\circ - \\frac{\\alpha}{2} \\]\n\\[ \\alpha = 20^\\circ \\]\n\nIn summary, \\( \\alpha = 40^\\circ \\) or \\( 20^\\circ \\).\n\n**Key Insight:** This problem examines the properties of rotation and the determination and properties of isosceles triangles. The key to solving the problem lies in utilizing the method of classification and discussion." }, { "problem_id": 1320, "question": "There is a right-angled triangle iron sheet with the lengths of the two legs being $\\mathrm{AC}=3 \\mathrm{~cm}$ and $\\mathrm{BC}=4 \\mathrm{~cm}$. To cut a square from it, there are two methods: one is to place one side of the square on the hypotenuse of the right-angled triangle, with the other two vertices on the two legs, as shown in Figure (1); the other is to place one pair of adjacent sides on the two legs of the right-angled triangle, with the other vertex on the hypotenuse, as shown in Figure (2). Calculate and explain which method results in a larger area for the square in both cases?\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch16-2024_06_15_a8c0d8712015e48b26a0g_0022_1.jpg", "batch16-2024_06_15_a8c0d8712015e48b26a0g_0022_2.jpg" ], "is_multi_img": true, "answer": "The area of ​​the square in the second case is larger.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) Since $\\triangle ABC$ is a right-angled triangle with side lengths of $3 \\mathrm{~cm}$ and $4 \\mathrm{~cm}$, then $AB=\\sqrt{3^{2}+4^{2}}=5$. Draw the altitude $CH$ on side $AB$, intersecting $DG$ at point $Q$.\n\nThus, $\\frac{5 CH}{2}=\\frac{3 \\times 4}{2}$, so $CH=\\frac{12}{5} \\mathrm{~cm}$.\n\nIt is easy to see that: $\\triangle DCG \\sim \\triangle ACB$, hence: $\\frac{CQ}{CH}=\\frac{DG}{AB}$.\n\nLet the side length of square $DEFG$ be $x \\mathrm{~cm}$, then: $\\frac{\\frac{12}{5}-x}{\\frac{12}{5}}=\\frac{x}{5}$, solving gives: $x=\\frac{60}{37}$.\n\n(2) Let $AC=3 \\mathrm{~cm}$, and set the side length of the square to $y \\mathrm{~cm}$.\n\nIt is easy to see that: $\\triangle ADE \\backsim \\triangle ACB$, thus: $\\frac{AD}{AC}=\\frac{DE}{CB}, \\frac{3-y}{3}=\\frac{y}{4}$, solving gives: $y=\\frac{12}{7}$.\n\n$\\because \\frac{60}{37}<\\frac{12}{7}, \\therefore$ the area of the square in the second scenario is larger.\n\n\n\n(1)\n\n\n\n(2)\n\n【Key Insight】 (1) The key to solving this problem is using the area method to find the altitude on the hypotenuse of the right-angled triangle;\n\n(2) It can be solved using $\\triangle ADE \\sim \\triangle ACB$ or $\\triangle BFE \\sim \\triangle BCA$." }, { "problem_id": 1321, "question": "A certain middle school installed wooden supports for cylindrical water barrels inside the school (as shown in Figure (1)). If the thickness of the wooden strips is not considered, the top view is shown in Figure (2). It is known that $\\mathrm{AD}$ bisects $\\mathrm{BC}$ at right angles, and $\\mathrm{AD}=\\mathrm{BC}=40 \\mathrm{~cm}$. The maximum radius of the base of the cylindrical water barrel is $\\mathrm{cm}$.\n\n\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch16-2024_06_15_aa818076c564b88bd64ag_0032_1.jpg", "batch16-2024_06_15_aa818076c564b88bd64ag_0032_2.jpg", "batch16-2024_06_15_aa818076c564b88bd64ag_0032_3.jpg" ], "is_multi_img": true, "answer": "25", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** \nTake any point **O** on **AD**. When **AO = BO**, the radius of the cylindrical water bucket reaches its maximum. Let **AO = BO = x**. Using the Pythagorean theorem, we can set up an equation to solve for **x**.\n\n\n\nLet **AO = BO = x**. According to the problem, we have:\n\n\\[\n\\left(\\frac{40}{2}\\right)^{2} + (40 - x)^{2} = x^{2}\n\\]\n\nSolving this equation gives:\n\n\\[\nx = 25\n\\]\n\nThus, the maximum radius of the cylindrical water bucket is **25 cm**.\n\n**Key Point:** Application of the Pythagorean theorem.\n\n**Comment:** The key to solving the problem lies in understanding the problem statement and the diagram, drawing the correct figure based on the given information, and accurately applying the Pythagorean theorem to find the solution." }, { "problem_id": 1322, "question": "As shown in Figure 1, the broken line segment \\( AOB \\) divides the circle \\( \\odot O \\) with area \\( S \\) into two sectors, with the areas of the larger and smaller sectors being \\( S_1 \\) and \\( S_2 \\) respectively. If \\( \\frac{S_1}{S} = \\frac{S_2}{S_1} = 0.618 \\), the smaller sector is called a \"golden sector.\" The folding fan in daily life (as shown in Figure 2) is roughly a \"golden sector.\" Therefore, the central angle of the \"golden sector\" is approximately \\(\\qquad\\) (accurate to 0.1). \n\nfigure 1\n\n\n\nfigure 2", "input_image": [ "batch16-2024_06_15_aa818076c564b88bd64ag_0067_1.jpg", "batch16-2024_06_15_aa818076c564b88bd64ag_0067_2.jpg" ], "is_multi_img": true, "answer": "$137.5^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since \\(\\frac{S_{1}}{S} = \\frac{S_{2}}{S_{1}} = 0.618\\),\n\nTherefore, the ratio of the central angles of the small sector to the large sector is 0.618.\nLet the degree of \\(\\angle AOB\\) be \\(0.618x\\), then the other angle is \\(x\\),\n\n\\(0.618x + x = 360\\)\n\nSolving gives \\(x \\approx 222.497\\),\n\n\\(0.618x \\approx 137.5^\\circ\\).\n\nHence, the answer is \\(137.5^\\circ\\).\n\nKey point: Concept of angles." }, { "problem_id": 1323, "question": "As shown in Figure 1, a revolving door consists of three rotating blades within a cylindrical space, dividing the cylindrical space into three equal sector spaces. Points $\\mathrm{AB}$ and $\\mathrm{CD}$ are the entrances and exits. During rotation, when one end of a rotating blade coincides with point $\\mathrm{B}$, one end of another blade among the remaining two coincides with point $\\mathrm{D}$; continuing to rotate, when one end of a rotating blade coincides with point $\\mathrm{A}$, one end of another blade among the remaining two coincides with point $\\mathrm{C}$. Figure 2 is a top-down schematic view, with point $\\mathrm{O}$ as the center of the circle. If the diameter of circle $\\mathrm{O}$ is 3 meters and the widths of the revolving door entrances and exits are equal, then the width of the revolving door entrances and exits is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_ae4516cb9d15abae975dg_0017_1.jpg", "batch16-2024_06_15_ae4516cb9d15abae975dg_0017_2.jpg" ], "is_multi_img": true, "answer": "1.5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Connect OA,\n\nFrom the problem statement, we have: $\\angle \\mathrm{AOB}=\\frac{1}{2} \\angle \\mathrm{BOD}=60^{\\circ}$,\n\nTherefore, $\\triangle \\mathrm{AOB}$ is an equilateral triangle,\n\nHence, $\\mathrm{AB}=\\mathrm{OB}=1.5$ meters,\n\nSo the answer is 1.5.\n\n\n\nFigure 2\n\n[Insight] This problem tests the basic properties of circles and the determination and properties of equilateral triangles. The key to solving the problem lies in correctly understanding the given condition that $\\angle \\mathrm{AOB}=\\frac{1}{2} \\angle \\mathrm{BOD}$." }, { "problem_id": 1324, "question": "As shown in the figure, there are four circles $\\mid A, B, C, D$, and the sum of the radii of circles $\\mathrm{A}$ and $\\mathrm{B}$ equals the radius of circle $\\mathrm{C}$, while the sum of the radii of circles $\\mathrm{B}$ and $\\mathrm{C}$ equals the radius of circle $\\mathrm{D}$. The circles $\\mathrm{A}, B, C$ are arranged as shown in Figure A, and the circles $\\mathrm{B}, C, D$ are arranged as shown in Figure B. If the shaded areas in Figure A and Figure B are $4 \\pi$ and $12 \\pi$ respectively, then the area of circle $\\mathrm{D}$ is $\\qquad$.\n\n\n\n\n\nFigure A\n\n", "input_image": [ "batch16-2024_06_15_ae4516cb9d15abae975dg_0022_1.jpg", "batch16-2024_06_15_ae4516cb9d15abae975dg_0022_2.jpg", "batch16-2024_06_15_ae4516cb9d15abae975dg_0022_3.jpg" ], "is_multi_img": true, "answer": "$28 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Since the area of the shaded part in Figure A is $4\\pi$, which means the area of circle A is $4\\pi$,\n\nTherefore, the radius of circle A is 2.\nLet the radius of circle B be $b$, then the radius of circle C is $b+2$, and thus the radius of circle D is $2b+2$.\n\nAccording to Figure B, we have $\\pi(2b+2)^{2}=12\\pi + \\pi b^{2} + \\pi(b+2)^{2}$.\n\nSimplifying this equation gives $b^{2} + 2b = 6$,\n\nTherefore, the area of circle D is $\\pi(2b+2)^{2} = 4\\pi(b^{2} + 2b) + 4\\pi = 28\\pi$,\n\nHence, the answer is: $28\\pi$\n\n[Insight] This problem mainly tests the calculation of the area of a circle. The key to solving it is to find the equivalent relationship based on the figure and set up the equation for solving." }, { "problem_id": 1325, "question": "Figure 1 shows an artistic arch, with a rectangular base \\(ABCD\\), where the lengths of \\(AB\\) and \\(AD\\) are \\(2\\sqrt{3} \\, \\text{m}\\) and \\(4 \\, \\text{m}\\) respectively. The upper part is an arc \\(CD\\) with center \\(O\\) and \\(\\angle COD = 120^\\circ\\). It is desired to topple the arch by using point \\(B\\) as the pivot. During the toppling process, the plane containing the rectangle \\(ABCD\\) remains perpendicular to the ground, as shown in Figure 2. Let \\(\\theta\\) be the angle formed by \\(BC\\) with the horizontal line of the ground, and let \\(h\\) be the distance from a point on the arch to the ground. When \\(h\\) reaches its maximum value, the corresponding \\(\\theta\\) is \\(\\qquad\\) \\(\\sim^\\circ\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_ae4516cb9d15abae975dg_0047_1.jpg", "batch16-2024_06_15_ae4516cb9d15abae975dg_0047_2.jpg" ], "is_multi_img": true, "answer": "$60^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $\\mathrm{OB}$,\n\nDraw a perpendicular from point $\\mathrm{O}$ to $\\mathrm{AB}$, intersecting $\\mathrm{DC}$ at point $\\mathrm{E}$ and $\\mathrm{AB}$ at point $\\mathrm{F}$.\n\nWhen $\\mathrm{OB}$ is perpendicular to the base, $\\mathrm{h}$ reaches its maximum value;\n\n$\\because \\angle \\mathrm{DOC}=120^{\\circ}$\n\n$\\therefore \\angle \\mathrm{EOC}=60^{\\circ}$\n\nUsing trigonometric functions,\n\n$\\mathrm{OC} \\times \\sin 60^{\\circ}=\\mathrm{EC}$\n\n$\\because \\mathrm{DC}=2 \\sqrt{3}$\n\n$\\therefore \\mathrm{EC}=\\sqrt{3}$\n\n$\\therefore \\mathrm{OC}=\\frac{\\sqrt{3}}{\\sin 60^{\\circ}}=2$\n\n$\\therefore \\mathrm{OE}=1$\n\nThen $\\mathrm{OF}=3$\n\n$\\because \\tan \\angle B O F=\\frac{B F}{O F}=\\frac{\\sqrt{3}}{3}$\n\n$\\therefore \\angle \\mathrm{BOF}=30^{\\circ}$\n\n$\\because \\mathrm{OF} / / \\mathrm{BC}$\n\n$\\therefore \\angle \\mathrm{OBC}=30^{\\circ}$\n\nWhen $\\mathrm{OB}$ rotates to the position of $\\mathrm{BC}$, $\\mathrm{h}$ reaches its maximum value,\n\nAt this time, $\\mathrm{BC}$ has also rotated by $30^{\\circ}$\n\nThus, $\\theta=90^{\\circ}-30^{\\circ}=60^{\\circ}$\n\nTherefore, the answer to this question is $60^{\\circ}$.\n\n\n\n【Highlight】This question tests knowledge points such as trigonometric functions and rotation, requiring analysis of the changes during the movement process." }, { "problem_id": 1326, "question": "As shown in Figure 1, the museum exhibits a wheel from the Warring States period. According to the \"Zhouli·Kaogongji\" (周礼·考工记), it is recorded that \"...the wheels of military chariots are six feet and six inches, and the wheels of farming carts are six feet and three inches...\" Based on this, to verify the type of wheel exhibited in the museum, we can infer by calculating the radius of the wheel. As shown in Figure 2, two points $A$ and $B$ are selected on the wheel, and the center of the circle where $AB$ lies is denoted as $O$ with a radius of $r \\mathrm{~cm}$. A perpendicular line $OC$ is drawn to the chord $AB$, with $D$ as the foot of the perpendicular. After measurement, $AB = 120 \\mathrm{~cm}$ and $CD = 30 \\mathrm{~cm}$, the radius of this wheel is $\\qquad$ $\\mathrm{cm}$. Through unit conversion (during the Warring States period, one foot was approximately $23 \\mathrm{~cm}$), the diameter of the wheel is approximately six feet and six inches, which verifies that this wheel is from a military chariot.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_b47dffe6cd34dd7f2e6eg_0066_1.jpg", "batch16-2024_06_15_b47dffe6cd34dd7f2e6eg_0066_2.jpg" ], "is_multi_img": true, "answer": "75", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since \\( OC \\perp AB \\) and \\( AB = 120 \\, \\text{cm} \\),\n\n\\[ AD = \\frac{1}{2} AB = 60 \\, \\text{cm}. \\]\n\nAccording to the problem, \\( OD = (r - 30) \\, \\text{cm} \\).\n\nIn the right triangle \\( \\triangle OAD \\), by the Pythagorean theorem:\n\n\\[ r^{2} = 60^{2} + (r - 30)^{2}, \\]\n\nSolving this equation yields:\n\n\\[ r = 75. \\]\n\nThus, the radius of the wheel is \\( 75 \\, \\text{cm} \\).\n\nThe answer is: 75.\n\n**Key Insight:** This problem tests the **Perpendicular Diameter Theorem**, which states that a diameter perpendicular to a chord bisects the chord and the arcs it subtends. It also involves the application of the **Pythagorean Theorem**." }, { "problem_id": 1327, "question": "Zisha is a unique handmade pottery craft in China, and its production process requires dozens of different tools. One of these tools is called the \"calibrated mouth frame,\" whose shape and usage are shown in Figure 1. When the potter aligns the arc part of the \"calibrated mouth frame\" perfectly with the boundary of the pot's mouth, it ensures that the centers of the pot's spout, handle, and mouth are on the same straight line. Figure 2 is a schematic diagram of the correct usage of this tool. As shown in Figure 3, $O$ is the mouth of a certain zisha pot, and it is known that points $A$ and $B$ lie on $O$. The line $l$ passes through point $O$ and is perpendicular to $AB$ at point $D$, intersecting $O$ at point $C$. If $AB = 30 \\mathrm{~mm}$ and $CD = 5 \\mathrm{~mm}$, then the radius $r$ of the mouth of this zisha pot is $\\qquad$ $\\mathrm{mm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0001_1.jpg", "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0001_2.jpg", "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0001_3.jpg" ], "is_multi_img": true, "answer": "25", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Given that \\( AB = 30 \\text{ mm} \\), \\( CD = 5 \\text{ mm} \\), radius \\( r \\), and \\( l \\perp AB \\),\n\nwe have:\n- \\( OD = (r - 5) \\text{ mm} \\),\n- \\( BD = \\frac{1}{2} AB = 15 \\text{ mm} \\),\n- \\( OB = r \\text{ mm} \\).\n\nAccording to the Pythagorean theorem:\n\\[\n(r - 5)^2 + 15^2 = r^2\n\\]\n\nSolving the equation:\n\\[\n(r - 5)^2 + 225 = r^2 \\\\\nr^2 - 10r + 25 + 225 = r^2 \\\\\n-10r + 250 = 0 \\\\\n10r = 250 \\\\\nr = 25 \\text{ mm}\n\\]\n\nTherefore, the answer is: 25.\n\n**Key Insight:** This problem tests the understanding of the Perpendicular Chord Bisector Theorem and the Pythagorean Theorem. Mastering these theorems is crucial for solving such problems." }, { "problem_id": 1328, "question": "The waterwheel is an ancient Chinese invention used for irrigation, showcasing the wisdom of ancient Chinese laborers. As shown in Figure 1, point $\\mathrm{P}$ represents a bucket on the waterwheel. In Figure 2, when the waterwheel is in operation, the path of the bucket is a circle with center $\\mathrm{O}$ and radius $5 \\mathrm{~m}$, with the center of the circle above the water surface. If the chord $\\mathrm{AB}$ formed by the intersection of the circle and the water surface has a length of $8 \\mathrm{~m}$, then the maximum depth of the bucket below the water surface when the waterwheel is operating is $\\qquad$ $\\mathrm{m}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0073_1.jpg", "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0073_2.jpg" ], "is_multi_img": true, "answer": "2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw a radius $\\mathrm{OD} \\perp \\mathrm{AB}$ at point $\\mathrm{E}$, as shown in the figure.\n\nThus, $\\mathrm{AE} = \\mathrm{BE} = \\frac{1}{2} \\mathrm{AB} = \\frac{1}{2} \\times 8 = 4$.\n\nIn the right triangle $\\triangle \\mathrm{AEO}$, $\\mathrm{OE} = \\sqrt{OA^{2} - AE^{2}} = \\sqrt{5^{2} - 4^{2}} = 3$.\n\nTherefore, $\\mathrm{ED} = \\mathrm{OD} - \\mathrm{OE} = 5 - 3 = 2$ (m).\n\nAnswer: When the waterwheel is operating, the maximum depth of the water bucket below the water surface is $2 \\mathrm{~m}$.\n\nHence, the answer is: 2.\n\n\n\nFigure 2\n\n[Key Insight] This problem examines the Perpendicular Chord Bisector Theorem, which states that a diameter perpendicular to a chord bisects the chord and the arcs it subtends. Proficiency in applying the Perpendicular Chord Bisector Theorem is crucial for solving this problem." }, { "problem_id": 1329, "question": "As shown in Figure 1, it is a baseball, and Figure 2 is its schematic diagram. $E$ is a point on the diameter $CD$, and points $C$ and $E$ are symmetric with respect to the chord $AB$. If $AB = 2CE = 4$, then the radius of $\\odot O$ is . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0084_1.jpg", "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0084_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{5}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $OA$,\n\n\n\nSince points $C$ and $E$ are symmetric with respect to chord $AB$,\n\n$\\therefore AB \\perp CD$, and $CF = EF = \\frac{1}{2} CE$.\n\nSince $CD$ is the diameter of $\\odot O$, and $AB = 2CE = 4$,\n\n$\\therefore AF = \\frac{1}{2} AB = 2$, and $CF = 1$.\n\nLet the radius of $\\odot O$ be $r$, i.e., $OA = OC = r$, then $OF = OC - CF = r - 1$.\n\nIn the right triangle $\\triangle AOF$, by the Pythagorean theorem: $OA^{2} = AF^{2} + OF^{2}$.\n\nThat is, $r^{2} = 2^{2} + (r - 1)^{2}$,\n\nSolving gives $r = \\frac{5}{2}$.\n\n$\\therefore$ The radius of $\\odot O$ is $\\frac{5}{2}$.\n\nHence, the answer is: $\\frac{5}{2}$.\n\n【Insight】This problem examines the application of the perpendicular chord theorem. Note the use of the right triangle formed by the radius of the circle, half the chord, and the distance from the center to the chord to solve practical problems. When tackling such problems, careful observation and analysis are essential to uncover the relationships between the segments." }, { "problem_id": 1330, "question": "As shown in Figure 1, the ancient wheel on display at the museum is recorded in the \"Zhouli・Kaogongji\" as: \"...Therefore, the wheels of military chariots are six feet and six inches, and the wheels of farming carts are six feet and three inches...\" Based on this, we can verify the type of wheel by calculating its radius. As shown in Figure 2, two points $A$ and $B$ are selected on the wheel, with the center of the circle passing through $A$ and $B$ being $O$ and the radius being $r \\mathrm{~cm}$. A perpendicular line $OC$ is drawn to the chord $AB$, with $D$ being the foot of the perpendicular. After measurement, $AB = 90 \\mathrm{~cm}$ and $CD = 15 \\mathrm{~cm}$, then $r = \\ldots \\mathrm{cm}$. Through unit conversion, the wheel diameter is approximately six feet and six inches, which verifies that this wheel is from a military chariot.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0088_1.jpg", "batch16-2024_06_15_cca9673171cfa1fd5ee0g_0088_2.jpg" ], "is_multi_img": true, "answer": "75", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, we have:\n$O D=(r-15) \\mathrm{cm}$,\n\nSince $O C \\perp A B$ and $A B=90 \\mathrm{~cm}$,\n\nTherefore, $A D=\\frac{1}{2} A B=45 \\mathrm{~cm}$,\n\nIn the right triangle $O A D$, by the Pythagorean theorem, we have: $O D^{2}+A D^{2}=O A^{2}$, which is $(r-15)^{2}+45^{2}=r^{2}$,\n\nSolving this gives $r=75(\\mathrm{~cm})$,\n\nHence, the answer is: 75.\n\n[Key Insight] This problem tests the understanding of the Perpendicular Chord Theorem and the Pythagorean theorem. Mastering the Perpendicular Chord Theorem is crucial for solving the problem." }, { "problem_id": 1331, "question": "As shown in Figure (1), point $\\mathrm{O}$ is on the straight line $A B$. A ray $O C$ is drawn such that $\\angle B O C = 60^{\\circ}$. A right-angled triangle ruler is placed as shown, with its right-angle vertex at point $\\mathrm{O}$ and one of its right-angle sides, $O P$, on ray $O A$. In Figure (1), the triangle ruler is rotated around point $\\mathrm{O}$ at a speed of $10^{\\circ}$ per second in the counterclockwise direction (as shown in Figure (2)). During the rotation, at the $\\mathrm{t}$-th second, the line $O Q$ exactly bisects $\\angle A O C$. The value of $\\mathrm{t}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_cef7e3f37d6cdcce06f4g_0035_1.jpg", "batch16-2024_06_15_cef7e3f37d6cdcce06f4g_0035_2.jpg" ], "is_multi_img": true, "answer": "3 or 21", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Draw a straight line $\\mathrm{DE}$ passing through point $\\mathrm{O}$ to bisect $\\angle \\mathrm{AOC}$, as shown in the figure.\n\nSince $\\angle BOC = 60^\\circ$,\n\nTherefore, $\\angle AOC = 120^\\circ$.\n\nBecause $\\mathrm{DE}$ bisects $\\angle \\mathrm{AOC}$,\n\nThus, $\\angle \\mathrm{AOE} = \\angle \\mathrm{BOD} = 60^\\circ$.\n\nWhen $\\mathrm{OQ}$ coincides with $\\mathrm{OD}$, the line where $OQ$ lies exactly bisects $\\angle AOC$,\n\nTherefore, $\\mathrm{t} = \\frac{180 - 90 - 60}{10} = 3$ (seconds);\n\nWhen $\\mathrm{OQ}$ coincides with $\\mathrm{OE}$, the line where $OQ$ lies exactly bisects $\\angle AOC$,\n\nTherefore, $t = \\frac{360 - 90 - 60}{10} = 21$,\n\nHence, the answer is: 3 or 21.\n\n\n\nFigure 2\n\n【Insight】This problem tests the calculation of rotation angles, the properties of angle bisectors, and the mixed operations of rational numbers. Correctly understanding the angles obtained by rotation in the figure and the position of $OQ$ is key to solving the problem." }, { "problem_id": 1332, "question": "As shown in Figure (1), $O$ is a point on the line $A B$. A ray $O C$ is drawn such that $\\angle A O C = 120^{\\circ}$. A right-angled triangle ruler is placed as shown, with its right-angle vertex at point $O$ and one of its right-angle sides, $O P$, on ray $O A$. In Figure (1), the triangle ruler is rotated around point $O$ at a speed of $5^{\\circ}$ per second in a counterclockwise direction (as shown in Figure (2)). During the rotation, at the $t$-th second, the line $O Q$ exactly bisects $\\angle B O C$. The value of $t$ is $\\qquad$.\n\n\n\n\n\nFigure (2)", "input_image": [ "batch16-2024_06_15_cef7e3f37d6cdcce06f4g_0039_1.jpg", "batch16-2024_06_15_cef7e3f37d6cdcce06f4g_0039_2.jpg" ], "is_multi_img": true, "answer": "24 or 60", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1, since \\(\\angle \\mathrm{AOC} = 120^\\circ\\),\n\n\\(\\therefore \\angle \\mathrm{BOC} = 60^\\circ\\),\n\nSince \\(\\mathrm{OQ}\\) bisects \\(\\angle \\mathrm{BOC}\\),\n\n\\(\\therefore \\angle \\mathrm{BOQ} = \\frac{1}{2} \\angle \\mathrm{BOC} = 30^\\circ\\),\n\n\\(\\therefore t = \\frac{90^\\circ + 30^\\circ}{5^\\circ} = 24 \\text{ s}\\);\n\nAs shown in Figure 2, since \\(\\angle \\mathrm{AOC} = 120^\\circ\\),\n\n\\(\\therefore \\angle \\mathrm{BOC} = 60^\\circ\\),\n\nSince \\(\\mathrm{OQ}'\\) bisects \\(\\angle \\mathrm{BOC}\\),\n\n\\(\\therefore \\angle \\mathrm{AOQ} = \\angle \\mathrm{BOQ}' = \\frac{1}{2} \\angle \\mathrm{BOC} = 30^\\circ\\),\n\n\\(\\therefore t = \\frac{180^\\circ + 30^\\circ + 90^\\circ}{5^\\circ} = 60 \\text{ s}\\),\n\nIn summary, if the line where \\(\\mathrm{OQ}\\) lies exactly bisects \\(\\angle \\mathrm{BOC}\\), then the value of \\(t\\) is either \\(24 \\text{ s}\\) or \\(60 \\text{ s}\\). Therefore, the answer is: 24 or 60.\n\n\n【Key Insight】This problem examines the definition of an angle bisector and the definition of a straight angle. Correctly drawing the diagram is key to solving the problem." }, { "problem_id": 1333, "question": "As shown in the figure, a set of triangular plates is arranged as shown in Figure A, with $\\angle A=45^{\\circ}, \\angle D=30^{\\circ}$, the hypotenuse $A B=5, C D=\\frac{17}{2}$. The triangle $\\triangle D C E$ is rotated $15^{\\circ}$ clockwise around point $C$ to obtain $\\triangle D_{1} C E_{1}$ (as shown in Figure B). At this time, $A B$ intersects $C D_{1}$ at point $O$. The length of segment $A D_{1}$ is $\\qquad$.\n\n\n\n(A)\n\n", "input_image": [ "batch16-2024_06_15_cef7e3f37d6cdcce06f4g_0062_1.jpg", "batch16-2024_06_15_cef7e3f37d6cdcce06f4g_0062_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{13}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since \\(\\angle ACB = \\angle DEC = 90^\\circ\\) and \\(\\angle D = 30^\\circ\\),\n\n\\(\\angle DCE = 90^\\circ - 30^\\circ = 60^\\circ\\),\n\n\\(\\angle ACD = 90^\\circ - 60^\\circ = 30^\\circ\\),\n\nGiven that the rotation angle is \\(15^\\circ\\),\n\n\\(\\angle ACD_{l} = 30^\\circ + 15^\\circ = 45^\\circ\\),\n\nAlso, since \\(\\angle CAB = 45^\\circ\\),\n\n\\(\\triangle ACO\\) is an isosceles right triangle,\n\n\\(\\angle ACO = \\angle BCO = 45^\\circ\\),\n\nSince \\(CA = CB\\),\n\n\\(AO = CO = \\frac{1}{2} AB = \\frac{1}{2} \\times 6 = \\frac{5}{2}\\),\n\nGiven \\(DC = \\frac{17}{2}\\),\n\n\\(D_{1}C = DC = \\frac{17}{2}\\),\n\nThus, \\(D_{I}O = \\frac{17}{2} - \\frac{5}{2} = 6\\),\n\nIn the right triangle \\(\\triangle AOD_{I}\\),\n\n\\(AD_{I} = \\sqrt{AO^2 + OD_{1}^2} = \\sqrt{\\left(\\frac{5}{2}\\right)^2 + 6^2} = \\frac{13}{2}\\).\n\nTherefore, the answer is: \\(\\frac{13}{2}\\).\n\n[Key Insight] This problem tests the properties of rotation and involves the application of the Pythagorean theorem. Determining that \\(AB \\perp CO\\) and mastering the Pythagorean theorem are crucial for solving the problem." }, { "problem_id": 1334, "question": "The Tai Chi push plate is a common fitness equipment (as shown in Figure 1), where turning two disks can exercise the body. Take the circles $A$ and $B$ with radii of 0.4 meters each (as shown in Figure 2) and $AB = 1$ meter, circle $A$ rotates counterclockwise around its center $A$ at a speed of $2^{\\circ}$ per second, and circle $B$ rotates clockwise around its center $B$ at a speed of $2^{\\circ}$ per second. When the rotation starts, the point $C$ on circle $A$ exactly falls on the line segment $AB$, and the point $D$ on circle $B$ is below $AB$ and satisfies $\\angle DBA = 60^{\\circ}$. Then, within 30 seconds of simultaneous rotation of both circles, the minimum value of $CD$ is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_e2cd64697b1ded5190edg_0018_1.jpg", "batch16-2024_06_15_e2cd64697b1ded5190edg_0018_2.jpg" ], "is_multi_img": true, "answer": "$0.5 \\# \\# \\frac{1}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Connect $CD$, and construct parallelogram $ACDE$ with adjacent sides $AC$ and $CD$. Draw a line $MN$ through $D$ parallel to $AB$,\n\n\n\nLet $\\angle CAB = 2t^\\circ$. According to the problem, $\\angle ABD = 60^\\circ - 2t^\\circ$,\n\n$\\therefore \\angle MDE = \\angle CAB = 2t^\\circ$, $\\angle BDM = 180^\\circ - \\angle ABD = 120^\\circ + 2t^\\circ$,\n\n$\\therefore \\angle BDE = 120^\\circ + 2t^\\circ + 2t^\\circ = 120^\\circ + 4t^\\circ$\n\nAlso, $DE = AC = DB$,\n\n$\\therefore \\angle EBD = \\angle BED = \\frac{180^\\circ - \\angle BDE}{2} = 30^\\circ - 2t^\\circ$,\n\n$\\therefore \\angle EBA = 30^\\circ$,\n\n$\\therefore$ When $AE \\perp EB$, the minimum length of $CD$ is $AE_{\\text{min}} = AB \\div 2 = 0.5$ meters.\n\nThus, the answer is: 0.5.\n\n【Insight】This problem examines the basic concepts of circles, the properties and determination of parallelograms, the properties of right-angled triangles with a 30-degree angle, and the properties of isosceles triangles. The key to solving this problem is to understand the given conditions and identify the necessary conditions to find the solution." }, { "problem_id": 1335, "question": "As shown in Figure 1, a circle with a radius of $7 \\mathrm{~cm}$ is divided into 20 equal parts, then cut out and rearranged into the shape shown in Figure 2. The perimeter of the resulting shape is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_e2cd64697b1ded5190edg_0028_1.jpg", "batch16-2024_06_15_e2cd64697b1ded5190edg_0028_2.jpg" ], "is_multi_img": true, "answer": "57.96", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "When a circle is approximated into a rectangle, the length of the rectangle equals half the circumference of the circle, and the width equals the radius of the circle.\n\nTherefore, the perimeter of this rectangle is longer than the original circumference of the circle by two radii, which is equivalent to the length of one diameter. That is: $3.14 \\times 2 \\times 7 + 7 \\times 2 = 57.96$.\n\nHence, the answer is: 57.96.\n\n[Highlight] This question tests the concept of shape transformation, with the main reasoning based on the derivation process of the area of a circle." }, { "problem_id": 1336, "question": "As shown in the figure, in $\\odot O$, the diameter $A B=6$, $B C$ is a chord, $\\angle A B C=30^{\\circ}$, point $P$ is on $B C$, point $Q$ is on $\\odot O$, and $O P \\perp P Q$. When point $P$ moves along $B C$, the maximum length of $P Q$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_e2f44fb13c36af2637acg_0077_1.jpg", "batch16-2024_06_15_e2f44fb13c36af2637acg_0077_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{3}{2} \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "As shown in Figure 3, connect $\\mathrm{OQ}$. Since in circle $\\odot O$, the diameter $\\mathrm{AB}=6$ and $\\mathrm{OP} \\perp \\mathrm{PQ}$,\n\n$\\therefore \\mathrm{OQ}=\\mathrm{OB}=3$, and $\\angle \\mathrm{OPQ}=90^{\\circ}$,\n\n$\\therefore \\mathrm{PQ}=\\sqrt{O Q^{2}-O P^{2}}=\\sqrt{9-O P^{2}}$,\n\n$\\therefore$ when $\\mathrm{OP}$ is the shortest, $\\mathrm{PQ}$ is the longest.\n\n$\\because$ point $\\mathrm{O}$ is a fixed point, and point $\\mathrm{P}$ is a moving point on segment $\\mathrm{BC}$,\n\n$\\therefore$ as shown in Figure 4, when $\\mathrm{OP} \\perp \\mathrm{BC}$ at point $\\mathrm{P}$, $\\mathrm{OP}$ is the shortest, and at this time, point $\\mathrm{Q}$ coincides with point $\\mathrm{C}$,\n\n$\\because \\mathrm{OP} \\perp \\mathrm{BC}$, and $\\angle \\mathrm{ABC}=30^{\\circ}$,\n\n$\\therefore \\mathrm{OP}=\\frac{1}{2} \\mathrm{OB}=\\frac{3}{2}$,\n\n$\\therefore$ at this time, $\\mathrm{PQ}=\\sqrt{9-O P^{2}}=\\sqrt{9-\\frac{9}{4}}=\\frac{3 \\sqrt{3}}{2}$.\n\nThat is, the maximum length of $\\mathrm{PQ}$ is: $\\frac{3 \\sqrt{3}}{2}$.\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nKey point: The key to solving this problem is to connect $\\mathrm{OQ}$ and use the Pythagorean theorem in right triangle $\\triangle \\mathrm{POQ}$ to obtain:\n\n$\\mathrm{PQ}=\\sqrt{O Q^{2}-O P^{2}}=\\sqrt{9-O P^{2}}$, from which it can be seen that when $\\mathrm{OP}$ is the shortest, $\\mathrm{PQ}$ is the longest; thus, when $\\mathrm{OP} \\perp \\mathrm{BC}$ and point $\\mathrm{Q}$ coincides with point $\\mathrm{C}$, $\\mathrm{PQ}$ is the longest. Combining with the given conditions, the maximum value of $\\mathrm{PQ}$ can be determined." }, { "problem_id": 1337, "question": "Figure 1 shows a toilet paper holder and a roll of toilet paper. As shown in Figure 2, the distance from the shelf to the outer edge of the roll is measured as \\( AB = 5 \\, \\text{cm} \\), and the distance to the outer edge of the core is \\( AC = 10 \\, \\text{cm} \\). The diameter of the core is \\( 10 \\, \\text{cm} \\), the thickness of one sheet of toilet paper is \\( 0.02 \\, \\text{cm} \\), the length of the hanging part of the roll is \\( 15 \\, \\text{cm} \\), and the part of the roll that is not hanging can be approximated as a ring, ignoring the seam. The total length of this roll of toilet paper is approximately \\(\\qquad\\) m. (The result should be accurate to \\( 1 \\, \\text{m} \\), and the value of \\(\\pi\\) is taken as 3.14.)\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch16-2024_06_15_ea6e762ba8e5e6688f6fg_0002_1.jpg", "batch16-2024_06_15_ea6e762ba8e5e6688f6fg_0002_2.jpg" ], "is_multi_img": true, "answer": "$118 m$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Given a cylindrical core with a diameter of $10 \\mathrm{~cm}$ and a length of $a \\mathrm{~cm}$, the total length of the rolled paper is approximately $x \\mathrm{~cm}$. According to the problem, we have:\n\n$\\pi\\left(10^{2}-5^{2}\\right) \\cdot a=0.02 \\cdot(x-15) \\cdot a$\n\nSolving the equation yields:\n\n$x \\approx 11790 \\mathrm{~cm}$\n\n$\\approx 118 \\mathrm{~m}$\n\nTherefore, the answer is: $118 \\mathrm{~m}$.\n\n[Key Insight] This problem tests the formula for the area of a circle, which is a fundamental concept. The difficulty level is easy, and mastering the relevant knowledge is key to solving the problem." }, { "problem_id": 1338, "question": "Figure 1 shows an image of the Xiaomi hanging chair, with its cross-sectional view depicted in Figure 2. The outer frame of the hanging chair is a parabola, with its highest point at point $E$. The inner frame consists of an arc $MN$ and two congruent right-angled triangles, with points $A, B, C, D$ lying on the same straight line. It is known that $BM \\perp MN, MN // AB$, the distance between points $A$ and $D$ is $80 \\text{ cm}$, and the distances from points $E$ and $N$ to the line $AB$ are $80 \\text{ cm}$ and $60 \\text{ cm}$, respectively. The triangle $MFN$ is isosceles, and by drawing $FH \\perp MN$ intersecting $MN$ at point $H$, it is found that $\\frac{FH}{HN} = \\frac{3}{4}$. The radius of the circle containing the arc $MN$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_f237e7d4a4dfea8e3947g_0054_1.jpg", "batch16-2024_06_15_f237e7d4a4dfea8e3947g_0054_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{125}{6} \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, extend $HF$, which must pass through point $E$. Extend $FH$ to intersect $BC$ at point $O$, which is the midpoint of $BC$. Taking $O$ as the origin of the coordinate system, the line passing through $A, B, C, D$ is the $x$-axis, and the line $OE$ is the $y$-axis, establishing a Cartesian coordinate system as shown in the figure.\n\n\n\nThus, $A(-40,0)$, $D(40,0)$, and $E(0,80)$.\n\nAssume the equation of the parabola is $y = a x^{2} + 80$.\n\nSubstituting $D(40,0)$ into the equation, we get $0 = a \\times 40^{2} + 80$,\n\nSolving for $a$, we find $a = -\\frac{1}{20}$,\n\nTherefore, the equation of the parabola is $y = -\\frac{1}{20} x^{2} + 80$.\n\nSince the $y$-coordinate of point $N$ is 60,\n\nFrom $60 = -\\frac{1}{20} x^{2} + 80$, we get: $x = 20$ or $x = -20$ (discarded).\n\nIn triangle $FHN$, $HN = 20$, and $\\frac{FH}{HN} = \\frac{3}{4}$,\n\nThus, $FH = \\frac{3}{4} \\times 20 = 15$.\n\nLet the center of the arc $MN$ be $P$ with radius $r$, then $HP = r - 15$.\n\nIn right triangle $PHN$, by the Pythagorean theorem: $r^{2} = 20^{2} + (r - 15)^{2}$,\n\nSolving for $r$, we get $r = \\frac{125}{6}$.\n\nTherefore, the answer is: $\\frac{125}{6} \\text{ cm}$.\n\n【Key Insight】This problem mainly examines the application of quadratic functions, basic concepts of circles, and the Pythagorean theorem. The key is to establish a Cartesian coordinate system based on the characteristics of the actual object and correctly determine the equation of the parabola." }, { "problem_id": 1339, "question": "As shown in Figure 1, a rectangular poster with the slogan \"Empty Plate Campaign, Start with Me\" is hung on the wall of a school cafeteria. On the left side of the poster is a depiction of a plate with a pair of chopsticks placed on it. The bottom edge of the poster is a horizontal line. Figure 2 is a schematic diagram of the poster. The point $\\mathrm{A}$ on the horizontal line $l$ is directly below the center $O$ of the plate. The chopsticks intersect with the plate at points $B$ and $C$ to the lower right of $O$. The segments $B D$ and $C E$ are perpendicular to $l$ at points $D$ and $E$ respectively. Measurements show that $A D = 3 D E = 15 \\mathrm{~cm}$ and $C E = \\frac{3}{2} B D = 15 \\mathrm{~cm}$. The radius of the plate $O$ is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_f237e7d4a4dfea8e3947g_0094_1.jpg", "batch16-2024_06_15_f237e7d4a4dfea8e3947g_0094_2.jpg" ], "is_multi_img": true, "answer": "25", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $OA$, and draw $OG \\perp CE$ at point $G$, intersecting $BD$ at point $F$.\n\n\n\nSince the segments $BD$ and $CE$ are perpendicular to line $l$ at points $D$ and $E$ respectively,\n$\\therefore BD \\parallel CE$,\n\n$\\therefore OF \\perp BD$,\n\nSince point $A$ on the horizontal line $l$ is directly below the center $O$ of the circle,\n\n$\\therefore OA \\perp AE$,\n\n$\\therefore$ Quadrilaterals $AOFD$ and $AOGE$ are both rectangles,\n\n$\\therefore AD = OF, \\quad AE = OG, OA = FD = GE$,\n\nGiven that $AD = 3DE = 15 \\text{ cm}, CE = \\frac{3}{2}BD = 15 \\text{ cm}$,\n\n$\\therefore OF = 15 \\text{ cm}, OG = AE = AD + DE = 20 \\text{ cm}$,\n\nLet $OA = x \\text{ cm}$, then $BF = (x - 10) \\text{ cm}, CG = (x - 15) \\text{ cm}$,\n\nIn right triangle $\\triangle BOF$, $BO^2 = OF^2 + BF^2 = 15^2 + (x - 10)^2$,\n\nIn right triangle $\\triangle COG$, $CO^2 = OG^2 + CG^2 = 20^2 + (x - 15)^2$,\n\nSince $BO = CO$,\n\n$\\therefore 15^2 + (x - 10)^2 = 20^2 + (x - 15)^2$,\n\n$\\therefore x = 30$\n\n$\\therefore BF = 20 \\text{ cm}$\n\n$\\therefore BO = \\sqrt{OF^2 + BF^2} = \\sqrt{15^2 + 20^2} = 25 \\text{ cm}$,\n\nThus, the radius of the disk $O$ is $25 \\text{ cm}$.\n\nTherefore, the answer is: 25.\n\n【Key Insight】This problem examines the knowledge of the perpendicular chord theorem and the Pythagorean theorem. Adding appropriate auxiliary lines to construct right triangles and using the Pythagorean theorem to establish the equation $15^2 + (x - 10)^2 = 20^2 + (x - 15)^2$ is the key to solving the problem." }, { "problem_id": 1340, "question": "After studying the chapter on \"Similar Figures,\" a middle school mathematics practice group decided to use their knowledge to measure the height of an ancient building \\( AB \\) (as shown in Figure 1). As shown in Figure 2, they took points \\( E \\) and \\( G \\) on the ground \\( BC \\), and erected two markers \\( EF \\) and \\( GH \\) each with a height of \\( 2 \\) meters. The distance between the markers \\( EG \\) is \\( 23 \\) meters. The ancient building \\( AB \\), the markers \\( EF \\) and \\( GH \\) are all in the same vertical plane. From the marker \\( EF \\), they stepped back \\( 2 \\) meters to point \\( D \\), where they observed point \\( A \\) such that \\( A, F, D \\) were collinear. From the marker \\( GH \\), they stepped back \\( 4 \\) meters to point \\( C \\), where they observed point \\( A \\) such that \\( A, H, C \\) were also collinear. Please use the above measurement data to help the practice group determine the height of the ancient building.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_f31f88b3fca43eb34c85g_0078_1.jpg", "batch16-2024_06_15_f31f88b3fca43eb34c85g_0078_2.jpg" ], "is_multi_img": true, "answer": "$25 \\mathrm{~m}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Since the benchmarks \\( EF \\) and \\( GH \\) are in the same vertical plane,\n\nTherefore, \\( EF \\parallel GH \\parallel AB \\),\n\nThus, \\( \\triangle DFE \\sim \\triangle DAB \\) and \\( \\triangle CHG \\sim \\triangle CAB \\),\n\nHence, \\( \\frac{EF}{AB} = \\frac{DE}{BD} \\) and \\( \\frac{GH}{AB} = \\frac{CG}{BC} \\),\n\nGiven that \\( EF = GH = 2 \\),\n\nTherefore, \\( \\frac{DE}{BD} = \\frac{CG}{BC} \\),\n\nGiven \\( DE = 2 \\), \\( EG = 23 \\), and \\( CG = 4 \\),\n\nThus, \\( \\frac{2}{BD} = \\frac{4}{BD - 2 + 23 + 4} \\),\n\nSolving gives: \\( BD = 25 \\),\n\nTherefore, \\( \\frac{2}{AB} = \\frac{2}{25} \\),\n\nSolving gives: \\( AB = 25 \\).\n\nAnswer: The height of the ancient building is \\( 25 \\mathrm{~m} \\).\n\n[Key Insight] This problem tests the application of similar triangles. Mastering the similarity theorem of triangles is crucial for solving it." }, { "problem_id": 1341, "question": "To promote the implementation of the \"Double Reduction\" policy, a school has introduced a variety of interest clubs during its after-school extended service hours. In the handcraft club, student Xiao Ming has crafted a special toy car. Figure 1 shows a side view of this toy car, where the rectangle \\( A B C D \\) represents the trunk of the car. When opening the trunk, the lid \\( A D E \\) can rotate counterclockwise around point \\( A \\). When the rotation angle is \\( 60^\\circ \\), the lid \\( A D E \\) reaches the position \\( A D' E' \\) (as shown in Figure 2, given \\( A D = 12 \\, \\text{cm}, D E = 4 \\, \\text{cm} \\)). Find the length of the path traveled by point \\( E \\) during the rotation. (The result should be expressed with square roots and \\( \\pi \\))\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_f5be70a8a25f75025825g_0022_1.jpg", "batch16-2024_06_15_f5be70a8a25f75025825g_0022_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{4 \\sqrt{10}}{3} \\pi$ cm", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Connect $AE$, $AE'$, and $EE'$ as shown in the figure.\n\n\n\nFrom the problem statement, we have: $AE' = AE$, $\\angle EAE' = 60^\\circ$,\n\n$\\therefore \\triangle AEE'$ is an equilateral triangle,\n\n$\\therefore EE' = AE$.\n\n$\\because$ Quadrilateral $ABCD$ is a rectangle,\n\n$\\therefore \\angle ADE = 90^\\circ$.\n\nIn the right triangle $\\triangle ADE$, $AD = 12$ cm, $DE = 4$ cm,\n\n$\\therefore AE = \\sqrt{AD^2 + DE^2} = \\sqrt{12^2 + 4^2} = 4\\sqrt{10}$ cm,\n\n$\\therefore$ The length of $EE'$ $= \\frac{60 \\pi \\cdot 4\\sqrt{10}}{180} = \\frac{4\\sqrt{10}}{3} \\pi$ cm.\n\nAnswer: The length of the path traced by point $E$ during rotation is $\\frac{4\\sqrt{10}}{3} \\pi$ cm.\n\n【Key Insight】This problem examines the properties of rectangles, the Pythagorean theorem, and the calculation of arc lengths. The key to solving the problem lies in using the Pythagorean theorem to find the length of $AE$ and applying the arc length formula." }, { "problem_id": 1342, "question": "The value of \\( n \\) is equal to \\(\\qquad\\) after cutting out a sector with a central angle of \\( n^{\\circ} \\) from a circle with a radius of 5 (as shown in Figure 1) and forming the cone shown in Figure 2.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_010e22232af4de0f2842g_0004_1.jpg", "batch17-2024_06_14_010e22232af4de0f2842g_0004_2.jpg" ], "is_multi_img": true, "answer": "144", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: The circumference of the base of the cone is: $2 \\pi \\times 3=6 \\pi$,\n\n$$\n\\text{Then } \\frac{(360-n) \\times \\pi \\times 5}{180}=6 \\pi \\text{,}\n$$\n\nSolving for $n$ gives: $\\mathrm{n}=144$.\n\nTherefore, the answer is 144.\n\n[Highlight] This problem mainly examines the relationship between cones and circles, and the arc length formula. The key to solving this problem lies in mastering the relevant knowledge points." }, { "problem_id": 1343, "question": "As shown in the figure, if an ant starts from point $B$ on the base of a cone and crawls along the surface to point $D$, the midpoint of the generatrix $A C$,\nthe shortest route length is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view\n\n", "input_image": [ "batch17-2024_06_14_010e22232af4de0f2842g_0038_1.jpg", "batch17-2024_06_14_010e22232af4de0f2842g_0038_2.jpg", "batch17-2024_06_14_010e22232af4de0f2842g_0038_3.jpg", "batch17-2024_06_14_010e22232af4de0f2842g_0038_4.jpg" ], "is_multi_img": true, "answer": "$3 \\sqrt{3}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By unfolding the lateral surface of the cone, we obtain the sector \\( A B B^{\\prime} \\), and the line segment \\( B F \\) represents the shortest path sought.\n\n\n\nLet \\( \\angle B A B^{\\prime} = n^{\\circ} \\).\n\nSince \\( \\frac{n \\pi \\cdot 6}{180} = 4 \\pi \\),\n\nIt follows that \\( n = 120 \\), meaning \\( \\angle B A B^{\\prime} = 120^{\\circ} \\).\n\nSince \\( E \\) is the midpoint of arc \\( B B^{\\prime} \\),\n\nTherefore, \\( \\angle A F B = 90^{\\circ} \\), \\( \\angle B A F = 60^{\\circ} \\),\n\nIn the right triangle \\( \\triangle A F B \\), \\( \\angle A B F = 30^{\\circ} \\), \\( A B = 6 \\)\n\nThus, \\( A F = 3 \\), \\( B F = \\sqrt{6^{2} - 3^{2}} = 3 \\sqrt{3} \\),\n\nTherefore, the length of the shortest path is \\( 3 \\sqrt{3} \\).\n\nHence, the answer is: \\( 3 \\sqrt{3} \\).\n\n【Highlight】This question examines the problem of finding the shortest path by \"transforming a curved surface into a plane,\" and it is of medium difficulty." }, { "problem_id": 1344, "question": "There is a rectangular paper piece $A B C D$, where $A D=4$, and a semicircle with $A D$ as its diameter, which is tangent to the opposite side $B C$, as shown in Figure (A). When it is folded along $D E$ so that point $\\mathrm{A}$ falls on $B C$, as shown in Figure (B), the area of the part of the semicircle that remains exposed (shaded area) is $\\qquad$.\n\n\n\n(A)\n\n\n\n(B)", "input_image": [ "batch17-2024_06_14_0556bc609848ade61821g_0075_1.jpg", "batch17-2024_06_14_0556bc609848ade61821g_0075_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{4}{3} \\pi-\\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in the figure, point $\\mathrm{O}$ is the center of the semicircle. A perpendicular $\\mathrm{OH}$ is drawn from $\\mathrm{O}$ to $\\mathrm{DK}$, intersecting at $\\mathrm{H}$.\n\nSince the semicircle with diameter $\\mathrm{AD}$ is tangent to the opposite side $\\mathrm{BC}$,\n\n$\\therefore \\mathrm{AD}=2 \\mathrm{CD}$.\n\nGiven that $\\angle \\mathrm{C}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{DAC}=30^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{ODK}=30^{\\circ}$.\n\nSince $\\mathrm{OD}=\\mathrm{OK}$,\n\n$\\therefore \\angle \\mathrm{DOK}=120^{\\circ}, \\angle \\mathrm{ODK}=\\angle \\mathrm{OKD}=30^{\\circ}$.\n\nThus, the area of sector ODK is $\\frac{120 \\pi \\times 4}{360}=\\frac{4}{3} \\pi$.\n\nGiven that $\\angle \\mathrm{ODK}=\\angle \\mathrm{OKD}=30^{\\circ}$ and $\\mathrm{OD}=2$,\n\n$\\therefore \\mathrm{OH}=1, \\quad \\mathrm{DH}=\\mathrm{KH}=\\sqrt{O D^{2}-O H^{2}}=\\sqrt{3}$,\n\n$\\therefore \\mathrm{DK}=2 \\sqrt{3}$.\n\nTherefore, the area of $\\triangle \\mathrm{ODK}$ is $\\frac{1}{2} \\times 1 \\times 2 \\sqrt{3}=\\sqrt{3}$.\n\nHence, the area of the part of the semicircle that remains exposed (the shaded part) is $\\left(\\frac{4}{3} \\pi-\\sqrt{3}\\right)$.\n\nThe final answer is: $\\left(\\frac{4}{3} \\pi-\\sqrt{3}\\right)$.\n\n\n\n(Part B)\n\n【Key Insight】This problem tests the properties of rectangles, folding problems, calculation of sector areas, properties of right-angled triangles with a 30-degree angle, properties of isosceles triangles, and the properties of tangents. Mastery of these concepts is crucial for solving the problem." }, { "problem_id": 1345, "question": "As shown in the figure, two types of square cards with side length 1 are illustrated in Figure (1), where the radii of the sectors on the cards are all 1; Figure (2) shows a pattern obtained by alternatingly placing cards of type $A$ and $B$; if a total of 2019 cards of both types are used to create this pattern, then this $\\qquad$. (The result should retain $\\pi$)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch17-2024_06_14_0556bc609848ade61821g_0077_1.jpg", "batch17-2024_06_14_0556bc609848ade61821g_0077_2.jpg" ], "is_multi_img": true, "answer": "$1010-\\frac{\\pi}{4}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the area of the square is 1.\n\nSince \\(2019 \\div 2 = 1009\\) with a remainder of 1,\n\nTherefore, the total area of the shaded parts in this pattern is: the area of 1009 squares plus the shaded area of A.\n\nThus, the area of the shaded parts is: \\(1009 + 1 - \\frac{\\pi}{4} = 1010 - \\frac{\\pi}{4}\\).\n\nHence, the answer is: \\(1010 - \\frac{\\pi}{4}\\).\n\n[Key Insight] This problem tests the pattern of graphical changes. The key to solving the problem lies in integrating the shaded areas of the two types of cards based on the given figure." }, { "problem_id": 1346, "question": "There is a rectangular paper piece $A B C D$, where $A D = 8 \\mathrm{~cm}$, and there is a semicircle with $A D$ as its diameter, which is tangent to the opposite side $B C$. As shown in Figure 1, folding it along $D E$ makes point $A$ fall on $B C$, as shown in Figure 2. At this time, the area of the part of the semicircle that is still exposed (the shaded part) is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_1c6ab25721766d50fc13g_0091_1.jpg", "batch17-2024_06_14_1c6ab25721766d50fc13g_0091_2.jpg" ], "is_multi_img": true, "answer": "$\\left(\\frac{16}{3} \\pi-4 \\sqrt{3}\\right)$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since the semicircle with diameter \\( AD \\) is exactly tangent to the opposite side \\( BC \\),\n\n\n\ntherefore \\( AD = 2CD \\).\n\nSince \\( \\angle C = 90^\\circ \\),\n\ntherefore \\( \\angle DA'C = 30^\\circ \\),\n\nand thus \\( \\angle A'DC = 60^\\circ \\).\n\nConsequently, \\( \\angle DOK = 120^\\circ \\),\n\nand the area of sector \\( ODK \\) is \\( \\frac{16}{3} \\pi \\, \\text{cm}^2 \\).\n\nDrawing \\( OH \\perp DK \\) at \\( H \\),\n\nsince \\( \\angle D = \\angle K = 30^\\circ \\) and \\( OD = 4 \\, \\text{cm} \\),\n\ntherefore \\( OH = 2 \\, \\text{cm} \\) and \\( DH = 2\\sqrt{3} \\, \\text{cm} \\).\n\nThus, the area of \\( ODK \\) is \\( 4\\sqrt{3} \\, \\text{cm}^2 \\),\n\nand the area of the part of the semicircle that remains exposed (the shaded part) is \\( \\left(\\frac{16}{3} \\pi - 4\\sqrt{3}\\right) \\, \\text{cm}^2 \\).\n\nHence, the answer is: \\( \\left(\\frac{16}{3} \\pi - 4\\sqrt{3}\\right) \\).\n\n【Insight】This problem examines folding issues, requiring attention to corresponding equivalent relationships. It also tests the properties of tangents to circles, where the radius is perpendicular to the tangent at the point of contact. Additionally, it involves the properties of right-angled triangles, specifically that if one leg is half the hypotenuse, the angle opposite that leg is \\( 30^\\circ \\)." }, { "problem_id": 1347, "question": "As shown in Figure (1), there is a circle with a radius of 1. By removing 2 circles with a radius of $\\frac{1}{2}$ from it, we obtain Figure (2). By removing $2^{2}$ circles with a radius of $\\left(\\frac{1}{2}\\right)^{2}$, we obtain Figure (3). The area of the shaded part in the nth (n > 1) figure is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch17-2024_06_14_1c6ab25721766d50fc13g_0100_1.jpg", "batch17-2024_06_14_1c6ab25721766d50fc13g_0100_2.jpg", "batch17-2024_06_14_1c6ab25721766d50fc13g_0100_3.jpg" ], "is_multi_img": true, "answer": "(1- $\\left.\\frac{1}{2^{n-1}}\\right) \\pi$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "The area of the shaded part in Figure (2) is: $\\pi \\times 1^{2}-\\pi \\times\\left(\\frac{1}{2}\\right)^{2} \\times 2=\\pi-\\frac{1}{2} \\pi=\\left(1-\\frac{1}{2}\\right) \\pi=\\frac{1}{2} \\pi$;\n\nThe area of the shaded part in Figure (3) is: $\\pi \\times 1^{2}-\\pi \\times\\left[\\left(\\frac{1}{2}\\right)^{2}\\right]^{2} \\times 2^{2}=\\pi-\\frac{1}{2^{2}} \\pi=\\left(1-\\frac{1}{2^{2}}\\right) \\pi=\\frac{3}{4} \\pi$;\n\nFigure (4) is a circle with a radius of 1, from which $2^{3}$ circles with a radius of $\\left(\\frac{1}{2}\\right)^{3}$ are removed. The area of the shaded part in Figure (4) is: $\\pi \\times 1^{2}-\\pi \\times\\left[\\left(\\frac{1}{2}\\right)^{3}\\right]^{2} \\times 2^{3}=\\pi-\\frac{1}{2^{3}} \\pi=\\left(1-\\frac{1}{2^{3}}\\right) \\pi=\\frac{7}{8} \\pi$;\n\nThus, the area of the shaded part in the $\\mathrm{n}$-th figure ($\\mathrm{n}>1$) is: $\\pi \\times 1^{2}-\\pi \\times\\left[\\left(\\frac{1}{2}\\right)^{n-1}\\right]^{2} \\times 2^{n-1}=\\pi-\\frac{1}{2^{n-1}} \\pi=\\left(1-\\frac{1}{2^{n-1}}\\right) \\pi$.\n\nTherefore, the answer is $\\left(1-\\frac{1}{2^{n-1}}\\right) \\pi$.\n\n【Key Insight】This question examines the formula for the area of a circle. The key to solving the problem lies in identifying the pattern based on the specific figures provided in the question." }, { "problem_id": 1348, "question": "Definition: A polygonal line formed by two chords with a common endpoint in a circle is called a broken chord of the circle. Archimedes' Theorem on Broken Chords: As shown in Figure 1, $\\mathrm{AB}$ and $\\mathrm{BC}$ form a broken chord of the circle, with $\\mathrm{AB}>\\mathrm{BC}$, $\\mathrm{M}$ being the midpoint of arc $\\mathrm{ABC}$, and $\\mathrm{MF} \\perp \\mathrm{AB}$ at $\\mathrm{F}$, then $\\mathrm{AF}=\\mathrm{FB}+\\mathrm{BC}$.\n\nAs shown in Figure 2, in $\\triangle \\mathrm{ABC}$, $\\angle \\mathrm{ABC}=60^{\\circ}$, $\\mathrm{AB}=8$, $\\mathrm{BC}=6$, $\\mathrm{D}$ is a point on $\\mathrm{AB}$ such that $\\mathrm{BD}=1$, draw $\\mathrm{DE} \\perp \\mathrm{AB}$ intersecting the circumcircle of $\\triangle \\mathrm{ABC}$ at $\\mathrm{E}$, and connect $\\mathrm{EA}$, then $\\angle \\mathrm{EAC}=$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_3472680bada5b9187c88g_0085_1.jpg", "batch17-2024_06_14_3472680bada5b9187c88g_0085_2.jpg" ], "is_multi_img": true, "answer": "$60^{\\circ}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: As shown in Figure 2, connect $\\mathrm{OA}$, $\\mathrm{OC}$, and $\\mathrm{OE}$.\n\nGiven:\n$$\n\\begin{aligned}\n& AB = 8, \\quad BC = 6, \\quad BD = 1, \\\\\n& \\therefore AD = 7, \\quad BD + BC = 7,\n\\end{aligned}\n$$\n\nThus, $\\mathrm{AD} = \\mathrm{BD} + \\mathrm{BC}$.\n\nSince $\\mathrm{ED} \\perp \\mathrm{AB}$,\n\nPoint $\\mathrm{E}$ is the midpoint of arc $\\mathrm{ABC}$, meaning $AE = CE$.\n\nTherefore, $\\angle \\mathrm{AOE} = \\angle \\mathrm{COE}$.\n\nGiven that $\\angle \\mathrm{AOC} = 2 \\angle \\mathrm{ABC} = 2 \\times 60^\\circ = 120^\\circ$,\n\nIt follows that $\\angle \\mathrm{AOE} = \\angle \\mathrm{COE} = 120^\\circ$.\n\nThus, $\\angle \\mathrm{CAE} = \\frac{1}{2} \\angle \\mathrm{COE} = 60^\\circ$.\n\nThe answer is $60^\\circ$.\n\n\n\nFigure 2\n\n**Key Insight:** This problem is a new definition type, testing the theorem of angles in a circle and its corollaries. The key to solving it lies in understanding and applying the given content related to the Archimedean Broken Chord Theorem." }, { "problem_id": 1349, "question": "As shown in Figure 1.4, in a right triangle with legs of lengths 3 and 4, each additional altitude drawn from the hypotenuse adds an incircle to the triangle. By continuing this process, Figure 10 contains 10 incircles of right triangles, whose areas are denoted as $\\mathrm{S}_{1}, \\mathrm{~S}_{2}, \\mathrm{~S}_{3}, \\ldots, \\mathrm{S}_{10}$. Then, $\\mathrm{S}_{1}+\\mathrm{S}_{2}+\\mathrm{S}_{3}+\\ldots+\\mathrm{S}_{10}=$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4", "input_image": [ "batch17-2024_06_14_350b5f90217d6d389c60g_0007_1.jpg", "batch17-2024_06_14_350b5f90217d6d389c60g_0007_2.jpg", "batch17-2024_06_14_350b5f90217d6d389c60g_0007_3.jpg", "batch17-2024_06_14_350b5f90217d6d389c60g_0007_4.jpg" ], "is_multi_img": true, "answer": "$\\pi$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Figure 1: Draw $OE \\perp AC$ and $OF \\perp BC$ from point $O$, with the feet of the perpendiculars at $E$ and $F$ respectively. Then, $\\angle OEC = \\angle OFC = 90^\\circ$.\nSince $\\angle C = 90^\\circ$,\nthe quadrilateral $OECF$ is a rectangle.\nBecause $OE = OF$,\nthe rectangle $OECF$ is a square.\nLet the radius of circle $O$ be $r$,\nthen $OE = OF = r$,\n$AD = AE = 3 - r$,\n$BD = 4 - r$.\nTherefore, $3 - r + 4 - r = 5$, so $r = \\frac{3 + 4 - 5}{2} = 1$.\nThus, $S_I = \\pi \\times 1^2 = \\pi$.\n\nFigure 2:\nFrom $S \\triangle ABC = \\frac{1}{2} \\times 3 \\times 4 = \\frac{1}{2} \\times 5 \\times CD$,\nwe get $CD = \\frac{12}{5}$.\nBy the Pythagorean theorem, $AD = \\sqrt{3^2 - \\left(\\frac{12}{5}\\right)^2} = \\frac{9}{5}$,\n$BD = 5 - \\frac{9}{5} = \\frac{16}{5}$.\nFrom (1), the radius of circle $O$ is $\\frac{\\frac{9}{5} + \\frac{12}{5} - 3}{2} = \\frac{3}{5}$,\nand the radius of circle $E$ is $\\frac{\\frac{12}{5} + \\frac{16}{5} - 4}{2} = \\frac{4}{5}$.\nTherefore, $S_1 + S_2 = \\pi \\times \\left(\\frac{3}{5}\\right)^2 + \\pi \\times \\left(\\frac{4}{5}\\right)^2 = \\pi$.\n\nFigure 3:\nFrom $S \\triangle CDB = \\frac{1}{2} \\times \\frac{12}{5} \\times \\frac{16}{5} = \\frac{1}{2} \\times 4 \\times MD$,\nwe get $MD = \\frac{48}{25}$.\nBy the Pythagorean theorem, $CM = \\sqrt{\\left(\\frac{12}{5}\\right)^2 - \\left(\\frac{48}{25}\\right)^2} = \\frac{36}{25}$,\n$MB = 4 - \\frac{36}{25} = \\frac{64}{25}$.\nFrom (1), the radius of circle $O$ is $\\frac{3}{5}$,\nthe radius of circle $E$ is $\\frac{12}{25}$,\nand the radius of circle $F$ is $\\frac{16}{25}$.\nTherefore, $S_1 + S_2 + S_3 = \\pi \\times \\left(\\frac{3}{5}\\right)^2 + \\pi \\times \\left(\\frac{12}{25}\\right)^2 + \\pi \\times \\left(\\frac{16}{25}\\right)^2 = \\pi$." }, { "problem_id": 1350, "question": "There is a rectangular paper piece $A B C D$, where $A B=2 \\sqrt{2}$ and $A D=4$. There is a semicircle with $A D$ as its diameter on the paper (as shown in Figure 1). Point $E$ is on side $A B$. When the paper is folded along $D E$, point $A$ exactly falls on $B C$. The area of the part of the semicircle that remains exposed (as shown in Figure 2, the shaded area) is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_3ad8cdbfee15d4c39edcg_0001_1.jpg", "batch17-2024_06_14_3ad8cdbfee15d4c39edcg_0001_2.jpg" ], "is_multi_img": true, "answer": "$\\pi-2$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the center of the circle where the shaded region lies be \\( O \\), and let \\( AD \\) intersect the semicircular arc at point \\( F \\), as shown in the figure. Connect \\( OF \\).\n\n\n\nFrom the properties of folding, we know: \\( AD = BC = 4 \\).\n\nFrom the properties of the rectangle, we know: \\( CD = AB = 2\\sqrt{2} \\), and \\( \\angle DCA = 90^\\circ \\).\n\nThus, \\( AC = \\sqrt{AD^2 - CD^2} = \\sqrt{4^2 - (2\\sqrt{2})^2} = 2\\sqrt{2} = CD \\).\n\nTherefore, \\( \\triangle ADC \\) is an isosceles right triangle.\n\nHence, \\( \\angle DAC = 45^\\circ \\).\n\nSince \\( OD \\parallel BC \\), \\( \\angle ODF = \\angle DAC = 45^\\circ \\).\n\nGiven that \\( OD = OF = 2 \\), \\( \\angle OFD = \\angle ODF = 45^\\circ \\).\n\nThus, \\( \\angle DOF = 180^\\circ - 45^\\circ - 45^\\circ = 90^\\circ \\).\n\nTherefore, the area of the shaded region \\( S_{\\text{shaded}} = S_{\\text{sector } DOF} - S_{\\triangle DOF} = \\frac{90 \\pi \\times 2^2}{360} - \\frac{1}{2} \\times 2 \\times 2 = \\pi - 2 \\).\n\nThe final answer is: \\( \\pi - 2 \\).\n\n【Insight】This problem examines the properties of folding, the Pythagorean theorem, the properties of parallel lines, the determination and properties of isosceles triangles, and the calculation of the area of a sector. The key to solving the problem lies in determining the degree measure of the central angle corresponding to the sector in which the shaded region lies." }, { "problem_id": 1351, "question": "There is a set of triangles (as shown in Figure 1), each with a circular hole of diameter $4 \\mathrm{~cm}$ in the center. Now, the $30^{\\circ}$ end of triangle $a$ is inserted into the circular hole of triangle $b$ (as shown in Figure 2). The maximum area of the part of triangle $a$ that passes through the circular hole of triangle $b$ is $\\qquad$ $\\mathrm{cm}^{2}$. (Neglecting the thickness of the triangles, the result should be accurate to $0.1 \\mathrm{~cm}^{2}$)\n\n\n\nFigure 1\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_43c28cf770c07f2dc4a0g_0097_1.jpg", "batch17-2024_06_14_43c28cf770c07f2dc4a0g_0097_2.jpg" ], "is_multi_img": true, "answer": "14.9", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Suppose the part of triangle $a$ that passes through the circular hole of triangle $b$ is $\\triangle ABC$, with $BC = 4 \\text{ cm}$ and $\\angle BAC = 30^\\circ$. Construct the circumcircle $\\odot P$ of $\\triangle ABC$, connect $PA$, $PB$, and $PC$, and draw $PD \\perp BC$ at point $D$. Then, $PB = PC = PA$.\n\nSince $\\angle BAC = 30^\\circ$,\n\n$\\angle BPC = 2 \\angle BAC = 60^\\circ$,\n\nThus, $\\triangle PBC$ is an equilateral triangle,\n\nTherefore, $BD = CD = 2$, $PD = 2\\sqrt{3}$, and $BP = BC = PA = 4$.\n\nConnecting $AD$, we have $AD \\leq AP + PD = 4 + 2\\sqrt{3}$.\n\nHence, when $A$, $P$, and $D$ are collinear, $AD$ reaches its maximum value,\nAt this point, $AD \\perp BC$,\n\nThus, the area of $\\triangle ABC$ is:\n\\[\nS_{\\triangle ABC} = \\frac{1}{2} \\times BC \\times AD = \\frac{1}{2} \\times 4 \\times (4 + 2\\sqrt{3}) = 8 + 4\\sqrt{3} \\approx 14.9 \\text{ cm}^2.\n\\]\n\nTherefore, the answer is: 14.9.\n\n\n\n【Key Insight】This problem examines the properties of the circumcircle of a triangle. The key to solving it lies in flexibly constructing the triangle to be solved based on the given conditions and skillfully applying the properties of the circumcircle to analyze and solve the problem." }, { "problem_id": 1352, "question": "Many families now use folding western-style dining tables to save space, which expand into a round tabletop when both sides are flipped open (as shown in Figure 1). The two sides of the table, $AD$ and $BC$, are parallel and equal in length (as shown in Figure 2). Xiaohua measured with a tape measure and found that $AC = 4$ meters and $AB = 2$ meters. After the tabletop is flipped into a round table, the area of the table will increase by __ square meters. (The result should be kept in terms of $\\pi$.)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_48a4d8adae62cb58f253g_0072_1.jpg", "batch17-2024_06_14_48a4d8adae62cb58f253g_0072_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{8 \\pi}{3}-2 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Complete the circle, let the center be $O$, connect $D O$, and draw $O E \\perp A D$ at point $E$. Since $A D$ and $B C$ are parallel and equal in length, quadrilateral $A B C D$ is a parallelogram.\n\n$\\because$ Parallelogram $A B C D$ is inscribed in the circle,\n\n$\\therefore \\angle D A B=\\angle A B C=90^{\\circ}$,\n\n$\\therefore A C$ is the diameter of the circle,\n\n$\\because A C=4$ meters, $A B=2$ meters,\n\n$\\therefore \\angle A C B=30^{\\circ}$,\n\n$\\because$ The two sides of the table $A B$ and $C D$ are parallel and equal,\n\n$\\therefore \\angle C=\\angle 1=30^{\\circ}$,\n\n$\\therefore E O=\\frac{1}{2} A O=1$,\n\n$\\therefore A E=\\sqrt{A O^{2}-O E^{2}}=\\sqrt{3}$,\n\n$\\therefore A D=2 \\sqrt{3}$,\n\n$\\because A O=D O$,\n\n$\\therefore \\angle 1=\\angle D=30^{\\circ}$,\n\n$\\therefore \\angle A O D=120^{\\circ}$,\n\n$\\therefore$ The area of the segment $A D$\n\n$=$ Area of sector $A O D -$ Area of triangle $A O D$\n\n$=\\frac{120 \\cdot \\pi \\times 2^{2}}{360}-\\frac{1}{2} \\times 1 \\times 2 \\sqrt{3}$,\n\n$=\\frac{4 \\pi}{3}-\\sqrt{3}$,\n$\\therefore$ After flipping the tabletop into a round table, the area of the table will increase by $\\left(\\frac{8 \\pi}{3}-2 \\sqrt{3}\\right)$ square meters.\n\nTherefore, the answer is: $\\frac{8 \\pi}{3}-2 \\sqrt{3}$.\n\n\n\n【Insight】This problem mainly examines comprehensive issues related to circles, including the calculation of sector areas and cyclic quadrilaterals. Mastering the relevant knowledge of circles is key to solving the problem." }, { "problem_id": 1353, "question": "When Xiao Ming was studying the properties of circles, he saw in the textbook that a T-square (as shown in Figure 1) could be used to find the center of a circle. He remembered that his father's toolbox had a T-square, so he wanted to use it to restore a damaged circle. Given that the head of the T-square $AB = 4 \\mathrm{~cm}$ and the scale line $l$ of the body is perpendicularly bisected by $AB$, the situation he set up is shown in Figure 2. He found that the body of the T-square intersected at the scale of $6 \\mathrm{~cm}$ in two measurements. Therefore, the diameter of this damaged circle is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_4ccb54cc871f23fb3a34g_0046_1.jpg", "batch17-2024_06_14_4ccb54cc871f23fb3a34g_0046_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{10}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure: Determine the center of the circle $O$, according to the problem:\n\n$\\because O C \\perp A B$\n\n$\\therefore A C=\\frac{1}{2} A B=2$\n\nIn the right triangle $O C A$:\n\n$O A^{2}=A C^{2}+O C^{2}=2^{2}+6^{2}=40$\n\n$O A=2 \\sqrt{10}$\n\nTherefore, the answer is $O A=2 \\sqrt{10}$.\n\n\n\n【Key Insight】This problem examines the combined application of the Perpendicular Diameter Theorem and the Pythagorean Theorem. The key is to establish a model of the circle based on the given conditions, use the Perpendicular Diameter Theorem to determine the length of the segment, and then solve the problem." }, { "problem_id": 1354, "question": "Given that the radius of circle $O$ is $a$, follow these steps to construct the diagram: (1) Construct the inscribed square $A B C D$ in circle $O$ (as shown in Figure 1); (2) Construct the inscribed circle within square $A B C D$, and then construct the inscribed square $A_{1} B_{1} C_{1} D_{1}$ within the smaller circle (as shown in Figure 2); (3) Construct the inscribed circle within square $A_{1} B_{1} C_{1} D_{1}$, and then construct the inscribed square $A_{2} B_{2} C_{2} D_{2}$ within it (as shown in Figure 3); ...; Continue this process, and the side length of square $A_{n} B_{n} C_{n} D_{n}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch17-2024_06_14_4ccb54cc871f23fb3a34g_0092_1.jpg", "batch17-2024_06_14_4ccb54cc871f23fb3a34g_0092_2.jpg", "batch17-2024_06_14_4ccb54cc871f23fb3a34g_0092_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{(\\sqrt{2})^{n+1}}{2^{n}} a$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem,\n\nIn Figure 1, the radius of circle \\( O \\) is \\( a \\), so the side length of square \\( ABCD \\) is \\( AB = \\sqrt{2}a \\).\n\nIn Figure 2, \\( A_{1}B = BB_{1} = \\frac{1}{2}AB = \\frac{\\sqrt{2}}{2}a \\),\n\nthus the side length of square \\( A_{1}B_{1}C_{1}D_{1} \\) is \\( A_{1}B_{1} = \\sqrt{2}A_{1}B = \\sqrt{2} \\times \\frac{\\sqrt{2}}{2}a = \\frac{(\\sqrt{2})^{2}}{2}a \\).\n\nIn Figure 3, \\( A_{1}A_{2} = A_{1}B_{2} = \\frac{1}{2}A_{1}B_{1} = \\frac{1}{2} \\times \\frac{(\\sqrt{2})^{2}}{2}a = \\left(\\frac{\\sqrt{2}}{2}\\right)^{2}a \\),\n\nthus the side length of square \\( A_{2}B_{2}C_{2}D_{2} \\) is \\( A_{2}B_{2} = \\sqrt{2}A_{1}A_{2} = \\sqrt{2} \\times \\left(\\frac{\\sqrt{2}}{2}\\right)^{2}a = \\frac{(\\sqrt{2})^{3}}{2^{2}}a \\).\n\n...\n\nFollowing this pattern, the side length of square \\( A_{n}B_{n}C_{n}D_{n} \\) is \\( A_{n}B_{n} = \\frac{(\\sqrt{2})^{n+1}}{2^{n}}a \\).\n\nTherefore, the answer is: \\( \\frac{(\\sqrt{2})^{n+1}}{2^{n}}a \\).\n\n【Key Insight】This problem examines the exploration of patterns in regular polygons inscribed in circles and the properties of squares. Observing the figures and correctly deducing the pattern of side length changes is crucial to solving the problem." }, { "problem_id": 1355, "question": "Figure 1 is an image of the Rainbow Bridge in Jilin City. Figure 2 is a schematic diagram of the Rainbow Bridge, where the arch $(A D B)$ can be approximated as an arc of a circle with a radius of $25 \\mathrm{~m}$. Nine steel cables connect the arch (arc $A B$) to the road surface (chord $A B$), with the steel cables perpendicular to the road surface (chord $A B$). The length of the road surface (chord $A B$) is $40 \\mathrm{~m}$. If these nine steel cables divide the road surface (chord $A B$) into ten equal parts, find the length of the longest steel cable $C D$.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch17-2024_06_14_5f4f95a2bc38a3c50875g_0089_1.jpg", "batch17-2024_06_14_5f4f95a2bc38a3c50875g_0089_2.jpg" ], "is_multi_img": true, "answer": "$10 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, $CD$ is the perpendicular bisector of chord $AB$. By the Perpendicular Bisector Theorem, the center $O$ of the circle on which $AB$ lies is on the extension of $DC$. Connect $AO$ and $BO$ as shown in the figure.\n\n\n\nSince $CD$ is the perpendicular bisector of chord $AB$ and $AB = 40 \\mathrm{~m}$,\n\n$\\therefore AC = \\frac{1}{2} AB = 20 \\mathrm{~m}$.\n\nGiven that the radius of the circle on which $AB$ lies is $25 \\mathrm{~m}$,\n\n$\\therefore DO = AO = 25 \\mathrm{~m}$.\n\nIn the right triangle $ACO$, $CO = \\sqrt{AO^2 - AC^2} = \\sqrt{25^2 - 20^2} = 15 \\mathrm{~m}$.\n\nTherefore, $CD = DO - CO = 10 \\mathrm{~m}$.\n\nAnswer: The length of the longest steel cable $CD$ is $10 \\mathrm{~m}$.\n\n【Key Insight】This problem primarily tests the application of the Perpendicular Bisector Theorem. Determining that the center $O$ of the circle on which $AB$ lies is on the extension of $DC$ is crucial for solving the problem." }, { "problem_id": 1356, "question": "The waterwheel is an ancient Chinese invention used for irrigation, showcasing the wisdom of ancient Chinese laborers. As shown in Figure 1, point $P$ represents a bucket on the waterwheel. As shown in Figure 2, when the waterwheel is in operation, the path of the bucket is a circle with center $O$ and radius $10 \\mathrm{~m}$, with the center of the circle above the water surface. If the chord $AB$ formed by the intersection of the circle and the water surface has a length of $16 \\mathrm{~m}$, then the maximum depth of the bucket below the water surface when the waterwheel is in operation is $\\qquad$ $m$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_66a00a94d67a94380be3g_0059_1.jpg", "batch17-2024_06_14_66a00a94d67a94380be3g_0059_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw the radius $O D \\perp A B$ at point $E$, as shown in the figure,\n\n\n\n$\\therefore A E=B E=\\frac{1}{2} A B=\\frac{1}{2} \\times 16=8$,\n\nIn the right triangle $\\triangle A E O$, $O E=\\sqrt{O A^{2}-A E^{2}}=\\sqrt{10^{2}-8^{2}}=6$,\n\n$\\therefore E D=O D-O E=10-6=4(\\mathrm{~m})$,\n\nTherefore, the answer is: 4\n\n【Key Insight】This problem examines the application of the Perpendicular Chord Theorem, which states that a diameter perpendicular to a chord bisects the chord and the arcs it subtends. Proficiency in applying the Perpendicular Chord Theorem is crucial for solving such problems." }, { "problem_id": 1357, "question": "The great ancient Chinese mathematician Liu Hui pointed out in his commentary on the \"Nine Chapters on the Mathematical Art\" written in 263 AD that \"the circumference is three times the diameter\" is not the value of pi, but actually the ratio of the perimeter of an inscribed regular hexagon to the diameter of the circle (Figure 1). Liu Hui discovered that as the number of sides of an inscribed regular polygon increases infinitely, the perimeter of the polygon approaches the circumference of the circle infinitely closely, thereby creating the \"polygon-cutting technique\" to establish a quite rigorous theory and a perfect algorithm for calculating pi, as shown in Figure 2, where hexagon \\(ABCDEF\\) is an inscribed regular hexagon. By bisecting each arc, an inscribed regular dodecagon is constructed, and lines \\(AG\\) and \\(CF\\) are drawn, intersecting at point \\(P\\). If \\(AP = 2\\sqrt{6}\\), then the length of \\(CG\\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_7a59a3fdb311209ac62bg_0046_1.jpg", "batch17-2024_06_14_7a59a3fdb311209ac62bg_0046_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{6}-2 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the center of the circumscribed circle of the regular hexagon be $O$.\n\nConnect $O G$, then $\\angle C O G=\\frac{360^{\\circ}}{12}=30^{\\circ}$.\n\nFrom the problem statement, $\\angle F A G=75^{\\circ}$ and $\\angle C F A=60^{\\circ}$.\n\nDraw $A H \\perp C F$ through point $A$, intersecting at $H$.\n\nThus, $\\angle A H F=90^{\\circ}$,\n\nand $\\angle F A H=30^{\\circ}$,\n\nso $\\angle H A P=45^{\\circ}$.\n\nTherefore, $\\triangle A H P$ is an isosceles right triangle,\n\nand $A H=\\frac{\\sqrt{2}}{2} A P=2 \\sqrt{3}$.\n\nThus, $A F=\\frac{A H}{\\sin 60^{\\circ}}=\\frac{2 \\sqrt{3}}{\\frac{\\sqrt{3}}{2}}=4$,\n\nand $O C=A F=4$.\n\nDraw $G Q \\perp O C$ through point $G$, with the foot of the perpendicular at $Q$.\n\nThus, $G Q=\\frac{1}{2} O G=2$,\n\nand $O Q=\\sqrt{4^{2}-2^{2}}=2 \\sqrt{3}$.\n\nTherefore, $Q C=O C-O Q=4-2 \\sqrt{3}$,\n\nand $C G=\\sqrt{G Q^{2}+Q C^{2}}=2 \\sqrt{6}-2 \\sqrt{2}$.\n\nHence, the answer is: $2 \\sqrt{6}-2 \\sqrt{2}$.\n\n\n\nFigure 2\n\n[Key Insight] This problem examines the properties of regular polygons and circles, specifically the properties of regular hexagons and dodecagons, solving right triangles, and calculating arc lengths. Correctly understanding the problem statement is key to solving it." }, { "problem_id": 1358, "question": "As shown in Figure (1), if $A_{1}, A_{2}, A_{3}, A_{4}$ divide the circumference of a circle into four equal parts, then there are 4 right-angled triangles with $A_{1}, A_{2}, A_{3}, A_{4}$ as vertices; as shown in Figure (2), if $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ divide the circumference of a circle into six equal parts, then there are 12 right-angled triangles with $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ as vertices; if $A_{1}, A_{2}, A_{3}, \\ldots, A_{2n}$ divide the circumference of a circle into $2n$ equal parts, then there are $\\qquad$ right-angled triangles with $A_{1}, A_{2}, A_{3}, \\ldots, A_{2n}$ as vertices.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch17-2024_06_14_7db72a59a831372d1996g_0097_1.jpg", "batch17-2024_06_14_7db72a59a831372d1996g_0097_2.jpg" ], "is_multi_img": true, "answer": "$2 \\mathrm{n}(\\mathrm{n}-1$ )", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: According to the Inscribed Angle Theorem, the angle subtended by a diameter on the circumference is a right angle.\n\n$\\therefore$ When points $\\mathrm{A}_{1}, \\mathrm{A}_{2}, \\mathrm{A}_{3}, \\mathrm{A}_{4}$ divide the circumference into four equal parts, the diameters in the circle are $\\mathrm{A}_{1}\\mathrm{A}_{3}$ and $\\mathrm{A}_{2}\\mathrm{A}_{4}$.\n\n$\\therefore$ (1) When $\\mathrm{A}_{1}\\mathrm{A}_{3}$ is the diameter, there are two right-angled triangles;\n\n(2) When $\\mathrm{A}_{2}\\mathrm{A}_{4}$ is the diameter, there are two right-angled triangles;\n\n$\\therefore$ If $\\mathrm{A}_{1}, \\mathrm{A}_{2}, \\mathrm{A}_{3}, \\mathrm{A}_{4}$ divide the circumference into four equal parts, then the number of right-angled triangles with vertices at $\\mathrm{A}_{1}, \\mathrm{A}_{2}, \\mathrm{A}_{3}, \\mathrm{A}_{4}$ is $(4 \\div 2) \\times (4 - 2) = 4$;\n\nWhen $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ divide the circumference into six equal parts, the number of right-angled triangles with vertices at $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ is $(6 \\div 2) \\times (6 - 2) = 12$;\n\nWhen $A_{1}, A_{2}, A_{3}, \\ldots, A_{2n}$ divide the circumference into $2n$ equal parts, the number of right-angled triangles with vertices at $A_{1}, A_{2}, A_{3}, \\ldots, A_{2n}$ is $(2n \\div 2) \\times (2n - 2) = 2n(n - 1)$.\n\nTherefore, the answer is: $2n(n - 1)$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n【Key Insight】This problem tests the Inscribed Angle Theorem: the angle subtended by a diameter on the circumference is a right angle. The key to solving the problem is to find the pattern based on the number of diameters and vertices." }, { "problem_id": 1359, "question": "Chinese cuisine emphasizes the aesthetics of color, aroma, and taste, and elegant plating can further enhance the beauty of the food. In the plating shown in Figure (1), the shape is a part of a sector. Figure (2) is its geometric diagram (the shaded area represents the plating). By measuring, it is found that $AC = BD = 12 \\text{ cm}$, the distance between points $C$ and $D$ is $4 \\text{ cm}$, and the central angle is $60^\\circ$. Then, the area of the plating in the figure is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch17-2024_06_14_80065ed648312b47a22eg_0087_1.jpg", "batch17-2024_06_14_80065ed648312b47a22eg_0087_2.jpg" ], "is_multi_img": true, "answer": "$40 \\pi \\mathrm{cm}^{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $CD$,\n\n\n\nGiven that $OC = OD$ and $\\angle O = 60^\\circ$,\n\n$\\therefore \\triangle OCD$ is an equilateral triangle,\n\n$\\therefore OC = OD = CD = 4 \\text{ cm}$,\n\n$\\because AC = BD = 12 \\text{ cm}$,\n\n$\\therefore OA = OB = 12 + 4 = 16 \\text{ cm}$,\n\n$$\n\\begin{aligned}\n& \\therefore S_{\\text{shaded}} = S_{\\text{sector } OAB} - S_{\\text{sector } OCD} \\\\\n& = \\frac{60 \\pi \\cdot OA^{2}}{360} - \\frac{60 \\pi \\cdot OC^{2}}{360} \\\\\n& = \\frac{60 \\pi \\cdot (16^{2} - 4^{2})}{360} \\\\\n& = 40 \\pi \\text{ cm}^{2}\n\\end{aligned}\n$$\n\nTherefore, the answer is: $40 \\pi \\text{ cm}^{2}$.\n\n【Key Insight】This problem tests the calculation of the area of a sector and the properties of an equilateral triangle. Mastering the sector area formula $S_{\\text{sector}} = \\frac{n \\cdot \\pi \\cdot R^{2}}{360}$ is crucial for solving the problem." }, { "problem_id": 1360, "question": "Traditional clothing has been gaining increasing attention. As shown in Figure 1, the horse-face skirt is one of the main skirt styles for women during the Ming and Qing dynasties. As shown in Figure 2, the horse-face skirt can be approximately considered as a sector ring, where the length of $A D$ is $\\frac{1}{3} \\pi$ meters, the length of $B C$ is $\\frac{3}{5} \\pi$ meters, and the central angle $\\angle A O D = 60^{\\circ}$. The length of the skirt $A B$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_a7b6e5eaee0a90438865g_0040_1.jpg", "batch17-2024_06_14_a7b6e5eaee0a90438865g_0040_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{4}{5}$ m $/ 0.8$ m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, we have:\n\n\\[\nl_{AD} = \\frac{60 \\pi \\cdot OA}{180} = \\frac{1}{3} \\pi, \\quad l_{BC} = \\frac{60 \\pi \\cdot OB}{180} = \\frac{3}{5} \\pi\n\\]\n\nSolving these equations, we find:\n\n\\[\nOA = 1, \\quad OB = \\frac{9}{5}\n\\]\n\nTherefore, the length of \\( AB \\) is:\n\n\\[\nAB = OB - OA = \\frac{4}{5}\n\\]\n\nThus, the answer is: \\(\\frac{4}{5}\\) meters or 0.8 meters.\n\n**Key Insight:** This problem tests the understanding of the arc length formula for a sector. The key to solving it lies in accurate calculations." }, { "problem_id": 1361, "question": "As shown in Figure 1, this is an ancient Chinese waterwheel. As shown in Figure 2, its schematic diagram is a circle with the center at the axis $O$ and a radius of $5 \\mathrm{~m}$. If the chord $A B$ obtained by intersecting $O$ with the water surface has a length of $8 \\mathrm{~m}$, find the maximum depth of the waterwheel below the water surface.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_a8583af8c8007361cd64g_0025_1.jpg", "batch17-2024_06_14_a8583af8c8007361cd64g_0025_2.jpg" ], "is_multi_img": true, "answer": "$2 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw the radius $O D \\perp A B$ at point $E$, as shown in the figure.\n\n$\\therefore A E = B E = \\frac{1}{2} A B = \\frac{1}{2} \\times 8 = 4 \\mathrm{~m}$.\n\nIn the right triangle $\\triangle A E O$, $O E = \\sqrt{O A^{2} - A E^{2}} = \\sqrt{5^{2} - 4^{2}} = 3 \\mathrm{~m}$.\n\n$\\therefore E D = O D - O E = 5 - 3 = 2 \\mathrm{~m}$.\n\nAnswer: The maximum depth of the water wheel below the water surface is $2 \\mathrm{~m}$.\n\n\n\nFigure 2\n\n[Key Insight] This problem examines the Perpendicular Chord Theorem: A diameter that is perpendicular to a chord bisects the chord and the arcs subtended by the chord." }, { "problem_id": 1362, "question": "As shown in Figure 1, a protractor is placed in an axisymmetric configuration with an equilateral triangle $(\\triangle A B C)$ paper. $C D \\perp A B$, with the foot of the perpendicular at $D$. The center of the semicircle (protractor) coincides with point $D$. At this moment, the distance from vertex $C$ to the highest point of the protractor is measured to be $C E=2 \\mathrm{~cm}$. The protractor is then translated along the direction of $D C$ by $1 \\mathrm{~cm}$, and the semicircle (protractor) just touches the sides $A C$ and $B C$ of $\\triangle A B C$, as shown in Figure 2. The length of $A B$ is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nReason 1\n\n\n\nReason 2", "input_image": [ "batch17-2024_06_14_a929c1e5dec734dac061g_0078_1.jpg", "batch17-2024_06_14_a929c1e5dec734dac061g_0078_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in the figure, let the center of the semicircle in Figure (2) be $O$, and the point of tangency with $BC$ be $M$. Connect $OM$, then $OM \\perp MC$,\n\n$\\therefore \\angle OMC = 90^{\\circ}$. According to the problem, $\\angle DCB = 30^{\\circ}$, and let $AB$ be $2x \\mathrm{~cm}$.\n\nSince $\\triangle ABC$ is an equilateral triangle, $\\therefore CD = \\sqrt{3}x \\mathrm{~cm}$, and $CE = 2 \\mathrm{~cm}$. Furthermore, the protractor is translated $1 \\mathrm{~cm}$ along the direction of $DC$,\n$\\therefore$ the radius of the semicircle is $OM = (\\sqrt{3}x - 2) \\mathrm{cm}$, and $OC = (\\sqrt{3}x - 1) \\mathrm{cm}$. Therefore, $\\sin \\angle DCB = \\frac{OM}{OC} = \\frac{1}{2}$,\n\n$\\therefore \\frac{\\sqrt{3}x - 2}{\\sqrt{3}x - 1} = \\frac{1}{2}, \\quad \\therefore x = \\sqrt{3}, \\quad \\therefore AB = 2x = 2\\sqrt{3} \\quad (\\mathrm{~cm})$.\n\nThus, the answer is $2\\sqrt{3}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Insight】This problem examines the properties of tangents to circles and the knowledge of solving right triangles. Using the properties of tangents for calculations or proofs often involves drawing auxiliary lines to connect the center of the circle and the point of tangency, utilizing perpendicularity to construct right triangles to solve related problems." }, { "problem_id": 1363, "question": "As shown in Figure 1, the nine-grid hot pot with Chongqing characteristics is divided into nine sections: four corner grids, a cross grid, and a central grid (the central grid is generally square). The design of the partitions has the following two types: (1) the horizontal and vertical partitions intersect perpendicularly at the three equal division points as shown in Figure 2; (2) the horizontal and vertical partitions intersect perpendicularly at the eight equal division points on the edge of the circular pot as shown in Figure 3. Given that the diameter of the circular pot is $40 \\mathrm{~cm}$, the difference between the areas of the central grids $S_{1}$ and $S_{2}$ in the two designs is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch17-2024_06_14_ab9db35038cd3f0b2280g_0003_1.jpg", "batch17-2024_06_14_ab9db35038cd3f0b2280g_0003_2.jpg", "batch17-2024_06_14_ab9db35038cd3f0b2280g_0003_3.jpg" ], "is_multi_img": true, "answer": "$640-400 \\sqrt{2} \\# \\#-400 \\sqrt{2}+640$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw \\( O B \\perp A P \\) at point \\( B \\), connect \\( O A \\), and draw \\( C F \\perp D G \\) with foot \\( F \\), then connect \\( C G \\) and \\( D E \\).\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nFrom the problem statement: \\( O A = C D = C E = 20 \\text{ cm} \\), \\( A H = H I = I P \\).\n\nSince the central grid is a square, it follows that: \\( O B = H B = B I = \\frac{1}{2} H I = \\frac{1}{2} A H \\), and \\( C F = F G \\).\n\nLet \\( O B = x \\text{ cm} \\), then \\( A B = 3x \\text{ cm} \\).\n\nIn the right triangle \\( \\triangle A B O \\), by the Pythagorean theorem: \\( x^{2} + 9x^{2} = 400 \\),\n\nThus, \\( x^{2} = 40 \\),\n\nTherefore, \\( S_{1} = 4x^{2} = 160 \\text{ cm}^{2} \\).\n\nSince the horizontal and vertical partitions intersect perpendicularly at the eight equal division points on the edge of the circular pot as shown in Figure 3,\n\nEach arc on the edge of the circular pot measures \\( 45^{\\circ} \\),\n\nThus, \\( \\angle C D F = \\frac{1}{2} \\times 45^{\\circ} = 22.5^{\\circ} \\),\n\nSince \\( \\angle C G F = 45^{\\circ} \\),\n\nTherefore, \\( \\angle D C G = \\angle C G F - \\angle C D F = 22.5^{\\circ} = \\angle C D F \\),\n\nHence, \\( C G = D G = \\sqrt{2} C F = \\sqrt{2} G F \\),\n\nThus, \\( F D = D G + F G = (\\sqrt{2} + 1) C F \\),\n\nIn the right triangle \\( \\triangle C D F \\), by the Pythagorean theorem: \\( C F^{2} + F D^{2} = C D^{2} \\), that is, \\( C F^{2} + (\\sqrt{2} + 1)^{2} \\cdot C F^{2} = 400 \\),\n\nThus, \\( C F^{2} = 200 - 100 \\sqrt{2} \\),\n\nTherefore, \\( S_{2} = 4 C F^{2} = (800 - 400 \\sqrt{2}) \\text{ cm}^{2} \\),\n\nHence, \\( S_{2} - S_{1} = 800 - 400 \\sqrt{2} - 160 = (640 - 400 \\sqrt{2}) \\text{ cm} \\).\n\nThe final answer is \\( 640 - 400 \\sqrt{2} \\).\n\n【Key Insight】This problem primarily tests the theorems of perpendicular chords, inscribed angles, and the Pythagorean theorem. Mastering these theorems is crucial for solving the problem." }, { "problem_id": 1364, "question": "As shown in Figure 1, the Gothic pointed arch is an axisymmetric figure composed of two arcs with different centers, known as the two-centered arch. As shown in Figure 2, given that $P$ and $Q$ are the centers of the circles containing $AC$ and $BC$ respectively, and both lie on $AB$, if $PQ = 2 \\text{ m}$ and $AB = 6 \\text{ m}$, then the height of the arch $CD$ is $\\qquad$ $\\text{m}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_ab9db35038cd3f0b2280g_0040_1.jpg", "batch17-2024_06_14_ab9db35038cd3f0b2280g_0040_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{15}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $C Q$.\n\n\n\nGiven that $C Q = C P$ and $C D \\perp P Q$,\n\n$\\therefore D Q = D P = \\frac{1}{2} P Q = 1(\\mathrm{~m})$,\n\n$\\because P A = Q B$,\n\n$\\therefore A Q = P B = \\frac{1}{2}(A B - P Q) = 2(\\mathrm{~m})$,\n\n$\\therefore P C = P A = 2 + 2 = 4(m)$,\n\n$\\therefore C D = \\sqrt{P C^{2} - P D^{2}} = \\sqrt{4^{2} - 1^{2}} = \\sqrt{15} \\quad(\\mathrm{~m})$,\n\nTherefore, the answer is: $\\sqrt{15}$.\n\n【Key Insight】This problem mainly examines the Pythagorean theorem and the perpendicular bisector theorem. Understanding that the diameter perpendicular to the chord bisects the chord is key to solving this problem." }, { "problem_id": 1365, "question": "As shown in Figure (1), point \\( P \\) is a moving point on the semicircle \\( O \\) with diameter \\( AB \\) (it can coincide with points \\( A \\) and \\( B \\)), and through point \\( P \\), \\( PD \\perp AB \\) at point \\( D \\), connecting \\( PA \\). Let \\( PA = x \\) and \\( PD = y \\). Figure (2) shows the graph of the functional relationship between \\( y \\) and \\( x \\) as point \\( P \\) moves, where \\( M \\) is the highest point of the curve. Then the value of \\( mn \\) is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch17-2024_06_14_ab9db35038cd3f0b2280g_0092_1.jpg", "batch17-2024_06_14_ab9db35038cd3f0b2280g_0092_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the image, we have $AB = 4$,\n\n$\\therefore$ the radius of circle $O$ is 2.\n\nWhen point $D$ coincides with point $O$, $PD$ is maximized,\nat this time, $PD$ is the radius of circle $O$,\n\n$\\therefore y = 2$,\n\nWhen $y = 2$, $x = \\sqrt{AD^{2} + PD^{2}} = \\sqrt{2^{2} + 2^{2}} = 2\\sqrt{2}$,\n\n$\\therefore$ the coordinates of point $M$ are $(2\\sqrt{2}, 2)$, i.e., $m = 2\\sqrt{2}$, $n = 2$,\n\n$\\therefore mn = 4\\sqrt{2}$,\n\nHence, the answer is: $4\\sqrt{2}$.\n\n【Insight】This problem mainly examines the circle, the Pythagorean theorem, and the function image of moving point problems. The key to solving the problem lies in mastering the basic knowledge of circles, using the Pythagorean theorem to solve right triangles, determining the radius of the circle from the function image properties, and identifying the position of point $D$ when $PD$ is maximized." }, { "problem_id": 1366, "question": "As shown in Figure 1, a sector of paper has a central angle of $90^{\\circ}$ and a radius of 6. As shown in Figure 2, this sector is folded such that point $\\mathrm{A}$ coincides with point $\\mathrm{O}$, with the fold line being $\\mathrm{CD}$. The shaded area in the figure represents the overlapping portion. The area of the shaded portion is $\\qquad$. (Answer in radical form)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_add63c90b341ad5ca0f4g_0010_1.jpg", "batch17-2024_06_14_add63c90b341ad5ca0f4g_0010_2.jpg" ], "is_multi_img": true, "answer": "$6 \\pi-\\frac{9 \\sqrt{3}}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Connect $\\mathrm{OD}$,\n\n$\\because$ The sector-shaped paper is folded such that point $\\mathrm{A}$ coincides exactly with point $\\mathrm{O}$, and the crease is $\\mathrm{CD}$,\n\n$\\therefore \\mathrm{AC}=\\mathrm{OC}, \\mathrm{OD}=2 \\mathrm{OC}=6$,\n\n$\\therefore \\mathrm{CD}=\\sqrt{0 \\mathrm{D}^{2}-0 \\mathrm{C}^{2}}=3 \\sqrt{3}$\n\n$\\therefore \\angle \\mathrm{CDO}=30^{\\circ}, \\angle \\mathrm{COD}=60^{\\circ}$,\n\n$\\therefore$ The area of the figure enclosed by arc $\\mathrm{AD}$, line segment $\\mathrm{AC}$, and $\\mathrm{CD}$ is $=\\mathrm{S}$ sector $\\mathrm{AOD}-\\mathrm{S} \\triangle \\mathrm{COD}=\\frac{60 \\pi \\times 6^{2}}{360}-\\frac{1}{2} \\times 3$ $\\sqrt{3} \\times 3=6 \\pi-\\frac{9 \\sqrt{3}}{2}$\n$\\therefore$ The area of the shaded part is $6 \\pi-\\frac{9 \\sqrt{3}}{2}$,\n\nHence, the answer is $6 \\pi-\\frac{9 \\sqrt{3}}{2}$.\n\n\n\nFigure 2\n\n【Highlight】This question examines the calculation of the area of a sector and the properties of folding, converting the area of an irregular figure into that of a regular figure. Remembering the formula for calculating the area of a sector is key to solving the problem." }, { "problem_id": 1367, "question": "A rectangular wooden board has a circular hole in its upper right corner. It is now intended to be transformed into a hot pot dining table surface, with the size of the wooden board remaining unchanged and the center of the circular hole positioned at the intersection of the diagonals of the rectangular tabletop. The carpenter came up with a clever solution. After measuring the distance from the tangent point \\( K \\) of \\( PQ \\) to point \\( B \\) and other relevant data (in cm), he cut from point \\( N \\) along the broken line \\( NF-FM \\) (where \\( NF \\parallel BC \\) and \\( FM \\parallel AB \\)), as shown in Figure 1. The rectangle \\( EFGH \\) in Figure 2 is a schematic diagram of the two pieces of wood joined together to form a rectangular tabletop that meets the requirements (no overlap, no gaps, no waste). The lengths of \\( CN \\) and \\( AM \\) are respectively \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_b097719b9cd28e7aa021g_0012_1.jpg", "batch17-2024_06_14_b097719b9cd28e7aa021g_0012_2.jpg" ], "is_multi_img": true, "answer": "$18 \\mathrm{~cm}, 31 \\mathrm{~cm}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Draw the auxiliary line as shown in the figure, and let the radius of the circular hole be \\( r \\).\n\n\n\nAccording to the Pythagorean theorem, we have \\( BH^{2} + KH^{2} = BK^{2} \\).\n\nThus, \\( (130 - 50)^{2} + (44 + r)^{2} = 100^{2} \\),\n\nTherefore, \\( r = 16 \\).\n\nAs required by the problem, after cutting, the distance from the center of circle \\( O \\) to the two opposite sides should be equal,\n\nThat is: \\( ON_{1} = KN_{2} = \\frac{1}{2} AB \\), \\( OM_{1} = KM_{1} + r = \\frac{1}{2} CB \\).\n\nSince \\( KN_{2} = \\frac{1}{2} AB = 42 \\),\nTherefore, \\( QN_{2} + r = 42 \\), which means \\( QN_{2} = 42 - 16 = 26 \\).\n\nThus, \\( CN = QH - QN_{2} = 44 - 26 = 18 \\).\n\nAlso, since \\( KM_{1} + r = \\frac{1}{2} CB \\), that is \\( KM_{1} + 16 = \\frac{1}{2} \\times 130 \\),\n\nTherefore, \\( KM_{1} = 49 \\).\n\nThus, \\( AM = BC - PD - KM_{1} = 130 - 50 - 49 = 31 \\).\n\nTherefore, \\( CN = 18 \\text{ cm} \\), \\( AM = 31 \\text{ cm} \\).\n\nThe final answer is: \\( 18 \\text{ cm}, 31 \\text{ cm} \\)\n\n【Key Insight】This problem examines knowledge related to rectangles, right triangles, and circles. The key to solving the problem lies in transforming practical problems into mathematical problems and utilizing the concept of graphical transformation, demonstrating the application value of mathematical thinking methods in real-world problems." }, { "problem_id": 1368, "question": "As shown in Figure (1), in an equilateral $\\triangle \\mathrm{ABC}$ with side length 8, $\\mathrm{CD} \\perp \\mathrm{AB}$, with the foot of the perpendicular at $\\mathrm{D}$, the center of $\\odot \\mathrm{O}$ coincides with point $\\mathrm{D}$, and $\\odot \\mathrm{O}$ intersects segment $\\mathrm{CD}$ at point $\\mathrm{E}$. If $\\odot \\mathrm{O}$ is translated upward along the direction of $\\mathrm{DC}$ by $1 \\mathrm{~cm}$, as shown in Figure (2), $\\odot \\mathrm{O}$ just touches the sides $\\mathrm{AC}$ and $\\mathrm{BC}$ of $\\triangle \\mathrm{ABC}$, then the length of $\\mathrm{CE}$ in Figure (1) is $\\qquad$ $\\mathrm{cm}$.\n\n\n(1)\n\n\n(2)", "input_image": [ "batch17-2024_06_14_b097719b9cd28e7aa021g_0087_1.jpg", "batch17-2024_06_14_b097719b9cd28e7aa021g_0087_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{3}-\\frac{1}{2}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** As shown in the figure, let the point of tangency between circle \\( \\mathrm{O} \\) and \\( \\mathrm{BC} \\) be \\( \\mathrm{M} \\). Connect \\( \\mathrm{OM} \\). According to the property of tangents, we have \\( \\angle \\mathrm{OMC} = 90^\\circ \\). Given the condition \\( \\angle \\mathrm{DCB} = 30^\\circ \\), let \\( \\mathrm{OM} = r \\, \\text{cm} \\). Since \\( \\triangle \\mathrm{ABC} \\) is an equilateral triangle with \\( \\mathrm{AB} = 8 \\), it follows that \\( \\mathrm{CD} = \\frac{\\sqrt{3}}{2} \\times 8 = 4\\sqrt{3} \\, \\text{cm} \\). As circle \\( \\mathrm{O} \\) is translated upward along \\( \\mathrm{DC} \\) by \\( 1 \\, \\text{cm} \\), \\( \\mathrm{OD} = 1 \\, \\text{cm} \\). Therefore, \\( \\sin \\angle \\mathrm{DCB} = \\frac{\\mathrm{OM}}{\\mathrm{CO}} = \\frac{1}{2} \\), which implies \\( \\mathrm{CO} = 2r \\). Thus, \\( 1 + 2r = 4\\sqrt{3} \\), leading to \\( r = \\frac{4\\sqrt{3} - 1}{2} \\). Consequently, \\( \\mathrm{CE} = \\mathrm{CD} - \\mathrm{OE} - \\mathrm{OD} = 4\\sqrt{3} - \\frac{4\\sqrt{3} - 1}{2} - 1 = \\left(2\\sqrt{3} - \\frac{1}{2}\\right) \\, \\text{cm} \\).\n\n**Final Answer:** \\( 2\\sqrt{3} - \\frac{1}{2} \\).\n\n**Key Concepts:** Properties of tangents; properties of equilateral triangles; properties of translations." }, { "problem_id": 1369, "question": "A wooden support for a cylindrical water barrel is placed at a construction site (as shown in Figure 1). If the thickness of the wooden strips is ignored, the top view (vertical projection from above) is shown in Figure 2. It is known that \\( AD \\) bisects \\( BC \\) perpendicularly, and \\( AD = BC = 48 \\text{ cm} \\). The maximum radius of the base of the cylindrical water barrel is \\(\\qquad\\) \\(\\text{cm}\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_b2ad361951525488114fg_0062_1.jpg", "batch17-2024_06_14_b2ad361951525488114fg_0062_2.jpg" ], "is_multi_img": true, "answer": "30 .", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Connect $OB$ as shown in the figure.\n\n\n\nWhen $\\odot O$ is the circumcircle of $\\triangle ABC$, the radius of the base of the cylindrical water bucket reaches its maximum.\n\nSince $AD$ is the perpendicular bisector of $BC$, and $AD = BC = 48 \\mathrm{~cm}$,\n\n$\\therefore$ Point $O$ lies on $AD$, and $BD = 24 \\mathrm{~cm}$; in the right triangle $\\triangle OBD$, let the radius be $r$, then $OB = r$, $OD = 48 - r$,\n\n$\\therefore r^{2} = (48 - r)^{2} + 24^{2}$,\n\nSolving gives $r = 30$.\n\nThus, the maximum radius of the base of the cylindrical water bucket is $30 \\mathrm{~cm}$.\n\n【Insight】This is a common exam question that requires students to first understand the problem, construct a right triangle, and then calculate the corresponding values. Mastery of this method is essential." }, { "problem_id": 1370, "question": "The egg-shaped nine-piece puzzle originates from the \"Columbus Egg Puzzle\" released in 1983 (as shown in Figure 1), which is a variation and extension of the traditional seven-piece puzzle. As shown in Figure 2, $A B$ and $C D$ are two mutually perpendicular diameters of circle $O$. Arcs are drawn with points $A$ and $B$ as centers and the length of $A B$ as the radius, intersecting the extensions of $A C$ and $B C$ at points $E$ and $F$ respectively. Then, with point $C$ as the center and the length of $C E$ as the radius, the arc $E F$ is drawn, forming the outline of the egg-shaped nine-piece puzzle. If $A B = 2$, the perimeter of the outer edge of the egg-shaped nine-piece puzzle is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_b82179b4d99cd6350ef8g_0038_1.jpg", "batch17-2024_06_14_b82179b4d99cd6350ef8g_0038_2.jpg" ], "is_multi_img": true, "answer": "$3 \\pi-\\frac{\\sqrt{2}}{2} \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, both $AOC$ and $BOC$ are isosceles right triangles,\n\n$\\therefore \\angle A = \\angle B = 45^{\\circ}$,\n\n$\\because AB = 2$,\n\n$\\therefore OA = OC = OB = 1$,\n\n$\\therefore AC = BC = \\sqrt{2}$.\n\n$\\therefore CF = CE = AE - AC = 2 - \\sqrt{2}$,\n\n$\\therefore$ The length of $AB$ $= \\frac{1}{2} \\times 2 \\pi \\times \\frac{2}{2} = \\pi$,\n\nThe length of $AF$ $=$ The length of $BE$ $= \\frac{45 \\times \\pi \\times 2}{180} = \\frac{\\pi}{2}$,\n\nThe length of $EF$ $= \\frac{90 \\times \\pi \\times (2 - \\sqrt{2})}{180} = \\frac{(2 - \\sqrt{2}) \\pi}{2}$,\n\nTherefore, the perimeter of the outer edge of the egg-shaped nine-piece puzzle $= \\pi + 2 \\times \\frac{\\pi}{2} + \\frac{(2 - \\sqrt{2}) \\pi}{2}$\n$= \\frac{(6 - \\sqrt{2}) \\pi}{2}$\n\n$= 3\\pi - \\frac{\\sqrt{2}}{2} \\pi$.\n\n【Insight】This problem mainly examines the calculation of arc lengths. When solving such problems, it is important to note that complex irregular shapes should be decomposed into a combination of regular shapes." }, { "problem_id": 1371, "question": "As shown in the figure, there are two circles, A and B, with their radii in the ratio of $3:8$. Each circle is divided into two sectors, black and white. The areas of the black and white sectors in circle A are in the ratio of $1:2$, and the areas of the black and white sectors in circle B are in the ratio of $1:3$. What is the ratio of the combined area of the two black sectors to the combined area of the two white sectors in the figure? \n\nA\n\n\n\nB", "input_image": [ "batch17-2024_06_14_b984483c3f49ac22b86bg_0017_1.jpg", "batch17-2024_06_14_b984483c3f49ac22b86bg_0017_2.jpg" ], "is_multi_img": true, "answer": "19:54", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the radius of circle A be \\(3a\\) and the radius of circle B be \\(8a\\).\n\nThen, the area of circle A is \\((3a)^2 \\pi = 9\\pi a^2\\), and the area of circle B is \\((8a)^2 \\pi = 64\\pi a^2\\).\n\nSince the ratio of the areas of the black and white sectors in circle A is \\(1:2\\), and in circle B it is \\(1:3\\),\n\nthe area of the black sector in circle A is \\(\\frac{1}{1+2} \\times 9\\pi a^2 = 3\\pi a^2\\), and the area of the white sector in circle A is \\(\\frac{2}{1+2} \\times 9\\pi a^2 = 6\\pi a^2\\).\n\nSimilarly, the area of the black sector in circle B is \\(\\frac{1}{1+3} \\times 64\\pi a^2 = 16\\pi a^2\\), and the area of the white sector in circle B is \\(\\frac{3}{1+3} \\times 64\\pi a^2 = 48\\pi a^2\\).\n\nTherefore, the ratio of the combined area of the two black sectors to the combined area of the two white sectors is \\(\\left(3\\pi a^2 + 16\\pi a^2\\right) \\div \\left(6\\pi a^2 + 48\\pi a^2\\right) = 19\\pi a^2 \\div 54\\pi a^2 = 19:54\\).\n\nHence, the answer is: \\(19:54\\).\n\n[Key Insight] This problem tests the understanding of the area of a circle and proportional distribution. Mastering the formula for the area of a circle and the method of proportional distribution is key to solving this problem." }, { "problem_id": 1372, "question": "As shown, a circle with a diameter of 4 cm is divided into several equal parts (as shown in Figure 1), and then it is cut and rearranged into the shape shown in Figure 2. The perimeter of the resulting shape is $\\qquad$ cm longer than the original circle's perimeter.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_b984483c3f49ac22b86bg_0060_1.jpg", "batch17-2024_06_14_b984483c3f49ac22b86bg_0060_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The radius of the circle is: $4 \\div 2=2$ (centimeters)\n\nBecause the perimeter of the assembled figure is longer than the original circle's perimeter by the length of two radii on the left and right,\n\nTherefore, the perimeter has increased by $2+2=4$ (centimeters)\n\nHence, the answer is: 4\n\n[Key Insight] This question tests the understanding of planar figures, and the key to solving the problem lies in comprehending the meaning of the perimeter of a figure and the relationship before and after the figure is assembled." }, { "problem_id": 1373, "question": "As shown in Figure 1, there is a fan-shaped promotional board. A partial schematic of the board is shown in Figure 2, which is a sector formed with $O$ as the center, $O A$ and $O B$ as the radii, and a central angle $\\angle O = 120^{\\circ}$. If $O D = 5 \\mathrm{~m}$ and $O C = 3 \\mathrm{~m}$, then the area of the shaded part is $\\qquad$ $\\mathrm{m}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_ba789f3124c6fd56797ag_0054_1.jpg", "batch17-2024_06_14_ba789f3124c6fd56797ag_0054_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{16}{3} \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: The area of the shaded region, \\( S_{\\text{shaded}} \\), is calculated as the difference between the area of sector \\( AOD \\) and the area of sector \\( OC \\).\n\n\\[\nS_{\\text{shaded}} = S_{\\text{sector } AOD} - S_{\\text{sector } OC}\n\\]\n\n\\[\n= \\frac{120 \\pi \\cdot OD^{2}}{360} - \\frac{120 \\pi \\cdot OC^{2}}{360}\n\\]\n\n\\[\n= \\frac{120 \\pi (OD^{2} - OC^{2})}{360}\n\\]\n\n\\[\n= \\frac{\\pi (5^{2} - 3^{2})}{3}\n\\]\n\n\\[\n= \\frac{16}{3} \\pi \\text{ m}^{2}\n\\]\n\nTherefore, the answer is: \\(\\frac{16}{3} \\pi\\).\n\n**Key Insight:** This problem tests the understanding of the area of a sector and the area of irregular shapes. Mastery of the sector area formula is crucial for solving such problems." }, { "problem_id": 1374, "question": "A square $\\mathrm{ABCD}$ with side length 2 and a square $\\mathrm{AEFG}$ with side length $2 \\sqrt{2}$ are placed as shown in Figure (1), with $\\mathrm{AD}$ and $\\mathrm{AE}$ on the same straight line, and $\\mathrm{AB}$ and $\\mathrm{AG}$ on the same straight line. The square $\\mathrm{ABCD}$ is rotated counterclockwise around point $\\mathrm{A}$ as shown in Figure (2). The segments $\\mathrm{DG}$ and $\\mathrm{BE}$ intersect at point $\\mathrm{H}$. The maximum value of the sum of the areas of $\\triangle \\mathrm{GHE}$ and $\\triangle \\mathrm{BHD}$ is\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch17-2024_06_14_bcac7156d2ebff2a659bg_0010_1.jpg", "batch17-2024_06_14_bcac7156d2ebff2a659bg_0010_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:**\n\nGiven that quadrilaterals \\(ABCD\\) and \\(AEFG\\) are both squares,\n\n\\[\n\\therefore AD = AB, \\quad AG = AE, \\quad \\angle DAB = \\angle EAG = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle DAB + \\angle BAG = \\angle EAG + \\angle BAG,\n\\]\n\n\\[\n\\therefore \\angle DAG = \\angle BAE,\n\\]\n\n\\[\n\\therefore \\triangle ADG \\cong \\triangle ABE \\quad (\\text{SAS}),\n\\]\n\n\\[\n\\therefore \\angle AGD = \\angle AEB.\n\\]\n\nIn square \\(AEFG\\), \\(\\angle AGE = \\angle AEG = 45^\\circ\\),\n\n\\[\n\\therefore \\angle HGE + \\angle HEG = 45^\\circ + \\angle AGD + 45^\\circ - \\angle AEB = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle GHE = 90^\\circ.\n\\]\n\nThus, for \\(\\triangle EGH\\), point \\(H\\) lies on the circle with diameter \\(EG\\),\n\n\\[\n\\therefore \\text{When point } H \\text{ coincides with point } A, \\text{ the height of } \\triangle EGH \\text{ is maximized.}\n\\]\n\nSimilarly, for \\(\\triangle BDH\\), point \\(H\\) lies on the circle with diameter \\(BD\\),\n\n\\[\n\\therefore \\text{When point } H \\text{ coincides with point } A, \\text{ the height of } \\triangle BDH \\text{ is maximized.}\n\\]\n\n\\[\n\\therefore \\text{The maximum sum of the areas of } \\triangle GHE \\text{ and } \\triangle BHD \\text{ is:}\n\\]\n\n\\[\n\\frac{1}{2} \\times 2^2 + \\frac{1}{2} \\times (2\\sqrt{2})^2 = 2 + 4 = 6.\n\\]\n\n**Final Answer:** 6.\n\n**Key Insight:** This problem tests the properties of squares, rotational properties, the criteria and properties of congruent triangles, and the inscribed angle theorem. Determining the position of point \\(H\\) that maximizes the sum of the areas of \\(\\triangle GHE\\) and \\(\\triangle BHD\\) is crucial to solving this problem. Note that an angle inscribed in a semicircle (or subtended by a diameter) is a right angle, and a \\(90^\\circ\\) inscribed angle subtends the diameter." }, { "problem_id": 1375, "question": "The purple clay pot is a unique handmade earthenware craft in China, whose production process requires dozens of different tools. Among them, there is a tool named \"calibrated mouth frame,\" the shape and usage of which are shown in Figure 1. When the artisan aligns the upper arc of the \"calibrated mouth frame\" perfectly with the boundary of the pot's mouth, it ensures that the pot spout, handle, and mouth center are on the same straight line. Figure 2 is a schematic diagram of the correct usage of this tool. As shown in Figure 3, $O$ is the mouth of a certain purple clay pot, and it is known that points $\\mathrm{A}$ and $\\mathrm{B}$ are on $O$. The line $l$ passes through point $O$ and is perpendicular to $A B$ at point $D$, intersecting $O$ at point $C$. If $A B=30 \\mathrm{~mm}$, $C D=5 \\mathrm{~mm}$, find the radius $r$ of the mouth of this purple clay pot.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch17-2024_06_14_c8d73b909e7b2d6a41cag_0040_1.jpg", "batch17-2024_06_14_c8d73b909e7b2d6a41cag_0040_2.jpg", "batch17-2024_06_14_c8d73b909e7b2d6a41cag_0040_3.jpg" ], "is_multi_img": true, "answer": "$25 \\mathrm{~mm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $O B$.\n\n\n\nSince line $l$ passes through the center $O$ and is perpendicular to $A B$, and $A B=30$,\n\n$\\therefore B D=\\frac{1}{2} A B=15$.\n\nGiven that $C D=5$,\n\n$\\therefore D O=r-5$.\n\nSince $B O^{2}=B D^{2}+D O^{2}$,\n\n$\\therefore r^{2}=15^{2}+(r-5)^{2}$.\n\nSolving this equation gives $r=25$.\n\n$\\therefore$ The radius $r$ of the mouth of this purple clay teapot is $25 \\mathrm{~mm}$.\n\n【Key Insight】This problem tests the understanding of the Perpendicular Chord Theorem and the Pythagorean Theorem. Mastering the Perpendicular Chord Theorem is crucial for solving this problem." }, { "problem_id": 1376, "question": "As shown in Figure 1, a spherical ball is placed in a V-shaped rack. Figure 2 is its plan view, where $CA$ and $CB$ are both tangents to the circle $O$, with tangent points $A$ and $B$ respectively. If the radius of $O$ is $2\\sqrt{3} \\text{ cm}$, and $AB = 6 \\text{ cm}$, then $\\angle ACB =$ \n\nfigure 1\n\n\n\nfigure 2", "input_image": [ "batch17-2024_06_14_e0bc0d293ee8823ad800g_0038_1.jpg", "batch17-2024_06_14_e0bc0d293ee8823ad800g_0038_2.jpg" ], "is_multi_img": true, "answer": "$60^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $OC$, $OA$, and $OC$ intersects $AB$ at point $D$,\n\n\n\nFigure 2\n\n$\\because CA$ and $CB$ are tangents to $\\odot O$\n\n$\\therefore CA = CB$,\n\n$\\because OA = OB, \\quad OC = OC$\n\n$\\therefore \\triangle ACO \\cong \\triangle BCO$\n\n$\\therefore \\angle ACO = \\angle BCO$, which means $OC$ bisects $\\angle ACB$\n$\\therefore OC \\perp AB$\n\n$\\because AB = 6$\n\n$\\therefore BD = \\frac{1}{2} AB = \\frac{1}{2} \\times 6 = 3$\n\nIn right triangle $\\triangle OBD$\n\n$\\because OB = 2 \\sqrt{3}$,\n\n$\\therefore \\sin \\angle BOD = \\frac{BD}{OB} = \\frac{3}{2 \\sqrt{3}} = \\frac{\\sqrt{3}}{2}$,\n\n$\\therefore \\angle BOD = 60^{\\circ}$\n\n$\\because B$ is the point of tangency\n\n$\\therefore OB \\perp BC$\n\n$\\therefore \\angle OCB = 30^{\\circ}$\n\n$\\therefore \\angle ACB = 60^{\\circ}$.\n\nTherefore, the answer is: $60^{\\circ}$.\n\n【Insight】This problem mainly examines the properties of tangents, the criteria and properties of congruent triangles, and solving right triangles. Constructing a right triangle to find the angle is a commonly used method." }, { "problem_id": 1377, "question": "Definition: A quadrilateral with a set of equal adjacent sides and complementary angles is called an equicomplementary quadrilateral. Example: As shown in Figure 1, the quadrilateral is inscribed in $\\odot O, A B=AD$. Then the quadrilateral $A B C D$ is an equicomplementary quadrilateral.\n\nExploration and application: As shown in Figure 2, in the equicomplementary quadrilateral $A B C D$, $AB=AD$, the bisector of its exterior angle $\\angle E A D$ intersects the extension of $CD$ at point $F$, if $CD=10, AF=5$, then the length of $D F$ is $\\_$. .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_e0bc0d293ee8823ad800g_0063_1.jpg", "batch17-2024_06_14_e0bc0d293ee8823ad800g_0063_2.jpg" ], "is_multi_img": true, "answer": "$5 \\sqrt{2}-5$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, connect $AC$,\n\n\n\nFigure 2\n\n$\\because$ Quadrilateral $ABCD$ is an equilateral supplementary quadrilateral,\n\n$\\therefore \\angle BAD + \\angle BCD = 180^\\circ$,\n\nAlso, $\\angle BAD + \\angle EAD = 180^\\circ$,\n\n$\\therefore \\angle EAD = \\angle BCD$,\n\n$\\because AF$ bisects $\\angle EAD$,\n\n$\\therefore \\angle FAD = \\frac{1}{2} \\angle EAD$,\n\n$\\because$ Quadrilateral $ABCD$ is an equilateral supplementary quadrilateral,\n\n$\\therefore$ Points $A, B, C, D$ are concyclic,\n\n$\\because AB = AD$,\n\n$\\therefore AB = AD$,\n\n$\\therefore \\angle ACD - \\angle ACB$\n\n$\\therefore \\angle FCA = \\frac{1}{2} \\angle BCD$,\n\n$\\therefore \\angle FCA = \\angle FAD$,\n\nAlso, $\\angle AFC = \\angle DFA$,\n\n$\\therefore \\triangle ACF \\sim \\triangle DAF$,\n\n$\\therefore \\frac{AF}{DF} = \\frac{CF}{AF}$,\n\nThat is, $\\frac{5}{DF} = \\frac{DF + 10}{5}$,\n\n$\\therefore DF = 5\\sqrt{2} - 5$.\n\nTherefore, the answer is: $5\\sqrt{2} - 5$.\n\n【Key Insight】This question examines the new definition of an equilateral supplementary quadrilateral, properties of circles, the determination of angle bisectors, and the properties and determination of similar triangles. The key to solving the problem lies in the ability to conduct independent study and exploration, and to apply these concepts effectively." }, { "problem_id": 1378, "question": "Chinese cuisine emphasizes the aesthetics of color, aroma, and taste, and elegant plating can further enhance the appeal of the dish. In Figure (1), the shape of the plating is a part of a sector. Figure (2) is its geometric diagram (the shaded area represents the plating). By measuring, it is found that $AC = BD = 12 \\text{ cm}$, the distance between points $C$ and $D$ is $3 \\text{ cm}$, and the central angle is $60^\\circ$. Then, the area of the plating in the figure is $\\qquad$. (Express in terms of $\\pi$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_e890efc93c61380c66b8g_0065_1.jpg", "batch17-2024_06_14_e890efc93c61380c66b8g_0065_2.jpg" ], "is_multi_img": true, "answer": "$36 \\pi \\mathrm{cm}^{2}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, connect $CD$,\n\n\n\nSince $OC = OD$ and $\\angle COD = 60^\\circ$,\n\nTherefore, $\\triangle OCD$ is an equilateral triangle,\n\nHence, $OC = OD = CD = 3 \\text{ cm}$,\n\nSince $AC = BD = 12 \\text{ cm}$,\n\nTherefore, $OA = OC + AC = 15 \\text{ cm}$,\n\nThus, the area of the sector in the figure is: $\\frac{60 \\times \\pi \\times 15^{2}}{360} - \\frac{60 \\times \\pi \\times 3^{2}}{360} = 36 \\pi \\text{ cm}^{2}$,\n\nSo the answer is: $36 \\pi \\text{ cm}^{2}$.\n\n【Key Insight】This problem examines the application of sector area to solve real-life problems, with the key being the calculation of the radii of the two sectors." }, { "problem_id": 1379, "question": "As shown in Figure 1, it is the entrance of an arched culvert. Figure 2 is a schematic diagram of the culvert. If the arch height $\\mathrm{CD}$ of the culvert is 6 meters, and the width of the ground at the entrance of the culvert $\\mathrm{AB}$ is 4 meters, please find the radius of the circle on which the arc of the culvert lies.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_eebeae68c3a0215b997eg_0083_1.jpg", "batch17-2024_06_14_eebeae68c3a0215b997eg_0083_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{10}{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "According to the problem, $CD$ passes through point $O$ and is perpendicular to $AB$. Connect $OA$, and let the radius be $x$ meters.\n\n\n\nThus, $AD = DB = 2$.\n\nIn the right triangle $\\triangle ADO$, by the Pythagorean theorem, we have $OA^{2} = OD^{2} + AD^{2}$,\n\nwhich is $x^{2} = (6 - x)^{2} + 2^{2}$,\n\nSolving this gives $x = \\frac{10}{3}$.\n\nAnswer: The radius is $\\frac{10}{3}$ meters." }, { "problem_id": 1380, "question": "As shown in Figure 1, the circular arch is a style commonly used in ancient Chinese architecture, both aesthetically pleasing and practical. Figure 2 is a schematic diagram of the arch. The base of the arch, $AB$, is 2.4 meters wide, and the height of the arch, $CD$, is 3.6 meters. Find the radius of the circle that the arch belongs to.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch17-2024_06_14_f2357a1129c784c0eecfg_0072_1.jpg", "batch17-2024_06_14_f2357a1129c784c0eecfg_0072_2.jpg" ], "is_multi_img": true, "answer": "2 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Connect $OA$ as shown in the diagram:\n\n\n\nSince $AB \\perp CD$ and $CD$ passes through the center $O$ of the circle, with $AB = 2.4 \\mathrm{~m}$,\n\n$\\therefore AD = \\frac{1}{2} AB = 1.2 \\mathrm{~m}$.\n\nLet $OA = OC = x \\mathrm{~m}$, then $OD = (3.6 - x) \\mathrm{m}$.\n\n$\\therefore$ In the right triangle $\\triangle ADO$, by the Pythagorean theorem: $1.2^{2} + (3.6 - x)^{2} = x^{2}$.\n\nSolving this equation yields: $x = 2$,\n\nwhich means the radius of the circle where the arch is located is 2 meters.\n\n【Key Insight】This problem primarily tests the theorem of perpendicular chords. Mastering this theorem is crucial for solving such problems." }, { "problem_id": 1381, "question": "The waterwheel is an ancient Chinese invention used for irrigation, showcasing the wisdom of ancient Chinese laborers. In Figure 1, point $P$ represents a bucket on the waterwheel. As shown in Figure 2, when the waterwheel operates, the path of the bucket is a circle with center $O$ and a radius of $5$ meters, with the center of the circle above the water surface. If the chord $AB$ formed by the intersection of the circle and the water surface has a length of $8$ meters, find the maximum depth of the bucket below the water surface when the waterwheel is in operation.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch18-2024_06_15_7621490088dc62e2752fg_0080_1.jpg", "batch18-2024_06_15_7621490088dc62e2752fg_0080_2.jpg" ], "is_multi_img": true, "answer": "$2 m$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw $OD \\perp AB$ at $E$, intersecting $O$ at point $D$.\n\n$\\therefore AE = \\frac{1}{2} AB$\n\n$\\because AB = 8$\n\n$\\therefore AE = 4$\n\nIn the right triangle $\\triangle AEO$, $AO = 5$\n\n$\\therefore OE = \\sqrt{OA^{2} - AE^{2}} = 3$\n$\\therefore ED = 2$\n\n$\\therefore$ When the waterwheel is operating, the maximum depth of the water bucket below the water surface is $2$ meters.\n\n\n\nFigure 2\n\n【Key Point】This problem tests the theorem of perpendicular chords, with the key to solving it lying in constructing auxiliary lines and using the Pythagorean theorem for calculations." }, { "problem_id": 1382, "question": "Exercise is beneficial for building physical strength. A school is holding a 200-meter race on the sports field, with each lane being 1.22 meters wide. The finish line is set at point $C$ as shown in the figure. Since athletes on different lanes will go through different curves, they should not start from the same starting line. How far apart should the starting lines for athletes in the first and second lanes be to ensure fairness? (Take $\\pi$ as 3.14, accurate to 0.01 meters)\n\n\n\nPage 62 of 121\n\n", "input_image": [ "batch18-2024_06_15_7634eea1cb0236f7e58cg_0045_1.jpg", "batch18-2024_06_15_7634eea1cb0236f7e58cg_0045_2.jpg" ], "is_multi_img": true, "answer": "7.66 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "The curves at both ends of the running track together form a perfect circle.\n\nAssuming the radius of the curve for the first lane is \\( r \\), then the radius of the curve for the second lane is \\( (r + 1.22) \\).\n\nTherefore, the difference in the total length of the two complete tracks is \\( 2\\pi(r + 1.22) - 2\\pi r = 2\\pi \\times 1.22 \\approx 7.66 \\) meters.\n\nHence, to ensure fairness, the starting lines for athletes on the first and second lanes should be spaced approximately 7.66 meters apart.\n\n[Key Insight] This question tests the application of the circumference of a circle. The key to solving the problem lies in understanding the mathematical model of the situation and being proficient in calculating the circumference of a circle. The curves at both ends of the running track form a circle, and the problem involves calculating the difference in the circumferences of the curves for the first and second lanes." }, { "problem_id": 1383, "question": "The size of an iron ball can be checked for compliance using a workpiece groove (as shown in Figure 1). It is known that both base angles of the workpiece groove are $90^{\\circ}$, and the dimensions are as shown (units: $\\mathrm{cm}$). When a regular-shaped iron ball is placed in the groove, if it has contact points $A$, $B$, and $E$ as shown in Figure 1, the size of the ball is considered compliant. Figure 2 is a cross-sectional diagram through the center $O$ of the ball and points $A$, $B$, and $E$. Find the diameter of this iron ball.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch18-2024_06_15_7dfe5be0474cbce67281g_0027_1.jpg", "batch18-2024_06_15_7dfe5be0474cbce67281g_0027_2.jpg" ], "is_multi_img": true, "answer": "$20 \\mathrm{~cm}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** \nConnect $\\mathrm{OA}$ and $\\mathrm{OE}$, and let $\\mathrm{OE}$ intersect $\\mathrm{AB}$ at point $\\mathrm{P}$. According to the problem, quadrilateral $\\mathrm{ABCD}$ is a rectangle. By the **Perpendicular Chord Theorem**, we deduce that $\\mathrm{PA} = 8 \\mathrm{~cm}$ and $\\mathrm{PE} = 4 \\mathrm{~cm}$. Then, using the **Pythagorean Theorem** in right triangle $\\triangle \\mathrm{AOP}$, we can find the value of $\\mathrm{OA}$, which will help determine the diameter of the circle.\n\n**Problem Solution:** \nConnect $\\mathrm{OA}$ and $\\mathrm{OE}$, and let $\\mathrm{OE}$ intersect $\\mathrm{AB}$ at point $\\mathrm{P}$, as shown in the figure.\n\n\n\nSince $\\mathrm{AC} = \\mathrm{BD}$, $\\mathrm{AC} \\perp \\mathrm{CD}$, and $\\mathrm{BD} \\perp \\mathrm{CD}$, quadrilateral $\\mathrm{ACDB}$ is a rectangle.\n\nGiven that $C D = 16 \\mathrm{~cm}$ and $\\mathrm{PE} = 4 \\mathrm{~cm}$, we have:\n$$\n\\mathrm{PA} = 8 \\mathrm{~cm}, \\quad \\mathrm{BP} = 8 \\mathrm{~cm}.\n$$\n\nIn right triangle $\\triangle \\mathrm{OAP}$, applying the **Pythagorean Theorem**:\n$$\n\\mathrm{OA}^{2} = \\mathrm{PA}^{2} + \\mathrm{OP}^{2}.\n$$\nSubstituting the known values:\n$$\n\\mathrm{OA}^{2} = 8^{2} + (\\mathrm{OA} - 4)^{2}.\n$$\nSolving this equation yields:\n$$\n\\mathrm{OA} = 10 \\mathrm{~cm}.\n$$\n\n**Answer:** \nThe diameter of the iron sphere is **$20 \\mathrm{~cm}$**." }, { "problem_id": 1384, "question": "To save energy and reduce emissions, more and more citizens are using shared electric bikes. Figure 1 shows a physical image of the electric bike, and Figure 2 shows a schematic diagram of the bike. $AB$ is parallel to the ground. Given that the radius of the wheels is $15 \\mathrm{~cm}$, $BE = 40 \\mathrm{~cm}$, and $\\angle ABE = 60^{\\circ}$, if the thickness of the seat is $EM = 12 \\mathrm{~cm}$, find the height of the seat $M$ above the ground. (The result should be accurate to $1 \\mathrm{~cm}$) (Reference data: $\\sqrt{3} \\approx 1.732$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch18-2024_06_15_e5d65fe06fcd3cf5c9beg_0038_1.jpg", "batch18-2024_06_15_e5d65fe06fcd3cf5c9beg_0038_2.jpg" ], "is_multi_img": true, "answer": "$62 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let the point where the rear wheel of the car touches the ground be point $O$. Connect $A O$, and draw a perpendicular from point $E$ to the ground, with the foot of the perpendicular being point $P$. The line $E P$ intersects $A B$ at point $Q$.\n\n\n\nSince $A B \\parallel O P$,\n\nTherefore, $E P \\perp A B$,\n\nFrom the property of the tangent to a circle: $A O \\perp O P$,\n\nThus, quadrilateral $A O P Q$ is a rectangle,\n\nTherefore, $P Q = A O = 15 \\mathrm{~cm}$,\n\nSince $B E = 40 \\mathrm{~cm}$ and $\\angle A B E = 60^{\\circ}$,\n\nTherefore, $E Q = B E \\cdot \\sin \\angle A B E = 40 \\times \\frac{\\sqrt{3}}{2} = 20 \\sqrt{3} \\mathrm{~cm}$,\n\nSince $E M = 12 \\mathrm{~cm}$,\nTherefore, $E M + E Q + P Q = 12 + 20 \\sqrt{3} + 15 \\approx 61.64 \\approx 62 \\mathrm{~cm}$,\n\nAnswer: The height of the seat $M$ from the ground is approximately $62 \\mathrm{~cm}$.\n\n【Key Insight】This problem examines the properties of tangents to circles, the determination and properties of rectangles, and the solution of right-angled triangles. The key to solving the problem lies in constructing auxiliary lines to form a rectangle and right-angled triangles." }, { "problem_id": 1385, "question": "As shown in Figure 1, a manufacturer designs and produces a traditional Chinese medicine (TCM) mortar. The bottom of the mortar is approximately arc-shaped (referred to as an arc in this problem), and a roller with an axle can be placed inside the mortar. When TCM is placed in the mortar and the roller is rolled, the TCM can be crushed. A simplified schematic diagram of the mortar is shown in Figure 2. After measurement, the distance between the two ends of the mortar, \\( CD \\), is \\( 32 \\, \\text{cm} \\), the diameter of the roller is \\( 20 \\, \\text{cm} \\), and the diameter of the axle of the roller is \\( 4 \\, \\text{cm} \\). Let the center of the cross-sectional circle of the roller's axle be \\( G \\). When the roller passes through the lowest point \\( E \\) of the mortar, \\( G \\) is tangent to \\( CD \\) at point \\( F \\). Please find the radius of the circle on which the arc \\( CED \\) at the bottom of the mortar lies.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch18-2024_06_15_e5d65fe06fcd3cf5c9beg_0042_1.jpg", "batch18-2024_06_15_e5d65fe06fcd3cf5c9beg_0042_2.jpg" ], "is_multi_img": true, "answer": "$20 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the center of the circle where the arc $C E D$ at the bottom of the grinding groove lies be $O$. Connect $O C$ and $O E$. According to the problem, it is known that $O$, $G$, $F$, and $E$ are collinear, with $G F=\\frac{1}{2} \\times 4=2 \\mathrm{~cm}$ and $G E=\\frac{1}{2} \\times 20=10 \\mathrm{~cm}$.\n\nSince circle $G$ is tangent to $C D$ at $F$,\n\n$\\therefore O E \\perp C D$,\n\n$\\therefore C F=D F=\\frac{1}{2} C D=16 \\mathrm{~cm}$,\n\n$\\therefore E F=G E-G F=8 \\mathrm{~cm}$,\n\nLet $O C=O E=x$, then $P F=x-8$,\n\nIn the right triangle $\\triangle O C F$, $O C^{2}=C F^{2}+O F^{2}$,\n\n$\\therefore x^{2}=16^{2}+(x-8)^{2}$,\n\nSolving for $x$ gives $x=20$,\n\n$\\therefore$ The radius of the circle where the arc $C E D$ at the bottom of the grinding groove lies is $20 \\mathrm{~cm}$.\n\n\n\nFigure 2\n\n【Key Insight】This problem primarily examines the properties of tangents, the chord perpendicularity theorem, and the Pythagorean theorem. Understanding the chord perpendicularity theorem is crucial for solving the problem." }, { "problem_id": 1386, "question": "Take a rectangular piece of paper and fold it according to the methods shown in Figure 1 and Figure 2. Then, cut along the diagonal line (dashed line) passing through the vertices of the rectangle as shown in Figure 3. Unfold and lay the cut-out portion (1) flat on the table. If the resulting shape is a regular hexagon, then the ratio of the width to the length of this rectangular piece of paper is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\nFigure 3", "input_image": [ "batch19-2024_05_24_99b873783029f03e0775g_0049_1.jpg", "batch19-2024_05_24_99b873783029f03e0775g_0049_2.jpg", "batch19-2024_05_24_99b873783029f03e0775g_0049_3.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{3}: 2$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Draw $\\mathrm{OB} \\perp \\mathrm{AD}$. Based on the given information, the following figure can be constructed:\n\n\n\n$\\because$ According to the folding method, we have:\n\n$\\mathrm{AB}=\\mathrm{AD}$, $\\mathrm{CD}=\\mathrm{CE}$, $\\angle \\mathrm{OAB}=60^{\\circ}$, and $\\mathrm{AO}$ is equal to the side length of the regular hexagon.\n\n$\\therefore \\angle \\mathrm{BOA}=30^{\\circ}$,\n$\\therefore 2 \\mathrm{AB}=\\mathrm{AO}$,\n\n$\\frac{\\mathrm{BO}}{\\mathrm{AB}}=\\tan 60^{\\circ}=\\sqrt{3}$\n\n$\\therefore \\mathrm{BO}: \\mathrm{AM}=\\sqrt{3}: 2$." }, { "problem_id": 1387, "question": "To promote indoor air circulation and maintain a dry indoor environment, a center-pivot window, as shown in Figure (1), is used for the attic skylight. The window is a square with a side length of $1.2 \\mathrm{~m}$. Figure (2) is a side view of the window when fully open. The pivot point of the casement window is at the center, and it is assumed that the wind direction is perpendicular to the window frame $B E$. The ventilation area is $0.48 \\mathrm{~m}^{2}$. Find the area swept by the window frame $B D$ as it rotates from $B E$. (The result should be accurate to $0.01 \\mathrm{~m}^{2}$. Reference data: $\\sqrt{3} \\approx 1.732, \\sqrt{2} \\approx 1.414$, $\\sin 48^{\\circ} \\approx 0.74, \\cos 48^{\\circ} \\approx 0.67, \\tan 48^{\\circ} \\approx 1.11, \\pi$ is taken as 3.14)\n\n\nFigure (1)\n\nFigure (2)", "input_image": [ "batch19-2024_05_24_9ac5dc8e2dc0bb6540e0g_0100_1.jpg", "batch19-2024_05_24_9ac5dc8e2dc0bb6540e0g_0100_2.jpg" ], "is_multi_img": true, "answer": "$0.15 \\mathrm{~m}^{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure below, draw a perpendicular line from point $D$ to $BE$, meeting at point $F$.\n\nGiven that the ventilation area is $0.48 \\mathrm{~m}^{2}$,\n\nThe ventilation area above and below point $B$ is each $0.24 \\mathrm{~m}^{2}$.\n\nSince the window is a square with a side length of $1.2 \\mathrm{~m}$, and the pivot point of the casement window is at the center,\n\nThus, $EF = \\frac{0.24}{1.2} = 0.2 \\mathrm{~m}$, and $BE = BD = 0.6 \\mathrm{~m}$.\n\nTherefore, $BF = BE - EF = 0.4 \\mathrm{~m}$.\n\nHence, $\\cos \\angle EBD = \\frac{BF}{BD} = \\frac{2}{3} \\approx 0.67$.\n\nThus, $\\angle EBD \\approx 48^{\\circ}$.\n\nTherefore, the area swept by the window frame $BD$ as it rotates from $BE$ is $\\frac{48 \\times \\pi \\times (0.6)^{2}}{360} = \\frac{6 \\pi}{125} \\approx 0.15 \\mathrm{~m}^{2}$.\n\n【Key Insight】This problem tests the formula for the area of a sector and the solution of right-angled triangles. The key to solving the problem lies in the comprehensive application of these knowledge points." }, { "problem_id": 1388, "question": "Figure 1 is a schematic diagram of a bicycle shed for students at a school (dimensions as shown), where the roof of the shed is part of the lateral surface of a cylinder, and its development is a rectangle. Figure 2 is a schematic diagram of the cross-section of the shed roof, with the circle on which \\( AB \\) lies having its center at \\( O \\). The roof of the shed is covered with a certain type of canvas. Find the area of the canvas covering the roof. (Neglecting factors such as seams, the result should retain \\(\\pi\\).)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch19-2024_05_24_c5e9f7cd37910effc40dg_0093_1.jpg", "batch19-2024_05_24_c5e9f7cd37910effc40dg_0093_2.jpg" ], "is_multi_img": true, "answer": "$160 \\pi$ 平方m.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure: Connect $\\mathrm{OB}$, and draw $\\mathrm{OE} \\perp \\mathrm{AB}$ from point $\\mathrm{O}$, with the foot of the perpendicular at $\\mathrm{E}$, intersecting $\\mathrm{AB}$ at $\\mathrm{F}$.\n\nSince $\\mathrm{E}$ is the midpoint of $\\mathrm{AB}$, and $\\mathrm{F}$ is also the midpoint of $\\mathrm{AB}$,\n\nTherefore, $\\mathrm{EF}$ is the height of the segment,\n\nThus, $\\mathrm{AE}=\\frac{1}{2} \\mathrm{AB}=2 \\sqrt{3}$, and $\\mathrm{EF}=2$.\n\nLet the radius be $\\mathrm{R}$ meters, then $\\mathrm{OE}=(\\mathbf{R}-2)$ meters.\n\nIn the right triangle $\\triangle A O E$, by the Pythagorean theorem, we have $R^{2}=(\\mathbf{R}-2)^{2}+(2 \\sqrt{3})^{2}$, solving which gives $R=4$.\n\nSince $\\sin \\angle \\mathrm{AOE}=\\frac{\\mathrm{AE}}{\\mathrm{OA}}=\\frac{\\sqrt{3}}{2}$,\n\nTherefore, $\\angle \\mathrm{AOE}=60^{\\circ}$,\n\nHence, $\\angle \\mathrm{AOB}=120$ degrees.\n\nThus, the length of arc $\\mathrm{AB}$ is $\\frac{120 \\times 4 \\pi}{180}=\\frac{8}{3} \\pi$ meters.\n\nTherefore, the area of the sailcloth is $\\frac{8}{3} \\pi \\times 60=160 \\pi$ square meters.\n\n\n\n【Key Insight】This problem mainly tests the perpendicular chord theorem of circles, the Pythagorean theorem, trigonometric functions of acute angles, the arc length formula, and the sector area formula. The key to solving this problem lies in the flexible application of these related concepts." }, { "problem_id": 1389, "question": "As shown in Figure 1, the diagram represents the bicycle shed for teachers at a certain experimental junior high school (dimensions as shown). The roof of the shed is part of the lateral surface of a cylinder, and its unfolded form is a rectangle. Figure 2 is a schematic diagram of the cross-section of the shed roof, where the circle containing $A B$ has its center at $O$. A line is drawn from $O$ perpendicular to $A B$, with the foot of the perpendicular at $C$, intersecting $A B$ at point $D$. Given $A B = 4 \\sqrt{3}$ and $C D = 2$, the roof of the shed is covered with a type of plastic steel sheet. Calculate the area of the plastic steel sheet covering the roof. (Neglecting factors such as seams, the result should retain $\\pi$.)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch19-2024_05_24_fc559d5bf2be98a30689g_0071_1.jpg", "batch19-2024_05_24_fc559d5bf2be98a30689g_0071_2.jpg" ], "is_multi_img": true, "answer": "$60 \\pi$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( OD \\perp AB \\),\n\nTherefore, \\( AC = \\frac{1}{2} AB = 2\\sqrt{3} \\),\n\nSince \\( CD = 2 \\),\n\nTherefore, \\( OC = OA - 2 \\),\n\nSince \\( OA^{2} - OC^{2} = AC^{2} \\),\n\nTherefore, \\( OA = 4 \\),\n\nThus, \\( \\sin \\angle AOD = \\frac{\\sqrt{3}}{2} \\),\n\nTherefore, \\( \\angle AOD = 60^{\\circ} \\),\n\nHence, \\( \\angle AOB = 2 \\angle AOD = 120^{\\circ} \\),\n\nTherefore, the area of the plastic steel sheet covering the shed roof \\( = \\frac{120 \\pi \\times 4}{180} \\times 60 = 160 \\pi \\).\n\n【Key Insight】This problem primarily examines the Perpendicular Chord Theorem, the arc length formula, and the Pythagorean Theorem. The key to solving this problem lies in utilizing the Pythagorean Theorem and the Perpendicular Chord Theorem to determine the lengths of relevant segments and angles." }, { "problem_id": 1390, "question": "As shown in Figure 1, $\\triangle A B_{1} C_{1}$ is an equilateral triangle with side length 2; as shown in Figure 2, take the midpoint $C_{2}$ of $A B_{1}$, and draw an equilateral $\\triangle A B_{2} C_{2}$,\n\nconnect $B_{1} B_{2}$; as shown in Figure 3, take the midpoint $C_{3}$ of $A B_{2}$, and draw an equilateral $\\triangle A B_{3} C_{3}$, connect $B_{2} B_{3}$; as shown in Figure 4, take the midpoint $C_{4}$ of $A B_{3}$,\n\ndraw an equilateral $\\triangle A B_{4} C_{4}$, connect $B_{3} B_{4}$, $\\qquad$ continue this pattern indefinitely, then the length of $B_{n} B_{n+1}$ is $\\qquad$ (expressed in terms of $n$).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4", "input_image": [ "batch1-2024_06_14_1a64532da5f7cd51dd61g_0011_1.jpg", "batch1-2024_06_14_1a64532da5f7cd51dd61g_0011_2.jpg", "batch1-2024_06_14_1a64532da5f7cd51dd61g_0011_3.jpg", "batch1-2024_06_14_1a64532da5f7cd51dd61g_0011_4.jpg" ], "is_multi_img": true, "answer": "$\\frac{\\sqrt{3}}{2^{n-1}}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in Figure (2), draw a perpendicular line from point $C_{2}$ to $B_{1} B_{2}$, intersecting at point $D$.\n\n\n\nFigure (2)\n\nSince $\\triangle A B_{1} C_{1}$ is an equilateral triangle with side length 2, and $C_{2}$ is the midpoint of $A B_{1}$,\n\n$\\therefore B_{1} C_{2}=B_{2} C_{2}=1$.\n\nSince $\\triangle A B_{2} C_{2}$ is also an equilateral triangle,\n\n$\\therefore \\angle B_{1} C_{2} B_{2}=120^{\\circ}, \\quad B_{1} C_{2}=B_{2} C_{2}$,\n\n$\\therefore \\angle D B_{1} C_{1}=\\angle D B_{2} C_{2}=30^{\\circ}$,\n\n$\\therefore C_{2} D=\\frac{1}{2} B_{1} C_{2}=\\frac{1}{2}$,\n\n$\\therefore B_{1} D=\\sqrt{B_{1} C_{2}{ }^{2}-C_{2} D^{2}}=\\sqrt{1^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\frac{\\sqrt{3}}{2}$\n\n$\\therefore B_{1} B_{2}=2 B_{1} D=\\sqrt{3}$,\n\nSimilarly, we can find that $B_{2} B_{3}=\\frac{\\sqrt{3}}{2}, B_{3} B_{4}=\\frac{\\sqrt{3}}{2^{2}}=\\frac{\\sqrt{3}}{4}$\n\n$B_{n} B_{n+1}=\\frac{\\sqrt{3}}{2^{n-1}}$.\n\nTherefore, the answer is: $\\frac{\\sqrt{3}}{2^{n-1}}$.\n\n【Key Insight】This problem tests the properties of equilateral triangles. The key to solving it lies in drawing the auxiliary line as per the given conditions, determining the length of $B_{1} B_{2}$, and identifying the pattern." }, { "problem_id": 1391, "question": "Liu Hui was a great mathematician during the Wei and Jin dynasties in China. In his commentary on the \"Nine Chapters on the Mathematical Art,\" he stated: \"The sum of the squares of the legs equals the square of the hypotenuse, clearly.\" That is, in the right triangle in Figure 1, the three sides \\(a\\), \\(b\\), and \\(c\\) satisfy the relationship \\(a^2 + b^2 = c^2\\). He constructed some basic figures in the book to solve problems. As shown in Figure 2, if a square with side length \\(a\\) and a square with side length \\(b\\) are placed in the lower left and upper right corners of a larger square with side length \\(c\\), respectively, then the area of the shaded part in the figure is equal to \\(\\qquad\\) (expressed as an algebraic expression containing the letter \\(a\\)); if \\((c-a)(c-b)=18\\), then \\(a+b-c=\\) \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch1-2024_06_14_1a64532da5f7cd51dd61g_0022_1.jpg", "batch1-2024_06_14_1a64532da5f7cd51dd61g_0022_2.jpg" ], "is_multi_img": true, "answer": "$a^{2} 6$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the problem, the area of the shaded region in the figure is $c^{2}-b^{2}$.\n\nSince $a^{2}+b^{2}=c^{2}$,\n\nIt follows that $a^{2}=c^{2}-b^{2}$,\n\nTherefore, the area of the shaded region in the figure is $a^{2}$.\n\nAs shown in the figure, $AB = c - b$, $AC = c - a$, $DE = a - (c - b) = a + b - c$.\n\nGiven that $(c - a)(c - b) = 18$,\n\nIt follows that $AB \\cdot AC = 18$, which means the area of rectangle $ACDB$ is 18.\n\nSince the area of the shaded region is $2 \\times \\text{Area of rectangle } ACDB + a^{2} - (a + b - c)^{2} = a^{2}$,\n\nIt follows that $(a + b - c)^{2} = 36$.\n\nMoreover, since $a + b > c$, which implies $a + b - c > 0$,\n\nTherefore, $a + b - c = 6$.\n\nHence, the answer is: 6.\n\n\n\nFigure 2\n\n【Key Insight】This problem mainly tests the Pythagorean theorem, the relationship between the sides of a triangle, and the extraction of square roots. The key to solving the problem lies in correctly understanding the problem and deriving $(a + b - c)^{2} = 36$." }, { "problem_id": 1392, "question": "Figure 1 shows the emblem pattern of the Seventh International Congress on Mathematical Education (ICME-7), which evolved from a series of right-angled triangles sharing a common vertex \\( O \\) (as shown in Figure 2). If in Figure 2, \\( OA_1 = A_1A_2 = A_2A_3 = \\cdots = A_7A_8 = 1 \\), then the length of \\( OA_8 \\) is \\(\\qquad\\).\n\n\n\nICME-7\n\nFigure 1\n\n\nFigure 2", "input_image": [ "batch1-2024_06_14_1a64532da5f7cd51dd61g_0045_1.jpg", "batch1-2024_06_14_1a64532da5f7cd51dd61g_0045_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since \\( O A_{1} = A_{1} A_{2} = A_{2} A_{3} = A_{3} A_{4} = \\ldots = 1 \\),\n\nTherefore, by the Pythagorean theorem, we have:\n\\[ O A_{2} = \\sqrt{O A_{1}^{2} + A_{1} A_{2}^{2}} = \\sqrt{2}, \\]\n\\[ O A_{3} = \\sqrt{O A_{2}^{2} + A_{2} A_{3}^{2}} = \\sqrt{2 + 1} = \\sqrt{3}, \\]\n\\[ O A_{4} = \\sqrt{O A_{3}^{2} + A_{3} A_{4}^{2}} = \\sqrt{3 + 1} = \\sqrt{4}, \\]\n\\[ \\ldots \\]\n\nIt can be deduced that \\( O A_{n} = \\sqrt{n} \\),\n\nThus, \\( O A_{8} = \\sqrt{8} = 2 \\sqrt{2} \\),\n\nHence, the answer is: \\( 2 \\sqrt{2} \\).\n\n**Key Insight:** This problem primarily examines the Pythagorean theorem and the properties of square roots. The key to solving the problem lies in deriving \\( O A_{n} = \\sqrt{n} \\) through calculation." }, { "problem_id": 1393, "question": "Figure 1 and Figure 2 are the actual scene and side view of a cliff swing at a scenic spot, respectively. When the swing is stationary, it is located on the vertical line $O C$, with the distance from the top pivot $O$ to the ground being $O C = 20 \\mathrm{~m}$. During the swing, a tourist reaches the highest point $\\mathrm{A}$ after swinging backward, where the distance from point $\\mathrm{A}$ to $O C$ is measured as $A E = 3 \\mathrm{~m}$, and the distance from point $\\mathrm{A}$ to the ground platform $D C$ is $A D = 2 \\mathrm{~m}$. The rope $O A$ swings from point $\\mathrm{A}$ to the farthest point $O B$ outside the cliff, where $O B \\perp O A$. Find the distance from $B$ to $O C$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch1-2024_06_14_3e09945285968e53508fg_0022_1.jpg", "batch1-2024_06_14_3e09945285968e53508fg_0022_2.jpg" ], "is_multi_img": true, "answer": " $18 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Construct $BF \\perp OC$, with the foot of the perpendicular at $F$.\n\nSince $AD \\perp CD$, $OC \\perp CD$, and $AE \\perp OC$,\n\n$\\therefore \\angle ADC = \\angle DCE = \\angle AEC = 90^{\\circ}$,\n\n$\\therefore$ quadrilateral $ADCE$ is a rectangle,\n\n$\\therefore AD = EC = 2 \\mathrm{~m}$.\n\nAlso, since $OA \\perp OB$,\n\n$\\therefore \\angle AOE + \\angle BOE = 90^{\\circ}$,\n\n$\\therefore BF \\perp OC$,\n\n$\\therefore \\angle BFO = 90^{\\circ}$,\n$\\therefore \\angle BOE + \\angle OBF = 90^{\\circ}$,\n\n$\\therefore \\angle AOE = \\angle OBF$,\n\nMoreover, $\\angle AEO = \\angle BFO = 90^{\\circ}$,\n\nand $OA = OB$,\n\n$\\therefore \\triangle OAE \\cong \\triangle BOF$ (by AAS),\n\n$\\therefore BF = OE = OC - EC = 20 - 2 = 18 (\\mathrm{~m})$\n\n$\\therefore$ The distance from $B$ to $OC$ is $18 \\mathrm{~m}$.\n\n\n\n【Insight】This problem examines the determination and properties of congruent triangles. Mastering the methods of determining congruent triangles and proving their congruence is key to solving the problem." }, { "problem_id": 1394, "question": "The following diagram shows the \"growth\" process of the \"Pythagoras tree\": As shown in Figure (1), a square with side length $a$ undergoes its first \"growth\" and two smaller squares grow above it, forming a right triangle with the three squares; after another \"growth,\" it becomes Figure (2); continuing this \"growth\" process, the sum of the areas of all the squares on the \"Pythagoras tree\" after the 2015th \"growth\" is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch1-2024_06_14_64f6306d23b67a7bde6dg_0045_1.jpg", "batch1-2024_06_14_64f6306d23b67a7bde6dg_0045_2.jpg" ], "is_multi_img": true, "answer": "$2016 a^{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: Let the initial area before growth be denoted as \\( S_{0} \\), and the area after the \\( n \\)-th \"growth\" be \\( S_{n} \\).\n\n\\[\n\\therefore S_{0} = a^{2},\n\\]\n\n\\[\nS_{1} = a^{2} + a^{2} = 2a^{2},\n\\]\n\n\\[\nS_{2} = 2a^{2} + a^{2} = 3a^{2},\n\\]\n\n\\[\nS_{n} = (n + 1)a^{2},\n\\]\n\nWhen \\( n = 2015 \\), \\( S_{2015} = (2015 + 1)a^{2} = 2016a^{2} \\).\n\nTherefore, the answer is: \\( 2016a^{2} \\).\n\n【Key Insight】This problem examines the pattern of graphical changes and the Pythagorean theorem. The key to solving it lies in correctly understanding the pattern of changes in the figure and accurately representing this pattern using algebraic expressions." }, { "problem_id": 1395, "question": "On a number line, there are three points $\\mathrm{A}$, $B$, and $C$. The numbers represented by points $\\mathrm{A}$ and $B$ are -10 and 7, respectively. Now, using point $C$ as the fold point, the number line is folded to the right (as shown in Figure 1). If the point corresponding to $\\mathrm{A}$, denoted as $A^{\\prime}$, falls to the right of point $B$ (as shown in Figure 2), and $A^{\\prime} B = 2$, then the number represented by point $C$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch1-2024_06_14_a061820b6ad51067d01eg_0007_1.jpg", "batch1-2024_06_14_a061820b6ad51067d01eg_0007_2.jpg" ], "is_multi_img": true, "answer": "$-0.5 \\# \\#-\\frac{1}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the number represented by point $C$ be $x$.\n\nSince the numbers represented by points $A$ and $B$ are $-10$ and $7$ respectively,\n\nwe have $AC = x - (-10) = x + 10$.\n\nGiven that $A'B = 2$ and the number represented by point $B$ is $7$,\n\nthe number represented by point $A'$ is $7 + 2 = 9$.\n\nAccording to the folding property, $AC = A'C$,\n\nthus $x + 10 = 9 - x$.\n\nSolving the equation gives $x = -0.5$.\n\nTherefore, the answer is: -0.5.\n\n【Key Insight】This problem examines the properties of folding, the meaning of numbers represented on a number line, and the formula for the distance between two points on a number line. Mastering the formula for the distance between two points on a number line is crucial for solving such problems. If points $A$ and $B$ on the number line represent numbers $a$ and $b$ respectively, then the distance between points $A$ and $B$ is $AB = |a - b|$." }, { "problem_id": 1396, "question": "In Volume 1 of \"Cold Cottage Miscellaneous Notes\" by Lu Yihuo of the Qing Dynasty, it is written: \"Recently, there is the 'Seven-Piece Puzzle,' which consists of five shapes and seven pieces, with over a thousand possible variations.\" The Seven-Piece Puzzle is an invention of ancient Chinese laborers and is an ancient traditional intellectual game in China. Its history can be traced back to at least the first century BC. The Seven-Piece Puzzle was one of the most popular puzzles in the 19th century. Construct a Seven-Piece Puzzle from a square $ABCD$ with side length $2\\sqrt{2}$ as shown in Figure 1, and arrange the pieces within square $EFGH$ to form the \"Cheerful Rabbit\" shape as shown in Figure 2. Then, the area of square $EFGH$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch1-2024_06_14_a061820b6ad51067d01eg_0039_1.jpg", "batch1-2024_06_14_a061820b6ad51067d01eg_0039_2.jpg" ], "is_multi_img": true, "answer": "20", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In Figure 2, connect $EG$, and draw $GM \\perp EM$ at point $G$.\n\nSince $ABCD$ is a square with side length $2\\sqrt{2}$,\n\n$\\therefore AB = AD = 2\\sqrt{2}$.\n\nLet the side length of the isosceles right triangle labeled 1 be $a$.\n\nBy the Pythagorean theorem: $a^{2} + a^{2} = (2\\sqrt{2})^{2}$,\n\nSolving gives: $a = 2$.\n\nAccording to the characteristics of the tangram, the hypotenuse of the isosceles right triangle labeled 3 is $\\frac{2\\sqrt{2}}{2} = \\sqrt{2}$.\n\nLet the legs of the isosceles right triangle labeled 3 be $b$.\n\nBy the Pythagorean theorem: $b^{2} + b^{2} = (\\sqrt{2})^{2}$,\n\nSolving gives: $b = 1$.\n\nAccording to the characteristics of the tangram, the legs of the isosceles right triangle labeled 7 are $\\frac{2\\sqrt{2}}{2} = \\sqrt{2}$.\n\nLet the hypotenuse of the isosceles right triangle labeled 7 be $c$.\n\nBy the Pythagorean theorem: $(\\sqrt{2})^{2} + (\\sqrt{2})^{2} = c^{2}$,\n\nSolving gives: $c = 2$.\n\nFrom Figure 1, the side length of the square labeled 4 is equal to the leg of the isosceles right triangle labeled 3.\n\nLet the side length of the square be $d$,\n\nThen $d = b = 1$.\n\nFrom Figure 2, in the right triangle $\\triangle EGM$,\n\n$EM = 1 + 1 + 2 + 2 = 6$,\n\n$MG = 2$.\n\nBy the Pythagorean theorem: $EG^{2} = EM^{2} + MG^{2} = 6^{2} + 2^{2} = 40$,\n\n$\\therefore EG = 2\\sqrt{10}$ or $EG = -2\\sqrt{10} < 0$ (discarded).\n\nLet the side length of square $EFGH$ be $e$.\n\nBy the Pythagorean theorem: $e^{2} + e^{2} = EG^{2} = 40$,\n\nSolving gives: $e = 2\\sqrt{5}$.\n\nThe area of square $EFGH$ is $e^{2} = (2\\sqrt{5})^{2} = 20$.\n\nTherefore, the answer is: 20.\n\n\n\n【Insight】This problem examines the Pythagorean theorem for right triangles and cleverly uses the relationships between the side lengths of the tangram pieces to solve for the lengths of the segments, thereby utilizing the right triangle to find the solution." }, { "problem_id": 1397, "question": "As shown in the figure, this is a three-dimensional view of a heavy-duty truck. The right image is a schematic diagram of a rectangular wooden crate loaded with goods being unloaded from the truck along a slope. It is known that the height of the truck body $A C = 3.6 \\mathrm{~m}$, and when the truck unloads, the rear support $A B$ folds down to the ground at point $A^{\\prime}$, with a measured distance of $A^{\\prime} C = 1.8 \\mathrm{~m}$. The wooden crate has a length of $E D = 4.5 \\mathrm{~m}$, a height of $E F = 2.25 \\mathrm{~m}$, and a width less than that of the truck body. When the bottom vertex $G$ of the crate coincides with the bottom point $A^{\\prime}$ of the slope, the distance from the top vertex $E$ of the crate to the ground $A^{\\prime} C$ is $\\qquad$ $\\mathrm{m}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch1-2024_06_14_a061820b6ad51067d01eg_0097_1.jpg", "batch1-2024_06_14_a061820b6ad51067d01eg_0097_2.jpg" ], "is_multi_img": true, "answer": "$4.5 \\# \\# \\frac{9}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Solution:** As shown in Figure 2, draw $EJ \\perp A^{\\prime}C$ at point $J$, $FH \\perp A^{\\prime}C$ at point $H$, and $FI \\perp FJ$ at point $I$.\n\n\n\nFigure 2\n\nSince $AC \\perp A^{\\prime}C$ and $GB \\perp AC$,\n\n$\\therefore FH \\parallel EJ \\parallel AB$, and $FI \\parallel BG \\parallel A^{\\prime}C$.\n\n$\\therefore IJ = FH$, $GH = BC$, $\\angle EIF = \\angle BGF = \\angle CHG = \\angle A^{\\prime}CB = 90^{\\circ}$, and $\\angle FBG = \\angle BA^{\\prime}C$.\n\nLet $A^{\\prime}B = AB = x$.\n\nSince $A^{\\prime}C^{2} + BC^{2} = A^{\\prime}B^{2}$, and $AC = 3.6 \\, \\text{m}$, $A^{\\prime}C = 1.8 \\, \\text{m}$,\n\n$\\therefore 1.8^{2} + (3.6 - x)^{2} = x^{2}$.\n\nSolving this equation gives $x = 2.25$.\n\n$\\therefore A^{\\prime}B = 2.25 \\, \\text{m}$, and $GH = BC = 3.6 - 2.25 = 1.35 \\, \\text{m}$.\n\nSince quadrilateral $A^{\\prime}DEF$ is a rectangle, $ED = 4.5 \\, \\text{m}$, and $EF = 2.25 \\, \\text{m}$,\n\n$\\therefore A^{\\prime}F = ED = 4.5 \\, \\text{m}$.\n\n$\\therefore EF = BF = A^{\\prime}B = 2.25 \\, \\text{m}$.\n\nIn triangles $\\triangle BGF$ and $\\triangle A^{\\prime}CB$,\n\n$\\left\\{\\begin{array}{l}\\angle BGF = \\angle A^{\\prime}CB \\\\ BF = A^{\\prime}B \\\\ \\angle FBG = \\angle BA^{\\prime}C\\end{array}\\right.$,\n\n$\\therefore \\triangle BGF \\cong \\triangle A^{\\prime}CB$ (ASA).\n\n$\\therefore FG = BC = 1.35 \\, \\text{m}$, and $BG = A^{\\prime}C = 1.8 \\, \\text{m}$.\n\n$\\therefore IJ = FH = FG + GH = 1.35 + 1.35 = 2.7 \\, \\text{m}$.\n\nSince $\\angle BFE = 90^{\\circ}$ and $\\angle IFH = 180^{\\circ} - \\angle CHG = 90^{\\circ}$,\n\n$\\therefore \\angle EFI = \\angle BFG = 90^{\\circ} - \\angle BFI$.\n\nIn triangles $\\triangle EIF$ and $\\triangle BGF$,\n\n$\\left\\{\\begin{array}{l}\\angle EIF = \\angle BGF \\\\ \\angle EFI = \\angle BFG \\\\ EF = BF\\end{array}\\right.$,\n\n$\\therefore \\triangle EIF \\cong \\triangle BGF$ (AAS).\n\n$\\therefore EI = BG = 1.8 \\, \\text{m}$.\n\n$\\therefore EJ = EI + IJ = 1.8 + 2.7 = 4.5 \\, \\text{m}$.\n\n$\\therefore$ The distance from the top vertex $E$ of the wooden box to the ground $A^{\\prime}C$ is $4.5 \\, \\text{m}$.\n\nHence, the answer is: **4.5**.\n\n**【Key Insight】** This problem focuses on testing knowledge and methods such as the equality of complementary angles, the Pythagorean theorem, the properties and criteria of congruent triangles, and the application of the Pythagorean theorem and properties of congruent triangles to solve practical problems. The key to solving the problem lies in correctly drawing the necessary auxiliary lines." }, { "problem_id": 1398, "question": "Connect four equal-length thin wooden sticks end to end to form a quadrilateral $A B C D$. Rotate this quadrilateral to change its shape. When $\\angle B=60^{\\circ}$, as shown in Figure 2, it is measured that $A C=\\sqrt{2}$. When $\\oplus B=90^{\\circ}$, as shown in Figure 1, the length of $A C$ is\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_00f733bcb43c6bcd8b5bg_0091_1.jpg", "batch20-2024_05_23_00f733bcb43c6bcd8b5bg_0091_2.jpg" ], "is_multi_img": true, "answer": "2 .", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, connect $AC$,\n\n\nSince $AB = BC = CD = DA$ and $\\angle B = 60^\\circ$,\n\nTherefore, quadrilateral $ABCD$ is a rhombus, and $\\triangle ABC$ is an equilateral triangle,\n\nHence, $AC = AB = BC = \\sqrt{2}$.\n\nAs shown in Figure 1, connect $AC$,\n\nSince $AB = BC = CD = DA$ and $\\angle B = 90^\\circ$,\n\nTherefore, quadrilateral $ABCD$ is a square,\n\nHence, $AB = BC = \\sqrt{2}$,\n\nThen, $AB^2 + BC^2 = AC^2$,\n\nTherefore, $AC = \\sqrt{(\\sqrt{2})^2 + (\\sqrt{2})^2} = 2$,\n\nThus, the answer is: 2.\n\n【Insight】This problem examines the determination and properties of equilateral triangles, the determination and properties of squares, and the Pythagorean theorem. Understanding the problem and determining the side lengths of the quadrilateral is key to solving it." }, { "problem_id": 1399, "question": "Connect four equal-length thin wooden sticks end to end, and nail them into a quadrilateral $A B C D$. Rotate this quadrilateral to change its shape. When $\\angle B=90^{\\circ}$, as shown in Figure 1, it is measured that $A C=2$. When $\\angle B=60^{\\circ}$, as shown in Figure 2, $A C=$ $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_00f733bcb43c6bcd8b5bg_0096_1.jpg", "batch20-2024_05_23_00f733bcb43c6bcd8b5bg_0096_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (1), since \\( AB = BC = CD = DA \\) and \\( \\angle B = 90^\\circ \\),\n\n\\(\\therefore\\) quadrilateral \\( ABCD \\) is a square.\n\nConnecting \\( AC \\), we have \\( AB^{2} + BC^{2} = AC^{2} \\),\n\n\\(\\therefore AB = BC = \\sqrt{\\frac{1}{2} AC^{2}} = \\sqrt{\\frac{1}{2} \\times 2^{2}} = \\sqrt{2} \\).\n\nAs shown in Figure (2), \\( \\angle B = 60^\\circ \\), connecting \\( AC \\), \\( AB = BC \\),\n\n\\(\\therefore ABC \\) is an equilateral triangle,\n\n\\(\\therefore AC = AB = BC = \\sqrt{2} \\).\n\nTherefore, the answer is: \\( \\sqrt{2} \\).\n\n\n\nFigure 1\n\n\n\nFigure (2)\n\n【Key Insight】This problem mainly examines the properties and determination of squares and equilateral triangles. The key to solving the problem lies in mastering the relevant knowledge and applying it effectively." }, { "problem_id": 1400, "question": "As shown in Figure 1, the circular arch is a style commonly used in ancient Chinese architecture, both aesthetically pleasing and practical. Figure 2 is a schematic diagram of the arch, with a base width of 1.8 meters and a height of 3 meters. Calculate the radius of the circle on which the arch is located.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_1a9fe8d403025e1c75a7g_0034_1.jpg", "batch20-2024_05_23_1a9fe8d403025e1c75a7g_0034_2.jpg" ], "is_multi_img": true, "answer": "1.635 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from the origin $O$ to $AB$ at point $D$, and connect $BO$ and $AO$, as shown in the figure:\n\n\n\nFigure 2\n\nGiven the conditions, $AB = 1.8$ and $CD = 3$.\n\nSince $BO$ and $AO$ are both radii of circle $O$,\n\n$\\therefore BO = AO$.\n\nMoreover, since $OD \\perp AB$,\n\n$\\therefore D$ is the midpoint of $AB$,\n\n$\\therefore DB = 0.9$.\n\nLet the radius of the circle be $r$, then $OB = r$, and $OD = CD - OC = 3 - r$.\n\nTherefore, in the right triangle $\\triangle ODB$, $OD^{2} + DB^{2} = OB^{2}$,\n\n$\\therefore (3 - r)^{2} + 0.9^{2} = r^{2}$.\n\nSolving this equation gives $r = 1.635$.\n\nThus, the radius of the circle in which the arch is inscribed is 1.635 meters.\n\n【Highlight】This problem tests the understanding of the Perpendicular Chord Bisector Theorem and the Pythagorean Theorem. The key to solving this problem lies in the flexible application of the learned knowledge." }, { "problem_id": 1401, "question": "There are three good friends who attended the same elementary school class. The first winter vacation of junior high school has arrived. One day, they gathered together, each attending different middle schools, and they excitedly discussed mathematics, each posing a mathematical problem. The problem posed by Jia is: What is the value of $\\left(\\frac{1}{1 \\times 3}+\\frac{1}{3 \\times 5}+\\frac{1}{5 \\times 7}+\\cdots+\\frac{1}{2017 \\times 2019}+\\frac{1}{2019 \\times 2021}\\right)$? The problem posed by Yi is: As shown in Figure 1, folding a corner of a rectangular paper such that vertex $\\mathrm{A}$ lands at $F$, with the crease being $B C$, and drawing the angle bisector $B E$ of $\\angle F B D$, folding $\\angle F B D$ along $B F$ such that $B E$ and $B D$ both lie within $\\angle F B C$, with $B E$ intersecting $C F$ at point $M$ and $B D$ intersecting $C F$ at $N$. If $B N$ bisects $\\angle C B M$, what is the degree measure of $\\angle A B C$?\n\nThe problem posed by Bing is: As shown in Figure 2, the segment $A B$ represents a rope folded in half, with point $P$ on $A B$ such that $A P=\\frac{1}{2} P B$. If the rope is cut at $P$, the longest of the three resulting segments is $8 \\mathrm{~cm}$. What was the length of the entire rope before it was cut? If we denote the value in Jia's problem as $a$, the degree measure of $\\angle A B C$ in Yi's problem as $b$, and the length of the rope in Bing's problem in centimeters as $c$.\n\nCan you surpass them and quickly calculate $a, b, c$ and connect them with \">\" (show the solving process)?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_21272d98feed83dbd443g_0046_1.jpg", "batch20-2024_05_23_21272d98feed83dbd443g_0046_2.jpg" ], "is_multi_img": true, "answer": "$b>c>a$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Arithmetic", "image_relavance": "0", "analysis": "Solution: Problem A:\n\n$\\left(\\frac{1}{1 \\times 3}+\\frac{1}{3 \\times 5}+\\frac{1}{5 \\times 7}+\\cdots+\\frac{1}{2017 \\times 2019}+\\frac{1}{2019 \\times 2021}\\right)$ $=\\frac{1}{2}\\left(1-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{5}+\\frac{1}{5}-\\frac{1}{7}+\\cdots+\\frac{1}{2017} \\frac{1}{2019}+\\frac{1}{2019} \\frac{1}{2021}\\right.$, $=\\frac{1}{2}\\left(1-\\frac{1}{2021}\\right)$\n\n$=\\frac{1}{2} \\times \\frac{2020}{2021}$\n\n$=\\frac{1010}{2021}$\n\n$\\therefore \\mathrm{a}=\\frac{1010}{2021}$\n\nProblem B: As shown in Figure 1, let $\\angle \\mathrm{DBE}=\\angle \\mathrm{EBF}=\\mathrm{x}$.\n\n\n\nFigure 1\n\n$\\because \\angle \\mathrm{FBD}^{\\prime}$ is obtained by folding $\\angle \\mathrm{FBD}$ along $\\mathrm{BF}$,\n\n$\\therefore \\angle \\mathrm{MBF}=\\angle \\mathrm{MBN}=\\mathrm{x}$,\n\n$\\because \\mathrm{BN}$ bisects $\\angle \\mathrm{CBM}$,\n\n$\\therefore \\angle \\mathrm{CBN}=\\angle \\mathrm{MBN}=\\mathrm{x}$,\n\n$\\therefore \\angle \\mathrm{CBF}=3 \\mathrm{x}$,\n\n$\\because \\triangle \\mathrm{CBF}$ is obtained by folding $\\triangle \\mathrm{CBA}$,\n\n$\\therefore \\angle \\mathrm{ABC}=\\angle \\mathrm{CBF}=3 \\mathrm{x}$,\n\n$\\because \\angle \\mathrm{ABF}+\\angle \\mathrm{FBD}=180^{\\circ}$,\n\n$\\therefore 8 \\mathrm{x}=180^{\\circ}$,\n\n$\\therefore \\mathrm{x}=22.5^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{ABC}=3 \\mathrm{x}=67.5^{\\circ}$,\n\n$\\therefore b=67.5$\n\nProblem C: According to the problem, $2 \\mathrm{~PB}=8 \\mathrm{~cm}$,\n\n$\\therefore \\mathrm{PB}=4(\\mathrm{~cm})$,\n\n$\\because P A=\\frac{1}{2} \\mathrm{~PB}=2(\\mathrm{~cm})$\n\n$\\therefore \\mathrm{AB}=\\mathrm{AP}+\\mathrm{PB}=6(\\mathrm{~cm})$,\n\n$\\therefore$ The length of the entire rope before cutting is $12 \\mathrm{~cm}$,\n\n$\\therefore \\mathrm{c}=12$,\n\n$\\because 67.5>12>\\frac{1010}{2021}$\n\n$\\therefore \\mathrm{b}>\\mathrm{c}>\\mathrm{a}$.\n\n【Highlight】This problem is a comprehensive geometry transformation question, examining the knowledge of folding transformations and rational number calculations. The key to solving the problem lies in understanding the question and flexibly applying the learned knowledge to solve it, making it a challenging problem for high school entrance exams." }, { "problem_id": 1402, "question": "(2016・Chengdu) As shown in the figure, in the parallelogram paper $\\mathrm{ABCD}$ with an area of 6, $\\mathrm{AB}=3, \\angle \\mathrm{BAD}=45^{\\circ}$, the following steps are taken for cutting and assembling.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\nStep 1: As shown in Figure (1), cut the parallelogram paper along the diagonal $\\mathrm{BD}$, resulting in paper pieces $\\triangle \\mathrm{ABD}$ and $\\triangle \\mathrm{BCD}$. Then cut the $\\triangle \\mathrm{ABD}$ paper along $\\mathrm{AE}$ (where $\\mathrm{E}$ is any point on $\\mathrm{BD}$), resulting in paper pieces $\\triangle \\mathrm{ABE}$ and $\\triangle \\mathrm{ADE}$.\n\nStep 2: As shown in Figure (2), translate the $\\triangle \\mathrm{ABE}$ paper to $\\triangle \\mathrm{DCF}$ and the $\\triangle \\mathrm{ADE}$ paper to $\\triangle \\mathrm{BCG}$.\n\nStep 3: As shown in Figure (3), flip the $\\triangle \\mathrm{DCF}$ paper over to place it on $\\triangle \\mathrm{PQM}$ (with side $\\mathrm{PQ}$ coinciding with $\\mathrm{DC}$, and $\\triangle \\mathrm{PQM}$ and $\\triangle \\mathrm{DCF}$ on the same side of $\\mathrm{DC}$). Similarly, flip the $\\triangle \\mathrm{BCG}$ paper over to place it on $\\triangle \\mathrm{PRN}$ (with side $\\mathrm{PR}$ coinciding with $\\mathrm{BC}$, and $\\triangle \\mathrm{PRN}$ and $\\triangle \\mathrm{BCG}$ on the same side of $\\mathrm{BC}$).\n\nIn the pentagon $\\mathrm{PMQRN}$ formed by the paper pieces, the minimum length of the diagonal $\\mathrm{MN}$ is $\\qquad$.", "input_image": [ "batch20-2024_05_23_2326e0ff8416e98e3b37g_0012_1.jpg", "batch20-2024_05_23_2326e0ff8416e98e3b37g_0012_2.jpg", "batch20-2024_05_23_2326e0ff8416e98e3b37g_0012_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{6 \\sqrt{10}}{5}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since triangles $\\triangle \\mathrm{ABE} \\cong \\triangle \\mathrm{CDF} \\cong \\triangle \\mathrm{PMQ}$,\n\nTherefore, $\\mathrm{AE}=\\mathrm{DF}=\\mathrm{PM}$, and $\\angle \\mathrm{EAB}=\\angle \\mathrm{FDC}=\\angle \\mathrm{MPQ}$.\n\nSince triangles $\\triangle \\mathrm{ADE} \\cong \\triangle \\mathrm{BCG} \\cong \\triangle \\mathrm{PNR}$,\n\nTherefore, $\\mathrm{AE}=\\mathrm{BG}=\\mathrm{PN}$, and $\\angle \\mathrm{DAE}=\\angle \\mathrm{CBG}=\\angle \\mathrm{RPN}$.\n\nHence, $\\mathrm{PM}=\\mathrm{PN}$.\n\nSince quadrilateral $\\mathrm{ABCD}$ is a parallelogram,\n\nTherefore, $\\angle \\mathrm{DAB}=\\angle \\mathrm{DCB}=45^{\\circ}$,\n\nThus, $\\angle \\mathrm{MPN}=90^{\\circ}$,\n\nTherefore, $\\triangle \\mathrm{MPN}$ is an isosceles right triangle.\n\nWhen $\\mathrm{PM}$ is at its minimum, the diagonal $\\mathrm{MN}$ is at its minimum, meaning $\\mathrm{AE}$ reaches its minimum value.\n\nTherefore, when $\\mathrm{AE} \\perp \\mathrm{BD}$, $\\mathrm{AE}$ reaches its minimum value.\n\nDrawing $\\mathrm{DF} \\perp \\mathrm{AB}$ at $\\mathrm{F}$,\n\nSince the area of parallelogram $\\mathrm{ABCD}$ is $6$ and $\\mathrm{AB}=3$,\nTherefore, $\\mathrm{DF}=2$.\n\nSince $\\angle \\mathrm{DAB}=45^{\\circ}$,\n\nTherefore, $\\mathrm{AF}=\\mathrm{DF}=2$,\n\nThus, $\\mathrm{BF}=1$,\n\nTherefore, $\\mathrm{BD}=\\sqrt{\\mathrm{DF}^{2}+\\mathrm{BF}^{2}}=\\sqrt{5}$,\n\nThus, $\\mathrm{AE}=\\frac{\\mathrm{DF} \\cdot \\mathrm{AB}}{\\mathrm{BD}}=\\frac{2 \\times 3}{\\sqrt{5}}=\\frac{6 \\sqrt{5}}{5}$,\n\nTherefore, $\\mathrm{MN}=\\sqrt{2} \\mathrm{AE}=\\frac{6 \\sqrt{10}}{5}$,\n\nHence, the answer is: $\\frac{6 \\sqrt{10}}{5}$.\n\n\n\n【Insight】This problem examines the properties of translation, reflection, the Pythagorean theorem, and the properties of parallelograms. Correctly identifying the figure is key to solving the problem." }, { "problem_id": 1403, "question": "As shown in the figure, there is a wooden stick $M N$ of fixed length placed inside the square $A B C D$ as shown in Figure 1, where the endpoint $M$ of the stick coincides exactly with point $\\mathrm{A}$, and point $N$ is on side $B C$, with $B N=2.5$. The stick is then slid downward along $A B$ by $a$ units to the position shown in Figure 2. Simultaneously, the other endpoint $N$ slides rightward along $B C$ by $b$ units to $N^{\\prime}$, with $a: b=7: 9$ and $B M^{\\prime}=3.9$. During the sliding process, the shortest distance from point $D$ to the midpoint $P$ of the stick is $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_2326e0ff8416e98e3b37g_0094_1.jpg", "batch20-2024_05_23_2326e0ff8416e98e3b37g_0094_2.jpg" ], "is_multi_img": true, "answer": "$6 \\sqrt{2}-\\frac{13}{4}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 2, we know that: $\\mathrm{AB}=\\mathrm{AM}^{\\prime}+\\mathrm{BM}^{\\prime}=\\mathrm{a}+3.9$,\n\nSince $\\mathrm{BN}=2.5$,\n\nTherefore, in Figure 1, $\\mathrm{MN}^{2}=\\mathrm{AB}^{2}+\\mathrm{BN}^{2}=(\\mathrm{a}+3.9)^{2}+2.5^{2}$,\n\nGiven that $\\mathrm{a}: \\mathrm{b}=7: 9$,\n\nThus, $b=\\frac{9}{7} a$,\n\nIn Figure 2, $\\mathrm{M}^{\\prime} \\mathrm{N}^{\\prime 2}=\\mathrm{BM}^{\\prime 2}+\\mathrm{BN}^{\\prime 2}=3.9^{2}+\\left(2.5+\\frac{9}{7} a\\right)^{2}$,\n\nSince $\\mathrm{MN}^{2}=\\mathrm{M}^{\\prime} \\mathrm{N}^{\\prime 2}$,\n\nTherefore, $(\\mathrm{a}+3.9)^{2}+2.5^{2}=3.9^{2}+\\left(2.5+\\frac{9}{7} a\\right)^{2}$.\n\nSolving this gives: $a=2.1$ or $a=0$ (discarded),\n\nThus, $\\mathrm{AB}=\\mathrm{a}+3.9=2.1+3.9=6$,\n\nTherefore, in Figure 1, $\\mathrm{MN}=\\sqrt{(a+3.9)^{2}+2.5^{2}}=\\sqrt{6^{2}+2.5^{2}}=6.5$,\n\nConnecting $\\mathrm{BP}, \\mathrm{BD}$, as shown in the figure,\n\n\n\nSince $\\angle \\mathrm{BAD}=90^{\\circ}, \\mathrm{AD}=\\mathrm{AB}=6$,\n\nThus, $\\mathrm{BD}=\\sqrt{A B^{2}+A D^{2}}=\\sqrt{6^{2}+6^{2}}=6 \\sqrt{2}$,\n\nSince $\\angle \\mathrm{M}^{\\prime} \\mathrm{BN}^{\\prime}=90^{\\circ}, \\mathrm{P}$ is the midpoint of $\\mathrm{M}^{\\prime} \\mathrm{N}^{\\prime}$,\n\nThus, $\\mathrm{BP}=\\frac{1}{2} \\mathrm{M}^{\\prime} \\mathrm{N}^{\\prime}=\\frac{1}{2} \\mathrm{MN}=\\frac{1}{2} \\times 6.5=\\frac{13}{4}$,\n\nSince $\\mathrm{DP} \\geq \\mathrm{BD}-\\mathrm{BP}$,\n\nTherefore, when points $\\mathrm{B}, \\mathrm{P}, \\mathrm{D}$ are collinear, $\\mathrm{DP}$ is the shortest, and at this time, $\\mathrm{DP}=\\mathrm{BD}-\\mathrm{BP}=6 \\sqrt{2}-\\frac{13}{4}$,\n\nThus, during the sliding process, the shortest distance from point $\\mathrm{D}$ to the midpoint $\\mathrm{P}$ of the stick is $6 \\sqrt{2}-\\frac{13}{4}$.\n\nHence, the answer is: $6 \\sqrt{2}-\\frac{13}{4}$.\n\n【Highlight】This problem examines the shortest distance in a square, the Pythagorean theorem, and belongs to the dynamic type of problem. The key to solving it is to use the learned knowledge and the actual situation to determine that when points $\\mathrm{B}, \\mathrm{P}, \\mathrm{D}$ are collinear, $\\mathrm{DP}$ is the shortest." }, { "problem_id": 1404, "question": "As shown in the figure, in triangle \\(ABC\\), points \\(A_1, B_1, C_1\\) are the midpoints of \\(BC, AC, AB\\) respectively, \\(A_2, B_2, C_2\\) are the midpoints of \\(B_1C_1, A_1C_1, A_1B_1\\) respectively, and so on. If the perimeter of \\(ABC\\) is 1, then the perimeter of \\(A_nB_nC_n\\) is \\(\\qquad\\).\n\n\n\n(a)\n\n\n\n(b)\n\n\n\n(c)", "input_image": [ "batch20-2024_05_23_2aea9e2b584677eadccfg_0077_1.jpg", "batch20-2024_05_23_2aea9e2b584677eadccfg_0077_2.jpg", "batch20-2024_05_23_2aea9e2b584677eadccfg_0077_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{1}{2^{n}}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since points \\( A_{1}, B_{1}, C_{1} \\) are the midpoints of \\( BC, AC, AB \\) respectively,\n\n\\[\n\\therefore A_{1} B_{1} = \\frac{1}{2} AB, \\quad A_{1} C_{1} = \\frac{1}{2} AC, \\quad B_{1} C_{1} = \\frac{1}{2} BC\n\\]\n\n\\[\n\\therefore C_{A_{1} B_{1} C_{1}} = \\frac{1}{2} C_{ABC} = \\frac{1}{2}\n\\]\n\nMoreover, since \\( A_{2}, B_{2}, C_{1} \\) are the midpoints of \\( B_{1} C_{1}, A_{1} C_{1}, A_{1} B_{1} \\) respectively,\n\n\\[\n\\therefore A_{2} B_{2} = \\frac{1}{2} A_{1} B_{1}, \\quad A_{2} C_{2} = \\frac{1}{2} A_{1} C_{1}, \\quad B_{2} C_{2} = \\frac{1}{2} B_{1} C_{1}\n\\]\n\n\\[\n\\therefore C_{A_{2} B_{2} C_{2}} = \\frac{1}{2} C_{A_{1} B_{1} C_{1}} = \\frac{1}{4} = \\frac{1}{2^{2}}\n\\]\n\n\\[\n\\therefore C_{A_{n} B_{n} C_{n}} = \\frac{1}{2} C_{A_{n-1} B_{n-1} C_{n-1}} = \\frac{1}{2^{n}}\n\\]\n\nHence, the answer is \\( \\frac{1}{2^{n}} \\).\n\n【Key Insight】This problem examines the Midsegment Theorem of a triangle: The midsegment of a triangle is parallel to the third side and is half as long." }, { "problem_id": 1405, "question": "Given: Connecting the midpoints of the sides of a rectangle in sequence yields a rhombus, as shown in Figure (1); then connecting the midpoints of the sides of the rhombus in sequence yields a new rectangle, as shown in Figure (2); then connecting the midpoints of the sides of the new rectangle in sequence yields a new rhombus, as shown in Figure (3); repeating this operation, the number of right-angled triangles in the 2014th figure is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0024_1.jpg", "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0024_2.jpg", "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0024_3.jpg" ], "is_multi_img": true, "answer": "4028", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "**Problem Analysis:** The number of right triangles in Figure (1) and Figure (2) is the same, both being \\(4\\), which can be expressed as \\(4 = 4 \\times 1\\). Similarly, the number of right triangles in Figure (3) and Figure (4) is the same, both being \\(8\\), which can be expressed as \\(8 = 4 \\times 2\\).\n\n\\(\\cdots\\)\n\nThe number of right triangles in Figure 2013 and Figure 2014 is the same, both being \\(4 \\times \\frac{2014}{2} = 4028\\).\n\nTherefore, the answer is **4028**.\n\n**Key Insight:** By observing several figures, determine the number of right triangles and identify the underlying pattern." }, { "problem_id": 1406, "question": "It is known that: by connecting the midpoints of each side of a rectangle in sequence, a rhombus is obtained, as shown in Figure (1). Then, by connecting the midpoints of each side of the rhombus in sequence, a new rectangle is obtained, as shown in Figure (2). Next, by connecting the midpoints of each side of the new rectangle in sequence, a new rhombus is obtained, as shown in Figure (3). By repeating this operation, the number of rhombuses in the 2017th figure is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0025_1.jpg", "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0025_2.jpg", "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0025_3.jpg" ], "is_multi_img": true, "answer": "1009", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: In the 1st figure, there is 1 rhombus,\n\nIn the 2nd figure, there is 1 rhombus,\n\nIn the 3rd figure, there are 2 rhombuses,\n\nIn the 4th figure, there are 2 rhombuses,\n\nand so on.\n\nFollowing this pattern,\n\nwhen \\( n \\) is odd, the number of rhombuses is \\( \\frac{n+1}{2} \\),\n\nwhen \\( n \\) is even, the number of rhombuses is \\( \\frac{n}{2} \\).\n\nTherefore, in the 2017th figure, the number of rhombuses is \\( \\frac{2017+1}{2} = 1009 \\).\n\nHence, the answer is 1009.\n\n[Key Insight] This problem primarily examines the changes in figures. Observing the number of rhombuses in the initial figures and deducing the relationship with their sequence numbers is the key to solving the problem." }, { "problem_id": 1407, "question": "When one vertex of square $\\mathrm{A}$ coincides with the intersection of the diagonals of square $\\mathrm{B}$, as shown in Figure 1, the area of the shaded part is $\\frac{1}{8}$ of the area of square $\\mathrm{A}$. When square $\\mathrm{A}$ is placed against square $\\mathrm{B}$ as shown in Figure 2, the area of the shaded part is $\\qquad$ of the area of square $\\mathrm{B}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0055_1.jpg", "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0055_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{1}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** Align one vertex of square A with the diagonal intersection of square B, as shown in Figure 1. The shaded area is exactly one-fourth of square B, and the area of the shaded region is one-eighth the area of square A, i.e., \\(\\frac{1}{4} a^{2} = \\frac{1}{8}\\). When squares A and B are arranged as in Figure 2, the shaded area is one-fourth of square A, so the area of the shaded region is one-fourth of the area of square B, calculated as \\(\\frac{1}{4}(2 a)^{2} = 4 \\times \\frac{1}{8} = \\frac{1}{2}\\).\n\n**Exam Focus:** Squares\n\n**Review:** This question tests the properties of squares. The key to solving the problem lies in the examinee's familiarity with the characteristics of squares and their ability to discern the relationship between the shaded area and the squares." }, { "problem_id": 1408, "question": "In rectangle \\( A B C D \\), point \\( P \\) is on \\( A D \\), \\( A B = \\sqrt{3} \\), \\( A P = 1 \\). Place the vertex of a right angle ruler at \\( P \\), and let its two sides intersect \\( A B \\) and \\( B C \\) at points \\( E \\) and \\( F \\) respectively, then connect \\( E F \\) (as shown in Figure 1). When point \\( E \\) coincides with point \\( B \\), point \\( F \\) coincides with point \\( C \\) (as shown in Figure 2). Starting from the position in Figure 2, rotate the right angle ruler clockwise around point \\( P \\) until point \\( E \\) coincides with point \\( A \\). During this process, the path length traversed by the midpoint of segment \\( E F \\) from start to stop is .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0071_1.jpg", "batch20-2024_05_23_3f82fe2c2a51c6c99425g_0071_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Figure 1\n\n\nFigure 2\n\nSolution: As shown in Figure 2,\n\nIn rectangle $\\mathrm{ABCD}$, $\\angle \\mathrm{A}=\\angle \\mathrm{D}=90^{\\circ}$,\n\nSince $\\mathrm{AP}=1$, $\\mathrm{AB}=\\sqrt{3}$,\n\nTherefore, $\\mathrm{PB}=\\sqrt{1^{2}+(\\sqrt{3})^{2}}=2$,\n\nSince $\\angle \\mathrm{ABP}+\\angle \\mathrm{APB}=90^{\\circ}$, $\\angle \\mathrm{BPC}=90^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{APB}+\\angle \\mathrm{DPC}=90^{\\circ}$,\n\nThus, $\\angle \\mathrm{ABP}=\\angle \\mathrm{DPC}$,\n\nTherefore, $\\triangle \\mathrm{APB} \\sim \\triangle \\mathrm{DCP}$,\n\nThus, $\\mathrm{AP}: \\mathrm{CD}=\\mathrm{PB}: \\mathrm{CP}$, that is, $1: \\sqrt{3}=2: \\mathrm{PC}$,\n\nTherefore, $\\mathrm{PC}=2 \\sqrt{3}$,\n\nLet the midpoint of segment $\\mathrm{EF}$ be $\\mathrm{O}$, connect $\\mathrm{OP}$, $\\mathrm{OB}$, as shown in Figure 1,\n\nIn right triangle $\\triangle \\mathrm{EPF}$, $\\mathrm{OP}=\\frac{1}{2} \\mathrm{EF}$,\n\nIn right triangle $\\triangle \\mathrm{EBF}$, $\\mathrm{OB}=\\frac{1}{2} \\mathrm{EF}$,\n\nTherefore, $\\mathrm{OP}=\\mathrm{OB}$,\n\nThus, point $\\mathrm{O}$ lies on the perpendicular bisector of segment $\\mathrm{BP}$,\n\nAs shown in Figure 2, when point $\\mathrm{E}$ coincides with point $\\mathrm{B}$, and point $\\mathrm{F}$ coincides with point $\\mathrm{C}$, the midpoint of $\\mathrm{EF}$ is the midpoint $\\mathrm{O}$ of $\\mathrm{BC}$,\n\nWhen point $\\mathrm{E}$ coincides with point $\\mathrm{A}$, the midpoint of $\\mathrm{EF}$ is the midpoint $\\mathrm{O}$ of $\\mathrm{PB}$,\nTherefore, $\\mathrm{OO}^{\\prime}$ is the midline of $\\triangle \\mathrm{PBC}$,\n\nThus, $\\mathrm{OO}^{\\prime}=\\frac{1}{2} \\mathrm{PC}=\\sqrt{3}$,\n\nTherefore, the length of the path traced by the midpoint of segment $\\mathrm{EF}$ is $\\sqrt{3}$.\n\nHence, the answer is $\\sqrt{3}$.\n\n【Insight】This problem examines the determination and properties of similar triangles, the combination of rectangles and triangles, and moving points on a rectangle. It is a complex problem that requires integrating information from both figures to arrive at the solution." }, { "problem_id": 1409, "question": "The stone arch bridge is a crystallization of the diligence and wisdom of the ancient Chinese people (as shown in Figure 1). The Zhaozhou Bridge, built during the Sui Dynasty, has a history of approximately 1400 years and is a representative of ancient Chinese stone arch bridges. Figure 2 is a geometric diagram drawn based on the actual image of a stone arch bridge, where the main arch of the bridge is an arc, denoted as $AB$. The span of the bridge (the chord length corresponding to the arc) $AB = 24 \\mathrm{~m}$. Let the center of the circle containing $AB$ be $O$, and the radius $OC \\perp AB$, with the perpendicular foot at $D$. The arch height (the distance from the midpoint of the arc to the chord) $CD = 5 \\mathrm{~m}$. Connect $OB$. Find the radius of the main arch of this stone arch bridge. (Round to the nearest $1 \\mathrm{~m}$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_435c21700c1d7dea955cg_0011_1.jpg", "batch20-2024_05_23_435c21700c1d7dea955cg_0011_2.jpg" ], "is_multi_img": true, "answer": "$17 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( OC \\perp AB \\),\n\nTherefore, \\( AD = BD \\).\n\nLet the radius of the main arch be \\( R \\). According to the problem, \\( AB = 24 \\) and \\( CD = 5 \\).\n\nThus, \\( BD = \\frac{1}{2} AB = 12 \\),\n\n\\( OD = OC - CD = R - 5 \\).\n\nSince \\( \\angle ODB = 90^\\circ \\),\n\nWe have \\( OD^2 + BD^2 = OB^2 \\),\n\nTherefore, \\( (R - 5)^2 + 12^2 = R^2 \\),\n\nSolving this gives \\( R = 16.9 \\approx 17 \\).\n\nAnswer: The radius of the main arch of this stone arch bridge is approximately \\( 17 \\mathrm{~m} \\).\n\n[Highlight] This problem tests the understanding of the Perpendicular Chord Bisector Theorem and the Pythagorean Theorem. The key to solving it lies in the application of equation-solving techniques. Combining the Perpendicular Chord Bisector Theorem with the Pythagorean Theorem to construct a right triangle can solve problems related to calculating chord lengths, radii, and the distance from the center to the chord. This type of problem generally employs the method of setting up equations, and this mathematical approach of using algebraic methods to solve geometric problems, known as geometric algebra, is essential to master." }, { "problem_id": 1410, "question": "As shown in the quadrilateral $A B C D$, it is formed by nailing four thin wooden sticks of equal length together at their ends, allowing it to rotate and change shape, as illustrated in Figure 1. When $\\angle B=90^{\\circ}$, $A C=6$; as shown in Figure 2, if $\\angle B A D=60^{\\circ}$, then $A C=$ . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_4e6da04e127696e73c75g_0056_1.jpg", "batch20-2024_05_23_4e6da04e127696e73c75g_0056_2.jpg" ], "is_multi_img": true, "answer": "$3 \\sqrt{6}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in Figure 1, connect $AC$.\n\nSince quadrilateral $ABCD$ is a square, let its side length be $x$.\n\nThen, $x^{2} + x^{2} = 36$, solving for $x$ gives $x = 3\\sqrt{2}$.\n\nThus, $AB = 3\\sqrt{2}$.\n\nAs shown in Figure 2, connect $AC$ and $BD$ intersecting at point $E$.\n\nSince quadrilateral $ABCD$ is a rhombus and $\\angle BAD = 60^\\circ$,\n\n$\\triangle ABD$ is an equilateral triangle.\n\nTherefore, $AE = \\frac{\\sqrt{3}}{2} \\times 3\\sqrt{2} = \\frac{3\\sqrt{6}}{2}$.\n\nHence, $AC = 2AE = 3\\sqrt{6}$.\n\nThe answer is: $3\\sqrt{6}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Insight】This problem tests the properties of squares, the properties of rhombuses, and the determination and properties of equilateral triangles. The key to solving this problem lies in mastering these concepts." }, { "problem_id": 1411, "question": "As shown in Figure 1, this is the \"Circular Cutting Eight Lines Diagram\" from the late Ming Dynasty in China. In Figure 2, points C, E, B, G, H, and D in Figure 1 are represented by $\\mathrm{A}, B, C, D, E, O$ respectively. The central angle of sector $A O D$ is $90^{\\circ}$, and its radius is $\\sqrt{3}$. Lines $D E$ and $A B$ are tangent to $A D$ at points $D$ and $\\mathrm{A}$ respectively. If $B C=A C$, then the length of $C E$ is $\\qquad$.\n\n\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_5b6a6dc4a2fb95543ee1g_0008_1.jpg", "batch20-2024_05_23_5b6a6dc4a2fb95543ee1g_0008_2.jpg", "batch20-2024_05_23_5b6a6dc4a2fb95543ee1g_0008_3.jpg" ], "is_multi_img": true, "answer": "$2-\\sqrt{3} \\# \\#-\\sqrt{3}+2$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AB \\) is a tangent,\n\n\\[\n\\therefore \\angle OAB = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle B + \\angle AOB = 90^\\circ, \\quad \\angle CAB + \\angle CAO = 90^\\circ,\n\\]\n\nSince \\( BC = AC \\),\n\n\\[\n\\therefore \\angle B = \\angle CAB\n\\]\n\n\\[\n\\therefore \\angle AOB = \\angle CAO\n\\]\n\n\\[\n\\therefore AC = OC,\n\\]\n\n\\[\n\\therefore AC = OC = BC\n\\]\n\nMoreover, since \\( OC = OA \\),\n\n\\[\n\\therefore OB = 2OA,\n\\]\n\n\\[\n\\therefore \\text{In right triangle } \\triangle ABO, \\quad \\cos \\angle AOB = \\frac{OA}{OB} = \\frac{1}{2},\n\\]\n\n\\[\n\\therefore \\angle AOB = 60^\\circ,\n\\]\n\nSince \\( \\angle AOD = 90^\\circ \\),\n\n\\[\n\\therefore \\angle EOD = 30^\\circ,\n\\]\n\nSince \\( DE \\) is a tangent,\n\n\\[\n\\therefore \\angle ODE = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\text{In right triangle } \\triangle ODE, \\quad OE = \\frac{OD}{\\cos \\angle EOD} = 2,\n\\]\n\n\\[\n\\therefore CE = OE - OC = 2 - \\sqrt{3},\n\\]\n\nHence, the answer is: \\( 2 - \\sqrt{3} \\).\n\n**Key Insight:** This problem primarily examines the properties of tangents, solving right triangles, and the properties and determination of isosceles triangles. The key to solving the problem lies in deducing that \\( \\angle AOB = 60^\\circ \\)." }, { "problem_id": 1412, "question": "As shown in Figure A, it consists of a square $A B C D$, a rectangle $C E F G$, and a circle, where points $B, C, E$ are on the same horizontal line, $B C=C E=15$, the circle is tangent to $A D$ at point $P$, and the circle is tangent to $C G$ at point $G$. According to the instability of quadrilaterals, Figure A transforms into Figure B, where the shapes of the two quadrilaterals change, but the side lengths remain the same. At this time, in Figure B, the diagonal $G E$ of quadrilateral $C E F G$ is perpendicular to $C E$. The size of the circle remains unchanged but its position shifts, still tangent to $A D$ at point $P$, and tangent to $C G$ at point $H$, with the distance from point $H$ to $A D$ being 4. The distance that the tangent point $P$ moves along $A D$ is $\\qquad$.\n\n\n\nFigure A\n\n\n\nFigure B", "input_image": [ "batch20-2024_05_23_5b6a6dc4a2fb95543ee1g_0097_1.jpg", "batch20-2024_05_23_5b6a6dc4a2fb95543ee1g_0097_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nFrom the problem statement, the distance moved by the tangent point \\( P \\) is the difference between \\( PD \\) in Figure A and \\( PD \\) in Figure B.\n\nLet the radius of the circle be \\( R \\), and the center of the circle be \\( O \\).\n\n**Figure A:** \nConnect \\( OG \\) and \\( OP \\), then \\( OG = OP = R \\). \nSince \\( \\angle OGD = \\angle OPD = \\angle ADG = 90^\\circ \\), \nthe quadrilateral \\( OGPD \\) is a square. \nThus, \\( PD = OG = R = GD \\), \nand \\( CG = 15 + R \\).\n\n**Figure B:** \nConnect \\( OP \\) and \\( OH \\), extend \\( PD \\), and draw \\( HM \\perp PD \\) with \\( M \\) as the foot of the perpendicular. \nThus, \\( OH = R \\), \\( HM = 4 \\). \nDraw \\( HN \\perp OP \\) with \\( N \\) as the foot of the perpendicular. \nThen, \\( ON = R - 4 \\), \nand \\( HN = \\sqrt{R^2 - (R - 4)^2} = \\sqrt{8R - 16} \\). \nConnect \\( GE \\). \nSince \\( CG = 15 + R \\), \\( CE = 15 \\), and \\( CG \\perp CE \\), \nwe have \\( GE = \\sqrt{(15 + R)^2 - 15^2} = \\sqrt{R^2 + 30R} \\). \nSince \\( \\triangle HMD \\sim \\triangle GEC \\), \nwe have \\( \\frac{GE}{HM} = \\frac{CE}{DM} = \\frac{CG}{HD} \\). \nThus, \\( DM = \\frac{60}{\\sqrt{R^2 + 30R}} \\), and \\( DH = \\frac{60 + 4R}{\\sqrt{R^2 + 30R}} \\). \nTherefore, \\( PD = PM - DM = HN - DM = \\sqrt{8R - 16} - \\frac{60}{\\sqrt{R^2 + 30R}} \\). \nSince \\( PD = DH \\), \nwe have \\( \\sqrt{8R - 16} - \\frac{60}{\\sqrt{R^2 + 30R}} = \\frac{60 + 4R}{\\sqrt{R^2 + 30R}} \\). \nSolving this, we find \\( R = 10 \\). \nThus, in Figure A, \\( PD = 10 \\), and in Figure B, \\( PD = 5 \\). \nTherefore, the distance moved by the tangent point \\( P \\) along \\( AD \\) is \\( 10 - 5 = 5 \\).\n\n**Final Answer:** \n\\(\\boxed{5}\\)\n\n**Key Insight:** \nThis problem tests the properties of tangents, the determination and properties of squares, the Pythagorean theorem, and the properties of similar triangles. Drawing auxiliary lines is crucial to solving this problem." }, { "problem_id": 1413, "question": "Using 4 congruent regular octagons, adjoin them such that each pair of adjacent octagons shares a common side, and form a ring where a square is formed in the center, as shown in Figure 1. Using $n$ congruent regular hexagons in this manner, as shown in Figure 2, if a ring is formed and a regular polygon is also formed in the center, then the value of $n$ is $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_771d9eeff12a9d0285e0g_0015_1.jpg", "batch20-2024_05_23_771d9eeff12a9d0285e0g_0015_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: When two regular hexagons are joined together, the angle formed at a common point is $240^{\\circ}$. Therefore, to achieve a tessellation, a regular polygon with an internal angle of $120^{\\circ}$ is needed in the middle. Since the internal angle of a regular hexagon is $120^{\\circ}$, the central polygon must also be a regular hexagon. Hence, $\\mathrm{n}=6$.\n\nThe answer is 6.\n\n[Key Insight] This problem tests knowledge of plane tessellation. The key to solving it lies in determining the required internal angle of the central regular polygon under tessellation conditions, thereby finding the value of $\\mathrm{n}$. The difficulty level is not high." }, { "problem_id": 1414, "question": "As shown in Figure (1), in quadrilateral $\\mathrm{ABCD}$, $\\mathrm{AD}=\\mathrm{CD}$, $\\mathrm{AB}=\\mathrm{CB}$. We call such a quadrilateral with equal adjacent sides on both pairs a \"kite\". The area of a kite is half the product of its diagonals. As shown in Figure (2), there is a right triangle $\\triangle A B C$ with known sides $A B=6$, $\\mathrm{AC}=8$, $\\mathrm{BC}=10$. Point $\\mathrm{P}$ is a moving point on side $\\mathrm{BC}$, and point $\\mathrm{N}$ is the midpoint of $\\mathrm{DE}$. If the area of kite $\\mathrm{ADPE}$ is 18, then the maximum value of $\\mathrm{AN}$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch20-2024_05_23_7e2c875fce25ca646505g_0022_1.jpg", "batch20-2024_05_23_7e2c875fce25ca646505g_0022_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{15}{4}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (2),\n\n\n\nFigure (2)\n\nSince ADPE is a kite,\n\nThe area of kite $\\mathrm{ADPE}$ $=\\frac{1}{2} \\times AP \\times DE=18$,\n\nTherefore, $AP \\times DE=36$,\n\nThus, when $\\mathrm{AP}$ reaches its minimum, $\\mathrm{DE}$ reaches its maximum,\n\nSince $\\mathrm{P}$ is a moving point on side $\\mathrm{BC}$,\n\nWhen $\\mathrm{AP} \\perp \\mathrm{BC}$, $\\mathrm{AP}$ reaches its minimum,\n\nTherefore, the minimum value of $\\mathrm{AP}$ $=\\frac{\\frac{1}{2} \\times AB \\times AC}{\\frac{1}{2} \\times BC}=\\frac{AB \\times AC}{BC}=\\frac{6 \\times 8}{10}=\\frac{24}{5}$,\n\nThus, $\\frac{24}{5} \\times DE=36$,\n\nTherefore, $\\mathrm{DE}=\\frac{15}{2}$,\n\nHence, the maximum value of $\\mathrm{DE}$ is $\\frac{15}{2}$,\n\nSince in right triangle $\\triangle \\mathrm{ADE}$, point $\\mathrm{N}$ is the midpoint of $\\mathrm{DE}$,\n\nTherefore, $\\mathrm{AN}=\\frac{1}{2} \\mathrm{DE}$,\n\nThus, when $\\mathrm{DE}$ reaches its maximum, $\\mathrm{AN}$ also reaches its maximum,\n\nTherefore, the maximum value of $\\mathrm{AN}$ is $\\frac{15}{4}$.\n\nThe final answer is: $\\frac{15}{4}$.\n\n【Key Insight】This problem examines the properties of the median to the hypotenuse in a right triangle and the area formula of a right triangle. Understanding the definition of a \"kite\" is crucial to solving the problem. The challenge lies in analyzing that when AP is minimized, DE is maximized." }, { "problem_id": 1415, "question": "The tangram is an intellectual game created by our ancestors, originating from the Pythagorean method. As shown in Figure (1), the entire tangram is composed of a square $\\mathrm{ABCD}$ divided into seven pieces (including five isosceles right-angled triangles, one square, and one parallelogram). Figure (2) shows a trapezoid formed by the tangram pieces. If the side length of the square $\\mathrm{ABCD}$ is $12 \\mathrm{~cm}$, then the perimeter of the trapezoid $\\mathrm{MNGH}$ is $\\qquad$ $\\mathrm{cm}$ (the result should be kept in radical form).\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch20-2024_05_23_7e2c875fce25ca646505g_0081_1.jpg", "batch20-2024_05_23_7e2c875fce25ca646505g_0081_2.jpg" ], "is_multi_img": true, "answer": "$24+24 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By observing the figure, we find that $\\mathrm{MH}=\\mathrm{GN}=\\mathrm{AD}=12$, and $\\mathrm{HG}=\\frac{1}{2} \\mathrm{AC}$.\n\nGiven that $\\mathrm{AD}=\\mathrm{DC}=12$,\n\nthen $\\mathrm{AC}=12 \\sqrt{2}$,\n\nand thus $\\mathrm{HG}=6 \\sqrt{2}$.\n\nThe perimeter of trapezoid $\\mathrm{MNGH}$ is calculated as $\\mathrm{HG}+\\mathrm{HM}+\\mathrm{MN}+\\mathrm{NG}=2 \\mathrm{HM}+4 \\mathrm{HG}=24+24 \\sqrt{2}$.\n\nTherefore, the answer is $24+24 \\sqrt{2}$.\n\n【Insight】This problem primarily tests the student's ability to apply the properties of isosceles trapezoids and squares, as well as their skill in analyzing and observing geometric figures." }, { "problem_id": 1416, "question": "In rectangle $\\mathrm{ABCD}$, point $\\mathrm{P}$ is on $\\mathrm{AD}$, $\\mathrm{AB}=\\sqrt{3}, \\mathrm{AP}=1$. Place the vertex of a right angle ruler at $\\mathrm{P}$, and let its two sides intersect $A B$ and $B C$ at points $E$ and $F$ respectively, then connect $E F$ (as shown in Figure 1). When point $E$ coincides with point $B$, point $\\mathrm{F}$ coincides with point $\\mathrm{C}$ (as shown in Figure 2). Starting from the position in Figure 2, rotate the right angle ruler clockwise around point $\\mathrm{P}$ until point $\\mathrm{E}$ coincides with point $\\mathrm{A}$. During this process, the path length traversed by the midpoint of segment $\\mathrm{EF}$ from start to stop is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_93a739bbe5659e20dd68g_0033_1.jpg", "batch20-2024_05_23_93a739bbe5659e20dd68g_0033_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\nLet the midpoint of segment $\\mathrm{EF}$ be $\\mathrm{O}$. Connect $\\mathrm{OP}$ and $\\mathrm{OB}$, as shown in the figure.\n\nIn the right triangle $\\triangle \\mathrm{EPF}$, $\\mathrm{OP} = \\frac{1}{2} \\mathrm{EF}$.\n\nIn the right triangle $\\triangle \\mathrm{EBF}$, $\\mathrm{OB} = \\frac{1}{2} \\mathrm{EF}$.\n\nTherefore, $\\mathrm{OP} = \\mathrm{OB}$.\n\nThus, point $\\mathrm{O}$ lies on the perpendicular bisector of segment $\\mathrm{BP}$.\n\nAs shown in the figure, when point $\\mathrm{E}$ coincides with point $\\mathrm{B}$, and point $\\mathrm{F}$ coincides with point $\\mathrm{C}$, the midpoint of $\\mathrm{EF}$ is the midpoint $\\mathrm{O}$ of $\\mathrm{BC}$.\n\nWhen point $\\mathrm{E}$ coincides with point $\\mathrm{A}$, the midpoint of $\\mathrm{EF}$ is the midpoint $\\mathrm{O}$ of $\\mathrm{PB}$.\n\nTherefore, $\\mathrm{OO}^{\\prime}$ is the midline of $\\triangle \\mathrm{PBC}$.\n\nThus, $\\mathrm{OO}^{\\prime} = \\frac{1}{2} \\mathrm{PC} = \\sqrt{3}$.\n\nThe length of the path traced by the midpoint of segment $\\mathrm{EF}$ is $\\sqrt{3}$.\n\nHence, the answer is $\\sqrt{3}$.\n\n【Key Insight】This problem primarily examines the properties of the median on the hypotenuse of a right triangle and the perpendicular bisector theorem. Finally, the solution is derived using the property of the midline of a triangle. Pay attention to the flexible application and transfer of knowledge." }, { "problem_id": 1417, "question": "As shown in Figure (1), the side length of square \\(ABCD\\) is 3, and the square is folded along the crease \\(MN\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\nAs shown in Figure (2), the square \\(ABCD\\) is unfolded, with points \\(E\\) and \\(F\\) located on sides \\(AB\\) and \\(BC\\) respectively, and \\(CE \\perp DF\\). Point \\(P\\) is a moving point on the crease \\(MN\\). If \\(CF = 1\\), the minimum value of \\(PB + PE\\) is \\(\\qquad\\).", "input_image": [ "batch20-2024_05_23_9809bc8556e8f99f5990g_0016_1.jpg", "batch20-2024_05_23_9809bc8556e8f99f5990g_0016_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{10}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: From the problem statement, we have $BN = NC = \\frac{3}{2}$.\n\nSince points $B$ and $C$ are symmetric with respect to line $MN$,\n\nwhen point $P$ coincides with the intersection of $CE$ and $NM$, the sum $PB + PE$ reaches its minimum value, which is equal to the length of $CE$. Let $DF$ intersect $CE$ at point $M$.\n\n\n\nGiven that $\\angle DFC + \\angle MCF = \\angle FDC + \\angle DFC = 90^\\circ$,\nit follows that $\\angle MCF = \\angle FDC$.\n\nSince $\\angle EBC = \\angle DCF = 90^\\circ$ and $BC = DC = 3$,\n\ntriangles $\\triangle EBC$ and $\\triangle FCD$ are congruent by the ASA criterion.\n\nTherefore, $EB = CF = 1$.\n\nBy the Pythagorean theorem, $CE = \\sqrt{3^2 + 1^2} = \\sqrt{10}$.\n\nThus, the minimum value of $PB + PE$ is $\\sqrt{10}$.\n\nThe answer is: $\\sqrt{10}$\n\n【Insight】This problem tests the properties of folding, the criteria and properties of congruent triangles, and the Pythagorean theorem. Mastering the criteria and properties of congruent triangles is key to solving this problem." }, { "problem_id": 1418, "question": "\"Doing math\" helps us accumulate mathematical activity experience. As shown in the figure, given a triangular paper $A B C$, the first fold makes point $B$ fall on point $B^{\\prime}$ of $B C$, with fold line $A D$ intersecting $B C$ at point $D$. The second fold makes point $A$ fall on point $D$, with fold line $M N$ intersecting $A B^{\\prime}$ at point $P$. If $B C=14, M P+M N=$\n\n\n\n(First fold)\n\n\n\n(Second fold)", "input_image": [ "batch20-2024_05_23_9809bc8556e8f99f5990g_0097_1.jpg", "batch20-2024_05_23_9809bc8556e8f99f5990g_0097_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Complete the diagram as shown:\n\n\n\n## (Second Fold)\n\nFrom the folding, we have: $AM = MD$, $MN \\perp AD$, $AD \\perp BC$,\n\n$\\therefore GN // BC$\n\n$\\therefore AG = BG$\n\n$\\therefore GN$ is the midline of triangle $ABC$,\n\n$\\therefore GN = \\frac{1}{2} BC = \\frac{1}{2} \\times 12 = 6$,\n\n$\\because PM = GM$\n\n$\\therefore MP + MN = GM + MN = GN = 6$,\n\nThus, the answer is: 6.\n\n【Highlight】This problem examines the midline theorem of a triangle and the properties of folding. The key to solving this problem lies in completing the diagram and proving that $GN$ is the midline of triangle $ABC$." }, { "problem_id": 1419, "question": "As shown in Figure 1, there is a mobile phone tablet stand, and Figure 2 is a schematic diagram of its side structure. The length of the tray is measured as \\( AB = 120 \\, \\text{mm} \\), the length of the support plate is \\( CD = 80 \\, \\text{mm} \\), and the length of the base is \\( DE = 90 \\, \\text{mm} \\). The tray \\( AB \\) is fixed at the top point \\( C \\) of the support plate, with \\( CB = 40 \\, \\text{mm} \\). The tray \\( AB \\) can rotate around point \\( C \\), and the support plate \\( CD \\) can rotate around point \\( D \\). As shown in Figure 2, if \\( \\angle DCB = 90^\\circ \\) and \\( \\angle CDE = 60^\\circ \\), find the distance from point \\( A \\) to the base \\( DE \\).\n\n(Reference data: \\( \\sqrt{2} \\approx 1.41 \\), \\( \\sqrt{3} \\approx 1.73 \\), \\( \\sqrt{5} \\approx 2.24 \\)) (The result should be accurate to \\( 0.1 \\, \\text{mm} \\))\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_9e9ee74f6a3d8ab5369cg_0060_1.jpg", "batch20-2024_05_23_9e9ee74f6a3d8ab5369cg_0060_2.jpg" ], "is_multi_img": true, "answer": "$109.2 \\mathrm{~mm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular from point $A$ to line $DE$, meeting at point $H$. From point $C$, draw a perpendicular to $AH$, meeting at point $N$, and a perpendicular to $DE$, meeting at point $M$.\n\n\n\n$\\therefore \\angle CNH = \\angle CMH = \\angle NHM = 90^{\\circ}$,\n\n$\\therefore$ quadrilateral $CMHN$ is a rectangle, and $AH \\parallel CM$,\n\n$\\because \\angle DCB = 90^{\\circ}, \\angle CDE = 60^{\\circ}$,\n\n$\\therefore \\angle DCM = 30^{\\circ}, \\angle BCM = 60^{\\circ}$,\n\n$\\therefore \\angle A = \\angle BCM = 60^{\\circ}$,\n\n$\\because AB = 120, \\quad CB = 40$,\n\n$\\therefore AC = 80$,\n\n$\\therefore AN = \\frac{1}{2} AC = 40, NC = \\sqrt{AC^{2} - AN^{2}} = 40 \\sqrt{3}$,\n\n$\\because CD = 80, \\angle CDE = 60^{\\circ}$,\n\n$\\therefore DM = \\frac{1}{2} CD = 40, MC = \\sqrt{CD^{2} - DM^{2}} = 40 \\sqrt{3}$,\n$\\therefore MC = NC$,\n\n$\\therefore$ quadrilateral $CMHN$ is a square,\n\n$\\therefore HN = 40 \\sqrt{3}$,\n\n$\\therefore AH = AN + HN = 40 + 40 \\sqrt{3} \\approx 109.2$,\n\nThat is, the distance from point $A$ to the base $DE$ is $109.2 \\mathrm{~mm}$.\n\n【Highlight】This problem examines the determination and properties of a square, the properties of parallel lines, the properties of a 30-degree right triangle, and the Pythagorean theorem. The key to solving the problem is constructing a 30-degree right triangle by drawing auxiliary lines." }, { "problem_id": 1420, "question": "Mathematician Moron discovered the world's first perfect rectangle in 1925 (as shown in Figure 1), which can be exactly divided into 10 squares of different sizes. Since then, people have become enthusiastic about the study of perfect graph divisions. $\\square E F G H$ is divided into 13 small equilateral triangles (as shown in Figure 2). It is known that the two smallest central equilateral triangles, $\\triangle A B C$ and $\\triangle A D C$, both have side lengths of $2$, and the perimeter of $\\square E F G H$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_a36c77a3f55629b66926g_0009_1.jpg", "batch20-2024_05_23_a36c77a3f55629b66926g_0009_2.jpg" ], "is_multi_img": true, "answer": "78", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure:\n\n\nLet the side length of $\\triangle B M N$ be $x$.\n\nSince the side lengths of $\\triangle A B C$ and $\\triangle A D C$ are both 2,\n\nTherefore, $P M = P K = A M = x + 2$,\n\n$F R = P N = x + 2 + x = 2x + 2$,\n\n$O H = O K = K D = x + 2 + 2 = x + 4$,\n\n$R G = R B = F R + x = 3x + 2$,\n\nThus, $E H = O E + O H = 2x + 6 + x + 4 = 3x + 10$, $F G = F R + R G = 2x + 2 + 3x + 2 = 5x + 4$,\n\nTherefore, $3x + 10 = 5x + 4$, solving gives: $x = 3$,\n\n$H G = H Q + Q G = x + 6 + 3x + 2 = 4x + 8 = 20$, $F G = 5x + 4 = 19$,\n\nThus, the perimeter of rectangle $E F G H$ is: $2 \\times (20 + 19) = 78$.\n\nHence, the answer is: 78.\n\n【Highlight】This question examines the properties of a parallelogram and the pattern of changes in a figure. Careful observation and step-by-step reasoning to express the lengths of the sides $E H$ and $F G$ of the parallelogram are key to solving the problem." }, { "problem_id": 1421, "question": "As shown, in $\\triangle \\mathrm{ABC}$, $\\mathrm{BC}=2$, points $P_{1}, M_{1}$ are the midpoints of sides $\\mathrm{AB}, \\mathrm{AC}$ respectively (Figure 1), points $P_{2}, M_{2}$ are the midpoints of sides $A P_{1}, A M_{1}$ respectively (Figure 2), points $P_{3}, M_{3}$ are the midpoints of sides $A P_{2}, A M_{2}$ respectively (Figure 3) $\\ldots$. Following this pattern, the length of $P_{5} M_{5}$ is $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch20-2024_05_23_b79c70140846dbac1f27g_0090_1.jpg", "batch20-2024_05_23_b79c70140846dbac1f27g_0090_2.jpg", "batch20-2024_05_23_b79c70140846dbac1f27g_0090_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{1}{16}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: In triangle \\( ABC \\), \\( BC = 2 \\). Points \\( P_1 \\) and \\( M_1 \\) are the midpoints of sides \\( AB \\) and \\( AC \\), respectively. Points \\( P_2 \\) and \\( M_2 \\) are the midpoints of \\( AP_1 \\) and \\( AM_1 \\), respectively. Points \\( P_3 \\) and \\( M_3 \\) are the midpoints of \\( AP_2 \\) and \\( AM_2 \\), respectively.\n\nWe can derive:\n\\[ P_1M_1 = 1, \\quad P_2M_2 = 1 \\times \\frac{1}{2} = \\frac{1}{2}, \\]\nand thus,\n\\[ P_nM_n = \\frac{1}{2^{(n-1)}}. \\]\n\nTherefore,\n\\[ P_5M_5 = \\frac{1}{2^4} = \\frac{1}{16}. \\]\n\nThe answer is: \\(\\frac{1}{16}\\).\n\n**Key Insight:** This problem tests the Midsegment Theorem of triangles. The key is to identify the pattern based on the midsegments to arrive at the solution." }, { "problem_id": 1422, "question": "The tangram is known as the \"Oriental Magic Square\" by Westerners. The two figures below are formed by the same set of tangram pieces. Given that the side length of the square formed by the tangram (as shown in Figure a) is 8, the area of the shaded part of \"Smooth Sailing\" (as shown in Figure b) is $\\qquad$\n\n\n\nFigure a\n\n\n\nFigure b", "input_image": [ "batch20-2024_05_23_d8f798e1105c2ee20feeg_0029_1.jpg", "batch20-2024_05_23_d8f798e1105c2ee20feeg_0029_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Question Analysis:**\n\n\n\nAs shown in the figure, the shaded area in the \"Smooth Sailing\" diagram is congruent to the two right-angled isosceles triangles labeled \"1\" and \"2\" within the square. In diagram $\\mathrm{b}$, the trapezoid has a side equal to the right-angled side of triangle 2, and the side length of the small square is equal to the right-angled side of triangle 2.\n\nThus, in diagram a, it can be deduced that the diagonal of the large square is equal to the sum of the side lengths of the four shaded right-angled triangles.\n\nGiven that the side length of the large square is 8, the diagonal $=\\sqrt{8^{2}+8^{2}}=\\sqrt{128}=8 \\sqrt{2}$.\n\nTherefore, the side length of the shaded triangle $=8 \\sqrt{2} \\div 4=2 \\sqrt{2}$.\n\nThe area is: $\\frac{1}{2} \\times 2 \\sqrt{2} \\times 2 \\sqrt{2}=4$.\n\n**Key Points:** Properties of right-angled isosceles triangles and squares.\n\n**Review:** This question is of medium difficulty, primarily testing students' understanding of the properties of right-angled isosceles triangles and squares in the context of a tangram. The key to solving the problem lies in recognizing the corresponding equal sides of the triangles in the two diagrams." }, { "problem_id": 1423, "question": "The tangram is an outstanding creation of our ancestors, praised by Westerners as the \"Oriental Magic Plate.\" The two figures below, the square (Figure 1) and the \"pinwheel\" (Figure 2), are both formed by the same set of tangram pieces. The ratio of the areas of the squares $A B C D$ and $E F G H$ in the figures is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_d8f798e1105c2ee20feeg_0042_1.jpg", "batch20-2024_05_23_d8f798e1105c2ee20feeg_0042_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{8}{13}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in the figure, let the side length of the small square in the tangram be a,\n\n$\\therefore \\mathrm{BD}=4 \\mathrm{a}$,\n\n$\\therefore \\mathrm{CD}=2 \\sqrt{2} \\mathrm{a}$,\n\n$\\therefore \\mathrm{S}$ square $\\mathrm{ABCD}=(2 \\sqrt{2} \\mathrm{a})^{2}=8 \\mathrm{a}^{2}$,\n\n$\\because \\mathrm{S}_{1}=\\mathrm{S}_{2}=\\mathrm{S}_{3}=\\mathrm{S}_{4}=\\frac{1}{2} \\mathrm{a} \\cdot 2 \\mathrm{a}=\\mathrm{a}^{2}, \\quad \\mathrm{~S}_{5}=\\mathrm{a}^{2}$\n\n$\\therefore \\mathrm{S}$ square $\\mathrm{EFGH}=8 \\mathrm{a}^{2}+4 \\mathrm{a}^{2}+\\mathrm{a}^{2}=13 \\mathrm{a}^{2}$,\n\n$\\therefore$ The area ratio of squares $\\mathrm{ABCD}$ and $\\mathrm{EFGH}$ is $\\frac{8 a^{2}}{13 a^{2}}=\\frac{8}{13}$,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nTherefore, the answer is $\\frac{8}{13}$\n\n【Insight】This problem uses the tangram to examine the properties of squares, isosceles right triangles, and the application of the Pythagorean theorem. Mastering these properties and theorems is crucial for solving the problem." }, { "problem_id": 1424, "question": "As shown in Figure 1, in square \\(ABCD\\), points \\(E\\), \\(F\\), \\(G\\), and \\(H\\) are located on sides \\(AB\\), \\(BC\\), \\(CD\\), and \\(DA\\) respectively, with \\(HA = EB = FC = GD\\). Lines \\(EG\\) and \\(FH\\) intersect at point \\(O\\). The square \\(ABCD\\) is cut along segments \\(EG\\) and \\(HF\\), and the resulting four quadrilaterals are rearranged into a single quadrilateral as shown in Figure 2. If the side length of square \\(ABCD\\) is \\(3 \\, \\text{cm}\\) and \\(HA = EB = FC = GD = 1 \\, \\text{cm}\\), then the area of the shaded part in Figure 2 is \\(\\qquad\\) \\(\\text{cm}^2\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_d8f798e1105c2ee20feeg_0060_1.jpg", "batch20-2024_05_23_d8f798e1105c2ee20feeg_0060_2.jpg" ], "is_multi_img": true, "answer": "1", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Connect $\\mathrm{EF}$, $\\mathrm{FG}$, $\\mathrm{GH}$, $\\mathrm{HE}$.\n\n\n\nSince quadrilateral $\\mathrm{ABCD}$ is a square,\n\n$\\therefore \\angle \\mathrm{A}=\\angle \\mathrm{B}=\\angle \\mathrm{C}=\\angle \\mathrm{D}=90^{\\circ}$, and $\\mathrm{AB}=\\mathrm{BC}=\\mathrm{CD}=\\mathrm{DA}$.\n\nSince $\\mathrm{HA}=\\mathrm{EB}=\\mathrm{FC}=\\mathrm{GD}$,\n\n$\\therefore \\mathrm{AE}=\\mathrm{BF}=\\mathrm{CG}=\\mathrm{DH}$,\n\n$\\therefore \\triangle \\mathrm{AEH} \\cong \\triangle \\mathrm{BFE} \\cong \\triangle \\mathrm{CGF} \\cong \\triangle \\mathrm{DHG}$,\n\n$\\therefore \\mathrm{EF}=\\mathrm{FG}=\\mathrm{GH}=\\mathrm{HE}$,\n\n$\\therefore$ quadrilateral $\\mathrm{EFGH}$ is a rhombus.\n\nSince $\\triangle \\mathrm{DHG} \\cong \\triangle \\mathrm{AEH}$,\n\n$\\therefore \\angle \\mathrm{DHG}=\\angle \\mathrm{AEH}$.\n\nSince $\\angle \\mathrm{AEH}+\\angle \\mathrm{AHE}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{DHG}+\\angle \\mathrm{AHE}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{GHE}=90^{\\circ}$,\n\n$\\therefore$ quadrilateral $\\mathrm{EFGH}$ is a square.\n\nSince $H A=E B=F C=G D=1$, and $A B=B C=C D=A D=3$,\n\n$\\therefore \\mathrm{GF}=\\mathrm{EF}=\\mathrm{EH}=\\mathrm{GH}=\\sqrt{5}$,\n\nSince quadrilateral $\\mathrm{EFGH}$ is a square,\n\n$\\therefore \\mathrm{GO}=\\mathrm{OF}$, and $\\angle \\mathrm{GOF}=90^{\\circ}$,\n\nBy the Pythagorean theorem: $\\mathrm{GO}=\\mathrm{OF}=\\frac{\\sqrt{10}}{2}$.\n\nSince the area of quadrilateral $\\mathrm{FCGO}=\\frac{1}{2} \\times 1 \\times 2+\\frac{1}{2} \\times \\frac{\\sqrt{10}}{2} \\times \\frac{\\sqrt{10}}{2}=\\frac{9}{4}$,\n\n$\\therefore$ the shaded area $=\\left(\\frac{\\sqrt{10}}{2}+\\frac{\\sqrt{10}}{2}\\right)^{2}-\\text{Area of quadrilateral } F C G O \\times 4=10-9=1$.\n\n【Insight】This problem examines the determination and properties of a square, with the key to solving it lying in drawing auxiliary lines and utilizing the properties of congruent triangles." }, { "problem_id": 1425, "question": "As shown in Figure 1, a rectangle \\( A B C D \\) and a square \\( E F G H \\) are cut along their diagonals \\( A C \\) and \\( E G \\) respectively, and rearranged to form the parallelogram \\( P Q M N \\) as shown in Figure 2, with the quadrilateral \\( K R S T \\) in the middle being a square. If the areas of the square \\( E F G H \\) and the square \\( K R S T \\) are 16 and 1 respectively, then the area of the rectangle \\( A B C D \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_d8f798e1105c2ee20feeg_0089_1.jpg", "batch20-2024_05_23_d8f798e1105c2ee20feeg_0089_2.jpg" ], "is_multi_img": true, "answer": "15", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since the areas of squares $\\mathrm{EFCH}$ and KRST are 16 and 1 respectively,\n\nTherefore, the side lengths of squares $\\mathrm{EFCH}$ and KRST are 4 and 1 respectively.\n\nThus, the area of rectangle $\\mathrm{ABCD}$ is $(4+1) \\times (4-1) = 15$.\n\nHence, the answer is: 15.\n\n[Highlight] This question tests knowledge on the properties of shapes, specifically the properties of rectangles and squares, and the key to solving it lies in applying the concept of combining numbers with shapes." }, { "problem_id": 1426, "question": "As shown in the figure, Figure 1 is a collapsible desk lamp, and Figure 2 is its plan view. The base $A O \\perp O E$ at point $O$, and the supports $A B, B C$ are fixed rods. The angle $\\angle B A O$ is twice the angle $\\angle C B A$, and the lamp body $C D$ can be adjusted by rotating around point $C$. Now, the lamp body $C D$ is rotated from the horizontal position to the position $C D^{\\prime}$ (shown as the dashed line in Figure 2). At this time, the line where the lamp body $C D^{\\prime}$ lies is exactly perpendicular to the support $A B$, and $\\angle B C D-\\angle D C D^{\\prime}=123^{\\circ}$. Then, $\\angle D C D^{\\prime}=$ \n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch20-2024_05_23_e6a1e11f9f80e6c7d7f1g_0003_1.jpg", "batch20-2024_05_23_e6a1e11f9f80e6c7d7f1g_0003_2.jpg" ], "is_multi_img": true, "answer": "$38^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Extend $O A$ to intersect $C D$ at point $F$, and extend $D^{\\prime} C$ to intersect $A B$ at point $G$.\n\n\n\nHorizontal tabletop\n\n$\\because A O \\perp O E$\n\n$\\therefore \\angle A O E=90^{\\circ}$,\n\n$\\because C D \\parallel O E$,\n\n$\\therefore \\angle A F C=\\angle A O E=90^{\\circ}$,\n\n$\\therefore O A \\perp C D$,\n\nSince $A O \\perp O E$ and $D^{\\prime} C \\perp A B$,\n\n$\\therefore \\angle A G C=\\angle A F C=90^{\\circ}$,\n\n$\\therefore \\angle G C F+\\angle G A F=180^{\\circ}$,\n\nSince $\\angle D C D^{\\prime}+\\angle G C F=180^{\\circ}$,\n\n$\\therefore \\angle D C D^{\\prime}=\\angle G A F$,\n\n$\\therefore \\angle B A O=180^{\\circ}-\\angle D C D^{\\prime}$,\n$\\therefore \\angle B=\\frac{1}{2}\\left(180^{\\circ}-\\angle D C D^{\\prime}\\right)$,\n\n$\\because \\angle B C D-\\angle D C D^{\\prime}=123^{\\circ}$,\n\n$\\therefore \\angle B C D=\\angle D C D^{\\prime}+123^{\\circ}$,\n\nIn quadrilateral $A B C F$, we have $\\angle G A F+\\angle B+\\angle B C D+\\angle A F C=360^{\\circ}$,\n\n$\\therefore \\angle D C D^{\\prime}+\\frac{1}{2}\\left(180^{\\circ}-\\angle D C D^{\\prime}\\right)+\\angle D C D^{\\prime}+123^{\\circ}+90^{\\circ}=360^{\\circ}$,\n\nSolving gives: $\\angle D C D^{\\prime}=38^{\\circ}$,\n\nTherefore, the answer is: $38^{\\circ}$.\n\n【Highlight】This problem tests the properties of parallel lines, with the key being a thorough understanding of the properties of parallel lines and the fact that the sum of the interior angles of a quadrilateral is 360 degrees." }, { "problem_id": 1427, "question": "Several rectangular paper pieces of the same shape and size are arranged to form a square. As shown in Figure (1), a square is formed by 4 rectangular paper pieces, with the area of the shaded part being 16; as shown in Figure (2), a square is formed by 8 rectangular paper pieces, with the area of the shaded part being 8; as shown in Figure (3), a square is formed by 12 rectangular paper pieces, then the perimeter of the shaded part is $\\qquad$.\n\n\n(1)\n\n\n(2)\n\n\n(3)", "input_image": [ "batch21-2024_06_15_09d88f35dee69465b1c7g_0098_1.jpg", "batch21-2024_06_15_09d88f35dee69465b1c7g_0098_2.jpg", "batch21-2024_06_15_09d88f35dee69465b1c7g_0098_3.jpg" ], "is_multi_img": true, "answer": "$16 \\sqrt{2}-16$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the problem statement, we have:\n\nIn Figure (1), the side length of the shaded part is $\\sqrt{16}=4$, and in Figure (2), the side length of the shaded part is $\\sqrt{8}=2 \\sqrt{2}$.\n\nLet the length of the rectangle be $a$ and the width be $b$. According to the problem, we have:\n\n$$\n\\left\\{\\begin{array}{l}\na - b = 4 \\\\\na - 2b = 2 \\sqrt{2}\n\\end{array}\\right.\n$$\n\nSolving the system of equations, we get:\n\n$$\n\\left\\{\\begin{array}{l}\na = 8 - 2 \\sqrt{2} \\\\\nb = 4 - 2 \\sqrt{2}\n\\end{array}\\right.\n$$\n\nTherefore, the side length of the shaded square in Figure (3) is:\n\n$$\na - 3b = 8 - 2 \\sqrt{2} - 3(4 - 2 \\sqrt{2}) = 4 \\sqrt{2} - 4\n$$\n\nThus, Figure (3) is a square formed by surrounding 12 rectangular pieces of paper, and the perimeter of the shaded part is:\n\n$$\n16 \\sqrt{2} - 16\n$$\n\nHence, the answer is: $16 \\sqrt{2} - 16$.\n\n【Insight】This problem mainly tests the properties of squares and the application of arithmetic square roots. Mastering these concepts is essential for solving such problems." }, { "problem_id": 1428, "question": "Figure (1) is a work by artist M.C. Escher, who perfectly combined mathematics with painting to create a three-dimensional effect on a flat surface. Figure (2) is a rhombus. By cutting off a rhombus with half the side length of Figure (2), we obtain Figure (3). Using Figure (3) to create a tessellation results in Figure (4). After coloring Figure (4), another tessellation yields Figure (1). The degree measure of $\\angle A B C$ in Figure (4) is $\\qquad$ $\\circ$.\n\n\n\n(Figure (1)\n\n\n\n(Figure (2))\n\n\n\n(Figure (3))\n\n\n\n(Figure (4)", "input_image": [ "batch21-2024_06_15_11ed3deff80f50053462g_0043_1.jpg", "batch21-2024_06_15_11ed3deff80f50053462g_0043_2.jpg", "batch21-2024_06_15_11ed3deff80f50053462g_0043_3.jpg", "batch21-2024_06_15_11ed3deff80f50053462g_0043_4.jpg" ], "is_multi_img": true, "answer": "60", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "As shown in the figure, since $\\angle BAD = \\angle BAE = \\angle DAE$ and $\\angle BAD + \\angle BAE + \\angle DAE = 360^\\circ$,\n\ntherefore, $\\angle BAD = \\angle BAE = \\angle DAE = 120^\\circ$.\n\nSince $BC \\parallel AD$,\n\ntherefore, $\\angle ABC = 180^\\circ - 120^\\circ = 60^\\circ$.\n\nHence, the answer is: 60.\n\n\n\n(Figure 4)\n\n[Insight] This question tests the properties of a rhombus and the student's ability to read and understand the problem. The key to solving it is to comprehend the question and determine the measure of $\\angle BAD$." }, { "problem_id": 1429, "question": "As shown in the figure, the school gate of a certain school is a retractable gate. Each row in the retractable gate has 20 rhombuses, with each rhombus having a side length of $25 \\mathrm{~cm}$. When the gate is closed, each rhombus has an obtuse angle of $120^{\\circ}$. When the gate is partially opened, the original $120^{\\circ}$ angle of each rhombus shrinks to $60^{\\circ}$. Then the gate opens by $\\qquad$ $\\mathrm{cm}$.\n\n\n", "input_image": [ "batch21-2024_06_15_11ed3deff80f50053462g_0059_1.jpg", "batch21-2024_06_15_11ed3deff80f50053462g_0059_2.jpg" ], "is_multi_img": true, "answer": "$(500 \\sqrt{3}-500) \\mathrm{cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, connect $AC$ intersecting $BD$ at $O$, and connect $B^{\\prime} D^{\\prime}$.\n\nSince quadrilateral $ABCD$ is a rhombus and $\\angle BAD = 120^\\circ$,\n\n$\\therefore \\angle BAO = \\frac{1}{2} \\angle BAD = 60^\\circ$, $\\angle AOB = 90^\\circ$, and $BD = 2OB$.\n\n$\\therefore \\angle ABO = 30^\\circ$,\n\n$\\therefore AO = \\frac{1}{2} AB = \\frac{25}{2} \\text{ cm}$,\n\n$\\therefore OB = \\sqrt{AB^2 - OA^2} = \\frac{25 \\sqrt{3}}{2} \\text{ cm}$,\n\n$\\therefore BD = 25 \\sqrt{3} \\text{ cm}$.\n\nSince quadrilateral $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ is a rhombus and $\\angle A^{\\prime} B^{\\prime} C^{\\prime} = 120^\\circ$,\n\n$\\therefore A^{\\prime} B^{\\prime} = A^{\\prime} D^{\\prime}$, and $A^{\\prime} D^{\\prime} \\parallel B^{\\prime} C^{\\prime}$,\n\n$\\therefore \\angle A^{\\prime} = 60^\\circ$,\n\n$\\therefore \\triangle A^{\\prime} B^{\\prime} D^{\\prime}$ is an equilateral triangle,\n\n$\\therefore B^{\\prime} D^{\\prime} = A^{\\prime} B^{\\prime} = 25 \\text{ cm}$.\n\n$\\therefore$ When the school gate is closed, the length of the gate is $25 \\sqrt{3} \\times 20 = 500 \\sqrt{3} \\text{ cm}$, and when the gate is open, the length is $25 \\times 20 = 500 \\text{ cm}$.\n\n$\\therefore$ The gate has opened by $(500 \\sqrt{3} - 500) \\text{ cm}$.\n\nHence, the answer is: $(500 \\sqrt{3} - 500)$.\n\n\n\n【Insight】This problem mainly examines the properties of a rhombus, the properties and determination of an equilateral triangle, the properties of a right triangle with a 30-degree angle, and the Pythagorean theorem. Understanding the properties of a rhombus is key to solving this problem." }, { "problem_id": 1430, "question": "As shown in Figure 1, the area of the rhombus paper $A B C D$ is $30 \\mathrm{~cm}^{2}$, and the length of the diagonal $A C$ is $6 \\mathrm{~cm}$. When this rhombus paper is cut along the diagonal, four congruent right-angled triangles are obtained. These four right-angled triangles are then arranged into a large square as shown in Figure 2. The area of the blank small square in the large square is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_11ed3deff80f50053462g_0099_1.jpg", "batch21-2024_06_15_11ed3deff80f50053462g_0099_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since \\( AC = 6 \\text{ cm} \\),\n\nTherefore, half of it is \\( 3 \\text{ cm} \\).\n\nGiven that the area of the rhombus \\( ABCD \\) is \\( 30 \\text{ cm}^2 \\),\n\nThus, the length of the other diagonal is \\( 30 \\times 2 \\div 6 = 10 \\text{ cm} \\),\n\nTherefore, half of it is \\( 5 \\text{ cm} \\).\n\nHence, the side length of the small blank square shown in Figure 2 is \\( 5 - 3 = 2 \\text{ cm} \\),\n\nTherefore, the area is \\( 2 \\times 2 = 4 \\text{ cm}^2 \\).\n\nThe answer is: 4.\n\n[Key Insight] This problem examines paper-cutting issues, properties of squares, properties of rhombuses, and congruent figures. The key to solving the problem lies in determining the side length of the small square in Figure 2." }, { "problem_id": 1431, "question": "\"Flowers shadow the wall, peaks overlap the window,\" the airy window lattices of Suzhou gardens contain many mathematical elements. The window lattice in Figure (1) is a frost crack pattern, and Figure (2) shows part of this pattern. If $\\angle 1 + \\angle 3 + \\angle 5 = 186^\\circ$, then\n\n$\\angle 2 + \\angle 4 + \\angle 6 =$ $\\qquad$ .\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_11ff388d4aac29d42a3eg_0006_1.jpg", "batch21-2024_06_15_11ff388d4aac29d42a3eg_0006_2.jpg" ], "is_multi_img": true, "answer": "366", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\n$\\because$ The sum of the exterior angles of a polygon is $360^{\\circ}$,\n\n$\\therefore \\angle 7+\\angle 8+\\angle 9=360^{\\circ}-(\\angle 1+\\angle 3+\\angle 5)=174^{\\circ}$,\n\n$\\because \\angle 2+\\angle 7=\\angle 4+\\angle 8=\\angle 6+\\angle 9=180^{\\circ}$,\n\n$\\therefore \\angle 2+\\angle 4+\\angle 6=180^{\\circ} \\times 3-(\\angle 7+\\angle 8+\\angle 9)=540^{\\circ}-174^{\\circ}=366^{\\circ}$;\nTherefore, the answer is: 366.\n\n【Insight】This question tests the application of the sum of the exterior angles of a polygon. Mastering the fact that the sum of the exterior angles of a polygon is $360^{\\circ}$ is key to solving the problem." }, { "problem_id": 1432, "question": "As shown in Figure 1, given a rectangular strip of paper \\( ABCD \\) with \\( AD \\parallel CD \\) and \\( AD \\parallel BC \\). After folding the strip along \\( EF \\), points \\( B \\) and \\( C \\) fall at positions \\( H \\) and \\( G \\) respectively. Then, folding along \\( GF \\) results in Figure 2, where points \\( A \\) and \\( D \\) fall at positions \\( Q \\) and \\( H \\) respectively. Given that \\( 2 \\angle QHG = 4 \\angle GFH - 108^\\circ \\), the angle \\( \\angle EFC \\) is:\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_16fcfef452df2c165789g_0040_1.jpg", "batch21-2024_06_15_16fcfef452df2c165789g_0040_2.jpg" ], "is_multi_img": true, "answer": "$63^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Let $\\angle Q H G = x$,\n\nBy rotation, we have: $\\angle Q H F = \\angle D = 90^{\\circ}$, $\\angle H G F = \\angle C = 90^{\\circ}$,\n\nThus, $\\angle G H F = 90^{\\circ} - x$,\n\nSince $2 \\angle Q H G = 4 \\angle G F H - 108^{\\circ}$,\n\nTherefore, $\\angle G F H = \\frac{1}{2} x + 27^{\\circ}$,\n\nIn right triangle $\\triangle G H F$, $\\angle G H F + \\angle G F H = 90^{\\circ}$,\n\nThus, $90 - x + \\frac{1}{2} x + 27 = 90$, solving gives $x = 54^{\\circ}$,\n\nBy rotation, $\\angle D F G = \\angle G F H = \\frac{1}{2} \\times 54^{\\circ} + 27^{\\circ} = 54^{\\circ}$,\n\nTherefore, $\\angle G F C = 180^{\\circ} - 54^{\\circ} = 126^{\\circ}$,\n\nHence, $\\angle E F C = \\frac{1}{2} \\angle G F C = 63^{\\circ}$,\n\nThe final answer is: $63^{\\circ}$\n\n【Highlight】This problem examines the properties of parallel lines: corresponding angles are equal; consecutive interior angles are supplementary; alternate interior angles are equal. It also tests the knowledge of the sum of angles in a triangle and the properties of folding." }, { "problem_id": 1433, "question": "As shown, a small square is cut out from a larger square, and then the shaded area in the figure is cut and rearranged into a rectangle, as shown in Figure 2. The length of this rectangle is 24, and the width is 16. Therefore, the area of the $\\mathrm{S}_{2}$ part in Figure 2 is\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0009_1.jpg", "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0009_2.jpg" ], "is_multi_img": true, "answer": "64", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the side length of the larger square be \\( a \\), and the side length of the smaller square be \\( b \\).\n\nAccording to the problem, we have the system of equations:\n\\[\n\\begin{cases}\nb + a = 24 \\\\\na - b = 16\n\\end{cases}\n\\]\n\nSolving the system, we find:\n\\[\n\\begin{cases}\na = 20 \\\\\nb = 4\n\\end{cases}\n\\]\n\nTherefore, the area of part \\( S_2 \\) in Figure 2 is:\n\\[\n4 \\times (20 - 4) = 64\n\\]\n\nThus, the answer is: 64.\n\n**Key Insight:** This problem primarily tests the properties of squares and the application of systems of linear equations. The key to solving the problem lies in deriving the equations \\( a + b = 24 \\) and \\( a - b = 16 \\)." }, { "problem_id": 1434, "question": "Figure 1 is the famous \"Zhao Shuang's Xian Diagram\" in China, which is formed by four congruent right-angled triangles. Extend the shorter sides of the four right-angled triangles (such as $A F$) outward by an equal length to obtain points $A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}$, resulting in Figure 2. It is known that the areas of the square $E F G H$ and the square $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ are $1 \\mathrm{~cm}^{2}$ and $85 \\mathrm{~cm}^{2}$ respectively, then the area of the shaded part is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0011_1.jpg", "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0011_2.jpg" ], "is_multi_img": true, "answer": "30", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since the areas of squares $\\mathrm{EFGH}$ and $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}$ are $1 \\mathrm{~cm}^{2}$ and $85 \\mathrm{~cm}^{2}$ respectively,\n\n$\\therefore \\mathrm{EF}=\\mathrm{FG}=\\mathrm{GH}=\\mathrm{HF}=1, \\quad \\mathrm{~A}^{\\prime} \\mathrm{B}^{\\prime}=\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}=\\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}=\\mathrm{A}^{\\prime} \\mathrm{D}^{\\prime}=\\sqrt{85}$.\n\nLet the shorter side of the four right-angled triangles be $\\mathrm{x}$. Then, in right-angled triangle $\\triangle \\mathrm{A}^{\\prime} \\mathrm{ED}^{\\prime}$,\n\n$D^{\\prime} E=2 x, A^{\\prime} E=2 x+1$, and according to the problem,\n\n$(2 \\mathrm{x})^{2}+(2 \\mathrm{x}+1)^{2}=85$,\n\nSimplifying gives\n\n$2 x^{2}+x-21=0$\n\n$\\therefore \\mathrm{x}_{1}=3, \\mathrm{x}_{2}=-3.5$ (discarded)\n\n$\\therefore \\mathrm{A}^{\\prime} \\mathrm{F}=\\mathrm{C}^{\\prime} \\mathrm{H}=6, \\quad \\mathrm{AE}=\\mathrm{CG}=4$\n\n$\\therefore$ The area of the shaded part in Figure 2 is $(3 \\times 6 \\div 2+3 \\times 4 \\div 2) \\times 2=30$.\n\nHence, the answer is: 30.\n\n[Key Insight] This problem tests the application of the Pythagorean theorem in the context of a chord diagram. Clearly understanding the length relationships of the relevant line segments in the diagram and setting up the equation based on the Pythagorean theorem are crucial for solving this problem." }, { "problem_id": 1435, "question": "As shown in Figure 1, given that the area of the small square \\(ABCD\\) is 1, extend each of its sides by one time to obtain the new square \\(A_{1}B_{1}C_{1}D_{1}\\): extend the sides of square \\(A_{1}B_{1}C_{1}D_{1}\\) by the same method to obtain square \\(A_{2}B_{2}C_{2}D_{2}\\), as shown in Figure 2; continue this process... then the area of square \\(A_{4}B_{4}C_{4}D_{4}\\) is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0041_1.jpg", "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0041_2.jpg" ], "is_multi_img": true, "answer": "625", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Initially, the side length is 1, and the area is 1. \nAfter the first extension, the side length becomes $\\sqrt{5}$, and the area becomes 5. \nAfter the second extension, the side length is $5^{1}=5$, and the area is $5^{2}=25$. \nAfter the next extension, the side length becomes $5 \\sqrt{5}$, and the area becomes $5^{3}=125$. \nFollowing this pattern, \nwhen $n=4$, the area of square $A_{4} B_{4} C_{4} D_{4}$ is: $5^{4}=625$. \n\nTherefore, the answer is: 625. \n\n[Insight] This problem primarily examines the properties of squares. In solving it, one must identify the pattern based on the given conditions to determine the area of the square." }, { "problem_id": 1436, "question": "Divide the rhombus with side length 10 in Figure 1 along its diagonals into four congruent right triangles, where one of the diagonals of the rhombus is 16. Arrange these four right triangles to form the square shown in Figure 2. Then the area of the shaded part in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0045_1.jpg", "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0045_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n$\\because$ Quadrilateral $A B C D$ is a rhombus, $A C=16$, $A D=10$,\n\n$\\therefore O A=O C=8$, $O B=O D$, $A C \\perp B D$,\n\n$O B=O D=\\sqrt{A D^{2}-O A^{2}}=6$,\n\n$\\therefore B D=2 O D=12$,\n\n$\\therefore$ The area of the rhombus $=\\frac{1}{2} \\times 12 \\times 16=96$,\n\nThe area of the square in Figure 2 $=10^{2}=100$,\n\n$\\therefore$ The area of the shaded region $=100-96=4$.\n\nTherefore, the answer is: 4.\n\n[Highlight] This question examines the properties of figure cutting and splicing, the properties of a rhombus, and the properties of a square. The key to solving this problem lies in understanding the properties of a rhombus." }, { "problem_id": 1437, "question": "In the $3 \\times 3$ grid shown in Figure 1, there is an octagon, where the side length of each small square is 1. Through investigation, it is found that this octagon can be divided into four congruent pentagons and a small square (1) as shown in Figure 2. Now, the four pentagons after division are reassembled (i.e., the shaded parts in Figure 2) to form a large square $A B C D$. It is found that the blank space in the middle of the large square (2) is also a square, and the area of square (2) is exactly twice the area of square (1). Then, the length of $A E$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0068_1.jpg", "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0068_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}-\\frac{1}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nFigure 2\n\nAccording to the problem, the side length of the octagon is $\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$,\n\nThe area of the octagon is $3 \\times 3 - 4 \\times \\frac{1}{2} \\times 1 \\times 1 = 7$,\n\n$\\therefore$ The area of square (2) is $(\\sqrt{2})^{2} = 2$,\n\n$\\because$ The area of square (2) is exactly twice the area of square (1),\n\n$\\therefore$ The area of square (1) is 1,\n\n$\\therefore$ The area of four congruent pentagons is $7 - 1 = 6$, and $AF = \\sqrt{1} = 1$,\n\n$\\therefore$ The area of the large square $ABCD$ is $6 + 2 = 8$,\n\n$\\therefore$ The side length of the large square is: $AB = \\sqrt{8} = 2 \\sqrt{2}$,\n\n$\\therefore BF = AB + AF = 2 \\sqrt{2} + 1$,\n\n$\\because$ The four pentagons are congruent,\n\n$\\therefore EF = BE = \\frac{1}{2} BF = \\frac{1}{2}(2 \\sqrt{2} + 1) = \\sqrt{2} + \\frac{1}{2}$,\n\n$\\therefore AE = EF = AF = \\sqrt{2} - \\frac{1}{2}$.\n\nTherefore, the answer is: $\\sqrt{2} - \\frac{1}{2}$.\n\n【Key Insight】This problem mainly examines the grid diagram and the Pythagorean theorem, as well as the area of squares. The key to solving the problem lies in determining the area of the four congruent pentagons and the area of the large square." }, { "problem_id": 1438, "question": "The tangram is known as the \"Oriental magic\" by Westerners. The two figures below are formed by the same set of tangram pieces. It is known that the side length of the square formed by the tangram (as shown in Figure 1) is $a(\\mathrm{~cm})$. If the shaded area in the \"little fox\" pattern of Figure 2 is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0086_1.jpg", "batch21-2024_06_15_1a48b6fa7bfae1347d11g_0086_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the side length of the shaded small square be \\( x \\mathrm{~cm} \\).\n\nAccording to the problem, we have:\n\\[\n\\frac{1}{2}(2x + 4x) \\cdot x = 6\n\\]\nSolving this equation, we find:\n\\[\nx = \\sqrt{2} \\quad \\text{or} \\quad x = -\\sqrt{2} \\quad (\\text{discarded})\n\\]\nTherefore, the side length of the small square is \\( \\sqrt{2} \\mathrm{~cm} \\).\n\nThe diagonal of the large square is then:\n\\[\n4 \\times \\sqrt{2} = 4\\sqrt{2} \\mathrm{~cm}\n\\]\nThus, the side length \\( a \\) of the large square is:\n\\[\na = \\frac{4\\sqrt{2}}{\\sqrt{2}} = 4 \\mathrm{~cm}\n\\]\nHence, the answer is: 4.\n\n**Key Insight:** This problem primarily tests knowledge of the tangram. Mastering the relationships between the sides of the tangram pieces is crucial for solving the problem." }, { "problem_id": 1439, "question": "Figure (1) shows an octagram-shaped cardboard with eight right angles and eight equal obtuse angles, and all sides are equal. As shown in Figure (2), the cardboard is cut along the dotted lines and seamlessly pieced together to form a large square as shown in Figure (3), with an area of $8 + 4\\sqrt{2}$. The length of line segment $AB$ in Figure (3) is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch21-2024_06_15_1c7aa28f72aa80a6bc06g_0009_1.jpg", "batch21-2024_06_15_1c7aa28f72aa80a6bc06g_0009_2.jpg", "batch21-2024_06_15_1c7aa28f72aa80a6bc06g_0009_3.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}+1 \\# \\# 1+\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Let the side length of the octagon be $a$.\n\nAccording to the problem, we have:\n\n\\[\n4 \\times \\frac{1}{2} \\cdot a \\cdot a + (2a + \\sqrt{2}a)^{2} = 8 + 4\\sqrt{2}\n\\]\n\nSimplifying, we get:\n\n\\[\n2a^{2} + (2a + \\sqrt{2}a)^{2} = 8 + 4\\sqrt{2}\n\\]\n\nExpanding the square:\n\n\\[\n2a^{2} + 4a^{2} + 4\\sqrt{2}a^{2} + 2a^{2} = 8 + 4\\sqrt{2}\n\\]\n\nCombining like terms:\n\n\\[\n8a^{2} + 4\\sqrt{2}a^{2} = 8 + 4\\sqrt{2}\n\\]\n\nFactoring out $a^{2}$:\n\n\\[\na^{2}(8 + 4\\sqrt{2}) = 8 + 4\\sqrt{2}\n\\]\n\nDividing both sides by $(8 + 4\\sqrt{2})$:\n\n\\[\na^{2} = 1\n\\]\n\nSince $a > 0$, we have:\n\n\\[\na = 1\n\\]\n\nTherefore, the length of $AB$ is:\n\n\\[\nAB = a + \\sqrt{2}a = 1 + \\sqrt{2}\n\\]\n\nThus, the answer is:\n\n\\[\n\\boxed{\\sqrt{2} + 1}\n\\]\n\n**Note:** This problem tests knowledge of geometric shapes and their properties, particularly the concept of area conservation when shapes are rearranged. The key to solving it is to set up an equation based on the area before and after the rearrangement, which is a common type of problem in middle school mathematics competitions." }, { "problem_id": 1440, "question": "As shown, in rectangle $A B C D(A D>A B)$, equilateral triangles are constructed inward from sides $A D$ and $B C$ (Figure 1); equilateral triangles are constructed inward from sides $A B$ and $C D$ (Figure 2). The overlapping parts of the two equilateral triangles are shaded. Let the area of the shaded part in Figure 1 be $S_{1}$, and the area of the shaded part in Figure 2 be $S_{2}$. If $\\frac{S_{1}}{S_{2}}=8$, then the value of $\\frac{A D}{A B}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_1c7aa28f72aa80a6bc06g_0061_1.jpg", "batch21-2024_06_15_1c7aa28f72aa80a6bc06g_0061_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{3 \\sqrt{3}}{4} \\# \\# \\frac{3}{4} \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( AD = BC = a \\) and \\( AB = CD = b \\), as shown in Figure 1.\n\n\n\nFigure 1\n\nGiven: \\( \\angle ADN = \\angle BCH = 60^\\circ \\),\n\nThus, \\( \\angle NDC = \\angle HCD = 30^\\circ \\).\n\nTherefore, \\( FD = FC \\).\n\nSince quadrilateral \\( ABCD \\) is a rectangle,\n\n\\( AD \\parallel BC \\),\n\nThus, \\( \\angle FNC = \\angle ADN = 60^\\circ \\).\n\nTherefore, \\( \\triangle FNC \\) is an equilateral triangle.\n\nThus, \\( FN = FC \\),\n\nHence, \\( FN = FD \\).\n\nTherefore, the area of \\( \\triangle FNC \\) equals the area of \\( \\triangle FDC \\), which is half the area of \\( \\triangle DCN \\).\n\nIn right triangle \\( \\triangle DNC \\), \\( DN = 2CN \\),\n\nBy the Pythagorean theorem, \\( NC = \\frac{\\sqrt{3}}{3} b \\).\n\nThus, the area of \\( \\triangle FNC \\) equals the area of \\( \\triangle FDC \\), which is half the area of \\( \\triangle DCN \\), calculated as \\( \\frac{1}{2} \\times \\frac{1}{2} NC \\cdot CD = \\frac{\\sqrt{3}}{12} b^2 \\).\n\nSimilarly: \\( S \\triangle DHF = S \\triangle AGE = S \\triangle ABE = S \\triangle BEM = \\frac{\\sqrt{3}}{12} b^2 \\).\n\nTherefore, \\( S_1 = S_{\\text{rectangle}} ABCD - S \\triangle NFC - S \\triangle DFC - S \\triangle DHF - S \\triangle MBE - S \\triangle ABE - S \\triangle AGE = ab - \\frac{\\sqrt{3}}{2} b^2 \\).\n\nAs shown in Figure 2, draw \\( HM \\perp AD \\) at point \\( M \\), and draw \\( GN \\perp AB \\) at point \\( N \\),\n\n\n\nFigure 2\n\nGiven: \\( \\angle HEF = \\angle HGF = \\angle GAB = \\angle EDC = 60^\\circ \\),\n\n\\( GA = AB = CD = ED = EC = GB \\).\n\nThus, \\( \\angle HAD = \\angle HDA = 30^\\circ \\),\n\nTherefore, \\( HA = HD \\).\n\nSince \\( HM \\perp AD \\),\n\nThus, \\( AM = \\frac{1}{2} AD = \\frac{1}{2} a \\).\n\nBy the Pythagorean theorem, \\( MH = \\frac{\\sqrt{3}}{6} a \\),\n\nTherefore, the area of \\( \\triangle HAD = \\frac{1}{2} \\cdot AD \\cdot MH = \\frac{\\sqrt{3}}{12} a^2 \\).\n\nSimilarly: \\( S \\triangle FBC = \\frac{\\sqrt{3}}{12} a^2 \\).\n\nSince \\( \\triangle GAB \\) is an equilateral triangle, and \\( GN \\perp AB \\),\n\nThus, \\( AN = \\frac{1}{2} AB = \\frac{1}{2} b \\),\n\nSince \\( AG = AB = b \\),\n\nThus, \\( GN = \\sqrt{AG^2 - AN^2} = \\frac{\\sqrt{3}}{2} b \\).\n\nTherefore, the area of \\( \\triangle ABG = \\frac{1}{2} AB \\cdot NG = \\frac{1}{2} b \\cdot \\frac{\\sqrt{3}}{2} b = \\frac{\\sqrt{3}}{4} b^2 \\).\n\nSimilarly: \\( S \\triangle CDE = \\frac{\\sqrt{3}}{4} b^2 \\).\n\nTherefore, \\( S_2 = S \\triangle ABG + S \\triangle CDE + S \\triangle ADH + S \\triangle BFC - \\text{Area of } ABCD = \\frac{\\sqrt{3}}{6} a^2 + \\frac{\\sqrt{3}}{2} b^2 - ab \\).\n\nGiven that \\( \\frac{S_1}{S_2} = 8 \\),\n\nThus, \\( \\frac{ab - \\frac{\\sqrt{3}}{2} b^2}{\\frac{\\sqrt{3}}{6} a^2 + \\frac{\\sqrt{3}}{2} b^2 - ab} = \\frac{8}{1} \\),\n\nTherefore, \\( 8 \\sqrt{3} a^2 - 54 ab + 27 \\sqrt{3} b^2 = 0 \\).\n\nSolving gives: \\( a = \\frac{3 \\sqrt{3}}{2} b \\) or \\( a = \\frac{3 \\sqrt{3}}{4} b \\).\n\nGiven that \\( a < \\sqrt{3} b \\),\n\nThus, \\( a = \\frac{3 \\sqrt{3}}{4} b \\).\n\nTherefore, \\( \\frac{AD}{AB} = \\frac{a}{b} = \\frac{3 \\sqrt{3}}{4} \\).\n\nThe answer is: \\( \\frac{3 \\sqrt{3}}{4} \\).\n\n【Highlight】This problem primarily examines the properties of rectangles, equilateral triangles, isosceles triangles, right triangles with a 30-degree angle, and the Pythagorean theorem. The key to solving the problem lies in expressing the area of the shaded region using the sum or difference of known areas." }, { "problem_id": 1441, "question": "As shown in the figure, the edge length of the cube wooden block is $3 \\mathrm{~cm}$. A corner is cut off along the diagonals of three adjacent faces (dashed lines in the figure), resulting in the geometric shape shown in Figure (2). The shortest distance for an ant to crawl along the surface of the geometric shape in Figure (2) from vertex $A$ to vertex $B$ is $\\mathrm{cm}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_250de0fcb6ff3e1054beg_0076_1.jpg", "batch21-2024_06_15_250de0fcb6ff3e1054beg_0076_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{3}{2}(\\sqrt{2}+\\sqrt{6})$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, the cross-section and the top surface are unfolded in the same plane, connecting $AB$ to intersect $CD$ at $E$. According to the principle that the shortest distance between two points is a straight line, the length of $AB$ is the required distance.\n\nGiven the conditions, $\\triangle ACD$ is an equilateral triangle, and $\\triangle BCD$ is an isosceles right triangle.\n\nThus, $BC = CD = 3 \\text{ cm}$, and $AC = AD = CD = \\sqrt{3^2 + 3^2} = 3\\sqrt{2} \\text{ cm}$.\n\nSince $BC = BD$, $AC = AD$, and $AB = AB$, it follows that $\\triangle ABC \\cong \\triangle ABD$ by the SSS (Side-Side-Side) congruence criterion.\n\nTherefore, $\\angle CBA = \\angle DBA$ and $\\angle CAB = \\angle DAB$, which implies that $AB \\perp CD$.\n\nHence, $CE = \\frac{1}{2} CD = \\frac{3}{2}\\sqrt{2} \\text{ cm}$, and $BE = \\frac{1}{2} CD = \\frac{3}{2}\\sqrt{2} \\text{ cm}$.\n\nThus, $AE = \\sqrt{AC^2 - CE^2} = \\frac{3}{2}\\sqrt{6} \\text{ cm}$.\n\nTherefore, $AB = AE + BE = \\frac{3}{2}(\\sqrt{2} + \\sqrt{6}) \\text{ cm}$.\n\nThe final answer is: $\\frac{3}{2}(\\sqrt{2} + \\sqrt{6})$.\n\n**Key Insight:** This problem primarily examines the principle that the shortest distance between two points is a straight line, the properties and congruence criteria of triangles, the property that the median to the hypotenuse of a right triangle is half the hypotenuse, the properties of squares, and the Pythagorean theorem. The key to solving the problem lies in a thorough understanding of the principle that the shortest distance between two points is a straight line." }, { "problem_id": 1442, "question": "The Pythagorean theorem holds an irreplaceable important position in plane geometry. In the ancient Chinese mathematical book Zhoubi Suanjing, there is a record stating \"if the gou is three, the gu is four, then the xian is five.\" As shown in Figure 1, it consists of small squares with side lengths of 1 and the right triangle Rt $\\triangle A B C$. The Pythagorean theorem can be verified using the area relationship. If Figure 1 is \"embedded\" into the rectangle $L M J K$ as shown in Figure 2, then the area of the rectangle is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_2abe4376579ee612d364g_0097_1.jpg", "batch21-2024_06_15_2abe4376579ee612d364g_0097_2.jpg" ], "is_multi_img": true, "answer": "110", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nExtend \\( AB \\) to intersect \\( KF \\) at point \\( O \\), and extend \\( AC \\) to intersect \\( GM \\) at point \\( P \\), as shown in Figure 3:\n\n\n\nThen, quadrilateral \\( OAPL \\) is a rectangle.\n\nSince \\( \\angle CBF = 90^\\circ \\),\n\\[ \\angle ABC + \\angle OBF = 90^\\circ. \\]\n\nAlso, in the right triangle \\( \\triangle ABC \\),\n\\[ \\angle ABC + \\angle ACB = 90^\\circ. \\]\n\nTherefore,\n\\[ \\angle OBF = \\angle ACB. \\]\n\nIn triangles \\( \\triangle OBF \\) and \\( \\triangle ACB \\),\n\\[\n\\begin{cases}\n\\angle BAC = \\angle BOF, \\\\\n\\angle ACB = \\angle OBF, \\\\\nBC = BF.\n\\end{cases}\n\\]\n\nThus,\n\\[ \\triangle OBF \\cong \\triangle ACB \\quad (\\text{by AAS}). \\]\n\nHence,\n\\[ AC = OB. \\]\n\nSimilarly,\n\\[ \\triangle ACB \\cong \\triangle PGC. \\]\n\nTherefore,\n\\[ PC = AB. \\]\n\nSo,\n\\[ OA = AP. \\]\n\nThus, rectangle \\( AOLP \\) is a square with side length\n\\[ AO = AB + AC = 3 + 4 = 7. \\]\n\nConsequently,\n\\[ KL = 3 + 7 = 10, \\quad LM = 4 + 7 = 11. \\]\n\nTherefore, the area of rectangle \\( KLMJ \\) is\n\\[ 10 \\times 11 = 110. \\]\n\n**Answer:** \\(\\boxed{110}\\).\n\n**Key Points:**\nThis problem examines the proof of the Pythagorean theorem, properties of rectangles, the determination and properties of squares, and the properties of congruent triangles. The key to solving the problem lies in constructing auxiliary lines to prove triangle congruence, leading to the identification of the square." }, { "problem_id": 1443, "question": "As shown in the figure, Figure (1) is a triangle. Connecting the midpoints of the three sides of this triangle yields the second figure (Figure (2)). Then, connecting the midpoints of the three sides of the smaller triangle in the center of Figure (2) results in the third figure (Figure (3)), and so on. According to this pattern, the $n$th figure ($n>1$) contains $\\qquad$ parallelograms.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch21-2024_06_15_324fa18eb13fa442f5f6g_0086_1.jpg", "batch21-2024_06_15_324fa18eb13fa442f5f6g_0086_2.jpg", "batch21-2024_06_15_324fa18eb13fa442f5f6g_0086_3.jpg" ], "is_multi_img": true, "answer": "$3 n-3$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Count the number of parallelograms in Figure (1), Figure (2), and Figure (3) respectively.\n\nThe number of parallelograms in Figure (1) is $0 = 3 \\times (1 - 1)$;\n\nThe number of parallelograms in Figure (2) is $3 = 3 \\times (2 - 1)$;\n\nThe number of parallelograms in Figure (3) is $6 = 3 \\times (3 - 1)$;\n\nIt can be observed that the number of parallelograms in the nth figure is the product of 3 and $(n - 1)$.\n\nFollowing this pattern, the total number of parallelograms in the nth figure is $3n - 3$.\n\nTherefore, the answer is: $3n - 3$.\n\n[Key Insight] This question primarily tests students' understanding and mastery of the knowledge point related to pattern recognition in graphical changes. The key to solving such problems lies in carefully analyzing the given figures and data, and then deducing the underlying pattern through thoughtful consideration and summarization." }, { "problem_id": 1444, "question": "Fold the rectangular paper $ABCD (AB < BC)$ along a line passing through point $B$ such that point $A$ falls on point $F$ on side $BC$, with the crease being $BE$ (as shown in Figure 1); then fold along a line passing through point $E$ such that point $D$ falls on point $D'$ on $BE$, with the crease being $EG$ (as shown in Figure 2); finally, unfold the paper (as shown in Figure 3), then the size of $\\angle FEG$ in Figure 3 is .\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_3795bbc3649f3c598283g_0057_1.jpg", "batch21-2024_06_15_3795bbc3649f3c598283g_0057_2.jpg", "batch21-2024_06_15_3795bbc3649f3c598283g_0057_3.jpg" ], "is_multi_img": true, "answer": "$22.5^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nFrom the folding, we know that $\\triangle AEB \\cong \\triangle FEB$.\n\nThus, $\\angle A = \\angle EFB = 90^\\circ$, and $AB = BF$.\n\nSince the sheet $ABCD$ is a rectangle,\n\nit follows that $AE \\parallel BF$.\n\nTherefore, $\\angle AEF = 180^\\circ - \\angle BFE = 90^\\circ$.\n\nGiven that $AB = BF$ and $\\angle A = \\angle AEF = \\angle EFB = 90^\\circ$,\n\nthe quadrilateral $ABFE$ is a square.\n\nHence, $\\angle AEB = 45^\\circ$,\n\nand $\\angle BED = 180^\\circ - 45^\\circ = 135^\\circ$.\n\nThus, $\\angle BEG = 135^\\circ \\div 2 = 67.5^\\circ$,\n\nand $\\angle FEG = 67.5^\\circ - 45^\\circ = 22.5^\\circ$.\n\n**Key Insight:** This problem examines the properties of folding, rectangles, and squares, as well as the properties of parallel lines. The key to solving this problem lies in combining the properties of squares and rectangles." }, { "problem_id": 1445, "question": "The tangram is an outstanding creation of our ancestors, known as the \"Magic Plate of the East.\" Xiao Ming used the tangram (as shown in Figure 1) to form a convex hexagon by leveraging the relationships between the side lengths of the pieces (as shown in Figure 2). The perimeter of this convex hexagon is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_47b0f811d85dbecc9dd5g_0078_1.jpg", "batch21-2024_06_15_47b0f811d85dbecc9dd5g_0078_2.jpg" ], "is_multi_img": true, "answer": "$32 \\sqrt{2}+16$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in the figure:\n\n- Figure 1: Side lengths are: $16, 8\\sqrt{2}, 8\\sqrt{2}$;\n- Figure 2: Side lengths are: $16, 8\\sqrt{2}, 8\\sqrt{2}$;\n- Figure 3: Side lengths are: $8, 4\\sqrt{2}, 4\\sqrt{2}$;\n- Figure 4: Side length is: $4\\sqrt{2}$;\n- Figure 5: Side lengths are: $8, 4\\sqrt{2}, 4\\sqrt{2}$;\n- Figure 6: Side lengths are: $4\\sqrt{2}, 8$;\n- Figure 7: Side lengths are: $8, 8, 8\\sqrt{2}$.\n\nTherefore, the perimeter of the convex hexagon is:\n\\[\n8 + 2 \\times 8\\sqrt{2} + 8 + 4\\sqrt{2} \\times 4 = 32\\sqrt{2} + 16 \\text{ cm}.\n\\]\n\n\n\nThus, the answer is \\(32\\sqrt{2} + 16\\).\n\n**Key Insight:** This problem tests the properties of squares, the Pythagorean theorem, and the properties of isosceles right triangles. Mastering the properties of squares and calculating the side lengths of each segment are crucial to solving the problem." }, { "problem_id": 1446, "question": "In the comprehensive practical activity class, Xiao Liang cut a sharp-angled triangle paper with an area of $6 \\mathrm{~cm}^{2}$ and one side $BC$ of $4 \\mathrm{~cm}$ (as shown in Figure 1) into two pieces, which were then seamlessly and without overlap assembled into a rectangle $BCDE$ (as shown in Figure 2). The perimeter of the rectangle is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_488728a8971d5024fb53g_0017_1.jpg", "batch21-2024_06_15_488728a8971d5024fb53g_0017_2.jpg" ], "is_multi_img": true, "answer": "$11 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "As shown in the figure, extend $AH$ to intersect $BC$ at point $M$.\n\nGiven: $\\triangle AGH \\cong \\triangle CGD$, $\\triangle AFH \\cong \\triangle BFE$.\n\nTherefore, $AG = CG$, $AH = CD$, $AF = BF$, $BE = AH$.\n\nSince quadrilateral $BCDE$ is a rectangle and $AM \\perp BC$,\n\nIt follows that $AH = CD = BE = HM$.\n\nThus, $AM = 2CD$.\n\nGiven that $\\frac{1}{2} BC \\cdot AM = 6$ and $BC = 4 \\mathrm{~cm}$,\n\nWe find $AM = 3 \\mathrm{~cm}$.\n\nTherefore, the perimeter of rectangle $BCDE$ is: $2BC + 2CD = 2 \\times 4 + 3 = 11 \\mathrm{~cm}$.\n\nHence, the answer is: $11 \\mathrm{~cm}$.\n\n\n\nFigure 2\n\n[Key Insight] This problem examines the concepts of graphical dissection and complementation, properties of rectangles, and properties of congruent triangles. The key is to deduce that $AH = CD = BE = HM$." }, { "problem_id": 1447, "question": "Divide the rhombus in Figure (1) along its diagonals into four congruent right triangles, and arrange these four right triangles into squares as shown in Figures (2) and (3) (the side length of the large square in Figure (2) is 5, and the side length of the small square in the middle of Figure (3) is 1). Then the area of the rhombus in Figure (1) is . $\\qquad$\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch21-2024_06_15_4ef222f18e483571ea8cg_0085_1.jpg", "batch21-2024_06_15_4ef222f18e483571ea8cg_0085_2.jpg", "batch21-2024_06_15_4ef222f18e483571ea8cg_0085_3.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Let the length of the longer diagonal of the rhombus be \\(2a\\), and the length of the shorter diagonal be \\(2b\\).\n\nAccording to the two types of puzzles, we have the system of equations:\n\\[\n\\begin{cases}\na + b = 5 \\\\\na - b = 1\n\\end{cases}\n\\]\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\nSolving the system, we find:\n\\[\n\\begin{cases}\na = 3 \\\\\nb = 2\n\\end{cases}\n\\]\n\nTherefore, the area of the rhombus is:\n\\[\n\\frac{1}{2} \\times 2a \\times 2b = 2ab = 2 \\times 3 \\times 2 = 12\n\\]\n\nHence, the answer is: 12.\n\n【Key Insight】This problem tests the properties of a rhombus and systems of equations. Mastering the properties of a rhombus and the method of solving systems of equations is crucial for solving the problem." }, { "problem_id": 1448, "question": "Divide the rhombus in Figure 1 along its diagonals into four congruent right triangles, and arrange these four right triangles to form the \"Zhao Shuang's Xian Diagram\" as shown in Figure 2. If the area of the small square in Figure 2 is 4, and the gōu (a) = 6, then the perimeter of the rhombus in Figure 1 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_5d36f644882161b318dbg_0067_1.jpg", "batch21-2024_06_15_5d36f644882161b318dbg_0067_2.jpg" ], "is_multi_img": true, "answer": "40", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since the area of the small square is 4,\n\nTherefore, the side length of the small square is 2,\n\nThus, \\( b = a + 2 = 6 + 2 = 8 \\),\n\nHence, \\( c = \\sqrt{a^{2} + b^{2}} = \\sqrt{6^{2} + 8^{2}} = 10 \\),\n\nTherefore, the perimeter of the rhombus is: \\( 4c = 4 \\times 10 = 40 \\),\n\nSo the answer is: 40.\n\n[Highlight] This question examines the properties of a rhombus and the Pythagorean theorem; mastering the properties of a rhombus is key to solving the problem." }, { "problem_id": 1449, "question": "As shown in Figure (1), Xiao Gang cuts along the lines connecting the midpoints of the sides of the rhombus paper $ABCD$ to obtain the quadrilateral paper $EFGH$. Then, according to Figure (2), the paper $EFGH$ is folded along $MN$ and $PQ$ respectively. When $PN \\parallel EF$, if the sum of the perimeters of the shaded areas is $16$, and the sum of the areas of $\\triangle AEH$ and $\\triangle CFG$ is $12$, then the length of one diagonal $BD$ of the rhombus paper $ABCD$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_5e1517d0f57673325b7eg_0021_1.jpg", "batch21-2024_06_15_5e1517d0f57673325b7eg_0021_2.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Connect $BD$ as shown in the figure:\n\n\n\nSince quadrilateral $ABCD$ is a rhombus,\n\n$\\therefore AB = AD$, and $AC$ is perpendicularly bisected by $BD$,\n\nSince $E$ is the midpoint of $AB$, and $H$ is the midpoint of $AD$,\n\n$\\therefore AE = AH$, and $EH$ is the midline of $\\triangle ABD$,\n\n$\\therefore EN = HN$, and $BD = 2EH = 4HN$,\n\nAccording to the problem, let $AN = PC = x$, and $EN = HN = PF = PG = y$.\n\nThen we have the system of equations:\n\\[\n\\left\\{\n\\begin{array}{c}\n2 \\times \\frac{1}{2} \\times 2y \\times x = 12 \\\\\n4y + 4(2x - y) = 16\n\\end{array}\n\\right.\n\\]\nSolving the system gives:\n\\[\n\\left\\{\n\\begin{array}{l}\nx = 2 \\\\\ny = 3\n\\end{array}\n\\right.\n\\]\n$\\therefore AN = 2$, and $HN = 3$,\n\n$\\therefore BD = 4HN = 12$;\n\nThus, the answer is: 12.\n\n【Insight】This problem examines the properties of a rhombus, the determination and properties of a rectangle, the midline theorem of a triangle, and the method of solving systems of equations. The key to solving the problem lies in constructing equations using parameters." }, { "problem_id": 1450, "question": "As shown in Figure (1), fold the rectangular paper along a line passing through point $\\mathrm{B}$, such that point $\\mathrm{A}$ falls onto point $\\mathrm{F}$ on side $\\mathrm{BC}$, with the crease being $\\mathrm{BE}$. Then, fold along a line passing through point $\\mathrm{E}$, such that point $\\mathrm{D}$ falls onto point $\\mathrm{M}$ on $\\mathrm{BE}$, with the crease being $\\mathrm{EG}$, as shown in Figure (2). In Figure (2), $\\angle \\mathrm{EGC} =$ _ degrees.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_64174d282920bed1586bg_0026_1.jpg", "batch21-2024_06_15_64174d282920bed1586bg_0026_2.jpg" ], "is_multi_img": true, "answer": "112.5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the folding, we can deduce that $\\angle \\mathrm{EBF}=\\frac{1}{2} \\angle \\mathrm{ABF}=45^{\\circ}$.\n\nSince $\\mathrm{AD} \\parallel \\mathrm{BC}$,\n\nit follows that $\\angle \\mathrm{BED}+\\angle \\mathrm{EBF}=180^{\\circ}$,\n\nhence $\\angle \\mathrm{BED}=135^{\\circ}$.\n\nFrom the folding, we can also deduce that $\\angle \\mathrm{BEG}=\\frac{1}{2} \\angle \\mathrm{BED}=67.5^{\\circ}$.\n\nTherefore, $\\angle \\mathrm{EGC}=\\angle \\mathrm{EBF}+\\angle \\mathrm{BEG}=45^{\\circ}+67.5^{\\circ}=112.5^{\\circ}$.\n\nThus, the answer is 112.5.\n\n【Highlight】This problem mainly examines the properties of parallel lines and the application of symmetry properties. When solving, note that for two parallel lines, the consecutive interior angles are supplementary." }, { "problem_id": 1451, "question": "As shown in Figure 1, in rectangle $\\mathrm{ABCD}$, point $\\mathrm{E}$ is on side $\\mathrm{AD}$, $A D / / B C, \\angle \\mathrm{A}=\\angle \\mathrm{D}=90^{\\circ}, \\angle \\mathrm{BEA}=60^{\\circ}$. Now, folding along lines $\\mathrm{BE}$ and $\\mathrm{CE}$, points $\\mathrm{A}$ and $\\mathrm{D}$ are folded towards $\\mathrm{BC}$, resulting in the position diagram of points $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}, \\mathrm{D}$, and $\\mathrm{E}$ on the same plane as shown in Figure 2. If $\\angle \\mathrm{AED}=15^{\\circ}$, then the degree measure of $\\angle \\mathrm{BCE}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_64174d282920bed1586bg_0092_1.jpg", "batch21-2024_06_15_64174d282920bed1586bg_0092_2.jpg" ], "is_multi_img": true, "answer": "$37.5^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\n(1)\n\n\n\n(2)\n\nIn rectangle $\\mathrm{ABCD}$, $\\angle \\mathrm{A}=90^{\\circ}$, $\\mathrm{AD} / / \\mathrm{BC}$,\n\nSince $\\mathrm{BE}=2 \\mathrm{AE}$,\n\nTherefore, $\\angle \\mathrm{ABE}=30^{\\circ}$,\n\nThus, $\\angle \\mathrm{AEB}=90^{\\circ}-\\angle \\mathrm{ABE}=90^{\\circ}-30^{\\circ}=60^{\\circ}$,\n\nSince $\\angle \\mathrm{AED}=15^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{BED}=\\angle \\mathrm{AEB}-\\angle \\mathrm{AED}=60^{\\circ}-15^{\\circ}=45^{\\circ}$,\n\nHence, $\\angle \\mathrm{DED}^{\\prime}=180^{\\circ}-60^{\\circ}-45^{\\circ}=75^{\\circ}$,\n\nAccording to the property of folding, $\\angle \\mathrm{CED}^{\\prime}=\\frac{1}{2} \\angle \\mathrm{DED}^{\\prime}=\\frac{1}{2} \\times 75^{\\circ}=37.5^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{BCE}=\\angle \\mathrm{CED}^{\\prime}=37.5^{\\circ}$.\n\nThe answer is: $37.5^{\\circ}$.\n\n【Highlight】This question examines the area of a rectangle, the properties of folding transformations, and the property that in a right-angled triangle, the side opposite the $30^{\\circ}$ angle is half the hypotenuse. Familiarity with these properties and accurate interpretation of the diagram are key to solving the problem." }, { "problem_id": 1452, "question": "As shown in Figure 1, in rectangle $A B C D$, point $E$ is on $A D$, and point $A$ is folded to the right along the fold line $B E$ as shown in Figure 2. Through point $A$, $A F \\perp D C$ at $F$. If $A B=6 \\sqrt{3}, B C=13, \\angle B E A=60^{\\circ}$, then in Figure 3, $A F=$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_65427ff9f2d9ed3cf7b3g_0046_1.jpg", "batch21-2024_06_15_65427ff9f2d9ed3cf7b3g_0046_2.jpg", "batch21-2024_06_15_65427ff9f2d9ed3cf7b3g_0046_3.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular from point $A$ to $BC$ at point $H$, as shown in the figure:\n\n\n\nFigure 3\n\nSince quadrilateral $ABCD$ is a rectangle,\n\n$\\therefore \\angle A=\\angle ABC=\\angle C=90^{\\circ}$,\n\nGiven that $\\angle BEA=60^{\\circ}$,\n$\\therefore \\angle EBA=30^{\\circ}$,\n\nFrom the properties of folding, we have: $\\angle ABH=90^{\\circ}-2 \\angle EBA=90^{\\circ}-60^{\\circ}=30^{\\circ}$,\n\nIn right triangle $\\triangle ABH$, $AB=6 \\sqrt{3}$, $\\angle AHB=90^{\\circ}$, $\\angle ABH=30^{\\circ}$,\n\n$\\therefore AH=3 \\sqrt{3}$,\n\n$\\therefore BH=\\sqrt{AB^{2}-AH^{2}}=9$,\n\n$\\therefore CH=BC-BH=13-9=4$,\n\nSince $AF \\perp CD$,\n\n$\\therefore \\angle AHC=\\angle C=\\angle AFC=90^{\\circ}$,\n\n$\\therefore$ quadrilateral $AFCH$ is a rectangle,\n\n$\\therefore AF=CH=4$.\n\nTherefore, the answer is: 4.\n\n[Highlight] This problem examines the knowledge of flip transformations, properties of rectangles, and the Pythagorean theorem. The key to solving the problem is to add common auxiliary lines and construct right triangles to solve the problem, which is a common type of question in middle school exams." }, { "problem_id": 1453, "question": "As shown in Figure 1, a line segment $M N$ is drawn on a rectangular paper $A B C D$, and the paper is folded along the line segment $M N$ (as shown in Figure 2). When $\\angle 1=70^{\\circ}$, $\\angle K N C=$ $\\qquad$ ${ }^{\\circ}$ (Note: The opposite sides of the rectangular paper are parallel, i.e., $C D / / A B, A D / / B C$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_65427ff9f2d9ed3cf7b3g_0091_1.jpg", "batch21-2024_06_15_65427ff9f2d9ed3cf7b3g_0091_2.jpg" ], "is_multi_img": true, "answer": "40", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since quadrilateral \\( ABCD \\) is a rectangle,\n\ntherefore \\( AM \\parallel DN \\).\n\nThus, \\( \\angle KNM = \\angle 1 \\).\n\nGiven that \\( \\angle 1 = 70^\\circ \\),\n\nit follows that \\( \\angle KNM = \\angle 1 = 70^\\circ \\).\n\nDue to the folding,\n\n\\( \\angle KNM = \\angle KMN = \\angle 1 = 70^\\circ \\),\n\nhence \\( \\angle MKN = 40^\\circ \\).\n\nSince \\( BM \\parallel CN \\),\n\n\\( \\angle KNC = \\angle MKN = 40^\\circ \\).\n\nTherefore, the answer is: 40.\n\n**Key Insight:** This problem tests the properties of rectangles and the transformations due to folding. The key to solving it lies in utilizing the invariance properties of folding, a common topic in middle school exams." }, { "problem_id": 1454, "question": "Tangram is an ancient Chinese traditional intellectual game. Xiao Ming used a tangram (as shown in Figure 1) to form the number \"7\" (as shown in Figure 2). If the area of the square $A B C D$ in Figure 1 is $32 \\mathrm{~cm}^{2}$, then the perimeter of Figure 2 is $\\qquad$ $\\mathrm{cm}$.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0056_1.jpg", "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0056_2.jpg" ], "is_multi_img": true, "answer": "36", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: $\\sqrt{32}=4 \\sqrt{2} \\mathrm{~cm}$,\n\n$4 \\sqrt{2} \\times \\frac{\\sqrt{2}}{2}=4 \\mathrm{~cm}$\n\n$4 \\div 2=2 \\mathrm{~cm}$,\n\n$4+4 \\sqrt{2}+2+2+2+2+(4-2 \\sqrt{2}) \\times 2+4+2+2 \\times 3+4=36 \\mathrm{~cm}$.\n\nAnswer: The perimeter of Figure 2 is $36 \\mathrm{~cm}$.\n\nTherefore, the answer is 36.\n\n[Key Insight] This problem mainly examines the relationship between the area and side length of a square, as well as the length relationships between the sides of an isosceles right-angled triangle." }, { "problem_id": 1455, "question": "As shown in Figure (1), a small square with side length $\\mathrm{b}$ is cut out from a large square with side length $\\mathrm{a}$. The shaded area in the figure is then cut and rearranged into a rectangle, as shown in Figure (2). The length of this rearranged rectangle is 24, and its width is 12. Therefore, the area of the part in Figure (2) is . $\\qquad$\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0066_1.jpg", "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0066_2.jpg" ], "is_multi_img": true, "answer": "72", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "According to the problem, we have the system of equations:\n\n\\[\n\\left\\{\n\\begin{array}{l}\na + b = 24 \\\\\na - b = 12\n\\end{array}\n\\right.\n\\]\n\nSolving the system, we find:\n\n\\[\n\\left\\{\n\\begin{array}{l}\na = 18 \\\\\nb = 6\n\\end{array}\n\\right.\n\\]\n\nTherefore, the area of part II in Figure (2) is:\n\n\\[\n6 \\times 12 = 72\n\\]\n\nThus, the answer is **72**.\n\n**Key Insight**: This problem tests the properties of squares, with the key to solving it lying in deriving the equations \\( a + b = 24 \\) and \\( a - b = 12 \\)." }, { "problem_id": 1456, "question": "As shown in Figure (1), it is given that the area of the small square \\(ABCD\\) is 1. Extend each of its sides by a factor of two to obtain the new square \\(A_1B_1C_1D_1\\); extend the sides of square \\(A_1B_1C_1D_1\\) by the same method to obtain square \\(A_2B_2C_2D_2\\) as shown in Figure (2); continue this process... The area of square \\(A_5B_5C_5D_5\\) is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0088_1.jpg", "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0088_2.jpg" ], "is_multi_img": true, "answer": "3125", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "In Figure (1), the area of square $\\mathrm{ABCD}$ is 1. After extending each side by double, the area of each small triangle is also 1. \n\nTherefore, the area of square $\\mathrm{A}_{1} \\mathrm{B}_{1} \\mathrm{C}_{1} \\mathrm{D}_{1}$ is 5. \n\nIn Figure (2), the area of square $\\mathrm{A}_{1} \\mathrm{B}_{1} \\mathrm{C}_{1} \\mathrm{D}_{1}$ is 5. After extending each side by double, the area of each small triangle is also 5. \n\nThus, the area of square $\\mathrm{A}_{2} \\mathrm{B}_{2} \\mathrm{C}_{2} \\mathrm{D}_{2}$ is $5^{2}=25$. \n\nFollowing this pattern, the area of square $\\mathrm{A}_{5} \\mathrm{B}_{5} \\mathrm{C}_{5} \\mathrm{D}_{5}$ is $5^{5}=3125$.\n\n【Insight】This problem examines the pattern in geometric shapes, focusing on identifying the rule by listing the areas of each figure." }, { "problem_id": 1457, "question": "As shown in Figure 1, cut along the diagonals $\\mathrm{AC}$ and $\\mathrm{EG}$ of the rectangular paper $\\mathrm{ABCD}$ and the square paper $\\mathrm{EFGH}$ respectively, and form the parallelogram $K L M N$ as shown in Figure 2. If the blank area in the middle is exactly a square $\\mathrm{OPQR}$, and the area of the parallelogram $\\mathrm{KLMN}$ is 50, then the area of the square $\\mathrm{EFGH}$ is $\\qquad$.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0092_1.jpg", "batch21-2024_06_15_6657dcbbf7a396cdcc28g_0092_2.jpg" ], "is_multi_img": true, "answer": "25", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let $\\mathrm{PM}=\\mathrm{PL}=\\mathrm{NR}=\\mathrm{KR}=\\mathrm{a}$, and the side length of square $\\mathrm{ORQP}$ be $\\mathrm{b}$.\n\nAccording to the problem: $a^{2}+b^{2}+(a+b)(a-b)=50$,\n\n$\\therefore \\mathrm{a}^{2}=25$,\n\n$\\therefore$ The area of square $\\mathrm{EFGH}$ $=\\mathrm{a}^{2}=25$,\n\nHence, the answer is: 25.\n\n[Insight] This problem mainly examines the properties of squares and rectangles, as well as the combination of shapes. The key is to use the properties of squares and rectangles to construct and solve the equation." }, { "problem_id": 1458, "question": "Xiao Ming made a movable rhombus tool using four wooden strips of equal length. He first adjusted the tool into the rhombus shown in Figure 1, measured that $\\angle B = 60^\\circ$, and found the length of diagonal $AC$ to be $30 \\text{ cm}$. Then, he adjusted the tool into the square shown in Figure 2. The length of diagonal $AC$ in Figure 2 is $\\qquad$ $\\text{cm}$.\n\n\n\n\\%1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0004_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0004_2.jpg" ], "is_multi_img": true, "answer": "$30 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in Figures 1 and 2, connect $AC$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nIn Figure 1, since $AB = BC$ and $\\angle B = 60^\\circ$,\n\n$\\therefore \\triangle ABC$ is an equilateral triangle,\n\n$\\therefore AB = BC = AC = 30$.\n\nIn Figure 2, since quadrilateral $ABCD$ is a square,\n\n$\\therefore AB = BC$ and $\\angle B = 90^\\circ$,\n\n$\\because AB = BC = 30 \\mathrm{~cm}$,\n\n$\\therefore AC = 30 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the answer is: $30 \\sqrt{2}$.\n\n【Insight】This question tests the properties of rhombuses, squares, and the Pythagorean theorem. The key to solving the problem is the flexible application of the learned knowledge, which is a common type of question in middle school exams." }, { "problem_id": 1459, "question": "A rectangular piece of paper with length 1 and width $\\mathrm{a}$ $(0<\\mathrm{a}<1)$ is folded as shown, cutting off a square with a side length equal to the width of the rectangle (referred to as the first operation); then the remaining rectangle is folded as shown again, cutting off a square with a side length equal to the width of the rectangle at that time (referred to as the second operation); this process is repeated. If after the $\\mathrm{n}$-th operation, the remaining rectangle is a square, the operation stops. When $n=3$, the value of $\\mathrm{a}$ may have $\\qquad$ possibilities.\n\n\n\nFirst fold\n\n\n\nSecond fold", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0012_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0012_2.jpg" ], "is_multi_img": true, "answer": "2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "**Solution:** According to the problem, when \\(0 < a < \\frac{1}{2}\\), after the first operation, the remaining rectangle has a length of \\(a\\) and a width of \\(1 - a\\). After the second operation, the sides of the remaining rectangle are \\(a\\) and \\(1 - 2a\\). At this point, there are two cases to consider:\n\n1. **Case 1:** If \\(1 - 2a > a\\), which implies \\(0 < a < \\frac{1}{3}\\), then the side length of the square after the third operation is \\(1 - 3a\\).\n\n Since the rectangle obtained after the third operation is a square, the width of the rectangle equals \\(a\\), i.e., \\(a = 1 - 3a\\). Solving this equation gives \\(a = \\frac{1}{4}\\).\n\n2. **Case 2:** If \\(1 - 2a < a\\), which implies \\(a > \\frac{1}{3}\\), then the side length of the square after the third operation is \\(1 - 2a\\).\n\n Then, \\(1 - 2a = a - (1 - 2a)\\), solving this equation gives \\(a = \\frac{2}{5}\\).\n\nTherefore, \\(a = \\frac{1}{4}\\) or \\(\\frac{2}{5}\\).\n\n**Answer:** There are 2 possible values.\n\n**Key Insight:** This problem tests the properties of folding and the properties of squares, as well as the application of linear equations. The key to solving the problem lies in considering the two cases separately to determine the sides of the remaining rectangle after the operations." }, { "problem_id": 1460, "question": "As shown in Figure (1), in the rectangular paper piece $ABCD$, $AB = 5, BC = 3$. First, as shown in Figure (2), fold the rectangle $ABCD$ along a line passing through point $A$, such that point $D$ falls on point $E$ on side $AB$, with the crease being $AF$. Then, as shown in Figure (3), fold along a line passing through point $F$, such that point $C$ falls on point $H$ on side $EF$, with the crease being $FG$. The area of the shaded part of the figure is $\\qquad$. \n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0026_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0026_2.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0026_3.jpg" ], "is_multi_img": true, "answer": "2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the folding transformation, we know that quadrilaterals $ADFE$ and $FHGC$ are both squares with side lengths of 3 and 2, respectively.\n\n$\\therefore$ The length of rectangle $BEHG$, $BE=5-3=2$, and its width, $EH=3-2=1$,\n\n$\\therefore$ The area of the shaded region $=2 \\times 1=2$,\n\nHence, the answer is: 2.\n\n[Insight] This problem examines the knowledge of folding transformations, properties of rectangles, and properties of squares. The key to solving the problem lies in the flexible application of the learned knowledge, which is a common type of question in middle school exams." }, { "problem_id": 1461, "question": "Xiao Ming made a movable rhombus tool using four wooden strips of equal length. He first transformed the tool into the rhombus shown in Figure 1, measured $\\angle A = 60^{\\circ}$, and the diagonal $BD = 10 \\mathrm{~cm}$. Then, he transformed the tool into the square shown in Figure 2. The length of the diagonal $AC$ in Figure 2 is . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0039_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0039_2.jpg" ], "is_multi_img": true, "answer": "$10 \\sqrt{2} \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1 and Figure 2, connect $BD$.\n\nIn Figure 1, since quadrilateral $ABCD$ is a rhombus,\n\n$\\therefore AD = AB$,\n\n$\\because \\angle A = 60^\\circ$,\n\n$\\therefore \\triangle ABD$ is an equilateral triangle,\n\n$\\therefore AB = BD = AD = 10 \\text{ cm}$.\n\nIn Figure 2, since quadrilateral $ABCD$ is a square,\n\n$\\therefore AB = AD$, $\\angle A = 90^\\circ$,\n\n$\\therefore \\triangle ABD$ is an isosceles right triangle,\n\n$\\therefore AC = BD = \\sqrt{2} \\cdot AD = 10\\sqrt{2} \\text{ cm}$.\n\nTherefore, the answer is: $10\\sqrt{2} \\text{ cm}$.\n\n\n\nFigure 1\n\nFigure 2\n\n【Insight】This question examines the properties of rhombuses, squares, and the Pythagorean theorem. The key to solving the problem is to be proficient in the properties of rhombuses and squares. This type of question is commonly seen in middle school exams." }, { "problem_id": 1462, "question": "As shown in Figure 1, the ratio of the lengths of the two legs of the right-angled triangle paper is $1: 2$. Four such right-angled triangle papers are cut and arranged into a square as shown in Figure 2. Given that the area of the small blank square in Figure 2 is 16, the side length of the large square in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0051_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0051_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{5}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the length of the shorter leg of the right-angled triangle be $x$, then the length of the longer leg is $2x$.\n\nSince the area of the small blank square is 16,\n\nThe side length of the small blank square is 4.\n\nFrom Figure 2, we have $2x - x = 4$.\n\nSolving for $x$, we get $x = 4$.\n\nTherefore, the length of the shorter leg of the right-angled triangle is 4, and the length of the longer leg is 8.\n\nThus, the side length of the large square is $\\sqrt{4^{2} + 8^{2}} = 4\\sqrt{5}$.\n\nHence, the answer is $4\\sqrt{5}$.\n[Insight] This problem examines the properties of squares, the Pythagorean theorem, and solving linear equations. The key to solving the problem lies in understanding the given information and applying the concept of combining numbers with shapes to find the solution." }, { "problem_id": 1463, "question": "The tangram shown in Figure 1 can be used to form the \"burning candle\" shown in Figure 2. Then, the \"flame height\" in Figure 2 is $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0056_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0056_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{10}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, according to the problem statement: Quadrilaterals $ABCD$ and $OGEF$ are both squares, $\\triangle COG$ is a right-angled triangle, $\\triangle AEF$ is an isosceles right-angled triangle, point $E$ is the midpoint of $AD$, the side length of square $ABCD$ is 4, and $CG$ is the required length.\n\n$\\therefore OC=\\frac{1}{2} AC=\\frac{1}{2} \\sqrt{AB^{2}+BC^{2}}=2 \\sqrt{2}, \\quad AE=\\frac{1}{2} AD=2, AF=EF=OG$,\n\n$\\therefore AE=\\sqrt{AF^{2}+EF^{2}}=\\sqrt{2} EF$, which implies $\\sqrt{2} EF=2$,\n\nSolving gives $EF=\\sqrt{2}$,\n\n$\\therefore OG=\\sqrt{2}$,\n\nIn right-angled triangle $\\triangle COG$, $CG=\\sqrt{OC^{2}+OG^{2}}=\\sqrt{(2 \\sqrt{2})^{2}+(\\sqrt{2})^{2}}=\\sqrt{10}$,\n\nThus, the \"flame height\" in Figure 2 is $\\sqrt{10}$,\n\nTherefore, the answer is: $\\sqrt{10}$.\n\n\n\n【Insight】This problem tests the properties of squares, isosceles right-angled triangles, and the Pythagorean theorem. Mastering the characteristics of the tangram is key to solving the problem." }, { "problem_id": 1464, "question": "Given a hydraulic lift as shown in Figure 1, Figures 2 and 3 are plan views of the hydraulic lift. The side length of the rhombus \\( C O D P \\) and the leg lengths of the isosceles triangles \\( O A B \\) and \\( P E F \\) are fixed and equal. As shown in Figure 2, the distance \\( h_1 \\) from the load platform \\( E F \\) to the horizontal base \\( A B \\) is \\( 60 \\text{ cm} \\), and at this time, \\( \\angle A O B = 120^\\circ \\); as shown in Figure 3, when \\( \\angle A O B = 90^\\circ \\), the distance \\( h_2 \\) from the load platform \\( E F \\) to the horizontal base \\( A B \\) is \\(\\qquad\\) \\(\\text{cm}\\) (the result is accurate to \\( 1 \\text{ cm} \\), reference data: \\( \\sqrt{2} \\approx 1.41 \\), \\( \\sqrt{3} \\approx 1.73 \\)).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0079_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0079_2.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0079_3.jpg" ], "is_multi_img": true, "answer": "85", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, connect $OP$ and extend it to intersect $AB$ and $EF$ at points $M$ and $N$ respectively. Connect $CD$, which intersects $MN$ at point $Q$.\n\n\n\nFigure 2\n\nSince quadrilateral $CODP$ is a rhombus, and the legs of the isosceles triangles $OAB$ and $PEF$ are of fixed and equal lengths,\n\n$\\therefore MN \\perp EF$, $MN \\perp AB$, and $QP = PN = QO = OM$.\n\nGiven that the distance $h_{1}$ from the loading platform $EF$ to the horizontal base $AB$ is $60 \\mathrm{~cm}$,\n\n$\\therefore MN = 60 \\mathrm{~cm}$.\n\n$$\n\\begin{aligned}\n& \\therefore QP = PN = QO = OM = 15 \\mathrm{~cm}, \\\\\n& \\because \\angle AOB = 120^{\\circ}, \\\\\n& \\therefore \\angle OAB = \\angle OBA = 30^{\\circ}, \\\\\n& \\therefore OA = 2OM = 30 \\mathrm{~cm}.\n\\end{aligned}\n$$\n\nAs shown in Figure 3, connect $OP$ and extend it to intersect $AB$ and $EF$ at points $G$ and $H$ respectively.\n\n\n\nFigure 3\n\nSimilarly, we can deduce that $GH = h_{2} = 4OG$.\n\nSince $\\angle AOB = 90^{\\circ}$,\n\n$\\therefore \\triangle OAB$ is an isosceles right triangle,\n\n$\\therefore OA = \\sqrt{2}OG$,\n\n$\\therefore OG = \\frac{30}{\\sqrt{2}} \\approx 21.3 \\mathrm{~cm}$,\n\n$\\therefore GH = h_{2} = 4 \\times 21.3 \\approx 85 \\mathrm{~cm}$.\n\nTherefore, the answer is 85.\n\n[Key Insight] This problem primarily examines the properties of rhombuses, squares, and the operations of square roots. Mastering these properties and operations is crucial for solving the problem." }, { "problem_id": 1465, "question": "As shown in Figure 1, this is a geometric design at a certain location in the park. Figure 2 is its schematic diagram. A part of the square is below the horizontal plane $E F$. It is measured that $D E = 2$ meters, $\\angle C D F = 45^{\\circ}$, and the total length of the material exposed above the horizontal plane is 140 meters (Note: There are 8 identical square designs, excluding wastage). The distance from point $B$ to the horizontal plane $E F$ is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_66c6b1a0e45f620e93deg_0088_1.jpg", "batch21-2024_06_15_66c6b1a0e45f620e93deg_0088_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{35 \\sqrt{2}}{8}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, extend $AE$ and $CD$ to intersect at point $M$, then connect $BM$ to intersect $EF$ at point $N$.\n\n\n\n$M$\n\nSince quadrilateral $ABCM$ is a square,\n\n$\\therefore \\angle A = \\angle AMC = 90^\\circ$, $\\angle BMA = \\angle BMC = 45^\\circ$, and $AB = BC = AM = CM$.\n\nGiven that $\\angle CDF = 45^\\circ$,\n\n$\\therefore \\angle MDN = 45^\\circ$,\n\n$\\therefore \\angle MED = 90^\\circ - \\angle MDN = 45^\\circ$,\n\n$\\therefore$ Right triangle $MDE$ is an isosceles right triangle,\n\n$\\therefore BM \\perp DE$, and $MN = EN = DN = \\frac{1}{2} DE = 1$ meter (since the three lines coincide in an isosceles triangle),\n\nIn right triangle $MDN$, $DM = \\sqrt{MN^2 + DN^2} = \\sqrt{2}$ meters,\n\nSimilarly, $EM = \\sqrt{2}$ meters.\n\nFrom the problem statement: $AB + BC + AE + CD = \\frac{140}{8} = \\frac{35}{2}$ meters.\n\nLet $AB = BC = AM = CM = x$ meters, then $AE = CD = (x - \\sqrt{2})$ meters.\n\n$\\therefore x + x + x - \\sqrt{2} + x - \\sqrt{2} = \\frac{35}{2}$,\n\nSolving for $x$ gives $x = \\frac{35 + 4\\sqrt{2}}{8}$ meters,\n\n$\\therefore AB = AM = \\frac{35 + 4\\sqrt{2}}{8}$ meters.\n\nIn right triangle $ABM$, $BM = \\sqrt{AB^2 + AM^2} = \\frac{35\\sqrt{2} + 8}{8}$ meters,\n\nThen $BN = BM - MN = \\frac{35\\sqrt{2} + 8}{8} - 1 = \\frac{35\\sqrt{2}}{8}$ meters,\n\nThus, the distance from point $B$ to the horizontal plane $EF$ is $\\frac{35\\sqrt{2}}{8}$ meters.\n\nTherefore, the answer is: $\\frac{35\\sqrt{2}}{8}$.\n\n【Highlight】This problem examines the properties of squares and the coincidence of three lines in an isosceles triangle. Mastering the properties of squares is crucial for solving the problem." }, { "problem_id": 1466, "question": "As shown in Figure 1, point \\( F \\) starts from vertex \\( A \\) of the rhombus \\( ABCD \\) and moves uniformly along the path \\( A \\rightarrow B \\rightarrow D \\) at a speed of \\( 1 \\, \\text{cm/s} \\) to point \\( D \\). Figure 2 shows the relationship between the area \\( y \\, (\\text{cm}^2) \\) of \\( \\triangle FDC \\) and the time \\( x \\, (\\text{s}) \\) as point \\( F \\) moves. The value of \\( a \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_6ee39017d17680334ff6g_0055_1.jpg", "batch21-2024_06_15_6ee39017d17680334ff6g_0055_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{25}{6}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from point $D$ to $AB$, intersecting at point $E$,\n\n\n\nFrom the graph, it can be observed that point $F$ moves from point $A$ to $B$ in $a$ seconds, and the area of triangle $FDC$ is $2a$ cm².\n\n$$\n\\therefore AB = a \\text{,}\n$$\n\n$\\therefore \\frac{1}{2} AB \\cdot DE = \\frac{1}{2} a \\cdot DE = 2a$,\n\n$\\therefore DE = 4$,\n\nWhen $F$ moves from $B$ to $D$, it takes $5$ seconds,\n\n$\\therefore BD = 5$\n\nIn right triangle $DBE$, $BE = \\sqrt{BD^2 - DE^2} = \\sqrt{5^2 - 4^2} = 3$,\n\n$\\because ABCD$ is a rhombus,\n\n$\\therefore AE = a - 3$, $AD = a$,\n\nIn right triangle $ADE$, $AD^2 = AE^2 + DE^2$\n\n$\\therefore a^2 = 4^2 + (a - 3)^2$,\n\nSolving gives $a = \\frac{25}{6}$.\n\nTherefore, the answer is: $\\frac{25}{6}$.\n\n【Highlight】This problem comprehensively examines the properties of a rhombus and the properties of a linear function graph. The solution process requires attention to the relationship between the changes in the function graph and the position of the moving point." }, { "problem_id": 1467, "question": "Divide the rhombus in Figure 1 along its diagonals into four congruent right triangles, and arrange these four right triangles into squares as shown in Figure 2 and Figure 3. Then the area of the rhombus in Figure 1 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_7c99ce7307c5c5b100afg_0003_1.jpg", "batch21-2024_06_15_7c99ce7307c5c5b100afg_0003_2.jpg", "batch21-2024_06_15_7c99ce7307c5c5b100afg_0003_3.jpg" ], "is_multi_img": true, "answer": "48", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Let the longer leg of the right triangle in Figure 1 be $a$, and the shorter leg be $b$. We have the system of equations:\n\\[\n\\left\\{\n\\begin{array}{l}\na + b = 10 \\\\\na - b = 2\n\\end{array}\n\\right.\n\\]\nSolving this system, we find:\n\\[\n\\left\\{\n\\begin{array}{l}\na = 6 \\\\\nb = 4\n\\end{array}\n\\right.\n\\]\n\nTherefore, the area of the rhombus in Figure 1 is:\n\\[\n\\frac{6 \\times 4}{2} \\times 4 = 48\n\\]\nHence, the answer is: 48\n\n【Key Insight】This problem examines the properties of a rhombus and a square, as well as the application of a system of linear equations. Understanding the properties of squares and rhombuses and setting up the system of equations based on the given conditions are crucial to solving the problem." }, { "problem_id": 1468, "question": "As shown, there is a square paper with a side length of 1. The first cut is made along the line segment $A P_{1}$, leaving the triangle $A B P_{1}$;\n\nFor the second cut, the midpoint $P_{2}$ of $B P_{1}$ is taken, and then the cut is made along $A P_{2}$, leaving the triangle $A B P_{2}$; for the third cut, the midpoint $P_{3}$ of $B P_{2}$ is taken,\n\nand then the cut is made along $A P_{3}$, leaving the triangle $A B P_{3}$ $\\qquad$ This process continues, and after the $n$-th cut, the area of the removed figure$\\qquad$ .\n\n\n\nFirst cut\n\n\n\nSecond cut\n\n\n\nThird cut", "input_image": [ "batch21-2024_06_15_7c99ce7307c5c5b100afg_0020_1.jpg", "batch21-2024_06_15_7c99ce7307c5c5b100afg_0020_2.jpg", "batch21-2024_06_15_7c99ce7307c5c5b100afg_0020_3.jpg" ], "is_multi_img": true, "answer": "$1-\\left(\\frac{1}{2}\\right)^{n}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement: After the first cut along segment $A P_{1}$, the remaining segment $B P_{1}=1=\\left(\\frac{1}{2}\\right)^{0}$;\n\nAfter the second cut along segment $A P_{2}$, the remaining segment $B P_{2}=\\frac{1}{2}=\\left(\\frac{1}{2}\\right)^{1}$;\n\nAfter the third cut along segment $A P_{3}$, the remaining segment $B P_{3}=\\left(\\frac{1}{2}\\right)^{2}$;\n\nFollowing this pattern: After the $n$th cut along segment $A P_{n}$, the remaining segment $B P_{n}=\\left(\\frac{1}{2}\\right)^{n-1}$;\n\n$\\therefore$ The area of the remaining figure is: $\\frac{1}{2} \\times 1 \\times\\left(\\frac{1}{2}\\right)^{n-1}=\\left(\\frac{1}{2}\\right)^{n}$,\n\n$\\because$ The area of the square is 1,\n\nThe area subtracted from the figure is: $1-\\left(\\frac{1}{2}\\right)^{n}$.\n\nHence, the answer is: $1-\\left(\\frac{1}{2}\\right)^{n}$\n\n【Insight】This problem examines the pattern recognition in geometric figures, specifically concerning the areas related to mid-segments. The key to solving the problem lies in identifying the pattern to determine the final remaining area $\\frac{1}{2} \\times 1 \\times\\left(\\frac{1}{2}\\right)^{n-1}=\\left(\\frac{1}{2}\\right)^{n}$." }, { "problem_id": 1469, "question": "As shown in Figure 1, a square piece of paper is folded in half to form a rectangle $\\mathrm{ABCD}$, then folded along the bisector $\\mathrm{DE}$ of $\\angle \\mathrm{ADC}$, as shown in Figure 2, with point $\\mathrm{C}$ landing at point $\\mathrm{C}^{\\prime}$. Finally, as shown in Figure 3, the paper is folded such that point $\\mathrm{A}$ lands at the midpoint $\\mathrm{A}^{\\prime}$ of $\\mathrm{DE}$, with the fold line being $\\mathrm{FG}$. If the original side length of the square paper is $9 \\mathrm{~cm}$, then $\\mathrm{FG}=$ $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_7ceb450d6a32d576e297g_0011_1.jpg", "batch21-2024_06_15_7ceb450d6a32d576e297g_0011_2.jpg", "batch21-2024_06_15_7ceb450d6a32d576e297g_0011_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{3}{2} \\sqrt{10}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Draw $\\mathrm{GM} \\perp \\mathrm{AC}^{\\prime}$ at $\\mathrm{M}$, $\\mathrm{A}^{\\prime} \\mathrm{N} \\perp \\mathrm{AD}$ at $\\mathrm{N}$, and let $\\mathrm{AA}^{\\prime}$ intersect $\\mathrm{EC}^{\\prime}$ at $\\mathrm{K}$. It is easy to see that $\\mathrm{MG}=\\mathrm{AB}=\\mathrm{AC}^{\\prime}$.\n\n\n\nFigure 3\n\nSince $\\mathrm{GF} \\perp \\mathrm{AA}^{\\prime}$,\n\n$\\therefore \\angle \\mathrm{AFG}+\\angle \\mathrm{FAK}=90^{\\circ}$, and $\\angle \\mathrm{MGF}+\\angle \\mathrm{MFG}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{MGF}=\\angle \\mathrm{KAC}^{\\prime}$,\n\n$\\therefore \\triangle \\mathrm{AKC}^{\\prime} \\cong \\triangle \\mathrm{GFM}$,\n\n$\\therefore \\mathrm{GF}=\\mathrm{AK}$.\n\nGiven that $\\mathrm{AN}=\\frac{27}{4} \\mathrm{~cm}$, $\\mathrm{A}^{\\prime} \\mathrm{N}=\\frac{9}{4} \\mathrm{~cm}$, and $\\mathrm{C}^{\\prime} \\mathrm{K} / / \\mathrm{A}^{\\prime} \\mathrm{N}$,\n\n$\\therefore \\frac{K C^{\\prime}}{A^{\\prime} N}=\\frac{A C^{\\prime}}{A N}$,\n\n$\\therefore \\frac{K C^{\\prime}}{\\frac{9}{4}}=\\frac{\\frac{9}{2}}{\\frac{27}{4}}$,\n\n$\\therefore \\mathrm{C}^{\\prime} \\mathrm{K}=1.5 \\mathrm{~cm}$.\n\nIn the right triangle $\\triangle \\mathrm{AC} \\mathrm{C}^{\\prime} \\mathrm{K}$, $\\mathrm{AK}=\\sqrt{A C^{\\prime 2}+C^{\\prime} K^{2}}=\\sqrt{4.5^{2}+1.5^{2}}=\\frac{3}{2} \\sqrt{10} \\mathrm{~cm}$,\n\n$\\therefore \\mathrm{FG}=\\mathrm{AK}=\\frac{3}{2} \\sqrt{10} \\mathrm{~cm}$.\n\nThus, the answer is $\\frac{3}{2} \\sqrt{10}$.\n\n【Insight】This problem examines the knowledge of flip transformations, properties of squares, properties of rectangles, and the determination and properties of congruent triangles. The key to solving the problem is to flexibly apply the learned knowledge, add common auxiliary lines, and construct congruent triangles to solve the problem. This type of problem is commonly seen in middle school exams." }, { "problem_id": 1470, "question": "Fold a square piece of paper as shown in the steps to obtain Figure (4), then cut off one corner along the dotted line, unfold it to obtain Figure (5), where $F M$ and $G N$ are the fold lines. If the area of the square $E F G H$ is equal to that of the pentagon $M C N G F$, the value of $\\frac{F M}{G F}$ is $\\qquad$\n\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)\n\n\n(5)", "input_image": [ "batch21-2024_06_15_7ceb450d6a32d576e297g_0036_1.jpg", "batch21-2024_06_15_7ceb450d6a32d576e297g_0036_2.jpg", "batch21-2024_06_15_7ceb450d6a32d576e297g_0036_3.jpg", "batch21-2024_06_15_7ceb450d6a32d576e297g_0036_4.jpg", "batch21-2024_06_15_7ceb450d6a32d576e297g_0036_5.jpg" ], "is_multi_img": true, "answer": "$\\frac{\\sqrt{5}-\\sqrt{2}}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Connect $\\mathrm{HF}$, and let the intersection point of line $\\mathrm{MH}$ with side $\\mathrm{AD}$ be $\\mathrm{P}$, as shown in the figure:\n\n\n\n(5)\nFrom the folding, it is known that points $P$, $H$, $F$, and $M$ are collinear, and $PH = MF$. Let the side length of square $ABCD$ be $2a$, then the area of square $\\mathrm{ABCD}$ is $4a^{2}$.\n\nSince the area of square EFGH is equal to the area of pentagon MCNGF,\n\nFrom the folding, the area of square $\\mathrm{EFGH}$ is $\\frac{1}{5}$ times the area of square $\\mathrm{ABCD}$, which is $\\frac{4}{5}a^{2}$.\n\nTherefore, the side length of square $\\mathrm{EFGH}$, $\\mathrm{GF}$, is $\\sqrt{\\frac{4}{5}a^{2}} = \\frac{2\\sqrt{5}}{5}a$.\n\nThus, $\\mathrm{HF} = \\sqrt{2} \\times GF = \\frac{2\\sqrt{10}}{5}a$.\n\nTherefore, $\\mathrm{MF} = \\mathrm{PH} = \\frac{2a - \\frac{2\\sqrt{10}}{5}a}{2} = \\frac{5 - \\sqrt{10}}{5}a$.\n\nHence, the ratio $\\frac{FM}{GF} = \\frac{\\frac{5 - \\sqrt{10}}{5}a}{\\frac{2\\sqrt{5}}{5}a} = \\frac{\\sqrt{5} - \\sqrt{2}}{2}$.\n\n【Key Point】This problem examines the properties of folding and the geometry of squares. The key to solving it lies in understanding the properties of folding." }, { "problem_id": 1471, "question": "As shown in Figure 1, $\\angle D E F=24^{\\circ}$, the rectangular paper $A B C D$ is folded along line $E F$ to form Figure 2, and then folded along line $G F$ to form Figure 3. In Figure 3, $\\angle C F E=$. \n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_811f63e0720dfa428709g_0045_1.jpg", "batch21-2024_06_15_811f63e0720dfa428709g_0045_2.jpg", "batch21-2024_06_15_811f63e0720dfa428709g_0045_3.jpg" ], "is_multi_img": true, "answer": "$108^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\angle DEF = 24^\\circ$ and $AE \\parallel BC$,\n\nIn Figure 2, $\\angle BFE = \\angle DEF = 24^\\circ$, and $\\angle EFC = 180^\\circ - \\angle DEF = 156^\\circ$.\n\nSince $\\angle GFC = \\angle EFC - \\angle EFG$,\n\nTherefore, $\\angle GFC = 132^\\circ$.\n\nIn Figure 3, $\\angle GFC = \\angle BFE + \\angle CFE$,\n\nThus, $\\angle CFE = \\angle GFC - \\angle BFE = 108^\\circ$.\n\nHence, the answer is: $108^\\circ$.\n\n【Key Insight】This problem examines the properties of parallel lines, the transformation of folding, and the properties of rectangles. The key is to identify the equal side and angle relationships based on the folding transformation." }, { "problem_id": 1472, "question": "As shown in Figure A, the dart-shaped floor drain is located at the center of a square tile, with each side parallel to the sides of the square tile. The design around it is cut along the solid lines in Figure A, forming 4 equal \"dart\" shaped tiles. Figure B is its structural diagram, where the cutting line $A F=4$, and $A B / / E F, \\angle A B G=60^{\\circ}$. Then the area of the square drain $E F G H$ is $\\qquad$.\n\n\n\n(Figure A)\n\n\n\n(Figure B)", "input_image": [ "batch21-2024_06_15_8438011e04c686eed18bg_0013_1.jpg", "batch21-2024_06_15_8438011e04c686eed18bg_0013_2.jpg" ], "is_multi_img": true, "answer": "$16-8 \\sqrt{3}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Answer:** \n**Solution:** \nAs shown in the figure, extend \\( GF \\) to intersect \\( AB \\) at point \\( M \\).\n\n\n\n(Diagram B)\n\nFrom the problem statement, we know that square \\( ABCD \\) is divided into 8 congruent right triangles and one square \\( EFGH \\).\n\nGiven that \\( AF = BG = 4 \\), and \\( AB \\parallel EF \\), with \\( \\angle ABG = 60^\\circ \\),\n\nwe have:\n\\[\nBM = \\frac{1}{2} BG = 2, \\quad MG = \\frac{\\sqrt{3}}{2} BG = 2\\sqrt{3}.\n\\]\nThus,\n\\[\nAB = 2 + 2\\sqrt{3}.\n\\]\nTherefore, the area of square \\( EFGH \\) is:\n\\[\n(2 + 2\\sqrt{3})^2 - 8 \\times \\frac{1}{2} \\times 2 \\times 2\\sqrt{3} = 16 - 8\\sqrt{3}.\n\\]\nHence, the answer is: \\( 16 - 8\\sqrt{3} \\).\n\n**Key Insight:** \nThis problem tests the properties of squares and the properties of \\( 30^\\circ \\) right triangles. The key to solving it lies in a solid understanding of these properties." }, { "problem_id": 1473, "question": "As shown in Figure 1, there is an octagram paperboard with eight right angles, eight equal obtuse angles, and all sides of equal length. As shown in Figure 2, the paperboard is cut along the dotted lines and seamlessly pieced together to form the large square shown in Figure 3, with an area of $8+4 \\sqrt{2}$. The length of line segment $A B$ in Figure 3 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_8438011e04c686eed18bg_0014_1.jpg", "batch21-2024_06_15_8438011e04c686eed18bg_0014_2.jpg", "batch21-2024_06_15_8438011e04c686eed18bg_0014_3.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the side length of the original octagon be \\( a \\). Then, the side length of the square in Figure 2 is \\( 2a + \\sqrt{2}a \\), and its area is \\( (2a + \\sqrt{2}a)^2 \\). The combined area of the four small triangles is \\( 2a^2 \\).\n\nSetting up the equation:\n\\[\n(2a + \\sqrt{2}a)^2 + 2a^2 = 8 + 4\\sqrt{2},\n\\]\nwe solve for \\( a \\) and find:\n\\[\na = 1.\n\\]\nThus, the length of \\( AB \\) is:\n\\[\nAB = \\sqrt{2}.\n\\]\nTherefore, the answer is:\n\\[\n\\boxed{\\sqrt{2}}\n\\]\n\n**Key Insight:** This problem tests the understanding of geometric transformations and the ability to identify how segment \\( AB \\) in Figure 3 is composed of segments in Figure 2." }, { "problem_id": 1474, "question": "On the occasion of the 100th anniversary of the founding of the Party, we should vigorously promote the \"Three Oxen Spirit.\" A set of tangram made from a square $ABCD$ with side length $2\\sqrt{2}$ is shown in Figure 1. When this set of tangram is arranged within a rectangle $EFGH$ to form the \"Ox\" pattern shown in Figure 2, the ratio of the area of the rectangle $EFGH$ to the area of the \"Ox\" pattern is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8438011e04c686eed18bg_0015_1.jpg", "batch21-2024_06_15_8438011e04c686eed18bg_0015_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{15+8 \\sqrt{2}}{8}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Since the \"Ruzi Niu\" is composed of tangram pieces,\n\nTherefore, the area of \"Ruzi Niu\" is $2 \\sqrt{2} \\times 2 \\sqrt{2}=8$,\n\nFrom the relationships between the sides of the tangram pieces, it can be deduced that the width of rectangle $EFGH$ is $2+\\sqrt{2}$, and its length is $7+\\frac{\\sqrt{2}}{2}$,\n\nThus, the area of rectangle $EFGH$ is $\\left(7+\\frac{\\sqrt{2}}{2}\\right)(2+\\sqrt{2})=15+8 \\sqrt{2}$,\n\nTherefore, the ratio of the area of rectangle $EFGH$ to that of \"Ruzi Niu\" is $\\frac{15+8 \\sqrt{2}}{8}$,\n\nHence, the answer is: $\\frac{15+8 \\sqrt{2}}{8}$.\n\n[Highlight] This problem mainly tests knowledge of tangram. Mastering the numerical relationships between the sides of tangram pieces is key to solving the problem." }, { "problem_id": 1475, "question": "Xiao Ming made a movable rhombus tool using four wooden strips of equal length. He first adjusted the tool to form the rhombus shown in Figure 1, measured $\\angle B = 60^\\circ$, and the diagonal $AC = 20 \\text{ cm}$. Then, he adjusted the tool to form the square shown in Figure 2. The length of the diagonal $AC$ in Figure 2 is $\\qquad$ $\\text{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8438011e04c686eed18bg_0034_1.jpg", "batch21-2024_06_15_8438011e04c686eed18bg_0034_2.jpg" ], "is_multi_img": true, "answer": "$20 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1 and Figure 2, connect $AC$,\n\n\n\nFigure 1\n\nFigure 2\n\nIn Figure 1, since quadrilateral $ABCD$ is a rhombus,\n\n$\\therefore AB = BC$,\n\nsince $\\angle B = 60^{\\circ}$,\n\n$\\therefore \\triangle ABC$ is an equilateral triangle,\n\n$\\therefore AB = BC = AC = 20 \\mathrm{~cm}$.\n\nIn Figure 2, since quadrilateral $ABCD$ is a square,\n\n$\\therefore AB = BC$, $\\angle B = 90^{\\circ}$,\n\n$\\therefore \\triangle ABC$ is an isosceles right triangle,\n\n$\\therefore AC = \\sqrt{AB^{2} + BC^{2}} = 20 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the answer is: $20 \\sqrt{2}$\n\n【Insight】This question tests the properties of rhombuses, squares, and the Pythagorean theorem. The key to solving it is to be proficient in the properties of rhombuses and squares. This type of question is commonly found in middle school exams." }, { "problem_id": 1476, "question": "As shown in Figure 1, the area of the rhombus paper $ABCD$ is $30 \\mathrm{~cm}^{2}$, and the length of the diagonal $AC$ is $6 \\mathrm{~cm}$. When this rhombus paper is cut along the diagonal, four congruent right-angled triangles are obtained. These four right-angled triangles are then arranged into a square as shown in Figure 2. The side length of the blank small square in the large square is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8438011e04c686eed18bg_0035_1.jpg", "batch21-2024_06_15_8438011e04c686eed18bg_0035_2.jpg" ], "is_multi_img": true, "answer": "2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let the diagonals $AC$ and $BD$ intersect at point $O$.\n\n\n\nIn rhombus $ABCD$, $AC \\perp BD$, $AO = OC$, and $OB = OD$.\n\nGiven that the area of rhombus $ABCD$ is $30 \\mathrm{~cm}^{2}$ and the length of diagonal $AC$ is $6 \\mathrm{~cm}$,\n\nwe have $\\frac{1}{2} \\cdot AC \\cdot BD = 30$, and $OA = 3 \\mathrm{~cm}$.\n\nThus, $BD = 10 \\mathrm{~cm}$,\n\nand $OB = 5 \\mathrm{~cm}$.\n\nTherefore, the side length of the blank small square in the larger square is equal to $OB - OA = 2 \\mathrm{~cm}$.\n\nHence, the answer is: 2\n\n【Key Insight】This problem primarily examines the properties of a rhombus. Mastering these properties is crucial for solving the problem." }, { "problem_id": 1477, "question": "The tangram is a great invention of the ancient Chinese working people and is known as the \"Magic Plate of the East.\" Xiao Luo arranged a set of tangram pieces (as shown in Figure 1) into a pattern of \"Tigers Roaring with Power\" and embedded this pattern into a rectangular frame, leaving a $5 \\mathrm{~cm}$ gap on all four sides (as shown in Figure 2). The perimeter of the rectangular frame is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8438011e04c686eed18bg_0100_1.jpg", "batch21-2024_06_15_8438011e04c686eed18bg_0100_2.jpg" ], "is_multi_img": true, "answer": "$160 \\sqrt{2}+40$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure below, mark the points in Figure 2.\n\nAccording to the problem, we know that $AB = FG = HE = KL = 5 \\mathrm{~cm}$, $BC = DE = EF = EI = \\frac{\\sqrt{40^{2} + 40^{2}}}{4} = 10 \\sqrt{2} \\mathrm{~cm}$, $CD = \\sqrt{\\left(\\frac{40}{2}\\right)^{2} + \\left(\\frac{40}{2}\\right)^{2}} = 20 \\sqrt{2} \\mathrm{~cm}$, $IK = \\frac{\\sqrt{40^{2} + 40^{2}}}{2} = 20 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, $AG = AB + BC + CD + DE + EF + FG = (50 \\sqrt{2} + 10) \\mathrm{cm}$,\n\n$HL = HE + EI + IK + KL = (30 \\sqrt{2} + 10) \\mathrm{cm}$.\n\nThus, the perimeter of the rectangle is $(50 \\sqrt{2} + 10) \\times 2 + (30 \\sqrt{2} + 10) \\times 2 = (160 \\sqrt{2} + 40) \\mathrm{cm}$.\n\nHence, the answer is: $160 \\sqrt{2} + 40$.\n\n\n\n【Highlight】This problem tests the properties of squares and the Pythagorean theorem. Mastering these concepts is key to solving the problem." }, { "problem_id": 1478, "question": "As shown in Figure 1, it is a three-section telescopic clothes rack. As shown in Figure 2, it is a side view of the clothes rack. MN is the fixed end of the rack on the wall, A is the fixed support point, B is the sliding support point, quadrilaterals DFGI and EIJH are rhombuses, and $\\mathrm{AF}=\\mathrm{BF}=\\mathrm{CH}=\\mathrm{DF}=\\mathrm{EH}$. When point $\\mathrm{B}$ slides on $\\mathrm{AN}$, the outer steel frame of the rack undergoes angular deformation, and its outer extension length (the distance between points $\\mathrm{A}$ and $\\mathrm{C}$) also changes, creating the telescopic effect of the rack. When the telescopic rack is in its initial state, the outer extension length is $42 \\mathrm{~cm}$. When point $\\mathrm{B}$ moves $8 \\mathrm{~cm}$ towards point $\\mathrm{A}$, the outer extension length is $90 \\mathrm{~cm}$. As shown in Figure 3, when the outer extension length is $120 \\mathrm{~cm}$, the distance $\\mathrm{PQ}$ between $\\mathrm{BD}$ and $\\mathrm{GE}$ is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_8611f6d433df8b9a1bd1g_0077_1.jpg", "batch21-2024_06_15_8611f6d433df8b9a1bd1g_0077_2.jpg", "batch21-2024_06_15_8611f6d433df8b9a1bd1g_0077_3.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "As shown in the figure, draw $\\mathrm{FK} \\perp \\mathrm{AB}$ at point $K$. Let $\\mathrm{AB} = 2x$ cm. According to the problem, $\\mathrm{FK} = 7$ cm. When $\\mathrm{AB} = (2x - 8)$ cm, $\\mathrm{FK} = 15$ cm.\n\n\n\nThen, we have $\\mathrm{AF}^{2} = x^{2} + 7^{2} = (x - 4)^{2} + 15^{2}$.\n\n$\\therefore x = 24$ cm,\n\n$\\therefore \\mathrm{AF} = \\sqrt{7^{2} + 24^{2}} = 25$ cm.\n\nAs shown in the figure, when $\\mathrm{OF} = 20$, in the right triangle $\\triangle \\mathrm{DFO}$, $\\mathrm{OD} = \\sqrt{25^{2} - 20^{2}} = 15$ cm.\n\n\n\n$\\because \\mathrm{PQ} \\perp \\mathrm{GI}$,\n\n$\\therefore \\frac{1}{2} \\cdot \\mathrm{FI} \\cdot \\mathrm{DG} = \\mathrm{DF} \\cdot \\mathrm{PQ}$,\n\n$\\therefore \\mathrm{PQ} = \\frac{\\frac{1}{2} \\times 40 \\times 30}{25} = 24$ cm.\n\nTherefore, the answer is 24.\n\n【Insight】This problem examines the properties of a rhombus and the Pythagorean theorem. The key to solving it is understanding the problem and using parameters to construct equations, which is a common type of question in middle school exams." }, { "problem_id": 1479, "question": "As shown in Figure 1, it is a \"Gou-Gu Circle-Square Diagram\" created by Zhao Shuang, a mathematician from the Three Kingdoms period. The rectangle in Figure 2 is divided into four congruent triangles and one square, which can be exactly arranged into such a \"Gou-Gu Circle-Square Diagram\". The ratio of the perimeter of the rectangle to the perimeter of the resulting square is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8d6a539c652873b8894bg_0056_1.jpg", "batch21-2024_06_15_8d6a539c652873b8894bg_0056_2.jpg" ], "is_multi_img": true, "answer": "$3: \\sqrt{5}($ or $3 \\sqrt{5}: 5$ )", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nLet the two perpendicular sides of the four congruent right triangles that divide the rectangle in Figure 2 be \\( a \\) and \\( b \\) (where \\( a > b \\)).\n\n- The side length of the larger square is \\( \\sqrt{a^{2} + b^{2}} \\).\n- The side length of the smaller square is \\( a - b \\).\n- The length of the rectangle is \\( 2a + a - b = 3a - b \\), and its width is \\( b \\).\n- Therefore, the perimeter of the rectangle is: \n \\[\n 2(3a - b + b) = 6a.\n \\]\n\nFrom Figure 2, the side length of the smaller square in the middle is \\( b \\). \nThus, \n\\[\na - b = b, \n\\]\nwhich implies \n\\[\na = 2b.\n\\]\n\nThe perimeter of the larger square is: \n\\[\n4\\sqrt{a^{2} + b^{2}} = 4\\sqrt{4b^{2} + b^{2}} = 4\\sqrt{5}b = 2\\sqrt{5}a.\n\\]\n\nThe ratio of the perimeter of the rectangle to the perimeter of the larger square is: \n\\[\n\\frac{6a}{2\\sqrt{5}a} = \\frac{3}{\\sqrt{5}} = \\frac{3\\sqrt{5}}{5}.\n\\]\n\n**Final Answer:** \nThe ratio is \\( 3 : \\sqrt{5} \\) (or \\( 3\\sqrt{5} : 5 \\)).\n\n**Key Insight:** \nThis problem primarily examines the Pythagorean theorem, the properties of rectangles, and the properties of squares. The key is to determine the relationship between the sides of the congruent right triangles and the sides of the rectangle and the larger square based on the given figure." }, { "problem_id": 1480, "question": "As shown in Figure 1, a right-angled triangle paper has one of its legs measuring 2 units. Four such right-angled triangle papers are cut and placed into a square with side length 3, as shown in Figure 2 (with no overlapping or gaps at the joints). The area of the shaded part in Figure 2 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8d6a539c652873b8894bg_0064_1.jpg", "batch21-2024_06_15_8d6a539c652873b8894bg_0064_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{5}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem,\n\nThe hypotenuse of the right triangle is 3, and one of the legs is 2.\n\nTherefore, the length of the other leg of the right triangle is: $\\sqrt{3^{2}-2^{2}}=\\sqrt{5}$.\n\nThus, the area of the shaded region is: $\\frac{2 \\times \\sqrt{5}}{2} \\times 4=4 \\sqrt{5}$.\n\nHence, the answer is: $4 \\sqrt{5}$.\n\n【Key Insight】This problem tests the application of the Pythagorean theorem in solving triangles and the properties of squares. Correctly understanding the relationship between the side length of the square, which is 3, and the right triangle is crucial for solving the problem." }, { "problem_id": 1481, "question": "Xiao Hui used a set of tangram pieces shown in Figure 1 to create the \"bowing figure\" as shown in Figure 2. Given that the side length of the square $A B C D$ is $4 d m$, the value of $h$ in Figure 2 is $\\qquad$ $d m$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8d6a539c652873b8894bg_0065_1.jpg", "batch21-2024_06_15_8d6a539c652873b8894bg_0065_2.jpg" ], "is_multi_img": true, "answer": "$4+\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since the side length of square \\( ABCD \\) is \\( 4 \\, \\text{dm} \\),\n\nTherefore, the height on the hypotenuse of (2) is \\( 2 \\, \\text{dm} \\), the height of (4) is \\( 1 \\, \\text{dm} \\), the height on the hypotenuse of (6) is \\( 1 \\, \\text{dm} \\), and the height on the hypotenuse of (7) is \\( \\sqrt{2} \\, \\text{dm} \\).\n\nThus, the value of \\( h \\) in Figure 2 is \\( (4 + \\sqrt{2}) \\, \\text{dm} \\).\n\nHence, the answer is: \\( (4 + \\sqrt{2}) \\).\n\n[Key Insight] This problem primarily examines the properties of a square, with the key to solving it being the calculation of the heights of (2), (4), (6), and (7)." }, { "problem_id": 1482, "question": "Cut the rhombus in Figure 1 along its diagonals to form four triangles, which are then rearranged without overlap to form the square shown in Figure 2. If the area of the larger square formed is 2 units greater than the area of the rhombus, then the difference between the longer and shorter diagonals of the rhombus is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_8d6a539c652873b8894bg_0098_1.jpg", "batch21-2024_06_15_8d6a539c652873b8894bg_0098_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, the difference between the longer diagonal and the shorter diagonal of the rhombus is $MN - PQ$. According to the problem, in the larger square $ABCD$, the area that exceeds the area of the rhombus is the area of square $EFGH$.\n\n$\\therefore$ The area of square $EFGH$ is 2,\n\n$\\therefore EF = \\sqrt{2}$,\n\n$\\because EF = AF - AE = AF - BF$,\n\n$\\therefore$ According to the problem, we have: $EF = NO - PO$,\n\nThat is: $NO - PO = \\sqrt{2}$,\n\nFrom the properties of a rhombus, we know: $PQ = 2PO$, $MN = 2NO$,\n\n$\\therefore MN - PQ = 2\\sqrt{2}$,\n\nHence, the answer is: $2\\sqrt{2}$.\n\n\n\n【Insight】This problem tests the properties of rhombuses and squares. Understanding the basic properties and applying them flexibly is key to solving the problem." }, { "problem_id": 1483, "question": "As shown in Figure (1), point $\\mathrm{F}$ starts from vertex $\\mathrm{A}$ of the rhombus $\\mathrm{ABCD}$ and moves uniformly along the path $\\mathrm{A} \\rightarrow \\mathrm{D} \\rightarrow \\mathrm{B}$ at a speed of $1 \\mathrm{~cm} / \\mathrm{s}$ to point B. Figure (2) shows the relationship between the area $\\mathrm{y}\\left(\\mathrm{cm}^{2}\\right)$ of $\\triangle \\mathrm{FBC}$ and time $\\mathrm{x}(\\mathrm{s})$ as point F moves. The value of $\\mathrm{a}$ is\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_8e52011320cf89d1949eg_0027_1.jpg", "batch21-2024_06_15_8e52011320cf89d1949eg_0027_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{5}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Through point $\\mathrm{D}$, draw $\\mathrm{DE} \\perp \\mathrm{BC}$ intersecting at point $\\mathrm{E}$.\n\n\n\nFrom the graph, it can be seen that the time taken for point $\\mathrm{F}$ to move from point $\\mathrm{A}$ to point $\\mathrm{D}$ is $a$ seconds, and the area of $\\triangle \\mathrm{FBC}$ is $\\mathrm{a} \\mathrm{cm}^{2}$.\n\n$\\therefore \\mathrm{AD}=\\mathrm{a}$,\n\n$\\therefore \\frac{1}{2} \\mathrm{DE} \\cdot \\mathrm{AD}=\\mathrm{a}$,\n\n$\\therefore \\mathrm{DE}=2$.\n\nWhen point $\\mathrm{F}$ moves from D to B, it takes $\\sqrt{5} \\mathrm{~s}$,\n\n$\\therefore \\mathrm{BD}=\\sqrt{5}$.\n\nIn right triangle $\\triangle \\mathrm{DBE}$,\n$\\mathrm{BE}=\\sqrt{B D^{2}-B E^{2}}=\\sqrt{(\\sqrt{5})^{2}-2^{2}}=1$.\n\n$\\because \\mathrm{ABCD}$ is a rhombus,\n\n$\\therefore \\mathrm{EC}=\\mathrm{a}-1, \\mathrm{DC}=\\mathrm{a}$,\n\nIn right triangle $\\triangle \\mathrm{DEC}$, $\\mathrm{a}^{2}=2^{2}+(\\mathrm{a}-1)^{2}$,\n\nSolving gives $\\mathrm{a}=\\frac{5}{2}$.\n\n【Insight】This problem examines the properties of a rhombus and the properties of linear function graphs. During the solution process, attention should be paid to the relationship between the changes in the function graph and the position of the moving point." }, { "problem_id": 1484, "question": "As shown in Figure 1, fold the square paper $ABCD$ so that $AB$ coincides with $CD$, with the fold line being $EF$. As shown in Figure 2, unfold and then fold again so that point $C$ coincides with point $E$, with the fold line being $GH$. The corresponding point of $B$ is point $M$, and $EM$ intersects $AB$ at $N$. If $AD = 2$, then $MN =$ $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_98d00b7828a9a1882e53g_0013_1.jpg", "batch21-2024_06_15_98d00b7828a9a1882e53g_0013_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{1}{3}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Problem Analysis: Let \\( DH = x \\), \\( CH = 2 - x \\). By the property of folding, \\( DE = 1 \\), \\( EH = CH = 2 - x \\). In the right triangle \\( \\triangle DEH \\), \\( DE^{2} + DH^{2} = EH^{2} \\), which gives \\( 1 + x^{2} = (2 - x)^{2} \\). Solving this equation yields \\( x = \\frac{3}{4} \\), and \\( EH = 2 - x = \\frac{5}{4} \\). Since \\( \\angle MEH = \\angle C = 90^{\\circ} \\), it follows that \\( \\angle AEN + \\angle DEH = 90^{\\circ} \\). Also, since \\( \\angle ANE + \\angle AEN = 90^{\\circ} \\), we have \\( \\angle ANE = \\angle DEH \\). Additionally, \\( \\angle A = \\angle D \\), so \\( \\triangle ANE \\sim \\triangle DEH \\). Therefore, \\( \\frac{AE}{DH} = \\frac{EN}{EH} \\), which translates to \\( \\frac{EN}{5} = \\frac{\\frac{1}{3}}{4} \\). Solving for \\( EN \\) gives \\( EN = \\frac{5}{3} \\), and \\( MN = ME - BC = 2 - \\frac{5}{3} = \\frac{1}{3} \\). Thus, the answer is \\( \\frac{1}{3} \\)." }, { "problem_id": 1485, "question": "Cut the rectangle along the dotted line (as shown in Figure (1)) to obtain quadrilateral $ABCD$. Then, fold quadrilateral $ABCD$ along diagonal $BD$, so that point $A$ falls on point $F$ of $DC$ (as shown in Figure (2)). Connect $AF$ to intersect $BD$ at point $E$, where $O$ is the midpoint of $BD$. The ray $OD$ rotates clockwise around point $O$ to intersect $DC$ at $N$. Through point $O$, draw $OM \\perp ON$ to intersect $BC$ at $M$, and connect $MN$. If $AD = 2$ and the area of quadrilateral $ABCD$ is $\\frac{15}{2}$, then the minimum perimeter of $\\triangle CMN$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_98d00b7828a9a1882e53g_0030_1.jpg", "batch21-2024_06_15_98d00b7828a9a1882e53g_0030_2.jpg" ], "is_multi_img": true, "answer": "$3+\\frac{3 \\sqrt{2}}{2}$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Test Question Analysis:" }, { "problem_id": 1486, "question": "Figure 1 is a Penrose tiling pattern, which is composed of two types of \"fat\" and \"thin\" rhombuses as shown in Figure 2 (without overlapping or gaps). Then, the $\\angle \\alpha$ in Figure 2 is $\\qquad$ degrees.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_9c37109890948e79dcc9g_0014_1.jpg", "batch21-2024_06_15_9c37109890948e79dcc9g_0014_2.jpg" ], "is_multi_img": true, "answer": "36", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the problem, we know that each side of the \"fat\" and \"thin\" rhombus is equal, so the figure is a regular decagon.\n\n$\\therefore$ The sum of the interior angles of a regular decagon is: $(10-2) \\times 180^{\\circ}=1440^{\\circ}$,\n\n$\\therefore$ One interior angle of a regular decagon is: $1440^{\\circ} \\div 10=144^{\\circ}$,\n\nIn the \"thin\" rhombus,\n\n$\\because$ the opposite sides are parallel, $\\therefore \\alpha$ is supplementary to one interior angle of the regular decagon,\n\n$\\therefore \\alpha=180^{\\circ}-144^{\\circ}=36^{\\circ}$,\n\nHence, the answer is: 36.\n\n【Insight】This problem examines the properties of a rhombus and the theorem on the sum of interior angles of a polygon. Mastering the properties of a rhombus is key to solving the problem." }, { "problem_id": 1487, "question": "As shown in Figure (1), the quadrilateral $A B C D$ obtained by overlapping two congruent rectangles is a rhombus. It is known that the length of the rectangle is 8 and the width is 4. When these two pieces of paper are overlapped as shown in Figure (2), the area of the rhombus $A B C D$ is maximized. At this time, the area of the rhombus is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch21-2024_06_15_9c37109890948e79dcc9g_0048_1.jpg", "batch21-2024_06_15_9c37109890948e79dcc9g_0048_2.jpg" ], "is_multi_img": true, "answer": "20", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since $ABCD$ is a rhombus,\n\n$\\therefore AB = BC = CD = AD$.\n\nAs shown in the figure, from rectangle $BEDK$, we have: $\\angle K = \\angle E = 90^{\\circ}$, $KD = BE = 8$, and $BK = DE = 4$.\n\n\n\nLet $BC = x$, then $CE = 8 - x$, and $CD = BC = x$.\n\nIn right triangle $\\triangle CDE$, $CE^{2} + DE^{2} = CD^{2}$,\n\n$\\therefore (8 - x)^{2} + 4^{2} = x^{2}$,\n\nSolving for $x$, we get $x = 5$,\n\n$\\therefore S = BC \\cdot DE = 20$.\n\nTherefore, the answer is: 20\n\n【Highlight】This problem examines the properties of a rhombus, the Pythagorean theorem, and the properties of a rectangle. The key to solving this problem is using the Pythagorean theorem to establish the equation." }, { "problem_id": 1488, "question": "In life, some people like to fold the note being transmitted into a shape.\n\n\n\nThe folding process is shown in the figure (the shaded part represents the back of the paper strip):\n\n\n\nIt is known that the rectangular paper strip made from letter paper (Figure (1) has a length of $25 \\mathrm{~cm}$ and a width of $x \\mathrm{~cm}$. If it can be folded into the shape shown in Figure (4), and for the sake of aesthetics, the lengths by which the ends of the strip extend beyond point $P$ are equal, i.e., the final shape is axisymmetric, then the distance between the starting point $M$ and point $\\mathrm{A}$ when folding begins (expressed in terms of $x$) is $\\qquad$ $\\mathrm{cm}$.", "input_image": [ "batch21-2024_06_15_9ea084d454a37577eddeg_0026_1.jpg", "batch21-2024_06_15_9ea084d454a37577eddeg_0026_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{25-5 x}{2}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: According to the folding process, it is observed that the middle length has 5 widths,\n\nTherefore, the distance between the starting point $M$ and point $\\mathrm{A}$ when beginning to fold is: $\\frac{25-5 x}{2}$,\n\nHence, the answer is: $\\frac{25-5 x}{2}$.\n\n[Highlight] This question examines the transformation of flipping (folding problem), which is an important topic. The difficulty is relatively easy, and mastering the relevant knowledge is key to solving the problem." }, { "problem_id": 1489, "question": "As shown in Figure 1, in rectangle \\( A B C D \\), diagonals \\( A C \\) and \\( B D \\) intersect at point \\( O \\). A moving point \\( P \\) starts from point \\( B \\) and moves at a constant speed along segment \\( B C \\), stopping when it reaches point \\( C \\). Let the distance traveled by point \\( P \\) be \\( x \\), and the length of segment \\( O P \\) be \\( y \\). If the graph of the relationship between \\( y \\) and \\( x \\) is as shown in Figure 2, then the perimeter of rectangle \\( A B C D \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_a042b32142ea66b4073dg_0056_1.jpg", "batch21-2024_06_15_a042b32142ea66b4073dg_0056_2.jpg" ], "is_multi_img": true, "answer": "28 .", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Since when \\( O P \\perp A B \\), \\( O P \\) is minimized,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nand at this time \\( B P = 4 \\), \\( O P = 3 \\),\n\ntherefore \\( B C = 2 B P = 8 \\),\n\n\\( A B = 2 O P = 6 \\),\n\nthus the perimeter of rectangle \\( A B C D \\) is \\( C_{\\text{rectangle } A B C D} = 2(A B + A D) = 2 \\times (8 + 6) = 28 \\).\n\nHence, the answer is: 28.\n\n【Highlight】This question examines the properties of rectangles and the shortest perpendicular segment property. Mastering the properties of rectangles and flexibly applying the shortest perpendicular segment to determine the problem-solving goal is key to solving the problem." }, { "problem_id": 1490, "question": "In the comprehensive practical activity class, Xiao Liang cut a sharp-angled triangle paper with an area of $24 \\mathrm{~cm}^{2}$ and one side $B C$ of $8 \\mathrm{~cm}$ (as shown in Figure 1) into two pieces, and then pasted them together to form a seamless and non-overlapping rectangle $D E F G$ as shown in Figure 2. The perimeter of the rectangle is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_a042b32142ea66b4073dg_0068_1.jpg", "batch21-2024_06_15_a042b32142ea66b4073dg_0068_2.jpg" ], "is_multi_img": true, "answer": "20", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, let the intersection of $AB$ and $DG$ be point $H$, and the intersection of $AC$ and $EF$ be point $I$. Draw $AJ \\perp BC$ through point $A$, intersecting $BC$ at point $J$.\n\n\n\nGiven that the area of $\\triangle ABC$ is $24 \\mathrm{~cm}^{2}$ and $BC = 8 \\mathrm{~cm}$,\n\nwe have $AJ = \\frac{2 \\times \\text{Area of } \\triangle ABC}{BC} = \\frac{2 \\times 24}{8} = 6 \\mathrm{~cm}$.\n\nSince $\\triangle ABC$ is cut and rearranged into a seamless, non-overlapping rectangle $DEFG$,\n\nit follows that $\\triangle HAG \\cong \\triangle HBD$ and $\\triangle IAF \\cong \\triangle ICE$.\n\nThus, $AG = BD$ and $AF = CE$.\n\nSince quadrilateral $DEFG$ is a rectangle,\n\nall its angles are right angles, i.e., $\\angle G = \\angle GDE = \\angle FED = \\angle F = 90^{\\circ}$.\n\nGiven that $AJ \\perp BC$,\n\nwe have $\\angle AJD = \\angle AJE = 90^{\\circ}$.\n\nTherefore, $\\angle G = \\angle GDE = \\angle AJD = 90^{\\circ}$ and $\\angle AJE = \\angle FED = \\angle F = 90^{\\circ}$.\n\nThis implies that quadrilaterals $GAJD$ and $AFEJ$ are both rectangles.\n\nHence, $AG = JD$, $AF = JE$, and $FE = AJ = 6 \\mathrm{~cm}$.\n\nThus, $BD = JD$ and $CE = JE$.\n\nGiven that $BC = 8 \\mathrm{~cm}$,\n\nwe have $BD + JD + CE + JE = 2JD + 2JE = 2(JD + JE) = 2DE = 8 \\mathrm{~cm}$.\n\nTherefore, $DE = 4 \\mathrm{~cm}$.\n\nThe perimeter of rectangle $DEFG$ is $2DE + 2FE = 2 \\times 6 + 2 \\times 4 = 20 \\mathrm{~cm}$.\n\nThe final answer is: 20.\n\n【Key Insight】This problem tests the properties and determination theorems of rectangles, the area formula of triangles, and the correct understanding of the cutting and rearrangement process is crucial for solving the problem." }, { "problem_id": 1491, "question": "As shown in the figure, in the rectangular paper $\\mathrm{ABCD}$, $\\mathrm{AB}=8 \\mathrm{~cm}$, $\\mathrm{AD}=6 \\mathrm{~cm}$. The following steps are taken for cutting and assembling:\n\n**Step 1:** As shown in Figure (1), take an arbitrary point $\\mathrm{E}$ on segment $\\mathrm{AD}$, and cut out a triangular paper piece $\\mathrm{EBC}$ along $\\mathrm{EB}$ and $\\mathrm{EC}$ (the remaining part is not used anymore).\n\n**Step 2:** As shown in Figure (2), cut the paper along the median line $\\mathrm{GH}$ of triangle $\\mathrm{EBC}$ into two parts. Take an arbitrary point $\\mathrm{M}$ on segment $\\mathrm{GH}$ and an arbitrary point $\\mathrm{N}$ on segment $\\mathrm{BC}$, and cut the trapezoidal paper piece $\\mathrm{GBCH}$ into two parts along $\\mathrm{MN}$.\n\n**Step 3:** As shown in Figure (3), rotate the paper piece to the left of $\\mathrm{MN}$ around point $\\mathrm{G}$ clockwise by $180^{\\circ}$, making segment $\\mathrm{GB}$ coincide with $\\mathrm{GE}$. Rotate the paper piece to the right of $\\mathrm{MN}$ around point $\\mathrm{H}$ counterclockwise by $180^{\\circ}$, making segment $\\mathrm{HC}$ coincide with $\\mathrm{HE}$. Assemble them into a quadrilateral paper piece with the same area as the triangular paper piece $\\mathrm{EBC}$ (the cutting and assembling processes are seamless and non-overlapping). The maximum perimeter of the assembled quadrilateral paper piece is $\\mathrm{cm}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch21-2024_06_15_a405fffcbbf108a15681g_0045_1.jpg", "batch21-2024_06_15_a405fffcbbf108a15681g_0045_2.jpg", "batch21-2024_06_15_a405fffcbbf108a15681g_0045_3.jpg" ], "is_multi_img": true, "answer": "$12+4 \\sqrt{13}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Draw a schematic diagram of the quadrilateral $\\mathrm{M}_{1} \\mathrm{~N}_{1} \\mathrm{~N}_{2} \\mathrm{M}_{2}$ after the third step of cutting and rearranging, as shown in Answer Figure 1.\n\n\n\nAnswer Figure 1\n\nIn the figure, $\\mathrm{N}_{1} \\mathrm{~N}_{2}=\\mathrm{EN}_{1}+\\mathrm{EN}_{2}=\\mathrm{NB}+\\mathrm{NC}=\\mathrm{BC}$,\n\n$\\mathrm{M}_{1} \\mathrm{M}_{2}=\\mathrm{M}_{1} \\mathrm{G}+\\mathrm{GM}+\\mathrm{MH}+\\mathrm{M}_{2} \\mathrm{H}=2(\\mathrm{GM}+\\mathrm{MH})=2 \\mathrm{GH}=\\mathrm{BC}$ (Midline Theorem of a Triangle),\n\nAlso, since $\\mathrm{M}_{1} \\mathrm{M}_{2} / / \\mathrm{N}_{1} \\mathrm{~N}_{2}$,\n\n$\\therefore$ the quadrilateral $\\mathrm{M}_{1} \\mathrm{~N}_{1} \\mathrm{~N}_{2} \\mathrm{M}_{2}$ is a parallelogram,\n\nIts perimeter is $2 \\mathrm{~N}_{1} \\mathrm{~N}_{2}+2 \\mathrm{M}_{1} \\mathrm{~N}_{1}=2 \\mathrm{BC}+2 \\mathrm{MN}$.\n\nSince $\\mathrm{BC}=6$ is a fixed value,\n\n$\\therefore$ the perimeter of the quadrilateral depends on the length of $\\mathrm{MN}$.\n\nAs shown in Answer Figure 2, which is the complete schematic before cutting and rearranging,\n\n\n\nAnswer Figure 2\n\nDraw lines parallel to side $\\mathrm{BC}$ through points $G$ and $H$, intersecting $\\mathrm{AB}$ and $C D$ at points $\\mathrm{P}$ and $Q$ respectively, then quadrilateral $\\mathrm{PBCQ}$ is a rectangle, which is half of rectangle $\\mathrm{ABCD}$,\n\nSince $\\mathrm{M}$ is any point on line segment $\\mathrm{PQ}$, and $\\mathrm{N}$ is any point on line segment $\\mathrm{BC}$,\n\nAccording to the shortest perpendicular segment, the minimum value of $\\mathrm{MN}$ is the distance between the parallel lines of $\\mathrm{PQ}$ and $\\mathrm{BC}$, i.e., the minimum value of $\\mathrm{MN}$ is\n\n4 ;\n\nWhile the maximum value of $\\mathrm{MN}$ equals the length of the rectangle's diagonal, i.e.,\n\n$\\sqrt{P B^{2}+B C^{2}}=\\sqrt{4^{2}+6^{2}}=2 \\sqrt{13}$,\n\nThe perimeter of quadrilateral $\\mathrm{M}_{1} \\mathrm{~N}_{1} \\mathrm{~N}_{2} \\mathrm{M}_{2}$ $=2 \\mathrm{BC}+2 \\mathrm{MN}=12+2 \\mathrm{MN}$,\n\n$\\therefore$ the maximum value is $12+2 \\times 2 \\sqrt{13}=12+4 \\sqrt{13}$.\n\nTherefore, the answer is: $12+4 \\sqrt{13}$.\n\n【Highlight】This problem tests manual dexterity and spatial imagination through the cutting and rearranging of shapes. Determining the shape after cutting and rearranging, and exploring the different positional relationships of MN to find the extremum of the quadrilateral's perimeter, is the key to solving the problem." }, { "problem_id": 1492, "question": "In the rectangle $A B C D$ shown in Figure (1), point $E$ is on $A D$, and by folding point $A$ to the right along $B E$, it forms the configuration shown in Figure (2). Then, $A F \\perp C D$ is drawn at point $F$, as shown in Figure (3). If $A B=2, B C=3, \\angle B E A=60^{\\circ}$, then the length of $A F$ in Figure (3) is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch21-2024_06_15_bdd83528d6ff58166c03g_0070_1.jpg", "batch21-2024_06_15_bdd83528d6ff58166c03g_0070_2.jpg", "batch21-2024_06_15_bdd83528d6ff58166c03g_0070_3.jpg" ], "is_multi_img": true, "answer": "$3-\\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure below, draw $\\mathrm{AH} \\perp \\mathrm{BC}$ at $\\mathrm{H}$. Then $\\angle \\mathrm{AHC}=90^{\\circ}$,\n\n\n\nSince quadrilateral $A B C D$ is a rectangle,\n\n$\\therefore \\angle \\mathrm{B}=\\angle \\mathrm{C}=\\angle \\mathrm{EAB}=90^{\\circ}$,\n\nSince $A F \\perp C D$,\n\n$\\therefore \\angle \\mathrm{AFC}=90^{\\circ}$,\n\n$\\therefore$ quadrilateral $\\mathrm{AFCH}$ is a rectangle, and $A F=C H$,\n\nSince $\\angle B E A=60^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{EAB}=30^{\\circ}$,\n\n$\\therefore$ According to the properties of folding, $\\angle \\mathrm{AEH}=90^{\\circ}-2 \\angle \\mathrm{EAB}=30^{\\circ}$,\n\nSince in right triangle $\\triangle \\mathrm{ABH}$, $\\mathrm{AB}=2$,\n\n$\\therefore A H=\\frac{1}{2} A B=1$,\n\nBy the Pythagorean theorem, $B H=\\sqrt{A B^{2}-A H^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}$\n\nSince $\\mathrm{BC}=3$,\n\n$\\therefore A F=H C=B C-B H=3-\\sqrt{3}$.\n\nTherefore, the answer is: $3-\\sqrt{3}$.\n\n【Key Insight】This problem examines the properties and determination of rectangles, folding transformations, the Pythagorean theorem, and right triangles with a $30^{\\circ}$ angle. The key to solving this problem is the ability to construct auxiliary lines to form right triangles." }, { "problem_id": 1493, "question": "As shown in the figure, Figure (1) is a quadrilateral paper strip \\(ABCD\\) where \\(AB \\parallel CD\\), and \\(E\\), \\(F\\) are two points on sides \\(AB\\) and \\(CD\\) respectively, with \\(\\angle BEF = 27^\\circ\\). The paper strip \\(ABCD\\) is folded along the line containing \\(EF\\) to obtain Figure (2). Then, the quadrilateral \\(BCFM\\) in Figure (2) is folded along the line containing \\(DF\\) to obtain Figure (3). The degree measure of \\(\\angle EFC\\) in Figure (3) is \\qquad. \n\nFigure (1)\n\n Figure 2\n\n\n\n Figure(3)", "input_image": [ "batch21-2024_06_15_bdd83528d6ff58166c03g_0083_1.jpg", "batch21-2024_06_15_bdd83528d6ff58166c03g_0083_2.jpg", "batch21-2024_06_15_bdd83528d6ff58166c03g_0083_3.jpg" ], "is_multi_img": true, "answer": "$99^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution:\n\n\n\nAs shown in Figure (1), draw EB' // DF. From the folding, we have: $\\angle B^{\\prime} E F=\\angle F E M=27^{\\circ}$,\n\n$\\therefore \\angle B^{\\prime} E M=\\angle B^{\\prime} E F+\\angle F E M=27^{\\circ}+27^{\\circ}=54^{\\circ}$\n\n$\\because A E / / D F$,\n\n$\\therefore \\angle E F M=\\angle B^{\\prime} E F=27^{\\circ}, \\angle D M E=\\angle B^{\\prime} E M=54^{\\circ}$,\n\n$\\therefore \\angle B M F=\\angle D M E=54^{\\circ}$\n\n$\\because B M / / C F$,\n\n$\\therefore \\angle C F M+\\angle B M F=180^{\\circ}$,\n\n$\\therefore \\angle C F M=180^{\\circ}-54^{\\circ}=126^{\\circ}$\n\n\n\nFigure (3)\n\nFrom the folding, we have:\n\nAs shown in Figure (3), $\\angle M F C=126^{\\circ}$,\n\n$\\therefore \\angle E F C=\\angle M F C-\\angle E F M=126^{\\circ}-27^{\\circ}=99^{\\circ}$,\n\nTherefore, the answer is: $99^{\\circ}$\n\n【Insight】This question examines the properties of parallel lines and the properties of folding transformations; mastering the properties of parallel lines and folding transformations to derive equal angles is key to solving the problem." }, { "problem_id": 1494, "question": "In the rectangle $A B C D$ shown in Figure (1), point $E$ is on $A D$, and $B E = 2 A E$. Now, using $B E$ and $C E$ as fold lines, we fold $A$ and $D$ towards $B C$. Figure (2) shows the position where all five points $A, B, C, D, E$ are on the same plane after folding. If in Figure (2), $\\angle A E D = 15^\\circ$, then the degree measure of $\\angle B C E$ is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch21-2024_06_15_bdd83528d6ff58166c03g_0099_1.jpg", "batch21-2024_06_15_bdd83528d6ff58166c03g_0099_2.jpg" ], "is_multi_img": true, "answer": "$37.5^{\\circ}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "In rectangle \\( A B C D \\), \\( \\angle A = 90^\\circ \\), and \\( AD \\parallel BC \\).\n\nGiven:\n\\[\n\\begin{aligned}\n& \\because BE = 2AE, \\\\\n& \\therefore \\angle ABE = 30^\\circ, \\\\\n& \\therefore \\angle AEB = 90^\\circ - \\angle ABE = 90^\\circ - 30^\\circ = 60^\\circ, \\\\\n& \\because \\angle AED = 15^\\circ, \\\\\n& \\therefore \\angle BED = \\angle AEB - \\angle AED = 60^\\circ - 15^\\circ = 45^\\circ,\n\\end{aligned}\n\\]\n\nThus,\n\\[\n\\angle DED' = 180^\\circ - 60^\\circ - 45^\\circ = 75^\\circ.\n\\]\n\nAccording to the property of folding,\n\\[\n\\angle CED' = \\frac{1}{2} \\angle DED' = \\frac{1}{2} \\times 75^\\circ = 37.5^\\circ,\n\\]\n\\[\n\\therefore \\angle BCE = \\angle CED' = 37.5^\\circ.\n\\]\n\nTherefore, the answer is: \\( 37.5^\\circ \\).\n\n\n\n(1)\n\n\n\n(2)\n\n**Key Insight:** This problem examines the properties of a right triangle with a \\( 30^\\circ \\) angle, the properties of folding, and the properties of a parallelogram. The key to solving the problem lies in understanding the properties of a right triangle with a \\( 30^\\circ \\) angle, the properties of folding, and the properties of a parallelogram." }, { "problem_id": 1495, "question": "As shown in the figure, the rectangular paper $A B C D$ (Figure (1)) is operated as follows: (1) Fold the paper along a line passing through point $\\mathrm{A}$, such that point $B$ exactly falls on the $A D$ side, and the crease intersects the $B C$ side at point $E$ (as shown in Figure (2)); (2) Fold the paper along a line passing through point $E$, such that point $\\mathrm{A}$ falls on the $B C$ side, and the crease $E F$ intersects the $A D$ side at point $F$ (as shown in Figure (3)); then the degree measure of $\\angle A E F$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (8)\n\n\n\nFigure (3)", "input_image": [ "batch21-2024_06_15_c455d5467ab3ddfddedfg_0077_1.jpg", "batch21-2024_06_15_c455d5467ab3ddfddedfg_0077_2.jpg", "batch21-2024_06_15_c455d5467ab3ddfddedfg_0077_3.jpg" ], "is_multi_img": true, "answer": "$67.5^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since \\( \\mathrm{AB} = \\mathrm{BE} \\) and \\( \\angle \\mathrm{ABE} = 90^\\circ \\),\n\nit follows that \\( \\angle \\mathrm{AEB} = \\angle \\mathrm{BAE} = 45^\\circ \\).\n\nIn Figure (3),\n\nsince \\( \\mathrm{DF} \\parallel \\mathrm{CE} \\),\n\nwe have \\( \\angle \\mathrm{DFA} = \\angle \\mathrm{EAF} = 45^\\circ \\).\n\nTherefore, \\( \\angle \\mathrm{AFE} = \\angle \\mathrm{EFA'} = \\frac{180^\\circ - 45^\\circ}{2} = 67.5^\\circ \\),\n\nand \\( \\angle \\mathrm{AEF} = \\angle \\mathrm{FEA'} = 180^\\circ - 67.5^\\circ - 45^\\circ = 67.5^\\circ \\).\n\nThus, the answer is: \\( 67.5^\\circ \\).\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n【Key Insight】This problem examines the properties of flip transformations. It is essential to recognize that flipping is a type of symmetry transformation, belonging to axial symmetry. The shape and size of the figure remain unchanged before and after the flip, but the position changes, and the corresponding sides and angles are equal. This understanding is crucial for solving the problem." }, { "problem_id": 1496, "question": "As shown in Figure 1, a rectangular paper piece $A B C D$ with $A B=a$ and $B C=b$ satisfies $\\frac{1}{2} b\n\n(Figure 1)\n\n\n\n(Figure 2)\n\n\n\n(Figure 3)\n\n\n\n(Figure 4)", "input_image": [ "batch21-2024_06_15_c455d5467ab3ddfddedfg_0095_1.jpg", "batch21-2024_06_15_c455d5467ab3ddfddedfg_0095_2.jpg", "batch21-2024_06_15_c455d5467ab3ddfddedfg_0095_3.jpg", "batch21-2024_06_15_c455d5467ab3ddfddedfg_0095_4.jpg" ], "is_multi_img": true, "answer": "$2 b-2 a$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "According to the problem, the figure is as follows:\n\n\n\n$\\because$ In rectangle $\\mathrm{ABCD}$, $AB = a$, $BC = b$\n\n$\\therefore \\mathrm{AE} = \\mathrm{GI} = \\mathrm{BF} = \\frac{b}{2}$, $\\angle \\mathrm{A} = 90^{\\circ}$\n\n$\\because \\triangle \\mathrm{GEI}$ is obtained by folding $\\triangle \\mathrm{GEA}$\n\n$\\therefore \\mathrm{EI} = \\mathrm{AE} = \\frac{b}{2} = \\mathrm{GI}$, $\\angle \\mathrm{EIG} = 90^{\\circ}$,\n\n$\\therefore \\mathrm{IF} = a - \\frac{b}{2}$\n\n$\\therefore \\triangle \\mathrm{GIE}$ is an isosceles right triangle,\n\n$\\therefore \\angle \\mathrm{IEG} = 45^{\\circ}$\n\n$\\because \\angle \\mathrm{EFB} = 90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{EJ} B_{1} = 90^{\\circ}$\n\n$\\therefore \\triangle \\mathrm{EJM}$ is an isosceles right triangle\n$\\because \\mathrm{IF} = a - \\frac{b}{2}$,\n\n$\\therefore \\mathrm{J} = a - \\frac{b}{2}$,\n\n$\\therefore \\mathrm{JE} = a - \\left(a - \\frac{b}{2}\\right) - \\left(a - \\frac{b}{2}\\right) = b - a$\n\n$\\therefore$ In the isosceles right triangle $\\triangle \\mathrm{EJM}$, $\\mathrm{JM} = b - a$\n\n$\\therefore \\mathrm{MN} = 2b - 2a$\n\nTherefore, the answer is: $2b - 2a$\n\n【Highlight】This problem examines the folding of a rectangle. The key to solving it lies in utilizing the characteristics of folding, which result in the figures before and after folding being completely identical." }, { "problem_id": 1497, "question": "As shown in the figure, Figure 1 is a packaging bottom box made of carbon cardboard, which is an open-top hexagonal prism box. The sides are rectangular or square. After measurement, the hexagon base has three sides of length $6 \\mathrm{~cm}$ and three sides of length $2 \\mathrm{~cm}$, with each internal angle being $120^{\\circ}$. The height of the hexagonal prism is $2 \\mathrm{~cm}$. Now, if it is cut along its lateral edges and flattened, a planar unfolding diagram as shown in Figure 2 is obtained. If a piece of equilateral triangular cardboard is cut along the dashed lines in Figure 3 to form the template shown in Figure 2, then the side length of this equilateral triangle should be at least $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_c485505caa9fed29584dg_0001_1.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0001_2.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0001_3.jpg" ], "is_multi_img": true, "answer": "$(4 \\sqrt{3}+10)$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure:\n\n\n\n$\\because$ Quadrilateral $DCFE$ is a rectangle, quadrilateral $FIKJ$ is a square, and $\\angle EFI = 120^\\circ$,\n\n$\\therefore CF = FI = FJ = 2 \\text{ cm}, \\quad \\angle CFJ = 360^\\circ - 120^\\circ - 90^\\circ - 90^\\circ = 60^\\circ$.\n\nExtend $FJ$ to intersect $AB$ at point $G$, and extend $IK$ to intersect $BH$ at point $H$,\n\nthen $\\angle CGF = 30^\\circ$,\n\n$\\therefore CG = 2\\sqrt{3} \\text{ cm}, GF = 2CF = 2FJ = 4 \\text{ cm}, \\angle BGH = 180^\\circ - 30^\\circ - 90^\\circ = 60^\\circ$,\n\n$\\therefore \\triangle BGH$ is an equilateral triangle, and quadrilateral $GHKJ$ is a square,\n\n$\\therefore BG = GH = FI = 2 \\text{ cm}$,\n\n$\\therefore AD = BC = CG + BG = 2\\sqrt{3} + 2 \\text{ cm}$,\n\n$\\therefore AB = AD + CB + CD = 4\\sqrt{3} + 4 + 6 = 4\\sqrt{3} + 10 \\text{ cm}$,\n\n$\\therefore$ The side length of this equilateral triangle should be at least $(4\\sqrt{3} + 10) \\text{ cm}$.\n\nHence, the answer is: $(4\\sqrt{3} + 10)$.\n\n【Insight】This problem examines the determination and properties of equilateral triangles and squares. The key to solving the problem lies in skillfully constructing a 30-degree right triangle, an equilateral triangle, and a square within the figure, and performing calculations based on their properties." }, { "problem_id": 1498, "question": "The tangram is an invention of the ancient Chinese working people and is an ancient traditional intellectual game in China. Xiao Hui used a set of tangrams shown in Figure 1 to create the \"bowing figure\" as shown in Figure 2. Given that the side length of the square $A B C D$ is $4 \\mathrm{dm}$, the value of $h$ in Figure 2 is $\\qquad$ $\\mathrm{dm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_c485505caa9fed29584dg_0028_1.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0028_2.jpg" ], "is_multi_img": true, "answer": "$4+\\sqrt{2} \\# \\# \\sqrt{2}+4$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the legs of the isosceles right triangle be $x$, then the hypotenuse $=\\sqrt{x^{2}+x^{2}}=\\sqrt{2} x$.\n\n$\\therefore$ The hypotenuse of the isosceles right triangle is $\\sqrt{2}$ times the length of its legs. The altitude to the hypotenuse coincides with the median, so the altitude equals half the length of the hypotenuse.\n\n(1)(2) are isosceles right triangles with a hypotenuse of $4 \\mathrm{dm}$, so the legs are $2 \\sqrt{2} \\mathrm{dm}$.\n\nThe legs of the isosceles right triangle (6) are equal in length to the sides of the square (5) and sum up to $2 \\sqrt{2} \\mathrm{dm}$, so the side length of the square is $\\sqrt{2} \\mathrm{dm}$.\n\nThe legs of the isosceles right triangle (6) are $\\sqrt{2} \\mathrm{dm}$, so the hypotenuse is $2 \\mathrm{dm}$.\n\n$\\therefore$ The legs of the isosceles right triangle (7) are $2 \\mathrm{dm}$.\n\n$\\therefore$ The altitude to the hypotenuse of the isosceles right triangle $(7)$ is $\\sqrt{2} \\mathrm{dm}$.\n\nIn Figure 1, the sum of the altitudes of the isosceles right triangles (2)(3) and the parallelogram (4) equals the side length of the square $A B C D$, which is $4 \\mathrm{dm}$.\n\n$\\because$ The legs of the two isosceles right triangles (3)(6) are equal to the side length of the square (5).\n\n$\\therefore$ (3)6 are two identical isosceles right triangles.\n\n$\\therefore$ In Figure 2, $h=$ the side length of the square $A B C D$ + the altitude to the hypotenuse of the isosceles right triangle (7).\n\n$\\therefore h=(4+\\sqrt{2}) \\mathrm{dm}$.\n\nTherefore, the answer is: $4+\\sqrt{2}$\n\n【Highlight】This problem examines the properties of squares, isosceles right triangles, and the Pythagorean theorem. Mastering the side length relationships of isosceles right triangles is key to solving the problem." }, { "problem_id": 1499, "question": "Given, as shown in Figure 1, a square cardboard with side length 4 is cut along the dividing lines to form a set of tangram pieces. Among them, Figure (1) is a square, Figure (2) is a parallelogram, and Figures (3)(4)(5)(6)(7) are all isosceles right-angled triangles. Now, using this set of tangram pieces, a new square is formed as shown in Figure 2. The gap in the figure is represented by a shaded arrow shape $A B C D E F G H$, which consists of an isosceles right-angled triangle $A B C$, an isosceles right-angled triangle $E F G$, and a rectangle $A C D H$. The quadrilateral $E F M N$ is Figure (1).\n\n\n\nFigure 1\n\n\n\n(1) The side length of the new square is $\\qquad$ ;\n\n(2) The perimeter of the arrow shape is $\\qquad$ .", "input_image": [ "batch21-2024_06_15_c485505caa9fed29584dg_0031_1.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0031_2.jpg" ], "is_multi_img": true, "answer": "$3 \\sqrt{2} \\quad 16 \\sqrt{2}-12 \\# \\#-12+16 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) The segmentation diagram in Figure 2 is as follows. Combined with Figure 1, it is known that $Q M=M N=N P=\\frac{1}{4} \\times 4 \\sqrt{2}=\\sqrt{2}$,\n\n$\\therefore P Q=Q M+M N+N P=3 \\sqrt{2}$\n\n$\\therefore$ The side length of the new square is $3 \\sqrt{2}$.\n\nTherefore, the answer is: $3 \\sqrt{2}$;\n\n\n\n(2) $\\because A B=B C=3 \\sqrt{2}-4$,\n\n$\\therefore H D=A C=\\sqrt{2} A B=\\sqrt{2}(3 \\sqrt{2}-4)=6-4 \\sqrt{2}, H A=D C=2 \\sqrt{2}, G E=\\sqrt{2} E F=2$,\n\n$\\therefore G H+D E=G E-H D=G E-A C=2-(6-4 \\sqrt{2})=4 \\sqrt{2}-4$,\n\n$\\therefore$ The perimeter of the arrow figure is: $A B+B C+C D+D E+E F+F G+G H+H A$\n\n$=2(3 \\sqrt{2}-4)+4 \\sqrt{2}+(4 \\sqrt{2}-4)+2 \\sqrt{2}=16 \\sqrt{2}-12$,\n\nTherefore, the answer is: $16 \\sqrt{2}-12$\n\n【Highlight】This problem tests the ability to calculate common figures through the tangram." }, { "problem_id": 1500, "question": "As shown in the figure, Xiao Ming made a movable rhombus tool using four wooden strips of equal length. He first adjusted the tool into the rhombus shown in Figure 1, measuring $\\angle A = 120^{\\circ}$. Then, he adjusted the tool into the square shown in Figure 2, measuring the diagonal $AC = 30 \\mathrm{~cm}$. The length of the diagonal $AC$ in Figure 1 is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_c485505caa9fed29584dg_0039_1.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0039_2.jpg" ], "is_multi_img": true, "answer": "$15 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in Figures 1 and 2, connect $AC$.\n\n\n\nFigure 2\n\nIn Figure 2, since quadrilateral $ABCD$ is a square,\n\n$\\therefore AB = BC$, and $\\angle B = 90^{\\circ}$.\n\nGiven that $AC = 30$,\n\n$\\therefore AB = BC = 15 \\sqrt{2}$.\n\nIn Figure 1, since $\\angle BAD = 120^{\\circ}$,\n\n$\\therefore \\angle B = 60^{\\circ}$, and $BA = BC$,\n\n$\\therefore \\triangle ABC$ is an equilateral triangle,\n\n$\\therefore AC = BC = 15 \\sqrt{2}$.\n\nHence, the answer is: $15 \\sqrt{2}$.\n\n【Insight】This problem examines the properties of rhombuses, squares, and the Pythagorean theorem. The key to solving it lies in the flexible application of the learned knowledge, which is a common type of question in middle school exams." }, { "problem_id": 1501, "question": "In the pentagon paper piece $A B C D E$, $B C=C D=E D, A B=A E, \\angle B C D=\\angle D=90^{\\circ}$. As shown in Figure 1, first cut it along the dotted lines $A F, C F$ into three pieces, then use these three small paper pieces to splice together, exactly forming a square without gaps or overlaps as shown in Figure 2. The value of $\\frac{A F}{A B}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_c485505caa9fed29584dg_0044_1.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0044_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{\\sqrt{10}}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "As shown in the figure, the assembled square is $AHCF$, and $BE$ is connected.\n\n\n\nFrom the cutting and pasting, we know: $\\triangle AEF \\cong \\triangle ABH$, $\\triangle CFD \\cong \\triangle CHB$.\n\nLet $CD = a$, $EF = b$, $AE = c$.\n\nSince $\\triangle AEF \\cong \\triangle ABH$,\n\n$\\therefore EF = BH = b$.\n\nSince $\\triangle CFD \\cong \\triangle CHB$,\n\n$\\therefore DF = BH = b$,\n\n$\\therefore EG = DF = b$.\n\nSince $BC = CD = ED$, and $\\angle BCD = \\angle D = 90^{\\circ}$,\n\n$\\therefore 2b = a$, and $BE = CD = a$.\n\nSince $\\angle HAF = \\angle BAH + \\angle BAF = \\angle EAF + \\angle BAF = \\angle BAE = 90^{\\circ}$,\n\n$\\therefore$ In the right triangle $\\triangle BAE$, $AB^{2} + AE^{2} = BE^{2}$,\n\n$\\therefore c^{2} + c^{2} = a^{2}$,\n\n$\\therefore a^{2} = 2c^{2}$,\n\n$\\therefore a = \\sqrt{2}c$.\n\nIn the right triangle $\\triangle CDF$, $CD^{2} + DF^{2} = CF^{2}$,\n\n$\\therefore a^{2} + \\left(\\frac{1}{2}a\\right)^{2} = CF^{2}$,\n\n$\\therefore CF = \\frac{\\sqrt{5}}{2}a$ or $CF = -\\frac{\\sqrt{5}}{2}a$ (discarded),\n\n$\\therefore CF = \\frac{\\sqrt{10}}{2}c$.\n\nSince $AE = AB = c$, and $CF = AF$,\n\n$\\therefore \\frac{AF}{AB} = \\frac{\\sqrt{10}}{2}$.\n\nTherefore, the answer is: $\\frac{\\sqrt{10}}{2}$.\n\n【Insight】This problem mainly examines the properties of figure assembly, the properties of squares, and solving right triangles. The key to solving this problem is learning to use parameters effectively." }, { "problem_id": 1502, "question": "【Reading Material】As shown in Figure (1), in quadrilateral $ABCD$, $AB=AD$, $\\angle B+\\angle D=180^{\\circ}$, points $E$ and $F$ are on $BC$ and $CD$ respectively. If $\\angle BAD=2 \\angle EAF$, then $EF=BE+DF$.\n\n\n\nFigure (1)\n\n【Solving the Problem】As shown in Figure (2), in a park on the same horizontal plane, four roads form quadrilateral $ABCD$. It is known that $CD=CB=100 \\mathrm{~m}$, $\\angle D=60^{\\circ}$, $\\angle ABC=120^{\\circ}$, $\\angle BCD=150^{\\circ}$, and there are scenic spots $M$ and $N$ on roads $AD$ and $AB$ respectively, with $DM=100 \\mathrm{~m}$ and $BN=50(\\sqrt{3}-1) \\mathrm{m}$. If a straight road is built between $M$ and $N$, the length of route $M \\rightarrow N$ is shorter than the length of route $M \\rightarrow A \\rightarrow N$ by $\\qquad$ $\\mathrm{m}$ (rounded to the nearest whole number, reference data: $\\sqrt{3} \\approx 1.7$).\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_c485505caa9fed29584dg_0051_1.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0051_2.jpg" ], "is_multi_img": true, "answer": "370", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "**Solution:** As shown in the figure, extend lines \\( AB \\) and \\( DC \\) to meet at point \\( E \\), and connect points \\( C M \\) and \\( C N \\).\n\n\n\nGiven:\n- \\( \\angle D = 60^\\circ \\),\n- \\( \\angle ABC = 120^\\circ \\),\n- \\( \\angle BCD = 150^\\circ \\),\n\nWe deduce:\n- \\( \\angle A = 30^\\circ \\),\n- \\( \\angle E = 90^\\circ \\).\n\nSince \\( DC = DM = 100 \\),\n- Triangle \\( DCM \\) is equilateral,\n- Therefore, \\( \\angle DCM = 60^\\circ \\),\n- Hence, \\( \\angle BCM = 90^\\circ \\).\n\nIn the right triangle \\( BCE \\):\n- \\( BC = 100 \\),\n- \\( \\angle ECB = 180^\\circ - \\angle BCD = 30^\\circ \\),\n- \\( EB = \\frac{1}{2} BC = 50 \\),\n- \\( EC = \\sqrt{3} \\cdot EB = 50\\sqrt{3} \\).\n\nThus:\n- \\( DE = DC + EC = 100 + 50\\sqrt{3} \\).\n\nIn the right triangle \\( ADE \\):\n- \\( AD = 2DE = 200 + 100\\sqrt{3} \\),\n- \\( AE = \\sqrt{3} \\cdot DE = 100\\sqrt{3} + 150 \\).\n\nTherefore:\n- \\( AM = AD - DM = 200 + 100\\sqrt{3} - 100 = 100 + 100\\sqrt{3} \\),\n- \\( AN = AB - BN = (AE - EB) - BN = (100\\sqrt{3} + 150 - 50) - 50(\\sqrt{3} - 1) = 50\\sqrt{3} + 150 \\).\n\nHence:\n- \\( AM + AN = 100 + 100\\sqrt{3} + 50\\sqrt{3} + 150 = 250 + 150\\sqrt{3} \\).\n\nIn the right triangle \\( CMB \\):\n- \\( BM = \\sqrt{BC^2 + CM^2} = 100\\sqrt{2} \\).\n\nSince:\n- \\( EN = EB + BN = 50 + 50(\\sqrt{3} - 1) = 50\\sqrt{3} = EC \\),\n- Triangle \\( ECN \\) is an isosceles right triangle.\n\nTherefore:\n- \\( \\angle NCM = \\angle BCM - \\angle NCB = \\angle BCM - (\\angle NCE - \\angle BCE) = 75^\\circ = \\frac{1}{2} \\angle DCB \\).\n\nFrom the given material:\n- \\( MN = DM + BN = 100 + 50(\\sqrt{3} - 1) = 50(\\sqrt{3} + 1) \\).\n\nThus:\n- The length of the path \\( M \\rightarrow N \\) is shorter than the path \\( M \\rightarrow A \\rightarrow N \\) by:\n \\[\n 250 + 150\\sqrt{3} - 50(\\sqrt{3} + 1) = 200 + 100\\sqrt{3} \\approx 370 \\text{ m}.\n \\]\n\n**Final Answer:** 370.\n\n**Key Insight:** This problem tests the properties of right triangles with a 30-degree angle and the Pythagorean theorem. Understanding the given conditions is crucial for solving the problem." }, { "problem_id": 1503, "question": "As shown, a square with side length $12 \\mathrm{~cm}$ is divided to make a set of tangram pieces. Figure 2 shows a \"little house\" formed by arranging these pieces, where the area of the shaded part is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_c485505caa9fed29584dg_0062_1.jpg", "batch21-2024_06_15_c485505caa9fed29584dg_0062_2.jpg" ], "is_multi_img": true, "answer": "72", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Extend the diagonal $AG$ to point $D$, as shown in the figure:\n\n\n\nFigure 1\n\n$\\therefore BE = EF = FG = GD$\n\n$\\therefore$ The area of the shaded part is equal to half the area of a square with a side length of $12 \\mathrm{~cm}$, that is, $S_{\\text{shadow}} = S_{\\triangle ACD}$.\n\nThus, according to the problem, $12 \\times 12 \\div 2 = 72\\left(\\mathrm{~cm}^{2}\\right)$.\n\nTherefore, the answer is: 72.\n\n【Insight】This problem mainly examines the tangram puzzle, as well as the methods for calculating the areas of squares and triangles, which should be thoroughly mastered." }, { "problem_id": 1504, "question": "As shown in Figure 1, fold the square paper $ABCD$ so that $AB$ coincides with $CD$, with the fold line being $EF$. As shown in Figure 2, unfold and then fold again so that point $C$ coincides with point $E$, with the fold line being $GH$. The corresponding point of $B$ is $M$, and $EM$ intersects $AB$ at $N$. If $AD = 8$, then the length of the fold line $GH$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_c704fad97646df3fdc39g_0038_1.jpg", "batch21-2024_06_15_c704fad97646df3fdc39g_0038_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{5}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Connect $CE$, and draw $GJ \\perp CD$ at point $J$. Let the intersection of $EC$ and $GH$ be point $O$.\n\n\n\nFigure 2\n\nSince quadrilateral $ABCD$ is a square,\n\n$\\therefore \\angle BCD = 90^\\circ$, $\\angle B = \\angle BCD = 90^\\circ$,\n\n$\\therefore$ quadrilateral $BCJG$ is a rectangle,\n\n$\\therefore GJ = BC = EF$.\n\nFrom the folding, we have: $E$ and $C$ are symmetric with respect to $GH$,\n\n$\\therefore EC \\perp GH$, and $AB = EF = CD$,\n\n$\\therefore \\angle OHC + \\angle OCH = 90^\\circ$,\n\nAlso, $\\angle OCH + \\angle ECF = 90^\\circ$,\n\n$\\therefore \\angle ECF = \\angle GHJ$.\n\nIn triangles $\\triangle EFC$ and $\\triangle GJH$,\n\n$\\left\\{\\begin{array}{l}\\angle ECF = \\angle GHJ \\\\ \\angle CFE = \\angle HJG \\\\ EF = GJ\\end{array}\\right.$,\n\n$\\therefore \\triangle EFC \\cong \\triangle GJH$ (by AAS),\n\n$\\therefore EC = GH$.\n\nGiven that $AD = 8$,\n\n$\\therefore EF = 8$, and $CF = 4$,\n\n$\\therefore GH = CE = \\sqrt{8^2 + 4^2} = 4\\sqrt{5}$.\n\nThus, the answer is: $4\\sqrt{5}$.\n\n【Insight】This problem examines the properties of squares, folding transformations, the Pythagorean theorem, and the criteria and properties of congruent triangles. The key to solving the problem lies in adding common auxiliary lines and constructing congruent triangles to find the solution, making it a challenging problem typical of high school entrance exams." }, { "problem_id": 1505, "question": "The tangram, also known as the \"wisdom board,\" is an outstanding creation of our ancient ancestors. Xiao Hua used a tangram (as shown in Figure 1) to assemble a model of a house (as shown in Figure 2). Given that the side length of the square $A B C D$ in Figure 1 is $4 \\mathrm{~cm}$, the perimeter of the hexagon EFGHIJ in Figure 2 is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_c704fad97646df3fdc39g_0053_1.jpg", "batch21-2024_06_15_c704fad97646df3fdc39g_0053_2.jpg" ], "is_multi_img": true, "answer": "$(8 \\sqrt{2}+4)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Add nodes $K$ and $L$ to Figure 2:\n\n\n\nFigure 2\n\nBy observing Figure 1 and Figure 2, we can see:\n\n$EK = EF = FL = HG = \\frac{1}{2} BD$,\n\n$JI = KH = LG = \\frac{1}{2} EK = \\frac{1}{4} BD$,\n\n$EJ = IH$,\n\nSince the side length of square $ABCD$ is $4 \\mathrm{~cm}$,\n\n$\\therefore BD = \\sqrt{4^{2} + 4^{2}} = 4 \\sqrt{2}$,\n\n$FL = EF = HG = \\frac{1}{2} \\times 4 \\sqrt{2} = 2 \\sqrt{2}$,\n\n$JI = KH = LG = \\frac{1}{2} EK = \\frac{1}{4} \\times 4 \\sqrt{2} = \\sqrt{2}$,\n\nThen $EJ = IH = 2$,\n\n$\\therefore$ The perimeter of hexagon $EFGKIJ$ is:\n\n$EJ + JI + IH + HG + (LG + FL) + EF = 2 + \\sqrt{2} + 2 + 2 \\sqrt{2} + \\sqrt{2} + 2 \\sqrt{2} + 2 \\sqrt{2} = 8 \\sqrt{2} + 4$.\n\nTherefore, the answer is: $(8 \\sqrt{2} + 4) \\mathrm{cm}$.\n\n【Highlight】This question tests the understanding of tangram diagrams, properties of isosceles right triangles, and the Pythagorean theorem. It requires careful observation of the figures and analysis of the length relationships between the segments." }, { "problem_id": 1506, "question": "$\\triangle A B C$ is an isosceles right triangle cardboard, $\\angle C=R t \\angle, A C=B C=4$. A largest possible square is cut out from this cardboard, which is called the 1st cut; in the remaining Rt $\\triangle A D E$ and Rt $\\triangle B D F$, squares are cut out, resulting in two identical squares, which is called the 2nd cut (as shown in Figure 2); the process continues...; after the 64th cut, the sum of the areas of all the remaining small triangles is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n", "input_image": [ "batch21-2024_06_15_c72ed8c65b301e6be3f5g_0068_1.jpg", "batch21-2024_06_15_c72ed8c65b301e6be3f5g_0068_2.jpg", "batch21-2024_06_15_c72ed8c65b301e6be3f5g_0068_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{1}{2^{61}}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: Since quadrilateral ECFD is a square,\n\nTherefore, DE = EC = CF = DF, and ∠AED = ∠DFB = 90°.\n\nSince triangle ABC is an isosceles right triangle,\n\nTherefore, ∠A = ∠B = 45°,\n\nHence, AE = DE = EC = DF = BF = EC = CF.\n\nGiven that AC = BC = 4,\n\nTherefore, DE = DF = 2.\n\nThus, the area of triangle AED plus the area of triangle DBF equals the area of square ECFD, which is S₁ = 4.\n\nSimilarly, S₂ represents the sum of the areas of the remaining triangles after the second cut, and Sₙ represents the sum of the areas of the remaining triangles after the nth cut.\n\nTherefore, after the first cut, the sum of the areas of the remaining triangles is: 8 - S₁ = 4 = 2² = S₁.\n\nAfter the second cut, the sum of the areas of the remaining triangles is: S₁ - S₂ = 4 - 2 = 2¹ = S₂.\n\nAfter the third cut, the sum of the areas of the remaining triangles is: S₂ - S₃ = 2⁰ = S₃.\n\nAfter the nth cut, the sum of the areas of the remaining triangles is: Sₙ₋₁ - Sₙ = Sₙ = 1 / 2ⁿ⁻³.\n\nThus, after the 64th cut, the sum of the areas of all the remaining small triangles is: 1 / 2⁶¹.\n\nTherefore, the answer is 1 / 2⁶¹.\n\n[Insight] This problem examines the properties of squares and isosceles right triangles. It is a challenging problem that belongs to the category of pattern recognition. The key to solving this problem lies in identifying the pattern: Sₙ represents the sum of the areas of the remaining triangles after the nth cut." }, { "problem_id": 1507, "question": "Mathematician Academician Wu Wenjun attached great importance to the proposition proposed by the ancient mathematician Jia Xian: \"From any point on the diagonal of a rectangle, draw two lines parallel to the adjacent sides, then the areas of the two rectangles thus formed are equal\" (as shown in Figure 1: \"$S_{\\text{rectangle } D N F G}=S_{\\text{rectangle } F E B M}$\"). Starting from this proposition, he used the principle of \"complementary subtraction and addition\" to restore the ancient proofs of the nine problems in the \"Sea Island Calculation\" (Hai Dao Suan Jing). Problem Solving: As shown in Figure 2, point $P$ is a point on the diagonal $B D$ of rectangle $A B C D$. A line $E F$ is drawn through point $P$ parallel to $B C$, intersecting $A B$ and $C D$ at points $E$ and $F$ respectively. Lines $A P$ and $C P$ are connected. If $D F=4$ and $E P=3$, then the sum of the areas of the shaded parts in the figure is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_c7983477ed5fa2a425a9g_0065_1.jpg", "batch21-2024_06_15_c7983477ed5fa2a425a9g_0065_2.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $P M \\perp A D$ through point $P$, intersecting $B C$ at $N$,\n\n\nSince quadrilateral $A B C D$ is a rectangle and $E F / / B C$,\n\nTherefore, quadrilaterals $A E P M$, $M P F D$, $B N P E$, and $P N C F$ are all rectangles,\n\nHence, $A E=D F=4$,\n\nThus, $S_{\\triangle A E P}=\\frac{1}{2} \\times A E \\times P E=\\frac{1}{2} \\times 4 \\times 3=6$,\n\n$S_{\\triangle A E P}=S_{\\triangle A P M}, \\quad S_{\\triangle P E B}=S_{\\triangle P B N}, \\quad S_{\\triangle P D M}=S_{\\triangle P F D}$,\n\n$S_{\\triangle P C N}=S_{\\triangle P C F}, \\quad S_{\\triangle A B D}=S_{\\triangle B C D}$,\n\nTherefore, $S_{\\text{rectangle } A E P M}=S_{\\text{rectangle } P N C F}$,\n\nHence, $S_{\\triangle A E P}=S_{\\triangle P C F}=6$,\n\nThus, $S_{\\triangle A E P}+S_{\\triangle P C F}=12$,\n\nThat is, the sum of the areas of the shaded parts in the figure is 12,\n\nTherefore, the answer is: 12.\n\n【Insight】This question tests the properties of rectangles and the area of triangles, among other knowledge. The key to solving the problem is to understand the given information and proceed with the proof based on the known conclusions." }, { "problem_id": 1508, "question": "As shown in the figure, while assembling a puzzle, Xiao Ming found that 8 identical rectangles could perfectly form a larger rectangle as shown in Figure (1). Seeing this, Xiao Hong said, \"I'll give it a try too.\" As a result, Xiao Hong managed to piece together the rectangles into a square as shown in Figure (2), with a hole left in the center, which happened to be a small square with a side length of $5 \\mathrm{~mm}$. The area of each small rectangle is __ $\\mathrm{mm}^{2}$.\n\n\n\n## Figure 1\n\n\n\nFigure 2\n\n##", "input_image": [ "batch21-2024_06_15_c7983477ed5fa2a425a9g_0074_1.jpg", "batch21-2024_06_15_c7983477ed5fa2a425a9g_0074_2.jpg" ], "is_multi_img": true, "answer": "375", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of each small rectangle be \\( x \\mathrm{~mm} \\) and the width be \\( y \\mathrm{~mm} \\).\n\nAccording to the problem, we have the system of equations:\n\\[\n\\left\\{\n\\begin{array}{l}\n3x = 5y \\\\\nx + 5 = 2y\n\\end{array}\n\\right.\n\\]\n\nSolving the system, we find:\n\\[\n\\left\\{\n\\begin{array}{l}\nx = 25 \\\\\ny = 15\n\\end{array}\n\\right.\n\\]\n\nThus, the area \\( xy \\) is:\n\\[\nxy = 25 \\times 15 = 375\n\\]\n\nTherefore, the answer is: 375.\n\n【Key Insight】This problem tests the application of a system of linear equations and the area of a rectangle. The key to solving it lies in correctly identifying the relationships and setting up the system of equations accurately." }, { "problem_id": 1509, "question": "As shown in the figure, in the rhombus paper $A B C D$, $A B=4, \\angle B A D=60^{\\circ}$, the rhombus paper is folded as follows:\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\nStep 1: As shown in Figure (1), fold the rhombus paper $A B C D$ so that the corresponding point $A^{\\prime}$ of point $A$ lies exactly on side $C D$. The fold line $E F$ intersects sides $A D$ and $A B$ at points $E$ and $F$ respectively. The fold line $E F$ intersects the line connecting the corresponding points $A$ and $A^{\\prime}$ at point $G$.\n\nStep 2: As shown in Figure (2), fold the quadrilateral paper $B C A^{\\prime} F$ so that the corresponding point $C^{\\prime}$ of point $C$ lies exactly on $A^{\\prime} F$. The fold line $M N$ intersects sides $C D$ and $B C$ at points $M$ and $N$ respectively.\n\nStep 3: Unfold the rhombus paper $A B C D$ and connect $G C^{\\prime}$. The minimum value of $G C^{\\prime}$ is $\\qquad$.", "input_image": [ "batch21-2024_06_15_c9b4e0196542f57ad332g_0043_1.jpg", "batch21-2024_06_15_c9b4e0196542f57ad332g_0043_2.jpg", "batch21-2024_06_15_c9b4e0196542f57ad332g_0043_3.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Solution:** As shown in the figure, draw \\( GH \\perp AB \\) at \\( H \\), \\( DR \\perp AB \\) at \\( R \\), \\( GP \\perp A'F \\) at \\( P \\), and \\( A'Q \\perp AB \\) at \\( Q \\).\n\n\n\nSince quadrilateral \\( ABCD \\) is a rhombus,\n\n\\[\nDA = AB = BC = CD = 4, \\quad AB \\parallel CD,\n\\]\n\n\\[\nA'Q = DR.\n\\]\n\nGiven that \\( \\angle BAD = 60^\\circ \\),\n\n\\[\nA'Q = DR = \\frac{\\sqrt{3}}{2} AD = 2\\sqrt{3}.\n\\]\n\nSince \\( A' \\) and \\( A \\) are symmetric with respect to \\( EF \\),\n\n\\[\nEF \\text{ is the perpendicular bisector of } AA',\n\\]\n\n\\[\nAG = A'G, \\quad \\angle AFE = \\angle A'FE,\n\\]\n\n\\[\nGP = PH.\n\\]\n\nMoreover, since \\( GH \\perp AB \\) and \\( A'Q \\perp AB \\),\n\n\\[\nGH \\parallel A'B,\n\\]\n\n\\[\nGH = \\frac{1}{2} A'Q = \\frac{1}{2} DR = \\sqrt{3}.\n\\]\n\nTherefore, \\( GC' \\geq GP = \\sqrt{3} \\), and the minimum value of \\( GC' \\) is \\( \\sqrt{3} \\) when \\( C' \\) coincides with \\( P \\).\n\nHence, the answer is: \\( \\sqrt{3} \\).\n\n**Key Insight:** A thorough understanding of the properties of rhombuses, the characteristics of folding, and the method for determining the shortest path is crucial for solving the problem." }, { "problem_id": 1510, "question": "Figure 1 is a pattern formed by a Penrose tiling using a \"kite\" and a \"dart\" as the basic components. The \"kite\" and \"dart\" in Figure 2 are made from special rhombuses as shown in Figure 3. In the rhombus \\(ABCD\\), \\(\\angle BAD = 72^\\circ\\). By taking \\(AE = AB\\) on the diagonal \\(AC\\) and connecting \\(BE\\) and \\(DE\\), the rhombus can be divided into a \"kite\" (convex quadrilateral \\(ABED\\)) and a \"dart\" (concave quadrilateral \\(BCDE\\)). Then, in Figure 2, \\(\\alpha = \\) \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_cac1185079aa2a9e3486g_0100_1.jpg", "batch21-2024_06_15_cac1185079aa2a9e3486g_0100_2.jpg", "batch21-2024_06_15_cac1185079aa2a9e3486g_0100_3.jpg" ], "is_multi_img": true, "answer": "144", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "In rhombus \\( A B C D \\), \\( \\angle D A E = \\angle B A E \\).\n\nGiven that \\( \\angle B A D = 72^\\circ \\),\n\nit follows that \\( \\angle D A E = \\angle B A E = \\frac{1}{2} \\times \\angle B A D = \\frac{1}{2} \\times 72^\\circ = 36^\\circ \\).\n\nSince \\( A E = A B \\),\n\nwe have \\( \\angle A E B = \\angle A B E = \\frac{1}{2} \\times (180^\\circ - 36^\\circ) = 72^\\circ \\).\n\nIn triangles \\( \\triangle A B E \\) and \\( \\triangle A D E \\),\n\n\\[\n\\begin{aligned}\n& \\left\\{\n\\begin{array}{c}\nA B = A D \\\\\n\\angle E A B = \\angle E A D \\\\\nA E = A E\n\\end{array}\n\\right. \\\\\n& \\therefore \\triangle A B E \\cong \\triangle A D E \\\\\n& \\therefore \\angle A E B = \\angle A E D = 72^\\circ \\\\\n& \\therefore \\alpha = 72^\\circ \\times 2 = 144^\\circ\n\\end{aligned}\n\\]\n\nThus, the answer is: 144\n\n【Key Insight】The problem primarily examines the properties of the diagonals of a rhombus, the properties of isosceles triangles, and the methods for proving triangle congruence, with a moderate level of difficulty." }, { "problem_id": 1511, "question": "As shown in Figure 1, this is a wooden movable coat rack. Figure 2 is a schematic diagram of the coat rack, which consists of 3 identical rhombuses, with the side length $AB$ being $30 \\mathrm{~cm}$. Xiaonan fixed the two endpoints $B$ and $E$ of the coat rack on the wall, making the distance between points $B$ and $E$ $108 \\mathrm{~cm}$. Without considering the width of the material, the distance between points $A$ and $C$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_ddaa4e8f04b94cbb8b1ag_0032_1.jpg", "batch21-2024_06_15_ddaa4e8f04b94cbb8b1ag_0032_2.jpg" ], "is_multi_img": true, "answer": "$48 \\mathrm{~cm}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $AC$ and $BD$ intersecting at point $O$,\n\n\n\nSince the distance between $BE$ is adjusted to $108 \\mathrm{~cm}$, and the wooden coat hanger is composed of three congruent rhombuses,\n\n$\\therefore BD=36 \\mathrm{~cm}$,\n\nSince quadrilateral $ABCD$ is a rhombus,\n\n$\\therefore BO=DO, AO=CO, AC \\perp BD$,\n\n$\\therefore BO=18 \\mathrm{~cm}$,\n\nSince the side length of the rhombus $AB=30 \\mathrm{~cm}$,\n\nIn right triangle $ABO$,\n\n$AO=\\sqrt{AB^{2}-BO^{2}}=\\sqrt{30^{2}-18^{2}}=24 \\mathrm{~cm}$,\n\n$\\therefore AC=48 \\mathrm{~cm}$.\n\nTherefore, the answer is: $48 \\mathrm{~cm}$\n\n【Key Insight】This problem mainly examines the properties of a rhombus and the Pythagorean theorem. Mastering the properties of a rhombus and the Pythagorean theorem is crucial for solving the problem." }, { "problem_id": 1512, "question": "As shown in Figure 1, in rhombus \\( A B C D \\), the line \\( l \\) is perpendicular to side \\( A B \\) and moves to the right from point \\( A \\) at a speed of \\( 1 \\mathrm{~cm} / \\mathrm{s} \\). If the length of the line segment \\( M N \\) intercepted by line \\( l \\) inside the rhombus \\( A B C D \\) is \\( y(\\mathrm{~cm}) \\), the graph reflecting the functional relationship between \\( y(\\mathrm{~cm}) \\) and the moving time \\( x(\\mathrm{~s}) \\) is shown in Figure 2. Then the area of rhombus \\( A B C D \\) is \\(\\qquad\\) \\(\\mathrm{cm}^{2}\\).\n\n\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_ddaa4e8f04b94cbb8b1ag_0070_1.jpg", "batch21-2024_06_15_ddaa4e8f04b94cbb8b1ag_0070_2.jpg" ], "is_multi_img": true, "answer": "20", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 2, it can be seen that when the line $l$ passes through point $D$, $x = AN = 3$ and $y = MN = 4$.\n\nTherefore, the height of rhombus $ABCD$ is equal to the length of segment $MN$.\n\nBy the Pythagorean theorem, $AD = \\sqrt{3^{2} + 4^{2}} = 5$.\n\nThus, the area of the rhombus is $5 \\times 4 = 20 \\left(\\mathrm{~cm}^{2}\\right)$.\n\nHence, the answer is: 20.\n\n[Insight] This problem involves the concept of a moving point on a function graph and the Pythagorean theorem. The key to solving such problems lies in combining the motion of the figure with the analysis of the function graph." }, { "problem_id": 1513, "question": "As shown in Figure 1, in rectangle \\( A B C D \\), diagonals \\( A C \\) and \\( B D \\) intersect at point \\( O \\). A moving point \\( P \\) starts from point \\( B \\), moves uniformly along segment \\( B C \\), and stops when it reaches point \\( C \\). Let the distance traveled by point \\( P \\) be \\( x \\), and the length of segment \\( O P \\) be \\( y \\). If the graph of the function of \\( y \\) with respect to \\( x \\) is as shown in Figure 2, then the area of rectangle \\( A B C D \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_de2762c0f79ea21fe223g_0072_1.jpg", "batch21-2024_06_15_de2762c0f79ea21fe223g_0072_2.jpg" ], "is_multi_img": true, "answer": "48", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in Figure 2, when $\\mathrm{OP} \\perp \\mathrm{BC}$, $\\mathrm{BP}=\\mathrm{CP}=4$, and $\\mathrm{OP}=3$,\n\ntherefore, $\\mathrm{AB}=2 \\mathrm{OP}=6$, and $\\mathrm{BC}=2 \\mathrm{BP}=8$,\n\nso the area of rectangle $\\mathrm{ABCD}$ is $6 \\times 8=48$.\nHence, the answer is: 48.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Insight】This problem examines the function graph of a moving point, with the key being to determine that $\\mathrm{BP}=\\mathrm{CP}=4$ and $\\mathrm{OP}=3$ based on the given function graph and the trajectory of the point." }, { "problem_id": 1514, "question": "Divide the rectangle $\\mathrm{ABCD}$ in Figure 1 into four congruent right triangles, and arrange these four right triangles into squares as shown in Figure 2 and Figure 3. According to the lengths indicated in the figures, the area of the rectangle $\\mathrm{ABCD}$ in Figure 1 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_e0c6cb68abd23015f7f5g_0026_1.jpg", "batch21-2024_06_15_e0c6cb68abd23015f7f5g_0026_2.jpg", "batch21-2024_06_15_e0c6cb68abd23015f7f5g_0026_3.jpg" ], "is_multi_img": true, "answer": "48", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nLet the length of the rectangle in Figure 1 be \\( \\mathrm{AB} = x \\), and the width be \\( \\mathrm{AD} = y \\).\n\nFrom Figure 2, we have:\n\\[\n\\frac{1}{2}x + y = 10 \\quad \\text{(1)}\n\\]\n\nFrom Figure 3, we have:\n\\[\n\\frac{1}{2}x - y = 2 \\quad \\text{(2)}\n\\]\n\nAdding equations (1) and (2), we get:\n\\[\nx = 12\n\\]\n\nSubstituting \\( x = 12 \\) into equation (1), we obtain:\n\\[\n6 + y = 10\n\\]\nSolving for \\( y \\), we find:\n\\[\ny = 4\n\\]\n\nTherefore, the area of rectangle \\( \\mathrm{ABCD} \\) in Figure 1 is:\n\\[\n12 \\times 4 = 48\n\\]\n\nThe final answer is: **48**.\n\n**Key Insight:**\nThis problem tests the understanding of systems of linear equations. Mastery of the properties of rectangles, congruent triangles, and the relationships between their sides is crucial for solving such problems." }, { "problem_id": 1515, "question": "\"Doing math\" helps us accumulate mathematical activity experience. As shown in the figure, given a triangular paper $A B C$, the first fold makes point $B$ fall on point $B^{\\prime}$ on side $B C$, with fold line $A D$ intersecting $B C$ at point $D$; the second fold makes point $A$ fall on point $D$, with fold line $M N$ intersecting $A B^{\\prime}$ at point $P$. If $B C=12$, then $M P+M N=$ $\\qquad$ .\n\n\n\nFirst fold\n\n\n\nSecond fold", "input_image": [ "batch21-2024_06_15_e5005ea2d67caff7f240g_0080_1.jpg", "batch21-2024_06_15_e5005ea2d67caff7f240g_0080_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Given a triangular piece of paper \\( ABC \\), the first fold places point \\( B \\) onto point \\( B' \\) on side \\( BC \\), with the crease \\( AD \\) intersecting \\( BC \\) at point \\( D \\).\n\nThus, \\( BD = DB' = \\frac{1}{2} BB' \\), and \\( AD \\perp BC \\).\n\nThe second fold places point \\( A \\) onto point \\( D \\), with the crease \\( MN \\) intersecting \\( AB' \\) at point \\( P \\).\n\nTherefore, \\( AM = DM \\), \\( AN = ND \\), and \\( MN \\perp AD \\), which implies \\( MN \\parallel BC \\).\n\nSince \\( AM = DM \\), \\( MN \\) is the midline of triangle \\( ADC \\).\n\nHence, \\( MP = \\frac{1}{2} DB' \\), and \\( MN = \\frac{1}{2} DC \\).\n\nGiven \\( BC = 12 \\), and \\( BD + DC = CB' + 2BD = BC \\), it follows that:\n\n\\[ MP + MN = \\frac{1}{2} DB' + \\frac{1}{2} DC = \\frac{1}{2}(DB' + DB' + B'C) = \\frac{1}{2} BC = 6 \\]\n\nTherefore, the answer is: 6.\n\n[Key Insight] This problem primarily examines the properties of folding and the midline of a triangle. Understanding the properties of folding and the midline of a triangle is crucial for solving the problem." }, { "problem_id": 1516, "question": "A square paper with side length $8 \\mathrm{~cm}$ is folded, unfolded, and drawn upon to create a tangram as shown in Figure 1. Then, Figure 1 is divided along the solid lines and rearranged to form a \"house\" shape as shown in Figure 2. The area of the small square (shaded part) in this shape is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_e66ab86fe71c7bffc431g_0082_1.jpg", "batch21-2024_06_15_e66ab86fe71c7bffc431g_0082_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 1, it can be seen that the area of the small square is $\\frac{1}{8}$ of the area of the large square.\n\nSince the area of the large square is $8^{2}=64 \\mathrm{~cm}^{2}$,\n\nthe area of the small square (shaded part) is $64 \\times \\frac{1}{8}=8 \\mathrm{~cm}^{2}$.\n\nTherefore, the answer is 8.\n\n[Highlight] This problem utilizes the properties of squares to find the solution. The area of each piece in a tangram can be determined using the properties of squares." }, { "problem_id": 1517, "question": "Connect four equal-length thin wooden sticks end to end, and nail them into a quadrilateral $\\mathrm{ABCD}$. Rotate this quadrilateral to change its shape. When $\\angle B=90^{\\circ}$, as shown in Figure 1, measure $\\mathrm{AC}=2$. When $\\angle B=60^{\\circ}$, as shown in Figure 2, then $\\mathrm{BD}=$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_ea0f37e97ad0f9e9013bg_0008_1.jpg", "batch21-2024_06_15_ea0f37e97ad0f9e9013bg_0008_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{6}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 1,\n\n$\\because \\mathrm{AB}=\\mathrm{BC}=\\mathrm{CD}=\\mathrm{DA}, \\angle \\mathrm{B}=90^{\\circ}$,\n\n$\\therefore$ quadrilateral $\\mathrm{ABCD}$ is a square.\n\nConnecting $\\mathrm{AC}$, we have $\\mathrm{AB}^{2}+\\mathrm{BC}^{2}=\\mathrm{AC}^{2}=4$,\n\n$\\therefore \\mathrm{AB}=\\mathrm{AD}=\\sqrt{2}$.\n\nAs shown in Figure 2, $\\angle \\mathrm{ABC}=60^{\\circ}$. Connecting $\\mathrm{BD}$ and drawing $\\mathrm{AE} \\perp \\mathrm{BD}$ at point $\\mathrm{E}$,\n\n$\\therefore \\triangle \\mathrm{ABD}$ is an isosceles triangle.\n\n$\\because A B=A D=\\sqrt{2}$, quadrilateral $\\mathrm{ABCD}$ is a rhombus,\n\n$\\therefore \\angle \\mathrm{ABE}=30^{\\circ}$,\n\n$$\n\\begin{aligned}\n& \\therefore \\mathrm{AE}=\\frac{1}{2} \\mathrm{AB}=\\frac{\\sqrt{2}}{2}, \\\\\n& \\therefore \\mathrm{BE}=\\sqrt{A B^{2}-A E^{2}}=\\frac{\\sqrt{6}}{2}, \\\\\n& \\therefore \\mathrm{BD}=2 \\mathrm{BE}=\\sqrt{6} .\n\\end{aligned}\n$$\n\nTherefore, the answer is: $\\sqrt{6}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Insight】This problem examines the properties of squares and rhombuses, as well as the application of the Pythagorean theorem. The key is to use the Pythagorean theorem to determine the side length of the square." }, { "problem_id": 1518, "question": "Xiao Ming made an adjustable rhombus tool using four wooden strips of equal length. He first adjusted the tool into the rhombus shown in Figure 1 and measured $\\angle B = 60^{\\circ}$. Then, he adjusted the tool into the square shown in Figure 2 and measured the diagonal of the square\n$A C = 40 \\mathrm{~cm}$. The length of the diagonal $A C$ in Figure 1 is $\\mathrm{cm}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_ea0f37e97ad0f9e9013bg_0017_1.jpg", "batch21-2024_06_15_ea0f37e97ad0f9e9013bg_0017_2.jpg" ], "is_multi_img": true, "answer": "$20 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in Figures 1 and 2, connect $AC$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nIn Figure 2, since quadrilateral $ABCD$ is a square,\n\n$\\therefore AB = BC$, and $\\angle B = 90^{\\circ}$.\n\nGiven that $AC = 40 \\mathrm{~cm}$,\n\n$\\therefore AB = BC = 20 \\sqrt{2} \\mathrm{~cm}$.\n\nIn Figure 1, quadrilateral $ABCD$ is a rhombus, and $BA = BC$.\n\nSince $\\angle B = 60^{\\circ}$,\n\n$\\therefore \\triangle ABC$ is an equilateral triangle,\n\n$\\therefore AC = BC = 20 \\sqrt{2} \\mathrm{~cm}$.\n\nThus, the answer is: $20 \\sqrt{2}$.\n\n【Highlight】This question examines the properties of rhombuses and squares, as well as the Pythagorean theorem. The key to solving the problem lies in the flexible application of the learned knowledge. This type of question is commonly encountered in middle school exams." }, { "problem_id": 1519, "question": "As shown in Figure 1, this is the class emblem designed by Xiao Yuan, where the \"$Z$\" shaped part is obtained by the following construction method: as shown in Figure 2, points $E$ and $F$ are taken on sides $AB$ and $CD$ of square $ABCD$ respectively, and points $G$ and $H$ are taken on the extensions of $CB$ and $AD$ respectively, such that $BE = BG = DF = DH$. Lines $AG$, $EG$, $AF$, $CE$, $FH$, and $CH$ are connected. Let the sum of the areas of $\\triangle AEG$ and $\\triangle CFH$ be $S_1$, and the area of quadrilateral $AECF$ be $S_2$. If $\\frac{S_1}{S_2} = \\frac{3}{7}$ and $S_1 + S_2 = 20$, then the area of square $ABCD$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_ea0f37e97ad0f9e9013bg_0070_1.jpg", "batch21-2024_06_15_ea0f37e97ad0f9e9013bg_0070_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{49}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let $\\mathrm{BE}=\\mathrm{BG}=\\mathrm{DF}=\\mathrm{DH}=\\mathrm{x}$, and $\\mathrm{AE}=\\mathrm{CF}=\\mathrm{y}$.\nSince quadrilateral $\\mathrm{ABCD}$ is a square,\n\n$\\therefore \\mathrm{AB}=\\mathrm{BC}=\\mathrm{CD}=\\mathrm{AD}=\\mathrm{x}+\\mathrm{y}$, $\\angle \\mathrm{ABC}=\\angle \\mathrm{ABG}=90^{\\circ}$, $\\angle \\mathrm{ADF}=\\angle \\mathrm{CDH}=90^{\\circ}$, and $\\mathrm{AE} \\parallel \\mathrm{CF}$,\n\n$\\therefore$ quadrilateral $\\mathrm{AECF}$ is a parallelogram.\n\nGiven that $\\frac{S_{1}}{S_{2}}=\\frac{3}{7}$ and $S_{1}+S_{2}=20$,\n\n$\\therefore \\frac{1}{2} x y+\\frac{1}{2} x y+y(x+y)=20$ (1),\n\n$\\frac{\\frac{1}{2} x y+\\frac{1}{2} x y}{y(x+y)}=\\frac{3}{7}$ (2).\n\nFrom (1) and (2), we can derive $\\left\\{\\begin{array}{l}x=\\frac{3 \\sqrt{2}}{2} \\\\ y=2 \\sqrt{2}\\end{array}\\right.$,\n\n$\\therefore$ the area of the square $=\\left(2 \\sqrt{2}+\\frac{3 \\sqrt{2}}{2}\\right)^{2}=\\frac{49}{2}$,\n\nHence, the answer is: $\\frac{49}{2}$.\n\n【Highlight】This problem examines the application and design of geometric constructions, the properties of squares, and the properties and determination of parallelograms. The key to solving the problem lies in using parameters to construct and solve a system of equations." }, { "problem_id": 1520, "question": "The Pythagorean theorem is an important theorem in geometry. In the ancient Chinese mathematical text \"Zhoubi Suanjing,\" there is a record stating \"if the shorter leg is three, the longer leg is four, then the hypotenuse is five.\" Figure 1 is composed of small squares with equal side lengths and right-angled triangles, and the Pythagorean theorem can be verified using their area relationships. Figure 2 is obtained by placing Figure 1 within a rectangle, where $\\angle B A C=90^{\\circ}, A B=3, A C=4$, and points $D, E, F, G, H, I$ are all on the sides of the rectangle $K L M J$. The area of the rectangle $K L M J$ is $\\qquad$.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch21-2024_06_15_ee2f1c1788f4783d1082g_0010_1.jpg", "batch21-2024_06_15_ee2f1c1788f4783d1082g_0010_2.jpg" ], "is_multi_img": true, "answer": "110", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:** As shown in the figure, extend \\( AB \\) to intersect \\( KL \\) at \\( O \\), and extend \\( AC \\) to intersect \\( ML \\) at \\( N \\). This forms the quadrilateral \\( AOLN \\), which is a rectangle.\n\nSince quadrilateral \\( BFGC \\) is a square,\n\n\\[ BC = BF = FG = CG, \\quad \\angle CBF = \\angle BFG = \\angle FGC = \\angle BCG = 90^\\circ. \\]\n\nThus,\n\n\\[ \\angle ABC + \\angle OBF = \\angle OFB + \\angle LFG = \\angle FGL + \\angle NGC = \\angle ACB + \\angle NCG = 90^\\circ. \\]\n\nSince\n\n\\[ \\angle BAC = \\angle AOL = \\angle L = \\angle ANL = 90^\\circ, \\]\n\nit follows that\n\n\\[ \\angle ACB + \\angle ABC = \\angle OBF + \\angle OFB = \\angle LFG + \\angle LGF = \\angle NCG + \\angle NGC = 90^\\circ. \\]\n\nTherefore,\n\n\\[ \\angle ABC = \\angle OFB = \\angle FGL = \\angle NCG, \\]\n\nand\n\n\\[ \\triangle CAB \\cong \\triangle BOF \\cong \\triangle FLG \\cong \\triangle GNC. \\]\n\nHence,\n\n\\[ BO = FL = GN = AC = 4, \\quad OF = LG = CN = AB = 3. \\]\n\nThus,\n\n\\[ JK = 4 + 3 + 4 = 11, \\quad KL = 3 + 3 + 4 = 10. \\]\n\nTherefore, the area of rectangle \\( KLMJ \\) is\n\n\\[ 11 \\times 10 = 110. \\]\n\nThe final answer is: \\(\\boxed{110}\\).\n\n**Key Insight:** This problem tests the application of the Pythagorean theorem. The key to solving it lies in constructing auxiliary lines to form a rectangle and identifying congruent triangles." }, { "problem_id": 1521, "question": "As shown in the figure, Figure 1 is a children's slide, and $AE, DF, MN$ are the three reinforcing supports of the slide (as shown in Figure 2). Both $AE$ and $DF$ are perpendicular to the ground $BC$, $N$ is the midpoint of the slide $DC$, and Xiao Zhou measured $FM=1$ meter, $MN=2$ meters, $MC=3$ meters. Through calculation, he found that the length of the slide $DC$ is $\\qquad$ meters.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_ee2f1c1788f4783d1082g_0040_1.jpg", "batch21-2024_06_15_ee2f1c1788f4783d1082g_0040_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{7}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, connect $FN$, and draw $NG \\perp CF$ at point $G$.\n\nGiven that $FM = 1$ meter and $MC = 3$ meters,\n\n$\\therefore CF = FM + MC = 4$ meters.\n\nSince $DF \\perp BC$,\n\n$\\therefore \\angle DFC = 90^\\circ$.\n\nSince $N$ is the midpoint of the slide $DC$,\n\n$\\therefore FN = \\frac{1}{2} DC = CN$.\n\nSince $NG \\perp CF$,\n\n$\\therefore FG = CG = \\frac{1}{2} CF = 2$ meters.\n\n$\\therefore MG = FG - FM = 2 - 1 = 1$ meter.\n\nIn the right triangle $\\triangle MNG$, by the Pythagorean theorem:\n\n$NG = \\sqrt{MN^2 - MG^2} = \\sqrt{2^2 - 1^2} = \\sqrt{3}$ meters.\n\nIn the right triangle $\\triangle CNG$, by the Pythagorean theorem:\n\n$CN = \\sqrt{NG^2 + CG^2} = \\sqrt{(\\sqrt{3})^2 + 2^2} = \\sqrt{7}$ meters.\n\n$\\therefore DC = 2CN = 2\\sqrt{7}$ meters.\n\nTherefore, the answer is: $2\\sqrt{7}$.\n\n\n\nFigure 2\n\n【Insight】This problem examines the application of the Pythagorean theorem, the property of the median to the hypotenuse in a right triangle, and the properties of isosceles triangles. Familiarity with these properties is key to solving the problem." }, { "problem_id": 1522, "question": "The \"Tangram\" is an outstanding creation of our ancestors, capable of forming many interesting shapes, and is known as the \"Magic Square of the East.\" Figure (1) is a \"Tangram\" made by dividing a square thin plate with a side length of $20 \\mathrm{~cm}$ into 7 pieces. Figure (2) is a \"home\" shape formed by assembling this \"Tangram.\" The side length of the square piece among the 7 pieces in this \"Tangram\" is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_f5940cec6216cb44f929g_0005_1.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0005_2.jpg" ], "is_multi_img": true, "answer": "$5 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, in the square \\( ABCD \\), \\( AB = BC = CD = AD = 20 \\, \\text{cm} \\), and \\( \\angle BAD = 90^\\circ \\).\n\nTherefore, the diagonal \\( BD = \\sqrt{20^2 + 20^2} = 20\\sqrt{2} \\, \\text{cm} \\).\n\nHence, \\( NB = ND = 10\\sqrt{2} \\, \\text{cm} \\).\n\n\n\nFrom figure (2), it can be deduced that \\( DG = MN \\), and triangle \\( GNM \\) is an isosceles right triangle.\n\nThus, \\( NG = NM = DG = 5\\sqrt{2} \\, \\text{cm} \\).\n\nThe answer is: \\( 5\\sqrt{2} \\).\n\n【Insight】This problem examines the properties of the tangram, the characteristics of a square, and the application of the Pythagorean theorem. The key is to understand the relationship between the side length of one of the seven pieces of the tangram and the side length of the square in the tangram." }, { "problem_id": 1523, "question": "In the comprehensive practical class, Xiao Ming cut a square paper with side length $2 \\mathrm{~cm}$ along its diagonal $A C$, as shown in Figure $l$. Then, fixing the paper $\\triangle A B C$, he translated the paper $\\triangle A D C$ along the direction of $A C$ to obtain $\\triangle A^{\\prime} D^{\\prime} C^{\\prime}$, connecting $A^{\\prime} B, D^{\\prime} B$, $D^{\\prime} C$. During the translation process: (1) The shape of quadrilateral $A^{\\prime} B C D^{\\prime}$ is always ; (2) The minimum value of $A^{\\prime} B+D^{\\prime} B$ is .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_f5940cec6216cb44f929g_0007_1.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0007_2.jpg" ], "is_multi_img": true, "answer": "parallelogram, $2 \\sqrt{5}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: (1) As shown in Figure 2, since \\( A^{\\prime} D^{\\prime} = BC \\) and \\( A^{\\prime} D^{\\prime} \\parallel BC \\), the quadrilateral \\( A^{\\prime} B C D^{\\prime} \\) is a parallelogram. Therefore, the answer is: parallelogram.\n\n(2) As shown in Figure 2, draw the line \\( DD^{\\prime} \\), and find the symmetric point \\( C^{\\prime \\prime} \\) of point \\( C \\) with respect to the line \\( DD^{\\prime} \\). Connect \\( D^{\\prime} C^{\\prime \\prime} \\) and \\( B C^{\\prime \\prime} \\), and draw a perpendicular \\( BH \\) from point \\( B \\) to \\( CC^{\\prime \\prime} \\), intersecting at point \\( H \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nSince the quadrilateral \\( ABCD \\) is a square, \\( AB = BC = 2 \\) and \\( \\angle ABC = 90^\\circ \\). Therefore, \\( AC = \\sqrt{2} \\cdot AB = 2\\sqrt{2} \\). Since \\( BJ \\perp AC \\), \\( AJ = JC \\), and thus \\( BJ = \\frac{1}{2} AC = \\sqrt{2} \\). Since \\( \\angle BJC = \\angle JCH = \\angle H = 90^\\circ \\), the quadrilateral \\( BHCJ \\) is a rectangle. Given that \\( BJ = CJ \\), the quadrilateral \\( BHCJ \\) is a square, and thus \\( BH = CH = \\sqrt{2} \\).\n\nIn the right triangle \\( \\triangle BHC^{\\prime \\prime} \\), \\( BH = \\sqrt{2} \\) and \\( HC^{\\prime \\prime} = 3\\sqrt{2} \\). Therefore, \\( BC^{\\prime \\prime} = \\sqrt{BH^2 + HC^{\\prime \\prime 2}} = \\sqrt{(\\sqrt{2})^2 + (3\\sqrt{2})^2} = 2\\sqrt{5} \\).\n\nSince the quadrilateral \\( A^{\\prime} B C D^{\\prime} \\) is a parallelogram, \\( A^{\\prime} B = CD^{\\prime} \\). Thus, \\( A^{\\prime} B + BD^{\\prime} = BD^{\\prime} + CD^{\\prime} = BD^{\\prime} + D^{\\prime} C^{\\prime \\prime} \\geq BC^{\\prime \\prime} \\). Therefore, \\( A^{\\prime} B + BD^{\\prime} \\geq 2\\sqrt{5} \\), and the minimum value of \\( A^{\\prime} B + D^{\\prime} B \\) is \\( 2\\sqrt{5} \\).\n\nHence, the answer is: \\( 2\\sqrt{5} \\).\n\n【Insight】This problem examines the concepts of translation transformation, the shortest path problem using symmetry, and the Pythagorean theorem. The key to solving the problem lies in utilizing symmetry to address the shortest path problem, which is a common type of question in middle school exams." }, { "problem_id": 1524, "question": "As shown in Figure (1), it is given that the area of the small square \\(ABCD\\) is 1. Extend each of its sides by a factor of 2 to obtain the new square \\(A_1B_1C_1D_1\\); extend the sides of square \\(A_1B_1C_1D_1\\) by the same method to obtain square \\(A_2B_2C_2D_2\\) (as shown in Figure (2))... and so on. The area of square \\(A_{2021}B_{2021}C_{2021}D_{2021}\\) is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_f5940cec6216cb44f929g_0008_1.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0008_2.jpg" ], "is_multi_img": true, "answer": "$5^{2021}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Since the square of the side length of square $A_{1} B_{1} C_{1} D_{1}$ is: $(1+1)^{2}+1^{2}=5$,\n\nTherefore, the square of the area of square $A_{1} B_{1} C_{1} D_{1}$ is: 5.\n\nMoreover, since the square of the side length of square $A_{2} B_{2} C_{2} D_{2}$ is: $(2 \\sqrt{5})^{2}+(\\sqrt{5})^{2}=25$,\n\nThus, the area of square $A_{2} B_{2} C_{2} D_{2}$ is: $25=5^{2}$.\n\nFollowing this pattern, the square of the side length of square $A_{3} B_{3} C_{3} D_{3}$ is: $(2 \\times 5)^{2}+5^{2}=125$.\n\nTherefore, the area of square $A_{3} B_{3} C_{3} D_{3}$ is: $125=5^{3}$.\n\nHence, the area of square $A_{n} B_{n} C_{n} D_{n}$ is $5^{n}$.\n\nConsequently, the area of square $A_{2021} B_{2021} C_{2021} D_{2021}$ is: $5^{2021}$.\n\nThe answer is: $5^{2021}$.\n\n[Highlight] This problem examines the pattern of changes in geometric figures, with the key to solving it being to identify the relationship between the area of the square and the index $n$." }, { "problem_id": 1525, "question": "As shown in Figure (1), it is given that the area of the small square \\( A B C D \\) is 1. Extend each of its sides by a factor of two to obtain the new square \\( A_{1} B_{1} C_{1} D_{1} \\); extend the sides of square \\( A_{1} B_{1} C_{1} D_{1} \\) by the same method to obtain square \\( A_{2} B_{2} C_{2} D_{2} \\) (as shown in Figure (2)); $\\qquad$ .\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch21-2024_06_15_f5940cec6216cb44f929g_0021_1.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0021_2.jpg" ], "is_multi_img": true, "answer": "625", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Connect $AC$ and $B_{1}C$ as shown in Figure (1).\n\nSince $AB = BB_{1}$ and $BC = CC_{1}$,\n\nWe have $S_{\\triangle ABC} = S_{\\triangle BB_{1}C}$ and $S_{\\triangle BB_{1}C} = S_{\\triangle CC_{1}B_{1}}$,\n\nTherefore, $S_{\\triangle BB_{1}C} = 2 S_{\\triangle ABC} = S_{\\text{square } ABCD} = 1$,\n\nHence, $S_{\\text{square } A_{1}B_{1}C_{1}D_{1}} = 5 S_{\\text{square } ABCD} = 5$,\n\nSimilarly, we can derive $S_{\\text{square } A_{2}B_{2}C_{2}D_{2}} = 5 S_{\\text{square } B_{1}C_{1}D_{1}} = 5^{2}$,\n\nThus, the area of square $A_{4}B_{4}C_{4}D_{4}$ is $5^{4} = 625$.\n\nThe final answer is: 625.\n\n\n\nFigure (1)\n\n[Highlight] This problem examines the properties of squares and also explores the method for solving pattern-based problems. It utilizes the principle that two triangles with the same base and height have equal areas to find the solution." }, { "problem_id": 1526, "question": "The Five-Piece Tangram is an intellectual puzzle similar to the Seven-Piece Tangram. It is formed by dividing a square in the manner shown in Figure 1, where figure (1) is a square. Xiao Ming discovered that the Five-Piece Tangram can be assembled into the \"triangle\" and \"airplane\" models shown in Figure 2. In the \"airplane\" model, the ratio of the width to the height is $\\frac{l}{h}=$ . $\\qquad$\n\n\n\nFigure 1\n\n\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_f5940cec6216cb44f929g_0023_1.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0023_2.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0023_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{6}{7}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the side length of the small square (1) in Figure 1 be $a$. According to the way the figures are assembled in the problem, we have $l=3a$ and $h=3.5a$.\n\nTherefore, the ratio $\\frac{l}{h} = \\frac{3a}{3.5a} = \\frac{6}{7}$.\n\nHence, the answer is: $\\frac{6}{7}$.\n\n[Highlight] This question examines the properties of squares and the area of figures. The key to solving the problem lies in observing the diagram and using the side length of the small square (1) in Figure 1 to represent the width and height of the \"airplane\" model." }, { "problem_id": 1527, "question": "As shown in the figure, the following are diagrams formed by connecting 1, 2, and $n$ squares respectively. In Figure 1, $x=70^{\\circ}$; in Figure 2, $y=28^{\\circ}$. Based on the above calculations, please write the expression for $a+b+c+\\ldots+d$ in Figure 3 $\\qquad$ (expressed in terms of $n$).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch21-2024_06_15_f5940cec6216cb44f929g_0026_1.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0026_2.jpg", "batch21-2024_06_15_f5940cec6216cb44f929g_0026_3.jpg" ], "is_multi_img": true, "answer": "$90 n$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Connect the diagonals of each small square as shown in the following figures:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nIn Figure 1, $61^{\\circ} + 119^{\\circ} + 20^{\\circ} + x + 45^{\\circ} \\times 2 = 360^{\\circ}$,\n\nwhich simplifies to $20^{\\circ} + x = 90^{\\circ}$.\n\nIn Figure 2, $61^{\\circ} + 119^{\\circ} + 31^{\\circ} + 121^{\\circ} + y + 45^{\\circ} \\times 4 = 360^{\\circ}$,\n\nwhich simplifies to $31^{\\circ} + 121^{\\circ} + y = 180^{\\circ} = 2 \\times 90^{\\circ}$.\n\nFollowing this pattern, $a + b + c + \\ldots + d = n \\times 90^{\\circ} = 90^{\\circ} n$.\n\nTherefore, the answer is: $90^{\\circ} n$.\n\n【Insight】The problem mainly tests the ability to derive corresponding algebraic expressions based on patterns and the properties of squares. Understanding the problem and exploring the patterns are key to solving it." }, { "problem_id": 1528, "question": "Xiao Ying was able to manipulate the rhombus learning tool into the shape shown in Figure 1, and measured that \\( AC = 5 \\) and \\( \\angle B = 60^\\circ \\). Then, she manipulated the tool again to form a square as shown in Figure 2, and the length of \\( A'C' \\) at this time is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_f70bf3c076d9c8ee25f4g_0022_1.jpg", "batch21-2024_06_15_f70bf3c076d9c8ee25f4g_0022_2.jpg" ], "is_multi_img": true, "answer": "$5 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "From the problem statement, we know that \\( A B C D \\) is a rhombus,\n\n\\[\n\\therefore A B = B C\n\\]\n\n\\[\n\\because \\angle B = 60^{\\circ},\n\\]\n\n\\[\n\\therefore \\triangle A B C \\text{ is an equilateral triangle},\n\\]\n\n\\[\n\\because A C = 5\n\\]\n\n\\[\nA B = B C = 5,\n\\]\n\n\\[\nA^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime} \\text{ is a square},\n\\]\n\n\\[\n\\therefore A^{\\prime} B^{\\prime} = B^{\\prime} C^{\\prime} = 5\n\\]\n\n\\[\nA^{\\prime} C^{\\prime} = \\sqrt{A^{\\prime} B^{\\prime 2} + B^{\\prime} C^{\\prime 2}} = 5 \\sqrt{2},\n\\]\n\nThus, the answer is: \\( 5 \\sqrt{2} \\).\n\n【Insight】This problem examines the properties of a rhombus, square, and equilateral triangle, as well as the Pythagorean theorem; the key to solving the problem lies in the flexible application of these properties to perform correct calculations." }, { "problem_id": 1529, "question": "\"Doing mathematics\" helps us accumulate mathematical activity experience. As shown in the figure, given a triangular paper $ABC$, the first fold brings point $B$ to point $B'$ on side $BC$, with fold line $AD$ intersecting $BC$ at point $D$; the second fold brings point $A$ to point $D$, with fold line $MN$ intersecting $AB'$ at point $P$. If $BC=16$, then $MP+MN=$ $\\qquad$ .\n\n\n\n(First fold)\n\n\n\n(Second fold)", "input_image": [ "batch21-2024_06_15_f70bf3c076d9c8ee25f4g_0069_1.jpg", "batch21-2024_06_15_f70bf3c076d9c8ee25f4g_0069_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "(Second Fold)\n\nSolution: As shown in the figure, from the folded diagram we obtain:\n\n$AM = MD$, $MN \\perp AD$, $AD \\perp BC$,\n\n$\\therefore GN // BC$,\n\n$\\therefore AG = BG$,\n\n$\\therefore GN$ is the midline of $\\triangle ABC$,\n\n$\\therefore GN = \\frac{1}{2} BC = \\frac{1}{2} \\times 16 = 8$,\n\n$\\because PM = GM$,\n\n$\\therefore MP + MN = GM + MN = GN = 8$.\n\nTherefore, the answer is: 8.\n\n[Highlight] This question examines the midline theorem of a triangle and the properties of folding. The key to solving this problem is to complete the figure and prove that $GN$ is the midline of $\\triangle ABC$." }, { "problem_id": 1530, "question": "As shown in Figure A, within a plane, there is a point \\( P \\) such that the distances from \\( P \\) to the three vertices of \\(\\triangle ABC\\) are \\( PA, PB, \\) and \\( PC \\). If the condition \\( PA^2 = PB^2 + PC^2 \\) is satisfied, then point \\( P \\) is called the Pythagorean point of \\(\\triangle ABC\\) with respect to point \\( A \\). As shown in Figure B, \\( E \\) is a point inside rectangle \\( ABCD \\), and point \\( C \\) is the Pythagorean point of \\(\\triangle ABE\\) with respect to point \\( A \\). Connect \\( DE \\). It is given that \\( AB = 5 \\), \\( BC = 6 \\), and \\( DA = DE \\). The length of \\( AE \\) is \\(\\qquad\\).\n\n\n\n(Figure A)\n\n\n\n(Figure B)", "input_image": [ "batch21-2024_06_15_f70bf3c076d9c8ee25f4g_0095_1.jpg", "batch21-2024_06_15_f70bf3c076d9c8ee25f4g_0095_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{6 \\sqrt{10}}{5}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since point \\( C \\) is the Pythagorean point of triangle \\( ABE \\) with respect to point \\( A \\),\n\n\\[ CA^{2} = CB^{2} + CE^{2}. \\]\n\nSince quadrilateral \\( ABCD \\) is a rectangle,\n\n\\[ \\angle ABC = 90^{\\circ}, \\quad AB = CD, \\]\n\n\\[ CA^{2} = AB^{2} + CB^{2} = CB^{2} + CD^{2}, \\]\n\n\\[ CE = CD. \\]\n\nAs shown in the figure, draw the altitudes \\( CF \\) and \\( EG \\) of triangle \\( ECD \\), and the altitude \\( EH \\) of triangle \\( AED \\).\n\n\n\nSince \\( CE = CD = AB = 5 \\), and \\( DE = 6 \\),\n\n\\[ EF = \\frac{1}{2} ED = 3, \\]\n\n\\[ CF = \\sqrt{CE^{2} - EF^{2}} = 4. \\]\n\nSince the area of triangle \\( ECD \\) is\n\n\\[ S_{\\triangle ECD} = \\frac{1}{2} ED \\cdot CF = \\frac{1}{2} CD \\cdot EG, \\]\n\n\\[ EG = \\frac{ED \\cdot CF}{CD} = \\frac{6 \\times 4}{5} = \\frac{24}{5}. \\]\n\n\\[ DH = \\frac{24}{5}, \\quad AH = 6 - \\frac{24}{5} = \\frac{6}{5}. \\]\n\nBy the Pythagorean theorem,\n\n\\[ AE^{2} - \\left(\\frac{6}{5}\\right)^{2} = 6^{2} - \\left(\\frac{24}{5}\\right)^{2}, \\]\n\nSolving gives:\n\n\\[ AE = \\frac{6 \\sqrt{10}}{5}. \\]\n\nTherefore, the answer is:\n\n\\[ \\boxed{\\frac{6 \\sqrt{10}}{5}}. \\]\n\n【Insight】This problem tests the properties of rectangles, with the key being to master the properties of rectangles and the Pythagorean theorem." }, { "problem_id": 1531, "question": "Theorem: The median to the hypotenuse of a right triangle is equal to half the hypotenuse. That is, as shown in Figure 1, in right $\\triangle ABC$ where $\\angle ACB = 90^\\circ$, if point $D$ is the midpoint of the hypotenuse $AB$, then $CD = \\frac{1}{2} AB$.\n\nApplication: As shown in Figure 2, in $\\triangle ABC$ where $\\angle BAC = 90^\\circ$, $AB = 2$, and $AC = 3$, point $D$ is the midpoint of $BC$. By folding $\\triangle ABD$ along $AD$ to form $\\triangle AED$, and connecting $BE$, $CE$, and $DE$, the length of $CE$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch21-2024_06_15_f760c267f00991d8fd72g_0022_1.jpg", "batch21-2024_06_15_f760c267f00991d8fd72g_0022_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{5 \\sqrt{13}}{13}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Let $BE$ intersect $AD$ at $O$, and draw $AH \\perp BC$ at $H$.\n\nIn the right triangle $\\triangle ABC$, $\\angle BAC = 90^\\circ$, $AB = 2$, $AC = 3$.\n\nBy the Pythagorean theorem, we have: $BC = \\sqrt{13}$.\n\nSince point $D$ is the midpoint of $BC$,\n\n$\\therefore AD = DC = DB = \\frac{\\sqrt{13}}{2}$.\n\nBecause $\\frac{1}{2} \\cdot BC \\cdot AH = \\frac{1}{2} \\cdot AB \\cdot AC$,\n\n$\\therefore AH = \\frac{6 \\sqrt{13}}{13}$.\n\nSince $AE = AB$, $DE = DB$,\n\n$\\therefore$ point $A$ lies on the perpendicular bisector of $BE$, and point $D$ lies on the perpendicular bisector of $BE$,\n\n$\\therefore AD$ is the perpendicular bisector of segment $BE$.\n\nBecause $\\frac{1}{2} AD \\cdot BO = \\frac{1}{2} BD \\cdot AH$,\n\n$\\therefore OB = \\frac{6 \\sqrt{13}}{13}$,\n\n$\\therefore BE = 2 OB = \\frac{12 \\sqrt{13}}{13}$.\n\nSince $DE = DB = CD$,\n\n$\\therefore \\angle DBE = \\angle DEB$, $\\angle DEC = \\angle DCE$,\n\n$\\therefore \\angle DEB + \\angle DEC = \\frac{1}{2} \\times 180^\\circ = 90^\\circ$, that is: $\\angle BEC = 90^\\circ$,\n\n$\\therefore$ in the right triangle $\\triangle BCE$, $EC = \\sqrt{BC^2 - BE^2} = \\sqrt{(\\sqrt{13})^2 - \\left(\\frac{12 \\sqrt{13}}{13}\\right)^2} = \\frac{5 \\sqrt{13}}{13}$.\n\nTherefore, the answer is: $\\frac{5 \\sqrt{13}}{13}$.\n\n\n\n【Key Insight】This problem mainly examines the properties of right triangles, the Pythagorean theorem, and the properties of folding. Mastering the concept that \"the median to the hypotenuse of a right triangle is half the hypotenuse\" and using the area method to find the height of a triangle are key to solving the problem." }, { "problem_id": 1532, "question": "As shown in Figure 1, a rectangular paper $\\mathrm{ABCD}$ with $\\mathrm{AB}=5$ and $\\mathrm{BC}=8$ is folded in the following sequence. The length of $\\mathrm{MN}$ in Figure 4 is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4", "input_image": [ "batch21-2024_06_15_f8cd00f9f6da9a42b30ag_0065_1.jpg", "batch21-2024_06_15_f8cd00f9f6da9a42b30ag_0065_2.jpg", "batch21-2024_06_15_f8cd00f9f6da9a42b30ag_0065_3.jpg", "batch21-2024_06_15_f8cd00f9f6da9a42b30ag_0065_4.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Since quadrilateral $\\mathrm{ABCD}$ is a rectangle, $AB=5$, $BC=8$.\n\nTherefore, $EF=AB=5$, $AD=BC=8$.\n\nAs shown in Figure 3, by the properties of folding, we have: $\\angle A_{1}EP=\\angle A_{1}EQ=45^{\\circ}$, $A_{1}E=AE=\\frac{1}{2}AD=4$, $EP=EQ$.\n\nThus, $\\triangle EPQ$ is an isosceles right triangle, and point $A_{1}$ is the midpoint of $\\mathrm{PQ}$, $A_{1}F=EF-A_{1}E=1$.\n\nAs shown in Figure 4, by the properties of folding, we have: $\\triangle EMN$ is an isosceles right triangle, $OE=A_{1}E-1=3$, and point $\\mathrm{O}$ is the midpoint of $\\mathrm{MN}$.\n\nTherefore, $MN=2OE=2 \\times 3=6$.\n\nThe answer is: 6.\n\n\n\nFigure 3\n\n\n\nFigure 4\n\n【Key Insight】This question examines the knowledge points of rectangles and folding problems, as well as the determination and properties of isosceles right triangles. Mastering the properties of folding is key to solving the problem." }, { "problem_id": 1533, "question": "As shown in Figure 1, in the rectangular paper $A B C D$, $A B=4, B C=4 \\sqrt{3}$. The rectangular paper is folded along the diagonal $A C$, with point $D$ landing at point $D^{\\prime}$. Line segment $B D^{\\prime}$ is connected, as shown in Figure 2. Find the length of segment $B D^{\\prime}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch22-2024_06_14_7aa24deed846fccaa898g_0074_1.jpg", "batch22-2024_06_14_7aa24deed846fccaa898g_0074_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Let $\\mathrm{AD}^{\\prime}$ intersect $\\mathrm{BC}$ at $\\mathrm{O}$.\n\nDraw $\\mathrm{BE} \\perp \\mathrm{AD}^{\\prime}$ at point $\\mathrm{E}$.\n\n\n\nFigure 1\n\n\n\nIn rectangle $\\mathrm{ABCD}$,\n\nsince $\\mathrm{AD} / / \\mathrm{BC}$ and $\\mathrm{AD}=\\mathrm{BC}$,\n\n$\\angle \\mathrm{B}=\\angle \\mathrm{D}=\\angle \\mathrm{BAD}=90^{\\circ}$.\n\nIn right triangle $\\triangle \\mathrm{ABC}$,\n\nsince $\\tan \\angle \\mathrm{BAC}=\\frac{B C}{A B}=\\frac{4 \\sqrt{3}}{4}=\\sqrt{3}$,\n\n$\\therefore \\angle \\mathrm{BAC}=60^{\\circ}$, and thus $\\angle \\mathrm{DAC}=90^{\\circ}-\\angle \\mathrm{BAC}=30^{\\circ}$.\n\nSince $\\triangle A C D$ is folded along diagonal $A C$ to obtain $\\triangle A C D^{\\prime}$,\n\n$\\therefore \\mathrm{AD}^{\\prime}=\\mathrm{AD}=\\mathrm{BC}=4 \\sqrt{3}$, and $\\angle 1=\\angle \\mathrm{DAC}=30^{\\circ}$.\n\n$\\therefore \\angle 4=\\angle \\mathrm{BAC}-\\angle 1=30^{\\circ}$.\n\nIn right triangle $\\triangle \\mathrm{ABE}$, $\\angle \\mathrm{AEB}=90^{\\circ}$, so $\\mathrm{BE}=2$.\n\n$\\therefore \\mathrm{AE}=\\sqrt{A B^{2}-B E^{2}}=2 \\sqrt{3}$.\n\n$\\therefore \\mathrm{D}^{\\prime} \\mathrm{E}=\\mathrm{AD}^{\\prime}-\\mathrm{AE}=2 \\sqrt{3}$.\n\n$\\therefore \\mathrm{AE}=\\mathrm{D}^{\\prime} \\mathrm{E}$, which means $\\mathrm{BE}$ is the perpendicular bisector of $\\mathrm{AD}^{\\prime}$.\n\n$\\therefore \\mathrm{BD}^{\\prime}=\\mathrm{AB}=4$.\n\n【Key Point】This problem utilizes the congruence of folded figures, the Pythagorean theorem, and the properties and criteria of congruent triangles." }, { "problem_id": 1534, "question": "As shown in Figure 1, $A F$ and $B E$ are medians of $\\triangle A B C$, with $A F \\perp B E$ intersecting at point $\\mathrm{P}$. Let $B C=a, A C=b, A B=c$. Then $a^{2}+b^{2}=5 c^{2}$. Use this property to calculate. As shown in Figure 2, in $\\square A B C D$, E, F, G are the midpoints of $A D, B C, C D$ respectively, with $E B \\perp E G$ at point $\\mathrm{E}, A D=8, A B=2 \\sqrt{5}$. Find the value of $A F$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch22-2024_06_14_890001fa003ceba2b0b2g_0027_1.jpg", "batch22-2024_06_14_890001fa003ceba2b0b2g_0027_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{15}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Connect $\\mathrm{AC}$ and $\\mathrm{EF}$ intersecting at point $\\mathrm{N}$, and connect $\\mathrm{BE}$ and $\\mathrm{AF}$ intersecting at point $\\mathrm{M}$. Then, connect $\\mathrm{EC}$ as shown in the figure:\n\n\n\nSince $\\mathrm{E}$ and $\\mathrm{G}$ are the midpoints of $\\mathrm{AD}$ and $\\mathrm{CD}$ respectively,\n\nin $\\triangle ACD$, $EG \\parallel AC$.\n\nGiven that $EB \\perp EG$,\n\nit follows that $EB \\perp AC$.\n\nSince $\\mathrm{E}$ and $\\mathrm{F}$ are the midpoints of $\\mathrm{AD}$ and $\\mathrm{BC}$ respectively,\n\n$\\mathrm{AE}$ is parallel and equal in length to $\\mathrm{BF}$, and $\\mathrm{AE}$ is also parallel and equal in length to $\\mathrm{FC}$.\n\nThus, $\\mathrm{ABFE}$ is a parallelogram, and $\\mathrm{AFCE}$ is also a parallelogram.\n\nTherefore, point $\\mathrm{M}$ is the midpoint of $\\mathrm{AF}$ (since the diagonals of a parallelogram bisect each other),\n\nand point $\\mathrm{N}$ is the midpoint of $\\mathrm{EF}$ (for the same reason).\n\nHence, $\\triangle AFE$ satisfies the property shown in Figure 1,\n\nso $AF^{2} + EF^{2} = 5AE^{2}$.\n\nGiven that $EF = AB = 2\\sqrt{5}$ and $AE = \\frac{1}{2}AD = 4$,\n\nwe have $AF^{2} + (2\\sqrt{5})^{2} = 5 \\times 4^{2}$.\n\nSolving this, we find: $AF = 2\\sqrt{15}$.\n\n【Insight】This problem primarily tests the Midsegment Theorem of triangles and the properties and determination of parallelograms. The key to solving the problem lies in the flexible application of the Midsegment Theorem of triangles." }, { "problem_id": 1535, "question": "【Initial Exploration】As shown in Figure 1, Xiao Ming arranged two congruent triangles with a $30^{\\circ}$ angle to form an equilateral triangle, finding that $BC = \\frac{1}{2} BD = \\frac{1}{2} AB$. Xiao Ming concluded that in a right triangle, if one of the acute angles is $30^{\\circ}$, then the hypotenuse is twice the length of the opposite leg. Please use Xiao Ming's conclusion to complete the following exploration problem.\n\n【In-depth Exploration】In parallelogram $ABCD$, where $AB // CD$, $AD // BC$, $AB = BC = 4$, and $\\angle B = 60^{\\circ}$, point $P$ starts from point $B$ and moves along segment $BA$ towards point $A$ at a speed of 1 unit per second. A perpendicular $PE$ is drawn from point $P$ to $BC$ at point $E$, and another perpendicular $PF$ is drawn from point $P$ intersecting line $CD$ at point $F$ and line $BC$ at point $Q$. The time for point $P$ to move is $t$ (seconds). Determine the value of $t$ when $\\triangle PBE$ is congruent to $\\triangle QCF$, and provide the reasoning.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure", "input_image": [ "batch22-2024_06_14_c785d82426b0cbf016a1g_0079_1.jpg", "batch22-2024_06_14_c785d82426b0cbf016a1g_0079_2.jpg", "batch22-2024_06_14_c785d82426b0cbf016a1g_0079_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{4}{3}$ or 4", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: When point $Q$ is on the segment $BC$,\n\n$\\because AB \\parallel CD$, and $PF \\perp AB$,\n\n$\\therefore \\angle F = 90^{\\circ}$,\n\n$\\because \\triangle PBE$ is congruent to $\\triangle QCF$,\n\n$\\therefore CQ = BP = t$,\n\n$\\because \\angle B = 60^{\\circ}$,\n\n$\\therefore \\angle BQP = 30^{\\circ}$,\n\n$\\therefore BQ = 2t$,\n$\\therefore 2t + t = 4$,\n\n$\\therefore t = \\frac{4}{3}$,\n\nWhen point $Q$ is on the extension of $BC$,\n\n\n\n$\\because \\triangle PBE$ is congruent to $\\triangle QCF$,\n\n$\\therefore BP = CQ$,\n\nSimilarly, $BQ = 2t$,\n\n$\\therefore BC = t = 4$,\n\nIn summary: when $t = \\frac{4}{3}$ or 4, $\\triangle PBE$ is congruent to $\\triangle QCF$.\n\n【Insight】This problem mainly examines the properties of congruent triangles, the properties of parallelograms, and the knowledge that the leg opposite a $30^{\\circ}$ angle in a right triangle is half the hypotenuse. Mastering the properties of congruent triangles and conducting classified discussions are key to solving the problem." }, { "problem_id": 1536, "question": "The Huizhuang City Government has introduced a shared bike service to facilitate green commuting for its citizens. Figure (1) is a physical image of a certain brand of shared bike placed on a horizontal ground, and Figure (2) is its schematic diagram. In the diagram, both $AB$ and $CD$ are parallel to the ground $l$, $\\angle BCD = 60^\\circ$, and $\\angle BAC = 54^\\circ$. At what degree does $\\angle MAC$ need to be for $AM$ to be parallel to $CB$?\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch23-2024_06_14_15786e07d91640923609g_0065_1.jpg", "batch23-2024_06_14_15786e07d91640923609g_0065_2.jpg" ], "is_multi_img": true, "answer": "when $\\angle M A C=66^{\\circ}$, $A M / / C B$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since both $AB$ and $CD$ are parallel to the ground line $l$,\n\n$\\therefore AB \\parallel CD$,\n\n$\\therefore \\angle BAC + \\angle ACD = 180^\\circ$,\n\n$\\therefore \\angle BAC + \\angle ACB + \\angle BCD = 180^\\circ$,\n\nGiven that $\\angle BCD = 60^\\circ$ and $\\angle BAC = 54^\\circ$,\n\n$\\therefore \\angle ACB = 66^\\circ$,\n\n$\\therefore$ When $\\angle MAC = \\angle ACB = 66^\\circ$, $AM \\parallel CB$.\n\n【Highlight】This problem examines the determination and properties of parallel lines. Familiarity with the theorems regarding the determination and properties of parallel lines is crucial for solving the problem." }, { "problem_id": 1537, "question": "The law of light reflection by a plane mirror: The angle between the light ray incident on the plane mirror and the reflected light ray is equal to the acute angle formed with the plane mirror. As shown in Figure (1), a light ray $m$ is incident on plane mirror $a$, and the reflected light ray is $n$. The acute angle between the incident ray $m$, the reflected ray $n$, and the plane mirror $a$ is equal, i.e., $\\angle 1 = \\angle 2$. As shown in Figure (2), $AB$ and $CD$ are two plane mirrors. After two reflections, the positional relationship between the incident ray $m$ and the reflected ray $n$ will change accordingly. Please calculate: in Figure (2), what should be the angle $\\angle ABC$ between the two plane mirrors $AB$ and $CD$ for the incident ray $m$ to be parallel but opposite in direction to the reflected ray $n$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch23-2024_06_14_22f118a9c0a4c542475bg_0004_1.jpg", "batch23-2024_06_14_22f118a9c0a4c542475bg_0004_2.jpg" ], "is_multi_img": true, "answer": "$\\angle A B C=90^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the problem statement, we know: $\\angle 1 = \\angle 2$, $\\angle 3 = \\angle 4$.\n\nFor the incident ray $m$ to be parallel to the reflected ray $n$, it must hold that $\\angle 5 + \\angle 6 = 180^{\\circ}$.\n\nSince $\\angle 1 + \\angle 2 + \\angle 5 = 180^{\\circ}$ and $\\angle 3 + \\angle 6 + \\angle 4 = 180^{\\circ}$,\n\nit follows that $\\angle 1 + \\angle 2 + \\angle 3 + \\angle 4 = 180^{\\circ}$,\n\nhence $\\angle 2 + \\angle 3 = 90^{\\circ}$,\n\ntherefore, in triangle $ABC$, $\\angle ABC = 90^{\\circ}$.\n\n\n\n【Insight】This problem tests the application of the criteria for parallel lines. The key to solving it lies in the ability to flexibly use theorems for reasoning." }, { "problem_id": 1538, "question": "As shown in the figure, the following are the three views of a geometric body. What is the lateral surface area of this geometric body?\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch24-2024_06_15_2e5d00e8ed7276384599g_0003_1.jpg", "batch24-2024_06_15_2e5d00e8ed7276384599g_0003_2.jpg", "batch24-2024_06_15_2e5d00e8ed7276384599g_0003_3.jpg" ], "is_multi_img": true, "answer": "$6 \\pi \\mathrm{cm}^{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By observing the three views, it is known that the geometric body is a cylinder with a height of $3 \\mathrm{~cm}$ and a base diameter of $2 \\mathrm{~cm}$. The lateral area is calculated as: $\\pi \\mathrm{dh} = 2 \\times 3 \\pi = 6 \\pi \\mathrm{cm}^{2}$. Therefore, the answer is: $6 \\pi \\mathrm{cm}^{2}$.\n\n【Highlight】This question tests the ability to determine a geometric body from its three views and to perform calculations related to a cylinder. The key to solving the problem lies in first identifying the geometric body." }, { "problem_id": 1539, "question": "Teacher Han made 4 identical cubes and placed them as shown in Figure 1, and then rearranged them as shown in Figure 2. The sum of the numbers on the bottom squares of the 4 cubes in Figure 2 is $\\qquad$ .\n\nFigure 1\n\n\nFigure 2\n\n", "input_image": [ "batch24-2024_06_15_2e5d00e8ed7276384599g_0019_1.jpg", "batch24-2024_06_15_2e5d00e8ed7276384599g_0019_2.jpg" ], "is_multi_img": true, "answer": "16", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From Figure 1, it can be deduced that the opposite of 1 dot is 5 dots, the opposite of 2 dots is 4 dots, and the opposite of 3 dots is 6 dots. Therefore, in Figure 2, the number of dots in the four bottom squares are $1, 3, 6, 6$ respectively, summing up to 16.\n\n【Highlight】This question mainly tests students' spatial imagination and reasoning abilities. One could also manually create a cube, mark the number of dots on each face according to the problem statement, and then determine the number of dots on the opposite faces. This exercise can enhance hands-on skills and spatial imagination. The key to solving the problem lies in determining that the face opposite to 5 dots is 1 dot, based on the rotation of Figure 1 and Figure 4." }, { "problem_id": 1540, "question": "As shown in Figure 1, it is the side expansion diagram of a small cube. The small cube flips to the 1st grid, the 2nd grid, and the 3rd grid in sequence from the position shown in Figure 2. The word on the top side of the small cube at this time is $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch24-2024_06_15_2e5d00e8ed7276384599g_0029_1.jpg", "batch24-2024_06_15_2e5d00e8ed7276384599g_0029_2.jpg" ], "is_multi_img": true, "answer": "会", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From Figure 1, it can be deduced that \"构\" is opposite to \"会\"; \"建\" is opposite to \"谐\"; and \"和\" is opposite to \"社\".\n\nFrom Figure 2, when the small cube flips from its position in Figure 2 to the third grid, \"构\" is on the bottom, which means the character on the top face of the small cube at this time is \"会\".\n\nTherefore, the answer is \"会\".\n\n[Highlight] This question tests the topic: Characters on opposite faces of a cube, with the key to solving the problem lying in understanding the surface development diagram of the cube." }, { "problem_id": 1541, "question": "There is a die, and its three orientations are shown in the figures. The sum of the dots on the bottom faces in these three orientations is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch24-2024_06_15_2e5d00e8ed7276384599g_0083_1.jpg", "batch24-2024_06_15_2e5d00e8ed7276384599g_0083_2.jpg", "batch24-2024_06_15_2e5d00e8ed7276384599g_0083_3.jpg" ], "is_multi_img": true, "answer": "14", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "1", "analysis": "Solution: According to the problem, face \"1\" is opposite to face \"6\", face \"4\" is opposite to face \" \", and face \"2\" is opposite to face \"3\". Therefore, the number of dots on the bottom faces for these three placements are $3$, $5$, and $6$ respectively, and their sum is 14.\n\nThus, the answer is: 14.\n\n[Key Insight] This question examines the spatial structural characteristics of a cube. Pay attention to the spatial figure of the cube and analyze and solve the problem by considering the opposite faces." }, { "problem_id": 1542, "question": "Figure (1) shows a square cardboard with a side length of $24 \\mathrm{~cm}$. After cutting out the shaded area, it is folded into a rectangular box as shown in Figure (2). Given that the width and height of the rectangular box are equal, its volume is $\\qquad$ $\\mathrm{cm}^{3}$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch24-2024_06_15_2e5d00e8ed7276384599g_0089_1.jpg", "batch24-2024_06_15_2e5d00e8ed7276384599g_0089_2.jpg" ], "is_multi_img": true, "answer": "432", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the height of the rectangular prism be \\( x \\) cm. Then, its width can be expressed as \\( \\frac{24 - 2x}{2} = 12 - x \\) cm.\n\nAccording to the problem, we have:\n\\[ 12 - x = x \\]\n\nSolving the equation:\n\\[ x = 6 \\]\n\nThus, the width of the rectangular prism is 6 cm, and the length is 12 cm.\n\nTherefore, the volume of the rectangular prism is:\n\\[ 6 \\times 6 \\times 12 = 432 \\text{ cm}^3 \\]\n\n**Key Insight**: This problem tests the application of linear equations in one variable and the understanding of geometric shapes. The key to solving it lies in identifying the relationship between the variables and setting up the correct equation." }, { "problem_id": 1543, "question": "A cube has six faces labeled with the numbers $1, 2, 3, 4, 5, 6$. The views from three different directions are shown in Figure 1, and Figure 2 is the side view of the cube's net. The number represented by $x$ in the figure is $\\qquad$.\n\n\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0001_1.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0001_2.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0001_3.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 1, it can be seen that the numbers adjacent to 1 are $2, 3, 4, 6$.\n\nTherefore, the number opposite to 1 is 5.\n\nSince the numbers adjacent to 4 are $1, 3, 5, 6$,\n\nthe number opposite to 4 is 2.\n\nThus, the number opposite to 3 is 6.\n\nFrom Figure 2, it is known that the number opposite to 6 is $x$.\n\nTherefore, the value of $x$ is 3.\n\nHence, the answer is: 3.\n\n[Highlight] This question examines the text on opposite faces of a cube. The key to solving the problem is determining the numbers on opposite faces based on the numbers on adjacent faces." }, { "problem_id": 1544, "question": "As shown in the figure, in rectangle $A B C D$, $A B=a, B C=b$, and $a>b$. When rectangle $A B C D$ is rotated around the line containing side $A B$, it forms cylinder A. When rectangle $A B C D$ is rotated around the line containing side $B C$, it forms cylinder B. Let the lateral areas of the two cylinders be $S_{\\text {A }}$ and $S_{\\text {B }}$, respectively. Then $S_{\\text {A }}$ $\\qquad$ $S_{\\text {B }}$. (Fill in the blank with >, <, or =)\n\n\n\nA\n\n", "input_image": [ "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0025_1.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0025_2.jpg" ], "is_multi_img": true, "answer": "$=$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Since \\( S_{\\text{ A}} = 2 \\pi \\cdot AD \\cdot AB \\) and \\( S_{\\text{ B}} = 2 \\pi \\cdot AB \\cdot AD \\),\n\nTherefore, \\( S_{\\text{ A}} = S_{\\text{ B}} \\).\n\nHence, the answer is: \\( = \\).\n\n【Insight】This question examines the formation of a cylinder and the calculation of its lateral surface area. Understanding the correct formula for calculating the lateral surface area is key to solving the problem." }, { "problem_id": 1545, "question": "Fold the unfolded cube diagram shown in Figure 1 into a cube (with the text facing outward), and then roll this cube sequentially to the 1st, 2nd, 3rd, and 4th positions according to Figure 2. The text on the top face of the cube at this point is . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0043_1.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0043_2.jpg" ], "is_multi_img": true, "answer": "富", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From Figure 1, it can be deduced that \"Fu\" (富) is opposite to \"Wen\" (文); \"Qiang\" (强) is opposite to \"Zhu\" (主); and \"Min\" (民) is opposite to \"Ming\" (明).\n\nFrom Figure 2, when the small cube is flipped sequentially from its position in Figure 2 to the 4th position, \"Wen\" is on the bottom. Therefore, the character on the top face of the small cube at this time is \"Fu\".\n\nHence, the answer is: Fu.\n\n[Key Insight] This question primarily tests the understanding of the unfolded diagram of a cube. The key to solving it lies in identifying the characters on opposite faces and determining the character on the bottom after rotation." }, { "problem_id": 1546, "question": "There is a regular hexahedron die placed on a tabletop. The die is rolled clockwise as shown in the figure, with each roll of $90^{\\circ}$ counting as one move. After the 2022nd roll, the number of dots on the face of the die that is facing down is $\\qquad$.\n\n\n\nFirst roll\n\n\n\nSecond roll\n\n\n\nThird roll", "input_image": [ "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0061_1.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0061_2.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0061_3.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: By observing the figure, we can see that the number of dots 3 and 4 are opposite each other, and the number of dots 2 and 5 are opposite each other. According to the problem, the number of dots on the bottom face cycles every four times, with the sequence of dots in each cycle being $2, 3, 5, 4$. Calculating $2022 \\div 4$ gives a quotient of 505 with a remainder of 2.\n\nTherefore, after the 2022nd roll, the number of dots on the bottom face of the die is 3.\n\nHence, the answer is: 3.\n\n[Key Insight] This problem examines the text on opposite faces of a cube and the pattern of changes in the figure. The key to solving it lies in identifying the underlying pattern." }, { "problem_id": 1547, "question": "A geometric solid, as seen from different directions, is shown in the figures below. The number of small cubes that make up this geometric solid is . $\\qquad$\n\n\n\nView from the front\n\n\n\nView from above\n\n\n\nView from the left", "input_image": [ "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0084_1.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0084_2.jpg", "batch24-2024_06_15_5e9fa8d04bd98b86738fg_0084_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the given diagram, it can be seen that the first layer of the geometric shape consists of 4 small cubes, and the second layer consists of 2 small cubes. Therefore, the total number of small cubes that make up this geometric shape is 6.\n\nHence, the answer is: 6.\n\n[Key Insight] This question primarily tests the ability to observe objects and geometric shapes from different perspectives, with a key focus on developing students' spatial imagination skills." }, { "problem_id": 1548, "question": "As shown in Figure A, for an angle $\\angle M O N$ on the plane that does not exceed $90^{\\circ}$, we define the following: If point $P$ is inside $\\angle M O N$, and perpendiculars $P E \\perp O M, P F \\perp O N$ are drawn with feet at points $E$ and $F$ respectively, then the value of $P E + P F$ is called the \"point-angle distance\" of point $P$ relative to $\\angle M O N$, denoted as $d(P, \\angle M O N)$. As shown in Figure B, in the Cartesian coordinate plane $x O y$, point $P$ is in the first quadrant, and the abscissa of point $P$ is 1 greater than its ordinate. For $\\angle x O y$, it satisfies $d(P, \\angle x O y) = 5$. The coordinates of point $P$ are $\\qquad$.\n\n\n\nFigure A\n\n", "input_image": [ "batch24-2024_06_15_73f98a27cfd6bfcaa53fg_0050_1.jpg", "batch24-2024_06_15_73f98a27cfd6bfcaa53fg_0050_2.jpg" ], "is_multi_img": true, "answer": "(3, 2)", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the abscissa (x-coordinate) of point \\( P \\) be \\( x \\), then the ordinate (y-coordinate) of point \\( P \\) is \\( x - 1 \\).\n\nAccording to the problem, we have:\n\\[\nx + (x - 1) = 5\n\\]\nSolving for \\( x \\):\n\\[\n2x - 1 = 5 \\\\\n2x = 6 \\\\\nx = 3\n\\]\nThus, the y-coordinate is:\n\\[\nx - 1 = 3 - 1 = 2\n\\]\nTherefore, the coordinates of point \\( P \\) are \\( (3, 2) \\).\n\nThe final answer is \\( (3, 2) \\).\n\n【Key Insight】This problem primarily tests the understanding of point coordinates. The key to solving it lies in identifying the relationship between the coordinates based on the given image." }, { "problem_id": 1549, "question": "As shown in Figure A, for an angle $\\angle \\mathrm{MON}$ on the plane that does not exceed $90^{\\circ}$, we provide the following definition: If point $\\mathrm{P}$ is inside $\\angle \\mathrm{MON}$, and lines $\\mathrm{PE} \\perp \\mathrm{OM}$ and $\\mathrm{PF} \\perp \\mathrm{ON}$ are drawn with feet at points $\\mathrm{E}$ and $\\mathrm{F}$ respectively, then the value of $\\mathrm{PE} + \\mathrm{PF}$ is referred to as the \"point-angle distance\" of point $\\mathrm{P}$ relative to $\\angle \\mathrm{MON}$, denoted as $\\mathrm{d}(\\mathrm{P}, \\angle \\mathrm{MON})$. As shown in Figure B, in the plane rectangular coordinate system $\\mathrm{xOy}$, point $\\mathrm{P}$ is within the coordinate plane, and the abscissa of point $\\mathrm{P}$ is 2 greater than its ordinate. For $\\angle \\mathrm{xOy}$, the condition $\\mathrm{d}(\\mathrm{P}, \\angle \\mathrm{xOy}) = 10$ is satisfied. The coordinates of point $\\mathrm{P}$ are $\\qquad$.\n\n\n\nFigure A\n\n", "input_image": [ "batch24-2024_06_15_a400b9562ba267452a35g_0048_1.jpg", "batch24-2024_06_15_a400b9562ba267452a35g_0048_2.jpg" ], "is_multi_img": true, "answer": "(6,4) or $(-4,-6)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the abscissa of point $\\mathrm{P}$ be $\\mathrm{x}$, then the ordinate of point $\\mathrm{P}$ is $\\mathrm{x}-2$. According to the problem statement,\n\nWhen point $\\mathrm{P}$ is in the first quadrant, $x + x - 2 = 10$,\n\nSolving gives $\\mathrm{x} = 6$,\n\n$\\therefore \\mathrm{x} - 2 = 4$,\n\n$\\therefore \\mathrm{P}(6, 4)$;\n\nWhen point $\\mathrm{P}$ is in the third quadrant, $-x - x + 2 = 10$,\n\nSolving gives $x = -4$,\n\n$\\therefore x - 2 = -6$,\n\n$\\therefore P(-4, -6)$.\n\nThus, the answers are $(6, 4)$ or $(-4, -6)$.\n\n[Key Insight] This problem primarily tests the understanding of point coordinates. The key to solving it lies in comprehending the given information, understanding the definition of \"point-angle distance,\" and setting up the equation accordingly." }, { "problem_id": 1550, "question": "Figure 1 shows the surface development of a cube. When it is folded back into its original cube (Figure 2), the point that coincides with point $P$ should be $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch24-2024_06_15_cbf9fb5e4e372bcb251cg_0040_1.jpg", "batch24-2024_06_15_cbf9fb5e4e372bcb251cg_0040_2.jpg" ], "is_multi_img": true, "answer": "$T$ and $V$", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the diagram, we can see that after folding into a cube, point $Q$ coincides with point $S$, and point $P$ coincides with point $T$. Additionally, point $T$ coincides with point $V$. Therefore, the two points that coincide with point $P$ should be $T$ and $V$.\n\nHence, the answer is: $T$ and $V$.\n\n【Insight】This question tests the concept of folding a net into a geometric solid. When solving, do not forget the characteristics of a quadrangular prism and the various scenarios of cube nets. You can also try hands-on manipulation to enhance your spatial imagination." }, { "problem_id": 1551, "question": "The cube wooden block shown in Figure (1) has a side length of $6 \\mathrm{~cm}$. A corner is cut off along the diagonals of three adjacent faces (shown as dashed lines in the figure), resulting in the geometric shape shown in Figure (2). The shortest distance for an ant to crawl on the surface of the geometric shape from vertex $\\mathrm{A}$ to vertex $\\mathrm{B}$ is $\\qquad$ $\\mathrm{cm}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch24-2024_06_15_cbf9fb5e4e372bcb251cg_0093_1.jpg", "batch24-2024_06_15_cbf9fb5e4e372bcb251cg_0093_2.jpg" ], "is_multi_img": true, "answer": "$(3 \\sqrt{2}+3 \\sqrt{6})$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "As shown in the figure:\n\n\n\n$\\triangle \\mathrm{BCD}$ is an isosceles right triangle, and $\\triangle \\mathrm{ACD}$ is an equilateral triangle.\n\nIn the right triangle $\\triangle \\mathrm{BCD}$, $\\mathrm{CD}=\\sqrt{B C^{2}+B D^{2}}=6 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, $\\mathrm{BE}=\\frac{1}{2} \\mathrm{CD}=3 \\sqrt{2} \\mathrm{~cm}$.\n\nIn the right triangle $\\triangle \\mathrm{ACE}$, $\\mathrm{AE}=\\sqrt{A C^{2}-C E^{2}}=3 \\sqrt{6} \\mathrm{~cm}$.\n\nThus, the shortest distance from vertex $\\mathrm{A}$ to vertex $\\mathrm{B}$ is $(3 \\sqrt{2}+3 \\sqrt{6}) \\mathrm{cm}$.\n\nThe answer is $(3 \\sqrt{2}+3 \\sqrt{6})$.\n\n【Key Insight】This problem examines the shortest path on a plane by unfolding the surface of the geometric shape shown in Figure (2) into a flat plane. The solution leverages the properties of isosceles right triangles and equilateral triangles." }, { "problem_id": 1552, "question": "As shown in the figure, the number of small cubes that make up this geometric solid is $n$. Then, all possible values of $n$ are $\\qquad$.\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0011_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0011_2.jpg" ], "is_multi_img": true, "answer": "$8,9,10,11$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "The top view shows 5 squares,\n\nso the bottom layer has 5 cubes.\n\nFrom the front view, the second layer has at least 2 cubes, and the third layer has at least 1 cube;\n\nFrom the front view, the second layer has at most 4 cubes, and the third layer has at most 2 cubes;\n\nTherefore, the possible values for $\\mathrm{n}$ are: $1+4+3=8, 1+5+3=9, 1+6+3=10, 1+4+4=9, 1+5+4=10, 1+6+4=11$. Hence, the answers are $8, 9, 10, 11$.\n\n【Highlight】This question tests the understanding and application of three-view drawings and spatial imagination. The front view helps distinguish the vertical layers of the object, while the top view helps identify the horizontal and depth positions. By combining these analyses, the number of small cubes can be determined." }, { "problem_id": 1553, "question": "As shown in the figure, a geometric body is formed by small cubic blocks with a side length of $1 \\mathrm{~cm}$. The figures obtained from the front view and the top view are shown in the figure. The minimum surface area of such a geometric body is $\\mathrm{cm}^{2}$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0027_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0027_2.jpg" ], "is_multi_img": true, "answer": "34", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "To construct such a geometric shape, a minimum of \\(6 + 2 + 1 = 9\\) small cubes is required,\n\nand a maximum of \\(6 + 5 + 2 = 13\\) small cubes is needed;\n\nTherefore, the maximum number of small cubes needed is 13, and the minimum is 9.\n\nThe surface area of the geometric shape constructed with the minimum number of small cubes is \\((6 + 6 + 5) \\times 2 = 34\\).\n\nHence, the answer is 34;\n\n【Insight】This question tests the ability to determine a geometric shape from its three views. When solving such problems, one should utilize the characteristics of the three views to represent the object: understand the top and bottom, left and right shapes from the front view; understand the left and right, front and back shapes from the top view; and understand the top and bottom, front and back shapes from the side view. Comprehensive analysis, reasonable conjecture, and combining life experience to sketch a draft, then verify if it meets the problem's requirements." }, { "problem_id": 1554, "question": "As shown in the figure, a rectangular prism is formed by arranging small cubes with edge length $\\mathrm{a}$. According to this arrangement, the surface area of the $\\mathrm{n}$-th rectangular prism is $\\qquad$\n\n\n\nThe 1st one\n\n\n\nThe 2nd one\n\n\n\nThe 3rd one", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0031_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0031_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0031_3.jpg" ], "is_multi_img": true, "answer": "(4n+6) a2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "Based on the analysis of the question stem, it can be deduced that the surface area of the nth rectangular prism is: the area of $4n + 6$ small squares;\n\nThe area of one face of a small square is: $a \\times a = a^{2}$,\n\nTherefore, the surface area of the nth rectangular prism is: $[(n + 1) \\times 4 + 2] a^{2} = (4n + 6) a^{2}$.\n\nHence, the answer is: $(4n + 6) a^{2}$.\n\n[Highlight] This question mainly examines the surface area of geometric solids, paying attention to identifying patterns based on the figures." }, { "problem_id": 1555, "question": "The three views of a cone are shown in the figure. The surface area of this cone is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0037_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0037_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0037_3.jpg" ], "is_multi_img": true, "answer": "$55 \\pi \\mathrm{cm}^{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three views, it is known that the radius is $5 \\mathrm{~cm}$ and the slant height of the cone is $6 \\mathrm{~cm}$.\n\n$\\therefore$ The surface area $=\\pi \\times 5 \\times 6 + \\pi \\times 5^{2} = 55 \\pi \\mathrm{cm}^{2}$. Therefore, the answer is: $55 \\pi \\mathrm{cm}^{2}$.\n\n【Highlight】This question examines the calculation of a cone. Determining the base diameter and slant height of the cone from the data in the three views is key to solving this problem. This question embodies the mathematical concept of combining numbers and shapes. If the base radius of the cone is $r$ and the slant height is $l$, then the surface area of the cone $=\\pi r l + \\pi r^{2}$." }, { "problem_id": 1556, "question": "As shown in the figure, the three views of a geometric body composed of identical small cubes are given. The maximum number of small cubes that can form this geometric body is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0039_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0039_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0039_3.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "According to the three-view drawing, the distribution of small cubes in this geometric figure is as shown below:\n\n$$\n\\begin{array}{|l|l|}\n\\hline 2 & 2 \\\\\n\\hline 3 & 2 \\\\\n\\hline\n\\end{array}\n$$\n\nTop View\n\nTherefore, the maximum number of small cubes that make up this geometric figure is 9.\n\nHence, the answer is 9.\n\n[Highlight] This question tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. If you master the mnemonic \"Use the top view to lay the foundation, the front view to build up, and the side view to remove violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1557, "question": "The area of the top view of a cuboid, as shown in the figures (in $\\mathrm{cm}$) from the front and left sides, is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0045_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0045_2.jpg" ], "is_multi_img": true, "answer": "$6 \\mathrm{~cm}^{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Based on the relevant data from the shape diagrams viewed from the left and the front, it can be deduced that:\n\nThe shape diagram viewed from above is a rectangle with a length of 3 and a width of 2.\n\nTherefore, the area of the shape diagram viewed from above is \\(2 \\times 3 = 6 \\mathrm{~cm}^{2}\\).\n\nHence, the answer is \\(6 \\mathrm{~cm}^{2}\\).\n\n【Key Point】This question tests the ability to determine a geometric shape from its three-view drawings. The crucial point is to deduce from the relevant data of the shape diagrams viewed from the left and the front that the shape diagram viewed from above is a rectangle with a length of 3 and a width of 2." }, { "problem_id": 1558, "question": "As shown in the figure below, the geometric shape is formed by some identical small cubes. The shape views from the front, left, and top are as follows. Then, the number of small cubes used to form this geometric shape is $\\qquad$ cubes.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from the top", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0049_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0049_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0049_3.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the front view, top view, and left view, we can deduce:\n\nThe base layer consists of 6 small cubes, the second layer has 2 small cubes, and the third layer contains 1 small cube.\n\nTherefore, the total number of small cubes used to construct this geometric figure is $6 + 2 + 1 = 9$.\nHence, the answer is: 9.\n\n[Key Insight] This question tests the ability to determine a geometric figure based on its three views." }, { "problem_id": 1559, "question": "A geometric solid is formed by arranging identical small cubes, as shown in the following diagrams, which represent the front, left, and top views of the geometric solid. The number of small cubes used to form this geometric solid is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0059_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0059_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0059_3.jpg" ], "is_multi_img": true, "answer": "8\n", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the number of small cubes on the bottom layer is 6. From the other views, it can be seen that there is one cube each in the second row, second column, and the second layer of the third column. Therefore, the total number of cubes is $6 + 2 = 8$.\n\nThus, the answer is: 8" }, { "problem_id": 1560, "question": "The main view and top view of a geometric shape formed by small cubes of the same size are shown in the figures below. The maximum number of small cubes that can form this geometric shape is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0060_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0060_2.jpg" ], "is_multi_img": true, "answer": "11", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the front view and the top view, the base of this geometric figure should have at most \\(3 + 2 = 5\\) small cubes. The second layer can have up to 3 small cubes, and the third layer can also have up to 3 small cubes.\n\nTherefore, the maximum number of small cubes that make up this geometric figure is \\(5 + 3 + 3 = 11\\).\n\nHence, the answer is 11.\n\nKey Insight: This question is designed to test students' mastery and flexible application of the three-view drawing concept, as well as their spatial imagination skills. Remembering the mnemonic \"Top view lays the foundation, front view builds up wildly, side view removes violations\" can make it easier to arrive at the answer." }, { "problem_id": 1561, "question": "As shown in the figure, the following are the three views of a geometric body, and the total surface area of this geometric body is . (Take $\\pi$ as 3.14)\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0067_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0067_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0067_3.jpg" ], "is_multi_img": true, "answer": "9.42", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "This geometric shape is a cone.\n\nThe lateral surface area of the cone is: $\\frac{1}{2} \\times 2 \\pi \\times 2=2 \\pi$;\n\nThe base area is: $\\pi$,\n\nThus, the total surface area is: $2 \\pi+\\pi=3 \\pi \\approx 9.42$.\n\nTherefore, the answer is: 9.42." }, { "problem_id": 1562, "question": "As shown in the figure, the front view and the left view of a geometric solid are both equilateral triangles with side lengths of $1 \\mathrm{~cm}$, and the top view is a circle. Therefore, the lateral surface area of this geometric solid is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0071_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0071_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{\\pi}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the front view, top view, and left view, it can be deduced that the geometric shape should be a cone with a base circle diameter of 1 and a generatrix length of 1. Therefore, the lateral surface area is calculated as $\\frac{1}{2} \\times (1 \\times \\pi) \\times 1 = \\frac{\\pi}{2} \\mathrm{~cm}^{2}$.\n\nHence, the answer is: $\\frac{\\pi}{2}$.\n\n[Insight] This question tests the lateral surface area of a cone and the understanding of three-view drawings. Mastering these knowledge points is crucial for solving the problem." }, { "problem_id": 1563, "question": "As shown in the figure, the main view and top view of a geometric shape composed of several identical small cubes are given. The geometric shape may be composed of $\\qquad$ cubes.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0073_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0073_2.jpg" ], "is_multi_img": true, "answer": "6or 7 or 8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in the figure, the numbers on the top view indicate the number of layers of small cubes at the corresponding positions. There are five different scenarios in total, and the number of small cubes in this geometric structure could be 6, 7, or 8.\n\n" }, { "problem_id": 1564, "question": "A warehouse manager needs to count the items in the warehouse, which are all identical cubic boxes of the same size. He cannot move the boxes to count them, but he comes up with an idea: by observing the three views of the items, he can determine the number of boxes in the warehouse. The three views of the stacked boxes he sees are shown on the right. Therefore, the number of boxes the warehouse manager counts is . $\\qquad$\n\n\n\nLeft View\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0076_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0076_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0076_3.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** From the left and top views of the three-view drawing, only the two square boxes in the very center have two layers, while the others have only one layer. There are 4 small square boxes in the very center. Considering the front and top views, apart from the small square in the center, there are four other small squares, all of which are single-layered. Therefore, there are a total of 8 boxes.\n\n**Exam Point:** Three-view drawing\n\n**Review:** This question tests the understanding of three-view drawings, requiring candidates to grasp the concept of three-view drawings and be able to interpret the three views of a geometric shape." }, { "problem_id": 1565, "question": "The main view and left view of a geometric shape composed of some identical small cubes are shown in the figures below. The number of small cubes that make up this geometric shape could be $\\qquad$ .\n\n\n\nMain View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0083_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0083_2.jpg" ], "is_multi_img": true, "answer": "4 or 5 or 6 or 7 .", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the front view provided in the problem, we know that the object consists of two columns, with the left column being two layers high and the right column being at most one layer high. From the side view, we can see that there are two rows on the left and one row on the right.\n\n$\\therefore$ It can be determined that there is only one small cube on the left, while on the right, there could be either one row with a single layer and one row with two layers, or both rows could have two layers.\n\n$\\therefore$ The minimum number of small cubes in the figure is 4, and the maximum is 7. That is, the number of small cubes that make up this geometric shape could be 4, 5, 6, or 7." }, { "problem_id": 1566, "question": "Eight small cubes of the same size are glued together to form a large cube as shown in Figure 1. The three-view diagram of the resulting geometric body is shown in Figure 2. If Xiao Ming removes several of the eight small cubes, leaving the remaining cubes in their original positions, and the three-view diagram of the new geometric body is still as shown in Figure 2, then the maximum number of small cubes he can remove is $\\qquad$.\n\n\n\nFigure 1\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0084_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0084_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since the layers are adhered, by alternately removing two cubes from diagonal positions in each layer, it is ensured that each row and each column in every layer retains one cube. Therefore, the maximum number of small cubes that can be removed is 4.\n\nThus, the answer is: 4." }, { "problem_id": 1567, "question": "As shown in the figure, the following are the three views of a packaging box. The volume of this packaging box is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0085_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0085_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0085_3.jpg" ], "is_multi_img": true, "answer": "$2000 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, we have:\n$\\pi \\times 10 \\times 10 \\times 20 = 2000 \\pi$.\n\nTherefore, the answer is: $2000 \\pi$." }, { "problem_id": 1568, "question": "As shown in the figure, the following are the three views and relevant dimensions (units: cm) of a geometric object. The lateral surface area of this geometric object is $\\mathrm{cm}^{2}$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0086_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0086_2.jpg" ], "is_multi_img": true, "answer": "$2 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three views, it is easy to determine that this geometric body is a cone. According to the problem, the diameter of the base is 2, and the length of the generatrix (slant height) is 2. \n\nTherefore, the lateral surface area of the geometric body is:\n\\[\n\\frac{1}{2} \\times 2 \\times 2 \\pi = 2 \\pi.\n\\]" }, { "problem_id": 1569, "question": "The three views of a geometric solid are shown in the figure, where the front view and the left view are two congruent isosceles triangles, and the top view is a circle. The lateral surface area of the geometric solid is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_03ceafbdd4e2455f177fg_0087_1.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0087_2.jpg", "batch25-2024_06_17_03ceafbdd4e2455f177fg_0087_3.jpg" ], "is_multi_img": true, "answer": "$65 \\pi$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** According to the problem statement, the geometric figure is a cone. The height of the cone is 12, and the diameter of the base circle is 10, which means the radius of the base circle is 5.\n\nTherefore, the slant height of the cone \\( l = \\sqrt{12^{2} + 5^{2}} = 13 \\).\n\nThus, the lateral surface area of the cone \\( = \\frac{1}{2} \\times 13 \\times 2 \\pi \\times 5 = 65 \\pi \\).\n\n**Key Points:** 1. Calculation of a cone; 2. Determining the geometric figure from its three views." }, { "problem_id": 1570, "question": "As shown in the figure, the three views of a simple geometric body are given. The lateral surface area of this geometric body is equal to $\\qquad$ .\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0003_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0003_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0003_3.jpg" ], "is_multi_img": true, "answer": "18", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three views of the geometric solid, it can be determined that the solid is a triangular prism with an equilateral triangle base of side length 2 and a height of 3.\n\n$\\therefore$ The lateral surface area of this geometric solid is equal to $3 \\times 2 \\times 3=18$,\n\nHence, the answer is 18.\n\n【Insight】This question tests the understanding of three views and the lateral surface area of a triangular prism, and examines the application of three views of simple geometric solids. The key to solving the problem lies in having spatial imagination and basic computational skills." }, { "problem_id": 1571, "question": "A cuboid's three views are shown in the figure. If its top view is a square, then the volume of this cuboid is $\\qquad$.\n\n\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nSide view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0004_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0004_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0004_3.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0004_4.jpg" ], "is_multi_img": true, "answer": "12.", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** Let the side length of the square in the top view be $a$.\n\n$\\because$ The top view is a square, and from the front view, it can be seen that the diagonal of the square is $2 \\sqrt{2}$,\n\n$\\therefore a^{2}+a^{2}=(2 \\sqrt{2})^{2}$,\n\nSolving gives $a^{2}=4$,\n\n$\\therefore$ The volume of this cuboid is $4 \\times 3=12$." }, { "problem_id": 1572, "question": "As shown in the figure, the three views of a geometric object constructed from several small squares are given. Therefore, this geometric object is composed of $\\qquad$ small squares.\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0005_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0005_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0005_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Test Question Analysis: By synthesizing the three views, we can deduce that the base layer of this geometric figure should consist of 4 small cubes, and the second layer should have 2 small cubes. Therefore, the total number of small cubes needed to construct this geometric figure is $2 + 4 = 6$." }, { "problem_id": 1573, "question": "The number of small cubes that make up the geometric object, as seen from three directions, is $\\qquad$.\n\n\n\nFrom the front view\n\n\n\nFrom the left view\n\n\n\nFrom the top view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0009_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0009_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0009_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Analysis:** From the front view, it can be seen that the figure consists of 4 columns, with the leftmost column having 2 layers and the other 3 columns having 1 layer each. From the side view, the figure is composed of two rows, with the front row having 1 layer and the back row having 2 layers. From the top view, it is evident that the left column at the back has 2 layers, and the front row has only one column. Based on this information, the problem can be solved.\n\n**Detailed Explanation:** From the three views, the number of small squares in the geometric figure can be determined as: 1 + 4 + 1 = 6. Therefore, the answer is 6.\n\n**Key Insight:** This question tests the ability to observe objects and geometric figures from different perspectives, enhancing students' spatial imagination and abstract thinking skills." }, { "problem_id": 1574, "question": "Given the three views of a geometric object as shown in the figure, where the top view is a regular hexagon, the sum of all the lateral areas of the geometric object is .\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0018_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0018_2.jpg" ], "is_multi_img": true, "answer": "48", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three-view drawing, it is known that the geometric body is a regular hexagonal prism with a base edge length of 2 and a height of 4. Therefore, the total lateral surface area is $2 \\times 4 \\times 6 = 48$.\n\nHence, the answer is 48.\n\n[Highlight] This question tests the knowledge of determining the geometric body from three-view drawings. The key to solving the problem lies in the ability to interpret the shape and dimensions of the geometric body from the three views, which is not difficult." }, { "problem_id": 1575, "question": "The front view and top view of a geometric object are shown in the figures below. If this geometric object is composed of at most $m$ small cubes and at least $n$ small cubes, then $m+n=$ $\\qquad$ .\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0020_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0020_2.jpg" ], "is_multi_img": true, "answer": "16", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "It is easy to see that the first layer has 4 cubes, the second layer has a maximum of 3 cubes and a minimum of 2 cubes, and the third layer has a maximum of 2 cubes and a minimum of 1 cube.\n\n$m=4+3+2=9, n=4+2+1=7$,\n\nTherefore, $m+n=9+7=16$.\n\nHence, the answer is: 16.\n\n【Insight】This question tests the students' mastery and flexible application of the three-view concept, and also reflects the assessment of spatial imagination skills." }, { "problem_id": 1576, "question": "Given that an object is composed of $x$ identical cubes stacked together, its three views are shown in the figures below. Then $x=$ . $\\qquad$\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0028_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0028_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0028_3.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By synthesizing the three views, the object has a total of 3 layers,\n\nThe first layer has 6 units,\n\nThe second layer has 2 units,\n\nTotaling to $6+2=8$ units,\n\nThus, $\\mathrm{x}=8$,\n\nTherefore, the answer is: 8.\n\n【Highlight】This examines the ability to determine geometric shapes from three views, tests the proficiency and flexible application of understanding three views, and also reflects the assessment of spatial imagination capabilities." }, { "problem_id": 1577, "question": "The main view and top view of a geometric body composed of some identical small cubes are shown in the figure. The maximum number of small cubes that can form this geometric body is $\\qquad$ .\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0029_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0029_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the combination of the front view and the top view, it can be determined that the upper left layer has a maximum of 2 blocks, the lower left layer also has a maximum of 2 blocks, and the right side has only one layer with a single block.\n\nTherefore, the maximum number of small cubes in the diagram is 5.\n\nHence, the answer is 5.\n\n[Highlight] This question tests the students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills." }, { "problem_id": 1578, "question": "The exhibition hall requires the construction of a display platform using identical cubic wooden blocks, with the three views as shown in the figures. The total number of such cubic blocks needed for this platform is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0042_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0042_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0042_3.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Comprehensive front view, top view, and left view show that there are $3+1+2=6$ small cubes on the bottom layer, 2 small cubes on the second layer, and 2 small cubes on the third layer. Therefore, the total number of small cubes used to build this geometric figure is $6+2+2=10$.\n\n【Highlight】This question tests students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills. If one masters the mnemonic \"Top view lays the foundation, front view builds up wildly, left view removes violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1579, "question": "A certain object is composed of several identical small cubes, with its front view and left view as shown. The maximum number of small cubes the object can contain is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0055_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0055_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the front view, the number of columns of the smallest cubes at the bottom layer is 3.\n\nFrom the side view, the number of rows of the smallest cubes at the bottom layer is 2.\n\n$\\therefore$ The maximum number of cubes at the bottom layer is $3 \\times 2=6$.\n\n$\\because$ There is only 1 cube in the second layer,\n\n$\\therefore$ The maximum number of small cubes contained in the object is 7.\n\nTherefore, the answer is 7.\n\n[Highlight] This problem tests the knowledge of judging geometric shapes from three views; determining the maximum number of small cubes at the bottom layer by multiplying the number of rows by the number of columns is the key to solving this problem." }, { "problem_id": 1580, "question": "The following is a three-view drawing of a geometric shape composed of identical small cubes. The number of these identical small cubes is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft side view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0056_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0056_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0056_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "The base of the geometric shape consists of 4 small cubes. The shape has two layers, with the second layer containing 1 small cube, making a total of 5 cubes.\n\nTherefore, the answer is:" }, { "problem_id": 1581, "question": "A geometric solid is formed by arranging identical small cubes, as shown below are the planar views obtained from the front, left, and top of the geometric solid. The number of small cubes used to form this geometric solid is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0059_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0059_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0059_3.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer contains 6 small cubes. From the other views, we can see that there is one cube each in the second row, second column, and the third column of the second layer. Therefore, the total number of cubes is $6 + 2 = 8$.\n\nThus, the answer is: 8." }, { "problem_id": 1582, "question": "As shown in the figure, the three views of a geometric structure formed by identical small cubes are given. The geometric structure consists of $\\qquad$ small cubes.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0060_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0060_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0060_3.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Test Question Analysis: From the front view and the side view, mark the number of small cubes that are common in the vertical direction at each position within the top view's small squares.\n\nTherefore, this geometric figure consists of a total of $1+1+1+1+2+3+1=10$ small cubes. Hence, the answer is 10.\n\n\n\nTop View\n\nExam Point: Determining the geometric figure from the three views." }, { "problem_id": 1583, "question": "Several identical squares are stacked together. The front view (i.e., the view from the front) and the top view (i.e., the view from above) of the combination are shown below. Then, the maximum number of cubes in the combination is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0061_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0061_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the front view, the first and third columns can have at most two small cubes stacked together, and the second column can have at most one small cube.\n\nAccording to the top view, the first column has one small cube, the second column has two small cubes, and the third column has three small cubes.\n\nTherefore, the maximum number of cubes in the combined structure is: $1 \\times 2 + 2 + 3 \\times 2 = 10$ cubes.\n\nThus, the answer is 10.\n\n[Key Insight] This problem mainly tests the understanding of the three views of a simple combined structure. Enhancing spatial imagination skills is crucial for solving such problems." }, { "problem_id": 1584, "question": "As shown in the figure, the three views of a geometric body are given. According to the data marked in the figure, the lateral surface area of the geometric body can be calculated as $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0064_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0064_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0064_3.jpg" ], "is_multi_img": true, "answer": "$2 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the main view and the left view being rectangles, it can be deduced that this geometric body is a prism. From the top view being a circle, it can be deduced that this geometric body is a cylinder.\n\nIt is easy to determine that the diameter of the base of the cylinder is [diameter], and the height is [height],\n\nThe lateral area is [lateral area].\n\nTherefore, the answer is: [answer].\n\n[Insight] This question tests the calculation formula for the lateral area of a cylinder, with the key point being to identify the shape of the geometric body." }, { "problem_id": 1585, "question": "As shown in the figure, the three views of a geometric object composed of several small cubes with a side length of 2 are given. The volume of this geometric object is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0069_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0069_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0069_3.jpg" ], "is_multi_img": true, "answer": "40", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By synthesizing the three views, we can deduce that the base layer of this geometric model consists of \\(3 + 1 = 4\\) small cubes, and the second layer has 1 small cube. \n\nTherefore, the total number of small cubes used to construct this geometric model is \\(4 + 1 = 5\\).\n\n\\(\\therefore\\) The volume of this geometric body is \\(2 \\times 2 \\times 2 \\times 5 = 40\\),\n\nHence, the answer is: 40.\n\n【Key Insight】This question tests the understanding of three-view drawings, and using these views to determine the number of cubes is crucial for solving the problem." }, { "problem_id": 1586, "question": "As shown in the figure, the following are the three views of a solid figure. Please identify the name of the solid figure based on the views $\\qquad$ .\n\n\nFront View\n\n\nLeft View\n\n\nTop View\n\n##", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0073_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0073_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0073_3.jpg" ], "is_multi_img": true, "answer": "cone", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Test Question Analysis: The characteristics of the three views can be used to determine the shape.\n\nFrom the diagram, it can be seen that this three-dimensional figure is a cone.\n\nExam Focus: Determining the geometric shape from the three views.\n\nComment: This question is a basic application problem involving the three views of a geometric shape, primarily testing students' understanding of the three views of geometric shapes. It is relatively common in middle school exams, usually appearing in the form of multiple-choice or fill-in-the-blank questions. It is a fundamental question with a low level of difficulty." }, { "problem_id": 1587, "question": "There is a regular six-sided die placed on a table. The die is rolled clockwise as shown in the figure, with each roll of $90^{\\circ}$ counting as one move. After the 2014th roll, the number of dots on the face of the die that is facing down is $\\qquad$.\n\n\n\n\n\nFirst roll\n\n\n\nSecond roll\n\n\n\nThird roll", "input_image": [ "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0084_1.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0084_2.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0084_3.jpg", "batch25-2024_06_17_09dc9edc9ad7bd47fbdeg_0084_4.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: Observing the figure, we know that the number three is opposite to the number four, and the number two is opposite to the number five.\n\n$\\therefore$ According to the problem, the sequence of numbers facing downward cycles through $2, 3, 5, 4, \\ldots$ in a repeating pattern every four numbers.\n\n$$\n\\therefore 2014 \\div 4 = 503 \\text{ with a remainder of } 2,\n$$\n\nTherefore, after the 2014th turn, the number facing downward is 3.\n\n[Highlight] This problem examines the text on opposite faces of a cube and the pattern of graphical changes. The key to solving it lies in identifying the underlying pattern." }, { "problem_id": 1588, "question": "Figure (1) shows a simple device for hoisting a heavy object. $\\mathrm{AB}$ is the hoisting pole. When it is tilted, the heavy object is lifted. As it gradually becomes upright, the heavy object can be gradually raised. In sunlight, when the inclination angle of the hoisting pole $\\angle \\mathrm{ABC}=60^{\\circ}$, the shadow length of the hoisting pole measured is $\\mathrm{BC}=11.5$ meters. The hoisting pole is quickly made upright (the time required for the upright process is negligible), as shown in Figure (2). When $\\mathrm{AB}$ is perpendicular to the ground, the shadow length of the hoisting pole $\\mathrm{AB}$ measured is $\\mathrm{BC}=4$ meters. Find the length of the hoisting pole $\\mathrm{AB}$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_1a0f7200b797c80a9fc9g_0044_1.jpg", "batch25-2024_06_17_1a0f7200b797c80a9fc9g_0044_2.jpg" ], "is_multi_img": true, "answer": "$23-4 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "In Figure (1), draw a line $\\mathrm{AD} \\perp \\mathrm{BC}$ through point $\\mathrm{A}$, intersecting at point $\\mathrm{D}$.\n\nLet $\\mathrm{BD} = \\mathrm{x}$ meters, then $\\mathrm{DC} = (11.5 - \\mathrm{x})$ meters.\n\nSince $\\angle \\mathrm{ABC} = 60^{\\circ}$,\n\nit follows that $\\mathrm{AD} = \\sqrt{3} \\mathrm{x}$ meters, and $\\mathrm{AB} = 2 \\mathrm{x}$ meters.\n\nBecause sunlight is parallel, at the same moment and location, the height of an object is proportional to the length of its shadow,\n\nthe ratio \"AD:DC\" in Figure (1) is equal to the ratio \"AB:BC\" in Figure (2),\n\nthat is, $\\sqrt{3} x : (11.5 - x) = 2 x : 4$,\n\nsolving this gives $x = 11.5 - 2 \\sqrt{3}$ (the solution $x = 0$ is discarded as it does not fit the context),\n\nthus the boom $\\mathrm{AB} = 2 \\mathrm{x} = (23 - 4 \\sqrt{3})$ meters.\n\n\n\n(1)\n\n【Insight】This problem primarily tests the understanding of the properties and determination of similar triangles, as well as trigonometric knowledge. The key to solving the problem is recognizing that $\\triangle \\mathrm{ACD} \\sim \\triangle \\mathrm{ACB}$." }, { "problem_id": 1589, "question": "The flagpole in Shenzhen Civic Center Plaza is shown in Figure (1). A school interest group measured the height of the flagpole, as shown in Figure (2). At a certain moment, the shadow of the flagpole $A B$ falls partly on the platform and partly on the slope. The length of the shadow on the platform, $B C$, is measured to be 16 meters, and the length of the shadow on the slope, $C D$, is 8 meters. $A B \\perp B C$; at the same time, the angle between the sunlight and the horizontal plane is $45^{\\circ}$. A 1-meter pole $E F$ standing on the slope casts a shadow $F G$ of 2 meters. Find the height of the flagpole.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch25-2024_06_17_1a0f7200b797c80a9fc9g_0047_1.jpg", "batch25-2024_06_17_1a0f7200b797c80a9fc9g_0047_2.jpg" ], "is_multi_img": true, "answer": "20 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "As shown in the figure, draw $CM // AB$ intersecting $AD$ at $M$, and $MN \\perp AB$ at $N$.\n\nSince $\\triangle MCD \\sim \\triangle PQR$,\n\nTherefore, $\\frac{CM}{CD} = \\frac{PQ}{QR}$, which means $\\frac{CM}{1} = \\frac{8}{2}$, so $CM = 4$ meters.\n\nMoreover, since $MN // BC$ and $AB // CM$,\n\nTherefore, quadrilateral $MNBC$ is a rectangle,\nHence, $MN = BC = 16$ meters, and $BN = CM = 4$ meters.\n\nSince in the right triangle $\\triangle AMN$, $\\angle AMN = 45^\\circ$,\n\nTherefore, $AN = MN = 16$ meters,\n\nThus, $AB = AN + BN = 20$ meters.\n\n\n\nFigure 2\n\n[Insight] This question examines the application of similar triangles and shadow length, etc. The key to solving the problem is to correctly add auxiliary lines and construct right triangles to solve the problem, which is a common type of question in middle school exams." }, { "problem_id": 1590, "question": "Place a set of triangles as shown in Figure 1, where $\\angle \\mathrm{ACB}=\\angle \\mathrm{DEC}=90^{\\circ}$, the hypotenuse $\\mathrm{AB}=6, \\mathrm{DC}=7$. Rotate triangle $\\mathrm{DCE}$ clockwise around point $\\mathrm{C}$ so that side $\\mathrm{CD}$ passes exactly through the midpoint $\\mathrm{O}$ of $\\mathrm{AB}$, resulting in $\\triangle \\mathrm{D}_{1} \\mathrm{C}_{1} \\mathrm{E}_{1}$, as shown in Figure 2. The length of segment $\\mathrm{AD}_{1}$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_24be905cd8e625b47629g_0031_1.jpg", "batch25-2024_06_17_24be905cd8e625b47629g_0031_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, draw $\\mathrm{D}_{1} \\mathrm{H} \\perp \\mathrm{CA}$ intersecting the extension of $\\mathrm{CA}$ at $\\mathrm{H}$.\n\n\n\nFigure 2\n\nSince $\\mathrm{CA}=\\mathrm{CB}$, $\\angle \\mathrm{ACB}=90^{\\circ}$, and $\\mathrm{AO}=\\mathrm{OB}$,\n\nit follows that $\\mathrm{OC} \\perp \\mathrm{AB}$, and $\\mathrm{OC}=\\mathrm{OA}=\\mathrm{OB}=3$,\n\nthus $\\mathrm{AC}=3 \\sqrt{2}$.\n\nSince $\\mathrm{D}_{1} \\mathrm{H} \\perp \\mathrm{CH}$,\n\nit follows that $\\angle \\mathrm{HCD}_{1}=90^{\\circ}$.\n\nGiven that $\\angle \\mathrm{HCD}_{1}=\\frac{1}{2} \\angle \\mathrm{ACB}=45^{\\circ}$ and $\\mathrm{CD}_{1}=7$,\n\nwe have $\\mathrm{CH}=\\mathrm{HD}_{1}=\\frac{7 \\sqrt{2}}{2}$.\n\nTherefore, $\\mathrm{AH}=\\mathrm{CH}-\\mathrm{AC}=\\frac{\\sqrt{2}}{2}$.\n\nIn the right triangle $\\triangle \\mathrm{AHD}_{1}$, $\\mathrm{AD}_{1}=\\sqrt{A H^{2}+D_{1} H^{2}}=\\sqrt{\\left(\\frac{\\sqrt{2}}{2}\\right)^{2}+\\left(\\frac{7 \\sqrt{2}}{2}\\right)^{2}}=5$.\n\nHence, the answer is: 5.\n\n【Insight】This problem tests knowledge of rotational transformations, the properties of the median to the hypotenuse of a right triangle, and the Pythagorean theorem. The key to solving it lies in adding common auxiliary lines to construct right triangles for problem-solving, a type of question frequently encountered in middle school exams." }, { "problem_id": 1591, "question": "As shown in the figure, the geometric structure is composed of several identical small cubes. The images from the front and left sides are provided. How many small cubes make up this geometric structure?\n\n\n\nView from the front\n\n\n\nView from the left", "input_image": [ "batch25-2024_06_17_26670771c405ed326f86g_0025_1.jpg", "batch25-2024_06_17_26670771c405ed326f86g_0025_2.jpg" ], "is_multi_img": true, "answer": " 4 or 5 or 6 or 7.", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "To construct such a geometric shape, a minimum of $3+1=4$ small cubes are required, and a maximum of $6+1=7$ small cubes are needed. Therefore, the number of small cubes could be 4, 5, 6, or 7.\n\n\n【Key Insight】This question tests the ability to determine a geometric shape using three-view drawings. The key to solving the problem lies in reconstructing the geometric shape based on the definitions provided by the three views." }, { "problem_id": 1592, "question": "As shown in Figure 1, there is a square cardboard with a side length of 6. Small squares of equal side length are cut out from each corner. The cardboard is then folded into an open-top square box with a base perimeter of 16, as shown in Figure 2. What is the height of this open-top box?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_26670771c405ed326f86g_0030_1.jpg", "batch25-2024_06_17_26670771c405ed326f86g_0030_2.jpg" ], "is_multi_img": true, "answer": " 1", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the side length of the base is $16 \\div 4=4$.\n\nLet the height of the lidless box be $x$. Then we have $4 + x + x = 6$,\n\nSolving gives: $\\mathrm{x}=1$.\n\nAnswer: The height of the lidless box is 1.\n\n[Highlight] This problem tests the application of linear functions and the net of a rectangular prism. Understanding the problem and identifying the relationship is key to solving it." }, { "problem_id": 1593, "question": "As shown in Figure (1), given that the area of the small square \\(ABCD\\) is 1, extend each of its sides by one time to obtain the new square \\(A_1B_1C_1D_1\\); extend the sides of square \\(A_1B_1C_1D_1\\) by the same method to obtain square \\(A_2B_2C_2D_2\\) (as shown in Figure (2)); continue this process, then the area of square \\(A_nB_nC_nD_n\\) is \\(\\qquad\\).\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_36e89897fe8f9d3ab063g_0009_1.jpg", "batch25-2024_06_17_36e89897fe8f9d3ab063g_0009_2.jpg" ], "is_multi_img": true, "answer": "$5 n$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "**Question Analysis:** According to the area formula of a triangle, it is known that each time the sides are doubled, the area of the resulting right triangle is equal to the area of the original square before the extension. That is, each time the sides are doubled, the area of the resulting figure is 5 times the area of the original square, thus solving the problem.\n\nAs shown in Figure (1), given that the area of the small square $\\mathrm{ABCD}$ is 1, then after doubling each of its sides, the area of triangle $\\mathrm{AA}_{1} \\mathrm{~B}_{1}$ is 1, and the area of the new square $\\mathrm{A}_{1} \\mathrm{~B}_{1} \\mathrm{C}_{1} \\mathrm{D}_{1}$ is 5. Consequently, the area of square $\\mathrm{A}_{2} \\mathrm{~B}_{2} \\mathrm{C}_{2} \\mathrm{D}_{2}$ is $5 \\times 5=25$, and the area of square $A_{n} B_{n} C_{n} D_{n}$ is $5^{n}$.\n\n**Key Point:** Identifying Patterns - Changes in Figures\n\n**Commentary:** The key to solving such problems lies in carefully analyzing the characteristics of the given figures to identify the pattern, and then applying this pattern to solve the problem." }, { "problem_id": 1594, "question": "Place a set of triangles as shown in Figure 1, such that the right sides $A C$ and $M D$ of the two triangles coincide. Given $A B=A C=8 \\mathrm{~cm}$, after rotating $\\triangle M E D$ counterclockwise by $60^{\\circ}$ around point $A(M)$ (Figure 2), the area of the overlapping (shaded) part of the two triangles is approximately $\\qquad$ $\\mathrm{cm}^{2}$ (the result is accurate to $0.1, \\sqrt{3} \\approx 1.73$).\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_36e89897fe8f9d3ab063g_0053_1.jpg", "batch25-2024_06_17_36e89897fe8f9d3ab063g_0053_2.jpg" ], "is_multi_img": true, "answer": "20.3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure\n\n\n\nLet lines $\\mathrm{BC}$ and $\\mathrm{AD}$ intersect at point $\\mathrm{G}$. From the intersection point $\\mathrm{G}$, draw $\\mathrm{GF} \\perp \\mathrm{AC}$ intersecting $\\mathrm{AC}$ at point $\\mathrm{F}$. Let $\\mathrm{FC}=\\mathrm{x}$, then $\\mathrm{GF}=\\mathrm{FC}=\\mathrm{x}$. With a rotation angle of $60^{\\circ}$, we can deduce that $\\angle \\mathrm{FAG}=60^{\\circ}$,\n\n$\\therefore \\mathrm{AF}=\\mathrm{GF} \\cot \\angle \\mathrm{FAG}=\\frac{\\sqrt{3}}{3} \\mathrm{x}$.\n\nThus, $x+\\frac{\\sqrt{3}}{3} x=8$, leading to $x=12-\\frac{\\sqrt{3}}{3}$.\n\nTherefore, $\\mathrm{S}_{\\triangle A G C}=\\frac{1}{2} \\times 8 \\times\\left(12-\\frac{\\sqrt{3}}{3}\\right) \\approx 20.3 \\mathrm{~cm}^{2}$.\n\nHence, the answer is: 20.3" }, { "problem_id": 1595, "question": "In ancient Mesopotamia, people made clay tablets for writing on. A clay tablet from the Old Babylonian period, BM15285 (Figure 1), records mathematical geometry exercises from a priestly school, featuring perfect equal circles. Inspired by the patterns on the tablet, a designer created a pattern resembling a semi-umbrella for plane tessellation (Figure 2). If the maximum distance between the umbrella top and the handle in the pattern is 2, then the area of one umbrella-shaped pattern is . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_36e89897fe8f9d3ab063g_0055_1.jpg", "batch25-2024_06_17_36e89897fe8f9d3ab063g_0055_2.jpg" ], "is_multi_img": true, "answer": "2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Observing the figure,\n\n\n\nThe area of one umbrella-shaped pattern is: Area of the rectangle - Area of the lower semicircle + Area of the upper semicircle = Area of the rectangle,\n\n$\\therefore$ The area of one umbrella-shaped pattern is: $2 \\times 1=2$.\n\nTherefore, the answer is: 2.\n\n[Insight] This question examines the translation, rotation, and central symmetry of figures. Combining numbers with shapes is the key to solving the problem." }, { "problem_id": 1596, "question": "Figure A is a border design created by Xiao Ming, which is obtained by symmetry and translation of the shape in Figure B. In Figure B, $\\triangle A E O \\cong \\triangle A D O \\cong \\triangle B C O \\cong \\triangle B F O$, with $E, O, F$ all lying on the line $M N$, $E F = 12$, and $A E = 14$. The length of $O A$ is $\\qquad$.\n\n\n\nA\n\n\n\nB", "input_image": [ "batch25-2024_06_17_36e89897fe8f9d3ab063g_0056_1.jpg", "batch25-2024_06_17_36e89897fe8f9d3ab063g_0056_2.jpg" ], "is_multi_img": true, "answer": "16", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular line from point $A$ to $EF$, meeting at point $H$.\n\n\n\nSince $\\triangle AEO \\cong \\triangle ADO \\cong \\triangle BCO \\cong \\triangle BFO$,\n\n$\\therefore \\angle AOE = \\angle AOB = \\angle BOF$, and $OF = OF = \\frac{1}{2} EF = 6$.\n\nSince $\\angle AOE + \\angle AOB + \\angle BOF = 180^{\\circ}$,\n\n$\\therefore \\angle AOE = \\angle AOB = \\angle BOF = 60^{\\circ}$.\n\nLet $OH = x$, then $AO = 2x$, and $AH = \\sqrt{3}x$.\n\nIn the right triangle $\\triangle AEH$, $AE^{2} = AH^{2} + EH^{2}$,\n\n$\\therefore 14^{2} = (\\sqrt{3}x)^{2} + (x - 6)^{2}$.\n\nSolving this equation gives $x = 8$ or $-5$ (discarding the negative root),\n\n$\\therefore OA = 16$.\n\nThus, the answer is: 16.\n\n【Highlight】This problem examines the design of patterns using translation, the properties of congruent triangles, the properties of right triangles with a 30-degree angle, and the Pythagorean theorem. The key to solving the problem lies in constructing equations using parameters." }, { "problem_id": 1597, "question": "Li Ming discovered a workpiece while visiting a factory lathe workshop. By observing and drawing the three views of this workpiece, and using a ruler to measure some of the lengths, as shown in the figure, what is the volume of the workpiece?\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_399abdf59eb2f45264f8g_0014_1.jpg", "batch25-2024_06_17_399abdf59eb2f45264f8g_0014_2.jpg", "batch25-2024_06_17_399abdf59eb2f45264f8g_0014_3.jpg" ], "is_multi_img": true, "answer": "$17 \\pi \\mathrm{cm}^{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the three-view drawing, the geometric body consists of two cylinders stacked together,\n\nwith base diameters of \\(2 \\mathrm{~cm}\\) and \\(4 \\mathrm{~cm}\\),\n\nand heights of \\(4 \\mathrm{~cm}\\) and \\(1 \\mathrm{~cm}\\) respectively.\n\nTherefore, the volume is calculated as: \\(4 \\pi \\times 2^{2} + \\pi \\times 1^{2} \\times 1 = 17 \\pi \\left(\\mathrm{cm}^{3}\\right)\\).\n\nAnswer: The volume of the workpiece is \\(17 \\pi \\mathrm{cm}^{3}\\).\n\n[Highlight] This question tests the ability to determine a geometric body from its three-view drawing and the calculation of cylinders. Correctly identifying the shape of the geometric body is key to solving the problem." }, { "problem_id": 1598, "question": "As shown in Figure (1), the Great Wind Pavilion is a landmark building of Xi'an Han City Lake, inspired by the poetic imagery of \"The Great Wind Song\" by Liu Bang, the first emperor of the Han Dynasty: \"When the great wind rises, the clouds fly; His power extends over the land as he returns to his hometown; How can he find brave warriors to guard the four corners of the world?\" On a sunny weekend, Xiaohua and Xiaoli went to measure the height of the Great Wind Pavilion, $A B$, as shown in Figure (2). First, they placed a plane mirror at point $C$. Xiaohua then backed away along the direction of $B C$ until she reached point $E$, where she could just see the image of the top of the Great Wind Pavilion, $A$, in the mirror. The distance from Xiaohua's eyes to the ground, $D E$, was 1.5 meters, and the distance from the mirror to point $E$, $C E$, was 1.2 meters. Next, at a certain moment, the shadow of the Great Wind Pavilion reached its tip at point $M$. At the same time, Xiaoli measured the shadow length of Xiaohua's height, $E G$, which was 0.8 meters, and Xiaohua's height, $E F$, was 1.6 meters. The distance from the base of the Great Wind Pavilion to the tip of its shadow, $M C$, was 19.2 meters. It is known that $A B \\perp B G$ and $E F \\perp B G$, and points $B, M, C, E, G$ are on the same horizontal line, while points $E, D, F$ are on a straight line. Please calculate the height of the Great Wind Pavilion, $A B$. (The size and thickness of the plane mirror are negligible.)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch25-2024_06_17_399abdf59eb2f45264f8g_0019_1.jpg", "batch25-2024_06_17_399abdf59eb2f45264f8g_0019_2.jpg" ], "is_multi_img": true, "answer": "64 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: From the problem, we have: $\\angle ABC = \\angle FEC = \\angle FEG = 90^{\\circ}$, $\\angle AMB = \\angle FGE$, $\\angle ACB = \\angle DCE$,\n\n$\\therefore \\triangle ABM \\sim \\triangle FEG$, $\\triangle ABC \\sim \\triangle DEC$,\n\n$\\therefore \\frac{AB}{EF} = \\frac{BM}{EG}$, $\\frac{AB}{DE} = \\frac{BC}{EC}$,\n\n$\\therefore \\frac{AB}{1.6} = \\frac{BM}{0.8}$, $\\frac{AB}{1.5} = \\frac{BM + 19.2}{1.2}$,\n\nSolving gives $AB = 64$.\n\n$\\therefore$ The height of the Great Wind Pavilion $AB$ is 64 meters.\n\n[Key Insight] This problem examines the practical application of the principle of light reflection, parallel projection, and the similarity of triangles. Mastering the principle of light reflection, the determination and properties of similar triangles are key to solving the problem." }, { "problem_id": 1599, "question": "As shown in Figure (1), $O$ is a point on the line $A B$. A ray $O C$ is drawn such that $\\angle A O C = 120^{\\circ}$. A right triangle ruler is placed as shown, with its right-angle vertex at point $O$ and one of its right-angle sides, $O P$, on ray $O A$. In Figure (1), the triangle ruler is rotated around point $O$ at a speed of $6^{\\circ}$ per second in a clockwise direction (as shown in Figure (2)). During the rotation, at the $t$-th second, the line $O P$ exactly bisects $\\angle B O C$. The value of $t$ is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch25-2024_06_17_3fe05a133b218ccbe5deg_0028_1.jpg", "batch25-2024_06_17_3fe05a133b218ccbe5deg_0028_2.jpg" ], "is_multi_img": true, "answer": "25 or 55", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Since \\(\\angle \\mathrm{AOC} = 120^\\circ\\),\n\n\\(\\therefore \\angle \\mathrm{BOC} = 60^\\circ\\),\n\nBecause the line where \\(\\mathrm{OP}\\) lies exactly bisects \\(\\angle \\mathrm{BOC}\\),\n\n\\(\\therefore \\angle \\mathrm{BOP} = \\frac{1}{2} \\angle \\mathrm{BOC} = 30^\\circ\\), or \\(\\angle \\mathrm{BOP} = 180^\\circ - 30^\\circ = 150^\\circ\\),\n\n\\(\\therefore 6t = 180 - 30\\) or \\(6t = 180 + 150\\),\n\n\\(\\therefore t = 25\\) or 55,\n\nHence, the answer is: 25 or 55.\n\n[Highlight] This problem examines the application of linear equations in one variable, the definition of an angle bisector, and the definition of a straight angle. Formulating the correct equation is key to solving this problem." }, { "problem_id": 1600, "question": "As shown, a set of triangular rulers is placed as in Figure 1, with $AB = CD = \\sqrt{6}$, and the vertices $E$ coincide. The triangle $\\triangle DEC$ is rotated around its vertex $E$, as shown in Figure 2. During the rotation, when $\\angle AED = 75^\\circ$, lines $AD$ and $BC$ are connected, and the area of the quadrilateral $ABCD$ at this moment is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_3fe05a133b218ccbe5deg_0094_1.jpg", "batch25-2024_06_17_3fe05a133b218ccbe5deg_0094_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{3}+3$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, extend $CE$ to intersect $AB$ at point $F$,\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n$\\because \\angle AED = 75^\\circ$\n\n$\\therefore \\angle EAD + \\angle ADE = 180^\\circ - \\angle AED = 105^\\circ$,\n\nAlso, $\\angle BAE + \\angle CDE = 45^\\circ + 30^\\circ = 75^\\circ$,\n\n$\\therefore \\angle BAD + \\angle CDA = \\angle BAE + \\angle CDE + \\angle EAD + \\angle ADE = 180^\\circ$,\n\n$\\therefore AB \\parallel CD$,\n\n$\\because AB = CD$,\n\n$\\therefore$ Quadrilateral $ABCD$ is a parallelogram,\n$\\because CE \\perp CD$,\n\n$\\therefore CE \\perp AB$, which means $EF \\perp AB$,\n\n$\\therefore EF = \\frac{1}{2} AB = \\frac{\\sqrt{6}}{2}, \\quad EC = \\frac{\\sqrt{3}}{3} CD = \\sqrt{2}$,\n\n$\\therefore CF = EC + EF = \\frac{\\sqrt{6} + 2\\sqrt{2}}{2}$,\n\n$\\therefore S_{\\text{Quadrilateral } ABCD} = AB \\times CF = \\frac{\\sqrt{6} + 2\\sqrt{2}}{2} \\times \\sqrt{6} = 2\\sqrt{3} + 3$.\n\nTherefore, the answer is: $2\\sqrt{3} + 3$.\n\n【Key Insight】This problem examines the properties of rotation, the determination of a parallelogram, and the calculation of the area of a parallelogram. The key to solving the problem is first proving that quadrilateral $ABCD$ is a parallelogram." }, { "problem_id": 1601, "question": "As shown in Figure 1, in the isosceles right $\\triangle ABC$, $\\angle ACB = 90^\\circ$, point $O$ is the midpoint of $AC$. In $\\triangle DEF$, $\\angle F = 90^\\circ$, $\\angle DEF = 30^\\circ$, and $DE = AC$. Align $DE$ with $AC$, as shown in Figure 2. Then, rotate $\\triangle DEF$ around point $O$ clockwise by $60^\\circ$. $AB$ intersects $EF$ at point $G$ and $DE$ at point $H$. If $AG = 2$, then the length of $GH$ is $\\qquad$.\n\n\n\n", "input_image": [ "batch25-2024_06_17_46a486f71bc6240f8f1ag_0007_1.jpg", "batch25-2024_06_17_46a486f71bc6240f8f1ag_0007_2.jpg" ], "is_multi_img": true, "answer": "$4-2 \\sqrt{3} \\# \\#-2 \\sqrt{3}+4$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:** As shown in the figure, let \\( EF \\) intersect \\( AC \\) at point \\( N \\). Draw \\( HP \\perp AC \\) at point \\( P \\).\n\n\n\nSince \\( \\triangle DEF \\) is rotated clockwise by \\( 60^\\circ \\) around point \\( O \\),\n\n\\[\n\\angle AOE = 60^\\circ, \\quad AO = OE.\n\\]\n\nGiven \\( \\angle DEF = 30^\\circ \\),\n\n\\[\n\\angle ONE = 180^\\circ - \\angle AOE - \\angle DEF = 180^\\circ - 60^\\circ - 30^\\circ = 90^\\circ.\n\\]\n\nThus,\n\n\\[\nON = \\frac{1}{2} OE = \\frac{1}{2} OA,\n\\]\n\nand\n\n\\[\nON = NA.\n\\]\n\nSince \\( \\angle BAC = 45^\\circ \\) and \\( \\angle ANE = 90^\\circ \\),\n\n\\[\n\\angle BAC = \\angle AGN = 45^\\circ,\n\\]\n\nso\n\n\\[\nAN = NG.\n\\]\n\nTherefore,\n\n\\[\nAG = \\sqrt{AN^2 + NG^2} = \\sqrt{2} AN = 2,\n\\]\n\nwhich implies\n\n\\[\nAN = \\sqrt{2}.\n\\]\n\nHence,\n\n\\[\nOA = 2\\sqrt{2}.\n\\]\n\nSince \\( \\angle BAC = 45^\\circ \\) and \\( PH \\perp AO \\),\n\n\\[\nAP = PH.\n\\]\n\nThus,\n\n\\[\nAH = \\sqrt{AP^2 + PH^2} = \\sqrt{2} AP.\n\\]\n\nGiven \\( \\angle AOE = 60^\\circ \\),\n\n\\[\n\\angle PHO = 30^\\circ.\n\\]\n\nTherefore,\n\n\\[\nOP = \\frac{1}{2} OH,\n\\]\n\nand\n\n\\[\nPH = \\sqrt{OH^2 - OP^2} = \\sqrt{3} OP.\n\\]\n\nHence,\n\n\\[\nAP = \\sqrt{3} OP.\n\\]\n\nSince \\( OP + AP = OA = 2\\sqrt{2} \\),\n\n\\[\nOP + \\sqrt{3} OP = 2\\sqrt{2},\n\\]\n\nwhich gives\n\n\\[\nOP = \\sqrt{6} - \\sqrt{2}.\n\\]\n\nThus,\n\n\\[\nAP = OA - OP = 2\\sqrt{2} - (\\sqrt{6} - \\sqrt{2}) = 3\\sqrt{2} - \\sqrt{6},\n\\]\n\nand\n\n\\[\nAH = \\sqrt{2} AP = \\sqrt{2}(3\\sqrt{2} - \\sqrt{6}) = 6 - 2\\sqrt{3}.\n\\]\n\nFinally,\n\n\\[\nGH = AH - AG = (6 - 2\\sqrt{3}) - 2 = 4 - 2\\sqrt{3}.\n\\]\n\n**Answer:** \\( 4 - 2\\sqrt{3} \\).\n\n**Key Insight:** This problem examines the properties of rotation, isosceles right triangles, the Pythagorean theorem, and the relationship between the sides of a 30-degree right triangle. Mastery of these concepts is essential for solving the problem." }, { "problem_id": 1602, "question": "As shown in the figure, the following are the three views of a geometric object, and the total surface area of this geometric object is $\\qquad$ .\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0002_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0002_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0002_3.jpg" ], "is_multi_img": true, "answer": "$3 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By examining the three views of the geometric figure, it is easily determined that the figure is a cone with a base diameter of 2 and a generatrix length of 2.\n\nTherefore, the lateral area of the cone \\(= \\pi r l = 2 \\times 1 \\times \\pi = 2\\pi\\),\n\nThe area of the base circle \\(= \\pi r^{2} = \\pi\\),\n\nThus, the total surface area of the geometric figure \\(= 2\\pi + \\pi = 3\\pi\\),\n\nHence, the answer is: \\(3\\pi\\).\n\n[Insight] This problem tests the knowledge of determining a geometric figure from its three views and calculating the dimensions of a cone. The key to solving the problem lies in identifying the shape of the geometric figure, and the difficulty level is not high." }, { "problem_id": 1603, "question": "The three views of a geometric object are shown in the figure. It is composed of identical cubic wooden blocks, each with an edge length of $1 \\mathrm{~cm}$. The surface area of the geometric object is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0003_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0003_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0003_3.jpg" ], "is_multi_img": true, "answer": "18", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, there are two layers. From the top view, the bottom layer consists of three cube blocks. From the side view, there is one cube in the middle of the upper layer, with none on either side.\n\nTherefore, the number of cube blocks in this geometric figure is 4, and its surface area is: $3 \\times 2 + 3 \\times 2 + 3 \\times 2 = 18$.\n\nThus, the answer is: 18.\n\n[Insight] This question aims to test students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. The key to solving the problem lies in mastering the mnemonic: \"The top view lays the foundation, the front view builds up, and the side view removes the excess,\" which makes it easier to arrive at the answer." }, { "problem_id": 1604, "question": "A geometric solid is shown from the front, from the left, and from above as depicted in the figures below. The geometric solid is $\\qquad$\n\n\n\nFrom the front\n\n\n\nFrom the left\n\n\n\nFrom above", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0005_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0005_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0005_3.jpg" ], "is_multi_img": true, "answer": "triangular pyramid", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the main view and the side view being triangles, it is determined that the shape is a cone. Considering the top view is also a triangle, it can be inferred that the geometric body in question is a triangular pyramid.\n\nTherefore, the answer is: triangular pyramid.\n\n[Insight] This question tests the ability to deduce the original geometric shape from the three views of an object. The key to solving it lies in having a spatial imagination capability for the three types of views of geometric bodies." }, { "problem_id": 1605, "question": "A certain model is formed by arranging identical cubes. The views from the front, left, and top are shown in the figures below. This model is composed of $\\qquad$ cubes.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0014_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0014_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0014_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By observing the figure, it can be seen that this model is composed of $4+1=5$ cubes arranged together.\n\nTherefore, the answer is: 5.\n\n[Highlight] This question tests the ability to determine the geometry from three-view drawings; one can distinguish the number of layers from top to bottom and left to right from the main view, and the left-right and front-back positions from the top view. By synthesizing the above analysis, the number of small cubes can be counted." }, { "problem_id": 1606, "question": "As shown, the main view, left view, and top view of a geometric shape composed of several small cubes are given. The number of small cubes in this geometric shape is $\\qquad$.\n\n\n\nMain View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0015_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0015_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0015_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, from the top view, we can see that there are 5 small squares on the bottom layer. From the front view, only the middle part has 2 layers. Combining this with the side view, we can see that the frontmost part in the middle has 2 layers, as illustrated. Therefore, the total number of small cubes in this geometric figure is 6.\n\nHence, the answer is: 6\n\n\n\nTop View\n\n[Insight] This question tests the understanding and flexible application of three-view drawings, as well as the ability to visualize spatial relationships. If you master the mnemonic \"Top view lays the foundation, front view builds up, side view removes the excess,\" it becomes easier to arrive at the answer." }, { "problem_id": 1607, "question": "The figure shows the views of a geometric structure made up of several identical small cubes from different directions. The maximum number of small cubes that can form this geometric structure is $\\qquad$.\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0017_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0017_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Indicate the maximum number of small cubes that can be placed at the corresponding positions in the top view, as shown in the figure:\n\n\n\n## From the top view\n\n$\\therefore$ The maximum number of small cubes that can form this geometric figure is 8.\n\n[Key Insight] This problem tests the ability to deduce the number of small cubes from the three views of a simple composite figure. The key to solving the problem lies in placing small cubes on the top view according to the characteristics of the three views and determining the maximum number of small cubes that can be placed at each position." }, { "problem_id": 1608, "question": "A geometric solid is formed by arranging identical small cubes. The planar views obtained from the front, top, and left sides of the geometric solid are shown in the figures below. The number of small cubes that make up this geometric solid is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from the top\n\n\n\nView from the left", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0019_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0019_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0019_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three views of the geometric body, the distribution of the number of small cubes is as follows:\n\n| 1 | 1 |\n| :--- | :--- |\n| 1 | 2 |\n\n## View from above\n\nTherefore, the number of small cubes that make up this geometric body is 5.\n\nHence, the answer is: 5\n\n[Highlight] This question tests the ability to determine the geometric body from three views; one can distinguish the layers of the object from top to bottom and left to right from the main view, and the left-right and front-back positions of the object from the top view. By synthesizing the above analysis, the minimum and maximum number of small cubes can be counted." }, { "problem_id": 1609, "question": "A geometric structure composed of 13 identical small cubes is arranged such that the bottom layer consists of 9 small cubes. Its front view and left view are shown in the figures below. The number of ways to construct this geometric structure is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0022_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0022_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem's top view: Except for the positions of $A, B, C$, which cannot be determined, the numbers of small cubes in the other positions are as shown in the figure.\n\n| $A$ | 1 | $B$ |\n| :--- | :--- | :--- |\n| 1 | 1 | 1 |\n| 3 | 1 | $C$ |\n\nFront view\n\n$\\because$ The geometric body is constructed from 13 identical small cubes, with the bottom layer consisting of 9 small cubes,\n\n$\\therefore$ $A$ is $1$, $B$ is $2$, $C$ is $2$; or $A$ is $2$, $B$ is $2$, $C$ is $1$; or $A$ is $2$, $B$ is $1$, $C$ is $2$. There are three possible scenarios in total.\n\nTherefore, the answer is: 3.\n\n【Insight】This problem tests the understanding of using three views to determine the geometry of a body. The key to solving it lies in comprehending the problem and flexibly applying the knowledge learned." }, { "problem_id": 1610, "question": "As shown in the figure, the left and front views of a certain three-dimensional object are identical, and the top view is a circle. According to the given length and height data, its surface area is (expressed with $\\pi$ allowed) $\\qquad$.\n\n\n\nLeft and front views\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0025_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0025_2.jpg" ], "is_multi_img": true, "answer": "$48 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be seen that the geometric body is a combination of a cone and a cylinder. Its axial cross-section is an isosceles triangle with a height of 4 and a base of 6, and a rectangle with a length of 6 and a width of 4.\n\n$\\therefore$ The legs of the isosceles triangle $=\\sqrt{3^{2}+4^{2}}=5$,\n\n$\\therefore$ The radius of the base circle of the cone is 3, and the length of the generatrix is 5,\n\n$\\therefore$ The circumference of the base is $6 \\pi$,\n\n$\\therefore$ The lateral area of the cone is $\\frac{1}{2} \\times 6 \\pi \\times 5=15 \\pi$,\n\n$\\because$ The area of the base circle is $\\pi r^{2}=9 \\pi$,\n\n$\\therefore$ The total surface area is $15 \\pi+9 \\pi+4 \\times 6 \\pi=48 \\pi$,\n\nTherefore, the answer is: $48 \\pi$.\n\n[Key Insight] This problem tests the understanding of three-view drawings and the calculation of a cone. The key to solving this problem is to correctly understand the relationship between the lateral development of the cone and the original sector, recognizing that the generatrix of the cone is the radius of the sector, and the circumference of the base circle of the cone is the arc length of the sector." }, { "problem_id": 1611, "question": "A geometric shape is constructed by stacking several identical small cubes. The shape of the geometric shape as seen from above and from the left is shown in the figures below. Please construct all geometric shapes that meet the conditions, then the constructed geometric shape consists of $\\qquad$ small cubes.\n\n\n\nView from above\n\n\n\nView from the left", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0028_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0028_2.jpg" ], "is_multi_img": true, "answer": "5 or 6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view and the left view of the geometric shape, it can be inferred that the number of columns in the front view is the same as the number of columns in the top view, and the number of columns in the left view is the same as the number of rows in the top view. Therefore, the number of small squares in that column in the top view is as shown in the figure:\n\n\nSolution: The constructed geometric shape is as shown in the figure:\n\n\nThe constructed geometric shape consists of 5 or 6 small cubes, hence the answer is: 5 or 6.\n\n[Highlight] This question tests the drawing method of the three views of a geometric shape. From the top view and the left view of the geometric shape, it can be inferred that the number of columns in the front view is the same as the number of columns in the top view, and the number of columns in the left view is the same as the number of rows in the top view. Moreover, the number of small squares in each column is the maximum number of squares in the corresponding row in the top view." }, { "problem_id": 1612, "question": "A geometric solid is constructed using several identical small cubes. The images below show the shapes viewed from the front and from above, respectively. At least $\\qquad$ small cubes are needed to construct this geometric solid.\n\n\n\nViewed from the front\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0029_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0029_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the front view, at least one position in the first column of the top view has 2 layers, so the geometric figure is composed of at least 6 small cubes.\n\nTherefore, the answer is: 6.\n\n[Highlight] This question tests the ability to determine the number of small cubes in a figure based on its three views. Correctly interpreting the three views is key to solving the problem." }, { "problem_id": 1613, "question": "An object is composed of multiple small cubes with edge lengths of 1. The shapes of the geometric body as seen from the front, left, and top are shown in the figures below. Then, the volume of the geometric body is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from the top", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0034_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0034_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0034_3.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the three views, it can be seen that the small squares of the geometric body are arranged as shown in the figure (the numbers on the small squares indicate the number of small squares at that position):\n\n\n\nThere are a total of 8 cubes.\n\nTherefore, the answer is: 8.\n\n[Key Insight] This question mainly tests the three views of a geometric body. Mastering the method of reconstructing the geometric body from the three views is the key to solving the problem." }, { "problem_id": 1614, "question": "A geometric object's three views are shown in the figure, so this geometric object is $\\qquad$ .\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0036_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0036_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0036_3.jpg" ], "is_multi_img": true, "answer": "hollow cylinder", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: In the three-view drawing of this geometric solid, both the front view and the side view are rectangles, and the top view is a ring. Therefore, the geometric solid is a hollow cylinder.\n\nHence, the answer is: hollow cylinder.\n\n[Key Insight] This question tests the understanding of three-view drawings of geometric solids. The key to solving the problem lies in the ability to determine the geometric solid that matches the given conditions through its three views." }, { "problem_id": 1615, "question": "As shown, the main view and top view of a simple geometric shape composed of small cubes of the same size are given. If the number of small cubes that make up this geometric shape is $n$, then the sum of all possible values of $n$ is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0037_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0037_2.jpg" ], "is_multi_img": true, "answer": "38", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the front view and the top view, we can determine that the simple geometric body consists of 3 layers. The bottom layer has 5 cubes; the middle layer could have 2, 3, or 4 cubes; and the top layer could have 1 or 2 cubes.\n\nTherefore, the possible total number of small cubes, \\( n \\), that make up this geometric body are:\n\n(1) \\(5 + 2 + 1 = 8\\)\n\n(2) \\(5 + 3 + 1 = 9\\)\n\n(3) \\(5 + 3 + 2 = 10\\)\n\n(4) \\(5 + 4 + 1 = 10\\)\n\n(5) \\(5 + 4 + 2 = 11\\)\n\nThus, the sum of all possible values of \\( n \\) is \\(8 + 9 + 10 + 10 + 11 = 38\\).\n\nThe final answer is: 38.\n\n[Key Insight] This problem tests the ability to determine the number of small cubes in a geometric body based on its three views. The key to solving it lies in accurately interpreting the three views to count the number of small cubes." }, { "problem_id": 1616, "question": "The teacher arranged 10 small cubes of $1 \\mathrm{~cm} \\times 1 \\mathrm{~cm} \\times 1 \\mathrm{~cm}$ to form a three-dimensional figure, whose front view is shown in Figure (1). In the figure, any two adjacent small cubes share at least one edge ($1 \\mathrm{~cm}$) or one face ($1 \\mathrm{~cm} \\times 1 \\mathrm{~cm}$). The teacher then provided a $3 \\mathrm{~cm} \\times 4 \\mathrm{~cm}$ grid paper (as shown in Figure (2)) and asked Xiao Liang to place these 10 small cubes according to the front view within the grid cells. The maximum surface area of the geometric shape after Xiao Liang's arrangement is $\\qquad$ $\\mathrm{cm}^{2}$. (When placing the small cubes, they must not be suspended in the air, and each cube's edges must be either perpendicular or parallel to the horizontal line.)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0040_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0040_2.jpg" ], "is_multi_img": true, "answer": "52", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, when 10 small cubes are arranged as in the top view, the surface area of the geometric shape is maximized.\n\n\n\nFigure (1)\n\n\n\nTop View\n\nFigure (2)\n\nThe maximum value is calculated as follows: $3 \\times 6 + 2 \\times 10 + 14 = 52\\left(\\mathrm{~cm}^{2}\\right)$.\n\nTherefore, the answer is:\n\n52.\n\n【Key Insight】This problem examines the calculation of the maximum surface area of a geometric shape given its front view. The key to solving the problem lies in understanding the question and accurately depicting the arrangement that maximizes the surface area." }, { "problem_id": 1617, "question": "As shown in the figure, the three views of a geometric body are given. If the volume of this geometric body is 6, then its surface area is $\\qquad$ .\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0041_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0041_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0041_3.jpg" ], "is_multi_img": true, "answer": "22", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the main view, the length of the cuboid is 3, and the width is 1. The volume of this geometric solid is 6.\n\nLet the height be \\( h \\), then \\( 3 \\times 1 \\times h = 6 \\).\n\nSolving for \\( h \\), we get: \\( h = 2 \\).\n\nIts surface area is:\n\n\\[ 2 \\times 3 \\times 2 + 2 \\times 3 \\times 1 + 2 \\times 1 \\times 2 = 22 \\]\n\nTherefore, the answer is: 22.\n\n【Key Point】This question mainly tests the ability to determine the edge lengths of a geometric solid using three views. Finding the height of the figure is the key to solving the problem." }, { "problem_id": 1618, "question": "A geometric shape is composed of several identical small cubes, as shown in the shapes viewed from three different directions.\n\n\n\nViewed from the left\n\n\n\nViewed from the front\n\n\n\nViewed from above\n\n##\n\nTo construct this geometric shape, at least $\\qquad$ small cubes are needed.", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0042_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0042_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0042_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: The distribution of the geometric shapes is as shown in the figure below:\n\n\n\n## View from above\n\nThe minimum number of small cubes is at least \\(2 + 1 + 1 + 1 = 5\\).\n\nTherefore, the answer is: 5.\n\n**Insight:** This question tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Use the top view as the foundation, the front view to build up, and the side view to remove obstructions\" can make it easier to arrive at the answer." }, { "problem_id": 1619, "question": "The geometric figure composed of several small cubes of the same size is shown as viewed from above and from the front. The minimum number of small cubes in this geometric figure is $\\qquad$.\n\n\n\nViewed from above\n\n\n\nViewed from the front", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0045_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0045_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, the minimum number of small cubes in this geometric figure is $2 + 1 + 1 + 1 + 1 + 3 = 9$ (units).\n\n\n\nView from above\n\n\n\nFront view\n\nTherefore, the answer is: 9.\n\n[Key Insight] This question tests the ability to determine a geometric figure from its three views. The key to solving it lies in understanding the definition of the three views, a common type of question in middle school exams." }, { "problem_id": 1620, "question": "The geometric solid, when viewed from three different directions, yields the planar figures shown below. Therefore, the geometric solid is $\\qquad$ .\n\n\n\nViewed from the front\n\n\n\nViewed from the left\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0060_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0060_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0060_3.jpg" ], "is_multi_img": true, "answer": "Cylinder", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: Since both the front view and the side view are rectangles,\n\nTherefore, this geometric body is a prism,\n\nSince the top view is a circle,\nTherefore, this geometric body is a cylinder.\n\nHence, the answer is: Cylinder.\n\n[Insight] This question tests the ability to determine a geometric body using three views. If two of the three views are identical, it can determine whether the geometric body is a prism, a pyramid, or a sphere, and the specific shape is determined by the third view." }, { "problem_id": 1621, "question": "The geometric figure formed by some small cubes of the same size is shown in the front view and the left view as follows. The minimum number of small cubes required to form this geometric figure is $\\qquad$\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0062_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0062_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the front view and the side view, the geometric figure is drawn as shown in the images:\n\n\n\n4 small cubes\n\n\n\n4 small cubes\n\n\n\n5 small cubes\n\nTherefore, the answer is: 4.\n\n[Insight] This question is designed to test students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills." }, { "problem_id": 1622, "question": "A geometric shape is constructed by stacking several identical small cubes. The images below show the shapes viewed from the front and from above, respectively. The geometric shape can be constructed with a maximum of $m$ small cubes and a minimum of $n$ small cubes. Therefore, $m+n=$ $\\qquad$ .\n\n\n\nViewed from the front\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0067_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0067_2.jpg" ], "is_multi_img": true, "answer": "17", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nViewed from the front\n\n\n\nViewed from the front\n\n| 2 | 2 |\n| :--- | :--- |\n| 2 | 2 |\n| 2 | |\n| | |\n\nViewed from above\n\n\n\n## Viewed from above\n\n$m=2+2+2+2+2=10$, $n=2+2+1+1+1=7$,\n\n$\\therefore m+n=10+7=17$,\n\nTherefore, the answer is: 17.\n\n【Insight】This question mainly tests the students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills. If one masters the mnemonic \"The top view lays the foundation, the front view builds wildly, and the side view removes violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1623, "question": "As shown in the figure, the top view and left view of a geometric structure formed by stacking identical small cubes are provided. The maximum number of small cubes used to construct the geometric structure is $\\qquad$.\n\nTop View\n\n\n\nLeft View\n\n", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0069_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0069_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer has 4 cubes. From the left view, it is easy to determine that the second layer has a maximum of 3 cubes and a minimum of 1 cube.\n\nTherefore, the total number of small cubes could be 5, 6, or 7.\n\nHence, the answer is: 7\n\n[Key Insight] This question tests the understanding of the three views of a geometric solid. The key to solving the problem lies in determining the number of cubes in each layer based on the three views of the geometric solid." }, { "problem_id": 1624, "question": "A geometric shape is constructed by stacking several identical small cubes. The images below show the shapes viewed from the front and from above, respectively. If the minimum number of small cubes required to form this geometric shape is $m$, and the maximum number is $n$, then $m-n=$ $\\qquad$ .\n\n\n\nViewed from the front\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0072_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0072_2.jpg" ], "is_multi_img": true, "answer": "-4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the main view and the top view, the top view when the maximum number of required cubes is determined as follows:\n\nFor the maximum number of small squares:\n\n| | 2 | |\n| :--- | :--- | :--- |\n| | 2 | 3 |\n| 1 | 2 | 3 |\n\n## Viewed from above\n\n$\\therefore n=1+2+2+2+3+3=13$,\n\nFor the minimum number of small squares:\n\n\n\nViewed from above\n\n$\\therefore m=1+1+1+2+1+3=9$,\n\n$\\therefore m-n=9-13=-4$,\n\nTherefore, the answer is: -4\n\n[Key Insight] This question mainly tests the ability to determine the geometry from three views. The key is to draw the top view with the maximum and minimum number of required cubes based on the main view and the top view." }, { "problem_id": 1625, "question": "A geometric solid is composed of some small cubes of the same size. As shown in the figure, write down its front view and left view. Then, the maximum number of small cubes required to form the geometric solid is $\\qquad$\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0076_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0076_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view provided in the problem, we can see that the object consists of three columns. The left column has two layers, the middle column has one layer, and the right column has up to two layers.\n\nFrom the side view, we can observe that there are two layers on the left and one layer on the right. Therefore, the maximum number of small cubes in the figure is $5 + 3 = 8$.\n\nThus, the answer is 8.\n\n[Key Insight] This problem primarily tests the understanding of three-view drawings. Clarifying the definitions of the three views and the rules for reconstructing the geometric shape from these views is crucial for solving the problem." }, { "problem_id": 1626, "question": "A geometric solid is composed of multiple identical small cubes, and its three views are shown in the figure. The number of small cubes that make up this geometric solid is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0080_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0080_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0080_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer consists of 3 small cubes. The front view indicates that there are 2 cubes on the right side of the second layer. The side view shows that there is only one row in the second layer. Therefore, the total number of cubes is \\(3 + 2 = 5\\).\n\nThus, the answer is: 5.\n\n[Key Insight] This problem tests the ability to determine the shape of a geometric object from its three views, while also assessing the student's spatial imagination and understanding of three-dimensional shapes." }, { "problem_id": 1627, "question": "As shown in the figure, the shapes of the given geometric object as viewed from three directions are provided. Therefore, this geometric object can be composed of a maximum of $\\qquad$ small cubes.\n\n\n\nViewed from the front\n\n\n\nViewed from the left\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0081_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0081_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0081_3.jpg" ], "is_multi_img": true, "answer": "11", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: To determine the maximum number of small squares that make up this geometric figure, we can start by analyzing the top view, which easily shows that the bottom layer consists of 5 small cubes. From other views, we can see that the second layer has 5 small cubes, and the third layer has 1 small cube, as illustrated below:\n\n| 2 | 3 | 2 |\n| :--- | :--- | :--- |\n| | 2 | 2 |\n| | | |\n\n## From the top view\n\nThus, the total maximum number of small cubes is $5 + 5 + 1 = 11$.\n\nTherefore, the answer is: 11.\n\n[Insight] This problem tests the student's grasp and flexible application of the three-view concept, as well as their spatial imagination skills. The key to solving the problem lies in mastering the mnemonic: \"The top view lays the foundation, the front view builds up wildly, and the side view removes violations,\" which makes it easier to arrive at the answer." }, { "problem_id": 1628, "question": "A geometric solid is constructed by stacking identical small cubes, as shown in the figures representing the shapes viewed from the front and from above. The geometric solid is composed of at least $\\qquad$ small cubes.\n\n\n\nViewed from the front\n\n\n\n## Viewed from above\n\n##", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0082_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0082_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, there are at least three small cubes arranged in two layers; from the top view, there are at least five small cubes arranged in two columns; $\\square$\n\n$\\therefore$ To satisfy the problem's requirements, it is only necessary to ensure that there is one cube in the top layer of the front view and five small cubes in the top view.\n\n$\\therefore$ The minimum number of small cubes needed to build this structure is 6.\n\nHence, the answer is: 6.\n\n[Insight] This problem mainly tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"foundation with the top view, building up with the front view, and removing violations with the side view\" can make it easier to find the answer." }, { "problem_id": 1629, "question": "Using some identical cubes to build a geometric shape, the shapes seen from the front and top are shown in the figures below. The minimum number of cubes required to build this geometric shape is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from the top", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0084_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0084_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the front view provided in the problem, we know that the object consists of 3 columns, with 2 columns having the highest two layers and 1 column having one layer. From the top view, we know there are a total of 5 columns.\n\nTherefore, the geometric figure with the least number of small cubes is: $2 + 2 + 1 + 1 + 1 = 7$ cubes.\n\nHence, the answer is: 7.\n\n[Insight] This problem tests students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills. If one masters the mnemonic \"top view lays the foundation, front view builds up wildly, side view removes violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1630, "question": "As shown in the figures, the shapes seen from three directions of the given geometric object are provided. Therefore, this geometric object is composed of at most $\\qquad$ small cubes.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0085_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0085_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0085_3.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, it can be seen that the first layer has 5 small cubes.\n\nFrom the given front view and side view, it is known that the second layer can have at most 5 small cubes.\n\nTherefore, the maximum number of small cubes in this geometric figure is $5+5=10$.\n\nHence, the answer is: 10\n\n[Key Insight] This problem tests knowledge of the three views of a geometric figure: the front view is the view from the front, the side view is the view from the side, and the top view is the view from above. Mastering this knowledge is key to solving the problem." }, { "problem_id": 1631, "question": "As shown in the figure, the shape diagrams of a cylinder are observed from different directions. Calculate the lateral area of this cylinder based on the data in the figure. $\\qquad$ (The result should retain $\\pi$.)\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0088_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0088_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0088_3.jpg" ], "is_multi_img": true, "answer": "$6 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, it can be seen that the base diameter of the cylinder is 2, and the height is 3.\n\nTherefore, the lateral area $= 2 \\cdot \\pi \\times 3 = 6\\pi$.\n\nHence, the answer is: $6\\pi$.\n\n[Highlight] This question tests the three-dimensional view of geometric shapes and the spatial imagination ability of students, as well as the formula for the lateral area of a cylinder. The key to solving the problem lies in determining the base diameter and height of the cylinder from the front view." }, { "problem_id": 1632, "question": "The main view and top view of a geometric body composed of several identical small cubes are shown in the figure. The minimum number of small cubes is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0092_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0092_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer has 4 cubes. From the front view, the second layer has at least 2 cubes, and the third layer has at least 1 cube.\n\nTherefore, the minimum number of cubes is $4 + 2 + 1 = 7$.\nHence, the answer is: 7.\n\n[Key Insight] This problem tests the ability to determine the geometry from three views. The number of small squares in the top view corresponds to the number of small cubes in the bottom layer, while the number of small squares in the second and third layers of the front view indicates the minimum number of cubes in those respective layers." }, { "problem_id": 1633, "question": "As shown, the main view and left view of a geometric shape composed of several identical small cubes are given. The minimum number of small cubes required to form this geometric shape is $\\qquad$.\n\n\n\nMain View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0096_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0096_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Given the front view and the left view, to determine the minimum number of stacked geometric solids: draw a simulated top view, label the columns for the front view and the rows for the left view, assign the same number if they are identical, and assign 0 if they are different. The specific illustration is as follows:\n\n\n\nTherefore, the answer is: 3.\n\n【Insight】This question tests students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills. If one masters the mnemonic \"top view lays the foundation, front view builds wildly, left view dismantles violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1634, "question": "The maximum value of $n$ for a geometric shape formed by stacking $n$ identical small cubes, with the front view and top view as shown, is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0098_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0098_2.jpg" ], "is_multi_img": true, "answer": "13", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By integrating the front view and the top view, counting from top to bottom, the bottom layer can have a maximum of \\(2 + 2 + 3 = 7\\) units, the second layer can have a maximum of \\(1 + 1 + 2 = 4\\) units, and the third layer can have a maximum of \\(1 + 0 + 1 = 2\\) units. Therefore, the maximum value of \\(n\\) is \\(7 + 4 + 2 = 13\\). Hence, the answer is: 13.\n\n【Insight】This question tests the understanding of the front view and top view in three-view drawings. Mastering the related concepts of three-view drawings is key to solving the problem." }, { "problem_id": 1635, "question": "The following are the views of a geometric object composed of identical small cubes from three directions. At least $\\qquad$ small cubes are needed to form this geometric object.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0099_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0099_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0099_3.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By deriving the following three-dimensional diagram, it becomes easy to count the number of small cubes, which totals 7.\n\n\n\nTherefore, the answer is 7.\n\n[Highlight] This question tests the ability to deduce a three-dimensional figure from its three views, assessing students' spatial imagination. The key to solving the problem lies in constructing the spatial three-dimensional figure by combining the diagrams from the three directions." }, { "problem_id": 1636, "question": "The three views of a solid figure are shown in the figure. According to the data in the figure, the lateral surface area of this solid figure is obtained as:\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0100_1.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0100_2.jpg", "batch25-2024_06_17_4dfbb5a57aa37617b8fbg_0100_3.jpg" ], "is_multi_img": true, "answer": "$15 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three-view drawing, it is known that the base radius of the cone is 3, and the height is 4.\n\nTherefore, the slant height is $\\sqrt{3^{2}+4^{2}}=5$.\n\nThus, the lateral area is $=\\pi r l=3 \\times 5 \\pi=15 \\pi$.\n\nHence, the answer is: $15 \\pi$.\n\n【Insight】This question mainly examines determining the geometry from the three-view drawing and calculating the lateral area of the cone, involving the Pythagorean theorem. Remembering the formula is key to solving the problem, and the difficulty is not high." }, { "problem_id": 1637, "question": "As shown, given the front view and top view of a geometric object, find the volume of the geometric object. (Unit: cm)\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4e0f2d61cbde4601ea59g_0034_1.jpg", "batch25-2024_06_17_4e0f2d61cbde4601ea59g_0034_2.jpg" ], "is_multi_img": true, "answer": "$(30000+3200 \\pi) \\mathrm{cm}^{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view and the top view, it can be determined that the lower part of this geometric figure is a rectangular prism with a length of $30 \\mathrm{~cm}$, a width of $25 \\mathrm{~cm}$, and a height of $40 \\mathrm{~cm}$. The upper part is a cylinder with a base diameter of $20 \\mathrm{~cm}$ and a height of $32 \\mathrm{~cm}$.\n\nTherefore, the volume of this geometric figure is calculated as $30 \\times 25 \\times 40 + \\pi \\times \\left(\\frac{20}{2}\\right)^{2} \\times 32 = 30000 + 3200 \\pi \\left(\\mathrm{cm}^{3}\\right)$.\n\nAnswer: The volume of this geometric figure is $(30000 + 3200 \\pi) \\mathrm{cm}^{3}$.\n\n[Highlight] This question tests the understanding of front and top views. Correctly identifying the composition of the geometric figure is key to solving the problem." }, { "problem_id": 1638, "question": "As shown in the figure, the following are the three views of a geometric body. Based on the data in the figure, find the volume of the geometric body (the result should retain $\\pi$).\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4e0f2d61cbde4601ea59g_0064_1.jpg", "batch25-2024_06_17_4e0f2d61cbde4601ea59g_0064_2.jpg", "batch25-2024_06_17_4e0f2d61cbde4601ea59g_0064_3.jpg" ], "is_multi_img": true, "answer": "$3 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: The geometric body is a cylinder.\n\n$\\because$ By combining the three views, we can determine that the diameter of the base circle of the cylinder is 2, and the height is 3.\n\n$\\therefore$ The volume of the geometric body is: $\\pi \\times\\left(\\frac{2}{2}\\right)^{2} \\times 3=3 \\pi$.\n\n【Highlight】This question examines the identification of a geometric body from its three views and the calculation of its volume. The key to solving the problem lies in understanding the method for calculating the volume of a cylinder." }, { "problem_id": 1639, "question": "Smiley $(2)$ is obtained from smiley $(1)$ through $\\qquad$ transformation.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_4eba79f8db69661e9076g_0036_1.jpg", "batch25-2024_06_17_4eba79f8db69661e9076g_0036_2.jpg" ], "is_multi_img": true, "answer": "rotation", "answer_type": "single-step", "difficulty": "Low", "grade": "Elementary", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: According to the diagram, it can be seen that the smiling face (2) is obtained by rotating the smiling face (1). Therefore, the answer is rotation.\n\n[Key Point] This question examines the properties of rotation." }, { "problem_id": 1640, "question": "The given text translates to:\n\nAs shown in the figure, the front view and the left view of a simple geometric structure composed of identical small cubes are provided. The maximum number of small cubes that can compose this geometric structure is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0002_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0002_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the front view provided in the problem, we know that the object consists of 2 columns, with only one column having a maximum of two layers.\n\nFrom the side view, we know there are 3 rows, so the maximum number of small cubes that make up this geometric figure is $6+1=7$. Therefore, the answer is: 7.\n\n[Insight] This question tests students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills. If one masters the mnemonic \"Plan view lays the foundation, front view builds up wildly, side view removes violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1641, "question": "As shown, the three views of a geometric structure formed by several identical small cubes are given. The number of small cubes used to construct this geometric structure is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0003_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0003_2.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0003_3.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it can be seen that the base layer of this geometric figure consists of 3 small cubes.\n\nCombining the front view and the side view, it is evident that there is 1 cube in the upper back left position.\n\nTherefore, the total number of small cubes that make up this geometric figure is 4.\n\nHence, the answer is: 4.\n\n[Insight] This question tests students' understanding and flexible application of the three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Top view lays the foundation, front view builds up, side view removes the excess\" can make it easier to arrive at the answer." }, { "problem_id": 1642, "question": "The given text translates to English as follows:\n\nAs shown, the main view and top view of a simple geometric shape composed of identical small cubes are provided. If the number of these small cubes that make up the geometric shape is $n$, then the sum of the minimum and maximum values of $n$ is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0007_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0007_2.jpg" ], "is_multi_img": true, "answer": "26", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the front view and the top view, the minimum number of small cubes in the geometric body is as follows:\n\n\n\nTop view\n\nThus, the minimum value of $\\mathrm{n}$ is $1+1+1+1+3+2+1=10$.\n\nThe maximum number of small cubes in the geometric body is as follows:\n\n\n\nTop view\n\nThe maximum number of small cubes in the geometric body is $3+3+3+2+2+2+1=16$.\n\nTherefore, the sum of the maximum and minimum values is $10+16=26$.\n\nHence, the answer is: 26\n\n[Insight] This problem mainly examines the issue of determining the number of small cubes in a geometric body from three views. The layers of the object from top to bottom and left to right can be distinguished from the front view, and the left-right and front-back positions of the object can be distinguished from the top view. By synthesizing the above analysis, the possible number of small cubes can be counted." }, { "problem_id": 1643, "question": "On the table is a geometric shape composed of several identical cubes, with its front view and left view shown in the figures. Let the number of small cubes that make up this geometric shape be $\\mathrm{n}$. Then the minimum value of $\\mathrm{n}$ is\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0008_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0008_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the front view and the left view, it can be deduced that this geometric body has three rows and three columns.\n\n$\\because$ The bottom layer has at least 3 small cubes, and the second layer has at least 2 cubes,\n\n$\\therefore$ The minimum number of small cubes that make up this geometric body is 5,\n\n$\\therefore$ The minimum value of $\\mathrm{n}$ is 5,\n\nHence, the answer is: 5\n\n【Highlight】This question tests the knowledge of judging geometric bodies from three views. The key to solving this problem is to use the principle of \"the front view covers extensively, and the left view removes obstructions\" to determine the required number of cubes." }, { "problem_id": 1644, "question": "The front view and left view of a geometric shape formed by some identical small cubes are shown in the figures below. The maximum number of small cubes that can form this geometric shape is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0009_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0009_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the front view provided in the problem, we know that the object consists of three columns, with the left column being two layers high and the right column being one layer high at most. From the side view, we can see that there are two rows on the left and one row on the right. Therefore, it can be determined that there is only one small cube on the left, while on the right, there could be either one row with two layers and one row with a single layer, or both rows could have two layers. Thus, the minimum number of small cubes in the figure is 4, and the maximum is 7. Hence, the answer is 7.\n\n【Key Point】This question mainly tests the understanding of three-view drawings. Mastering the three-view drawings of geometric shapes is crucial for solving such problems." }, { "problem_id": 1645, "question": "As shown in the figure, this is the front view and top view of a cuboid. From the given data (unit: $\\mathrm{cm}$), the volume of the cuboid can be determined to be $\\qquad$ $\\mathrm{cm}^{3}$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0011_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0011_2.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, we can determine that the length of the cuboid is 4 and the height is 3. From the top view, we can determine that the width of the cuboid is 2.\n\nTherefore, the volume of the cuboid $= 4 \\times 3 \\times 2 = 24 \\mathrm{~cm}^{3}$.\n\n[Key Insight] This problem tests the ability to calculate the volume of a three-dimensional figure using its three views. It is a simple problem, and the key to solving it lies in understanding the three views correctly." }, { "problem_id": 1646, "question": "Several identical cubes are stacked together. The front view and top view of the combined structure are shown on the right. Therefore, the maximum number of cubes in the combined structure is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0013_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0013_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The first layer consists of \\(1 + 2 + 3 = 6\\) cubes, and the second layer can have up to 4 cubes. Therefore, the maximum number of cubes in this geometric figure is \\(6 + 4 = 10\\).\n\nThus, the answer is: 10.\n\n[Insight] This problem involves determining a geometric figure based on its three-view drawings, testing students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills." }, { "problem_id": 1647, "question": "As shown in the figure, the following are the three views of a geometric body. Based on the data given in the figure, calculate the lateral surface area of this geometric body is $\\qquad$ .\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0019_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0019_2.jpg" ], "is_multi_img": true, "answer": "$65 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be determined that the original geometric figure is a cone. Let the generatrix of the cone be of length \\( l \\), and the radius of the base circle be \\( r \\). We have \\( l = 13 \\) and \\( r = 5 \\).\n\nThe lateral surface area \\( S \\) of the cone is calculated by the formula:\n\\[\nS = \\pi r l = \\pi \\times 5 \\times 13 = 65\\pi.\n\\]\n\nTherefore, the answer is: \\( 65\\pi \\).\n\n[Highlight] This question tests the understanding of three-view drawings and the formula for the lateral surface area of a cone. The key to solving the problem lies in identifying the original geometric figure from the three-view drawing, and then applying the formula to find the answer." }, { "problem_id": 1648, "question": "A transparent open-topped cube container is filled with some liquid. The edge $\\mathrm{AB}$ always remains on the horizontal tabletop, and the inclination angle of the bottom of the container is $\\alpha$ ($\\angle CBE = \\alpha$, as shown in Figure 1). At this point, the liquid surface just reaches the edge $\\mathrm{CD}$ and intersects the edge $BB'$ at point $\\mathrm{Q}$. The shape of the liquid at this moment is a right triangular prism, with its three views and dimensions shown in Figure 2. When the cube is placed flat (with square $\\mathrm{ABCD}$ on the tabletop), the depth of the liquid is $\\qquad$ $d m$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0027_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0027_2.jpg" ], "is_multi_img": true, "answer": "1.5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the diagram, we know that $\\mathrm{CQ} \\parallel \\mathrm{BE}$, with $\\mathrm{BQ}=4$ and $\\mathrm{CQ}=5$.\n\nAccording to the Pythagorean theorem: $B Q=\\sqrt{5^{2}-4^{2}}=3(\\mathrm{dm})$.\n\nThe volume of the liquid is: $\\frac{1}{2} \\times 3 \\times 4 \\times 4=24\\left(\\mathrm{dm}^{3}\\right)$.\n\nThe depth of the liquid is: $24 \\div(4 \\times 4)=1.5(\\mathrm{dm})$.\n\nTherefore, the answer is: 1.5.\n\n**Key Insight:** This problem primarily examines the calculation of the volume of a quadrilateral and the understanding of three-dimensional views. Correctly interpreting the volume calculation of a prism is crucial to solving the problem." }, { "problem_id": 1649, "question": "A geometric shape is constructed by stacking several identical small cubes. The front view and the left view of the shape are shown in the figures below. The minimum number of small cubes required to construct this geometric shape is $\\qquad$ .\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0039_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0039_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the front view and the side view, the top view that requires the minimum number of cubes is determined to be:\n\n\n\nTop View\nThus, the minimum number of small cubes needed to form this geometric shape is 5.\n\nTherefore, the answer is: 5.\n\n[Key Insight] This question primarily tests the ability to determine a geometric shape from its three views. The key is to draw the top view that requires the least number of cubes based on the front and side views." }, { "problem_id": 1650, "question": "As shown, the geometric figure is composed of small cubes of the same size, and the plane figures obtained from viewing it from the left and from above are as follows. The minimum number of small cubes required to form this geometric figure is $\\qquad$.\n\n\n\nViewed from the left\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0040_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0040_2.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer consists of 5 small cubes. The second layer has at least 1 small cube, and the third layer also has at least 1 small cube.\n\nTherefore, the minimum number of small cubes needed to construct this geometric figure is \\(5 + 1 + 1 = 7\\).\n\nHence, the answer is: 7.\n\n[Insight] This question tests students' understanding and flexible application of the three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Top view lays the foundation, front view builds up, side view removes the excess\" can make it easier to arrive at the answer." }, { "problem_id": 1651, "question": "The following are the left view and top view of a simple geometric structure composed of small cubes of the same size. The number of small cubes that make up this geometric structure could be $\\qquad$.\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0046_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0046_2.jpg" ], "is_multi_img": true, "answer": "5 or 6 or 7 ", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the left view, it is easy to determine that this geometric figure has 2 layers. The top view shows that the base of this geometric figure is composed of 4 small cubes. From the left view, it can be inferred that the second layer may have 1, 2, or 3 small cubes. Therefore, the total number of small cubes that make up this geometric figure is: \n\n$4 + 1 = 5$ (cubes),\n\n$4 + 2 = 6$ (cubes),\n\n$4 + 3 = 7$ (cubes),\n\nHence, the answer is: 5 cubes, or 6 cubes, or 7 cubes.\n\n[Insight] This question mainly tests the students' mastery and flexible application of the three-view concept, and also reflects the assessment of spatial imagination ability. When solving such problems, it is essential to remember the mnemonic: \"The top view lays the foundation, the front view builds up, and the left view removes the excess.\"" }, { "problem_id": 1652, "question": "As shown in the figure, a transparent open cubical container $\\mathrm{ABCD}-\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}$ is filled with some liquid. The edge $\\mathrm{AB}$ always remains on the horizontal tabletop. As shown in Figure 1, the liquid surface just reaches the edge $\\mathrm{CD}$ and intersects the edge $\\mathrm{BB}^{\\prime}$ at point $\\mathrm{Q}$. At this moment, the shape of the liquid is a right triangular prism, and its three views and dimensions are shown in Figure 2. The volume of the liquid is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0056_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0056_2.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Given that \\( \\mathrm{CQ} = 5 \\, \\mathrm{dm} \\) and \\( \\mathrm{BC} = 4 \\, \\mathrm{dm} \\),\n\n\\[\n\\mathrm{BQ} = \\sqrt{5^{2} - 4^{2}} = 3 \\, \\mathrm{dm},\n\\]\n\nTherefore, the volume of the liquid is:\n\n\\[\n\\mathrm{V}_{\\text{liquid}} = \\frac{1}{2} \\times 3 \\times 4 \\times 4 = 24 \\, \\mathrm{dm}^{3}.\n\\]\n\nHence, the answer is: \\( 24 \\, \\mathrm{dm}^{3} \\).\n\n**Key Insight:** This problem tests the calculation of the volume of a quadrilateral and the understanding of three-dimensional views, as well as the application of the Pythagorean theorem. Correctly understanding the calculation of the volume of a prism is crucial." }, { "problem_id": 1653, "question": "The main view and top view of a geometric shape composed of several identical small cubes are shown in the figures below. The maximum number of small cubes that can form this geometric shape is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0058_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0058_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, it is easy to see that the bottom layer has 4 small cubes, and the second layer can have at most 2 small cubes. Therefore, the maximum number of small cubes needed to build this geometric figure is \\(4 + 2 = 6\\).\n\nThus, the answer is: 6\n\n【Key Point】This question tests the ability to determine a geometric figure based on its three views. The key to solving it lies in mastering the mnemonic: \"Use the top view to lay the foundation, the front view to build up, and the side view to remove excess,\" which makes it easier to arrive at the answer." }, { "problem_id": 1654, "question": "The maximum value of $n$ is $\\qquad$ for a geometric shape formed by stacking $n$ identical small cubes, as shown in the following views:\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0061_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0061_2.jpg" ], "is_multi_img": true, "answer": "18", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By synthesizing the front view and the top view, the maximum number of units on the bottom layer is \\(2 + 3 + 2 = 7\\), on the second layer is also \\(2 + 3 + 2 = 7\\), and on the third layer is \\(2 + 0 + 2 = 4\\). Therefore, the maximum value of \\(\\mathrm{n}\\) is \\(7 + 7 + 4 = 18\\).\n\nHence, the answer is: 18.\n\n[Key Insight] This question tests the understanding of the front view and top view in three-view drawings. Mastering the related concepts of three-view drawings is crucial for solving the problem." }, { "problem_id": 1655, "question": "The number of small cubes used to construct the geometric figure, whose three views are shown in the images, is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0062_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0062_2.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0062_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the three views, we can deduce that this geometric figure consists of 3 rows, 2 columns, and 2 layers.\n\nThe bottom layer should have 4 small cubes;\n\nThe second layer should have 1 small cube;\n\nTherefore, the total number of small cubes that make up this geometric figure is $4+1=5$.\n\nHence, the answer is: 5.\n\n[Highlight] This question tests the ability to reconstruct a physical object from its three views, which is key to solving the problem." }, { "problem_id": 1656, "question": "The three views of a geometric object are shown in the figure below. What is this geometric object?\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0070_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0070_2.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0070_3.jpg" ], "is_multi_img": true, "answer": "cone", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three views (front view, left view, and top view), the shape of this geometric body can be determined as follows:\n\n\n\nThat is, this geometric body is a cone.\n\nTherefore, the answer is: cone.\n\n[Key Insight] This question tests the ability to determine the shape of a geometric body from its three views. Mastering the relevant concepts is crucial for solving the problem." }, { "problem_id": 1657, "question": "A geometric solid is composed of several identical small cubes. The shape diagrams seen from the front and from above are shown in the figures below. The maximum number of small cubes in this geometric solid is $\\qquad$.\n\n\n\nSeen from the front\n\n\n\nSeen from above", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0075_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0075_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, the geometric figure has two layers. From the top view, the bottom layer of the geometric figure consists of 3 small cubes. Combining the front view and the top view, the top layer of the geometric figure has either 1 or 2 small cubes. Therefore, the maximum number of small cubes in this geometric figure is \\(3 + 2 = 5\\). Hence, the answer is: 5.\n\n[Key Insight] This problem tests the ability to reconstruct a geometric figure based on its front and top views. The key to solving this problem lies in analyzing the number of layers and the number of small cubes in each layer from the front and top views." }, { "problem_id": 1658, "question": "Construct a geometric shape using small cubes such that its front view and top view are as shown in the figures. The minimum number of small cubes required for this geometric shape is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0078_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0078_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By observing the front view and the top view, we can determine:\n\nThe minimum number of such geometric bodies required is $(2+1+1)+(3+1)+1=9$ (units), hence the answer is 9.\n\n[Highlight] This question tests the understanding of simple spatial graphics through three views, assesses spatial imagination ability, and is a basic question of moderate difficulty. The shape of the geometric body is examined from the front view and the side view, while the top view reveals the minimum and maximum number of small cubes in the geometric body." }, { "problem_id": 1659, "question": "As shown in the figure, the following are the three views of a workpiece, with dimensions labeled. The volume of this workpiece is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0079_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0079_2.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0079_3.jpg" ], "is_multi_img": true, "answer": "$17 \\pi \\mathrm{cm}^{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three-view drawing, it can be seen that this is a geometric body composed of two cylinders, one large and one small. The diameter of the large cylinder is $4 \\mathrm{~cm}$, and its height is $4 \\mathrm{~cm}$. The diameter of the small cylinder is $2 \\mathrm{~cm}$, and its height is $1 \\mathrm{~cm}$.\n\n$\\therefore$ The volume of this workpiece $=\\left(\\frac{4}{2}\\right)^{2} \\pi \\times 4+\\left(\\frac{2}{2}\\right)^{2} \\pi \\times 1=17 \\pi\\left(\\mathrm{cm}^{3}\\right)$. Therefore, the answer is: $17 \\pi \\mathrm{cm}^{3}$.\n\n【Insight】This question tests the composition of the three views of a geometric body. It requires obtaining the corresponding data of the geometric body from the three views and using the formula to calculate the volume of the cylinder." }, { "problem_id": 1660, "question": "As shown in the figure, the three views of a geometric solid formed by small cubes with a side length of 3 are given. Please determine the volume of the geometric solid.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0088_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0088_2.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0088_3.jpg" ], "is_multi_img": true, "answer": "135", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By examining the three views, it is found that the geometric body is composed of a total of $1+4=5$ cubes.\n\n$\\because$ The edge length of each cube is 3,\n\n$\\therefore$ The volume of each cube is 27,\n\n$\\therefore$ The volume of the geometric body is $5 \\times 27=135$.\n\nTherefore, the answer is: 135.\n\n【Insight】This question tests the ability to determine the geometric body from three views. The key to solving the problem is to ascertain the number of cubes that make up the geometric body." }, { "problem_id": 1661, "question": "As shown in the figure, the geometric shape is composed of identical small cubes and is viewed from three different directions. The number of small cubes that make up this geometric shape is $\\qquad$.\n\n\n\nViewed from the front\n\n\n\nViewed from the left\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0096_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0096_2.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0096_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "According to the three views of the geometric solid, the distribution of the cubes that make up the solid is as follows:\n\n\n\n$\\therefore$ There are 6 small cubes that constitute this geometric solid.\n\nTherefore, the answer is: 6.\n\n[Key Insight] This question mainly examines the three views of a geometric solid. The key to solving the problem is to visualize the shape of the geometric solid based on its three views." }, { "problem_id": 1662, "question": "As shown, the main view, top view, and left view of a geometric shape composed of identical small cubes are given. The number of small cubes that make up this geometric shape is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0100_1.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0100_2.jpg", "batch25-2024_06_17_4f92a1be4ee0234e0f7bg_0100_3.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the front view and the left view, it can be determined that the geometric body has two layers. The top view indicates that the bottom layer consists of 3 small cubes. Combining the front view and the left view, it is known that the second layer has 1 small cube. Therefore, there are a total of 4 small cubes.\n\nThus, the answer is 4.\n\n[Key Insight] This question primarily tests the ability to determine the number of small cubes that make up a geometric body based on three views. Mastering the knowledge of three views is crucial for solving such problems." }, { "problem_id": 1663, "question": "As shown in Figure 1, there are two completely overlapping rectangular sheets of paper. One of them is rotated $90^{\\circ}$ clockwise around point $\\mathrm{A}$ to obtain rectangle $A M E F$. Lines $B D$ and $M F$ are connected, and $\\angle A D B=30^{\\circ}$. The triangles $\\triangle B C D$ and $\\triangle M E F$ are cut out, and $\\triangle A B D$ is rotated clockwise around point $\\mathrm{A}$ to form $\\triangle A B_{1} D_{1}$. The side $A D_{1}$ intersects $F M$ at point $K$ (as shown in Figure 2). Let the rotation angle be $\\alpha\\left(0^{\\circ}<\\alpha<90^{\\circ}\\right)$. When $\\triangle A F K$ is an isosceles triangle, the degree measure of $\\alpha$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_51148592e5c9b3ed67e4g_0016_1.jpg", "batch25-2024_06_17_51148592e5c9b3ed67e4g_0016_2.jpg" ], "is_multi_img": true, "answer": "$60^{\\circ}$ or $15^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From the problem statement, we have: $\\angle F=\\angle A D B=30^{\\circ}$, $\\angle B A D=\\angle F A M=90^{\\circ}=\\angle B_{1} A D_{l}$, $\\angle B A B_{I}=\\alpha$. When $A K=K F$, triangle $A F K$ is isosceles,\n\n$\\therefore \\angle K A F=\\angle F=30^{\\circ}$,\n\n$\\therefore \\angle B A B_{l}=\\alpha=180^{\\circ}-\\angle B_{l} A D_{1}-\\angle K A F=60^{\\circ}$;\n\nWhen $A F=K F$, triangle $A F K$ is isosceles,\n\n$\\therefore \\angle F A K=\\angle F K A$\n\n$\\because \\angle F A K+\\angle F K A+\\angle F=180^{\\circ}$,\n\n$\\therefore 2 \\angle F A K=180^{\\circ}-\\angle F=150^{\\circ}$,\n\n$\\therefore \\angle F A K=75^{\\circ}$,\n\n$\\therefore \\angle B A B_{l}=\\alpha=180^{\\circ}-\\angle B_{l} A D_{1-} \\angle K A F=15^{\\circ}$;\nThus, the answer is: $60^{\\circ}$ or $15^{\\circ}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n[Insight] This problem primarily examines the properties of isosceles triangles, the triangle angle sum theorem, and the properties of rotation. The key to solving it lies in a thorough understanding and application of these concepts." }, { "problem_id": 1664, "question": "A set of right-angled triangles are stacked as shown in Figure (1). Now, the triangle \\( ADE \\) with a \\( 45^\\circ \\) angle is fixed in place, and the triangle \\( ABC \\) with a \\( 30^\\circ \\) angle is rotated clockwise around the vertex \\( A \\) (with the rotation angle not exceeding \\( 180^\\circ \\)), so that the two triangles have one set of sides parallel to each other. For example, in Figure (2), when \\( \\angle BAD = 15^\\circ \\), \\( BC \\parallel DE \\). When \\( 90^\\circ < \\angle BAD < 180^\\circ \\), the degree measure of \\( \\angle BAD \\) is \\qquad. \n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_51148592e5c9b3ed67e4g_0084_1.jpg", "batch25-2024_06_17_51148592e5c9b3ed67e4g_0084_2.jpg" ], "is_multi_img": true, "answer": "$105^{\\circ}$ or $135^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure (1), when \\( AC \\parallel DE \\), \\( \\angle BAD = \\angle DAE = 45^\\circ \\);\n\nAs shown in Figure (2), when \\( BC \\parallel AD \\), \\( \\angle DAB = \\angle B = 60^\\circ \\);\n\nAs shown in Figure (3), when \\( BC \\parallel AE \\),\n\n\\[\n\\begin{aligned}\n& \\because \\angle EAB = \\angle B = 60^\\circ, \\\\\n& \\therefore \\angle BAD = \\angle DAE + \\angle EAB = 45^\\circ + 60^\\circ = 105^\\circ;\n\\end{aligned}\n\\]\n\n\n\nFigure (1)\n\n\n\nFigure (3)\n\n\n\nFigure (2)\n\n\n\nAs shown in Figure (4), when \\( AB \\parallel DE \\),\n\n\\[\n\\because \\angle E = \\angle EAB = 90^\\circ,\n\\]\n\\[\n\\therefore \\angle BAD = \\angle DAE + \\angle EAB = 45^\\circ + 90^\\circ = 135^\\circ.\n\\]\n\\[\n\\therefore \\text{When } 90^\\circ < \\angle BAD < 180^\\circ,\n\\]\n\\[\n\\angle BAD = 105^\\circ \\text{ or } 135^\\circ.\n\\]\n\nTherefore, the answer is: \\( 105^\\circ \\) or \\( 135^\\circ \\).\n\n[Key Insight] This problem examines the properties of rotation, the determination and properties of parallel lines. Drawing the figures according to the given conditions and utilizing the properties of parallel lines and the characteristics of right-angled triangles are crucial for solving this problem." }, { "problem_id": 1665, "question": "According to the views shown in the figure (unit: $\\mathrm{mm}$), find the volume of the object.\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_538b70e6602da668df5ag_0010_1.jpg", "batch25-2024_06_17_538b70e6602da668df5ag_0010_2.jpg" ], "is_multi_img": true, "answer": "$1088 \\pi \\mathrm{mm} 3$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Question Analysis:** \nBy examining the front view and the top view, it can be determined that the object is formed by stacking two different cylinders vertically. Based on the data provided in the front view and the side view, the volumes of the two cylinders can be calculated separately, and then summed to obtain the total volume of the object.\n\n**Question Explanation:** \nThis is a composite shape consisting of two cylinders stacked vertically.\n\n\\[\nV = 16 \\times \\pi \\times \\left(\\frac{16}{2}\\right)^{2} + 4 \\times \\pi \\times \\left(\\frac{8}{2}\\right)^{2} = 1088\\pi \\, (\\text{mm}^{3})\n\\]\n\nTherefore, the volume of the object is \\(1088\\pi \\, \\text{mm}^{3}\\).\n\n**Key Insight:** \nThe volume of a cylinder is calculated as the base area multiplied by the height." }, { "problem_id": 1666, "question": "A workpiece is made by removing a smaller rectangular prism from the center top of a larger rectangular prism, as shown in the figure. Its front view is an axisymmetric concave shape. According to the dimensions in the figure (unit: cm), calculate the volume of the workpiece.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_538b70e6602da668df5ag_0013_1.jpg", "batch25-2024_06_17_538b70e6602da668df5ag_0013_2.jpg" ], "is_multi_img": true, "answer": "$558 \\mathrm{~cm} 3$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since the volume of the large cuboid $=9 \\times 11 \\times 7=693\\left(\\mathrm{~cm}^{3}\\right)$,\n\nand the volume of the small cuboid $=9 \\times 5 \\times 3=135\\left(\\mathrm{~cm}^{3}\\right)$,\n\ntherefore, the volume of the workpiece $=693-135=558\\left(\\mathrm{~cm}^{3}\\right)$.\n\nHence, the answer is $558 \\mathrm{~cm}^{3}$.\n[Insight] This question primarily tests the knowledge of three-view drawings." }, { "problem_id": 1667, "question": "The front view and top view of a cuboid are shown in the figures below. What is the volume of this cuboid?\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_538b70e6602da668df5ag_0020_1.jpg", "batch25-2024_06_17_538b70e6602da668df5ag_0020_2.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, we can determine that the length and height of the cuboid are 4 and 3, respectively. From the top view, we can determine that the length and width of the cuboid are 4 and 2, respectively.\n\nTherefore, the length, width, and height of the cuboid are $4$, $2$, and $3$, respectively.\n\nThus, the volume of the cuboid is $4 \\times 2 \\times 3 = 24$.\n\nAnswer: The volume of the cuboid is 24.\n\n[Key Insight] Determining the geometric shape from the three views and being able to synthesize the information from these views is crucial for solving the problem." }, { "problem_id": 1668, "question": "As shown in the figure, the three views of a workpiece are given, with dimensions labeled. Find the volume of this workpiece. (The result should retain $\\pi$.)\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_538b70e6602da668df5ag_0022_1.jpg", "batch25-2024_06_17_538b70e6602da668df5ag_0022_2.jpg", "batch25-2024_06_17_538b70e6602da668df5ag_0022_3.jpg" ], "is_multi_img": true, "answer": "$17 \\pi\\left(\\mathrm{cm}^{3}\\right)$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the given views, it can be determined that this workpiece is composed of two cylinders stacked together. The diameters of the base circles are $4 \\mathrm{~cm}$ and $2 \\mathrm{~cm}$ respectively, and their heights are $4 \\mathrm{~cm}$ and $1 \\mathrm{~cm}$ respectively.\n\n$\\therefore$ The volume of this workpiece is $\\pi \\times(4 \\div 2)^{2} \\times 4+\\pi \\times(2 \\div 2)^{2} \\times 1=17 \\pi\\left(\\mathrm{cm}^{3}\\right)$.\n\n【Highlight】This question tests the students' mastery and flexible application of three-view drawings, as well as their spatial imagination skills. Correctly identifying the shape of the geometric body is key to solving this problem." }, { "problem_id": 1669, "question": "As shown in the figure, the three views of a geometric body are given, where the front view and the left view are both isosceles triangles with a waist length of $13 \\mathrm{~cm}$ and a base of $10 \\mathrm{~cm}$. Find the volume of this geometric body.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_538b70e6602da668df5ag_0024_1.jpg", "batch25-2024_06_17_538b70e6602da668df5ag_0024_2.jpg", "batch25-2024_06_17_538b70e6602da668df5ag_0024_3.jpg" ], "is_multi_img": true, "answer": "$100 \\pi$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three-view drawing, it can be determined that this geometric body is a cone.\n\nAccording to the problem, the length of the generatrix (slant height) \\( l = 13 \\), and the radius of the base \\( r = 5 \\).\n\nTherefore, the height \\( h \\) on the base can be calculated as:\n\\[\nh = \\sqrt{l^2 - r^2} = \\sqrt{13^2 - 5^2} = 12\n\\]\n\nThus, the volume of the cone is:\n\\[\n\\text{Volume} = \\frac{1}{3} \\pi r^2 \\cdot h = \\frac{1}{3} \\pi \\times 5^2 \\times 12 = 100\\pi\n\\]\n\n**Key Insight:** This problem primarily tests knowledge of three-view drawings and the calculation of a cone's volume. The key to solving such problems lies in deriving the three-dimensional shape from the three-view drawing. Students often make mistakes due to insufficient spatial imagination, such as failing to identify the cone's base radius or not being proficient in applying the cone's volume formula." }, { "problem_id": 1670, "question": "Place a set of triangles as shown in Figure 1, where $\\angle ACB = \\angle DEC = 90^\\circ, \\angle A = 45^\\circ, \\angle D = 30^\\circ$, the hypotenuse $AB = 6$, and $CD = 8$. Rotate triangle $DCE$ around point $C$ clockwise by $15^\\circ$ to obtain triangle $D_1CE$ (as shown in Figure 2). At this point, $AB$ intersects $CD_1$ at point $O$. The length of segment $AD_1$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_541d2a6cd3628ed9c20eg_0001_1.jpg", "batch25-2024_06_17_541d2a6cd3628ed9c20eg_0001_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{34}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure,\n\n\n\nGiven that $\\angle A=45^{\\circ}$ and $\\angle D=30^{\\circ}$,\n\nIf the rotation angle is $15^{\\circ}$, then $\\angle \\mathrm{ACO}=30^{\\circ}+15^{\\circ}=45^{\\circ}$.\n\nTherefore, $\\angle \\mathrm{AOC}=180^{\\circ}-\\angle \\mathrm{ACO}-\\angle \\mathrm{CAO}=90^{\\circ}$.\n\nIn the isosceles right triangle $\\triangle \\mathrm{ABC}$, with $\\mathrm{AB}=6$, then $\\mathrm{AC}=\\mathrm{BC}=3 \\sqrt{2}$.\n\nSimilarly, we can find: $\\mathrm{AO}=\\mathrm{OC}=3$.\n\nIn the right triangle $\\triangle \\mathrm{AOD}_{1}$, $\\mathrm{OA}=3$ and $\\mathrm{OD}_{1}=\\mathrm{CD}_{1}-\\mathrm{OC}=5$.\n\nUsing the Pythagorean theorem, we find: $\\mathrm{AD}_{1}=\\sqrt{3^{2}+5^{2}}=\\sqrt{34}$.\n\nThus, the answer is $\\sqrt{34}$.\n\n【Insight】This problem mainly tests the properties of rotation and the comprehensive application of solving right triangles. The key to solving this problem is recognizing that $\\mathrm{AO} \\perp \\mathrm{OC}$." }, { "problem_id": 1671, "question": "Two isosceles right-angled triangular paper pieces $AOB$ and $COD$ are placed as shown in Figure 1, with their right-angled vertices coinciding at point $O$, $AB=13, CD=7$. Keeping paper piece $AOB$ stationary, rotate paper piece $COD$ around point $O$ counterclockwise by $\\alpha \\left(0<\\alpha<90^{\\circ}\\right)$, as shown in Figure 2. When $BD$ and $CD$ are collinear (as shown in Figure 3), the area of $\\triangle ABC$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch25-2024_06_17_541d2a6cd3628ed9c20eg_0082_1.jpg", "batch25-2024_06_17_541d2a6cd3628ed9c20eg_0082_2.jpg", "batch25-2024_06_17_541d2a6cd3628ed9c20eg_0082_3.jpg" ], "is_multi_img": true, "answer": "30", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Let the intersection point of $\\mathrm{AO}$ and $\\mathrm{BC}$ be point $\\mathrm{G}$,\n\n\n\nSince $\\angle \\mathrm{AOB} = \\angle \\mathrm{COD} = 90^{\\circ}$,\n\nTherefore, $\\angle \\mathrm{AOC} = \\angle \\mathrm{DOB}$,\n\nIn triangles $\\triangle \\mathrm{AOC}$ and $\\triangle \\mathrm{BOD}$,\n\n$$\n\\left\\{\\begin{array}{l}\nOA = OB \\\\\n\\angle AOC = \\angle BOD \\\\\nOC = OD\n\\end{array}\\right.\n$$\n\nThus, $\\triangle \\mathrm{AOC} \\cong \\triangle \\mathrm{BOD}$ (SAS),\nTherefore, $\\mathrm{AC} = \\mathrm{BD}$, and $\\angle \\mathrm{CAO} = \\angle \\mathrm{DBO}$,\n\nSince $\\angle \\mathrm{DBO} + \\angle \\mathrm{OGB} = 90^{\\circ}$,\n\nAnd $\\angle \\mathrm{OGB} = \\angle \\mathrm{AGC}$,\n\nTherefore, $\\angle \\mathrm{CAO} + \\angle \\mathrm{AGC} = 90^{\\circ}$,\n\nThus, $\\angle \\mathrm{ACG} = 90^{\\circ}$,\n\nTherefore, $\\mathrm{CG} \\perp \\mathrm{AC}$,\n\nLet $\\mathrm{AC} = x$, then $\\mathrm{BD} = \\mathrm{AC} = x$, and $\\mathrm{BC} = x + 7$,\n\nSince $\\mathrm{BD}$ and $\\mathrm{CD}$ are on the same straight line, and $\\mathrm{BD} \\perp \\mathrm{AC}$,\n\nTherefore, $\\triangle \\mathrm{ABC}$ is a right-angled triangle,\n\nThus, $\\mathrm{AC}^{2} + \\mathrm{BC}^{2} = \\mathrm{AB}^{2}$,\n\n$x^{2} + (x + 7)^{2} = 13^{2}$\n\nSolving gives $x = 5$, that is, $\\mathrm{AC} = 5$, and $\\mathrm{BC} = 5 + 7 = 12$,\n\nIn the right-angled triangle $\\mathrm{ABC}$, the area $\\mathrm{S} = \\frac{1}{2} \\times 5 \\times 12 = 30$,\n\nTherefore, the answer is: 30.\n\n【Key Insight】This problem examines the properties of rotation, the criteria and properties of congruent triangles, the Pythagorean theorem, and the properties of isosceles right-angled triangles. The key to solving the problem lies in correctly identifying congruent triangles and utilizing their properties to find the solution." }, { "problem_id": 1672, "question": "The following images show the front and top views of a geometric object. Calculate the volume of the geometric object. ( $\\pi \\approx 3.14)$\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_59559f556bf9f589781dg_0025_1.jpg", "batch25-2024_06_17_59559f556bf9f589781dg_0025_2.jpg" ], "is_multi_img": true, "answer": "$45420 \\mathrm{~cm}^{3}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: \n\n$\\because$ The front view shows a rectangle on top and a rectangle at the bottom, and the top view shows a circle in the middle with a square around it,\n\n$\\therefore$ The geometric shape consists of a cylinder on top and a rectangular prism at the bottom.\n\nFrom the diagram, we can see: the length, width, and height of the rectangular prism are $30$, $30$, and $40$ respectively, and the diameter of the cylinder's base is $20$ with a height of $30$.\n\n$\\therefore$ The volume of the geometric shape $=30 \\times 30 \\times 40 + 3.14 \\times (20 \\div 2)^{2} \\times 30$\n\n$=36000 + 3.14 \\times 100 \\times 30$\n\n$=36000 + 9420$\n\n$=45420\\left(\\mathrm{~cm}^{3}\\right)$.\n\nAnswer: The volume of the geometric shape is $45420 \\mathrm{~cm}^{3}$.\n\n【Key Insight】This question tests the understanding of three-view drawings. Correctly identifying the shape of the geometric figure and mastering the volume formulas of various geometric shapes are key to solving the problem." }, { "problem_id": 1673, "question": "Two rectangles $\\mathrm{ABCD}$ and $\\mathrm{EFGH}$ with length 2 and width 1 are placed on a straight line 1 as shown in Figure 1, with $\\mathrm{DE}=2$. The rectangle $\\mathrm{ABCD}$ is rotated clockwise around point $\\mathrm{D}$ by an angle $\\alpha$ $\\left(0^{\\circ}<\\alpha<90^{\\circ}\\right)$, and the rectangle $\\mathrm{EFGH}$ is rotated counterclockwise around point $\\mathrm{E}$ by the same angle. During the rotation, consider Figure 2: When the overlapping part of rectangles $\\mathrm{ABCD}$ and $\\mathrm{EFGH}$ forms a square, $\\alpha=$ $\\qquad$ ${ }^{\\circ}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_5cd77505a794128d8ae6g_0069_1.jpg", "batch25-2024_06_17_5cd77505a794128d8ae6g_0069_2.jpg" ], "is_multi_img": true, "answer": "45 .", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Since quadrilateral MFNC is a square, and rectangle $\\mathrm{ABCD}$ is rotated clockwise around point $\\mathrm{D}$ while rectangle $\\mathrm{EFGH}$ is rotated counterclockwise around point $\\mathrm{E}$ by the same angle,\n\n$\\therefore \\mathrm{NF}=\\mathrm{NC}, \\angle \\mathrm{FNC}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{DNE}=90^{\\circ}, \\mathrm{ND}=\\mathrm{NE}$,\n\n$\\therefore \\angle \\mathrm{NDE}=\\angle \\mathrm{NED}=45^{\\circ}$,\n\n$\\therefore \\angle \\alpha=180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$,\n\n$\\therefore \\alpha=45^{\\circ}$.\n\nTherefore, the answer is: 45\n\n【Key Insight】This question examines the properties of rotation: two figures are congruent before and after rotation, the distance from corresponding points to the center of rotation is equal, and the angle between the lines connecting corresponding points and the center of rotation equals the rotation angle. Mastering the properties of rotation is crucial for solving such problems." }, { "problem_id": 1674, "question": "As shown in the figure, the top and bottom surfaces of the gift box are congruent regular hexagons. Both the front view and the side view are composed of rectangles. The dimensions of the large rectangle in the front view are shown in the figure. The side view includes two congruent rectangles. If the gift box is wrapped with colored tape as shown, the minimum length of tape required is $\\qquad$ centimeters.\n\n\n\nPhysical Diagram\n\n\n\nFront View\n\n\n\nSide View", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0006_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0006_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0006_3.jpg" ], "is_multi_img": true, "answer": "$180 \\sqrt{3}+120$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "According to the problem, the top base of the actual figure is constructed as shown in the figure: $AC$ and $CD$ are the two sides of the top surface.\n\n\n\nSince the diameter of the regular hexagon is $60 \\mathrm{~cm}$,\n\nthen $AC = 60 \\div 2 = 30 (\\mathrm{~cm})$, and $\\angle ACD = 120^\\circ$.\n\nDraw $CB \\perp AD$ at point $B$,\n\nthen $AB = AC \\times \\sin 60^\\circ = 30 \\times \\frac{\\sqrt{3}}{2} = 15 \\sqrt{3} (\\mathrm{~cm})$,\n\nso $AD = 2AB = 30 \\sqrt{3} (\\mathrm{~cm})$.\n\nThe minimum length of the tape required is $= 6 \\times 30 \\sqrt{3} + 6 \\times 20 = 180 \\sqrt{3} + 120 \\quad (\\mathrm{~cm})$.\n\nTherefore, the answer is: $180 \\sqrt{3} + 120$.\n\n【Insight】This problem tests the properties of a regular hexagon, the three-dimensional view of a geometric figure, and the student's spatial imagination. Note that to find the distance between opposite sides of a regular hexagon given the maximum distance between two vertices, one needs to construct a right triangle and use the corresponding trigonometric functions to solve." }, { "problem_id": 1675, "question": "A geometric shape is constructed using identical small cubes. The shape of the geometric shape as seen from the front and from above is shown below. To construct such a geometric shape, the minimum number of these small cubes required is $a$, and the maximum number is $b$. Therefore, $a-b=$ $\\qquad$ .\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0007_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0007_2.jpg" ], "is_multi_img": true, "answer": "-2", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By synthesizing the front view and the top view, the base layer of this geometric figure consists of 4 small cubes.\n\nThe second layer has a minimum of 2 and a maximum of 4 cubes.\n\nTherefore, the minimum number of small cube blocks needed to construct such a geometric figure is: $4 + 2 = 6$ blocks, i.e., $a = 6$;\n\nThe maximum number of small cube blocks needed is: $4 + 4 = 8$ blocks, i.e., $b = 8$.\n\nThus, $a - b = -2$.\n\nHence, the answer is: -2.\n\n【Insight】This problem examines the three views of a geometric figure. The key to solving it lies in remembering the mnemonic: \"The top view lays the foundation, the front view builds up wildly, and the side view removes violations,\" which helps in determining the values of $a$ and $b$." }, { "problem_id": 1676, "question": "As shown in the figure, the following are the three views of a geometric body. The volume of the geometric body is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0011_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0011_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0011_3.jpg" ], "is_multi_img": true, "answer": "$4 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be determined that the three-dimensional figure is a cylinder with a base circle diameter of 2 and a height of 4.\n\nTherefore, the volume \\( V = S_{\\text{base}} \\times h = \\pi \\times \\left(\\frac{2}{2}\\right)^2 \\times 4 = 4\\pi \\).\n\nHence, the answer is: \\( 4\\pi \\).\n\n[Key Insight] This problem primarily examines the three-view drawing of a three-dimensional figure and the calculation of the volume of a cylinder. The key to solving the problem lies in determining the shape of the three-dimensional figure based on the three-view drawing." }, { "problem_id": 1677, "question": "A geometric shape composed of 8 identical small cubes is shown in Figure 1. After removing $\\qquad$ small cubes, the front view and left view of the resulting geometric shape are both as shown in Figure 2.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0014_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0014_2.jpg" ], "is_multi_img": true, "answer": "3、4、5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "According to the problem, after removing several small cubes, the shapes viewed from the front and left sides must be as shown in Figure 2. Therefore, the bottom layer must have at least 2 small cubes, and the layer above must retain 1 cube, as illustrated.\n\n\n\nRemove 5 cubes\n\n\n\nRemove 4 cubes\n\n\n\nRemove 3 cubes\n\nThus, the answer is: $3, 4, 5$.\n\n【Key Insight】This question tests the ability to determine a geometric shape from its three views. Understanding the three types of views of a geometric shape is crucial. The key to solving such problems lies in deriving the three views from the three-dimensional shape. Students often make mistakes due to insufficient spatial imagination skills." }, { "problem_id": 1678, "question": "The three views of a geometric figure composed of several small cubes are shown in the figure below. The number of small cubes that make up this geometric figure is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0024_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0024_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0024_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, there are at least 4 small cubes, and from the front view, there are at least 6 small cubes. Combining this with the side view, there are only 6 small cubes.\n\nTherefore, the answer is: 6.\n\n[Highlight] This question tests the students' mastery and flexible application of the three-view drawing, determining the shape of an object based on the three views, and also assesses the students' spatial imagination skills." }, { "problem_id": 1679, "question": "As shown in the figure, a geometric body is constructed using small cubes. The views from the front and from above are shown in the figure. The minimum number of cubes required for such a geometric body is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0026_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0026_2.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the top view, the bottom layer of this geometric figure consists of 6 small cubes.\n\nFrom the front view, it is known that the top layer requires at least 3 small cubes, and the middle layer also requires at least 3 small cubes. Therefore, the minimum number of small cubes needed for such a geometric figure is $3 + 3 + 6 = 12$.\n\nThus, the answer is:\n\n12.\n\n[Highlight] This problem tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Top view lays the foundation, front view builds up\" can make it easier to arrive at the answer." }, { "problem_id": 1680, "question": "The following figure shows the three views of a geometric shape composed of identical small cubes. The number of small cubes that make up this geometric shape is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0033_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0033_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0033_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the three views, we can deduce that the base layer of this geometric model consists of \\(3 + 1 = 4\\) small cubes, and the second layer has 1 small cube. \n\nTherefore, the total number of small cubes used to construct this geometric model is \\(4 + 1 = 5\\).\n\nHence, the answer is: 5.\n\n[Key Insight] This question tests the students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills. Mastering the mnemonic \"Use the top view to lay the foundation, the front view to build up, and the side view to correct mistakes\" is crucial for solving such problems." }, { "problem_id": 1681, "question": "As shown in the figure, the main view and top view of a geometric structure composed of several small cubes are given. Let the maximum number of small cube blocks required to build such a geometric structure be $m$, and the minimum number be $n$. Then $m+n=$ $\\qquad$ .\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0037_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0037_2.jpg" ], "is_multi_img": true, "answer": "15", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: There are two possibilities;\n\nFrom the front view, we can deduce that this geometric body has a total of 3 layers.\n\nFrom the top view, we can see that the first layer consists of 4 cubes. From the front view, the second layer has a minimum of 2 cubes and a maximum of 3 cubes.\n\nThe third layer consists of only one cube.\n\nTherefore, the maximum number of small cubes, \\( m \\), is \\( 3 + 4 + 1 = 8 \\), and the minimum number, \\( n \\), is \\( 2 + 4 + 1 = 7 \\).\n\nThus, \\( m + n = 15 \\).\n\nThe answer is: 15\n\n[Key Insight] This problem mainly tests the ability to determine the geometric body from its three views. The key is to master the mnemonic: \"The top view lays the foundation, the front view builds up wildly, and the side view removes the violations,\" which makes it easy to arrive at the answer." }, { "problem_id": 1682, "question": "As shown in the figure, the three views of a geometric object composed of several identical small cubes are given. The number of small cubes that make up this geometric object is . $\\qquad$\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0043_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0043_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0043_3.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the three-view drawing, the figure can be sketched as follows:\n\n\n\nThe number of small cubes that make up this geometric figure is: $1+3+1+1+1+2=9$;\n\nTherefore, the answer is: 9.\n\n[Key Insight] This problem primarily tests the understanding of the three-view drawing of a geometric figure. Mastering the three-view drawing of geometric figures is crucial for solving such problems." }, { "problem_id": 1683, "question": "A geometric solid is constructed by stacking several identical small cubes, as shown in the figures representing the shapes viewed from the front and from the left. If the total number of small cubes used in the geometric solid is $n$, then the number of all possible values for $n$ is $\\qquad$.\n\n\n\nViewed from the front\n\n\n\nViewed from the left", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0045_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0045_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the front view and the side view, the top view corresponding to the minimum and maximum number of required small cubes is determined as follows:\n\nTop view with the maximum number of cubes:\n\n| 1 | 1 | 1 |\n| :--- | :--- | :--- |\n| 1 | 2 | 2 |\n| 1 | 3 | 2 |\n\nTop view with the minimum number of cubes:\n\n| 1 | 0 | 0 |\n| :--- | :--- | :--- |\n| 0 | 0 | 2 |\n| 0 | 3 | 0 |\n\nThus, the geometric figure is composed of a minimum of 6 small cubes and a maximum of 14 small cubes. Therefore, the possible values for \\( n \\) are \\( 6, 7, 8, 9, 10, 11, 12, 13, 14 \\).\n\nHence, the answer is: 9.\n\n[Key Insight] This problem primarily tests the ability to determine the geometry from three views. The key is to draw the top view corresponding to the minimum and maximum number of cubes based on the front and side views." }, { "problem_id": 1684, "question": "The front view and left view of a geometric shape formed by small cubes of equal size are shown in the figures below. The minimum number of small cubes required to form this geometric shape is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0046_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0046_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By carefully observing the front view and the side view of the object, it can be determined that the lower part of the geometric shape requires at least 2 small cubes, and the upper part requires at least 1 small cube.\n\nTherefore, the geometric shape is composed of at least 3 small cubes.\n\nHence, the answer is: 3.\n\n[Insight] This question primarily tests students' mastery and flexible application of the three-view drawing, and also reflects the assessment of spatial imagination skills. If one masters the mnemonic \"the top view lays the foundation, the front view builds wildly, and the side view removes violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1685, "question": "A geometric solid is composed of multiple identical cubes, with its front view and top view shown in the figures below. The maximum number of cubes that can compose this geometric solid is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0050_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0050_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By combining the front view and the top view, it can be determined that the upper left layer has a maximum of 2 cubes, the lower left layer also has a maximum of 2 cubes, and the right side has only one layer with 2 cubes.\n\nTherefore, the maximum number of cubes in this geometric figure is \\(2 + 2 + 2 = 6\\).\n\nHence, the answer is: 6.\n\n[Highlight] This question primarily examines the ability to deduce a geometric figure from its three views, testing the student's mastery and flexible application of three-view concepts, as well as their spatial imagination skills." }, { "problem_id": 1686, "question": "Given the three views of a geometric object as shown in the figure, where the top view is an equilateral triangle, the length of the front view is $1 \\mathrm{~cm}$, and the height is $2 \\mathrm{~cm}$, the area of the left view of the geometric object is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0054_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0054_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0054_3.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the three-view drawing, it can be determined that the geometric body is a triangular prism with an equilateral triangle as its base.\n\n$\\because$ The length of the front view is 1, and the height is 2.\n\n$\\therefore$ The side length of the equilateral triangle base of the prism is 1, and the height is 2.\n\n$\\therefore$ The length of the side view $=\\sqrt{1^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\frac{\\sqrt{3}}{2}$, and the height of the side view is 2.\n\n$\\therefore$ The area of the side view $=\\frac{\\sqrt{3}}{2} \\times 2=\\sqrt{3}$.\n\nTherefore, the answer is: $\\sqrt{3}$.\n\n【Insight】This question tests the understanding of three-view drawings, the properties of equilateral triangles, and the Pythagorean theorem. Mastering the three-view drawings of common geometric bodies is key to solving this problem." }, { "problem_id": 1687, "question": "As shown in the figure, these are the shape diagrams of a geometric body formed by some small cubic blocks as seen from different directions. If, on the basis of the constructed geometric body (without changing the positions of the small cubic blocks in the original geometric body), we continue to add the same small cubic blocks to form a large cube, then at least $\\qquad$ more small cubic blocks are needed.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0058_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0058_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0058_3.jpg" ], "is_multi_img": true, "answer": "19", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the main view, it can be seen that the original geometry has three layers and 3 columns;\n\nFrom the left view, it can be seen that the constructed geometry has a total of 3 rows;\nFrom the top view, it is easy to determine that the bottom layer has 5 small cubes, the second layer has 2 small cubes, and the third layer has 1 small cube, totaling $5+2+1=8$ small cubes.\n\n$\\because$ The large cube to be constructed consists of $3 \\times 3 \\times 3=27$ small cubes,\n\n$\\therefore$ At least $27-8=19$ more small cubes are needed.\n\nTherefore, the answer is: 19.\n\n[Highlight] This question tests the understanding of three-view drawings, focusing on developing students' spatial imagination. The key to solving the problem lies in determining the total number of small cubes in the original geometry and the large cube to be constructed." }, { "problem_id": 1688, "question": "A geometric shape is composed of several identical small cubes. The top view and left view of this geometric shape are shown in the figures below. The minimum number of small cubes in this geometric shape is $\\qquad$.\n\n\n\nTop View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0060_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0060_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view and the left view, it can be determined that this geometric figure has 2 layers. The first layer consists of 4 small cubes, and the second layer has at least 1 small cube.\n\n$\\therefore$ The minimum number of small cubes in this geometric figure is 5. Hence, the answer is: 5\n\n【Insight】This question tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills." }, { "problem_id": 1689, "question": "The front view and top view of a cuboid are shown in the figures below. The volume of this cuboid is $\\qquad$.\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0061_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0061_2.jpg" ], "is_multi_img": true, "answer": "36", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the front view, we can determine that the length and height of this rectangular prism are 4 and 3, respectively.\n\nFrom the top view, we can determine that the length and width of this rectangular prism are 4 and 3, respectively.\n\nTherefore, the length, width, and height of this rectangular prism are $4$, $3$, and $3$, respectively.\n\nThus, the volume of this rectangular prism is $4 \\times 3 \\times 3 = 36$.\n\nHence, the answer is: 36.\n\n[Key Insight] This question tests the ability to determine the geometry of an object from its three views. Note: The front view primarily reflects the length and height of the object, the side view primarily reflects the width and height, and the top view primarily reflects the length and width." }, { "problem_id": 1690, "question": "As shown in the figure, the three views of a geometric object formed by several small cubes with a side length of 1 are given. The surface area of this geometric object is $\\qquad$\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0071_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0071_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0071_3.jpg" ], "is_multi_img": true, "answer": "22", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the three views, we can deduce that the base layer of this geometric model consists of $3+1=4$ small cubes, and the second layer has 1 small cube.\n\nTherefore, the total number of small cubes used to construct this geometric model is $4+1=5$.\n\n$\\therefore$ The surface area of this geometric body is $5 \\times 6 - 8 = 22$,\n\nHence, the answer is: 22.\n\n[Insight] This question tests the students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills. Mastering the mnemonic \"Use the top view to lay the foundation, the front view to build up, and the side view to remove obstructions\" is key to solving the problem." }, { "problem_id": 1691, "question": "As shown in the figure, the geometric shape is formed by stacking identical small cubes, and the shape diagrams of the geometric shape as seen from the front, left, and top are provided. Therefore, the number of small squares that make up this three-dimensional figure is $\\qquad$.\n\n\n\nViewed from the front\n\n\n\nViewed from the left\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0076_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0076_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0076_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer consists of 4 cubes, and the second layer has 1 cube. Therefore, the total number of cubes is $4 + 1 = 5$.\n\nThus, the answer is 5.\n\n[Highlight] This question tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Use the top view to lay the foundation, the front view to build up, and the side view to correct mistakes\" can make it easier to arrive at the answer." }, { "problem_id": 1692, "question": "As shown, these are the shape diagrams of a geometric object viewed from three different directions. Based on the data in the diagrams (unit: cm), its surface area can be calculated as $\\qquad$ $\\mathrm{cm}^{2}$. (The result should retain $\\pi$)\n\n\n\nView from above\n\n\n\nView from the front\n\n\n\nView from the left", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0080_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0080_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0080_3.jpg" ], "is_multi_img": true, "answer": "$8 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the three-view drawing, the geometric figure is a cylinder.\n\nThe surface area of the cylinder is calculated as \\(2 \\times \\pi \\times 1^{2} + 2 \\pi \\times 3 = 8 \\pi \\, (\\mathrm{cm}^{2})\\).\n\nTherefore, the answer is: \\(8 \\pi\\)\n\n[Highlight] This question tests knowledge of three-view drawings, surface area, and volume of geometric figures. The key to solving it lies in mastering the fundamental concepts, and it is a common type of question in middle school exams." }, { "problem_id": 1693, "question": "The following three diagrams are the front view, top view, and left view of a structure built with small wooden cubes. Please observe and determine how many small wooden cubes it consists of.\n\n\n\nTop view\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0084_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0084_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0084_3.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Since there are 6 squares in the top view,\n\nTherefore, the bottom layer has 6 small cube blocks.\n\nFrom the front view and the side view, it can be deduced that the second layer has 3 small cube blocks, and the third layer has 1 small cube block. Thus, the total number of small cube blocks is $6 + 3 + 1 = 10$ (blocks).\n\nHence, the answer is: 10.\n\n[Insight] This question tests the ability to determine the geometry from three views. Mastering the three views is crucial for solving such problems." }, { "problem_id": 1694, "question": "A geometric structure is built with several identical small cubes, as shown in the figures representing the front view and the top view respectively. If the total number of small cubes used in the structure is $n$, then the sum of the maximum and minimum values of $n$ is $\\qquad$.\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0085_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0085_2.jpg" ], "is_multi_img": true, "answer": "22", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the front view and the top view, the minimum and maximum number of items that can be placed in the corresponding positions on the top view are as follows:\n\n\n\nA minimum of 9 items are required, and a maximum of 13 items are needed.\n\nTherefore, $n = 9 + 13 = 22$,\n\nHence, the answer is: 22.\n\n[Highlight] This question tests the understanding of the three views of a simple composite object. Marking the number of items placed in the corresponding positions on the top view is key to solving the problem." }, { "problem_id": 1695, "question": "The geometric shapes shown are all formed by arranging unit cubes with a side length of 1 unit. After calculation, it is found that the surface area of the (1) geometric shape is 6 square units, the surface area of the (2) geometric shape is 18 square units, the surface area of the (3) geometric shape is 36 square units, ... Following this pattern, the surface area of the (20) geometric shape is $\\qquad$ square units.\n\n\n(1)\n\n\n(2)\n\n\n(3)\n\n\n(4)", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0091_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0091_2.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0091_3.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0091_4.jpg" ], "is_multi_img": true, "answer": "1260", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By analyzing the figures, we observe the following: (1) In the first figure, there are \\(1 \\times 6 = 6\\) square units. (2) In the second figure, there are \\((1 + 2) \\times 6 = 18\\) square units. Extending this pattern, the surface area of the 20th figure can be calculated as \\((1 + 2 + \\cdots + 20) \\times 6 = 1260\\) square units. Therefore, the answer is: 1260.\n\n**Key Insight:** This problem primarily examines pattern recognition related to geometric figures. Calculating the surface area based on the figures is crucial for solving the problem." }, { "problem_id": 1696, "question": "The geometric figure shown in Figure 1 is composed of 8 identical small cubes, and its three views are all $2 \\times 2$ squares. If several small cubes are removed, and the views from the front and left sides are as shown in Figure 2, then the number of small cubes that can be removed is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_5daa69e8a399ace673c6g_0098_1.jpg", "batch25-2024_06_17_5daa69e8a399ace673c6g_0098_2.jpg" ], "is_multi_img": true, "answer": "4 or 5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, after removing several small cubes, the shapes seen from the front and left sides are as shown in Figure 2. Therefore, the number of small cubes that can be removed is either 5 or 4.\n\nHence, the answer is: 4 or 5.\n\n[Highlight] This question tests the three-view representation of simple composite solids, primarily assessing students' spatial imagination skills." }, { "problem_id": 1697, "question": "Describe the shape of the object based on the three views, and find the volume of the geometric body.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_65f655dbf2e12f0c83fcg_0094_1.jpg", "batch25-2024_06_17_65f655dbf2e12f0c83fcg_0094_2.jpg", "batch25-2024_06_17_65f655dbf2e12f0c83fcg_0094_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{35}{4} \\pi$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** Based on the three views of the geometric figure, it is determined that the figure is a hollow cylinder. The radius of the larger base circle is 4, the radius of the smaller base circle is 3, and the height is 5. The volume of the cylinder can be calculated using the cylinder volume formula.\n\n**Problem Solution:**\n\nFrom the three views, it can be deduced that the geometric figure is a hollow cylinder.\n\nThe volume of the inner cylinder is calculated as $\\pi\\left(\\frac{3}{2}\\right)^{2} \\times 5=\\frac{45}{4} \\pi$, and the volume of the outer cylinder is $\\pi\\left(\\frac{4}{2}\\right)^{2} \\times 5=20 \\pi$.\n\nTherefore, the volume of the geometric figure is $20 \\pi-\\frac{45}{4} \\pi=\\frac{35}{4} \\pi$.\n\n**Key Insight:** This problem tests the ability to determine the volume of a geometric figure from its three views, assessing students' computational skills and spatial imagination. The key to solving the problem lies in reconstructing the geometric figure from the three views." }, { "problem_id": 1698, "question": "To stack a geometric shape using several small cubes, the shape viewed from the front and from above is shown in the figures below. The maximum number of small cubes required is $\\qquad$.\n\n\n\nViewed from the front\n\n\n\nViewed from above", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0002_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0002_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, we can determine that the bottom layer consists of 6 small cubes.\n\nFrom the front view, we can see that the first, second, and third columns can each have 2 cubes.\n\nTherefore, the maximum number of cubes required is $6 + 3 = 9$.\n\nHence, the answer is: 9.\n\n[Key Insight] This problem tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. The key to solving the problem lies in starting with the top view for consideration." }, { "problem_id": 1699, "question": "A geometric solid is constructed by stacking several small cubes with edge lengths of 1. The figures below show the shapes viewed from the front, left, and top of the geometric solid, respectively. What is the surface area of this geometric solid?\n\n\n\nViewed from the front\n\n\n\nViewed from the left\n\n\n\nViewed from the top", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0003_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0003_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0003_3.jpg" ], "is_multi_img": true, "answer": "22", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: $(4+3+4) \\times 2$\n\n$=11 \\times 2$\n\n$=22$.\n\nTherefore, the surface area of this geometric solid is 22.\n\nHence, the answer is: 22.\n\n[Key Insight] This problem primarily examines the ability to determine a geometric solid from its three-view drawings and the calculation of its surface area. Note: The surface area of the geometric solid is equal to twice the sum of the areas of the three views." }, { "problem_id": 1700, "question": "The following are the three views of a geometric body. It is known that the front view and the left view are two congruent rectangles. If the adjacent sides of the front view are 4 and 2, respectively, and the top view is a circle with a diameter of 2, then the total surface area of this geometric body is . $\\qquad$\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0017_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0017_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0017_3.jpg" ], "is_multi_img": true, "answer": "$10 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, the geometric body is identified as a cylinder,\n\nwith the radius of the base circle being 1 and the height being 4.\n\n$\\therefore$ The volume of this geometric body is: $2 \\pi \\times 1^{2} + 2 \\pi \\times 1 \\times 4 = 10 \\pi$.\n\nTherefore, the answer is: $10 \\pi$.\n\n【Key Insight】This question tests the ability to determine the area from three-view drawings. Mastering the drawing rules of three-view drawings and restoring the geometric features of the object from the three-view drawings is crucial." }, { "problem_id": 1701, "question": "Given several small cubes with each side length of $1 \\mathrm{~cm}$, they are arranged on the table according to the following pattern. The exposed area (excluding the bottom surface) of the 5th figure is equal to $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0019_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0019_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0019_3.jpg" ], "is_multi_img": true, "answer": "75", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: According to the problem, the surface area of each figure is the sum of the surface areas of the cubes in the bottommost layer.\n\nIn the 5th figure, there are 5 layers in total; from top to bottom, the number of cubes in each layer is $1, 3, 6, 10, 15$, totaling 35 cubes.\n\nThe area of the exposed parts (excluding the base) of the 5th figure is equal to $15 \\times 5 = 75\\left(\\mathrm{~cm}^{2}\\right)$.\n\nTherefore, the answer is: 75.\n\n[Key Insight] This problem primarily examines the pattern of graphical changes: identifying changes through specific images, summarizing the patterns, and then using these patterns to solve the problem. It also tests the understanding of three-dimensional views." }, { "problem_id": 1702, "question": "There are some identical cubic wooden blocks on the table. The planar view from the front is shown in Figure (1), and the planar view from the left is shown in Figure (2). To form such a shape, the maximum number of cubic wooden blocks required is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0021_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0021_2.jpg" ], "is_multi_img": true, "answer": "20", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "$\\because$ The planar figure obtained from the front view is as shown in Figure (1), and the planar figure obtained from the left view is as shown in Figure (2),\n\n\n\n(1)\n\n\n\n(2)\n\n$\\therefore$ The original geometric structure is a $4 \\times 4$ grid, with the maximum number of blocks in each layer as shown below:\n\n| 2 | 2 | 1 | 1 |\n| :--- | :--- | :--- | :--- |\n| 1 | 1 | 1 | 1 |\n| 1 | 1 | 1 | 1 |\n| 2 | 2 | 1 | 1 |\n\nTo construct such a figure, a maximum of 20 square wooden blocks are required.\n\nTherefore, the answer is: 20.\n\n【Key Insight】This question tests the ability to view from different directions. Correctly understanding the significance of the figures obtained from different perspectives is key to solving the problem." }, { "problem_id": 1703, "question": "The side area of an object is shown in the figure below, which is equal to $\\qquad$\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0032_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0032_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0032_3.jpg" ], "is_multi_img": true, "answer": "$15 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be determined that the geometric body is a cone, whose axial cross-section is an isosceles triangle with a height of 4,\n\nand the radius of the base circle of the cone is 3, with a generatrix length of 5,\n\n$\\therefore$ the circumference of the base is $6 \\pi$,\n\n$\\therefore$ the lateral area is $\\frac{1}{2} \\times 6 \\pi \\times 5=15 \\pi$.\n\nTherefore, the answer is: $15 \\pi$.\n\n[Insight] This question examines the determination of a geometric body from a three-view drawing and the calculation of a cone. The key to solving this problem lies in correctly understanding the relationship between the lateral development of the cone and the original sector." }, { "problem_id": 1704, "question": "As shown in the figure, the three views of a geometric object are given, with both the front view and the left view being equilateral triangles. Based on the information in the figure, the surface area of the geometric object can be determined to be $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0033_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0033_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0033_3.jpg" ], "is_multi_img": true, "answer": "$3 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be determined that the geometric body is a cone with a base diameter of 2 and a height of $\\sqrt{3}$.\n\n$\\therefore$ The radius of the base is 1,\n\n$\\therefore$ The circumference of the base is $2 \\pi$, and the area is $\\pi$,\n\n$\\because$ The height is $\\sqrt{3}$,\n\n$\\therefore$ The slant height of the cone is $\\sqrt{(\\sqrt{3})^{2}+1^{2}}=2$,\n\n$\\therefore$ The lateral surface area of the cone is $\\frac{1}{2} \\times 2 \\pi \\times 2=2 \\pi$,\n\n$\\therefore$ The total surface area of the cone is $2 \\pi+\\pi=3 \\pi$,\n\nHence, the answer is: $3 \\pi$.\n【Insight】This question examines the identification of common geometric bodies from their three-view drawings and the determination of dimensions such as length, width, and height based on the data in the three-view drawings. The key to solving the problem lies in calculating the surface area of the cone." }, { "problem_id": 1705, "question": "As shown in the figure, the left view and top view of a geometric structure composed of identical small cubes are given. The maximum number of small cubes required is $\\qquad$.\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0035_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0035_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it can be observed that the geometric figure has three rows and two columns. From the side view, it can be seen that the back row has two layers, the middle row has at most two layers, and the front row has a maximum of one layer.\n\n$\\therefore$ The maximum number of small cubes required is $2+2+2+1+1=8$. Therefore, the answer is: 8.\n\n【Insight】This question tests the understanding of three-view drawings. The key to solving it lies in visualizing the geometric figure from two views." }, { "problem_id": 1706, "question": "The side surface area of a geometric body is shown in the figure below.\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0039_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0039_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0039_3.jpg" ], "is_multi_img": true, "answer": "$2 \\pi+4 \\# \\# 4+2 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By observing the three views of the geometric solid, it is identified as a half-cylinder with a diameter of 2 and a height of 1. Therefore, its lateral surface area is calculated as: $\\frac{1}{2} \\times 2 \\pi \\times 2 + 2 \\times 2 = 2 \\pi + 4$.\n\nThus, the answer is: $2 \\pi + 4$.\n\n[Insight] This question tests the knowledge of determining a geometric solid from its three views and calculating the lateral surface area of the solid. The key to solving the problem lies in correctly identifying the solid as a half-cylinder." }, { "problem_id": 1707, "question": "The following figure shows the three views of a geometric shape built with identical rectangular blocks. The geometric shape is composed of $\\qquad$ rectangular blocks in total.\n\n\nFront view\n\n\nLeft view\n\n\nTop view", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0042_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0042_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0042_3.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, we know that the bottom layer consists of 3 rectangular blocks. From the front view and the side view, we understand the following arrangement:\n\n| 2 | |\n| :--- | :--- |\n| 1 | 1 |\n\nTop View\n\nThis diagram shows two layers, with the top layer having 1 rectangular block. Therefore, this geometric shape is constructed from a total of 4 rectangular blocks. Hence, the answer is: 4.\n\n【Highlight】This question tests the understanding of three-view drawings, and mastering spatial imagination is key to solving the problem." }, { "problem_id": 1708, "question": "The three views of a geometric object are shown in the figure. The lateral surface area of the geometric object is $\\qquad$ (the result should retain $\\pi$).\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0046_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0046_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0046_3.jpg" ], "is_multi_img": true, "answer": "$6 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be determined that the geometric body is a cylinder with a base radius of 1 and a height of 3. The lateral development of the cylinder is a rectangle with length and width of $2\\pi$ and $3$, respectively.\n\n$\\therefore$ The lateral area is $2\\pi \\times 3 = 6\\pi$,\n\nHence, the answer is: $6\\pi$.\n\n[Insight] This question examines the three-view drawing of a simple geometric body and the lateral area of a cylinder. The key to solving the problem lies in determining the geometric body based on the three-view drawing." }, { "problem_id": 1709, "question": "The three views of a certain triangular prism are shown in the figure. The front view is a triangle, while the left view and the top view are both rectangles. The area of the top view is twice that of the left view. In the left view, the rectangle $A B C D$ has side lengths $A B=3$. The area of the front view is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0047_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0047_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0047_3.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since the lengths of the front view, top view, and left view are equal, and if the side length $AB=3$ of rectangle $ABCD$ in the left view is given, and the area of the top view is twice that of the left view,\n\nTherefore, the width of the front view is $2AB=6$,\n\nSince the relationship between the front view and the left view indicates that the height of the triangle in the front view is $AB=3$,\n\nTherefore, the area of the front view is $\\frac{1}{2} \\times 6 \\times 3=9$,\n\nHence, the answer is: 9.\n\n[Key Insight] This problem tests the understanding of the relationships between the dimensions in three-view drawings. Mastering the principles of \"length alignment, height leveling, and width equality\" and accurately determining the corresponding side lengths from the three views are crucial for solving the problem." }, { "problem_id": 1710, "question": "To construct a geometric shape using small cubes, the front view and the left view are shown in the figures below. At least $\\qquad$ small cubes are needed to form such a geometric shape.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0050_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0050_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the front view and the side view, the base layer of this geometric figure has at least 3 small cubes, and the second layer has at least 2 small cubes. Therefore, to construct such a geometric figure, a minimum of \\(3 + 2 = 5\\) small cubes are required. Hence, the answer is 5.\n\n[Highlight] This question tests the understanding of three-view drawings. By following the principle of \"using the top view to lay the foundation, the front view to build up, and the side view to adjust for discrepancies,\" one can deduce the number of small cubes needed." }, { "problem_id": 1711, "question": "A geometric structure composed of several identical small cubes is arranged such that the bottom layer has 9 small cubes. Its front view and left view are shown in the figures below. The number of ways to construct this geometric structure is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0052_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0052_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the top view is as follows: Except for the positions of $A, B, C$, which cannot be determined, the numbers of small cubes in the other positions are as shown in the figure.\n\n| $A$ | 1 | $B$ |\n| :--- | :--- | :--- |\n| 1 | 1 | 1 |\n| 3 | 1 | $C$ |\n\nTop View\n\n$\\because$ The geometric body is constructed from several identical small cubes, with the bottom layer consisting of 9 small cubes,\n\n$\\therefore$ $A$ is 1, $B$ is 2, $C$ is 2; or $A$ is 2, $B$ is 2, $C$ is 1; or $A$ is 2, $B$ is 1, $C$ is 2; or $A$ is 2, $B$ is 2, $C$ is 2.\n\nThere are four possible scenarios in total,\n\nHence, the answer is 4.\n\n[Insight] This question tests the understanding of geometric bodies through three-view drawings. The key to solving the problem lies in comprehending the given information and flexibly applying the learned knowledge." }, { "problem_id": 1712, "question": "As shown, the main view and top view of a simple geometric body composed of small cubes of the same size are given. If the maximum number of small cubes that make up this geometric body is $m$, and the minimum number is $n$, then the value of $m-n$ is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0053_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0053_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Since the top view shows 5 squares,\n\nTherefore, the bottom layer has 5 cubes.\n\nFrom the front view, the second layer has at least 2 cubes, and the third layer has at least 1 cube. From the front view, the second layer can have up to 4 cubes, and the third layer can have up to 2 cubes.\n\nTherefore, the minimum number of cubes in this composite geometric figure is \\( n = 5 + 2 + 1 = 8 \\), and the maximum number is \\( m = 5 + 4 + 2 = 11 \\).\n\nThus, \\( m - n = 3 \\).\n\nThe answer is: 3\n\n[Key Insight] The problem involves determining the geometry from views. The key points are: the number of squares in the top view equals the number of cubes in the bottom layer; the minimum number of cubes in the composite figure is the sum of the cubes in the bottom layer and the number of squares in the second and third layers from the front view." }, { "problem_id": 1713, "question": "As shown in the figure, the three views of a geometric structure made up of small cubes are given. If we continue to add the same small cubes on top of the existing structure (without changing the positions of the small cubes in the original structure) to form a large cube, at least $\\qquad$ more small cubes are needed.\n\n\nFront View\n\n\nLeft View\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0054_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0054_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0054_3.jpg" ], "is_multi_img": true, "answer": "54", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it is easy to determine that the bottom layer consists of 7 small cubes, the second layer has 2 small cubes, and the third layer has 1 small cube. Therefore, the total number of geometric units is $7 + 2 + 1 = 10$.\n\nIf these are to be assembled into a large cube, a total of $4 \\times 4 \\times 4 = 64$ small cubes are needed.\n\nThus, an additional $64 - 10 = 54$ small cubes are required.\n\nHence, the answer is: 54.\n\n[Key Insight] This problem tests the understanding of three-view drawings. Mastering the three-view drawing technique is crucial for solving such problems." }, { "problem_id": 1714, "question": "The following shape diagrams are obtained by observing a geometric body composed of small cubes from the front, left, and top views, respectively. Then, the number of small cubes that constitute this geometric body is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from the top", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0055_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0055_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0055_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "According to the problem, the number of small cubes at each position of the constructed geometric figure (viewed from above) is as shown in the figure,\n\n\n\nThe total number of small cubes in this geometric figure is \\(2 + 1 + 1 + 1 = 5\\). Therefore, the answer is: 5.\n\n【Insight】This question primarily tests the students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills. If one masters the mnemonic \"the top view lays the foundation, the front view builds up, and the side view removes the excess,\" it becomes easier to arrive at the answer." }, { "problem_id": 1715, "question": "As shown in the figure, the following are the three views of a geometric body, the volume of which is (expressed as an algebraic expression containing $\\pi$) $\\qquad$ .\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0058_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0058_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0058_3.jpg" ], "is_multi_img": true, "answer": "$4 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, we know that the geometric body is a cylinder with a base diameter of 2 and a height of 4. Therefore, the volume of this geometric body is $\\pi \\times\\left(\\frac{2}{2}\\right)^{2} \\times 4=4 \\pi$.\n\nHence, the answer is: $4 \\pi$.\n\n[Key Insight] This problem primarily tests the ability to determine a geometric body from its three-view drawing. The key to solving it lies in understanding the three-view representations of common geometric bodies." }, { "problem_id": 1716, "question": "If a geometric solid is constructed by stacking several identical small cubes, as shown in the left view and top view, the number of small cubes used in the geometric solid is $m$. Then the minimum value of $m$ is $\\qquad$.\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0064_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0064_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the left view and the top view, the minimum and maximum number of small cubes required can be determined as follows:\n\n\n\nTop view\n\n\n\nTop view\n\n\n\nTop view\n\nThus, the minimum number of small cubes needed to form this geometric shape is 9, and the maximum is 11. Therefore, the minimum value of \\( m \\) is 9.\n\nHence, the answer is: 9.\n\n[Key Insight] This problem primarily tests the ability to determine the geometric shape from its three views. The key is to draw the top view that requires the minimum and maximum number of cubes based on the front and left views." }, { "problem_id": 1717, "question": "As shown are the top view and front view of a geometric object, the maximum number of small cubes that make up this geometric object is $\\qquad$.\n\n\n\nTop View\n\n\n\nFront View", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0067_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0067_2.jpg" ], "is_multi_img": true, "answer": "13", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the top view and the front view, the left column of this geometric figure has 1 small cube, the middle column can have up to 6 small cubes, and the right column can also have up to 6 small cubes.\n\nTherefore, the maximum number of small cubes that make up this geometric figure is 13.\n\nHence, the answer is: 13\n\n[Highlight] This question tests the ability to determine the number of small cubes in a geometric figure constructed from small cubes based on three views. Mastering the use of the top view to determine position and the front or side view to determine quantity is key to solving the problem." }, { "problem_id": 1718, "question": "A geometric structure built with identical small wooden cubes is shown from the front, from the left, and from above in the figures below. The structure is composed of $\\qquad$ small wooden cubes.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0070_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0070_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0070_3.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the top view, with the front view and side view as aids, the total number of small cubes in the first layer and the cubes directly above them can be determined, as shown in the figure:\n\n\n\n## View from above\n\nTherefore: The number of cubes in this geometric figure is: $2+3+2+1+1+1=10$ (cubes)\n\nThus, the answer is: 10.\n\n[Key Insight] This question tests the knowledge of determining geometric figures from three views. The key to solving it lies in understanding the definition of three views, and it is a common type of question in middle school exams." }, { "problem_id": 1719, "question": "As shown, the following images are the views of a geometric structure built with cubic blocks from three different directions, and the number of cubic blocks is $\\qquad$.\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0084_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0084_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0084_3.jpg" ], "is_multi_img": true, "answer": "7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to see that the bottom layer has 5 small cubes. From the front view and the side view, it is known that the second layer has 2 small cubes. Therefore, the total number of small cubes used to build this geometric figure is $5 + 2 = 7$.\n\n\n\nThus, the answer is:" }, { "problem_id": 1720, "question": "A table has several plates of the same size and shape placed on it. When viewed from three directions, the three views are shown in the figures below. The total number of plates on the table is $\\qquad$.\n\n\n\nTop view\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0097_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0097_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0097_3.jpg" ], "is_multi_img": true, "answer": "12", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the problem statement, we have,\n\n$3 + 4 + 5 = 12$.\n\nTherefore, the answer is: 12.\n\n[Key Insight] This question tests the ability to determine a geometric shape from its three views. The key to solving it lies in abstracting the actual shape of the geometric body from different perspectives." }, { "problem_id": 1721, "question": "A geometric structure built with small cubic blocks is shown from the front, left, and top views in the figures below. This geometric structure is composed of $\\qquad$ small cubic blocks.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_6aff45ce53b1253d001bg_0099_1.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0099_2.jpg", "batch25-2024_06_17_6aff45ce53b1253d001bg_0099_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Based on the three-view drawing, it can be determined that there are 2 layers in total. The first layer consists of 4 small squares, and the second layer has 2 small cubes on the left, as shown in the figure below:\n\n| 2 | 1 | 1 |\n| :--- | :--- | :--- |\n| 2 | | |\n| | | |\n| | | |\n| | | |\n\nTherefore, this geometric shape is constructed from 6 small cubes.\n\nHence, the answer is: 6.\n\n【Key Insight】This question tests the ability to reconstruct a geometric shape from its three-view drawing. Accurate interpretation of the drawing is crucial for solving the problem." }, { "problem_id": 1722, "question": "A set of triangles is placed as shown in Figure 1, where $O$ is the midpoint of side $B C(D F)$, and $B C = 20 \\mathrm{~cm}$. As shown in Figure 2, $\\triangle A B C$ is rotated $60^{\\circ}$ clockwise around point $O$, and $A C$ intersects $E F$ at point $G$. The length of $F G$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_703b497dc835336ab828g_0066_1.jpg", "batch25-2024_06_17_703b497dc835336ab828g_0066_2.jpg" ], "is_multi_img": true, "answer": "$(5 \\sqrt{3}-5) \\mathrm{cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, $BC$ intersects $EF$ at point $N$,\n\n\n\nGiven the conditions, $\\angle EDF = \\angle BAC = 90^\\circ$, $\\angle DEF = 60^\\circ$, $\\angle DFE = 30^\\circ$, $\\angle ABC = \\angle ACB = 45^\\circ$, and $BC = DF = 20 \\text{ cm}$.\n\nSince $O$ is the midpoint of side $BC$ (or $DF$), we have: $BO = OC = DO = OF = 10 \\text{ cm}$.\n\nBecause $\\triangle ABC$ is rotated $60^\\circ$ clockwise around point $O$, and $\\angle DFE = 30^\\circ$,\n\n$\\therefore \\angle BOD = \\angle NOF = 60^\\circ$,\n\n$\\therefore \\angle NOF + \\angle F = 90^\\circ$,\n\n$\\therefore \\angle FNO = 180^\\circ - \\angle NOF - \\angle F = 90^\\circ$,\n\n$\\therefore \\triangle ONF$ is a right-angled triangle,\n\n$\\therefore ON = \\frac{1}{2} OF = 5 \\text{ cm}$,\n\n$\\therefore FN = \\sqrt{OF^2 - ON^2} = 5\\sqrt{3} \\text{ cm}$, and $NC = OC - ON = 5 \\text{ cm}$.\n\nSince $\\angle FNO = 90^\\circ$ and $\\angle ACB = 45^\\circ$,\n\n$\\therefore \\angle GNC = 180^\\circ - \\angle FNO = 90^\\circ$,\n\n$\\therefore \\triangle CNG$ is a right-angled triangle,\n\n$\\therefore \\angle NGC = 180^\\circ - \\angle GNC - \\angle ACB = 45^\\circ$,\n\n$\\therefore \\triangle CNG$ is an isosceles right-angled triangle,\n\n$\\therefore NG = NC = 5 \\text{ cm}$,\n$\\therefore FG = FN - NG = (5\\sqrt{3} - 5) \\text{ cm}$.\n\nThus, the answer is: $(5\\sqrt{3} - 5)$.\n\n【Highlight】This problem examines the properties of a right-angled triangle with a 30-degree angle, the determination and properties of an isosceles right-angled triangle, the properties of rotation, and the Pythagorean theorem. The key to solving the problem lies in understanding the properties of a right-angled triangle with a 30-degree angle and the inherent angles in the triangle." }, { "problem_id": 1723, "question": "First, place a rectangle $A B C D$ (with $A B=4$ and $B C=3$) in the Cartesian coordinate system such that point $\\mathrm{A}$ coincides with the origin of the coordinate system, and sides $A B$ and $A D$ lie on the $x$-axis and $y$-axis, respectively (as shown in Figure 1). Then, rotate this rectangle counterclockwise around the origin by $30^{\\circ}$ in the coordinate plane (as shown in Figure 2). The coordinates of point $C$ in Figure 2 are $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_75f7cd293cd2246e2faeg_0012_1.jpg", "batch25-2024_06_17_75f7cd293cd2246e2faeg_0012_2.jpg" ], "is_multi_img": true, "answer": "$\\left(\\frac{4 \\sqrt{3}-3}{2}, \\frac{3 \\sqrt{3}+4}{2}\\right)$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Given that \\( AB = 4 \\) and \\( BC = 3 \\),\n\nTherefore, the coordinates of point \\( C \\) in Figure 1 are \\( (4, 3) \\).\n\nIn Figure 2, let \\( CD \\) intersect the \\( y \\)-axis at point \\( M \\), and draw \\( CN \\perp y \\)-axis at point \\( N \\). Then, \\( \\angle DOM = 30^\\circ \\) and \\( OD = 3 \\),\n\\[\n\\therefore DM = 3 \\cdot \\tan 30^\\circ = \\sqrt{3}, \\quad OM = 3 \\div \\cos 30^\\circ = 2\\sqrt{3},\n\\]\n\\[\n\\text{Thus, } CM = 4 - \\sqrt{3}.\n\\]\nIt is easy to see that \\( \\angle NCM = 30^\\circ \\),\n\\[\n\\therefore MN = CM \\cdot \\sin 30^\\circ = \\frac{4 - \\sqrt{3}}{2}, \\quad CN = CM \\cdot \\cos 30^\\circ = \\frac{4\\sqrt{3} - 3}{2},\n\\]\n\\[\n\\text{Then, } ON = OM + MN = \\frac{3\\sqrt{3} + 4}{2},\n\\]\n\\[\n\\therefore \\text{The coordinates of point } C \\text{ in Figure 2 are: } \\left( \\frac{4\\sqrt{3} - 3}{2}, \\frac{3\\sqrt{3} + 4}{2} \\right).\n\\]\nHence, the answer is \\( \\left( \\frac{4\\sqrt{3} - 3}{2}, \\frac{3\\sqrt{3} + 4}{2} \\right) \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Key Insight】This problem examines the properties of rectangles and rotation issues. The key is to recognize that the degrees of corresponding angles and the lengths of corresponding segments remain unchanged before and after rotation. It is important to construct right triangles for solving the problem." }, { "problem_id": 1724, "question": "As shown in Figure 1, in the right triangle $\\triangle \\mathrm{ABC}$, $\\angle \\mathrm{ACB}=90^{\\circ}$, $\\mathrm{AC}=1$, $\\mathrm{BC}=2$. Place $\\triangle \\mathrm{ABC}$ in the plane rectangular coordinate system such that point $\\mathrm{A}$ coincides with the origin and point $\\mathrm{C}$ lies on the positive $x$-axis. Roll $\\triangle \\mathrm{ABC}$ clockwise (without slipping) as shown in Figure 2. After rolling 2013 times, the coordinates of point $\\mathrm{B}$ are . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_75f7cd293cd2246e2faeg_0069_1.jpg", "batch25-2024_06_17_75f7cd293cd2246e2faeg_0069_2.jpg" ], "is_multi_img": true, "answer": "$(2014+671 \\sqrt{5}, 2)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "According to the triangle rolling pattern, every 3 rolls form a cycle.\n\n$\\because 2013 \\div 3=671$,\n\n$\\therefore$ After rolling 2013 times, the vertical coordinate of point $\\mathrm{B}$ is the same as after the 3rd roll, which is 2.\n\n$\\because \\angle \\mathrm{ACB}=90^{\\circ}, \\mathrm{AC}=1, \\mathrm{BC}=2$,\n\n$\\therefore \\mathrm{OB}=\\sqrt{1^{2}+2^{2}}=\\sqrt{5}$,\n\n$\\therefore$ The sum of the lengths of the triangle's sides is: $1+2+\\sqrt{5}=3+\\sqrt{5}$,\n\nThen, after rolling 2013 times, the horizontal coordinate of point $\\mathrm{B}$ is: $1+671(3+\\sqrt{5})=2014+671 \\sqrt{5}$.\n\nTherefore, the coordinates of point B are: $(2014+671 \\sqrt{5}, 2)$.\n\nHence, the answer is $(2014+671 \\sqrt{5}, 2)$.\n\n【Highlight】This question mainly tests the pattern of point coordinates, and understanding the pattern of point changes is key to solving the problem." }, { "problem_id": 1725, "question": "The main view, left view, and top view of a geometric object are shown in the figures below. The volume of the geometric object is:\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_77dae53a40120b5475d5g_0004_1.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0004_2.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0004_3.jpg" ], "is_multi_img": true, "answer": "$3 \\pi$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the three-view drawing, it can be determined that the geometric shape is a cylinder with a base radius of 1 and a height of 3. Therefore, the volume is calculated as: $\\pi r^{2} h = \\pi \\times 1 \\times 3 = 3\\pi$.\n\nThus, the answer is $3\\pi$.\n\n[Key Insight] This question tests the knowledge of identifying geometric shapes from three-view drawings. The key to solving it lies in understanding the three views of a cylinder and being familiar with the method for calculating its volume." }, { "problem_id": 1726, "question": "As shown in the figure, the main view, left view, and top view of a geometric object composed of several small cubes with equal length, width, and height are given. The total number of small cubes used to form this geometric object is $\\qquad$.\n\n\n\nMain View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_77dae53a40120b5475d5g_0027_1.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0027_2.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0027_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the main view, it is observed that the left and right sides of the combined body are one layer high, while the middle part is two layers high.\n\nFrom the left view, it is seen that the front and back rows are one layer high, with the middle part being two layers high.\n\nFrom the top view, it is noted that there are no small squares at the four corners of the base.\n\nThis leads to the conclusion that the actual object is a small square placed in the middle on the basis of the top view.\n\nTherefore, there are a total of six.\n\n[Key Point] This question tests the application of simple combined body three-view drawings, and the key to solving the problem lies in identifying the connections between the three views." }, { "problem_id": 1727, "question": "As shown in the figure, given the three views of a geometric object, the geometric object is $\\qquad$ .\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_77dae53a40120b5475d5g_0037_1.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0037_2.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0037_3.jpg" ], "is_multi_img": true, "answer": "Four-sided pyramid", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Given that the front view and side view are isosceles triangles, and the top view is a rectangle, it can be deduced that the geometric body is a rectangular pyramid with a rectangular base.\n\n[Key Point] This question tests the knowledge point of determining the shape of a geometric body from its three views, which is a fundamental problem." }, { "problem_id": 1728, "question": "A geometric shape is formed by stacking many small cubes of the same size. The front view and the left view are shown in the figures. To form such a shape, the minimum number of cubes required is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_77dae53a40120b5475d5g_0080_1.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0080_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "According to the three-view drawing, it can be deduced that the second layer has 2 small cubes. Based on the front view and the side view, the first layer must have at least 4 cubes. Therefore, a minimum of 6 cubes are required. (The top view is as follows)\n\n\n\n## Top View\n\nThus, the answer is 6.\n\n[Highlight] This question tests the students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills. If one masters the mnemonic \"Front view covers wildly, side view removes violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1729, "question": "As shown in the figure, the following are the three views of a geometric body. What is this geometric body?\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_77dae53a40120b5475d5g_0095_1.jpg", "batch25-2024_06_17_77dae53a40120b5475d5g_0095_2.jpg" ], "is_multi_img": true, "answer": "hollow cylinder", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "As shown in the figure, two of the three views of this geometric solid are rectangles, and one view is a circular ring, indicating that the geometric solid is a hollow cylinder.\n\nTherefore, the answer is a hollow cylinder.\n\n[Key Insight] This question tests the ability to determine the shape of a geometric solid from its three views, primarily assessing students' spatial imagination and understanding of three-dimensional shapes." }, { "problem_id": 1730, "question": "As shown in the figure, these are the three views of a geometric object. The geometric object is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0003_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0003_2.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0003_3.jpg" ], "is_multi_img": true, "answer": "cube with an edge length of 2", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be determined that the geometric body is a cube with an edge length of 2.\n\nTherefore, the answer is: a cube with an edge length of 2.\n\n[Key Point] This question tests the ability to determine the shape of a geometric body based on its three-view drawing. Mastering the three-view characteristics of common geometric bodies is crucial for solving such problems." }, { "problem_id": 1731, "question": "A geometric figure composed of several identical cubes is placed on the table. The shapes seen from the front and from above are shown in the figures below. Let the number of small cubes that make up this geometric figure be $\\mathrm{n}$. Then the maximum value of $\\mathrm{n}$ is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0010_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0010_2.jpg" ], "is_multi_img": true, "answer": "11 .", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in the figure: the cube is composed of three layers, with the bottom layer having 6 units, the second layer having a maximum of 3 units, and the top layer having a maximum of 2 units. Therefore, the maximum number of units is $n=6+3+2=11$.\n\nHence, the answer is: 11.\n\n[Key Insight] This question tests the composition of a cube, and understanding the positional relationships in the figure is crucial for solving the problem." }, { "problem_id": 1732, "question": "Stack several identical small cubes on a horizontal table to form a large three-dimensional shape. The images below show the views from the front and from above. The minimum number of small cubes required to form this shape is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0014_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0014_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By integrating the top view and the front view, the base layer of this geometric figure consists of 6 small cubes, the second layer has at least 2 small cubes, and the third layer has 1 small cube.\n\nTherefore, the minimum number of small cubes required to form this geometric figure is $6 + 2 + 1 = 9$.\n\nHence, the answer is: 9.\n\n[Insight] This question tests the composition of three-dimensional figures and assesses spatial awareness. Such problems require a comprehensive observation of both the top and front views to determine the number of small cubes in each layer, thereby finding the minimum value." }, { "problem_id": 1733, "question": "The main view and top view of a geometric object composed of several identical small cubes are shown in the figures. The letters and numbers in the grid of the top view represent the number of small cubes at that position. Therefore, $x+y=$ $\\qquad$ .\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0015_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0015_2.jpg" ], "is_multi_img": true, "answer": "4 or 5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it can be seen that the combined structure has two rows and two columns.\n\nIn the front row of the left column, there are two cubes. Combining this with the front view, it is evident that the left column has a maximum stack of 2 cubes, hence $\\mathrm{x}=$ 1 or 2.\n\nFrom the front view of the right column, it is clear that the right column has a maximum stack of 3 cubes, so $\\mathrm{y}=3$.\n\nTherefore, $x+y=4$ or $x+y=5$.\n\nThus, the answer is: 4 or 5.\n\n【Key Insight】This problem tests the ability to determine the composition of a geometric body based on its three views and the spatial imagination of the three views of the geometric body. Pay attention to identifying the maximum number of small cubes in each column of the geometric body's front view." }, { "problem_id": 1734, "question": "A geometric figure is constructed using identical small cubes, and the views from three different directions are shown in the figures below. The total number of such small cubes needed to construct this geometric figure is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0019_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0019_2.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0019_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By synthesizing the front view, top view, and left view, it is determined that there are 4 cubes on the bottom layer and 1 cube on the second layer. Therefore, the total number of small cubes used to construct this geometric figure is 5.\n\nHence, the answer is: 5.\n\n[Insight] This question tests students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills. If one masters the mnemonic \"the top view lays the foundation, the front view builds wildly, and the left view removes violations,\" it becomes easier to arrive at the answer." }, { "problem_id": 1735, "question": "In a warehouse, there are several identical small cubic crates stacked together. The warehouse manager has drawn the three views of this stack of crates, as shown in the figures below. How many cubic crates are there in total?\n\n\n\nLeft View\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0022_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0022_2.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0022_3.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the three-view drawing, it can be seen that the geometric body has 3 layers. The top view indicates that there are 6 cubes in the first layer. The front view and side view show that there are 2 cubes in the second layer and 1 cube in the third layer, totaling 9 cubes. Therefore, there are 9 small cube containers in this stack.\n\n[Key Point] This question mainly examines the three-view drawing of a geometric body. Generally, starting from the top view, the number of small cubes is determined based on the principles of \"length alignment, height leveling, and width equality.\"" }, { "problem_id": 1736, "question": "The main view and top view of a geometric shape formed by small cubes of the same size are shown in the figure below. The minimum number of small cubes required to form this geometric shape is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0028_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0028_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "According to the problem statement:\n\n\n\nFront view\n\n\n\nSide view or\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nFront view\n\n\n\nTop view\n\nThe minimum number of small cubes needed to construct this geometric figure is \\(1 + 1 + 1 + 2 + 1 = 6\\) (cubes).\n\nTherefore, the answer is 6.\n\n**Insight:** This question tests the ability to determine a geometric figure from its three views. The key to solving this problem lies in correctly representing the numbers on the top view." }, { "problem_id": 1737, "question": "A geometric shape is formed by several small cubes of the same size. The shapes of the geometric shape as seen from above and from the left are shown in the figures below. The minimum number of small cubes in this geometric shape is $\\qquad$.\n\n\n\nView from above\n\n\n\nView from the left", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0034_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0034_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "To construct such a geometric shape, a minimum of $4+1=5$ small cubes is required, and a maximum of $4+2=6$ small cubes is needed. Therefore, the answer is 5.\n\n【Highlight】This question tests students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Use the top view as the foundation, build wildly with the front view, and dismantle violations with the side view\" can make it easier to arrive at the answer." }, { "problem_id": 1738, "question": "As shown in the figure, the following are the three views of a geometric body, then the lateral area of this geometric body is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0039_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0039_2.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0039_3.jpg" ], "is_multi_img": true, "answer": "36", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By observing the three views of the geometric solid, it can be determined that the solid is a triangular prism. A triangular prism has three lateral faces, each of which is a rectangle. Therefore, the lateral surface area of this geometric solid is: $3 \\times 4 \\times 3 = 36 \\mathrm{~cm}^{2}$.\n\nThus, the answer is 36.\n\n[Key Insight] This question tests the ability to identify a geometric solid through its three views. Mastering the three views of geometric solids and understanding the characteristics of prisms are crucial for solving such problems." }, { "problem_id": 1739, "question": "A geometric shape composed of several small cubes of the same size is shown from the front and from above as illustrated. The maximum number of small cubes that can form this geometric shape is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0046_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0046_2.jpg" ], "is_multi_img": true, "answer": "11", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer consists of 6 small cubes, and the second layer can have up to 5 small cubes. Therefore, the maximum number of small cubes needed to construct this geometric figure is $6 + 5 = 11$. Hence, the answer is 11.\n\n[Highlight] This problem tests students' mastery and flexible application of the three-view drawing, as well as their spatial imagination skills. Understanding the mnemonic \"top view lays the foundation, front view builds up, and side view removes the excess\" can make it easier to arrive at the answer." }, { "problem_id": 1740, "question": "The front view and left view of a cuboid are shown in the figure (units: $\\mathrm{cm}$), then the area of the top view of the cuboid is equal to $\\mathrm{cm}^{2}$.\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0053_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0053_2.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the top view of the cube is a rectangle with a length of \\(6 \\mathrm{~cm}\\) and a width of \\(4 \\mathrm{~cm}\\). The area is calculated as \\(6 \\times 4 = 24 \\mathrm{~cm}^2\\).\n\nTherefore, the answer is 24.\n[Key Insight] This problem tests the knowledge of determining geometric shapes from three-view drawings. The key to solving this problem lies in obtaining the side lengths of the rectangle in the top view from the given views." }, { "problem_id": 1741, "question": "A geometric shape is constructed using several identical small cubes. The shape of the geometric shape as seen from the front and from above is shown in the figures below. The geometric shape is composed of at least $\\qquad$ small cubes.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0058_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0058_2.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, it can be determined that there are 5 blocks. From the front view, it can be determined that there are two layers on the left side, with at least one block, so there are 6 blocks in total.\n\nTherefore, the answer is 6 blocks.\n\n[Key Point] This question tests the knowledge of the three views of a geometric solid. By analyzing and imagining the original figure from two or three views of the geometric solid, it is more challenging than drawing the \"three views\" from the original figure, but it enhances the understanding of spatial graphics. If there are only two views, it is possible to uniquely determine the arrangement of the figure." }, { "problem_id": 1742, "question": "Below is the three views of a geometric solid (Note: views seen from the front, top, and left sides), the number of small cubes that make up this solid is $\\qquad$ .\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0060_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0060_2.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0060_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the first layer consists of 4 small cubes. Based on the front view and the side view, it is known that the second layer has 1 small cube, making a total of 5 cubes.\n\nTherefore, the answer is 5.\n\n[Highlight] This question tests students' understanding and flexible application of the three-view drawings, as well as their spatial imagination skills. Mastering the mnemonic \"Top view lays the foundation, front view builds up, side view removes the excess\" can make it easier to arrive at the answer." }, { "problem_id": 1743, "question": "In a warehouse, there are several identical cubic crates stacked together. The warehouse manager observes the stack of crates from different directions, as shown in Figure 6. The total number of crates in the stack is $\\qquad$.\n\n\n\nView from the left\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0067_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0067_2.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0067_3.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the comprehensive three-view drawing, it can be deduced that the bottom layer of this stack of cargo boxes has \\(3 + 2 + 1 = 6\\) boxes, the second layer has 2 boxes, and the third layer should have 1 box.\n\nTherefore, this stack of cubic cargo boxes has a total of \\(6 + 2 + 1 = 9\\) boxes.\n\n【Highlight】This question tests the ability to determine the geometry from three-view drawings, and the key to solving it lies in mastering the skill of interpreting three-view drawings to understand the geometric structure." }, { "problem_id": 1744, "question": "As shown, the three views of a geometric structure formed by several identical small cubes are given. The number of small cubes used to construct this geometric structure is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_7f2203197b4adb04186cg_0091_1.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0091_2.jpg", "batch25-2024_06_17_7f2203197b4adb04186cg_0091_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the front view and the left view, the numbers can be annotated on the top view as follows:\n\nThe front view has three columns, with the number of blocks in each column being: 1, 2, 2;\n\nThe left view has two columns, with the number of blocks in each column being: 2, 1;\n\nThe top view has three columns, with the number of blocks in each column being: 1, 2, 2;\n\nTherefore, the total number of blocks is $1+2+2=5$.\n\nHence, the answer is 5.\n\n[Highlight] This problem tests the ability to determine the geometry from three views. By \"building the foundation with the top view\" and combining the front and left views to determine the height of each column, it becomes easy to ascertain the number of small cubes." }, { "problem_id": 1745, "question": "Two right-angled triangles are overlapped, with $\\angle \\mathrm{A}=30^{\\circ}, \\angle \\mathrm{C}=45^{\\circ}$. $\\triangle \\mathrm{COD}$ is fixed, while $\\triangle \\mathrm{AOB}$ rotates counterclockwise around point $\\mathrm{O}$ by $\\alpha^{\\circ}\\left(0^{\\circ}<\\alpha<180^{\\circ}\\right)$, such that at least one pair of sides of the two triangles are perpendicular to each other. Then $\\alpha=$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_8a3c6106076a39639699g_0019_1.jpg", "batch25-2024_06_17_8a3c6106076a39639699g_0019_2.jpg" ], "is_multi_img": true, "answer": "$45^{\\circ}$ or $60^{\\circ}$ or $90^{\\circ}$ or $105^{\\circ}$ or $135^{\\circ}$ or $150^{\\circ}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "As shown in the figure:\n\n(1) When $\\alpha$ is $45^{\\circ}$, $\\mathrm{BO} \\perp \\mathrm{CD}$;\n\n\n\n(2) When $\\alpha$ is $60^{\\circ}$, $\\mathrm{AB} \\perp \\mathrm{OD}$;\n\n\n\n(3) When $\\alpha$ is $90^{\\circ}$, $\\mathrm{AO} \\perp \\mathrm{DO}$;\n\n\n\n(4) When $\\alpha$ is $105^{\\circ}$, $\\mathrm{AB} \\perp \\mathrm{CD}$;\n\n\n\n(5) When $\\alpha$ is $135^{\\circ}$, $\\mathrm{AO} \\perp \\mathrm{CD}$;\n\n\n\n(6) When $\\alpha$ is $150^{\\circ}$, $\\mathrm{AB} \\perp \\mathrm{CO}$.\n\n\n\nIn summary, when rotating $\\triangle \\mathrm{AOB}$ counterclockwise around point $\\mathrm{O}$ by $\\alpha$ degrees $\\left(0^{\\circ}<\\alpha<180^{\\circ}\\right)$, and when $\\alpha$ is $45^{\\circ}$, $60^{\\circ}$, $90^{\\circ}$, $105^{\\circ}$, $135^{\\circ}$, or $150^{\\circ}$, at least one pair of sides of the two triangles are perpendicular.\n\nTherefore, the answer is $45^{\\circ}$, $60^{\\circ}$, $90^{\\circ}$, $105^{\\circ}$, $135^{\\circ}$, or $150^{\\circ}$.\n\n【Key Insight】This problem primarily examines the properties of rotation: the angle between corresponding points and the rotation center equals the rotation angle. Applying a categorical approach to graphical analysis is crucial for solving the problem." }, { "problem_id": 1746, "question": "Figure 2 shows the three views of the cuboid in Figure 1. If $\\mathrm{S}$ represents the area, $S_{\\text{front}}=x^{2}+2 x, S_{\\text{left}}=x^{2}+x$, find $S_{\\text{top}}$.\n\n\n\nFront View\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_8c2fde83e26b65584ff2g_0085_1.jpg", "batch25-2024_06_17_8c2fde83e26b65584ff2g_0085_2.jpg" ], "is_multi_img": true, "answer": "$\\mathrm{S}_{\\text {㹸 }}=x^{2}+3 x+2$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Since \\( \\mathrm{S}_{\\text{front}} = x^{2} + 2x = x(x + 2) \\), and \\( \\mathrm{S}_{\\text{left}} = x^{2} + x = x(x + 1) \\),\n\nTherefore, the length of the top view is: \\( x + 2 \\), and the width is: \\( x + 1 \\),\n\nThus, the area of the top view is: \\( \\mathrm{S}_{\\text{top}} = (x + 2)(x + 1) = x^{2} + 3x + 2 \\).\n\n【Insight】This problem mainly examines the calculation of side lengths and areas from three views and the mixed operations of polynomials. The key to solving the problem lies in visualizing the shapes of the front, top, and left side views of the geometric body, as well as its length, width, and height." }, { "problem_id": 1747, "question": "As shown in Figure 1, the right triangle $Rt \\triangle ABC$ is rotated counterclockwise by $180^{\\circ}$ around point $A$. During this process, the corresponding points of $B$ and $C$ are $B'$ and $C'$ respectively. Connect $B'C'$, and let the rotation angle be $x^{\\circ} (0 \\leq x \\leq 180)$, $y = B'C'^2$. The graph of the function relationship between $y$ and $x$ is shown in Figure 2. When $x = 150$, the value of $y$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_8dcf73293476a9a79a1dg_0070_1.jpg", "batch25-2024_06_17_8dcf73293476a9a79a1dg_0070_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: From Figures 1 and 2, it can be observed that when \\( x = 0 \\),\n\n\\[ y = B'C^2 = BC^2 = 5, \\]\nat which point \\( B' \\) coincides with point \\( B \\);\n\nwhen \\( x = 90 \\), \n\\[ y = B'C^2 = 1, \\]\nat which point \\( B' \\) lies on \\( AC \\),\n\n\\[ \\therefore \\angle BAC = 90^\\circ, \\quad B'C = 1, \\]\nor \\( B'C = -1 \\) (which is not feasible and thus discarded),\n\nBy rotation, \\( AB' = AB \\),\n\n\\[ \\therefore AC - AB = AC - AB' = 1, \\]\n\\[ \\therefore AC = AB + 1, \\]\n\\[ \\because AB^2 + AC^2 = BC^2, \\]\n\\[ \\therefore AB^2 + (AB + 1)^2 = 5, \\]\nSolving gives \\( AB = 1 \\) or \\( AB = -2 \\) (which is not feasible and thus discarded),\n\n\\[ \\therefore AC = 2, \\]\n\nWhen \\( x = 150 \\), as shown in Figure 3, then \\( \\angle BAB' = 150^\\circ \\),\n\n\\[ \\therefore \\angle CAB' = \\angle BAB' - \\angle BAC = 150^\\circ - 90^\\circ = 60^\\circ, \\]\n\nDrawing \\( B'D \\perp AC \\) at point \\( D \\), then \\( \\angle ADB' = \\angle CDB' = 90^\\circ \\),\n\n\n\nFigure 3\n\n\\[ \\therefore \\angle AB'D = 90^\\circ - \\angle CAB' = 90^\\circ - 60^\\circ = 30^\\circ, \\]\n\\[ \\because AB' = AB = 1, \\]\n\\[ \\therefore AD = \\frac{1}{2} AB' = \\frac{1}{2}, \\]\n\\[ \\therefore CD = AC - AD = 2 - \\frac{1}{2} = \\frac{3}{2}, \\quad B'D^2 = AB'^2 - AD^2 = 1^2 - \\left(\\frac{1}{2}\\right)^2 = \\frac{3}{4}, \\]\n\\[ \\therefore y = B'C^2 = CD^2 + B'D^2 = \\left(\\frac{3}{2}\\right)^2 + \\frac{3}{4} = 3 \\]\n\nHence, the answer is: 3.\n\n【Insight】This problem emphasizes the examination of coordinates and graphics, function graphs of moving points, properties of rotation, the complementary nature of the two acute angles in a right triangle, properties of a 30°-60°-90° triangle, the Pythagorean theorem, and methods for solving quadratic equations. The key to solving the problem lies in using the Pythagorean theorem to derive the equations \\( AB = 1 \\) and \\( AC = 2 \\)." }, { "problem_id": 1748, "question": "As shown in the figure, there are three views of a three-dimensional figure composed of identical small cubes. How many small cubes are there in this three-dimensional figure?\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0001_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0001_2.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0001_3.jpg" ], "is_multi_img": true, "answer": "5.", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Question Analysis: As shown in the figure: From the left view, it can be determined that the figure has 3 rows. From the top view, it can be determined that the figure has 2 columns.\n\nFrom the main view, it can be determined that the highest part of the figure consists of two cubes.\n\nTherefore, the small cubes that make up this three-dimensional figure total 5.\n\nKey Point: Determining the geometric shape from the three views." }, { "problem_id": 1749, "question": "The lateral surface area of the geometric object, given its three views and relevant dimensions, is $\\qquad$.\n\n\n\nLeft View\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0005_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0005_2.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0005_3.jpg" ], "is_multi_img": true, "answer": "$15 \\pi$.", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "Question Analysis: From the three views, it can be determined that this geometric figure is a cone with a generatrix length of 5 and a height of 4.\n\n$\\therefore \\mathrm{a}=2 \\sqrt{5^{2}-4^{2}}=6, \\therefore$ the radius of the base is $3, \\therefore$ the lateral area is: $\\pi \\times 5 \\times 3=15 \\pi$.\nExam Point:" }, { "problem_id": 1750, "question": "The front view and top view of a cuboid are shown in the figure, then the area of its left view is $\\qquad$\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0008_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0008_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Based on the three-view drawing, the length of the cuboid is 4, the width is 3, and the height is 1.\n\nTherefore, the area of the left view $=$ width $\\times$ height $=3$.\n\nHence, the answer is: 3\n\n[Key Insight] This question tests the ability to calculate area using three-view drawings. Understanding the relationships between the three views is crucial for solving the problem." }, { "problem_id": 1751, "question": "A geometric object's three views are shown in the figure, so this geometric object is $\\qquad$ .\n\n\n\nFront view\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0013_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0013_2.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0013_3.jpg" ], "is_multi_img": true, "answer": "triangular prism", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Question Analysis: As shown in the figure, the method of elimination can be used to solve the problem based on the knowledge of three-view drawings.\n\nSolution: Given that the top view is a triangle, and both the front view and the side view are rectangles, it can be deduced that the geometric figure is a triangular prism. Therefore, the answer is a triangular prism.\n\nExam Point: Determining the geometric figure from three-view drawings." }, { "problem_id": 1752, "question": "The three views of a geometric structure built with several small cubes are as follows. Therefore, the number of small cubes used to build this geometric structure is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0015_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0015_2.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0015_3.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Test Question Analysis: Based on the three-view drawing, the small cubes in this geometric figure total 8.\n\nKey Point: Three-view Drawing" }, { "problem_id": 1753, "question": "To form a geometric structure using identical small cubes such that the shape of the structure as seen from the front and from above matches the figures shown, the minimum number of small cubes required to construct this geometric structure is $\\qquad$ .\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0027_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0027_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** From the top view, the number and shape of the smallest cubes at the bottom layer can be determined. From the front view, the number of layers and the number of small cubes in each layer can be observed, allowing the calculation of the total number of cubes.\n\n**Solution:** \n$\\because$ The top view shows 5 squares, \n$\\therefore$ The bottom layer consists of 5 cubes. \n\nFrom the front view, the second layer has at least 2 cubes, and the third layer has at least 1 cube. \nFrom the front view, the second layer has at most 4 cubes, and the third layer has at most 2 cubes. \n\n$\\therefore$ The composite geometric figure has a minimum of $5 + 2 + 1 = 8$ cubes and a maximum of $5 + 4 + 2 = 11$ cubes. \nThus, the answer is **8**. \n\n**Exam Focus:** Determining the geometric figure from its three views." }, { "problem_id": 1754, "question": "As shown in the figure, the main view and top view of a geometric object formed by small cubes with equal edge lengths are given. Based on these views, it can be determined that at least $\\qquad$ small cubes are required to form this geometric object.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0035_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0035_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it can be seen that there are 5 small cubes forming the base of this geometric figure. From the front view, it is known that the second layer has at least 2 cubes, and the third layer has at least 1 cube. Therefore, the minimum number of small cubes that make up this geometric figure is \\(5 + 2 + 1 = 8\\).\n\n**Key Insight:** This problem primarily tests the student's ability to determine the geometric figure from its three views, and also reflects the assessment of spatial imagination skills. To solve such problems, one should remember the mnemonic: \"The top view lays the foundation, the front view builds up wildly, and the side view removes the violations.\"" }, { "problem_id": 1755, "question": "As shown in the figure, the main view and top view of a geometric shape constructed with several small cubes are provided. The minimum number of small cubes required to construct this geometric shape is $\\mathrm{m}$, and the maximum number is $\\mathrm{n}$. Therefore, $2 \\mathrm{~m}-\\mathrm{n}=$ $\\qquad$\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0036_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0036_2.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Question Analysis: Based on the three-view drawing, we have: $\\mathrm{m}=6, \\mathrm{n}=8$, thus $2 \\mathrm{~m}-\\mathrm{n}=12-8=4$.\n\nKey Point: Three-view Drawing" }, { "problem_id": 1756, "question": "The front view and top view of a geometric shape composed of several small cubes are shown in the figure below. The number of small cubes used to construct this geometric shape could be $\\qquad$.\n\n\n\nFront view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0045_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0045_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By synthesizing the front view and the top view, the minimum number of small cubes on the bottom layer is 5, on the second layer is 3, and on the third layer is 1. Therefore, the minimum number of small cubes required to construct this geometric figure is 9. Hence, the answer is 9." }, { "problem_id": 1757, "question": "The main view, left view, and top view of a geometric shape composed of several small cubes with equal side lengths are shown in the figure. The number of small cubes that make up this geometric shape is $\\qquad$.\n\n\n\nLeft view\n\n\n\nMain view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0053_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0053_2.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0053_3.jpg" ], "is_multi_img": true, "answer": "6", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the comprehensive three-view drawing, it can be deduced that the base layer of this geometric shape should consist of \\(2 + 1 + 1 + 1 = 5\\) small cubes.\n\nThe second layer should have 1 small cube.\n\nTherefore, the total number of small cubes used to construct this geometric shape is \\(5 + 1 = 6\\).\n\nHence, the answer is 6.\n\n[Highlight] This question tests the knowledge of determining a geometric shape from its three-view drawings. Based on the three-view drawings, the front view and the top view of the geometric shape can determine that it consists of two rows and three columns, thus allowing us to deduce the number of small cubes in the geometric shape." }, { "problem_id": 1758, "question": "The number of small cubes that make up the geometric figure, as seen from different directions, is shown in the images below. Therefore, the number of small cubes that make up this geometric figure could be $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0062_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0062_2.jpg" ], "is_multi_img": true, "answer": "8 or 9 or 10", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Since the top view consists of 4 squares, the bottom layer must have 4 cubes. From the front view, the second layer has at least 2 cubes, and the third layer also has at least 2 cubes. Additionally, from the front view, the second layer can have at most 3 cubes, and the third layer can also have at most 3 cubes. Therefore, the composite geometric figure has a minimum of \\(4 + 2 + 2 = 8\\) cubes and a maximum of \\(4 + 3 + 3 = 10\\) cubes. Thus, the possible number of cubes is 8, 9, or 10.\n\nKey Insight: This problem tests the ability to determine the geometry of an object from its three views. The front view helps distinguish the number of layers from top to bottom and left to right, while the top view helps determine the left-right and front-back positions of the object. By synthesizing this information, one can calculate the minimum and maximum number of small cubes." }, { "problem_id": 1759, "question": "As shown in the figure, the top view and left view of a geometric structure formed by several identical small cubes are given. The number of small cubes is . $\\qquad$\n\n\n\nTop View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0068_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0068_2.jpg" ], "is_multi_img": true, "answer": "5 or 6 or 7", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer has 4 cubes. From the left view, it is easy to determine that the second layer has a maximum of 3 cubes and a minimum of 1 cube.\n\nTherefore, the number of small cubes could be 5, 6, or 7.\n\nHence, the answer is 5, 6, or 7." }, { "problem_id": 1760, "question": "The front view and top view of a cuboid are shown in the figures below. The surface area of this cuboid is:\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0079_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0079_2.jpg" ], "is_multi_img": true, "answer": "88", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the front view, we can see that the length of the cuboid is 6, and the height is 4. From the top view, we can see that the width of the cuboid is 2. Therefore, the surface area of the cuboid is: $2 \\times (6 \\times 4 + 6 \\times 2 + 4 \\times 2) = 88$. Hence, the answer is 88." }, { "problem_id": 1761, "question": "As shown in the figure, the front view and the top view of a cuboid are given. Based on the given data (in cm), the volume of the cuboid is $\\qquad$ $\\mathrm{cm}^{3}$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0086_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0086_2.jpg" ], "is_multi_img": true, "answer": "75", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: By observing its views, we can determine that the geometric body is a cube with a length of 5, a width of 3, and a height of 5. Therefore, its volume is calculated as: $5 \\times 5 \\times 3 = 75$.\n\nThus, the answer is: 75.\n\n[Highlight] This question tests the ability to determine a geometric body from its three views, and the key to solving it lies in remembering the method for calculating the volume of a cube." }, { "problem_id": 1762, "question": "Given that an object is composed of $\\mathrm{x}$ identical cubes stacked together, its front view and left view are shown in the figures below. Then the maximum value of $\\mathrm{x}$ is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0093_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0093_2.jpg" ], "is_multi_img": true, "answer": "11", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "By synthesizing the front view and the left view, the base layer can have a maximum of \\(3 \\times 3 = 9\\) small cubes, and the second layer can have a maximum of 2 small cubes. Therefore, the maximum value of \\(\\mathrm{x}\\) should be \\(9 + 2 = 11\\).\n\nThus, the answer is 11.\n\nKey Point: This question tests the understanding and application of three-view drawings and spatial imagination. Although the top view is not provided in this problem, the condition that \\(\\mathrm{x}\\) takes the maximum value indirectly indicates the scenario of the top view." }, { "problem_id": 1763, "question": "As shown in Figure (1), there is a small cube with an edge length of $a$. Figures (2) and (3) are composed of several such identical small cubes arranged in a similar manner. Continuing to arrange them in this way, the layers from top to bottom are called the first layer, the second layer, ..., the $n$-th layer. The number of small cubes in the $n$-th layer is denoted as $s$ (Hint: In the first layer, $s=1$; in the second layer, $s=3$). Then in the $n$-th layer, $s=$ $\\qquad$ . (Express in terms of $n$)\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch25-2024_06_17_9131a80dd800029c5b30g_0098_1.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0098_2.jpg", "batch25-2024_06_17_9131a80dd800029c5b30g_0098_3.jpg" ], "is_multi_img": true, "answer": "$\\frac{1}{2} n(n+1)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: \n\n$\\because$ The first figure has 1 layer, totaling 1 small cube,\n\nThe second figure has 2 layers, with the number of cubes in the second layer being $1+2$,\n\nThe third figure has 3 layers, with the number of cubes in the third layer being $1+2+3$,\n\n$\\therefore$ For the $n$th layer, $s=1+2+3+\\ldots+n=\\frac{1}{2} n(n+1)$.\n\nHence, the answer is $\\frac{1}{2} n(n+1)$.\n\n【Insight】This question examines the regularity of changes in figures; understanding the pattern of the number of cubes in the $n$th layer is key to solving this problem." }, { "problem_id": 1764, "question": "The front view and the left view of a cuboid are shown in the figures below. The surface area of this cuboid is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFront View\n\n\n\nLeft View", "input_image": [ "batch25-2024_06_17_9e565a461277c7b1b67cg_0001_1.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0001_2.jpg" ], "is_multi_img": true, "answer": "94", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "From the front view, we can determine that the length and height of this rectangular prism are 5 and 3, respectively.\n\nFrom the side view, we can determine that the width and height of this rectangular prism are 4 and 3, respectively.\n\nTherefore, the length, width, and height of this rectangular prism are 5, 4, and 3, respectively.\n\nThus, the surface area of this rectangular prism is calculated as \\(5 \\times 3 \\times 2 + 4 \\times 3 \\times 2 + 5 \\times 4 \\times 2 = 94 \\text{ cm}^2\\).\n\nHence, the answer is 94.\n\n[Insight] This problem examines the characteristics of a rectangular prism through two types of views. This type of question frequently appears in middle school exam papers. The knowledge used in this problem is: the front view primarily reflects the length and height of the object, and the side view primarily reflects the width and height of the object." }, { "problem_id": 1765, "question": "Figure 1 shows a tall building in the shape of a regular hexagonal prism. The shaded area in Figure 2 represents the top view of the building. $P$, $Q$, $M$, and $N$ indicate the activity areas on the ground where Xiaoming is located. Xiaoming wants to see three sides of the building at the same time. He should be in the $\\qquad$ area (fill in the area code).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch25-2024_06_17_9e565a461277c7b1b67cg_0053_1.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0053_2.jpg" ], "is_multi_img": true, "answer": "$Q$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "When Xiao Ming is in the $\\mathrm{P}$ and $\\mathrm{N}$ regions, he can only see one side of the building.\n\nWhen Xiao Ming is in the $M$ region, he can see two sides of the building.\n\nSince the $\\mathrm{Q}$ region is the common area for viewing three sides of the building, Xiao Ming can see three sides of the building when he is in the $\\mathrm{Q}$ region.\n\nTherefore, the answer is: $\\mathrm{Q}$.\n\n[Insight] This question examines the three-view drawing of a geometric solid and the definition of a blind spot." }, { "problem_id": 1766, "question": "As shown in Figure (1), there is a building in the shape of a regular pentagonal prism. Figure (2) is its top view. Xiao Ming observes the building from the ground. When he can only see one side of the building, his activity area has $\\qquad$ sides.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_9e565a461277c7b1b67cg_0058_1.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0058_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Based on the shape of a regular pentagonal prism building and its top view, it can be determined that:\n\nWhen only one side of the building is visible, the activity area is an equilateral triangle with each side of the regular pentagon as one of its sides.\n\n$\\therefore$ When only one side of the building is visible, there are 5 activity areas, each forming an equilateral triangle with one side of the pentagon.\n\nTherefore, the answer is 5.\n\n[Key Insight] This question primarily examines the concepts of viewpoint, viewing angle, and blind spots. The key to solving the problem lies in understanding that when only one side of the building is visible, the blind spot is an equilateral triangle formed by one side of the regular pentagon." }, { "problem_id": 1767, "question": "As shown in the images, the left view and top view of a simple geometric object composed of small cubes of the same size are provided. The minimum number of small cubes required to form this geometric object is $\\qquad$.\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9e565a461277c7b1b67cg_0075_1.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0075_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, it can be seen that there are 4 small cubes forming the base of this geometric figure. From the side view, it is known that the second layer has at least 1 cube.\n\nTherefore, the minimum number of small cubes that make up this geometric figure is: $4 + 1 = 5$ (cubes). Hence, the answer is 5." }, { "problem_id": 1768, "question": "As shown, the main view and top view of a geometric body composed of several identical small cubes are given. The minimum number of small cubes required to form this geometric body is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_9e565a461277c7b1b67cg_0080_1.jpg", "batch25-2024_06_17_9e565a461277c7b1b67cg_0080_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer consists of 6 cubes. From the front view, the second layer has at least 2 cubes, and the third layer has at least 1 cube. Therefore, the total number of cubes is 9.\n\nHence, the answer is: 9.\n\n[Key Insight] This question tests the understanding that the number of small squares in the top view corresponds to the number of cubes in the bottom layer, and the number of small squares in the second and third layers of the front view represents the minimum number of cubes in those respective layers." }, { "problem_id": 1769, "question": "A certain geometric object is composed of a cylinder and a rectangular prism. The views from the front and top are shown in the figures below. Calculate the volume of the geometric object. (Take $\\pi$ as 3.14, units: $\\mathrm{cm}$)\n\n\n\nView from the front\n\n\n\nView from the top", "input_image": [ "batch25-2024_06_17_9f2339989f7dc88d1ba7g_0031_1.jpg", "batch25-2024_06_17_9f2339989f7dc88d1ba7g_0031_2.jpg" ], "is_multi_img": true, "answer": "$40048 \\mathrm{~cm}^{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: The upper part of the geometric figure is a cylinder with a base diameter of $20 \\mathrm{~cm}$ and a height of $32 \\mathrm{~cm}$; the lower part is a rectangular prism with length, width, and height of $30 \\mathrm{~cm}$, $25 \\mathrm{~cm}$, and $40 \\mathrm{~cm}$ respectively. Therefore, the volume of the geometric figure is\n\n$3.14 \\times\\left(\\frac{20}{2}\\right)^{2} \\times 32 + 30 \\times 25 \\times 40 = 40048\\left(\\mathrm{~cm}^{3}\\right)$\n\nAnswer: The volume of the geometric figure is $40048 \\mathrm{~cm}^{3}$.\n\n【Key Insight】This problem examines the three-dimensional views of geometric figures and the calculation of volumes of common geometric shapes. The key to solving the problem lies in determining the diameter and height of the cylinder and the length, width, and height of the rectangular prism from the given figure." }, { "problem_id": 1770, "question": "When the large square in Figure (1) is rotated clockwise around its center by at least $\\qquad$ degrees, it can become Figure (2).\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch25-2024_06_17_b198ac3e13736d7a3336g_0066_1.jpg", "batch25-2024_06_17_b198ac3e13736d7a3336g_0066_2.jpg" ], "is_multi_img": true, "answer": "270", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure: When the large square in Figure (1) is rotated clockwise around its center by at least 270 degrees, it can transform into Figure (2).\n\n\n\n(1)\n\n\n\n(2)\n\nTherefore, the answer is 270." }, { "problem_id": 1771, "question": "As shown, the first pattern is composed of square tiles of the same size in black and white colors. The second and third patterns can be seen as the first pattern obtained through translation. Then, in the $\\mathrm{n}$th pattern, the number of black square tiles required is $\\qquad$ (expressed in terms of $\\mathrm{n}$).\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch25-2024_06_17_b198ac3e13736d7a3336g_0074_1.jpg", "batch25-2024_06_17_b198ac3e13736d7a3336g_0074_2.jpg", "batch25-2024_06_17_b198ac3e13736d7a3336g_0074_3.jpg" ], "is_multi_img": true, "answer": "$(3 \\mathrm{n}+1)$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "The first figure has black tiles totaling \\(3 + 1 = 4\\),\n\nthe second figure has black tiles totaling \\(3 \\times 2 + 1 = 7\\),\n\nthe third figure has black tiles totaling \\(3 \\times 3 + 1 = 10\\),\n\nand the \\(n\\)th figure requires black tiles totaling \\((3n + 1)\\).\n\nTherefore, the answer is \\((3n + 1)\\).\n\n【Insight】This problem primarily examines the changes in patterns, with the key being to deduce the underlying rule through induction and summarization." }, { "problem_id": 1772, "question": "A set of triangles is placed as shown in Figure 1, where $O$ is the midpoint of side $BC(DF)$, and $BC = 8 \\mathrm{~cm}$. As shown in Figure 2, $\\triangle ABC$ is rotated $60^{\\circ}$ clockwise around point $O$, and $AC$ intersects $EF$ at point $G$. The length of $FG$ is $\\qquad$ $\\mathrm{cm}$.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch25-2024_06_17_ca52227114d14ef65cffg_0058_1.jpg", "batch25-2024_06_17_ca52227114d14ef65cffg_0058_2.jpg" ], "is_multi_img": true, "answer": "$(2 \\sqrt{3}-2) \\# \\#(-2+2 \\sqrt{3})$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, $BC$ intersects $EF$ at point $N$,\n\n\n\nGiven the conditions, $\\angle EDF = \\angle BAC = 90^\\circ$, $\\angle DEF = 60^\\circ$, $\\angle DFE = 30^\\circ$, $\\angle ABC = \\angle ACB = 45^\\circ$, and $BC = DF = 8 \\text{ cm}$.\n\nSince $O$ is the midpoint of side $BC$ (or $DF$), we have: $BO = OC = DO = OF = 4 \\text{ cm}$.\n\nBecause $\\triangle ABC$ is rotated $60^\\circ$ clockwise around point $O$, and $\\angle DFE = 30^\\circ$,\n\nThus, $\\angle BOD = \\angle NOF = 60^\\circ$,\nTherefore, $\\angle NOF + \\angle F = 90^\\circ$,\n\nHence, $\\angle FNO = 180^\\circ - \\angle NOF - \\angle F = 90^\\circ$,\n\nSo, $\\triangle ONF$ is a right-angled triangle,\n\nThus, $ON = \\frac{1}{2} OF = 2 \\text{ cm}$, and $FN = \\sqrt{OF^2 - ON^2} = 2\\sqrt{3} \\text{ cm}$,\n\nTherefore, $NC = OC - ON = 2 \\text{ cm}$,\n\nSince $\\angle FNO = 90^\\circ$ and $\\angle ACB = 45^\\circ$,\n\nThus, $\\angle GNC = 180^\\circ - \\angle FNO = 90^\\circ$,\n\nSo, $\\triangle CNG$ is a right-angled triangle,\n\nHence, $\\angle NGC = 180^\\circ - \\angle GNC - \\angle ACB = 45^\\circ$,\n\nTherefore, $\\triangle CNG$ is an isosceles right-angled triangle,\n\nThus, $NG = NC = 2 \\text{ cm}$,\n\nTherefore, $FG = FN - NG = 2\\sqrt{3} - 2 \\text{ cm}$,\n\nThe final answer is: $(2\\sqrt{3} - 2)$.\n\n【Highlight】This problem examines the properties of a right-angled triangle with a 30-degree angle, the determination and properties of an isosceles right-angled triangle, the properties of rotation, and the Pythagorean theorem. The key to solving the problem lies in understanding the properties of a right-angled triangle with a 30-degree angle and the inherent angles in the set square." }, { "problem_id": 1773, "question": "The front view, left view, and top view of a geometric object are shown in the figures below, respectively. The surface area of the geometric object is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0004_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0004_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0004_3.jpg" ], "is_multi_img": true, "answer": "$8 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "From the three-view drawing, it can be determined that the geometric body is a cylinder with a base radius of 1 and a height of 3. Therefore, the surface area \\( S_{\\text{surface}} \\) is calculated as the sum of the lateral area \\( S_{\\text{lateral}} \\) and twice the base area \\( S_{\\text{base}} \\):\n\n\\[\nS_{\\text{surface}} = S_{\\text{lateral}} + 2S_{\\text{base}} = \\pi \\times 2 \\times 3 + 2 \\times \\pi \\times (2 \\div 2)^2 = 8\\pi\n\\]\n\nThus, the answer is \\( 8\\pi \\).\n\n**Key Insight:** This problem primarily tests the ability to identify a geometric body from its three-view drawing and the calculation of a cylinder's surface area. Mastery of the formula for calculating the surface area of a cylinder is crucial for solving this problem." }, { "problem_id": 1774, "question": "There is a right triangular prism with a base in the shape of an equilateral triangle. The three views are shown in the figure below. The volume of this right prism is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0005_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0005_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0005_3.jpg" ], "is_multi_img": true, "answer": "$8 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, we can determine:\n\nThe side length of the equilateral triangle base is 4, and the height of the equilateral triangle is: $2 \\sqrt{3}$. The height of the right triangular prism is 2.\n\n$\\therefore$ The area of the base $=\\frac{1}{2} \\times 2 \\sqrt{3} \\times 4=4 \\sqrt{3}$,\n\n$\\therefore$ The volume of the right prism $=4 \\sqrt{3} \\times 2=8 \\sqrt{3}$.\n\nTherefore, the answer is: $8 \\sqrt{3}$.\n\n【Key Insight】This problem tests the ability to calculate the volume of a geometric solid based on its three-view drawing. Mastering the principles of alignment in length, width, and height in three-view drawings is crucial for solving such problems." }, { "problem_id": 1775, "question": "As shown in the figure, the main view and left view of a geometric body composed of several small cubes with a side length of 1 are given. When the number of cubes in this geometric body is maximized, the surface area of the geometric body at this time is $\\qquad$.\n\n\n\nFront view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0006_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0006_2.jpg" ], "is_multi_img": true, "answer": "46", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "By synthesizing the views from the front and the left, it can be concluded that when the number of cubes in this geometric figure is at its maximum,\n\n\n\nTherefore, the number of small cubes used to construct this geometric figure is \\(5 + 3 + 6 = 14\\).\n\n\\(\\therefore\\) The surface area of this geometric figure is \\(14 \\times 6 - 38 = 46\\), so the answer is: 46.\n\n【Insight】This question tests the students' mastery and flexible application of the three-view concept, as well as their spatial imagination skills. Mastering the mnemonic \"Use the top view to lay the foundation, the front view to build wildly, and the side view to dismantle violations\" is key to solving the problem." }, { "problem_id": 1776, "question": "The area of the front view of a cuboid, given the views from the left and top along with their respective data as shown in the figures, is $\\qquad$.\n\n\n\nView from the left\n\n\n\nView from the top", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0010_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0010_2.jpg" ], "is_multi_img": true, "answer": "15", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the left view and the top view, we can determine that the length of the cuboid is 5, the width is 4, and the height is 3. Therefore, the shape seen from the front is a rectangle with a length of 5 and a width of 3.\n\nThus, the area of the shape seen from the front is \\(5 \\times 3 = 15\\).\n\nHence, the answer is: 15.\n\n[Key Insight] This question tests the understanding of three-view drawings. Mastering the concept of three-view drawings is crucial for solving the problem." }, { "problem_id": 1777, "question": "As shown in the figure, a geometric object is formed by stacking several identical small cubes. Its front view and top view are shown in the figure. If the minimum number of small cubes forming this geometric object is $a$, and the maximum number is $b$, then $a+b=$ $\\qquad$ .\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0021_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0021_2.jpg" ], "is_multi_img": true, "answer": "14", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "From the top view, it can be seen that the bottom layer has 5 small cubes.\n\nFrom the front view, the second layer has at least 1 small cube on the left and at most 3 small cubes on the left.\n\nTherefore, the minimum number of small cubes in this geometric structure is \\( a = 5 + 1 = 6 \\), and the maximum is \\( b = 5 + 3 = 8 \\).\n\nThus, \\( a + b = 6 + 8 = 14 \\).\n\nThe answer is: 14.\n\n【Key Point】This question tests the understanding of the three views of a geometric structure made of cubes and the evaluation of algebraic expressions. The key to solving the problem is to determine the values of \\( a \\) and \\( b \\) based on the given information." }, { "problem_id": 1778, "question": "A geometric shape is constructed using several small identical cubes. The shapes of the geometric shape as seen from the front and from above are shown in the figures below. If the number of small cubes used to construct this geometric shape is $n$, then the sum of the minimum and maximum values of $n$ is $\\qquad$.\n\n\n\nView from the front\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0027_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0027_2.jpg" ], "is_multi_img": true, "answer": "26", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The maximum number is: $3+3+3+2+2+2+1=16$ (units), and the minimum number is: $3+1+1+2+1+1+1=10$ (units).\n\n\n\nView from above showing the maximum scenario.\n\n\n\nView from above showing the minimum scenario.\n\n$16+10=26$ (units),\n\n$\\therefore$ The sum of the minimum and maximum values of $n$ is 26.\n\nTherefore, the answer is: 26.\n\n[Insight] This question tests the understanding of three-view drawings. The key to solving it lies in comprehending the definition of three-view drawings, which is a common type of question in middle school exams." }, { "problem_id": 1779, "question": "The surface area of a geometric object is shown in the figure below. (Take $\\pi$ as 3)\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nLeft view", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0037_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0037_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0037_3.jpg" ], "is_multi_img": true, "answer": "13", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Observing the three views of the geometric figure, it is identified as a half-cylinder.\n\nThe diameter of the half-cylinder is 2, and its height is 1.\nTherefore, its surface area is: $\\pi \\times 1^{2} + (\\pi + 2) \\times 2 = 3\\pi + 4$.\n\nWhen $\\pi$ is taken as 3, the original expression becomes $3 \\times 3 + 4 = 13$.\n\nHence, the answer is: 13.\n\n[Highlight] This question tests the knowledge of determining the geometric shape from its three views. The key to solving the problem lies in first identifying the shape of the geometric figure based on its three views." }, { "problem_id": 1780, "question": "The main view and top view of a geometric body composed of several small cubes are shown in the figures below. The maximum number of small cubes required to form such a geometric body is $\\qquad$.\n\n\n\nMain View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0042_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0042_2.jpg" ], "is_multi_img": true, "answer": "10", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that the bottom layer consists of 4 cubes. From the front view, the second layer can have up to 4 cubes, and the third layer can have up to 2 cubes.\n\nTherefore, the maximum number of cubes needed to form such a geometric structure is 10.\n\nHence, the answer is: 10.\n\n[Key Insight] This problem tests the ability to deduce the geometry from three views. Mastering the mnemonic \"Top view lays the foundation, front view builds up wildly, side view removes violations\" is crucial for solving this problem." }, { "problem_id": 1781, "question": "The figure shows the shape of a geometric structure made up of identical small cubes as seen from different directions. A total of $\\qquad$ small cubes were used to build this geometric structure.\n\n\n\nView from the front\n\n\n\nView from the left\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0043_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0043_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0043_3.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, it is easy to determine that there are 3 small cubes on the bottom layer. From the front view and the side view, we know there is 1 small cube on the second layer. Therefore, the total number of small cubes used to build this geometric figure is \\(3 + 1 = 4\\).\n\nHence, the answer is: 4.\n\n[Key Insight] This problem tests the understanding of the three views of a geometric figure and spatial imagination. The key to solving the problem lies in determining the number of small cubes in each layer based on the principle: \"Use the top view to lay the foundation, the front view to build up, and the side view to adjust for discrepancies.\"" }, { "problem_id": 1782, "question": "There are some identical small cubes forming a geometric structure, and the three views of this structure are shown below. How many small cubes are used to form this geometric structure?\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0044_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0044_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0044_3.jpg" ], "is_multi_img": true, "answer": "4", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the top view, we can see that the bottom layer consists of 3 small cubes. From the front view, there are 2 layers: the upper layer has 1 small cube, and the lower layer has 2 small cubes. From the side view, the back layer has 2 small cubes, and the front has 1 small cube.\n\nTherefore, this geometric figure consists of a total of 4 cubes.\n\nHence, the answer is: 4.\n\n[Key Insight] This problem tests the ability to reconstruct a geometric figure from its three views. Understanding that the front view is the shape seen from the front, the side view is the shape seen from the side, and the top view is the shape seen from above is crucial for solving the problem." }, { "problem_id": 1783, "question": "As shown in the figure, the three views of a geometric body are given (the top view is an equilateral triangle). The lateral surface area of this geometric body is:\n\n\n\nFront view\n\n\n\nLeft view\n\n\n\nTop view", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0051_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0051_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0051_3.jpg" ], "is_multi_img": true, "answer": "$18 \\mathrm{~cm}^{2} \\# \\# 18$ cm ${ }^{2}$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: The surface development diagram is as follows:\n\n\n\n$\\therefore$ The lateral surface area of this geometric solid is $3 \\times(2 \\times 3)=18 \\mathrm{~cm}^{2}$.\n\nTherefore, the answer is: $18 \\mathrm{~cm}^{2}$.\n\n[Key Insight] This question tests knowledge of three-view drawings and the lateral surface development of geometric solids. The key to solving the problem lies in understanding and interpreting the three-view drawings correctly." }, { "problem_id": 1784, "question": "Construct a geometric shape using small cubes such that the views from the left and from above are as shown in the figures. The maximum number of cubes required for such a geometric shape is $\\qquad$.\n\n\n\nView from the left\n\n\n\nView from above", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0058_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0058_2.jpg" ], "is_multi_img": true, "answer": "13", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: From the front view and the top view, we can determine:\n\n\n\n## Front View\n\nTop View\n\nAt least: $1+1+1+2+3+1=9$ small cubes are needed.\n\nAt most: $2+2+2+3+1=13$ small cubes are needed.\n\nTherefore, the answer is: 13.\n\n[Key Insight] This question tests the knowledge of three-view drawings. The front view is obtained by looking at the object from the front, and the top view is obtained by looking at the object from above. Note that the front view mainly informs about the number of layers and columns that make up the geometric shape." }, { "problem_id": 1785, "question": "A geometric solid is formed by stacking several small cubes with edge lengths of $1 \\mathrm{~cm}$, as shown in the figures below, which represent the shapes viewed from the front, left, and top respectively. The surface area of this geometric solid is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nViewed from the front\n\n\n\nViewed from the left\n\n\n\nViewed from the top", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0059_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0059_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0059_3.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be seen that the number of small cubes at each position of this geometric figure is as shown in the image:\n\n\n\nSince the area of each face of the small cube is $1 \\mathrm{~cm}^{2}$,\n\nthe surface area of this geometric figure is: $4+4+3+3+5+5=24 \\mathrm{~cm}^{2}$.\n\nTherefore, the answer is: 24.\n\n[Key Insight] This question primarily tests the understanding of three-view drawings. The key to solving the problem lies in reconstructing the geometric figure based on the three-view drawing." }, { "problem_id": 1786, "question": "Arrange 10 small cubes with edge length $1 \\mathrm{~cm}$ to form a three-dimensional figure, whose front view is shown in Figure (1), and any two adjacent small cubes share at least one edge or one face. Now, there is a grid paper of size $3 \\mathrm{~cm} \\times 4 \\mathrm{~cm}$ (as shown in Figure (2)) to place these 10 small cubes according to the front view in the grid cells. The maximum surface area of the resulting geometric body is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0067_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0067_2.jpg" ], "is_multi_img": true, "answer": "52", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, when 10 small cubes are arranged as in the top view, the surface area of the geometric shape is maximized.\n\n\n\nFigure (1)\n\n\n\nTop View\n\nFigure (2)\n\nThe maximum value is calculated as follows: $3 \\times 6 + 2 \\times 10 + 14 = 52 \\left(\\mathrm{~cm}^{2}\\right)$.\n\nTherefore, the answer is: 52.\n\n[Key Insight] This question tests knowledge of three-view drawings and the surface area of geometric shapes. The key to solving the problem lies in understanding the scenario and flexibly applying the learned knowledge to find the solution." }, { "problem_id": 1787, "question": "Several identical cubes are stacked together. The front view and top view of the combined structure are shown in the figures below. What is the minimum number of bottom cubes in the combined structure?\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0072_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0072_2.jpg" ], "is_multi_img": true, "answer": "8", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The first layer consists of \\(1 + 2 + 3 = 6\\) cubes, and the second layer has at least 2 cubes. Therefore, the minimum number of cubes that make up this geometric figure is \\(6 + 2 = 8\\).\n\nThus, the answer is: 8.\n\n[Insight] This question is designed to test students' understanding and flexible application of three-view drawings, as well as their spatial imagination skills." }, { "problem_id": 1788, "question": "Construct a geometric shape using small cubes such that its front view and top view are as shown in the figures. The minimum number of small cubes required for this geometric shape is $\\qquad$.\n\n\n\nFront View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0092_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0092_2.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, the distribution of the small cubes that make up the geometric body is as shown in the following diagram,\n\n\n\nTop view\n\n\n\nTop view or\n\n\n\nTop view\nTherefore, such a geometric body requires at least 5 small cubes.\n\nHence, the answer is: 5.\n\n[Key Insight] This problem mainly tests the understanding of three-view drawings. Mastering the relationships between the three views is the key to solving the problem." }, { "problem_id": 1789, "question": "As shown in the figure, the following are the three views of an object formed by stacking several identical small cubes. The number of small cubes used to form this object is . $\\qquad$\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0093_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0093_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0093_3.jpg" ], "is_multi_img": true, "answer": "5", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Based on the three-view drawing, the figure can be illustrated as follows:\n\n\n\nTop View\n\nThe number of small cubes that make up this geometric figure is: $1 + 2 + 1 + 1 = 5$.\n\nTherefore, the answer is: 5\n\n[Key Insight] This question primarily tests the understanding of the three-view drawing of a geometric figure. Mastering the three-view drawing of geometric figures is crucial for solving such problems." }, { "problem_id": 1790, "question": "As shown, the following are the three views of a geometric body. Based on the data in the figure, the lateral surface area of this geometric body is $\\qquad$ $\\mathrm{cm}^{2}$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0095_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0095_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0095_3.jpg" ], "is_multi_img": true, "answer": "$60 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be determined that the original geometric body is a cone.\n\n$\\because l=\\sqrt{\\left(\\frac{12}{2}\\right)^{2}+8^{2}}=10$,\n\n$\\therefore S_{\\text {solution }}=\\frac{1}{2} \\times \\pi d \\times l=\\frac{1}{2} \\times \\pi \\times 12 \\times 10=60 \\pi$.\n\nTherefore, the answer is: $60 \\pi$.\n\n[Highlight] This question examines the identification of a geometric body from its three-view drawing, the calculation of a cone, and the Pythagorean theorem. The key to solving the problem lies in recognizing that the original geometric body is a cone based on its three-view drawing." }, { "problem_id": 1791, "question": "The three views of a cuboid are shown in the figure. If its top view is a square, then the volume of this cuboid is $\\qquad$ $\\mathrm{cm}^{3}$.\n\n\n\nFront view\n\n\n\nTop view\n\n\n\nSide view", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0097_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0097_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0097_3.jpg" ], "is_multi_img": true, "answer": "144", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since the top view is a square, according to the front view, the diagonal of the square is $6 \\mathrm{~cm}$, and the height of the cuboid is $8 \\mathrm{~cm}$.\n\nTherefore, the volume of the cuboid is: $6 \\times 6 \\div 2 \\times 8=144\\left(\\mathrm{~cm}^{3}\\right)$.\n\nHence, the answer is: 144.\n\n[Highlight] This question tests the ability to determine the geometry from three-view drawings, using basic knowledge of three views and the formula for calculating the volume of a cuboid." }, { "problem_id": 1792, "question": "Given the three views of a geometric object as shown in the figure, the volume of the geometric object is $\\qquad$.\n\n\n\nFront View\n\n\n\nLeft View\n\n\n\nTop View", "input_image": [ "batch25-2024_06_17_da7d4874c13c5db0429eg_0100_1.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0100_2.jpg", "batch25-2024_06_17_da7d4874c13c5db0429eg_0100_3.jpg" ], "is_multi_img": true, "answer": "$48-3 \\pi$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From the three-view drawing, it can be seen that the original geometric body is a cube with a cylinder removed from the middle. The side length of the cube is \\(1 + 2 + 1 = 4\\), the diameter of the cylinder is 2, and both have a height of 3.\n\nTherefore, the volume of the geometric body is:\n\\[\n4^{2} \\times 3 - \\pi \\times \\left(\\frac{2}{2}\\right)^{2} \\times 3 = 48 - 3\\pi\n\\]\nHence, the answer is: \\(48 - 3\\pi\\).\n\n【Key Insight】This problem tests the understanding of the three-view drawing of a geometric body and the calculation of its volume. Correctly interpreting the three-view drawing of the geometric body is crucial for solving the problem." }, { "problem_id": 1793, "question": "In rectangle $\\mathrm{ABCD}$, point $\\mathrm{P}$ lies on $\\mathrm{AD}$, with $A B=\\sqrt{3}$ and $A P=1$. Place the vertex of a right angle ruler at point $\\mathrm{P}$, and let its sides intersect $\\mathrm{AB}$ and $\\mathrm{BC}$ at points $\\mathrm{E}$ and $\\mathrm{F}$ respectively. Connect $\\mathrm{EF}$ (as shown in Figure (1)). When point $\\mathrm{E}$ coincides with point $\\mathrm{B}$, point $\\mathrm{F}$ exactly coincides with point $\\mathrm{C}$ (as shown in Figure (2)). Starting from the position in Figure (2), rotate the ruler clockwise around point $\\mathrm{P}$ until point $\\mathrm{E}$ coincides with point $\\mathrm{A}$. During this process, the path (segment) traversed by the midpoint of segment $\\mathrm{EF}$ from start to stop is $\\qquad$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch25-2024_06_17_f95e8c4a1f4fffa24fc6g_0095_1.jpg", "batch25-2024_06_17_f95e8c4a1f4fffa24fc6g_0095_2.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "As shown in Figure (2),\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\nIn rectangle \\( A B C D \\), \\( \\angle A = \\angle D = 90^{\\circ} \\).\n\nSince \\( A P = 1 \\) and \\( A B = \\sqrt{3} \\),\n\nTherefore, \\( P B = 2 \\).\n\nSince \\( \\angle A B P + \\angle A P B = 90^{\\circ} \\) and \\( \\angle B P C = 90^{\\circ} \\),\n\nTherefore, \\( \\angle A P B + \\angle D P C = 90^{\\circ} \\),\n\nThus, \\( \\angle A B P = \\angle D P C \\),\n\nHence, \\( \\triangle A P B \\sim \\triangle C P D \\),\n\nTherefore, \\( A P : C D = P B : C P \\).\n\nThat is, \\( 1 : \\sqrt{3} = 2 : C P \\),\n\nThus, \\( C P = 2 \\sqrt{3} \\).\n\nAs shown in Figure (1), let the midpoint of segment \\( E F \\) be \\( O \\), and connect \\( O P \\) and \\( O B \\).\n\nIn right triangle \\( \\triangle E P F \\), \\( O P = \\frac{1}{2} E F \\). In right triangle \\( \\triangle E B F \\), \\( O B = \\frac{1}{2} E F \\),\n\nTherefore, \\( O P = O B \\),\n\nThus, point \\( O \\) lies on the perpendicular bisector of segment \\( B P \\).\n\nAs shown in Figure (2), when point \\( E \\) coincides with point \\( B \\) and point \\( F \\) coincides with point \\( C \\), the midpoint of \\( E F \\) is the midpoint \\( O \\) of \\( B C \\).\n\nWhen point \\( E \\) coincides with point \\( A \\), the midpoint of \\( E F \\) is the midpoint \\( O^{\\prime} \\) of \\( P B \\),\n\nTherefore, \\( O O^{\\prime} \\) is the midline of \\( \\triangle P B C \\),\n\nThus, \\( O O^{\\prime} = \\frac{1}{2} C P = \\sqrt{3} \\),\n\nTherefore, the length of the path traced by the midpoint of segment \\( E F \\) is \\( \\sqrt{3} \\).\n\n【Highlight】This problem examines the properties of rectangles and rotation. The key to solving part (2) is determining that point \\( O \\) lies on the perpendicular bisector of segment \\( B P \\)." }, { "problem_id": 1794, "question": "Xiaoming's study is equipped with a space-saving flip-up hidden bed (as shown in Figure 1). When not in use, it can be flipped upright and hidden inside the bed box (cabinet). Figure 2 is a side view schematic of the headboard. The bed rotates around the fixed point $O$ (pivot) on the bed box, with $AC$ and $BC$ being retractable gas struts that connect the bed and the bed box ($A$ and $B$ are fixed points on the bed frame, $C$ is a fixed point on the bed box), $AB = 8 \\text{ cm}$. When the bed is horizontally positioned, the gas strut $AC$ is perpendicular to the bed frame and measures $8 \\text{ cm}$ ($AC = 8 \\text{ cm}$); when the bed is upright, this gas strut extends to $40 \\text{ cm}$ ($A'C = 40 \\text{ cm}$). Then the other gas strut $B'C =$ . $\\qquad$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch26-2024_06_17_0a1559e37e8e7d7f2d24g_0036_1.jpg", "batch26-2024_06_17_0a1559e37e8e7d7f2d24g_0036_2.jpg" ], "is_multi_img": true, "answer": "$24 \\sqrt{2} \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Through point $C$, draw $CN \\perp A'B'$ intersecting at point $N$, as shown in the figure.\n\n\n\nAccording to the properties of rotation: $\\angle AOB' = 90^\\circ$, $A'B' = AB$, $OB' = OB$,\n\nThus, $\\triangle BOB'$ is an isosceles right triangle,\n\n$\\therefore \\angle OBB' = 45^\\circ$,\n\n$\\because AC \\perp AB$, $AB = AC = 8$ cm,\n\n$\\therefore \\triangle ABC$ is an isosceles right triangle,\n\n$\\therefore \\angle ABC = 45^\\circ$,\n\n$\\therefore \\angle OBB' = 45^\\circ = \\angle ABC$,\n\n$\\therefore$ points $C$, $B$, $B'$ are collinear,\n\n$\\because AC \\perp AO$, $CN \\perp NO$,\nAccording to the problem, $\\angle AON = 90^\\circ$,\n\n$\\therefore$ quadrilateral $ACNO$ is a rectangle,\n\n$\\therefore AO = CN$, $AC = NO = 8$,\n\n$\\because AO = AB + BO$, $NB' = NO + OB'$, $NA' = A'B' + NO + OB'$, $OB' = OB$,\n\n$\\therefore NC = AO = 8 + OB$, $NA' = 16 + OB$, $NB' = 8 + OB$,\n\n$\\because A'C = 40$,\n\n$\\therefore$ in right triangle $\\triangle A'CN$, $A'C^2 = A'N^2 + CN^2$,\n\n$\\therefore 40^2 = (OB + 16)^2 + (OB + 8)^2$,\n\nSolving gives $OB = 16$, (discarding the negative value)\n\n$\\therefore NC = 8 + OB = 24$, $NB' = 8 + OB = 24$,\n\n$\\therefore$ in right triangle $\\triangle B'CN$, $B'C = \\sqrt{B'N^2 + CN^2} = \\sqrt{24^2 + 24^2} = 24\\sqrt{2}$ cm,\n\nTherefore, the answer is: $24\\sqrt{2}$ cm.\n\n【Highlight】This problem mainly tests the Pythagorean theorem, the determination of isosceles right triangles, solving quadratic equations, and the properties of rotation. Understanding the problem and correctly drawing the diagram are key to solving this problem." }, { "problem_id": 1795, "question": "As shown, it is given that $\\triangle A C B$ and $\\triangle D F E$ are two congruent right triangles, with their hypotenuses measuring $10 \\mathrm{~cm}$ and the smaller acute angle being $30^{\\circ}$. The two triangles are arranged in the shape shown in Figure (1), such that points $B, C, F, D$ lie on the same straight line, and points $C$ and $F$ coincide. In Figure (1), $\\triangle A C B$ is rotated clockwise around point $C$ to the position shown in Figure (2), where point $E$ is on side $A B$, and $A C$ intersects $D E$ at point $G$. The length of segment $F G$ is $\\qquad$ cm (express in surd form).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch26-2024_06_17_62ae783bf5f126e06a07g_0009_1.jpg", "batch26-2024_06_17_62ae783bf5f126e06a07g_0009_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{5 \\sqrt{3}}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "From the problem statement, in right triangle $\\triangle ABC$, $\\angle A = 30^\\circ$ and $\\angle ABC = 60^\\circ$, thus $BC = \\frac{1}{2} AB = 5 \\text{ cm}$. By the properties of rotation, in figure (2), $CB = CE$.\n\nTherefore, $\\triangle BCE$ is an equilateral triangle.\n\nHence, $\\angle ECB = 60^\\circ$ and $CE = BC = 5 \\text{ cm}$.\n\nThus, $\\angle ECG = 30^\\circ$.\n\nSince $\\angle FED = 60^\\circ$,\n\nTherefore, $\\angle EGC = 90^\\circ$.\n\nHence, $EG = \\frac{1}{2} CE = \\frac{5}{2} \\text{ cm}$,\n\nAnd $CG = \\sqrt{CE^2 - EG^2} = \\frac{5 \\sqrt{3}}{2} \\text{ cm}$.\n\nThe final answer is: $\\frac{5 \\sqrt{3}}{2}$.\n\n【Key Insight】This problem primarily examines the properties and determination of equilateral triangles, the properties of right triangles containing a 30-degree angle, the properties of rotation, and the Pythagorean theorem. Understanding the properties and determination of equilateral triangles is crucial for solving the problem." }, { "problem_id": 1796, "question": "As shown in the figure, each pattern is formed by arranging black and white squares of equal side length according to a certain rule. Following this rule, the number of white squares in the $n$th pattern is $\\qquad$ more than the number of black squares. (Express in algebraic terms involving $n$)\n\n\n\n1st pattern\n\n\n\n2nd pattern\n\n\n\n3rd pattern", "input_image": [ "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0020_1.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0020_2.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0020_3.jpg" ], "is_multi_img": true, "answer": "$4 \\mathrm{n}+3$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Method One:\n\nThe first figure consists of a total of $3 \\times 3$ black and white squares, with 1 black and $3 \\times 3 - 1$ white squares. The second figure has $3 \\times 5$ black and white squares, with 2 black and $3 \\times 5 - 2$ white squares. The third figure contains $3 \\times 7$ black and white squares, with 3 black and $3 \\times 7 - 3$ white squares, and so on.\n\nThe nth figure consists of $3 \\times (2n + 1)$ black and white squares, with n black and $3 \\times (2n + 1) - n$ white squares, which means there are $5n + 3$ white squares and n black squares.\n\nTherefore, in the nth pattern, there are $4n + 3$ more white squares than black squares.\n\nMethod Two:\n\nThe first figure has 8 white squares and 1 black square, with 7 more white squares than black squares.\n\nThe second figure has 4 more white squares than black squares compared to the first figure, totaling $(7 + 4)$ more white squares.\n\nThe third figure has 4 more white squares than black squares compared to the second figure, totaling $(7 + 4 \\times 2)$ more white squares, and so on. Thus, in the nth pattern, there are $[7 + 4(n - 1)]$ more white squares than black squares, which is $(4n + 3)$ more.\n\nTherefore, in the nth pattern, there are $4n + 3$ more white squares than black squares.\n\n[Highlight] This question examines the changing patterns of geometric figures, which is an exploratory problem. The key to solving the problem lies in understanding the changing patterns in the figures." }, { "problem_id": 1797, "question": "As shown in Figure 3-3, a floor is paved with gray and white square tiles. In the 6th pattern, there are $\\qquad$ gray tiles.\n\n\n\n1st pattern\n\n\n\n2nd pattern\n\n\n\n3rd pattern", "input_image": [ "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0024_1.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0024_2.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0024_3.jpg" ], "is_multi_img": true, "answer": "14", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: When $\\mathrm{n}=1$, the number of gray tiles is: 4;\n\nWhen $\\mathrm{n}=2$, the number of gray tiles is: 6;\n\nWhen $\\mathrm{n}=3$, the number of gray tiles is: 8;\n$\\cdots ;$\n\nWhen $n=n$, the number of gray tiles is: $2(n+1)$.\n\n$\\therefore$ When $\\mathrm{n}=6$, the number of gray tiles is: $2 \\times 7=14$.\n\nThus, the number of gray tiles in the 6th pattern is 14.\n\n[Insight] This problem is about identifying patterns, a type of question frequently encountered in middle school exams. For pattern-finding questions, the first step is to determine which parts have changed and understand the rule governing these changes." }, { "problem_id": 1798, "question": "As shown in Figure (1), it represents 1 table and 6 chairs (each small semicircle represents 1 chair). If $n$ tables are arranged in this way, the number of chairs required is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0027_1.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0027_2.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0027_3.jpg" ], "is_multi_img": true, "answer": "$4 n+2 / 2+4 n$", "answer_type": "single-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: By analyzing the diagram, we observe the following pattern: With 1 dining table, there are 6 chairs. For each additional table beyond the first, 4 more chairs are added. Therefore, with a total of \\( n \\) dining tables, the number of chairs is calculated as \\( 6 + 4(n - 1) = 4n + 2 \\). Hence, the answer is \\( 4n + 2 \\).\n\n[Key Insight] This problem tests the ability to identify patterns in graphical representations. The key to solving it lies in recognizing the underlying rule governing the relationship between the number of tables and chairs." }, { "problem_id": 1799, "question": "Given two identical large rectangles with length $a$, each is placed with four identical small rectangles as shown in the figure, the difference in the perimeters of the shaded areas in Figure 1 and Figure 2 is $\\qquad$. (Express in algebraic form containing $a$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0032_1.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0032_2.jpg" ], "is_multi_img": true, "answer": "$a$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the width of the large rectangle in the figure be \\( b \\), the length of the small rectangle be \\( x \\), and the width of the small rectangle be \\( y \\). From the figure, we know that \\( x + 2y = a \\) and \\( x = 2y \\).\n\n\\[\n\\therefore a = 4y\n\\]\n\nThe perimeter of the shaded region in Figure 1 is:\n\n\\[\n2b + 2(a - x) + 2x = 2a + 2b\n\\]\n\nThe perimeter of the shaded region in Figure 2 is:\n\n\\[\n2(a + b - 2y) = 2a + 2b - 4y\n\\]\n\n\\[\n\\therefore \\text{The difference in the perimeters of the shaded regions in Figure 1 and Figure 2 is:}\n\\]\n\n\\[\n(2a + 2b) - (2a + 2b - 4y) = 4y = a\n\\]\n\nHence, the answer is: \\( a \\).\n\n【Insight】This problem mainly tests the application of algebraic expressions and integer addition and subtraction. The key to solving the problem lies in combining numbers and shapes to derive the perimeters of the shaded regions in Figure 1 and Figure 2." }, { "problem_id": 1800, "question": "On November 3, 2022, the basic configuration of the Chinese Space Station in the shape of the letter \"T\" was completed in orbit. The letter \"T\" symbolizes wisdom and excellence. Figure 1 shows a tangram made from a rectangular cardboard (with the given line lengths as shown), which is used to form the \"T\" shape in Figure 2. The perimeter of the \"T\" shape is $\\qquad$. (expressed in terms of $m$ and $n$)\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0033_1.jpg", "batch26-2024_06_17_70b1fe4582aa6a3c238ag_0033_2.jpg" ], "is_multi_img": true, "answer": "$2(m+4 n)$", "answer_type": "single-step", "difficulty": "Low", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: The perimeter of the \"T\"-shaped figure is $(m + 2n + 2n) \\times 2 = 2(m + 4n)$.\n\nTherefore, the answer is: $2(m + 4n)$.\n\n[Key Insight] This question tests the ability to formulate algebraic expressions, with a key focus on mastering the perimeter formula for rectangles and the concept of translating shapes." }, { "problem_id": 1801, "question": "As shown in Figure 1, in the right triangle $Rt \\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = 1$, $BC = 2$. The triangle $ABC$ is placed in the plane rectangular coordinate system such that point $A$ coincides with the origin, and point $C$ lies on the positive $x$-axis. The triangle $ABC$ is rolled clockwise (without slipping) as shown in Figure 2. After rolling 2021 times, the abscissa of point $B$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch26-2024_06_17_7b8d9aa2743c4fcf3338g_0059_1.jpg", "batch26-2024_06_17_7b8d9aa2743c4fcf3338g_0059_2.jpg" ], "is_multi_img": true, "answer": "$2022+673 \\sqrt{5}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: According to the rolling pattern of the triangle, it completes a cycle every 3 rolls,\n\n$\\therefore 2021 \\div 3=673 \\cdots 2$,\n\n$\\because \\angle A C B=90^{\\circ}, A C=1, B C=2$,\n\n$\\therefore A B=\\sqrt{A C^{2}+B C^{2}}=\\sqrt{1+4}=\\sqrt{5}$,\n\n$\\therefore$ The sum of the lengths of the three sides of the triangle is: $1+2+\\sqrt{5}=3+\\sqrt{5}$,\n\nThen, after rolling 2021 times, the horizontal coordinate of point $B$ is: $1+2+673(3+\\sqrt{5})=2022+673 \\sqrt{5}$.\n\n【Insight】This problem examines the exploration of point patterns under a coordinate system. The key to solving the problem lies in correctly abstracting and summarizing the numerical pattern from the figure." }, { "problem_id": 1802, "question": "Xiao Ming used the \"tangram\" in Figure 1 during a math interest activity class to design and form the spaceship in Figure 2. The ratio of the spaceship model's area to the area of the rectangle frame $A B C D$ is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch26-2024_06_17_7bf0fb9f48fee2e44f30g_0003_1.jpg", "batch26-2024_06_17_7bf0fb9f48fee2e44f30g_0003_2.jpg" ], "is_multi_img": true, "answer": "$8: 25 / \\frac{8}{25}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Elementary", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Let the side length of square (4) be \\( x \\). Then, the lengths of the right-angled sides of (1) and (2) are \\( 2x \\), the right-angled sides of (3) and (5) are \\( x \\), and the short side of (6) is \\( x \\).\n\nTherefore, the area of the tangram in Figure 1 is \\( 4S_{\\oplus} = 4 \\times \\frac{1}{2} \\times 2x \\times 2x = 8x^{2} \\), which means the area of the spaceship model is \\( 8x^{2} \\).\n\nIn the rectangular frame \\( ABCD \\), the length of \\( AD \\) can be considered as the sum of the right-angled sides of (1) and (2) plus the short side of (6).\n\nThus, \\( AD = 2x + 2x + x = 5x \\).\n\nTherefore, the area of the rectangular frame \\( ABCD \\) is \\( 5x \\times 5x = 25x^{2} \\).\n\nHence, the ratio of the area of the spaceship model to the area of the rectangular frame \\( ABCD \\) is \\( 8x^{2} : 25x^{2} = 8 : 25 \\).\n\nThe final answer is: \\( 8 : 25 \\).\n\n[Key Insight] This problem examines the concepts of shape cutting and assembly, tangram, and the area of rectangles. Understanding the quantitative relationships between the side lengths of the various shapes in the tangram is crucial for solving the problem." }, { "problem_id": 1803, "question": "As shown in Figure 1, in response to the national new energy construction, a city's bus stop shelters have been equipped with solar panels. In a certain season, the maximum angle between the sunlight (parallel rays) and the horizontal line is $62^{\\circ}$, as shown in Figure 2. The solar panel $AB$ is perpendicular to the sunlight at the moment of the maximum angle, and the angle between the solar panel $CD$ and the horizontal line is $48^{\\circ}$. To make $AB$ parallel to $CD$, the solar panel $CD$ needs to be rotated counterclockwise by $\\alpha$ degrees, then $\\alpha$ is $\\qquad$. $\\left(0<\\alpha<90^{\\circ}\\right)$\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch26-2024_06_17_9d67ef82a4fa63964076g_0013_1.jpg", "batch26-2024_06_17_9d67ef82a4fa63964076g_0013_2.jpg" ], "is_multi_img": true, "answer": "$20^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Since \\( EF \\perp AB \\), it follows that \\( \\angle EFO = 90^\\circ \\).\n\nGiven that \\( \\angle OEF = 62^\\circ \\), we can determine \\( \\angle EOF \\) as follows:\n\\[\n\\angle EOF = 180^\\circ - 90^\\circ - 62^\\circ = 28^\\circ\n\\]\n\nSince \\( AB \\parallel CD \\), the angle \\( \\angle MQD \\) is equal to \\( \\angle EOF \\), which is \\( 28^\\circ \\).\n\nTo ensure that \\( AB \\parallel CD \\), the panel \\( CD \\) needs to be rotated counterclockwise by an angle \\( \\alpha \\). Therefore:\n\\[\n\\alpha^\\circ = 48^\\circ - 28^\\circ = 20^\\circ\n\\]\n\nThus, the answer is: 20.\n\n**Key Insight:** This problem examines the properties of parallel lines, rotation, the triangle angle sum theorem, and the definition of perpendicularity. The crucial step is determining the measure of \\( \\angle MQD \\)." }, { "problem_id": 1804, "question": "As shown in Figure 1, for points \\( A \\) and \\( P \\) in the plane, if the segment \\( PA \\) is rotated counterclockwise by \\( 90^\\circ \\) around point \\( P \\) to obtain segment \\( PB \\), then point \\( B \\) is called the \"orthogonal reflection point\" of point \\( A \\) with respect to point \\( P \\). As shown in Figure 2, given point \\( A(4,0) \\), point \\( P \\) is a point on the \\( y \\)-axis, and point \\( B \\) is the \"orthogonal reflection point\" of point \\( A \\) with respect to point \\( P \\). Connecting \\( AB \\) and \\( OB \\), the minimum value of \\( OB \\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch26-2024_06_17_9d67ef82a4fa63964076g_0049_1.jpg", "batch26-2024_06_17_9d67ef82a4fa63964076g_0049_2.jpg" ], "is_multi_img": true, "answer": "$2 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) When the ordinate of point $P$ is $\\geq 0$, as shown in the figure, draw $BC \\perp y$-axis at point $C$,\n\n\n\n$\\angle CBP + \\angle CPB = 90^{\\circ}$, $\\angle OPA + \\angle CPB = 90^{\\circ}$, thus $\\angle CBP = \\angle OPA$,\n\nFrom the property of rotation: $PB = PA$,\n\nIn triangles $\\triangle BPC$ and $\\triangle PAO$: $\\angle PBC = \\angle APO$, $\\angle BCP = \\angle POA = 90^{\\circ}$, $BP = PA$,\n\n$\\therefore \\triangle BPC \\cong \\triangle PAO$ (AAS),\n\n$\\therefore BC = PO$, $PC = AO$,\n\nLet the length of $OP$ be $x$, then $PC = AO = 4$, $BC = x$, $B(x, x+4)$\n\n$$\n\\therefore OB = \\sqrt{x^{2} + (x+4)^{2}} = \\sqrt{2(x+2)^{2} + 8}\n$$\n\n$\\because x \\geq 0$,\n\n$\\therefore$ when $x = 0$, $OB$ is minimized, with a minimum value of 4,\n\n(2) When the ordinate of point $P$ is $< 0$, as shown in the figure, draw $BC \\perp y$-axis at point $C$,\n\n\n\nSimilarly, we can derive $\\triangle BPC \\cong \\triangle PAO$ (AAS), $BC = PO$, $PC = AO$,\n\nLet the length of $OP$ be $x$, then $PC = AO = 4$, $BC = x$, $B(-x, 4-x)$\n\n$\\therefore OB = \\sqrt{x^{2} + (4-x)^{2}} = \\sqrt{2(x-2)^{2} + 8}$\n\n$\\because x > 0$,\n\n$\\therefore$ when $x = 2$, $OB$ is minimized, with a minimum value of $2 \\sqrt{2}$,\n\nIn summary: The minimum value of $OB$ is $2 \\sqrt{2}$,\n\nTherefore, the answer is: $2 \\sqrt{2}$;\n\n【Insight】This problem examines the properties of rotation, the criteria and properties of congruent triangles, the Pythagorean theorem, and the properties of quadratic functions; the key to solving the problem lies in discussing the position of point $P$ by cases." }, { "problem_id": 1805, "question": "The history of magic squares is long and fascinating. As shown in Figure 1, by filling a 3x3 grid with 9 consecutive positive integers such that the sum of the numbers in each row, column, and diagonal is equal, a magic square can be obtained. In Figure 2, another set of 9 consecutive positive integers is filled into a 3x3 grid, where the sum of the numbers in each row, column, and diagonal is 2019. What is the smallest number among these 9 integers?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch26-2024_06_17_b5aa8c7dac408b10b0acg_0048_1.jpg", "batch26-2024_06_17_b5aa8c7dac408b10b0acg_0048_2.jpg" ], "is_multi_img": true, "answer": "669", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Because in Figure 1, the sum of the three numbers in each row, each column, and each diagonal is $9+5+1=15$, the middle number is $15 \\div 3=5$, and the smallest number is 1. In Figure 2, the sum of the three numbers in each row, each column, and each diagonal is 2019.\n\nTherefore, the middle number in Figure 2 is $2019 \\div 3=673$.\n\nThus, the smallest number is 673-4=669.\n\nHence, the answer is: 669.\n\n[Insight] This question mainly tests the application of rational number operations, and the key to solving it lies in understanding the characteristics of magic squares." }, { "problem_id": 1806, "question": "Place the same rectangular cards in the corner of a right angle as shown in the figure, each rectangular card has a length of 2 and a width of 1, and so on. When 2014 cards are placed, the length of the solid line part is . $\\qquad$\n\n\n\nWhen 1 card is placed, the length of the solid line part is 3.\n\n\n\nWhen 2 cards are placed, the length of the solid line part is 5.\n\n\n\nWhen 3 cards are placed, the length of the solid line part is 8.", "input_image": [ "batch26-2024_06_17_f75b7773378712736c4bg_0045_1.jpg", "batch26-2024_06_17_f75b7773378712736c4bg_0045_2.jpg", "batch26-2024_06_17_f75b7773378712736c4bg_0045_3.jpg" ], "is_multi_img": true, "answer": "5035", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Arithmetic", "image_relavance": "1", "analysis": "Solution: From the figure, we can deduce:\n\n- Placing 1 rectangle results in a solid line length of 3.\n- Placing 2 rectangles results in a solid line length of 5.\n- Placing 3 rectangles results in a solid line length of 8.\n- Placing 4 rectangles results in a solid line length of 10.\n- Placing 5 rectangles results in a solid line length of 13.\n\nThat is, for every even-numbered rectangle, the solid line length increases by 2 from the previous one, and for every odd-numbered rectangle, it increases by 3 from the previous one.\n\nSince placing 2014 rectangles is equivalent to adding 1007 increments of 2 and 1006 increments of 3 to the first rectangle,\n\nTherefore, when placing 2014 rectangles, the solid line length is: \n\\[ 3 + 1007 \\times 2 + 1006 \\times 3 = 5035. \\]\n\nHence, the answer is: 5035.\n\n[Insight] This problem tests the pattern of graphical changes. Understanding the problem and identifying the pattern is key to solving it." }, { "problem_id": 1807, "question": "The school gate of a certain school is a retractable electric gate (as shown in Figure 1). Each row of rhombuses in the retractable electric gate has 20 rhombuses, with each rhombus having a side length of $30 \\mathrm{~cm}$. When the internal angle of each rhombus is $60^{\\circ}$ (as shown in Figure 2), the gate opens to $5 \\mathrm{~m}$. How many meters does the gate open when the internal angle of each rhombus is $90^{\\circ}$?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch27-2024_06_17_0e23c84cef5a4b11ba4dg_0019_1.jpg", "batch27-2024_06_17_0e23c84cef5a4b11ba4dg_0019_2.jpg" ], "is_multi_img": true, "answer": "$11-6 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Connect $BD$,\n\nSince quadrilateral $ABCD$ is a rhombus,\n\nTherefore, $AB = AD$,\n\nGiven that $\\angle A = 60^\\circ$,\n\nThus, triangle $ABD$ is an equilateral triangle,\n\nHence, $BD = AB = 30 \\text{ cm} = 0.3 \\text{ m}$,\n\nTherefore, the width of the retractable door at this time is: $0.3 \\times 20 = 6 (\\text{m})$,\n\nWhen each internal angle of the rhombus is $90^\\circ$,\n\nSince $BD = \\sqrt{2} AB = \\frac{3 \\sqrt{2}}{10} (\\text{m})$,\nTherefore, the width of the retractable door at this time is: $\\frac{3 \\sqrt{2}}{10} \\times 20 = 6 \\sqrt{2} (\\text{m})$,\n\nThus, the school gate has opened: $6 + 5 - 6 \\sqrt{2} = 11 - 6 \\sqrt{2} (\\text{m})$.\n\nAnswer: The school gate has opened $(11 - 6 \\sqrt{2})$ meters.\n\n\n\n【Insight】This problem examines the properties of a rhombus and the determination and properties of an equilateral triangle. The key to solving the problem is to accurately draw auxiliary lines." }, { "problem_id": 1808, "question": "Xiao Ming tries to fold the rectangular paper $A B C D$ (as shown in Figure (1), $A D>C D$ ) along a line passing through point $A$, such that point $B$ falls on point $F$ on side $A D$, with the crease being $A E$ (as shown in Figure (2)); then fold along a line passing through point $D$, such that point $C$ falls on point $N$ on side $D A$, point $E$ falls on point $M$ on side $A E$, and the crease being $D G$ (as shown in Figure (3). If after the second fold, point $M$ is exactly on the bisector of $\\angle N D G$. Find the ratio of the length to the width of the rectangle $A B C D$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch27-2024_06_17_1fc79bd211a758b64a62g_0040_1.jpg", "batch27-2024_06_17_1fc79bd211a758b64a62g_0040_2.jpg", "batch27-2024_06_17_1fc79bd211a758b64a62g_0040_3.jpg" ], "is_multi_img": true, "answer": "$\\sqrt{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Connect $DE$, as shown in the figure,\n\n\nSince the figure is folded along the line passing through point $A$, causing point $B$ to land on point $F$ on side $AD$,\n\nTherefore, quadrilateral $ABEF$ is a square,\n\nHence, $\\angle EAD = 45^{\\circ}$, and similarly, it can be proven that $\\angle ADG = 45^{\\circ}$,\n\nThus, $\\angle AGD = 90^{\\circ}$,\n\nFrom the second fold, we know that point $M$ lies exactly on the angle bisector of $\\angle NDG$,\n\nTherefore, $DE$ bisects $\\angle GDC$, and right triangle $DGE$ is congruent to right triangle $DCE$,\n\nHence, $DC = DG$,\n\nMoreover, since $\\triangle AGD$ is an isosceles right triangle,\n\nTherefore, $AD = \\sqrt{2} DG = \\sqrt{2} CD$,\n\nThus, $BC = \\sqrt{2} AB$,\n\nTherefore, $\\frac{BC}{AB} = \\sqrt{2}$.\n\n[Insight] This problem examines the properties of folding: the two figures before and after folding are congruent. It also tests the properties of squares, angle bisectors, and isosceles right triangles; remembering the properties of folding transformations and squares is key to solving the problem." }, { "problem_id": 1809, "question": "As shown in Figure 1, the central flower bed $A B C D E$ in the square is known to have $A E=D E=2 A B=2 B C=2 C D=2$ meters, and $\\angle A=\\angle B=\\angle C=\\angle D=120^{\\circ}$. Now, extend $A B$ and $C D$ to intersect at point $F$, and complete Figure 1 into Figure 2. What is the distance between points $E$ and $F$ at this time?\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch27-2024_06_17_3d22a4f05aa46d81977bg_0025_1.jpg", "batch27-2024_06_17_3d22a4f05aa46d81977bg_0025_2.jpg" ], "is_multi_img": true, "answer": " $2 \\sqrt{3}$ m", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AE = DE = 2AB = 2BC = 2CD = 2 \\) meters,\n\nTherefore, \\( AB = BC = CD = 1 \\) meter,\n\nSince \\( \\angle B = \\angle C = 120^\\circ \\),\n\nTherefore, \\( \\angle FBC = \\angle FCB = 60^\\circ \\),\n\nThus, triangle \\( BFC \\) is an equilateral triangle,\n\nTherefore, \\( BF = CF = BC = 1 \\) meter, and \\( \\angle F = 60^\\circ \\),\n\nSince \\( \\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ \\),\n\nTherefore, \\( \\angle E = 60^\\circ \\),\n\nThus, \\( \\angle F = \\angle E \\),\n\nTherefore, quadrilateral \\( AFDE \\) is a parallelogram,\n\nSince \\( AE = DE \\),\n\nTherefore, quadrilateral \\( AFDE \\) is a rhombus,\n\nConnecting \\( AD \\) and \\( EF \\) intersecting at \\( O \\),\n\n\n\nFigure 2\n\nTherefore, \\( \\angle AOF = 90^\\circ \\), \\( \\angle AFO = 30^\\circ \\),\n\nThus, \\( AO = \\frac{1}{2} AF = 1 \\) meter,\n\nTherefore, \\( OF = \\sqrt{AF^2 - AO^2} = \\sqrt{2^2 - 1^2} = \\sqrt{3} \\) meters,\n\nThus, \\( EF = 2OF = 2\\sqrt{3} \\) meters,\n\nAnswer: The distance between points \\( E \\) and \\( F \\) is \\( 2\\sqrt{3} \\) meters.\n\n【Highlight】This problem examines the determination and properties of a rhombus, the Pythagorean theorem, and the determination and properties of an equilateral triangle. Mastering the theorem for determining a rhombus is key to solving the problem." }, { "problem_id": 1810, "question": "As shown in Figure 1, during the creation of a civilized city, a propaganda sign was set up on the roadside. Figure 2 is a mathematical model abstracted from this scene. The top of the sign $(A B)$ has a rope $(A C)$, which naturally hangs down, and the bottom of the rope is $0.7 \\mathrm{~m}$ above the ground (i.e., $B C=0.7$). The staff pulls the bottom of the rope to a point $3 \\mathrm{~m}$ away from the sign (i.e., the distance from point $E$ to $A B$ is $3 \\mathrm{~m}$), and the rope is just taut. Given that the staff's height $(D E)$ is $1.7 \\mathrm{~m}$, find the height of the sign $(A B)$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch27-2024_06_17_9ccecbba83dd0302531bg_0084_1.jpg", "batch27-2024_06_17_9ccecbba83dd0302531bg_0084_2.jpg" ], "is_multi_img": true, "answer": "$5.7 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular line $EF \\perp AB$ at point $F$.\n\nFrom the problem statement, we have $AC = AE$, $CB = 0.7$, $BF = DE = 1.7$, and $EF = BD = 3$.\n\nThus, $CF = BF - BC = DE - BC = 1.7 - 0.7 = 1 \\text{ m}$.\n\nLet $AF = x \\text{ m}$, then $AE = AC = (x + 1) \\text{ m}$.\n\nIn the right triangle $AEF$, $\\angle AFE = 90^\\circ$.\n\nBy the Pythagorean theorem, we have $AE^2 = AF^2 + EF^2$,\n\nwhich is $(x + 1)^2 = x^2 + 3^2$,\n\nsolving for $x$ gives $x = 4$.\n\nTherefore, $AB = AF + BF = 4 + 1.7 = 5.7 \\text{ m}$.\n\nAnswer: The height of the billboard $(AB)$ is $5.7 \\text{ m}$.\n\n\n\n【Key Insight】This problem tests the application of the Pythagorean theorem and the properties of rectangles. The key to solving the problem lies in appropriately drawing auxiliary lines to construct a right triangle and applying the Pythagorean theorem to establish an equation." }, { "problem_id": 1811, "question": "In class, Xiao Qiang performed the following practical operation:\n\nStep 1: Fold a rectangular piece of paper into a square using the method shown in Figure 1, then unfold the paper.\n\nStep 2: As shown in Figure 2, fold the square into two congruent rectangles, then unfold the paper.\n\nStep 3: As shown in Figure 3, fold the diagonal \\( AB \\) of the inner rectangle \\( AFBC \\), and then fold \\( AB \\) along \\( AQ \\) to \\( AD \\).\n\nStep 4: As shown in Figure 4, unfold the paper and fold along the point \\( D \\) to create \\( DE \\), resulting in the rectangle \\( BCDE \\).\n\nThen, \\(\\frac{BE}{DE}=\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nFigure 4", "input_image": [ "batch27-2024_06_17_ed60fb49715af1dfabddg_0004_1.jpg", "batch27-2024_06_17_ed60fb49715af1dfabddg_0004_2.jpg", "batch27-2024_06_17_ed60fb49715af1dfabddg_0004_3.jpg", "batch27-2024_06_17_ed60fb49715af1dfabddg_0004_4.jpg" ], "is_multi_img": true, "answer": "$\\frac{\\sqrt{5}-1}{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: Let the side length of rectangle \\( M N C B \\) be \\( a \\).\n\nFrom the second step, we know:\n\\[ N A = A C = \\frac{1}{2} N C = \\frac{1}{2} a \\]\n\nTherefore:\n\\[ A B = \\sqrt{B C^{2} + A C^{2}} = \\sqrt{a^{2} + \\left(\\frac{a}{2}\\right)^{2}} = \\frac{\\sqrt{5} a}{2} \\]\n\nFrom the third step, we have:\n\\[ A D = A B = \\frac{\\sqrt{5} a}{2} \\]\n\nThus:\n\\[ C D = A D - A C = \\frac{\\sqrt{5} a}{2} - \\frac{1}{2} a = \\frac{(\\sqrt{5} - 1) a}{2} \\]\n\nFrom the fourth step, we obtain:\n\\[ B E = C D, \\quad D E = B C \\]\n\nTherefore:\n\\[ \\frac{B E}{D E} = \\frac{C D}{B C} = \\frac{(\\sqrt{5} - 1) a}{2} = \\frac{\\sqrt{5} - 1}{2} \\]\n\nHence, the answer is:\n\\[ \\frac{\\sqrt{5} - 1}{2} \\]\n\n【Highlight】This problem primarily examines the combination of rectangles and the Pythagorean theorem. The key to solving it lies in the application of the Pythagorean theorem." }, { "problem_id": 1812, "question": "As shown in the figure, there is a streetlight between two teaching buildings of a middle school. Person A and Person B observe the top of the streetlight from their respective buildings, with their lines of sight as shown in Figure (1). Based on the actual situation, a planar diagram is drawn as shown in Figure (2), where $C D \\perp D F, A B \\perp D F$, and $E F \\perp D F$. Person A can see point $G$ from point $C$, and Person B can see point $D$ from point $E$. Point $B$ is the midpoint of $D F$, the streetlight $A B$ is 8 meters high, $D F = 120$ meters, and $\\tan \\angle A G B = \\frac{1}{3}$. Find the difference in the heights from the ground of the observation points of Person A and Person B.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch28-2024_06_17_283e09c392bc19054970g_0025_1.jpg", "batch28-2024_06_17_283e09c392bc19054970g_0025_2.jpg" ], "is_multi_img": true, "answer": "12 m.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AB \\perp DF \\) and \\( EF \\perp DF \\),\n\n\\(\\triangle ABD \\sim \\triangle EFD\\),\n\nTherefore, \\(\\frac{AB}{EF} = \\frac{DB}{DF}\\).\n\nSince point \\( B \\) is the midpoint of \\( DF \\),\n\n\\( DB = BF = 60 \\) meters,\n\nHence, \\( EF = 2AB = 16 \\) meters.\n\nSince \\( \\tan \\angle AGB = \\frac{DC}{DG} = \\frac{AB}{BG} = \\frac{1}{3} \\),\n\n\\( BG = 3AB = 24 \\) meters,\n\nTherefore, \\( DG = DB + BG = 84 \\) meters,\n\nAnd \\( CD = \\frac{1}{3} DG = 28 \\) meters,\n\nThus, \\( CD - EF = 28 - 16 = 12 \\) meters,\n\nTherefore, the difference in the heights of the two observers' points from the ground is 12 meters.\n\n[Highlight] This problem involves the application of similar triangles, primarily testing the properties of trigonometric functions and similar triangles. The key to solving this problem lies in determining the length of \\( CD \\)." }, { "problem_id": 1813, "question": "Figures (1) and (2) show the physical and schematic diagrams of a certain model of treadmill. It is known that the handles $\\mathrm{AB}$ of the treadmill are parallel to the ground and the height $h$ from the ground is approximately $1.05 \\mathrm{~m}$. The angle $\\angle C D E$ between the treadmill deck $C D$ and the ground $D E$ is $10^{\\circ}$, the length of the support (segment $A C$) is $0.8 \\mathrm{~m}$, and the angle $\\angle A C D$ is $82^{\\circ}$. Find the length of the treadmill deck $C D$ (accurate to $0.1 \\mathrm{~m}$).\n\n(Reference data: $\\sin 10^{\\circ}=\\cos 80^{\\circ} \\approx 0.17, \\sin 72^{\\circ}=\\cos 18^{\\circ} \\approx 0.95, \\tan 72^{\\circ} \\approx 3.1$)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch28-2024_06_17_283e09c392bc19054970g_0096_1.jpg", "batch28-2024_06_17_283e09c392bc19054970g_0096_2.jpg" ], "is_multi_img": true, "answer": " $1.7 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw a line $FG$ perpendicular to $AB$ at point $C$, intersecting $DE$ at point $G$.\n\n\n\nSince $AB$ is parallel to $DE$, it follows that $FG$ is perpendicular to $DE$, and thus $\\angle CGE = 90^\\circ$.\n\nGiven that $\\angle CDE = 10^\\circ$, we find that $\\angle GCD = 90^\\circ - 10^\\circ = 80^\\circ$.\n\nAlso, since $\\angle ACD = 82^\\circ$, we have $\\angle ACF = 180^\\circ - \\angle ACD - \\angle GCD = 180^\\circ - 80^\\circ - 82^\\circ = 18^\\circ$.\n\nTherefore, in the right triangle $\\triangle ACF$, $CF = AC \\cdot \\cos \\angle ACF = 0.8 \\cdot \\cos 18^\\circ \\approx 0.76$ meters.\n\nThen, $CG = h - CF = 1.05 - 0.76 = 0.29$ meters.\n\nThus, in the right triangle $\\triangle CDG$,\n\n$CD = \\frac{CG}{\\sin \\angle CDE} = \\frac{0.29}{\\sin 10^\\circ} \\approx \\frac{0.29}{0.17} \\approx 1.7$ meters.\n\nTherefore, the length of the treadmill pedal $CD$ is approximately $1.7$ meters.\n\n[Insight] This problem tests the understanding of solving right triangles and trigonometric functions. The key to solving it lies in mastering the application of solving right triangles and the calculation of trigonometric functions." }, { "problem_id": 1814, "question": "As shown in Figure 1, a spherical ball is placed in a $V$-shaped stand. Figure 2 is its planar schematic diagram, where $CA$ and $CB$ are both tangents to the circle $\\odot O$, with tangent points at $A$ and $B$ respectively. If the radius of $\\odot O$ is $2\\sqrt{3} \\, \\text{cm}$, and $AB = 6 \\, \\text{cm}$, find $\\angle ACB$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_2917f2173709a7a1a4d6g_0056_1.jpg", "batch28-2024_06_17_2917f2173709a7a1a4d6g_0056_2.jpg" ], "is_multi_img": true, "answer": "$60^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $OC$ intersecting $AB$ at point $D$.\n\nSince $CA$ and $CB$ are tangents to the circle $O$,\n\n$\\therefore CA = CB$, and $OC$ bisects $\\angle ACB$.\n\n$\\therefore OC \\perp AB$.\n\nGiven that $AB = 6$,\n\n$\\therefore BD = 3$.\n\nIn the right triangle $\\triangle OBD$,\n\nSince $OB = 2\\sqrt{3}$,\n\n$\\therefore \\sin \\angle BOD = \\frac{BD}{OB} = \\frac{3}{2\\sqrt{3}} = \\frac{\\sqrt{3}}{2}$.\n\n$\\therefore \\angle BOD = 60^\\circ$.\n\nSince $B$ is the point of tangency,\n\n$\\therefore OB \\perp BC$.\n\n$\\therefore \\angle OCB = 30^\\circ$.\n\n$\\therefore \\angle ACB = 60^\\circ$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n[Key Insight] This problem mainly examines the properties of tangents and solving right triangles. Constructing a right triangle to find the angle is a commonly used method." }, { "problem_id": 1815, "question": "As shown in the figure, in the rectangular paper $ABCD$, $AB = 6, AD = 8$. The folding process is carried out as follows:\n\nStep 1: First, fold the rectangle $ABCD$ in half, with the crease line being $MN$, as shown in Figure (1);\n\nStep 2: Then, fold point $B$ onto the crease line $MN$, with the crease being $AE$, and the corresponding point of $B$ on $MN$ being $B'$, resulting in the right triangle $\\triangle AB'E$, as shown in Figure (2);\n\nStep 3: Fold along $EB'$ with the crease being $EF$, and $AF$ intersecting the extension of $B'N$ at point $G$, as shown in Figure (3);\n\nThen, in the figure formed by the folded paper, the area $S_{\\triangle ABG}$ is $\\qquad$\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch28-2024_06_17_2f0ea24d8b1926806dd3g_0032_1.jpg", "batch28-2024_06_17_2f0ea24d8b1926806dd3g_0032_2.jpg", "batch28-2024_06_17_2f0ea24d8b1926806dd3g_0032_3.jpg" ], "is_multi_img": true, "answer": "$3 \\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: As shown in Figure 2, extend $EB'$ to intersect $AD$ at point $F$.\n\n\n\nFigure (2)\n\nGiven that $AM = \\frac{1}{2} AB'$,\n\nSince $\\angle M = 90^\\circ$,\n\nTherefore, $\\angle AB'M = 30^\\circ$,\n\nSince $MB' \\parallel AF$,\n\nTherefore, $\\angle AB'M = \\angle B'AF = 30^\\circ$,\n\nThus, $B'F = AB' \\cdot \\tan 30^\\circ = 2\\sqrt{3}$,\n\nTherefore, the area of $\\triangle AFB$ is $\\frac{1}{2} \\cdot AB' \\cdot FB' = \\frac{1}{2} \\times 6 \\times 2\\sqrt{3} = 6\\sqrt{3}$,\n\nSince $AG = GF$,\n\nTherefore, the area of $\\triangle AGB$ is $\\frac{1}{2} S_{\\triangle AB'F} = 3\\sqrt{3}$.\n\nHence, the answer is: $3\\sqrt{3}$.\n\n【Highlight】This problem examines the concepts of reflection transformation, properties of rectangles, and solving right triangles. The key to solving the problem lies in understanding the given conditions and flexibly applying the learned knowledge to find the solution." }, { "problem_id": 1816, "question": "As shown in Figure (1), $\\triangle A B C$ and $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ are two equilateral triangles with unequal side lengths, with points $B^{\\prime}, C^{\\prime}, B, C$ all lying on line $l$. $\\triangle A B C$ remains stationary, while $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is translated from left to right along line $l$. Initially, point $C^{\\prime}$ coincides with point $B$, and the translation stops when point $B^{\\prime}$ coincides with point $C$. Let the distance of translation of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ be $x$, and the area of overlap between the two triangles be $y$. The functional relationship between $y$ and $x$ is shown in Figure (2). The side length of $\\triangle A B C$ is $\\qquad$.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch28-2024_06_17_2f0ea24d8b1926806dd3g_0037_1.jpg", "batch28-2024_06_17_2f0ea24d8b1926806dd3g_0037_2.jpg" ], "is_multi_img": true, "answer": "5\n", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: When point $B^{\\prime}$ moves to point $B$, the area of the overlapping part no longer changes.\n\nAccording to the image, we know that $B^{\\prime} C^{\\prime} = a$ and $S_{\\triangle A^{\\prime} B^{\\prime} C^{\\prime}} = \\sqrt{3}$.\n\nDrawing $A^{\\prime} H \\perp B^{\\prime} C^{\\prime}$ through point $A^{\\prime}$,\n\nthen $A^{\\prime} H$ is the height of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$.\n\n\n\nSince $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is an equilateral triangle,\n\n$\\angle A^{\\prime} B^{\\prime} H = 60^{\\circ}$,\n\nthus $\\sin 60^{\\circ} = \\frac{A^{\\prime} H}{A^{\\prime} B^{\\prime}} = \\frac{\\sqrt{3}}{2}$,\n\nso $A^{\\prime} H = \\frac{\\sqrt{3}}{2} a$,\n\ntherefore $S_{\\triangle A^{\\prime} B^{\\prime} C^{\\prime}} = \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} a \\cdot a$, which is $\\frac{\\sqrt{3}}{4} a^{2} = \\sqrt{3}$,\n\nsolving gives $a = -2$ (discarded) or $a = 2$.\n\nWhen point $C$ moves to point $C$, the area of the overlapping part starts to decrease.\n\nAccording to the image, $B C = a + 3 = 2 + 3 = 5$,\n\nthus the side length of $\\triangle A B C$ is 5.\n\nTherefore, the answer is 5.\n\n【Insight】This problem mainly examines the function images and trigonometric functions in moving point problems. The key is to analyze clearly into which stages the moving process can be divided, how each stage changes, first a stage before point $B^{\\prime}$ reaches $B$, then a stage when point $C$ reaches $C$, and finally another stage when $B^{\\prime}$ reaches $C$. By distinguishing these stages and listing the equations based on the image information, the problem can be solved." }, { "problem_id": 1817, "question": "Congcong came across the following information in an extracurricular mathematics book: In 1925, mathematician Moron discovered the world's first perfect rectangle, as shown in Figure 1, which can be exactly divided into 10 differently sized squares. After carefully studying this diagram, Congcong designed a \"quasi-perfect rectangle\" as shown in Figure 2, where the squares labeled \"3 and 4\" are identical. If the side length of the square labeled \"1\" in the middle is $1 \\mathrm{~cm}$, find the area of this \"quasi-perfect rectangle\".\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_429fa9fcdc6e9005853dg_0083_1.jpg", "batch28-2024_06_17_429fa9fcdc6e9005853dg_0083_2.jpg" ], "is_multi_img": true, "answer": "$143 \\mathrm{~cm}^{2}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Let the side length of square 3 be \\( x \\) cm,\n\nthen the side length of square 2 is \\( (2x - 1) \\) cm,\n\nthe side length of square 5 is \\( (x + 1) \\) cm,\n\nand the side length of square 6 is \\( (x + 2) \\) cm.\n\nAccording to the problem, we have the equation:\n\\[ x + 1 + x + 2 = 2x - 1 + x \\]\n\nSolving the equation, we find:\n\\[ x = 4 \\]\n\nTherefore, the width of the rectangle is:\n\\[ x + 1 + x + 2 = 11 \\text{ cm} \\]\n\nAnd the length of the rectangle is:\n\\[ x + x + x + 1 = 13 \\text{ cm} \\]\n\nThus, the area of this \"quasi-perfect rectangle\" is:\n\\[ 13 \\times 11 = 143 \\text{ cm}^2 \\]\n\n[Highlight] This problem examines the application of linear equations, properties of squares, and the area of rectangles. The key to solving the problem is to understand the given conditions and establish the appropriate equation based on the relationships provided, then solve for the unknown." }, { "problem_id": 1818, "question": "As shown in Figure 1, Dalian Sightseeing Tower, originally named Dalian Radio and Television Tower, is located at the summit of Lushan Mountain south of Labor Park in downtown Dalian. It is a landmark building of Dalian and a national AAA-rated tourist attraction, serving as the first height and the best viewpoint for tourists to appreciate the scenery of Dalian. As shown in Figure 2, a person named Xiaopang (AB) with a height of $1.6 \\mathrm{~m}$ stands at point B, measuring the elevation angles of points N and M relative to point A as $30^{\\circ}$ and $50.86^{\\circ}$ respectively. After consulting reference materials, he learned that the height of CN is $170 \\mathrm{~m}$. Please help Xiaopang calculate the height of the tower MN.\n\n(The result should be rounded to the nearest whole number. Reference data: $\\sin 50.86^{\\circ} \\approx 0.78, \\cos 50.86^{\\circ} \\approx 0.63, \\tan 50.86^{\\circ} \\approx 1.23$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_53f13b9e7296b25323e4g_0078_1.jpg", "batch28-2024_06_17_53f13b9e7296b25323e4g_0078_2.jpg" ], "is_multi_img": true, "answer": "$190 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Draw $AD \\perp MC$ at point $D$,\n\n\n\nFigure 2\n\nSince $AB \\perp BC$ and $DC \\perp BC$,\n\nTherefore, quadrilateral $ABCD$ is a rectangle,\n\nHence, $CD = AB = 1.6 \\mathrm{~m}$,\n\nThus, $DN = CN - CD = 170 - 1.6 = 168.4 (\\mathrm{~m})$,\n\nIn right triangle $\\triangle ADN$, $\\angle NAD = 30^{\\circ}$, $\\tan \\angle NAD = \\frac{DN}{AD}$,\n\nTherefore, $AD = \\frac{DN}{\\tan 30^{\\circ}} = \\frac{842}{5} \\sqrt{3}$,\n\nIn right triangle $\\triangle ADM$, $\\angle MAD = 50.86^{\\circ}$,\n\n$DM = AD \\cdot \\tan \\angle MAD \\approx \\frac{842}{5} \\sqrt{3} \\times 1.23 \\approx 358.3 (\\mathrm{~m})$,\n\nHence, $MN = DM - DN = 358.3 - 168.4 \\approx 190 (\\mathrm{~m})$,\n\nAnswer: The height of tower $MN$ is $190 \\mathrm{~m}$.\n\n【Key Insight】This problem tests the application of acute angle trigonometric functions. Correctly understanding the problem and establishing a right triangle is the key to solving it." }, { "problem_id": 1819, "question": "Vanke Plaza has become an important place for people to relax and entertain on weekends. There is an escalator from the first floor to the second floor (as shown in Figure 1). Figure 2 is a side view schematic. It is known that the slope (or slope ratio) of the escalator $AC$ is $i=1:2$, $AC=6\\sqrt{5}$ meters, $BE$ is the ceiling of the second floor, $EF // MN$, point $B$ is on $EF$ and directly above the top of the escalator $C$. If $BC \\perp EF$, and the angle of elevation to point $B$ measured from the bottom of the escalator $A$ is $40^{\\circ}$, find the height of the second floor $BC$. (Round to the nearest 0.1 meter, reference data: $\\sin 40^{\\circ} \\approx 0.64, \\cos 40^{\\circ} \\approx 0.77, \\tan 40^{\\circ} \\approx 0.84$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_6745b66e3e336a9ea3eeg_0025_1.jpg", "batch28-2024_06_17_6745b66e3e336a9ea3eeg_0025_2.jpg" ], "is_multi_img": true, "answer": "4.1 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, extend $BC$ to intersect $MN$ at point $D$.\n\nSince $EF \\parallel MN$ and $BC \\perp EF$,\n\nIt follows that $BD \\perp MN$, which means $\\angle ADB = 90^\\circ$.\n\nGiven that the slope (or gradient) of the escalator $AC$ is $i = 1:2$,\n\nWe have $CD: AD = 1:2$,\n\nThus, $CD = \\frac{1}{2} AD$.\n\nIn the right triangle $\\triangle ADC$, by the Pythagorean theorem: $AC^2 = CD^2 + AD^2$,\n\nTherefore, $(6\\sqrt{5})^2 = \\left(\\frac{1}{2} AD\\right)^2 + AD^2$,\n\nSolving this, we find $AD = 12$ meters (discarding the negative value),\n\nHence, $CD = 6$ meters.\n\nIn the right triangle $\\triangle ABD$, with $\\angle ADB = 90^\\circ$ and $\\angle BAD = 40^\\circ$,\n\nWe calculate $BD = AD \\cdot \\tan \\angle BAD \\approx 10.1$ meters,\n\nThus, $BC = BD - CD = 4.1$ meters,\n\nTherefore, the height of the second floor $BC$ is approximately 4.1 meters.\n\n\n\nFigure 2\n\n[Key Insight] This problem primarily examines the practical application of solving right triangles and the Pythagorean theorem. The key to solving the problem lies in correctly constructing auxiliary lines to form right triangles." }, { "problem_id": 1820, "question": "Figure 1 shows an adjustable desk lamp, and Figure 2 is a schematic diagram of its plan view. The fixed base $O A$ is perpendicular to the horizontal plane $O E$, $A B$ is the fixed support rod, $B C$ is the adjustable rod that can rotate around point $B$, and the lamp body $C D$ always remains perpendicular to $B C$. $M N$ is the area illuminated by the desk lamp on the table surface. As shown in Figure 2, by rotating the adjustable rod so that $B C$ is parallel to the horizontal plane $O E$, the triangle $\\triangle D M N$ becomes an isosceles triangle with $D$ as its vertex. Given $A B=5 \\mathrm{dm}, O M=2 \\mathrm{dm}, B C=6 \\mathrm{dm}, \\tan B=\\frac{4}{3}$, find the length of the illuminated area $M N$ on the table surface.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_6745b66e3e336a9ea3eeg_0080_1.jpg", "batch28-2024_06_17_6745b66e3e336a9ea3eeg_0080_2.jpg" ], "is_multi_img": true, "answer": "$2 \\mathrm{dm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in the figure, draw $AP \\perp BC$ at point $P$ from point $A$, and draw $DQ \\perp MN$ at point $Q$ from point $D$.\n\nSince $OA \\perp OE$,\n\nTherefore, quadrilateral $OPCQ$ is a rectangle,\n\nHence, $PC = OQ$.\n\n\n\nFigure 2\n\nIn the right triangle $\\triangle ABP$,\n\nSince $AB = 5 \\text{ dm}$ and $\\tan B = \\frac{4}{3} = \\frac{AP}{BP}$,\n\nLet $AP = 4x$, $BP = 3x$,\n\nSince $(3x)^2 + (4x)^2 = 5^2$,\n\nTherefore, $x = 1$,\n\nHence, $AP = 4 \\text{ dm}$, $BP = 3 \\text{ dm}$,\n\nAlso, since $BC = 6 \\text{ dm}$,\n\nTherefore, $PC = 6 - 3 = 3 \\text{ dm} = OQ$,\n\nSince $OM = 2 \\text{ dm}$,\n\n$MQ = 3 - 2 = 1 \\text{ dm}$,\n\nSince $DM = DN$ and $DQ \\perp MN$,\n\nTherefore, $MN = 2MQ = 2 \\text{ dm}$.\n\n【Insight】This problem examines the application of solving right triangles, the determination and properties of rectangles, the properties of isosceles triangles, and the Pythagorean theorem. Mastering the relationship between the sides and angles of right triangles is essential for correct solutions, and constructing right triangles is key to solving the problem." }, { "problem_id": 1821, "question": "In the chapter \"Simple Machines and Work\" of the ninth-grade physics textbook by Suke Publishing, there is a problem that states: \"As shown in Figure 1, a uniform rod \\( AB \\) with a length of \\( 8 \\mathrm{dm} \\) can rotate freely in a vertical plane around a pivot point \\( A \\). A small fixed pulley is fixed at a height of \\( 10 \\mathrm{dm} \\) directly above point \\( A \\). A thin rope passes through the pulley and is connected to the other end \\( B \\) of the rod. The rod \\( AB \\) is slowly pulled upwards from a horizontal position. When the angle between the rod \\( AB \\) and the horizontal plane is \\( 30^\\circ \\), find the length of the lever arm.\" From a mathematical perspective, this problem can be interpreted as follows: As shown in Figure 2, given that \\( \\angle BAD = 30^\\circ \\), \\( AB = 8 \\mathrm{dm} \\), \\( CA \\perp AD \\) at point \\( D \\) and \\( CA = 10 \\mathrm{dm} \\), connect \\( CB \\) and find the distance from point \\( A \\) to \\( BC \\). Please write out the solution process to find the distance from point \\( A \\) to \\( BC \\). (The result should be kept in radical form)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_6745b66e3e336a9ea3eeg_0100_1.jpg", "batch28-2024_06_17_6745b66e3e336a9ea3eeg_0100_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{20 \\sqrt{7}}{7} \\mathrm{dm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, draw a perpendicular from point $B$ to $AC$ at point $F$, and draw a perpendicular from point $A$ to $BC$ at point $E$.\n\n\n\nFigure 2\n\nGiven that $\\angle BAD = 30^\\circ$, $AB = 8$, and $CA \\perp AD$,\n\nwe have $\\angle FAB = 90^\\circ - 30^\\circ = 60^\\circ$,\n\nthus $AF = AB \\cdot \\cos \\angle FAB = \\frac{1}{2} AB = 4$, and $BF = AB \\cdot \\sin \\angle FAB = \\frac{\\sqrt{3}}{2} \\times 8 = 4\\sqrt{3}$,\n\nso $FC = AC - AF = 10 - 4 = 6$,\n\nand $BC = \\sqrt{BF^2 + CF^2} = \\sqrt{(4\\sqrt{3})^2 + 6^2} = \\sqrt{84} = 2\\sqrt{21}$.\n\nSince $\\frac{1}{2} AE \\cdot BC = \\frac{1}{2} AC \\cdot BF$,\n\nwe have $AE = \\frac{AC \\cdot BF}{BC} = \\frac{10 \\times 4\\sqrt{3}}{2\\sqrt{21}} = \\frac{20\\sqrt{7}}{7}$,\n\nthus $AE = \\frac{20\\sqrt{7}}{7}$,\n\nwhich means the distance from point $A$ to $BC$ is $\\frac{20\\sqrt{7}}{7}$ dm.\n\n【Insight】This problem examines the application of solving right triangles, and constructing a right triangle is the key to solving it." }, { "problem_id": 1822, "question": "On October 23, 2021, China's first 10,000-ton maritime patrol vessel, the \"Haixun 09\" (Figure 1), was commissioned in Nansha, Guangzhou, joining the Chinese maritime law enforcement fleet. This marks the official entry into service of the country's largest, most advanced, and versatile public law enforcement vessel, which is at a world-leading level in terms of tonnage, equipment, and comprehensive capabilities. As shown in Figure 2, at 9:00 AM, the \"Haixun 09\" was at point A, which is 25° south of east from City B and 58° north of east from Island C. City B is 28° north of east from Island C and is 372 km away from it. At that time, the \"Haixun 09\" was sailing at a speed of 30 km/h along the direction of AC. The question is: approximately how long will it take for the \"Haixun 09\" to reach Island C? (Reference data: $\\sqrt{3} \\approx 1.73, \\sin 53^{\\circ} \\approx \\frac{4}{5}, \\cos 53^{\\circ} \\approx \\frac{3}{5}, \\tan 53^{\\circ} \\approx \\frac{4}{3}$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0041_1.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0041_2.jpg" ], "is_multi_img": true, "answer": "10 hours", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular from point $A$ to $BC$ at point $D$,\n\n\nGiven the conditions, $\\angle ABC = 28^\\circ + 25^\\circ = 53^\\circ$, $\\angle ACB = 58^\\circ - 28^\\circ = 30^\\circ$, and $BC = 372 \\text{ km}$. Let $AD = x \\text{ km}$.\n\nIn right triangle $\\triangle ABD$, since $\\angle ABD = 53^\\circ$,\n\n$\\therefore BD = \\frac{AD}{\\tan 53^\\circ} \\approx \\frac{3}{4} x \\text{ km}$.\n\nIn right triangle $\\triangle ACD$, since $\\angle ACD = 30^\\circ$,\n\n$\\therefore CD = \\frac{AD}{\\tan 30^\\circ} = \\sqrt{3} x \\text{ km}$.\n\nSince $BD + CD = BC$,\n\n$\\therefore \\frac{3}{4} x + \\sqrt{3} x = 372$,\n\nSolving for $x$, we get: $x \\approx 150$,\n\n$\\therefore AD = 150 \\text{ km}$, and $AC = 2 AD = 300 \\text{ km}$,\n\n$\\therefore 300 \\div 30 = 10 (\\text{ hours})$,\n\nAnswer: The patrol ship \"Haixun 09\" will take approximately 10 hours to reach island $C$.\n\n【Key Insight】This problem tests the application of solving right triangles and the properties of right triangles. Drawing the auxiliary line is crucial to solving this problem." }, { "problem_id": 1823, "question": "With the popularization of information-based teaching, many teaching scenarios have introduced projectors (as shown in Figure (1)). Figure (2) is a cross-sectional view of the projector installation. Given that the angle between the projector's light beams $\\angle B A C = 45^{\\circ}$, $\\angle A C B = 30^{\\circ}$, the length of the boom $A D = 0.5 \\mathrm{~m}$, the height of the projection screen $B C = 1.6 \\mathrm{~m}$, and $A D \\perp D E, D E \\perp E F$. Find the height from the bottom edge of the screen $C$ to the top of the classroom $C E$. (The result should be rounded to one decimal place, reference data: $\\sqrt{3} \\approx 1.73$)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0076_1.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0076_2.jpg" ], "is_multi_img": true, "answer": "2.4 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Through point $B$, draw $BG \\perp AC$ to meet at $G$, and through point $A$, draw $AH \\perp EF$ to meet at $H$.\n\n\n\nIn the right triangle $\\triangle BCG$, $BG = BC \\sin 30^\\circ = 1.6 \\times \\frac{1}{2} = 0.8 \\text{ m}$, and $CG = BC \\cos 30^\\circ = 1.6 \\times \\frac{\\sqrt{3}}{2} = 0.8 \\sqrt{3} \\text{ m}$.\n\nIn the right triangle $\\triangle ABG$, $AG = \\frac{BG}{\\tan 45^\\circ} = \\frac{0.8}{1} = 0.8 \\text{ m}$,\n\n$\\therefore AC = AG + CG = 0.8 + 0.8 \\sqrt{3} \\text{ m}$.\n\nIn the right triangle $\\triangle ACH$, $CH = AC \\cos 30^\\circ = (0.8 + 0.8 \\sqrt{3}) \\times \\frac{\\sqrt{3}}{2} = 0.4 \\sqrt{3} + 1.2 \\text{ m}$.\n\n$\\therefore CE = CH + EH = CH + AD = 0.4 \\sqrt{3} + 1.2 + 0.5 \\approx 2.4 \\text{ m}$.\n\nAnswer: The height $CE$ from the lower edge $C$ of the screen to the top of the classroom is approximately 2.4 meters.\n\n【Highlight】This question examines the practical application of trigonometric functions. The key to solving this problem lies in drawing the correct auxiliary lines." }, { "problem_id": 1824, "question": "Perform a paper folding game on an equilateral $\\triangle A B C$ paper with side length 2, where point $D$ is the midpoint of $A C$, as shown in Figure (1). Take any point $E$ on side $A B$, fold the paper along $D E$, and let point $A$ fall at $A^{\\prime}$, then fold the paper along $A^{\\prime} D$, and let point $E$ fall at $E^{\\prime}$, as shown in Figure (2); when point $E^{\\prime}$ exactly falls on one of the sides of the original equilateral triangle paper (not coinciding with the vertices), the length of segment $A E$ is . $\\qquad$\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch28-2024_06_17_c3cbaa3f7f75e1a27feag_0054_1.jpg", "batch28-2024_06_17_c3cbaa3f7f75e1a27feag_0054_2.jpg" ], "is_multi_img": true, "answer": "$2-\\sqrt{3}$ or $\\sqrt{3}-1 / \\sqrt{3}-1$ or $2-\\sqrt{3}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: Solve the problem by considering two scenarios:\n\n(1) As shown in Figure (1), when point \\( E^{\\prime} \\) lies on side \\( AB \\), the intersection of \\( DA^{\\prime} \\) and \\( AB \\) is point \\( M \\).\n\n\n\n(1)\n\nGiven the conditions: \\( EE^{\\prime} \\perp DA^{\\prime} \\), \\( AE = A^{\\prime}E \\), \\( \\angle A = 60^{\\circ} = \\angle EA^{\\prime}D \\), and \\( AD = \\frac{1}{2} AC = 1 \\).\n\nThus, \\( \\angle ADM = 30^{\\circ} \\), \\( \\angle A^{\\prime}EM = 30^{\\circ} \\).\n\nTherefore, \\( AM = \\frac{1}{2} AD = \\frac{1}{2} \\), and \\( EM = A^{\\prime}E \\cos 30^{\\circ} = \\frac{\\sqrt{3}}{2} AE \\).\n\nSince \\( AM = AE + EM \\),\n\\[ \\frac{1}{2} = AE + \\frac{\\sqrt{3}}{2} AE \\]\n\nSolving gives \\( AE = 2 - \\sqrt{3} \\);\n\n(2) As shown in Figure (2), when \\( E^{\\prime} \\) lies on side \\( BC \\), the intersection of \\( DA^{\\prime} \\) and \\( EE^{\\prime} \\) is point \\( M \\), and a perpendicular \\( EN \\) is drawn from \\( E \\) to \\( AC \\).\n\n\n\n(2)\n\nGiven the conditions: \\( EE^{\\prime} \\perp DA^{\\prime} \\), \\( DA^{\\prime} \\perp AC \\), \\( AE = A^{\\prime}E \\), \\( \\angle A = 60^{\\circ} = \\angle EA^{\\prime}D \\).\n\nThus, \\( \\angle EDA = \\angle EDA^{\\prime} = \\frac{1}{2} \\angle ADA^{\\prime} = 45^{\\circ} \\).\n\nTherefore, \\( EN = DN \\), \\( \\angle AEN = 30^{\\circ} \\), and \\( AN = \\frac{EN}{\\tan 60^{\\circ}} = \\frac{\\sqrt{3}}{3} EN \\).\n\nSince \\( AN + DN = AD \\),\n\\[ \\frac{\\sqrt{3}}{3} DN + DN = 1 \\]\n\nSolving gives \\( DN = \\frac{3 - \\sqrt{3}}{2} \\)\n\nThus, \\( AN = \\frac{\\sqrt{3} - 1}{2} \\)\n\nTherefore, \\( AE = 2 AN = \\sqrt{3} - 1 \\);\n\nThe final answers are: \\( 2 - \\sqrt{3} \\) or \\( \\sqrt{3} - 1 \\).\n\n【Highlight】This problem examines the properties of folding, equilateral triangles, right-angled triangles with a 30-degree angle, isosceles right-angled triangles, and trigonometric values. The key to solving the problem lies in considering different scenarios." }, { "problem_id": 1825, "question": "Using a tangram as shown in Figure 1, form a \"pinwheel pattern\" with a vacant square in the center as shown in Figure 2. Then, in Figure 2, $\\tan \\angle A B C=$ $\\qquad$ .\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_c3cbaa3f7f75e1a27feag_0090_1.jpg", "batch28-2024_06_17_c3cbaa3f7f75e1a27feag_0090_2.jpg" ], "is_multi_img": true, "answer": "3", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $AG$ and extend it to intersect $BC$ at point $H$,\n\n\n\nFigure 2\n\nLet $AK = a$,\n\nThen $EK = EC = EF = KG = AN = BG = a$,\n\nSince $BN = BF$ and $NG = FG$,\n\nTherefore, $BG$ is the perpendicular bisector of $NF$,\n\nSince $GK \\perp AE$,\n\nTherefore, points $B$, $G$, and $K$ lie on the same straight line,\n\nThus, $AB = \\sqrt{5}a$,\n\nSince $\\angle C = \\angle HAC = 45^\\circ$,\n\nTherefore, $\\angle AHC = 90^\\circ$,\n\nThus, $AH \\perp BC$,\n\nTherefore, $AH = \\frac{\\sqrt{2}}{2} AC = \\frac{3\\sqrt{2}}{2}a$,\n\nThus, $BH = \\sqrt{AB^2 - AH^2} = \\frac{\\sqrt{2}}{2}a$,\n\nTherefore, $\\tan \\angle ABC = \\frac{AH}{BH} = \\frac{\\frac{3\\sqrt{2}}{2}a}{\\frac{\\sqrt{2}}{2}a} = 3$.\n\nHence, the answer is: 3.\n【Key Insight】This problem examines the application of solving right triangles, the properties of tangram, and the properties of squares. The key to solving this problem lies in mastering the properties of squares." }, { "problem_id": 1826, "question": "Figure 1 shows a 3D model of a robotic arm. The base of the robotic arm, \\( AB \\), is fixed, with a height of \\( 40 \\, \\text{cm} \\). The connecting rod \\( BC \\) has a length of \\( 60 \\, \\text{cm} \\), the arm \\( CD \\) has a length of \\( 50 \\, \\text{cm} \\), and the manipulator \\( DE \\) has a length of \\( 30 \\, \\text{cm} \\). Points \\( B \\), \\( C \\), and \\( D \\) are rotational joints, and \\( AB \\), \\( BC \\), \\( CD \\), and \\( DE \\) always lie in the same plane. By rotating the connecting rod \\( BC \\), the arm \\( CD \\), and the manipulator \\( DE \\), we achieve \\( \\angle ABC = 150^\\circ \\), \\( CD \\parallel AF \\), and \\( \\angle CDE = 120^\\circ \\), as shown in Figure 2. Determine the height of the manipulator's endpoint \\( E \\) above the ground (round the result to the nearest \\( 1 \\, \\text{cm} \\)).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_c5b69eb29251fa396613g_0032_1.jpg", "batch28-2024_06_17_c5b69eb29251fa396613g_0032_2.jpg" ], "is_multi_img": true, "answer": "$66 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Through point $B$, draw $BG \\perp CD$, intersecting the extension of $DC$ at point $G$. Through point $E$, draw $EH \\perp CD$, intersecting the extension of $CD$ at point $H$.\n\nAs shown in the figure,\n\n\n\n$\\because \\angle ABC = 150^\\circ$,\n\n$\\therefore \\angle CBG = 30^\\circ$,\n\n$BG = BC \\cdot \\cos 30^\\circ = 60 \\times \\frac{\\sqrt{3}}{2} = 30\\sqrt{3}$,\n\n$\\because \\angle CDE = 120^\\circ$,\n\n$\\therefore \\angle EDH = 60^\\circ$,\n\n$HE = DE \\cdot \\sin 60^\\circ = 30 \\times \\frac{\\sqrt{3}}{2} = 15\\sqrt{3}$,\n\n$AG = AB + BG = 40 + 30\\sqrt{3}$,\n\n$AG - HE = 40 + 30\\sqrt{3} - 15\\sqrt{3} = 40 + 15\\sqrt{3} \\approx 66$,\n\n$\\therefore$ The height of point $E$ from the ground is approximately $66 \\mathrm{~cm}$.\n\n【Key Insight】This problem mainly tests the calculation of trigonometric functions. The key to solving it lies in clearly understanding the relationships between the line segments and correctly using trigonometric functions for calculations." }, { "problem_id": 1827, "question": "The earliest periscope in the world was made in ancient China as early as the 2nd century BC. The \"Huainan Wanbiji,\" a book compiled in the early Western Han Dynasty, contains the following record: \"Hang a large mirror high up, and place a water basin beneath it, and you will see your neighbors,\" as shown in Figure 1. Its working principle mainly utilizes the law of light reflection. In Figure 2, $AB$ is horizontal, the angle of incidence $\\angle BOC = 30^\\circ$, and $\\angle OAD = 15^\\circ$ (the angle of incidence equals the angle of reflection, with $OC$ and $AD$ as the normal lines). If $AB = 10\\sqrt{6}$ meters, find the distance the light travels from $O$ to $A$ (i.e., the length of $OA$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_e04d978f06aca8cfd83cg_0031_1.jpg", "batch28-2024_06_17_e04d978f06aca8cfd83cg_0031_2.jpg" ], "is_multi_img": true, "answer": "20 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, draw a perpendicular line $AE \\perp OB$ from point $\\mathrm{A}$ to point $E$.\n\n\n\nGiven that the angle of incidence $\\angle BOC = 30^{\\circ}$, and the angle of incidence equals the angle of reflection,\n\n$\\therefore \\angle AOC = \\angle BOC = 30^{\\circ}$,\n\n$\\therefore \\angle AOB = \\angle AOC + \\angle BOC = 60^{\\circ}$,\n\nSince $AD \\perp AB$ and $\\angle OAD = 15^{\\circ}$,\n\n$\\therefore \\angle OAB = 75^{\\circ}$,\n\n$\\therefore \\angle B = 180^{\\circ} - \\angle AOB - \\angle OAB = 45^{\\circ}$,\n\nSince $AE \\perp OB$,\n\n$\\therefore \\triangle AEB$ is an isosceles right triangle,\n\nIn the right triangle $\\triangle ABE$, $AB = 10 \\sqrt{6}$ meters, $AE = AB \\cdot \\sin \\angle B = 10 \\sqrt{6} \\times \\frac{\\sqrt{2}}{2} = 10 \\sqrt{3}$ meters,\n\nIn the right triangle $\\triangle AOE$, $OA = \\frac{AE}{\\sin \\angle AOE} = \\frac{10 \\sqrt{3}}{\\frac{\\sqrt{3}}{2}} = 20$ meters.\n\n$\\therefore$ The distance the light travels from $O$ to $\\mathrm{A}$ is 20 meters.\n\n【Key Insight】This problem mainly tests the application of solving right triangles. The key to solving the problem is to correctly construct the right triangle based on the given conditions and to determine $\\angle B = 45^{\\circ}$ using the triangle angle sum theorem." }, { "problem_id": 1828, "question": "As shown in Figure 1, there is a solar street light, which consists of a lamp pole and a lamp tube bracket as shown in Figure 2. $AB$ is the lamp pole, $CD$ is the lamp tube bracket, and the angle between the lamp tube bracket $CD$ and the lamp pole is $\\angle BDC = 60^{\\circ}$. The comprehensive practice group wants to know the length of the lamp tube bracket $CD$. They measured an elevation angle of $60^{\\circ}$ to the bottom of the lamp tube bracket $D$ from point $E$ on the ground, and an elevation angle of $30^{\\circ}$ to the top of the lamp tube bracket $C$ from point $F$. They also measured $AE = 3 \\mathrm{~m}$ and $EF = 8 \\mathrm{~m}$ ($\\mathrm{A}, E, F$ are collinear). Based on the above data, calculate the length of the lamp tube bracket $CD$ (the result should be accurate to $0.1 \\mathrm{~m}$, reference data: $\\sqrt{3} \\approx 1.73$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_e04d978f06aca8cfd83cg_0052_1.jpg", "batch28-2024_06_17_e04d978f06aca8cfd83cg_0052_2.jpg" ], "is_multi_img": true, "answer": "$1.2 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Extend $FC$ to intersect $AB$ at point $G$.\n\nIn the right triangle $\\triangle ADE$, $\\tan \\angle AED = \\frac{AD}{AE} = \\tan 60^\\circ = \\sqrt{3}$.\n\nGiven that $AE = 3\\,\\text{m}$,\n\nTherefore, $AD = \\sqrt{3} \\cdot AE = 3\\sqrt{3}\\,\\text{m}$.\n\nGiven that $AE = 3\\,\\text{m}$ and $EF = 8\\,\\text{m}$,\n\nTherefore, $AF = AE + EF = 11\\,\\text{m}$.\n\nIn the right triangle $\\triangle AFG$, $\\tan F = \\frac{AG}{AF} = \\tan 30^\\circ = \\frac{\\sqrt{3}}{3}$.\n\nTherefore, $AG = \\frac{11\\sqrt{3}}{3}\\,\\text{m}$.\n\n\n\nFigure 2\n\nIn the right triangle $\\triangle AFG$, $\\angle A = 90^\\circ$ and $\\angle F = 30^\\circ$.\n\nTherefore, $\\angle AGF = 60^\\circ$.\n\nSince $\\angle BDC = \\angle AGF = 60^\\circ$,\n\nTherefore, $\\triangle DGC$ is an equilateral triangle.\n\nThus, $DC = DG = AG - AD = \\frac{11}{3}\\sqrt{3} - 3\\sqrt{3} = \\frac{2}{3}\\sqrt{3} \\approx 1.2\\,\\text{m}$.\n\nAnswer: The length of the lamp holder $CD$ is approximately $1.2\\,\\text{m}$.\n\n【Key Insight】This problem primarily examines the practical application of solving right triangles, the properties and determination of equilateral triangles, and the importance of correctly constructing auxiliary lines to form right triangles in solving the problem." }, { "problem_id": 1829, "question": "The correct grip is important for students' learning and growth. Figure 1 shows a student's correct grip, and its schematic diagram is shown in Figure 2. The angle $\\alpha$ formed by the pen shaft and the paper surface is $53^{\\circ}$, and the length of the pen shaft $A B$ is $20 \\mathrm{~cm}$. Find the vertical height $B C$ of the top of the pen shaft from the paper surface. (Reference data: $\\sin 53^{\\circ} \\approx 0.80, \\cos 53^{\\circ} \\approx 0.60, \\tan 53^{\\circ} \\approx 1.33$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_e04d978f06aca8cfd83cg_0060_1.jpg", "batch28-2024_06_17_e04d978f06aca8cfd83cg_0060_2.jpg" ], "is_multi_img": true, "answer": "$16 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement: $BC \\perp AC$, and $BC = AB \\sin \\alpha = 20 \\times 0.80 = 16 \\mathrm{~cm}$.\n\n\n\nAnswer: The vertical height $BC$ from the top of the pen holder to the paper surface is $16 \\mathrm{~cm}$.\n\n[Key Insight] This problem tests the definition of acute angle trigonometric functions and the knowledge of solving right-angled triangles; mastering the definition of acute angle trigonometric functions is crucial for solving the problem." }, { "problem_id": 1830, "question": "As shown in Figure 1, the vehicle is a dump truck for unloading debris. The process of unloading debris is primarily accomplished by the hydraulic cylinder on the frame pushing the cargo box upward, allowing the debris inside to automatically pour out. Figure 2 is a side view schematic, where $OB$ represents the frame, $AO$ represents the cargo box, $OB = OA$, and $DC$ represents the hydraulic cylinder. It is known that $OA = 8$ meters. When the hydraulic cylinder $DC$ pushes the cargo box $OA$ to an angle $\\angle AOB = 40^{\\circ}$ and $\\angle CDO = 76^{\\circ}$ with $OB$, point $C$ is exactly the midpoint of $OA$. At this point, all the debris in the cargo box can be completely poured out. Calculate the distance $BD$ from the frame end point $B$ to the hydraulic cylinder end point $D$ at this moment. (The result should be accurate to 0.1 meters. Reference data: $\\sin 40^{\\circ} \\approx 0.64, \\cos 40^{\\circ} \\approx 0.77, \\tan 40^{\\circ} \\approx 0.84, \\sin 14^{\\circ} \\approx 0.24, \\cos 14^{\\circ} \\approx 0.97, \\tan 14^{\\circ} \\approx 0.25$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_e04d978f06aca8cfd83cg_0064_1.jpg", "batch28-2024_06_17_e04d978f06aca8cfd83cg_0064_2.jpg" ], "is_multi_img": true, "answer": "4.3 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Through point \\( C \\), draw \\( CE \\perp OD \\) intersecting at point \\( E \\),\n\n\\[\n\\therefore \\angle CEO = \\angle CED = 90^\\circ,\n\\]\n\n\\[\n\\because OA = 8, \\text{ and } C \\text{ is exactly the midpoint of } OA,\n\\]\n\n\\[\n\\begin{aligned}\n& \\therefore OC = 4, \\\\\n& \\because \\sin 40^\\circ = \\frac{CE}{CO} \\approx 0.64, \\\\\n& \\therefore CE \\approx 4 \\times 0.64 = 2.56 \\quad \\text{(meters)}, \\\\\n& \\because \\cos 40^\\circ = \\frac{OE}{CO} \\approx 0.77, \\\\\n& \\therefore OE = 3.08 \\quad \\text{(meters)}, \\\\\n& \\because \\tan 14^\\circ = \\frac{DE}{CE} \\approx 0.25, \\\\\n& \\therefore DE = 0.64 \\quad \\text{(meters)}, \\\\\n& \\therefore DO = DE + OE = 0.64 + 3.08 = 3.72 \\quad \\text{(meters)}, \\\\\n& \\therefore BD = OB - OD = 8 - 3.72 = 4.28 \\approx 4.3 \\quad \\text{(meters)}.\n\\end{aligned}\n\\]\n\nAnswer: At this moment, the distance \\( BD \\) from the frame endpoint \\( B \\) to the hydraulic cylinder endpoint \\( D \\) is approximately: 4.3 meters.\n\n\n\n【Highlight】This problem tests the knowledge of solving right triangles, and the key to solving it lies in the proficient application of solving right triangles." }, { "problem_id": 1831, "question": "As shown in Figure 1, this is a physical diagram of a crane, and Figure 2 is a schematic diagram of the crane in operation. When the crane is working, it rotates the boom \\( AB \\) around point \\( B \\) by extending and retracting the hydraulic rod \\( CD \\), thereby enabling the boom to lift and lower (the length of the boom \\( AB \\) can also be adjusted). During a lifting operation, the learning interest group measured and consulted with the workers and obtained the following information: as shown in Figure 3, the boom \\( AB = 10 \\) meters, the distance from point \\( B \\) to the ground \\( BE = 1.8 \\) meters, the line \\( AF \\) where the steel cable is located is perpendicular to the ground at point \\( F \\), and the distance from point \\( B \\) to \\( AF \\) is \\( BG = 8 \\) meters. Find the distance \\( AF \\) from point \\( A \\) to the ground in meters.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch28-2024_06_17_f38e72aa7f5f6ba2b1c8g_0076_1.jpg", "batch28-2024_06_17_f38e72aa7f5f6ba2b1c8g_0076_2.jpg", "batch28-2024_06_17_f38e72aa7f5f6ba2b1c8g_0076_3.jpg" ], "is_multi_img": true, "answer": "7.8 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle ABG$,\n\nBy the Pythagorean theorem, we have $AG = \\sqrt{AB^{2} - BG^{2}} = \\sqrt{10^{2} - 8^{2}} = 6$ meters.\n\nSince $BE \\perp EF$, $AF \\perp EF$, and $BG \\perp AF$,\n\nIt follows that $\\angle BEF = \\angle EFG = \\angle BGF = 90^{\\circ}$.\n\nBecause quadrilateral $BEFG$ is a rectangle,\n\nTherefore, $GF = BE = 1.8$ meters.\n\nThus, $AF = AG + GF = 6 + 1.8 = 7.8$ meters.\n\nAnswer: The distance from $A$ to the ground $AF$ is 7.8 meters.\n\n[Key Insight] The problem primarily tests the application of the Pythagorean theorem and the determination of a rectangle. Understanding the problem and comprehensively applying these concepts is crucial for solving it." }, { "problem_id": 1832, "question": "As shown in Figure 1, this is the \"Timed and Fixed-Location Waste Sorting and Disposal Point\" in a residential community in Shengzhou City. The intelligent key-operated method for opening the disposal door makes waste disposal in Shengzhou smarter and more environmentally friendly. Figure 2 is a side view of the disposal door after it has been opened. The length of the disposal flap $AB$ is $45 \\mathrm{~cm}$, the height of the bottom of the flap from the ground $BD$ is $125 \\mathrm{~cm}$, and the maximum opening angle of the flap is $57.6^{\\circ}$. Calculate the maximum distance $CF$ that the front end $C$ of the disposal door can be away from and the maximum height from the ground of the front end $C$ of the disposal door (reference data: $\\sin 57.6^{\\circ} \\approx 0.844, \\cos 57.6^{\\circ} \\approx 0.536, \\tan 57.6^{\\circ} \\approx 1.58$, results are accurate to $1 \\mathrm{~cm}$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_fa36e529dc60d6d0ff56g_0002_1.jpg", "batch28-2024_06_17_fa36e529dc60d6d0ff56g_0002_2.jpg" ], "is_multi_img": true, "answer": "The maximum distance $C$ from the front end of the delivery door is $C F=38 \\mathrm{~cm}$, and the maximum distance $C$ from the front end of the delivery door to the ground is $146 \\mathrm{~cm}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle ACF$, $\\angle A = 57.6^\\circ$, and $AC = AB = 45 \\text{ cm}$.\n\nTherefore, \n\\[ CF = AC \\cdot \\sin 57.6^\\circ \\approx 45 \\times 0.844 \\approx 38 \\text{ cm}, \\]\n\\[ AF = AC \\cdot \\cos 57.6^\\circ \\approx 45 \\times 0.536 \\approx 24 \\text{ cm}. \\]\n\nThus,\n\\[ FD = AB + BD - AF = 45 + 125 - 24 = 146 \\text{ cm}. \\]\n\nHence, the maximum horizontal distance from the front end $C$ of the delivery door is $CF = 38 \\text{ cm}$, and the maximum vertical distance from the front end $C$ to the ground is $146 \\text{ cm}$.\n\n**Key Insight**: This problem tests the application of solving right triangles. The key to solving it lies in proficiently applying the relationships of trigonometric functions of acute angles based on the given conditions." }, { "problem_id": 1833, "question": "A seesaw $A B$ with unequal arms in a children's playground is 2.8 meters long, and the support column $O C$ is perpendicular to the ground. As shown in Figure 1, when one end of $A B$, point $\\mathrm{A}$, touches the ground, the tangent of the angle between $A B$ and the ground is $\\frac{3}{4}$; as shown in Figure 2, when the other end of $A B$, point $B$, touches the ground, the sine of the angle between $A B$ and the ground is $\\frac{1}{3}$. Find the height of the support column $O C$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_fa36e529dc60d6d0ff56g_0021_1.jpg", "batch28-2024_06_17_fa36e529dc60d6d0ff56g_0021_2.jpg" ], "is_multi_img": true, "answer": "$0.6 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( OC = 3x \\) meters.\n\nAs shown in Figure 1, in the right triangle \\( \\triangle AOC \\), \\( \\angle ACO = 90^\\circ \\), and \\( \\tan A = \\frac{3}{4} \\).\n\nTherefore, \\( \\frac{OC}{AC} = \\frac{3}{4} \\),\n\nSo, \\( AC = 4x \\) meters,\n\nThus, \\( OA = \\sqrt{OC^2 + AC^2} = 5x \\) meters.\n\nAs shown in Figure 2, in the right triangle \\( \\triangle OBC \\), \\( \\sin B = \\frac{1}{3} \\).\n\nTherefore, \\( \\frac{OC}{OB} = \\frac{1}{3} \\),\n\nSo, \\( OB = 9x \\) meters,\n\nGiven that \\( AB = 2.8 \\) meters,\n\nWe have \\( 5x + 9x = 2.8 \\),\n\nSolving for \\( x \\), we get \\( x = 0.2 \\),\n\nThus, \\( OC = 0.6 \\) meters.\n\n【Key Insight】This problem primarily examines the practical application of solving right triangles and the Pythagorean theorem. Accurate calculation is crucial for solving the problem." }, { "problem_id": 1834, "question": "As shown in Figure 1 and Figure 2, these are the physical and schematic diagrams of a certain basketball hoop. It is known that the base \\( BC = 0.7 \\) meters, the angle formed by the base \\( BC \\) and the support \\( AC \\) is \\( \\angle ACB = 75^\\circ \\), the length of the support \\( AF \\) is 2.0 meters, the distance from the top of the backboard \\( F \\) to the rim \\( D \\) is \\( FD = 0.9 \\) meters, and the angle formed by the backboard bottom support \\( GE \\) and the support \\( AF \\) is \\( \\angle FGE = 42^\\circ \\). Find the distance from the rim \\( D \\) to the ground (accurate to 0.01 meters) (reference data: \\( \\cos 75^\\circ \\approx 0.2588 \\), \\( \\sin 75^\\circ \\approx 0.9659 \\), \\( \\tan 75^\\circ \\approx 3.732 \\), \\( \\sin 42^\\circ \\approx 0.6691 \\), \\( \\cos 42^\\circ \\approx 0.7431 \\), \\( \\tan 42^\\circ \\approx 0.9004 \\)).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch28-2024_06_17_fa36e529dc60d6d0ff56g_0041_1.jpg", "batch28-2024_06_17_fa36e529dc60d6d0ff56g_0041_2.jpg" ], "is_multi_img": true, "answer": "3.05 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "**Solution:** Extend $FE$ to intersect the extension of $CB$ at point $H$. Draw $AM \\perp FH$ passing through point $A$ to meet at point $M$.\n\n\n\nIn the right triangle $\\triangle ABC$, $\\tan \\angle ACB = \\frac{AB}{BC}$.\n\nThus, $AB = BC \\cdot \\tan 75^\\circ \\approx 0.7 \\times 3.732 = 2.6124$.\n\nTherefore, $MH = AB = 2.6124$.\n\nIn the right triangle $\\triangle AMF$,\n\nSince $\\angle FAM = \\angle FGE = 42^\\circ$, $\\sin \\angle FAM = \\frac{FM}{AF}$.\n\nThus, $FM = 2 \\times \\sin 42^\\circ \\approx 2 \\times 0.669 = 1.338$.\n\nHence, $DH = FM + MH - DF = 2.6124 + 1.338 - 0.9 = 3.05$ meters.\n\n**Answer:** The distance from the basketball hoop $D$ to the ground is 3.05 meters.\n\n**Key Insight:** This problem primarily tests the ability to solve right triangles. By constructing auxiliary lines, the problem is transformed into solving right triangles, and proficiency in trigonometric calculations is essential for solving it." }, { "problem_id": 1835, "question": "The Sangti is an ancient Chinese invention used for picking mulberry leaves. Figure (1) shows a depiction of the Sangti in the book \"Nong Zheng Quan Shu\" by the Ming Dynasty scientist Xu Guangqi. Given the diagram in Figure (2), where \\( AB = AC \\), \\( BC = 1 \\) meter, \\( AD = 1.2 \\) meters, and \\( \\angle CAB = 40^\\circ \\), find the length of \\( CD \\). (The result should be accurate to 0.1 meters. Reference data: \\( \\sin 70^\\circ = 0.94 \\), \\( \\cos 70^\\circ = 0.34 \\), \\( \\tan 70^\\circ = 2.75 \\))\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch29-2024_06_14_18230415f187b765f1d7g_0055_1.jpg", "batch29-2024_06_14_18230415f187b765f1d7g_0055_2.jpg" ], "is_multi_img": true, "answer": "$C D \\approx 2.7$ (m).", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Draw $AE \\perp BC$ at point $E$.\n\nSince $AB = AC$,\n\n$\\angle ABC = \\angle C$.\n\nGiven that $\\angle DAB = \\angle ABC + \\angle C = 140^\\circ$,\n\nit follows that $\\angle ABC = \\angle C = 70^\\circ$.\n\nSince $AE \\perp BC$,\n\n$CE = \\frac{1}{2} BC = \\frac{1}{2}$ meter.\n\nIn right triangle $AEC$, $\\cos \\angle C = \\cos 70^\\circ = \\frac{CE}{AC} = \\frac{\\frac{1}{2}}{AC} = 0.34$,\n\nthus $AC = \\frac{25}{17} \\approx 1.47$ meters.\n\nGiven that $AD = 1.2$ meters,\n\n$CD = AD + AC = 1.2 + 1.47 = 2.67 \\approx 2.7$ meters.\n\nTherefore, $CD \\approx 2.7$ meters.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n[Key Insight] This problem tests the understanding of solving right triangles and the properties of isosceles triangles. The key to solving the problem lies in correctly constructing the auxiliary line to form a right triangle." }, { "problem_id": 1836, "question": "Xiaoming's sink is equipped with a lift-type faucet (as shown in Figure (1)). When fully opened, the angle between the handle $AM$ and the horizontal line is $37^{\\circ}$. At this point, the handle end $A$, the water outlet point $B$, and the water drop point $C$ are collinear. The sink and faucet are illustrated in the diagram. The relevant data are $AM = 10 \\mathrm{~cm}$, $MD = 6 \\mathrm{~cm}$, $DE = 22 \\mathrm{~cm}$, $EH = 38 \\mathrm{~cm}$. Find the distance $HC$ from the water drop point $C$ to the basin wall (reference data: $\\sin 37^{\\circ} = \\frac{3}{5}$, $\\cos 37^{\\circ} = \\frac{4}{5}$, $\\tan 37^{\\circ} = \\frac{3}{4}$, $\\sqrt{3} \\approx 1.7$).\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0006_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0006_2.jpg" ], "is_multi_img": true, "answer": " $10.3 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $AG \\perp EH$ at point $G$, and draw $MN \\perp AG$ at point $N$.\n\n\n\nThen, $\\angle ANM = \\angle AGC = 90^\\circ$.\n\nSince $ME \\perp EC$,\n\nTherefore, quadrilateral $MNGE$ is a rectangle.\n\nHence, $EG = MN$, and $NG = ME = MD + DE = 28 \\text{ cm}$.\n\nGiven that $\\sin \\angle AMN = \\frac{AN}{AM}$, and $\\cos \\angle AMN = \\frac{MN}{AM}$,\n\nWe find $AN = AM \\cdot \\sin 37^\\circ = 6 \\text{ cm}$, and $MN = AM \\cdot \\cos 37^\\circ = 8 \\text{ cm}$.\n\nThus, $EG = 8 \\text{ cm}$, and $AG = AN + NG = 34 \\text{ cm}$.\n\nSince $\\angle ACG = 60^\\circ$,\n\n$CG = \\frac{AG}{\\cos \\angle ACG} \\approx \\frac{34}{1.73} = 19.7 \\text{ cm}$.\n\nTherefore, $EC = EG + GC = 27.7 \\text{ cm}$.\n\nHence, $CH = EH - EC = 10.3 \\text{ m}$.\n\nAnswer: The distance $HC$ from the falling point $C$ to the basin wall is $10.3 \\text{ m}$.\n\n【Highlight】This problem examines the solution of right triangles, the properties and determination of rectangles, and emphasizes the application of mathematical knowledge in solving real-world problems. It involves measuring, calculating, and modeling as part of the mathematical practice, highlighting the application awareness and problem-solving capabilities in mathematics. It incorporates mathematical modeling and encourages students to focus on real-life scenarios, using mathematical methods to solve practical problems." }, { "problem_id": 1837, "question": "The Yellow River is the main birthplace of Chinese civilization, and the Chinese people refer to it as the \"Mother River.\" To promote the inheritance and development of Yellow River culture, a school organized a study tour for students to a certain section of the Yellow River basin. A certain interest group, with only a meter ruler and an angle measuring instrument, wanted to determine the width of a certain part of the Yellow River in Henan Province (without reaching the opposite bank). As shown in the figure, it is known that there is a point $A$ on the opposite bank of the river. The interest group took two arbitrary points $B$ and $C$ along the riverbank on this side, with $A$ as the reference point, and measured $\\angle ABC = 65^\\circ$, $\\angle ACB = 45^\\circ$, and the length of $BC$ as $300 \\mathrm{~m}$. Find the width of the river. (The result should be accurate to $1 \\mathrm{~m}$, with reference values $\\sin 65^\\circ \\approx 0.91$, $\\cos 65^\\circ \\approx 0.42$, $\\tan 65^\\circ \\approx 2.14$.)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0018_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0018_2.jpg" ], "is_multi_img": true, "answer": " 204 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "As shown in the figure, draw a perpendicular line from point $A$ to $BC$ at point $D$.\n\nLet $AD = x$.\n\nFrom the figure, we can see that $\\angle ABD = 65^\\circ$ and $\\angle ACB = 45^\\circ$.\n\nIn the right triangle $\\triangle ABD$,\n\nSince $\\tan \\angle ABD = \\frac{AD}{BD}$,\n\nTherefore, $BD = \\frac{AD}{\\tan \\angle ABD} = \\frac{x}{\\tan 65^\\circ} \\approx 0.47x$.\n\nIn the right triangle $\\triangle ACD$,\n\nSince $\\angle ACD = 45^\\circ$,\n\nTherefore, $AD = CD = x$.\n\nSince $BD + DC = BC$,\n\nTherefore, $0.47x + x = 300$.\n\nSolving for $x$, we get $x \\approx 204$.\n\nThat is, $AD \\approx 204 \\mathrm{~m}$.\n\nThus, the width of the river is approximately 204 meters.\n\n【Insight】This problem tests the application of solving right triangles. The key to solving the problem is to use parameters to construct equations." }, { "problem_id": 1838, "question": "As shown in Figure 1, a new batch of new energy buses has been introduced in Taiyuan City. Figures 2 and 3 are top-view schematic diagrams of the bus's double doors when closed and open, respectively. \\( ME \\), \\( EF \\), and \\( FN \\) are the sliding tracks for the door hinges, with \\(\\angle E = \\angle F = 90^\\circ\\). The hinges \\( A, B, C, D \\) of the two doors \\( AB \\) and \\( CD \\) are all on the sliding tracks. When the doors are closed (Figure 2), \\( A \\) and \\( D \\) are at \\( E \\) and \\( F \\) respectively, with the door seam negligible (i.e., \\( B \\) and \\( C \\) coincide). Both doors open simultaneously, with points \\( A \\) and \\( D \\) sliding uniformly along the directions \\( E \\rightarrow M \\) and \\( F \\rightarrow N \\) respectively (Figure 3). When \\( B \\) reaches \\( E \\), \\( C \\) reaches \\( F \\), at which point the doors are fully open. During the opening process, when \\( BC = EB + CF \\), find the degree measure of \\(\\angle ABE\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0022_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0022_2.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0022_3.jpg" ], "is_multi_img": true, "answer": "$60^{\\circ}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since points $\\mathrm{A}$ and $\\mathrm{D}$ slide uniformly in the directions $E \\rightarrow M$ and $F \\rightarrow N$ respectively, when $B$ reaches $E$, $C$ just reaches $F$.\n\nTherefore, $EA = DF$.\n\nSince $\\angle E = \\angle F = 90^{\\circ}$,\n\nIn right triangles $\\triangle AEB$ and $\\triangle DFC$,\n\n\\[\n\\left\\{\n\\begin{array}{l}\nAB = DC \\\\\nAE = DF\n\\end{array}\n\\right.\n\\]\n\nThus, $\\triangle AEB \\cong \\triangle DFC$ (by the Hypotenuse-Leg theorem).\n\nTherefore, $EB = CF$.\n\nSince $BC = EB + CF$,\n\nIt follows that $EB = \\frac{1}{2} BC = \\frac{1}{4} EF$.\n\nSince $AB + CD = EF$,\n\nTherefore, $EB = \\frac{1}{2} AB$.\n\nIn right triangle $\\triangle ABE$, $\\cos \\angle ABE = \\frac{EB}{AB} = \\frac{1}{2}$.\n\nThus, $\\angle ABE = 60^{\\circ}$.\n\nTherefore, the measure of $\\angle ABE$ is $60^{\\circ}$.\n\n[Insight] This problem applies mathematical knowledge to real-life situations. Accurately understanding the problem and selecting the appropriate mathematical concepts for the solution is key to solving it." }, { "problem_id": 1839, "question": "As shown in Figure 1, there is a phone stand, and Figure 2 is a side view of it. $A B$ and $B C$ can rotate around points $A$ and $B$ respectively. It is measured that $B C = 10 \\mathrm{~cm}$ and $A B = 20 \\mathrm{~cm}$. When $A B$ and $B C$ rotate to $\\angle B A E = 60^{\\circ}$ and $\\angle A B C = 50^{\\circ}$, find the distance from point $C$ to $A E$. (The result should be rounded to one decimal place. Reference data: $\\sin 70^{\\circ} \\approx 0.94, \\sqrt{3} \\approx 1.73$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0031_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0031_2.jpg" ], "is_multi_img": true, "answer": "$7.9 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Construct $BG \\perp AE$ and $CF \\perp AE$, intersecting $AE$ at points $G$ and $F$ respectively. Then, construct $CD \\perp BG$, intersecting $BG$ at point $D$, as shown in the figure:\n\n\n\nFigure 2\n\nGiven that $\\angle BAE = 60^\\circ$ and $AB = 20 \\text{ cm}$,\n\n$\\therefore \\angle ABG = 30^\\circ$.\n\nThus, $AG = 10 \\text{ cm}$ and $BG = 10\\sqrt{3} \\text{ cm}$.\n\nSince $\\angle ABC = 50^\\circ$ and $\\angle ABG = 30^\\circ$,\n\n$\\therefore \\angle CBD = 20^\\circ$ and $\\angle BCD = 70^\\circ$.\n\nIn right triangle $DBC$,\n\nGiven that $BC = 10 \\text{ cm}$ and $\\sin 70^\\circ = \\frac{BD}{BC}$,\n\n$\\therefore BD = \\sin 70^\\circ \\times BC = 9.4 \\text{ cm}$.\n\nTherefore, $CF = DG = BG - BD = 10\\sqrt{3} - 9.4 \\approx 7.9 \\text{ cm}$.\n\n【Insight】This problem examines the solution of non-right-angled triangles. The key to solving it is to construct right-angled triangles and then proceed with the solution." }, { "problem_id": 1840, "question": "Due to the impact of the COVID-19 epidemic, some counties and cities have shifted their classroom teaching from \"offline\" to \"online.\" The Education Bureau of our city is responsible for organizing the recording of outstanding lesson examples for the \"Air Classroom\" across the region, with mobile phones becoming the primary tool for students' online learning. As shown in Figure 1, a mobile phone stand is depicted, and Figure 2 is a side view of it. The segments \\( AB \\) and \\( BC \\) can rotate around points \\( A \\) and \\( B \\) respectively. Measurements show that \\( BC = 8 \\text{ cm} \\) and \\( AB = 16 \\text{ cm} \\). When \\( AB \\) and \\( BC \\) are rotated to angles \\( \\angle BAE = 60^\\circ \\) and \\( \\angle ABC = 50^\\circ \\), the viewing experience is considered optimal. Calculate the distance from point \\( C \\) to \\( AE \\) at this position. (The result should be rounded to one decimal place. Reference data: \\( \\sin 50^\\circ \\approx 0.766 \\), \\( \\cos 50^\\circ \\approx 0.64 \\), \\( \\sin 70^\\circ \\approx 0.94 \\), \\( \\cos 70^\\circ \\approx 0.34 \\), \\( \\tan 70^\\circ \\approx 2.75 \\), \\( \\sqrt{3} \\approx 1.73 \\))\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0040_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0040_2.jpg" ], "is_multi_img": true, "answer": " $6.3 \\mathrm{~cm}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, perpendiculars are drawn from points $B$ and $C$ to $AE$, with feet $M$ and $N$ respectively. From point $C$, a perpendicular $CD$ is drawn to $BM$ at point $D$.\n\n\n\nFigure 2\n\nIn the right triangle $\\triangle ABM$, $\\angle A = 60^\\circ$ and $AB = 16 \\text{ cm}$,\n\n$\\therefore BM = AB \\cdot \\sin A$\n\n$= 16 \\times \\frac{\\sqrt{3}}{2}$\n\n$= 8\\sqrt{3} \\text{ cm}$.\n\nSince $\\angle ABM = 90^\\circ - 60^\\circ = 30^\\circ$ and $\\angle ABC = 50^\\circ$,\n\n$\\therefore \\angle CBD = 50^\\circ - 30^\\circ = 20^\\circ$,\n\n$\\therefore \\angle BCD = 90^\\circ - 20^\\circ = 70^\\circ$.\n\nIn the right triangle $\\triangle BCD$, $BC = 8 \\text{ cm}$ and $\\angle BCD = 70^\\circ$,\n\n$\\therefore BD = BC \\cdot \\sin 70^\\circ$\n\n$\\approx 8 \\times 0.94$\n\n$= 7.52 \\text{ cm}$.\n\n$CN = DM = BM - BD$\n\n$= 8\\sqrt{3} - 7.52$\n\n$\\approx 6.3 \\text{ cm}$.\n\nAnswer: The distance from point $C$ to $AE$ is approximately $6.3 \\text{ cm}$.\n\n【Insight】This problem examines the application of solving right triangles, and mastering the relationship between the sides and angles of a right triangle is essential for correctly answering the question." }, { "problem_id": 1841, "question": "As shown in Figure (1), a wall-mounted air conditioner is installed on the wall of a room according to customer requirements. Figure (2) is a side view of the installation. The air conditioner's blade $AF$ rotates from top to bottom around point $A$ to sweep the air. The installation requirements stipulate that when the blade just reaches the outer edge of the bed, the angle $\\alpha$ between the blade and the vertical line is $42^{\\circ}$. The bottom of the air conditioner $BC$ is perpendicular to the wall $CD$, with the foot at point $C$. Given $AB = 0.02 \\, \\text{m}$, $BC = 0.2 \\, \\text{m}$, and the length of the bed $DE = 2.5 \\, \\text{m}$, find the height $CD$ from the bottom of the installed air conditioner to the bed. (The result should be accurate to $0.1 \\, \\text{m}$, with $\\sin 42^{\\circ} \\approx 0.67$, $\\cos 42^{\\circ} \\approx 0.74$, and $\\tan 42^{\\circ} \\approx 0.90$.)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0041_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0041_2.jpg" ], "is_multi_img": true, "answer": "2.5 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Given that \\( BC = 0.2 \\, \\text{m} \\) and \\( DE = 2.5 \\, \\text{m} \\),\n\n\\[ EF = DE - DF = 2.5 - 0.2 = 2.3 \\, \\text{m}. \\]\n\nIn the right triangle \\( \\triangle AEF \\), since \\( \\tan 42^\\circ = \\frac{EF}{AF} \\),\n\n\\[ AF = \\frac{EF}{\\tan 42^\\circ} \\approx \\frac{2.3}{0.9} \\approx 2.56 \\, \\text{m}. \\]\n\nThus,\n\n\\[ CD = AF - AB = 2.56 - 0.02 = 2.54 \\approx 2.5 \\, \\text{m}. \\]\n\nTherefore, the height \\( CD \\) from the bottom of the installed air conditioner to the bed is \\( 2.5 \\, \\text{m} \\).\n\n**Key Insight:** This problem primarily examines the application of solving right triangles. The key to solving the problem lies in deriving \\( \\tan 42^\\circ = \\frac{EF}{AF} \\) from the given conditions, and attention should be paid to the precision requirements of the result." }, { "problem_id": 1842, "question": "As shown in Figure (1), an online teaching device consists of a base, a support arm \\( AB \\), a connecting rod \\( BC \\), a cantilever \\( CD \\), and a camera installed at \\( D \\). Figure (2) is a schematic diagram of the device placed on a horizontal tabletop \\( l \\). It is known that the support arm \\( AB \\perp l \\), \\( AB = 15 \\, \\text{cm} \\), \\( BC = 30 \\, \\text{cm} \\), and measurements yield \\( \\angle ABC = 148^\\circ \\), \\( \\angle BCD = 28^\\circ \\), \\( AE = 9 \\, \\text{cm} \\). Find the distance \\( DE \\) from the camera to the tabletop \\( l \\) (the result should be accurate to \\( 0.1 \\, \\text{cm} \\)). (Reference data:\n\n\\( \\sin 58^\\circ \\approx 0.85 \\), \\( \\cos 58^\\circ = 0.53 \\), \\( \\tan 58^\\circ \\approx 1.60 \\), \\( \\sqrt{3} \\approx 1.73 \\))\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0052_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0052_2.jpg" ], "is_multi_img": true, "answer": "$26.1 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "As shown in the figure, draw a perpendicular line $CF \\perp l$ from point $C$, with the foot of the perpendicular at $F$. From points $D$ and $B$, draw perpendicular lines to $CF$, with the feet of the perpendiculars at $H$ and $G$, respectively.\n\n\n\nIn the right triangle $\\triangle CBG$, $\\angle CBG = 148^\\circ - 90^\\circ = 58^\\circ$.\n\n$\\sin \\angle CBG = \\frac{CG}{CB}$, $\\cos \\angle CBG = \\frac{BG}{CB}$, and $BC = 30 \\mathrm{~cm}$.\n\nTherefore, $CG = 25.5 \\mathrm{~cm}$, $BG = 15.9 \\mathrm{~cm}$.\n\nAlso, since $AE = 9 \\mathrm{~cm}$,\n\n$DH = AE + BG = 24.9 \\mathrm{~cm}$.\n\nSince $\\angle CBG = 58^\\circ$,\n\n$\\angle BCG = 32^\\circ$.\n\nThus, $\\angle DCH = \\angle DCB + \\angle BCG = 60^\\circ$.\n\nIn the right triangle $\\triangle DCH$, $\\angle DCH = 60^\\circ$.\n\n$\\tan \\angle DCH = \\frac{DH}{CH}$,\n\nTherefore, $CH = \\frac{24.9}{\\sqrt{3}} \\approx 14.4 \\mathrm{~cm}$.\n\nHence, $DE = CG + AB - CH = 26.1 \\mathrm{~cm}$.\n\n【Insight】This problem examines the application of solving right triangles. The key to solving the problem lies in adding auxiliary lines to construct right triangles." }, { "problem_id": 1843, "question": "Figure 1 shows a tablet stand, and Figure 2 is a side view of it. $A B$ and $B C$ can rotate around points $A$ and $B$ respectively. It is measured that $B C = 10 \\mathrm{~cm}, A B = 20 \\mathrm{~cm}$. When $A B$ and $B C$ rotate to $\\angle B A E = 60^{\\circ}, \\angle A B C = 45^{\\circ}$, find the distance from point $C$ to $A E$ (the result should be accurate to $0.1, \\sin 15^{\\circ} \\approx 0.26, \\cos 15^{\\circ} \\approx 0.97, \\tan 15^{\\circ} \\approx 0.27, \\sqrt{2} \\approx 1.41, \\sqrt{3} \\approx 1.73$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0057_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0057_2.jpg" ], "is_multi_img": true, "answer": "$7.6 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $BM \\perp AE$ and $CN \\perp AE$, with the feet of the perpendiculars at points $M$ and $N$ respectively. Then draw $CD \\perp BM$, with the foot of the perpendicular at $D$.\n\n\n\n$\\therefore CDMN$ is a rectangle.\n\nIn the right triangle $\\triangle ABM$, $\\angle AMB = 90^{\\circ}$.\n\n$\\because \\angle BAM = \\angle BAE = 60^{\\circ}$, and $AB = 20$,\n\n$\\therefore BM = \\sin \\angle BAM \\cdot AB = \\sin 60^{\\circ} \\cdot AB = \\frac{\\sqrt{3}}{2} \\times 20 = 10\\sqrt{3} \\approx 17.3 \\text{ cm}$.\n\n$\\because$ in the right triangle $\\triangle ABM$, $\\angle ABD = 90^{\\circ} - \\angle BAE = 90^{\\circ} - 60^{\\circ} = 30^{\\circ}$,\n\n$\\therefore \\angle DBC = \\angle ABC - \\angle ABD = 45^{\\circ} - 30^{\\circ} = 15^{\\circ}$.\n\n$\\because$ in the right triangle $\\triangle BCD$, $\\angle DBC = 15^{\\circ}$, and $BC = 10$,\n\n$\\therefore BD = \\cos \\angle DBC \\cdot BC = \\cos 15^{\\circ} \\cdot BC \\approx 0.97 \\times 10 = 9.7 \\text{ cm}$.\n\n$\\therefore CN = MD = BM - BD \\approx 17.3 - 9.7 = 7.6 \\text{ cm}$.\n\n$\\therefore$ The distance from point $C$ to $AE$ is approximately $7.6 \\text{ cm}$.\n\n【Insight】This problem examines the application of solving right triangles and the properties of rectangles. The key to solving the problem lies in clarifying the quantitative relationships between the line segments." }, { "problem_id": 1844, "question": "Basketball is one of the additional test items in the physical education section of the high school entrance examination. To meet the daily needs of students, a school has purchased a set of basketball facilities. Figure 1 shows a basketball hoop purchased by the school that meets the latest international standards. It not only aligns with public aesthetics but also maximizes the safety of students during exercise. Figure 2 is a side view of the basketball hoop, which consists of the backboard \\(EF\\), the hoop point \\(E\\), the balance rod \\(CE\\), the support arm \\(BC\\), the auxiliary arm \\(BD\\), and the main support \\(OA\\). Among them, \\(MN\\) is the horizontal ground, the main support \\(OA \\perp MN\\) at \\(O\\), the support arm \\(BC\\) passes through point \\(A\\), the balance rod \\(CE\\) is parallel to the horizontal ground \\(MN\\), the backboard \\(EF \\perp CE\\) at point \\(E\\), and the other end \\(D\\) of the auxiliary arm \\(BD\\) is fixed on the ground \\(MN\\).\n\nMeasurements show that the auxiliary arm \\(BD\\) is 1.5 meters long with an inclination angle of \\(60^\\circ\\) to the horizontal plane, and the support arm \\(BC\\) is 2.9 meters long with an inclination angle of \\(37^\\circ\\) to the horizontal plane. Based on the above data, please calculate the distance from the hoop point \\(E\\) to the ground (the final result should be accurate to 0.01 meters, where \\(\\sin 37^\\circ \\approx 0.6\\), \\(\\cos 37^\\circ \\approx 0.8\\), \\(\\tan 37^\\circ \\approx 0.75\\), and \\(\\sqrt{3} = 1.732\\)).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0061_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0061_2.jpg" ], "is_multi_img": true, "answer": " 3.04 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure: Draw a perpendicular from point $B$ to $MN$ at point $H$, from point $C$ to $MN$ at point $T$, and from point $B$ to $CT$ at point $K$.\n\nIn right triangle $\\triangle BDH$, $BD = 1.5$ meters, $\\angle BDH = 60^\\circ$,\n\n$\\therefore BH = BD \\cdot \\sin 60^\\circ = 1.5 \\times \\frac{\\sqrt{3}}{2} \\approx 1.299$ meters.\n\nIn right triangle $\\triangle BCK$, $BC = 2.9$, $\\angle CBK = 37^\\circ$,\n\n$\\therefore CK = BC \\cdot \\sin 37^\\circ \\approx 2.9 \\times 0.6 \\times \\frac{\\sqrt{3}}{2} = 1.74$ meters.\n\nSince $\\angle BHT = \\angle HTK = \\angle BKT = 90^\\circ$,\n\n$\\therefore$ quadrilateral $BHTK$ is a rectangle,\n\n$\\therefore KT = BH = 1.299$ meters,\n\n$\\therefore CT = KT + CK = 1.299 + 1.74 \\approx 3.04$ meters,\n\n$\\therefore$ the distance from the basket point $E$ to the ground is approximately 3.04 meters.\n\n\n\nFigure 2\n\n[Insight] This problem examines solving right triangle problems involving slopes and angles. The key to solving it is to add auxiliary lines to construct right triangles." }, { "problem_id": 1845, "question": "As shown in Figure 1 and Figure 2, these are the actual photo and side view diagram of a certain basketball hoop respectively. It is known that the height of the rectangular base $B C L K$ is $B K=19 \\mathrm{~cm}$, the width is $B C=40 \\mathrm{~cm}$, the angle formed by the base $B C$ and the support $A C$ is $\\angle A C B=76^{\\circ}$, the length of the support $A F$ is $240 \\mathrm{~cm}$, the distance from the top of the backboard $F$ to the rim $D$ is $F D=90 \\mathrm{~cm}$ ($F E$ is perpendicular to the ground $L K$, the support $A K$ is perpendicular to the ground $L K$, the support $H E$ is perpendicular to $F E$), the angle formed by the bottom support $H E$ of the backboard and the support $A F$ is $\\angle F H E=66^{\\circ}$. Find the distance from the rim $D$ to the ground (accurate to $1 \\mathrm{~cm}$). (Reference data: $\\sin 66^{\\circ} \\approx \\frac{9}{10}, \\cos 66^{\\circ} \\approx \\frac{2}{5}, \\tan 66^{\\circ} \\approx \\frac{9}{4}$, $\\sin 76^{\\circ} \\approx 0.97, \\cos 76^{\\circ} \\approx 0.24, \\tan 76^{\\circ} \\approx 4.0$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0062_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0062_2.jpg" ], "is_multi_img": true, "answer": "$305 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Extend $FE$ to intersect the ground $LK$ at point $N$. Draw a perpendicular from point $A$ to $FN$, intersecting at point $M$, and draw a perpendicular from point $B$ to $FN$, intersecting at point $G$, as shown in the figure.\n\n\n\nSince $CB \\perp AB$, $\\angle ACB = 76^\\circ$, and $BC = 40 \\text{ cm}$,\n\nwe have $AB = BC \\times \\tan 76^\\circ \\approx 160 \\text{ cm}$,\n\nthus $MG \\approx 160 \\text{ cm}$.\n\nGiven $AF = 240 \\text{ cm}$, $AM \\perp FN$, and $\\angle FAM = \\angle FHE = 66^\\circ$,\n\nwe find $FM = AF \\times \\sin 66^\\circ \\approx 216 \\text{ cm}$.\n\nSince $DN = NG + MG + FM - FD$,\n\nand given $NG = BK = 19 \\text{ cm}$, $FD = 90 \\text{ cm}$,\n\nwe have $DN = 19 + 160 + 216 - 90 = 305 \\text{ cm}$.\n\nAnswer: The distance from the basket $D$ to the ground is $305 \\text{ cm}$.\n\n【Key Insight】This problem tests the ability to find the length of a side in a triangle using trigonometric values. It is essential to master the meanings of tangent, sine, and cosine, and correctly drawing auxiliary lines is key to solving this problem." }, { "problem_id": 1846, "question": "Figure 1 shows a wall-mounted air conditioner installed on the wall of a room. Figure 2 is a side view schematic of the air conditioner installation. The air conditioner's fan blade \\( AF \\) rotates from top to bottom around point \\( A \\) to sweep the air. The installation requires that when the fan blade just reaches the outer edge of the bed, the angle \\( \\alpha \\) between the fan blade and the vertical line is \\( 48^\\circ \\). The bottom of the air conditioner \\( BC \\) is perpendicular to the wall \\( CD \\), with \\( AB = 0.02 \\) meters and \\( BC = 0.1 \\) meters. The length of the bed is \\( DE = 2 \\) meters. Calculate the height \\( CD \\) from the bottom of the installed air conditioner to the bed, in meters, to the nearest 0.1 meter.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0063_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0063_2.jpg" ], "is_multi_img": true, "answer": "1.7 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem statement, we have:\n\n$\\because AB=0.02 \\mathrm{~m}, \\quad BC=0.1 \\mathrm{~m}, \\quad DE=2 \\mathrm{~m}, \\quad EM=ED-BC=1.9 \\mathrm{~m}, \\alpha=48^{\\circ}$, $\\therefore \\tan \\alpha=\\tan 48^{\\circ}=\\frac{EM}{AM}=\\frac{1.9}{0.02+BM} \\approx 1.11$,\n\nSolving gives: $BM \\approx 1.7(\\mathrm{~m})$,\n\nThus, $CD=1.7(\\mathrm{~m})$.\n\nAnswer: The height $CD$ from the bottom of the installed air conditioner to the bed is 1.7 meters.\n\n\n\nFigure 2\n\n[Insight] This problem mainly tests the application of solving right triangles, and the key to solving it is deriving $\\tan 48^{\\circ}=\\frac{EM}{AM}$ from the given information." }, { "problem_id": 1847, "question": "Figure 1 shows a solar street light, with a solar panel at its top. During the day, it absorbs sunlight to charge the battery pack inside the lamp post. At night, the battery pack provides power to the LED light source, enabling the lighting function. Figure 2 is a schematic diagram of the upper part of the street light. The lamp arm \\( AD \\), the bracket \\( BC \\), and the vertical pole intersect at points \\( A \\) and \\( B \\), respectively. The lamp arm \\( AD \\) and the bracket \\( BC \\) intersect at point \\( C \\). Given that \\( \\angle MAC = 75^\\circ \\), \\( \\angle ACB = 15^\\circ \\), \\( BC = 60 \\, \\text{cm} \\), \\( CD = 80 \\, \\text{cm} \\), and the distance from point \\( B \\) to the ground is \\( 5.14 \\, \\text{m} \\), find the distance from the highest point \\( D \\) of the street light to the ground. (The result should be accurate to \\( 0.01 \\, \\text{m} \\). Reference data:\n\n\\[\n\\sin 15^\\circ \\approx 0.26, \\quad \\cos 15^\\circ \\approx 0.97, \\quad \\tan 15^\\circ \\approx 0.27, \\quad \\sin 75^\\circ \\approx 0.97, \\quad \\cos 75^\\circ \\approx 0.26, \\quad \\tan 75^\\circ \\approx 3.73, \\quad \\sqrt{3} \\approx 1.73\n\\]\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0064_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0064_2.jpg" ], "is_multi_img": true, "answer": "$5.65 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $CE \\perp MB$ at point $E$, draw $DF \\perp MB$ at point $F$, and draw $CG \\perp DF$ at point $G$.\n\n$\\therefore$ Quadrilateral $EFGC$ is a rectangle.\n\n$\\therefore FE = CG$, and $CE \\parallel DF$.\n\n$\\because \\angle MAC = 75^\\circ$, $\\angle ACB = 15^\\circ$,\n\n$\\therefore \\angle EBC = \\angle MAC - \\angle ACB = 75^\\circ - 15^\\circ = 60^\\circ$,\n\n$\\angle ECA = 180^\\circ - \\angle AEC - \\angle EAC = 180^\\circ - 90^\\circ - 75^\\circ = 15^\\circ$.\n\n$\\because CE \\parallel DF$,\n\n$\\therefore \\angle D = \\angle ECA = 15^\\circ$.\n\nIn right triangle $\\triangle ECB$, $\\angle BEC = 90^\\circ$, $\\angle EBC = 60^\\circ$.\n\n$\\because \\cos \\angle EBC = \\frac{EB}{BC}$,\n\n$\\therefore EB = BC \\cdot \\cos 60^\\circ = 60 \\times \\frac{1}{2} = 30(\\mathrm{cm}) = 0.3 \\mathrm{m}$.\n\nIn right triangle $\\triangle CGD$, $\\angle CGD = 90^\\circ$, $\\angle D = 15^\\circ$.\n\n$\\because \\sin D = \\frac{GC}{CD}$,\n\n$\\therefore EF = GC = CD \\cdot \\sin 15^\\circ \\approx 80 \\times 0.26 = 20.8(\\mathrm{cm}) = 0.208 \\mathrm{m}$.\n\n$\\therefore BF = EB + EF = 0.3 + 0.208 = 0.508(\\mathrm{m})$.\n\n$\\therefore$ The distance from point $D$ to the ground: $0.508 + 5.14 = 5.648 \\approx 5.65(\\mathrm{m})$.\n\nAnswer: The distance from point $D$ to the ground is approximately $5.65 \\mathrm{m}$.\n\n\n\n【Highlight】This problem examines the application of solving right triangles. The key to solving it is the proficient use of the definitions of trigonometric functions for acute angles. This problem is of medium difficulty." }, { "problem_id": 1848, "question": "Figure (1) is a physical image of a disinfectant spray bottle, and its schematic diagram is shown in Figure (2). Given that \\( AB = 6 \\text{ cm} \\), \\( BC = 4 \\text{ cm} \\), \\( \\angle ABC = 85^\\circ \\), and \\( \\angle BCD = 120^\\circ \\), find the distance from point \\( A \\) to \\( CD \\). (Reference data: \\( \\sin 65^\\circ \\approx 0.905 \\),\n\n\\( \\cos 65^\\circ \\approx 0.423 \\), \\( \\tan 65^\\circ \\approx 2.144 \\), \\( \\sqrt{3} \\approx 1.732 \\).)\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0068_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0068_2.jpg" ], "is_multi_img": true, "answer": " $9.466 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $AE \\perp CD$ at point $E$, $BF \\perp AE$ at point $F$, and $CG \\perp BF$ at point $G$. Then, quadrilateral $CEFG$ is a rectangle.\n\n$\\therefore EF = CG$, and $\\angle AFB = \\angle BGC = \\angle AEC = 90^{\\circ}$.\n\nGiven that $\\angle ABC = 85^{\\circ}$ and $\\angle BCD = 120^{\\circ}$,\n\n$\\therefore \\angle BAE = 360^{\\circ} - \\angle ABC - \\angle BCD - \\angle AEC = 65^{\\circ}$,\n$\\therefore \\angle ABF = 90^{\\circ} - \\angle BAF = 25^{\\circ}$,\n\n$\\therefore \\angle CBG = \\angle ABC - \\angle ABF = 60^{\\circ}$.\n\nIn right triangle $\\triangle ABF$, $AF = AB \\cdot \\cos \\angle BAF \\approx 2.538 \\mathrm{~cm}$.\n\nIn right triangle $\\triangle BCG$, $CG = BC \\cdot \\sin \\angle CBG \\approx 6.928 \\mathrm{~cm}$.\n\n$\\therefore EF = CG = 6.928 \\mathrm{~cm}$,\n\n$\\therefore AE = AF + EF = 9.466 \\mathrm{~cm}$,\n\n$\\therefore$ The distance from point $A$ to $CD$ is approximately $9.466 \\mathrm{~cm}$.\n\n\n\n【Key Insight】This problem primarily examines the understanding of solving right triangles, the properties and determination of rectangles, the interior angle sum theorem of quadrilaterals, the complementary nature of acute angles in right triangles, and the distance from a point to a line. The key to solving the problem lies in correctly constructing auxiliary lines to form right triangles." }, { "problem_id": 1849, "question": "To strengthen epidemic prevention and control measures and avoid the phenomenon of personnel gathering during temperature measurement, a school in Yunyan District installed an infrared temperature detection device that quickly measures the temperature of individuals entering the detection area by detecting the infrared radiation energy emitted by the human body (as shown in Figure (1). It is known that the installation height of the device, $OC$, is $1.70 \\mathrm{~m}$ (as shown in Figure (2)). If the maximum detection angle ($\\angle OBC$) is $64.5^{\\circ}$ and the minimum detection angle ($\\angle OAC$) is $26.6^{\\circ}$, please calculate the length of the school's temperature measurement area, $AB$. (The result should be accurate to $0.1 \\mathrm{~m}$). Reference data: $\\sin 26.6^{\\circ} \\approx 0.45, \\cos 26.6^{\\circ} \\approx 0.89, \\tan 26.6^{\\circ} \\approx 0.50, \\sin 64.5^{\\circ} \\approx 0.90, \\cos 64.5^{\\circ} \\approx 0.43, \\tan 64.5^{\\circ} \\approx 2.10$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0075_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0075_2.jpg" ], "is_multi_img": true, "answer": "$2.4 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle BCO$,\n\nSince $\\tan \\angle OBC = \\frac{OC}{BC}$,\n\nTherefore, $BC = \\frac{OC}{\\tan \\angle OBC} = \\frac{OC}{\\tan 60^\\circ} = \\frac{1.7}{\\sqrt{3}} \\approx 0.99 \\quad (\\text{m})$.\n\nIn the right triangle $\\triangle ACO$,\n\nSince $\\tan \\angle OAC = \\frac{OC}{AC}$,\n\nTherefore, $AC = \\frac{OC}{\\tan 26.6^\\circ} \\approx \\frac{1.70}{0.50} \\approx 3.40 \\quad (\\text{m})$.\n\nThus, $AB = AC - BC$\n\n$\\approx 3.40 - 0.99$\n\n$= 2.41$\n\n$\\approx 2.4 \\quad (\\text{m})$.\n\nAnswer: The length of the school's temperature measurement area $AB$ is 2.4 meters.\n\n[Highlight] This problem primarily examines the solution of right triangles, and mastering the relationship between the sides and angles of a right triangle is key to solving this problem." }, { "problem_id": 1850, "question": "As shown in Figure 1, the Golden Pagoda is situated on the western riverbank. It was built in the first year of the Xianping era of the Song Dynasty, i.e., 998 AD. This pagoda is the oldest existing ancient tower structure in our province and is a protected national cultural relic under the seventh batch of national key protection units. The pagoda has nine stories and a hexagonal shape. The ninth-grade mathematics interest group conducted a practical activity to measure the height of the \"Golden Pagoda,\" with the following process:\n\n**Design of the Plan:** As shown in Figure 2, the Golden Pagoda \\( CD \\) is perpendicular to the ground. Two points \\( A \\) and \\( B \\) are selected on the ground to measure the angles \\( \\angle CAD \\) and \\( \\angle CBD \\) respectively (with \\( A, B, \\) and \\( D \\) on the same straight line).\n\n**Data Collection:** Through field measurements: The distance between points \\( A \\) and \\( B \\) on the ground is \\( 55 \\) meters, \\( \\angle CAD = 37^\\circ \\), and \\( \\angle CBD = 63.4^\\circ \\).\n\n**Problem Solving:** Determine the height of the Golden Pagoda \\( CD \\). Reference data: \\( \\sin 37^\\circ \\approx 0.6 \\), \\( \\cos 37^\\circ \\approx 0.8 \\), \\( \\tan 37^\\circ \\approx 0.75 \\); \\( \\sin 63.4^\\circ \\approx 0.89 \\), \\( \\cos 63.4^\\circ \\approx 0.45 \\), \\( \\tan 63.4^\\circ \\approx 2.00 \\). Based on the above plan and data, please complete the solution process.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0077_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0077_2.jpg" ], "is_multi_img": true, "answer": "30 m", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Let \\( CD = x \\) meters,\n\nIn the right triangle \\( \\triangle ACD \\), \\( \\angle CAD = 37^\\circ \\),\n\nThus, \\( \\tan \\angle CAD = \\frac{CD}{AD} \\), so \\( AD = \\frac{CD}{\\tan \\angle CAD} = \\frac{x}{\\tan 37^\\circ} \\),\n\nIn the right triangle \\( \\triangle BCD \\), \\( \\angle CBD = 63.4^\\circ \\),\n\nThus, \\( \\tan \\angle CBD = \\frac{CD}{BD} \\), so \\( BD = \\frac{CD}{\\tan \\angle CBD} = \\frac{x}{\\tan 63.4^\\circ} \\),\n\nSince \\( AD + BD = AB \\),\n\nTherefore, \\( \\frac{x}{\\tan 37^\\circ} + \\frac{x}{\\tan 63.4^\\circ} = 55 \\),\n\nSolving for \\( x \\): \\( x = \\frac{55 \\times \\tan 37^\\circ \\times \\tan 63.4^\\circ}{\\tan 37^\\circ + \\tan 63.4^\\circ} \\approx \\frac{55 \\times 0.75 \\times 2}{0.75 + 2} = 30 \\),\n\nAnswer: The height of the Golden Tower \\( CD \\) is approximately 30 meters.\n\n[Key Insight] This problem tests the understanding of solving right triangles. Mastering the definition and calculation methods of the tangent function is crucial for solving such problems." }, { "problem_id": 1851, "question": "The folding ladder, when fully extended, is shown in Figure 1. Points \\( B \\) and \\( C \\) are the contact points of the ladder on the ground, and point \\( D \\) is the fixed point of the highest step. Figure 2 is a schematic diagram of the ladder. It is known that \\( AB = AC \\), \\( BD = 140 \\, \\text{cm} \\), and \\( \\angle BAC = 42^\\circ \\). Find the height \\( DE \\) of point \\( D \\) from the ground. (Round the result to the nearest whole number. Reference data: \\( \\sin 69^\\circ \\approx 0.93 \\), \\( \\cos 69^\\circ \\approx 0.36 \\), \\( \\tan 69^\\circ \\approx 2.61 \\))\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0078_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0078_2.jpg" ], "is_multi_img": true, "answer": "$130 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AB = AC \\) and \\( \\angle BAC = 42^\\circ \\),\n\nit follows that \\( \\angle B = \\angle C = 69^\\circ \\).\n\nGiven that \\( DE \\perp BC \\),\n\nwe have \\( \\angle DEB = 90^\\circ \\).\n\nThus, \\( \\sin B = \\frac{DE}{BD} \\).\n\nSince \\( BD = 140 \\, \\text{cm} \\),\n\nit follows that \\( DE = BD \\cdot \\sin B \\approx 140 \\times 0.93 \\approx 130 \\, \\text{cm} \\).\n\nAnswer: The height \\( DE \\) of point \\( D \\) from the ground is \\( 130 \\, \\text{cm} \\).\n\n[Key Insight] This problem tests the application of solving right triangles. Mastering the relevant knowledge and accurately understanding the problem are crucial for solving it." }, { "problem_id": 1852, "question": "The fire tongs are a tool made of iron for picking up firewood. They are also used by cleaners to pick up litter on the ground. Figure 1 is a photograph of the actual tool, and Figure 2 is a schematic diagram of it. It is known that when the fire tongs are fully opened, the angle between the two tong arms $OC$ and $OD$ is $\\angle COD = 40^{\\circ}$. If $OC = OD = 40 \\, \\text{cm}$, find the distance between the ends of the tong arms $C$ and $D$. (The result should be accurate to $1 \\, \\text{cm}$. Reference data: $\\sin 70^{\\circ} \\approx 0.94, \\cos 70^{\\circ} \\approx 0.34, \\tan 70^{\\circ} \\approx 2.75$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0081_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0081_2.jpg" ], "is_multi_img": true, "answer": "$27 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, connect $CD$, and draw $OH \\perp CD$ at point $H$.\n\n\n\nSince $OC = OD = 40 \\mathrm{~cm}$, $OH \\perp CD$, and $\\angle COD = 40^{\\circ}$,\n\n$\\therefore \\angle OHD = 90^{\\circ}$, $CD = 2DH$, and $\\angle ODC = \\frac{1}{2}\\left(180^{\\circ} - \\angle COD\\right) = 70^{\\circ}$.\n\n$\\therefore DH = OD \\cdot \\cos \\angle ODC = OD \\cdot \\cos 70^{\\circ} \\approx 40 \\times 0.34 = 13.6 \\mathrm{~cm}$.\n\n$\\therefore CD = 2DH = 2 \\times 13.6 = 27.2 \\mathrm{~cm} \\approx 27 \\mathrm{~cm}$.\n\nAnswer: The distance between the endpoints $C$ and $D$ of the two arms is approximately $27 \\mathrm{~cm}$.\n\n【Highlight】This problem examines the application of trigonometric functions, the properties of isosceles triangles, and the concept of approximate numbers. The key to solving this problem lies in the flexible use of trigonometric functions and correctly drawing auxiliary lines." }, { "problem_id": 1853, "question": "As shown in Figure 1, a mobile phone tablet stand consists of a tray, a support plate, and a base. The phone is placed on the tray. Figure 2 is a schematic diagram of its side structure. The measurements are as follows: the length of the tray \\( AB = 120 \\text{ mm} \\), the length of the support plate \\( CD = 80 \\text{ mm} \\), and the length of the base \\( DE = 90 \\text{ mm} \\). The tray \\( AB \\) is fixed at point \\( C \\) at the top of the support plate, with \\( CB = 40 \\text{ mm} \\). The tray \\( AB \\) can rotate around point \\( C \\), and the support plate \\( CD \\) can rotate around point \\( D \\). When \\( \\angle DCB = 75^\\circ \\) and \\( \\angle CDE = 60^\\circ \\), find the distance from point \\( A \\) to the line \\( DE \\) (the result should be in surd form).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0083_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0083_2.jpg" ], "is_multi_img": true, "answer": "$40 \\sqrt{3}+40 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Through point \\( A \\), draw \\( AM \\perp DE \\), intersecting the extension of \\( ED \\) at point \\( M \\). Through point \\( C \\), draw \\( CF \\perp AM \\), with the foot of the perpendicular at \\( F \\). Through point \\( C \\), draw \\( CN \\perp DE \\), with the foot of the perpendicular at \\( N \\).\n\n\n\nFrom the problem statement, we know that \\( AC = AB - BC = 80 \\), \\( CD = 80 \\), \\( \\angle DCB = 75^\\circ \\), and \\( \\angle CDE = 60^\\circ \\).\n\nIn the right triangle \\( \\triangle CDN \\), \\( CN = CD \\cdot \\sin \\angle CDE = 80 \\times \\frac{\\sqrt{3}}{2} = 40\\sqrt{3} = FM \\).\n\n\\( \\angle DCN = 90^\\circ - 60^\\circ = 30^\\circ \\).\n\nSince \\( \\angle DCB = 75^\\circ \\), it follows that \\( \\angle BCN = 75^\\circ - 30^\\circ = 45^\\circ \\).\n\nBecause \\( AM \\perp DE \\) and \\( CN \\perp DE \\), \\( AM \\parallel CN \\).\n\nThus, \\( \\angle A = \\angle BCN = 45^\\circ \\).\n\nIn the right triangle \\( \\triangle AFC \\), \\( AF = AC \\cdot \\sin 45^\\circ = 80 \\times \\frac{\\sqrt{2}}{2} = 40\\sqrt{2} \\) mm.\n\nTherefore, \\( AM = AF + FM = 40\\sqrt{3} + 40\\sqrt{2} \\) mm.\n\nAnswer: The distance from point \\( A \\) to the line \\( DE \\) is approximately \\( (40\\sqrt{3} + 40\\sqrt{2}) \\) mm.\n\n【Key Insight】This problem tests the application of right triangles and the meaning of trigonometric functions of acute angles. Constructing auxiliary lines to form right triangles is a common and fundamental method." }, { "problem_id": 1854, "question": "A certain large bridge adopts the structure of a cable-stayed bridge with an H-shaped tower (as shown in Figure A). Figure B is the abstracted plan view derived from Figure A. It is measured that the angle between the cable $A B$ and the horizontal bridge deck is $45^{\\circ}$, and the angle between the cable $C D$ and the horizontal bridge deck is $65^{\\circ}$. The distance between the tops of the two cables $A C$ is 2 meters, and the distance between the bottoms of the two cables $B D$ is 10 meters. Please calculate the length of the pillar $A H$ (the result should be accurate to 1 meter).\n\n(Reference data: $\\sin 65^{\\circ} \\approx 0.91, \\cos 65^{\\circ} \\approx 0.42, \\tan 65^{\\circ} \\approx 2.14$)\n\n\n\nFigure A\n\n\n\nFigure B", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0086_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0086_2.jpg" ], "is_multi_img": true, "answer": "17 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of $AH$ be $x$ meters, then the length of $CH$ is $(x-2)$ meters.\n\nIn the right triangle $\\triangle ABH$, $AH = BH \\cdot \\tan 45^\\circ$,\n\n$\\therefore BH = x$,\n\n$\\therefore DH = BH - BD = x - 10$.\n\nIn the right triangle $\\triangle CDH$, $CH = DH \\cdot \\tan 65^\\circ$,\n\n$\\therefore x - 2 = 2.14(x - 10)$,\n\nSolving the equation: $x = 17.01 \\approx 17$.\n\nAnswer: The length of the pillar $AH$ is approximately 17 meters.\n\n[Key Insight] This problem examines the solution of right triangles. Mastering the function values of special angles and flexibly selecting trigonometric functions are key to solving the problem." }, { "problem_id": 1855, "question": "As shown in Figure 1, it is a partial diagram of the cable frame of a certain bridge, and Figure 2 is the plan view of the cable frame of the bridge. If the angle between the cable $AB$ and the horizontal bridge deck is measured to be $40^{\\circ}$, and the angle between the cable $CD$ and the horizontal bridge deck is $60^{\\circ}$, the distance between the tops of the two cables $BC$ is 2 meters, the distance between the bottoms of the two cables $AD$ is 20 meters, $A, D, H$ are collinear, $B, C, H$ are collinear, and $AH \\perp BH$, please calculate the length of the vertical column $BH$. (The result should be accurate to 1 meter; $\\sin 40^{\\circ} \\approx 0.64$,\n\n$\\cos 40^{\\circ} \\approx 0.7, \\quad \\tan 40^{\\circ} \\approx 0.84, \\sqrt{3} \\approx 1.73$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0097_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0097_2.jpg" ], "is_multi_img": true, "answer": " 31 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( CH = x \\) meters, then \\( BH = (x + 2) \\) meters.\n\nIn the right triangle \\( \\triangle CDH \\), \\( \\angle CDH = 60^\\circ \\), and \\( \\tan \\angle CDH = \\frac{CH}{DH} = \\sqrt{3} \\),\n\nThus, \\( DH = \\left(\\frac{\\sqrt{3}}{3} x\\right) \\) meters.\n\nIn the right triangle \\( \\triangle ABH \\), \\( \\angle A = 40^\\circ \\), and \\( \\tan A = \\frac{BH}{AH} \\approx 0.84 \\),\n\nTherefore, \\( AH \\approx \\frac{x + 2}{0.84} = \\frac{25}{21}(x + 2) \\) meters.\n\nGiven that \\( AD \\) is 20 meters, we have:\n\\[\n\\frac{25}{21}(x + 2) - \\frac{\\sqrt{3}}{3} x = 20\n\\]\n\nSolving for \\( x \\), we find \\( x \\approx 29 \\).\n\nHence, \\( BH = CH + BC \\approx 31 \\) meters.\n\nAnswer: The length of the pillar \\( BH \\) is approximately 31 meters.\n\n[Highlight] This problem tests the application of solving right triangles. The key to solving it lies in understanding the problem and mastering the definitions of trigonometric functions." }, { "problem_id": 1856, "question": "An alcohol disinfectant bottle is shown in Figure 1, where $AB$ is the nozzle, $\\triangle BCD$ is the pressing handle, $CE$ is the telescopic linkage, and $BE$ and $EF$ are the conduits. The schematic diagram is shown in Figure 2, with $\\angle DBE = \\angle BEF = 108^\\circ$, $BD = 6 \\text{ cm}$, $BE = 4 \\text{ cm}$, and $BC \\parallel EF$. Find the distance from point $D$ to the line $EF$ (the result should be accurate to $0.1 \\text{ cm}$). (Reference data:\n\n$\\sin 36^\\circ \\approx 0.59, \\cos 36^\\circ \\approx 0.81, \\tan 36^\\circ \\approx 0.73, \\sin 72^\\circ \\approx 0.95, \\cos 72^\\circ \\approx 0.31, \\tan 72^\\circ \\approx 3.08$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_198169e355ab3da4787bg_0099_1.jpg", "batch29-2024_06_14_198169e355ab3da4787bg_0099_2.jpg" ], "is_multi_img": true, "answer": "$7.3 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular from point $D$ to the extension of line $BC$, intersecting at point $N$, and draw a perpendicular from point $E$ to line $BC$ at point $M$.\n\n\n\nSince $BC \\parallel EF$,\n\n$\\angle BEF + \\angle MBE = 180^\\circ$,\n\n$\\angle MBE = 180^\\circ - \\angle BEF = 180^\\circ - 108^\\circ = 72^\\circ$,\n\n$\\angle DBN = \\angle DBE - \\angle MBE = 108^\\circ - 72^\\circ = 36^\\circ$.\n\nIn right triangle $\\triangle DBN$,\n\n$\\sin \\angle DBN = \\sin 36^\\circ = \\frac{DN}{DB}$,\n\n$DN = 6 \\times \\sin 36^\\circ \\approx 6 \\times 0.59 = 3.54$.\n\nIn right triangle $\\triangle EBM$,\n\n$\\sin \\angle MBE = \\sin 72^\\circ = \\frac{EM}{BE}$,\n\n$EM = 4 \\times \\sin 72^\\circ \\approx 4 \\times 0.95 = 3.80$.\n\nSince $BC \\parallel EF$,\n\nthe distance from point $D$ to line $EF$ is $EM + DN = 3.80 + 3.54 \\approx 7.3$.\n\nAnswer: The distance from point $D$ to line $EF$ is $7.3 \\mathrm{~cm}$.\n\n[Highlight] This problem mainly examines the practical application of solving right triangles, involving properties of parallel lines and trigonometric functions of acute angles. Mastering the definitions of trigonometric functions is key to solving the problem." }, { "problem_id": 1857, "question": "Qin Yang Village used a floating crane as shown in Figure 1 during the cleaning of the reservoir, significantly improving work efficiency. A mathematical research group, in order to calculate the length of the suspension cable $A C$ of the floating crane shown in Figure 2, conducted a series of measurements and obtained the following data: $\\angle A = 30^\\circ, \\angle A B C = 105^\\circ, A B = 20 \\text{ m}$. Please help them find the length of the suspension cable $A C$.\n\n(The result should be accurate to $0.1 \\text{ m}, \\sqrt{2} \\approx 1.414, \\sqrt{3} \\approx 1.732$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0006_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0006_2.jpg" ], "is_multi_img": true, "answer": "$27.3 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular line from point $B$ to $AC$, denoted as $BD \\perp AC$.\n\n\n\nGiven that $\\angle A = 30^\\circ$, $\\angle ABC = 105^\\circ$, and $AB = 20 \\mathrm{~m}$,\n\nwe have $BD = \\frac{1}{2} AB = 10 \\mathrm{~m}$, and $\\angle C = 180^\\circ - \\angle A - \\angle ABC = 45^\\circ$.\n\nThus, $\\triangle BDC$ is an isosceles right triangle.\n\nTherefore, $CD = BD = 10 \\mathrm{~m}$.\n\nIn the right triangle $\\triangle ABD$, $AD = \\sqrt{3} \\cdot BD = 10\\sqrt{3} \\approx 17.32 \\mathrm{~m}$.\n\nHence, $AC = AD + DC = 17.32 + 10 = 27.32 \\approx 27.3 \\mathrm{~m}$.\n\n【Key Insight】This problem tests the application of the Pythagorean theorem in solving right triangles, and the key to solving it lies in adding auxiliary lines." }, { "problem_id": 1858, "question": "An advertising design company plans to install a starry sky pattern neon light (as shown in Figure 2 at $AE$) on the street-facing wall of a high-rise building (as shown in Figure 2 at $AB$). They will use a ladder delivery vehicle as shown in Figure 1. It is known that the base of the ladder is 10 meters away from the wall. The ladder is raised from the bottom to install the neon lights. After measurement, when the top of the ladder reaches point $A$, the angle between the ladder and the horizontal plane is $65^{\\circ}$, and when the top of the ladder reaches point $E$, the angle between the ladder and the horizontal plane is $45^{\\circ}$. Find the height of this starry sky pattern neon light. (The result should be accurate to 1 meter. Reference data: $\\sin 65^{\\circ} \\approx 0.91, \\cos 65^{\\circ} \\approx 0.42, \\tan 65^{\\circ} \\approx 2.14$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0024_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0024_2.jpg" ], "is_multi_img": true, "answer": "$11 \\mathrm{~m}$\n", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from point $D$ to $AB$, meeting at point $F$, as shown in the figure. Then, quadrilateral $DFBC$ is a rectangle, with $\\angle EDF = 45^\\circ$ and $\\angle ADF = 65^\\circ$.\n\n$\\therefore BC = DF = 10 \\mathrm{~m}$.\n\n\n\nIn right triangle $\\triangle DEF$, $\\tan \\angle EDF = \\tan 45^\\circ = \\frac{EF}{DF} = 1$.\n\n$\\therefore EF = DF = 10 \\mathrm{~m}$.\n\nIn right triangle $\\triangle ADF$, $\\tan \\angle ADF = \\tan 65^\\circ = \\frac{AF}{DF} = \\frac{AF}{10} \\approx 2.14$.\n\n$\\therefore AF \\approx 21.4 \\mathrm{~m}$.\n\n$\\therefore AE = AF - EF \\approx 21.4 - 10 \\approx 11 \\mathrm{~m}$.\n\nAnswer: The height of this starry neon sign is approximately $11 \\mathrm{~m}$.\n\n【Key Insight】This problem tests the practical application of using trigonometric functions to measure distances. A thorough understanding of trigonometric concepts is crucial for solving such problems." }, { "problem_id": 1859, "question": "Figure 1 is a model of a certain type of stealth fighter in China. One of the highlights of this fighter is its all-moving swept-back vertical tail. Figure 2 shows the axial section of the vertical tail model, and the following data was obtained from the periphery of the vertical tail model: $BC = 8, DC = 2, \\angle D = 135^{\\circ}, \\angle C = 60^{\\circ}$, and $AB \\parallel CD$. Calculate the area of the vertical tail model $ABCD$. (The result should be kept in radical form)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0032_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0032_2.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw $CF \\perp AB$ at point $C$ and $DE \\perp AB$ at point $D$, as shown in the diagram:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nSince $AB \\parallel CD$,\n\n$\\therefore \\angle CBF = \\angle BCD = 60^\\circ$, and $\\angle A = 180^\\circ - \\angle ADC = 45^\\circ$.\n\nIn the right triangle $\\triangle BCF$, $\\sin 60^\\circ = \\frac{CF}{BC} = \\frac{CF}{8}$, and $\\cos 60^\\circ = \\frac{BF}{BC} = \\frac{BF}{8}$.\n\n$\\therefore CF = 4\\sqrt{3}$, and $BF = 4$.\n\nSince $CF \\perp AB$ and $DE \\perp AB$,\n\n$\\therefore CF \\parallel DE$.\n\nSince $AB \\parallel CD$,\n\n$\\therefore$ quadrilateral $DCFE$ is a rectangle.\n\n$\\therefore EF = CD = 2$, and $DE = CF = 4\\sqrt{3}$.\n\nIn the right triangle $\\triangle ADE$, $\\tan 45^\\circ = \\frac{DE}{AE} = \\frac{4\\sqrt{3}}{AE}$.\n\n$\\therefore AE = 4\\sqrt{3}$.\n\n$\\therefore AB = AE + EF - BF = 4\\sqrt{3} + 2 - 4 = 4\\sqrt{3} - 2$.\n\n$\\therefore$ The area of trapezoid $ABCD$ is $\\frac{DC + AB}{2} \\cdot DE = \\frac{2 + 4\\sqrt{3} - 2}{2} \\times 4\\sqrt{3} = 24$.\n\n【Key Insight】This problem examines the solution of right triangles, the determination and properties of rectangles, the properties and determination of parallel lines, and the mastery of trigonometric definitions as a prerequisite for correct calculations. Constructing right triangles is key to solving the problem." }, { "problem_id": 1860, "question": "The A-frame folding ladder, when fully extended, is shown in Figure 1. Points \\( B \\) and \\( C \\) are the contact points of the ladder on the ground, and point \\( D \\) is the fixed point of the highest step of the ladder. Figure 2 is a schematic diagram of the ladder. It is known that \\( AB = AC \\), \\( BD = 150 \\, \\text{cm} \\), and \\( \\angle BAC = 42^\\circ \\). Find the height \\( DE \\) of point \\( D \\) from the ground. (Round the result to the nearest whole number. Reference data: \\( \\sin 69^\\circ \\approx 0.93 \\), \\( \\cos 69^\\circ \\approx 0.36 \\), \\( \\tan 69^\\circ \\approx 2.61 \\))\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0034_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0034_2.jpg" ], "is_multi_img": true, "answer": "$140 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AB = AC \\) and \\( \\angle BAC = 42^\\circ \\),\n\nTherefore, \\( \\angle B = \\angle C = 69^\\circ \\).\n\nSince \\( DE \\perp BC \\),\n\nTherefore, \\( \\angle DEB = 90^\\circ \\).\n\nThus, \\( \\sin B = \\frac{DE}{BD} \\).\n\nGiven that \\( BD = 150 \\, \\text{cm} \\),\n\nTherefore, \\( DE = BD \\cdot \\sin B \\approx 150 \\times 0.93 \\approx 140 \\, \\text{cm} \\).\n\nAnswer: The height \\( DE \\) of point \\( D \\) from the ground is approximately \\( 140 \\, \\text{cm} \\).\n\n[Highlight] This problem examines the application of solving right triangles. Mastering the knowledge points and accurately understanding the problem are key to solving it." }, { "problem_id": 1861, "question": "Solar photovoltaic power generation, characterized by its cleanliness, safety, and efficiency, has become a key new energy industry for countries around the world. Figure (1) shows a physical image of a solar panel, with its cross-sectional diagram shown in Figure (2). $A B$ represents the solar panel, with one end $A$ fixed at an angle $\\angle D A B = 22^{\\circ}$ to the horizontal plane, and the other end $B$ connected to the support frame $B C$. The base of the frame $C D$ is perpendicular to the horizontal plane, and $\\angle B C D = 135^{\\circ}$. Given that $A D = 3 \\mathrm{~m}$ and $C D = 0.5 \\mathrm{~m}$, find the length of $A B$. (Reference data:\n\n$\\sin 22^{\\circ} \\approx 0.37, \\cos 22^{\\circ} \\approx 0.93, \\tan 22^{\\circ} \\approx 0.40$, with the result accurate to $0.01 \\mathrm{~m}$.)\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0043_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0043_2.jpg" ], "is_multi_img": true, "answer": " $2.70 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $BF \\perp AD$ at point $F$, and draw $CE \\perp BF$ at point $E$. Then, quadrilateral $CDFE$ is a rectangle.\n\n$\\therefore EF = CD = 0.5 \\mathrm{~m}, \\quad DF = CE, \\angle ECD = \\angle BEC = \\angle EFD = \\angle AFB = \\angle CEF = 90^{\\circ}$.\n\nSince $\\angle BCD = 135^{\\circ}$,\n\n$\\therefore \\angle BCE = 45^{\\circ}$,\n\n$\\therefore \\angle CBE = 45^{\\circ} = \\angle BCE$,\n\n$\\therefore BE = CE$.\n\nLet $BE = CE = DF = x \\mathrm{~m}$, then $BF = BE + EF = (x + 0.5) \\mathrm{~m}$, and $AF = AD - DF = (3 - x) \\mathrm{~m}$.\n\nIn right triangle $\\triangle AFB$, $\\tan \\angle BAF = \\frac{BF}{AF} = \\tan 22^{\\circ} \\approx 0.40$,\n\n$\\therefore \\frac{x + 0.5}{3 - x} \\approx 0.40$,\n\nSolving for $x$, we get $x \\approx 0.50$,\n\n$\\therefore BF \\approx 1 \\mathrm{~m}$,\n\n$\\therefore AB = \\frac{BF}{\\sin \\angle BAF} \\approx 2.70 \\mathrm{~m}$,\n\n$\\therefore$ The length of $AB$ is approximately $2.70 \\mathrm{~m}$.\n\n\n\n【Key Insight】This problem primarily examines the properties and determination of rectangles, solving right triangles, and the properties and determination of isosceles right triangles. The key to solving the problem lies in correctly constructing auxiliary lines to form right triangles." }, { "problem_id": 1862, "question": "The entrance of the Happy Home Community garage is equipped with a \"curved arm\" as shown in Figure 1. $O A \\perp A B, O A=1$ meter, where point $O$ is the pivot point for the arm's rotation, and point $C$ is the connection point of the two segments of the curved arm. The segment $C D$ of the curved arm always remains parallel to $A B$. As shown in Figure 2, when the curved arm is in its initial position, points $O, C, D$ are collinear. When the curved arm rises to position $O E$, $\\angle A O E=120^{\\circ}$, and the distance from point $E$ to $A B$ is 1.7 meters. When the curved arm rises to position $O F$, $\\angle C O F=66^{\\circ}$, what is the distance from point $F$ to $A B$ in meters? (The result should be accurate to 0.1. Reference data: $\\sqrt{3} \\approx 1.73, \\sin 66^{\\circ} \\approx 0.9$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0058_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0058_2.jpg" ], "is_multi_img": true, "answer": "2.3 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in the figure, through points $E$ and $F$, draw $EG \\perp OD$ and $FH \\perp OD$, intersecting at points $G$ and $H$ respectively.\n\n\n\nSince $OA \\perp AB$ and $OD // AB$,\n\n$\\therefore OA \\perp OD$\n\n$\\therefore \\angle AOD = 90^{\\circ}$\n\nGiven $\\angle AOE = 120^{\\circ}$ and $\\angle COF = 66^{\\circ}$,\n\n$\\therefore \\angle EOG = 120^{\\circ} - 90^{\\circ} = 30^{\\circ}$, and $\\angle FOH = \\angle COF = 66^{\\circ}$\n\nSince the distance from point $E$ to $AB$ is 1.7 meters and $OA = 1$ meter,\n\n$\\therefore EG = 1.7 - 1 = 0.7$ meters\n\nIn right triangle $\\triangle OEG$,\n\nSince $EG = OE \\times \\sin \\angle EOG$,\n$\\therefore OE = \\frac{0.7}{\\sin 30^{\\circ}} = \\frac{0.7}{\\frac{1}{2}} = 1.4$ meters\n\nGiven $OE = OF$,\n\nIn right triangle $\\triangle OFH$,\n\nSince $FH = OF \\times \\sin \\angle FOH = 1.4 \\times \\sin 66^{\\circ} \\approx 1.4 \\times 0.9 = 1.26$ meters\n\n$\\therefore FH + OA = 1.26 + 1 = 2.26 \\approx 2.3$ meters\n\n$\\therefore$ The distance from point $F$ to $AB$ is approximately 2.3 meters.\n\n【Insight】This problem tests the properties of right triangles, and the key to solving it is to be proficient in the sine function related to right triangles." }, { "problem_id": 1863, "question": "As shown in the figure, Figure 1 is a blood-red desk lamp, and Figure 2 is a side view of it (ignoring the height of the lamp base). The lamp arm \\( AC = 40 \\, \\text{cm} \\), the lamp shade \\( CD = 30 \\, \\text{cm} \\), and the angle formed by the lamp arm and the base is \\( \\angle CAB = 60^\\circ \\). The lamp shade \\( CD \\) can be adjusted up or down by a certain angle around point \\( C \\). During use, it is found that when \\( CD \\) forms an angle of \\( 30^\\circ \\) with the horizontal line, the lamp light is optimal. Find the distance from point \\( D \\) to the desktop at this time. (The result should be accurate to \\( 1 \\, \\text{cm} \\), and \\( \\sqrt{3} \\) is taken as 1.732)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0062_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0062_2.jpg" ], "is_multi_img": true, "answer": "$50 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Through point \\( D \\), draw \\( DH \\perp AB \\), intersecting the extension of \\( AB \\) at point \\( H \\). Through point \\( C \\), draw \\( CF \\perp AH \\) at \\( F \\), and through point \\( C \\), draw \\( CE \\perp DH \\) at \\( E \\).\n\n\n\nIn right triangle \\( \\triangle ACF \\), \\( \\angle A = 60^\\circ \\), \\( AC = 40 \\, \\text{cm} \\).\n\nSince \\( \\sin A = \\frac{CF}{AC} \\),\n\nTherefore, \\( CF = AC \\sin 60^\\circ = 20\\sqrt{3} \\, \\text{cm} \\).\n\nIn right triangle \\( \\triangle CDE \\), \\( \\angle DCE = 30^\\circ \\), \\( CD = 30 \\, \\text{cm} \\).\n\nSince \\( \\sin \\angle DCE = \\frac{DE}{CD} \\),\n\nTherefore, \\( DE = CD \\sin 30^\\circ = 15 \\, \\text{cm} \\).\n\nSince \\( DH \\perp AB \\), \\( CF \\perp AH \\), and \\( CE \\perp DH \\),\n\nTherefore, quadrilateral \\( CFHE \\) is a rectangle,\n\nHence, \\( CF = EH \\).\n\nSince \\( DH = DE + EH \\),\n\nTherefore, \\( DH = DE + EH = 20\\sqrt{3} + 15 \\approx 50 \\, \\text{cm} \\).\n\nAnswer: The distance from point \\( D \\) to the table surface is approximately \\( 50 \\, \\text{cm} \\).\n\n【Key Insight】This problem primarily examines the properties and determination of rectangles, the application of trigonometric functions of acute angles, and the key to solving the problem lies in constructing right triangles by drawing auxiliary lines." }, { "problem_id": 1864, "question": "As shown in Figure 1, a worker is using a lift to repair a street lamp. Figure 2 is a schematic diagram of the lift's operation. The interest study group plans to calculate the height of the street lamp \\( AB \\) using this schematic diagram. Through measurements and consulting with the worker, they have gathered the following information: the street lamp \\( AB \\) is perpendicular to the ground, the mechanical arm \\( DE = 2 \\) meters, \\( CD = 4 \\) meters, the distance from the top of the street lamp \\( A \\) to the work platform \\( AC = 1.5 \\) meters, the distance from the top of the carriage \\( EF \\) to the ground is 1.5 meters, \\( \\angle CDE = 75^\\circ \\), and \\( \\angle DEF = 55^\\circ \\). Based on this information, please calculate the height of the street lamp \\( AB \\). (The result should be accurate to 0.1 meters. Reference values: \\( \\sin 55^\\circ \\approx 0.82 \\), \\( \\cos 35^\\circ \\approx 0.82 \\), \\( \\sin 20^\\circ \\approx 0.34 \\), \\( \\sin 75^\\circ \\approx 0.97 \\)).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0065_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0065_2.jpg" ], "is_multi_img": true, "answer": "6 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular from point $D$ to $AB$ at point $H$, and draw a perpendicular from point $D$ to $EF$ at point $I$.\n\nIn the right triangle $\\triangle DEI$,\n\n$\\because \\sin \\angle DEI = \\frac{DI}{DE}$, and $\\angle DEF = 55^\\circ$, $DE = 2$ meters,\n\n$\\therefore DI = DE \\cdot \\sin 55^\\circ \\approx 0.82 \\times 2 = 1.64$ meters.\n\nFrom the construction, $DH \\parallel EF$,\n\n$\\therefore \\angle HDE = \\angle DEI = 55^\\circ$,\n\n$\\because \\angle CDE = 75^\\circ$,\n\n$\\therefore \\angle CDH = \\angle CDE - \\angle HDE = 75^\\circ - 55^\\circ = 20^\\circ$.\n\nIn the right triangle $\\triangle CDH$,\n\n$\\because \\sin \\angle CDH = \\frac{CH}{CD}$, and $CD = 4$ meters,\n\n$\\therefore CH = CD \\cdot \\sin 20^\\circ \\approx 0.34 \\times 4 = 1.36$ meters.\n\n$\\therefore AB = 1.5 + 1.36 + 1.64 + 1.5 = 6$ meters.\n\nAnswer: The height of the street lamp $AB$ is 6 meters.\n\n\n\n【Key Insight】This problem tests the application of solving right triangles. Understanding the problem and constructing the appropriate right triangle is the key to solving it." }, { "problem_id": 1865, "question": "A harrow is a traditional farming tool used to remove weeds between rice plants during their growth period. It comes in various sizes and styles. Figure 1 shows one type, and Figure 2 is a schematic diagram of it. Measurements have been taken: $\\mathrm{AC} = 40 \\mathrm{~cm}$, $\\angle \\mathrm{C} = 30^{\\circ}$, and $\\angle \\mathrm{BAC} = 45^{\\circ}$. To make the harrow more sturdy, a metal rod is commonly used at $\\mathrm{AB}$. How long of a metal rod is needed to make this harrow? (The result should be kept in radical form.)\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)", "input_image": [ "batch29-2024_06_14_1b144947cc87d7b4dfafg_0089_1.jpg", "batch29-2024_06_14_1b144947cc87d7b4dfafg_0089_2.jpg" ], "is_multi_img": true, "answer": "$20 \\sqrt{6}-20 \\sqrt{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Problem Analysis:** \nGiven that $\\angle \\mathrm{C}=30^{\\circ}$ and $\\angle \\mathrm{BAC}=45^{\\circ}$, we draw a perpendicular line from point $\\mathrm{B}$ to $\\mathrm{AC}$, intersecting at point $\\mathrm{D}$. This transforms the problem into solving two right triangles. Let $\\mathrm{BD}=x$. \n\nIn right triangle $\\triangle \\mathrm{ABD}$, using trigonometric relationships, we express $\\mathrm{BD}=\\mathrm{AD}=x$ and $\\mathrm{AB}=\\sqrt{2} x$. \n\nIn right triangle $\\triangle \\mathrm{CBD}$, using trigonometric relationships, we express $\\mathrm{CD}=\\sqrt{3} x$. \n\nBased on the relationship $\\mathrm{AD}+\\mathrm{CD}=\\mathrm{AC}$, we set up an equation and solve it. \n\n---\n\n**Problem Solution:** \nDraw a perpendicular line from point $\\mathrm{B}$ to $\\mathrm{AC}$, intersecting at point $\\mathrm{D}$, and let $\\mathrm{BD}=x$. \n\n1. In right triangle $\\triangle \\mathrm{ABD}$, since $\\angle \\mathrm{BAC}=45^{\\circ}$, we have $\\mathrm{BD}=\\mathrm{AD}=x$ and $\\mathrm{AB}=\\sqrt{2} x$. (1 point) \n\n2. In right triangle $\\triangle \\mathrm{CBD}$, since $\\angle \\mathrm{ACB}=30^{\\circ}$, we have $\\mathrm{CD}=\\sqrt{3} x$. (2 points) \n\n3. Since $\\mathrm{AD}+\\mathrm{CD}=\\mathrm{AC}$, we have: \n $$x + \\sqrt{3} x = 40$$ (4 points) \n\n4. Solving the equation: \n $$x = 20 \\sqrt{3} - 20$$ (6 points) \n\n5. Therefore, $\\mathrm{AB}=\\sqrt{2} x = \\sqrt{2}(20 \\sqrt{3} - 20) = 20 \\sqrt{6} - 20 \\sqrt{2}$. (7 points) \n\n**Answer:** \nTo make this tool, a metal strip of length $(20 \\sqrt{6} - 20 \\sqrt{2}) \\, \\mathrm{cm}$ is required. (8 points) \n\n**Key Point:** \nApplication of solving right triangles." }, { "problem_id": 1866, "question": "Figure 1 is a physical image of a crane, and Figure 2 is a schematic diagram of its operation. $A C$ is the telescopic boom, with its pivot point $A$ at a height $A H$ of 3.5 meters above the ground $B D$. When the length of the boom $A C$ is 8 meters and the angle $\\angle H A C$ is $118^{\\circ}$, find the height of the operating platform $C$ above the ground. (The result should be rounded to one decimal place)\n\n**Reference data:** $\\sin 28^{\\circ} \\approx 0.47$, $\\cos 28^{\\circ} \\approx 0.88, \\tan 28^{\\circ} \\approx 0.53$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1fd7f73b9b948dcb04d5g_0063_1.jpg", "batch29-2024_06_14_1fd7f73b9b948dcb04d5g_0063_2.jpg" ], "is_multi_img": true, "answer": " $7.3 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw $\\mathrm{CE} \\perp \\mathrm{BD}$ at $\\mathrm{E}$, and $\\mathrm{AF} \\perp \\mathrm{CE}$ at $\\mathrm{F}$, as shown in Figure 2. It is easy to see that quadrilateral $\\mathrm{AHEF}$ is a rectangle.\n\n$\\therefore \\mathrm{EF}=\\mathrm{AH}=3.5 \\mathrm{~m}, \\angle \\mathrm{HAF}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{CAF}=\\angle \\mathrm{CAH}-\\angle \\mathrm{HAF}=118^{\\circ}-90^{\\circ}=28^{\\circ}$,\n\nIn right triangle $\\triangle \\mathrm{ACF}$, since $\\sin \\angle \\mathrm{CAF}=\\frac{C F}{A C}$,\n\n$\\therefore \\mathrm{CF}=8 \\sin 28^{\\circ}=8 \\times 0.47=3.76$,\n\n$\\therefore \\mathrm{CE}=\\mathrm{CF}+\\mathrm{EF}=3.76+3.5 \\approx 7.3(\\mathrm{~m})$,\n\nAnswer: The height of the operating platform $\\mathrm{C}$ from the ground is $7.3 \\mathrm{~m}$.\n\n\n\nFigure 2\n\n[Insight] This problem examines the application of solving right triangles: first abstracting the practical problem into a mathematical problem (drawing a plane figure, constructing a right triangle to transform it into a problem of solving a right triangle), and then using the Pythagorean theorem and trigonometric definitions to perform calculations." }, { "problem_id": 1867, "question": "Figure (1) is a physical diagram of a crane, and Figure (2) is a schematic diagram of its operation. $A C$ is the telescopic boom, with its pivot point $\\mathrm{A}$ at a height $A H$ of $3.6 \\mathrm{~m}$ from the ground $B D$. When the length of the boom $A C$ is $12 m$ and the angle $\\angle H A C$ is $118^{\\circ}$, find the height of the operating platform $C$ from the ground. (The result should be rounded to one decimal place) [Reference data: $\\sin 28^{\\circ} \\approx 0.47$, $\\cos 28 \\approx 0.88, \\tan 28^{\\circ} \\approx 0.53$ ]\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_1fd7f73b9b948dcb04d5g_0081_1.jpg", "batch29-2024_06_14_1fd7f73b9b948dcb04d5g_0081_2.jpg" ], "is_multi_img": true, "answer": "$9.2 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw $CE \\perp BD$ at $E$, and $AF \\perp CE$ at $F$, as shown in Figure 2.\n\nIt is easy to see that quadrilateral $AHEF$ is a rectangle,\n\n$\\therefore EF = AH = 3.6 \\mathrm{~m}, \\angle HAF = 90^{\\circ}$,\n\n$\\therefore \\angle CAF = \\angle CAH - \\angle HAF = 118^{\\circ} - 90^{\\circ} = 28^{\\circ}$,\n\nIn right triangle $\\triangle ACF$, since $\\sin \\angle CAF = \\frac{CF}{AC}$,\n\n$\\therefore CF = 12 \\sin 28^{\\circ} = 12 \\times 0.47 = 5.64$,\n\n$\\therefore CE = CF + EF = 5.64 + 3.6 \\approx 9.2(m)$,\n\nAnswer: The height of the operation platform $C$ from the ground is $9.2 m$.\n\n\n\n## 【Key Insight】\n\nThis problem examines the application of solving right triangles: first, abstract the practical problem into a mathematical problem (drawing a plane figure and constructing a right triangle to transform it into a problem of solving a right triangle), then use the Pythagorean theorem and trigonometric definitions for geometric calculations." }, { "problem_id": 1868, "question": "In the cluster of peaks known as Gua Mountain northwest of Jiaocheng County, a white pagoda stands prominently on a central hilltop, making it the most striking feature among the many scenic spots on Gua Mountain (see Figure 1). A math group aimed to measure the height of the pagoda. From point $\\mathrm{A}$ (see Figure 2), they measured an elevation angle of $45^{\\circ}$ to the top of the pagoda at $C$. They then proceeded 13 meters along the slope $A B$ to reach point $B$, where they measured the distance to the base of the pagoda at $B D = 15$ meters. Given that $\\tan \\angle B A F = \\frac{5}{12}$ and $\\angle E = 90^{\\circ}$, they sought to determine the height of the white pagoda, $C D$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_215bf0351670f84c4debg_0067_1.jpg", "batch29-2024_06_14_215bf0351670f84c4debg_0067_2.jpg" ], "is_multi_img": true, "answer": "22 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: According to the problem, $AB=13$, quadrilateral $BDEF$ is a rectangle, and $\\angle CAE=45^{\\circ}$. In the right triangle $\\triangle BAF$, $AB=13$, and $\\tan \\angle BAF=\\frac{BF}{AF}=\\frac{5}{12}$.\n\nLet $BF=5k$, then $AF=12k$.\n\nThus, $AB=\\sqrt{AF^{2}+BF^{2}}=13k$.\n\nSince $AB=13$,\n\n$\\therefore k=1$.\n\nTherefore, $AF=12$, $BF=5$.\n\nGiven that $EF=BD=15$ and $DE=BF=5$,\n\n$\\therefore AE=AF+EF=12+15=27$.\n\nIn the right triangle $\\triangle ACE$, $AE=\\frac{CE}{\\tan CAE}=27$,\n\n$\\therefore CD=CE-DE=27-5=22$.\n\nAnswer: The height of the white pagoda $CD$ is 22 meters.\n\n【Highlight】This problem tests the application of solving right triangles, and mastering the relationship between the sides and angles in a right triangle is key to solving it." }, { "problem_id": 1869, "question": "Figure 1 shows a children's bicycle, with its U-shaped frame as shown in Figure 2. Given that \\( AB \\parallel CD \\), \\(\\angle ABE = 110^\\circ\\), and \\(\\angle CDE = 120^\\circ\\), find the measure of \\(\\angle BED\\).\n\n\nFigure 1\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_298ac09fc44b8d3eb665g_0060_1.jpg", "batch29-2024_06_14_298ac09fc44b8d3eb665g_0060_2.jpg" ], "is_multi_img": true, "answer": "$130^{\\circ}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a line $EF$ through point $E$ such that $EF \\parallel AB$,\n\nthen $AB \\parallel EF \\parallel CD$,\n\n$\\therefore \\angle ABE + \\angle BEF = 180^\\circ$, $\\angle CDE + \\angle DEF = 180^\\circ$,\n\n$\\because \\angle ABE = 110^\\circ$, $\\angle CDE = 120^\\circ$,\n\n$\\therefore \\angle BEF = 70^\\circ$, $\\angle DEF = 60^\\circ$,\n\n$\\therefore \\angle BED = \\angle BEF + \\angle DEF = 130^\\circ$.\n\n\n【Key Insight】This problem examines the properties of parallel lines. The key to solving it lies in understanding that when two lines are parallel, the consecutive interior angles are supplementary." }, { "problem_id": 1870, "question": "Figure 1 is a model of a certain type of stealth fighter in China. One of the highlights of this fighter is its all-moving swept-back vertical tail. Figure 2 shows the axial section of the vertical tail model, and the following data was obtained from the periphery of the vertical tail model: $BC = 8, DC = 2, \\angle D = 135^{\\circ}, \\angle C = 60^{\\circ}$, and $AB \\parallel CD$. Calculate the length of the axis $AB$ where the vertical tail is attached to the fuselage.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0002_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0002_2.jpg" ], "is_multi_img": true, "answer": "$4 \\sqrt{3}-2$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Draw $CF \\perp AB$ through point $C$, and draw $DE \\perp AB$ through point $D$, with the feet of the perpendiculars being $F$ and $E$ respectively.\n\nSince $AB \\parallel CD$,\n\n$\\angle CBF = \\angle BCD = 60^\\circ$, and $\\angle A = 180^\\circ - \\angle ADC = 45^\\circ$.\n\nIn $\\triangle BCF$, $\\sin 60^\\circ = \\frac{CF}{BC} = \\frac{CF}{8}$, and $\\cos 60^\\circ = \\frac{BF}{BC} = \\frac{BF}{8}$.\n\nThus, $CF = 4\\sqrt{3}$, and $BF = 4$.\n\nSince $CF \\perp AB$ and $DE \\perp AB$,\n\n$CF \\parallel DE$.\n\nSince $AB \\parallel CD$,\n\nQuadrilateral $DCFE$ is a parallelogram.\n\nSince $CF \\perp AB$,\n\nQuadrilateral $DCFE$ is a rectangle.\n\nTherefore, $EF = CD = 2$, and $DE = CF = 4\\sqrt{3}$.\n\nIn $\\triangle ADE$, $\\tan 45^\\circ = \\frac{DE}{AE} = \\frac{4\\sqrt{3}}{AE}$.\n\nThus, $AE = 4\\sqrt{3}$.\n\nTherefore, $AB = AE + EF - BF = 4\\sqrt{3} + 2 - 4 = 4\\sqrt{3} - 2$.\n\nHence, the length of the axis $AB$ at the attachment point of the tail fin to the fuselage is $4\\sqrt{3} - 2$.\n\n\n\n【Highlight】This problem tests the understanding of solving right-angled triangles. Mastering the definitions of trigonometric functions is essential for correct calculations, and constructing right-angled triangles is key to solving the problem." }, { "problem_id": 1871, "question": "As shown in Figure 1, to further enhance the teaching effect, a school replaced a batch of the latest projector equipment. There was a small episode during the installation of the projectors by the workers. As shown in Figure 2, $AB$ represents the height of the classroom, and $CD$ represents the width of the blackboard (points $C$ and $D$ are both on $AB$). To ensure that the students' view of the screen is not obstructed, the projector must be installed at point $E$, where $CE \\perp AB$ and $CE = 1.2$ meters. Due to the excessive weight of the projector, the horizontal axis $(CE)$ is insufficient to support it, so the workers decided to use some iron wire $(EF)$ for reinforcement, with point $F$ on the height of the classroom $AB$. During installation, the workers found that when $\\angle CEF = 30^\\circ$, the fixing coefficient at point $F$ was poor. Through practice, it was discovered that when point $F$ is moved upwards to point $G$ and $\\angle CEG = 40^\\circ$, the fixing coefficient is optimal. Please calculate how far the workers should move the fixing point $F$ upwards to reach point $G$ (i.e., find $FG$). (Given: $\\sin 40^\\circ \\approx 0.64, \\cos 40^\\circ \\approx 0.77, \\tan 40^\\circ \\approx 0.84, \\sqrt{3} \\approx 1.73$, the final result should be accurate to 0.1.)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0003_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0003_2.jpg" ], "is_multi_img": true, "answer": "$0.3 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle FCE$, $\\tan 30^{\\circ} = \\frac{CF}{CE}$,\n\nTherefore, $CF = CE \\cdot \\tan 30^{\\circ}$.\n\nIn the right triangle $\\triangle CGE$, $\\tan 40^{\\circ} = \\frac{CG}{CE}$,\n\nTherefore, $CG = CE \\cdot \\tan 40^{\\circ}$.\n\nThus, $FG = CG - CF$\n\n$= CE \\cdot \\tan 40^{\\circ} - CE \\cdot \\tan 30^{\\circ}$\n\n$= CE\\left(\\tan 40^{\\circ} - \\tan 30^{\\circ}\\right)$\n\n$\\approx 1.2 \\times \\left(0.84 - 1.73 \\times \\frac{1}{3}\\right)$\n\n$\\approx 0.3 \\mathrm{~m}$.\n\n【Insight】This problem examines the application of solving right triangles. The key to solving this problem lies in mastering the process of solving right triangles." }, { "problem_id": 1872, "question": "Figure 1 shows a showerhead, and Figure 2 is a schematic diagram of Figure 1. If the showerhead is fixed at point $\\mathrm{A}$ using a bracket, the handle length $A B = 20 \\mathrm{~cm}$, the angle between $A B$ and the wall $A D$ is $\\angle \\alpha = 30^{\\circ}$, and the angle formed by the water flow $B C$ and $A B$ is $\\angle A B C = 80^{\\circ}$. The resident now requires that when the person stands at point $E$ to shower, the water flow precisely hits point $C$ on the body, with $D E = 50 \\mathrm{~cm}$ and $C E = 150 \\mathrm{~cm}$. The question is: At what height from the ground should the installer fix the bracket? (The result should be accurate to $1 \\mathrm{~cm}$. Reference data:\n\n$\\left.\\sin 40^{\\circ} \\approx 0.64, \\cos 40^{\\circ} \\approx 0.77, \\tan 40^{\\circ} \\approx 0.84, \\sqrt{3} \\approx 1.73, \\sqrt{2} \\approx 1.41\\right)$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0010_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0010_2.jpg" ], "is_multi_img": true, "answer": "$166 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: From point $B$, draw $BF \\perp CE$, with the foot of the perpendicular at point $F$.\n\nFrom point $B$, draw $BG \\perp AD$, with the foot of the perpendicular at point $G$.\n\nIn right triangle $\\triangle ABG$, $\\angle a = 30^\\circ$.\n\nThus, $GB = \\frac{1}{2} AB = 10 \\text{ cm}$, and $AG = AB \\cos a = 10 \\sqrt{3} \\text{ cm} \\approx 17.3 \\text{ cm}$.\n\nIn right triangle $\\triangle BCF$, $\\angle FBC = 180^\\circ - 60^\\circ - 80^\\circ = 40^\\circ$.\n\n$BF = DE - BG = 40 \\text{ cm}$.\n\nTherefore, $CF = BF \\tan \\angle FBC = 40 \\tan 40^\\circ \\approx 33.6 \\text{ cm}$.\n\nHence, $AD = CE + CF - AG = 150 + 33.6 - 17.3 \\approx 166 \\text{ cm}$.\n\n\n\nFigure 2\n\nAnswer: The installation technician should fix the bracket at a height of $166 \\text{ cm}$ above the ground.\n\n[Highlight] This problem tests the application of solving right triangles, involving tangent, cosine, and the properties of a 30-degree right triangle. It is an important topic and relatively easy, with mastery of the relevant knowledge being key to solving the problem." }, { "problem_id": 1873, "question": "The \"Science and Technology Innovation Team\" at the school designed an intelligent lighting home (as shown in the design plan in Figure (2)): $M N$ is the base of the desk lamp, the bracket $A B$ forms an angle of $60^{\\circ}$ with $M N$. The angle between the brackets $A B$ and $B C$ is adjustable. After testing, it was found that when the angle between the brackets $A B$ and $B C$ is $108^{\\circ}$, a better lighting effect can be achieved. If $A B = 21 \\mathrm{~cm}$ and $B C = 28 \\mathrm{~cm}$, how far is point $C$ from the base $M N$? (The result should be accurate to $0.1 \\mathrm{~cm}$, reference data: $\\sqrt{2} \\approx 1.41$; $\\sqrt{3} \\approx 1.73$; $\\sin 48^{\\circ} \\approx 0.74$; $\\cos 48^{\\circ} \\approx 0.67$; $\\tan 48^{\\circ} \\approx 1.11$)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0019_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0019_2.jpg" ], "is_multi_img": true, "answer": "$38.9 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular from point $C$ to $MN$ at point $E$, and from point $B$ to $MN$ at point $F$. Then, draw a perpendicular from $B$ to $CE$ at point $G$, forming the rectangle $EGBF$.\n\n\n\nFigure (2)\n\nIn the right triangle $\\triangle ABF$,\n\nSince $\\angle BAF = 60^\\circ$ and $AB = 21 \\text{ cm}$,\n\n$\\angle ABF = 30^\\circ$,\n\nThus, $AF = \\frac{1}{2} AB = \\frac{21}{2} \\text{ cm}$,\n\nAnd $BF = \\sqrt{3} \\times AF = \\frac{21 \\sqrt{3}}{2} \\approx 18.165 \\text{ cm}$,\n\nTherefore, $GE = BF \\approx 18.165 \\text{ cm}$.\n\nIn the right triangle $\\triangle CGB$,\n\nSince $\\angle CBG = 108^\\circ - 60^\\circ = 48^\\circ$ and $BC = 28 \\text{ cm}$,\n\n$CG = BC \\times \\sin 48^\\circ \\approx 28 \\times 0.74 \\approx 20.72 \\text{ cm}$,\n\nThus, $CE = CG + GE = 20.72 + 18.165 \\approx 38.9 \\text{ cm}$.\n\nAnswer: The distance from point $C$ to the base $MN$ is approximately $38.9 \\text{ cm}$.\n\n【Highlight】This problem tests the application of solving right triangles. The key to solving this problem lies in mastering the methods of solving right triangles." }, { "problem_id": 1874, "question": "The world's longest cross-sea bridge, the Hong Kong-Zhuhai-Macao Bridge, connects the three cities of Hong Kong, Macao, and Zhuhai, with a total length of 55 kilometers. A section of the bridge adopts a low-tower cable-stayed bridge design, and Figure 2 is a plan view derived from Figure 1. Suppose you are standing on the bridge and measure that the angle between the cable $AB$ and the horizontal bridge deck is $30^{\\circ}$, and the angle between the cable $CD$ and the horizontal bridge deck is $60^{\\circ}$. The distance between the tops of the two cables, $BC$, is 2 meters, and the distance between the bottoms of the two cables, $AD$, is 20 meters. Calculate the length of the pillar $BH$. (The result should be accurate to 0.1 meters,\n\n$\\sqrt{3}=1.732$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0038_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0038_2.jpg" ], "is_multi_img": true, "answer": " 16.3 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of $DH$ be $x$ meters. According to the problem, $\\angle AHB = 90^\\circ$.\n\nSince $\\angle CDH = 60^\\circ$ and $\\angle AHB = 90^\\circ$,\n\nwe have $CH = DH \\cdot \\tan \\angle CDH = DH \\cdot \\tan 60^\\circ = \\sqrt{3}x$ meters.\n\nTherefore, $BH = CH + BC = (2 + \\sqrt{3}x)$ meters.\n\nGiven that $\\angle A = 30^\\circ$,\n\nwe find $AH = BH \\cdot \\tan A = BH \\cdot \\tan 60^\\circ = (2\\sqrt{3} + 3x)$ meters.\n\nSince $AH = AD + DH$,\n\nwe have $2\\sqrt{3} + 3x = x + 20$,\n\nwhich simplifies to $x = 10 - \\sqrt{3}$.\n\nThus, $BH = 2 + \\sqrt{3}(10 - \\sqrt{3}) = 10\\sqrt{3} - 1 \\approx 16.3$ meters.\n\nAnswer: The length of the column $BH$ is approximately 16.3 meters.\n\n\n\nFigure 2\n\n[Key Insight] This problem primarily tests the understanding of solving right-angled triangles. The key to solving it lies in mastering the methods of solving right-angled triangles." }, { "problem_id": 1875, "question": "The Xianren Street Scenic Area in Shiqian County is located at the summit of Zhongling Mountain in the western part of the city, at an altitude of 1300 meters, 15 kilometers from the county seat. It is now known as the \"back garden\" of Shiqian County. The attractions include: the world's longest aerial ship-shaped glass suspension bridge (Figure $a$), the world's only natural Xianren stone street, a vast original ecological stone and azalea flower sea, and more. Figure $b$ is a schematic diagram of the aerial ship-shaped glass suspension bridge. Given: $\\angle B = 37^{\\circ}, \\angle C = 30^{\\circ}, CD = 120 \\mathrm{~m}, AD \\perp BC$, find the length of the suspension bridge $BD$. (The result should be rounded to the nearest whole number, $\\sqrt{3} \\approx 1.73, \\sin 37^{\\circ} \\approx 0.60, \\cos 37^{\\circ} \\approx 0.80, \\tan 37^{\\circ} \\approx 0.75$)\n\n\n\nFigure a\n\n\n\nFigure b", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0043_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0043_2.jpg" ], "is_multi_img": true, "answer": " $92 \\mathrm{~m}$.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Since \\( AD \\perp BC \\),\n\nTherefore, \\( \\angle ADB = \\angle ADC = 90^{\\circ} \\).\n\nIn the right triangle \\( \\triangle ACD \\), \\( \\angle C = 30^{\\circ} \\) and \\( CD = 120 \\, \\text{m} \\),\n\n\\[\n\\tan 30^{\\circ} = \\frac{AD}{CD} = \\frac{AD}{120} = \\frac{\\sqrt{3}}{3},\n\\]\n\nSolving for \\( AD \\), we get \\( AD = 40\\sqrt{3} \\, \\text{m} \\).\n\nIn the right triangle \\( \\triangle ADB \\), \\( \\tan B = \\frac{AD}{BD} = \\frac{40\\sqrt{3}}{BD} \\approx 0.75 \\),\n\n\\[\n\\therefore BD \\approx \\frac{40 \\times 1.73}{0.75} = 92 \\frac{4}{15} \\approx 92 \\, \\text{m}.\n\\]\n\nThus, the length of the overhang \\( BD \\) is approximately \\( 92 \\, \\text{m} \\).\n\n[Key Insight] This problem tests the application of solving right triangles. Mastering the methods of solving right triangles and memorizing the definitions of the trigonometric functions of acute angles are crucial. Pay attention to approximate calculations as they are key to solving the problem." }, { "problem_id": 1876, "question": "The bridge type in Figure 1 is a low-tower cable-stayed bridge, and Figure 2 is an abstracted plan view derived from Figure 1. It is measured that the angle between the cable $A B$ and the horizontal bridge deck is $30^{\\circ}$, and the angle between the cable $C D$ and the horizontal bridge deck is $60^{\\circ}$. The distance between the tops of the two cables $A C$ is 4 meters, and the distance between the bottoms of the two cables $B D$ is 20 meters. Calculate the length of the pylon $A E$ (the result should be accurate to 0.1 meters, $\\sqrt{3} \\approx 1.732$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0048_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0048_2.jpg" ], "is_multi_img": true, "answer": " 15.3 m.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( CE = x \\), then \\( AE = x + 4 \\).\n\nFrom the problem statement, we have: \\( \\angle CDE = 60^\\circ \\), \\( \\angle B = 30^\\circ \\), and \\( BD = BE - DE \\).\n\nIn the right triangle \\( \\triangle CDE \\), \\( \\tan \\angle CDE = \\tan 60^\\circ = \\frac{CE}{DE} = \\sqrt{3} \\), solving for \\( DE \\) gives \\( DE = \\frac{\\sqrt{3}}{3} x \\).\n\nIn the right triangle \\( \\triangle ABE \\), \\( \\tan B = \\tan 30^\\circ = \\frac{AE}{BE} = \\frac{\\sqrt{3}}{3} \\), solving for \\( BE \\) gives \\( BE = \\sqrt{3} AE = \\sqrt{3}(x + 4) \\).\n\nSince \\( BD = BE - DE \\), we have:\n\\[\n\\sqrt{3}(x + 4) - \\frac{\\sqrt{3}}{3} x = 20\n\\]\nSolving this equation for \\( x \\) yields:\n\\[\nx = 10\\sqrt{3} - 6 \\approx 11.32 \\text{ m}\n\\]\nThus, \\( AE = x + 4 \\approx 15.3 \\text{ m} \\).\n\nAnswer: The length of the pillar \\( AE \\) is approximately 15.3 meters.\n\n【Key Insight】This problem tests the practical application of trigonometric functions. The key is to use trigonometric relationships in two right triangles separately and then set up an equation based on the given conditions to solve for the unknown." }, { "problem_id": 1877, "question": "To provide passengers with a comfortable waiting environment, a city has installed numerous bus shelters near the bus stops (as shown in Figure (1)), where the length of the support post \\( CD \\) is \\( 2.1 \\, \\text{m} \\), and the support post \\( DC \\) is perpendicular to the ground \\( DG \\). The canopy beam \\( AE \\) is \\( 1.5 \\, \\text{m} \\) long, and \\( BC \\) is the connecting post. The angle between the connecting post and the support post is \\( \\angle BCD = 150^\\circ \\), and the angle between the connecting post and the canopy beam is \\( \\angle ABC = 135^\\circ \\). It is required that one end of the beam \\( E \\) lies on the extension of the support post \\( DC \\), and at this point, the distance between the connecting point \\( B \\) and point \\( E \\) is measured to be \\( 0.35 \\, \\text{m} \\). Based on the above measurements, find the distance from point \\( A \\) to the ground \\( DG \\) (the result should be accurate to \\( 0.1 \\, \\text{m} \\), with reference data: \\( \\sqrt{2} \\approx 1.41 \\), \\( \\sin 15^\\circ \\approx 0.26 \\), \\( \\cos 15^\\circ \\approx 0.97 \\), \\( \\tan 15^\\circ \\approx 0.27 \\)).\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0056_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0056_2.jpg" ], "is_multi_img": true, "answer": " $3.0 m$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: As shown in the figure, connect $EC$. According to the problem statement:\n\n$\\angle EBC = 180^\\circ - \\angle ABC = 45^\\circ$, $\\angle ECB = 180^\\circ - \\angle BCD = 30^\\circ$.\n\nDraw a perpendicular from point $E$ to $BC$, denoted as $EP$.\n\n$\\therefore EP = BE \\times \\sin 45^\\circ = 0.35 \\times \\frac{\\sqrt{2}}{2} \\approx 0.25 \\text{ m}$.\n\n$\\therefore CE = 2EP = 0.5 \\text{ m}$;\n\nDraw a perpendicular from point $A$ to $DG$, denoted as $AF$, and from point $E$ to $AF$, denoted as $EM$,\n\n$\\angle BEC = \\angle BCD - \\angle EBC = 150^\\circ - 45^\\circ = 105^\\circ$, $\\angle ABM = \\angle BEC - \\angle MED = 105^\\circ - 90^\\circ = 15^\\circ$,\n\n$\\therefore AM = AE \\times \\sin 15^\\circ$.\n\n$\\therefore AF = AM + CE + DC$\n\n$= AE \\times \\sin 15^\\circ + 2BE \\times \\sin 45^\\circ + 2.1$\n\n$\\approx 0.39 + 0.50 + 2.1$\n\n$= 2.99$\n\n$\\approx 3.0$ (m).\n\nThus, the distance from point $A$ to the ground is $3.0 \\text{ m}$.\n\n【Insight】This problem examines the application of right-angled triangles, and using mathematical knowledge to solve practical problems is an important part of middle school mathematics. The key to solving this problem lies in correctly understanding the problem statement and establishing a mathematical model, thereby transforming the practical problem into a mathematical one." }, { "problem_id": 1878, "question": "The dump truck can automatically unload its cargo by adjusting the tilt angle of the cargo compartment through the extension and retraction of the hydraulic arm, as shown in Figure 1 and Figure 2, which are cross-sectional diagrams of a certain dump truck during unloading. The distance between the hydraulic arm base $A$ and the cargo compartment pivot $O$ is $AO = 2.4 \\mathrm{~m}$, and the distance between the extension arm pivot $B$ and the cargo compartment pivot $O$ is $BO = 2 \\mathrm{~m}$. When the angle between the cargo compartment base and the truck frame base is $\\angle AOB = 37^{\\circ}$, find the length of the hydraulic arm $AB$. (The result should be kept in radical form, with reference data $\\sin 37^{\\circ} = \\frac{3}{5}$, $\\cos 37^{\\circ} = \\frac{4}{5}$, $\\tan 37^{\\circ} = \\frac{3}{4}$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0060_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0060_2.jpg" ], "is_multi_img": true, "answer": "$\\frac{2 \\sqrt{13}}{5} \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular from point $B$ to $OA$, meeting at point $C$, as shown in the figure.\n\n\n\nIn the right triangle $\\triangle OBC$, $\\angle OCB = 90^\\circ$, $\\angle BOC = 37^\\circ$, and $BO = 2$.\n\nThus, $BC = OB \\cdot \\sin \\angle BOC = 2 \\times \\frac{3}{5} = \\frac{6}{5}$,\n\n$OC = OB \\cdot \\cos \\angle BOC = 2 \\times \\frac{4}{5} = \\frac{8}{5}$,\n\nTherefore, $AC = OA - OC = 2.4 - \\frac{8}{5} = \\frac{4}{5}$.\n\nIn the right triangle $\\triangle ABC$, since $\\angle ACB = 90^\\circ$,\n\n$AB = \\sqrt{AC^2 + BC^2} = \\sqrt{\\left(\\frac{4}{5}\\right)^2 + \\left(\\frac{6}{5}\\right)^2} = \\frac{2\\sqrt{13}}{5}$.\n\nHence, the length of the hydraulic arm $AB$ is $\\frac{2\\sqrt{13}}{5} \\text{ m}$.\n\n【Insight】This problem tests the understanding of solving right triangles and the Pythagorean theorem. The key to solving this problem lies in correctly interpreting the problem statement and establishing a mathematical model, thereby transforming the practical problem into a mathematical one." }, { "problem_id": 1879, "question": "Shared bicycles provide convenience for public travel. Figure 1 shows a physical image of the bicycle, and Figure 2 shows a schematic diagram of the bicycle. $AB$ is parallel to the ground, points $A, B, D$ are collinear, and points $D, F, G$ are collinear. The saddle $C$ can be adjusted along the direction of ray $BE$. It is known that $\\angle ABE = 70^\\circ$, $\\angle EAB = 45^\\circ$, the radius of the wheel is $30 \\, \\text{cm}$, and $BE = 40 \\, \\text{cm}$. After experiencing the ride, Xiao Ming feels that it is more comfortable when the height of the saddle $C$ from the ground is $90 \\, \\text{cm}$. Find the length of $CE$ at this time. (The result should be accurate to $1 \\, \\text{cm}$) (Reference data: $\\sin 70^\\circ \\approx 0.94$, $\\cos 70^\\circ \\approx 0.34$, $\\tan 70^\\circ \\approx 2.75$, $\\sqrt{2} \\approx 1.41$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0069_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0069_2.jpg" ], "is_multi_img": true, "answer": "$24 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: Draw a perpendicular line from point $C$ to $AB$, intersecting $AB$ at point $M$ and the ground at point $N$. According to the problem, $MN = 30 \\text{ cm}$.\n\nWhen $CN = 90 \\text{ cm}$, $CM = 60 \\text{ cm}$.\n\nTherefore, in the right triangle $\\triangle BCM$, $\\angle ABE = 70^\\circ$.\n\nThus, $\\sin \\angle ABE = \\sin 70^\\circ = \\frac{CM}{CB} = 0.94$.\n\n## Therefore, $BC \\approx 64 \\text{ cm}$,\n\nHence, $CE = BC - BE = 64 - 40 = 24 \\text{ cm}$.\n\n\n\nFigure 2\n\n[Insight] This problem tests the application of solving right triangles. Constructing a right triangle and placing the given angle within it is the key to solving the problem." }, { "problem_id": 1880, "question": "As shown in Figure 1, this is the state diagram of an automatic dump truck during unloading, and Figure 2 is its schematic diagram. The truck's cargo compartment uses a hydraulic mechanism, with the bottom support point $B$ of the support rod $AB$ of the compartment located below the horizontal line $OP$. The angle between $OB$ and the horizontal line $OP$ is $15^{\\circ}$, and when unloading, the compartment $OA$ forms a $40^{\\circ}$ angle with the horizontal line $OP$. At this time, the angle between $OB$ and the support rod $AB$ is $45^{\\circ}$. If $OA = 3$ meters, find the length of $AB$. (The result should be accurate to one decimal place, $\\sin 80^{\\circ} \\approx 0.9848, \\cos 80^{\\circ} \\approx 0.1736$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0086_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0086_2.jpg" ], "is_multi_img": true, "answer": "3.5 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from point $O$ to $AB$, meeting at point $M$.\n\nGiven the conditions, $\\angle ABO = 45^\\circ$, $\\angle AOP = 40^\\circ$, $\\angle BOP = 15^\\circ$, and $OA = 3$ meters.\n\nIn triangle $AOB$, $\\angle A = 180^\\circ - \\angle ABO - \\angle AOB = 180^\\circ - 45^\\circ - 40^\\circ - 15^\\circ = 80^\\circ$.\n\nIn right triangle $AOM$, $\\angle A = 80^\\circ$ and $OA = 3$ meters.\n\nThus, $OM = OA \\cdot \\sin 80^\\circ \\approx 3 \\times 0.9848 \\approx 2.95$ meters,\n\n$AM = OA \\cdot \\cos 80^\\circ \\approx 3 \\times 0.1736 \\approx 0.52$ meters.\n\nIn right triangle $BOM$, $\\angle ABO = 45^\\circ$,\n\nTherefore, $BM = OM = 2.95$ meters,\n\nHence, $AB = AM + BM = 2.95 + 0.52 \\approx 3.5$ meters.\n\nAnswer: The length of $AB$ is approximately 3.5 meters.\n\n\n\n[Highlight] This problem tests the practical application of solving right triangles and the properties of isosceles right triangles. The key to solving the problem lies in correctly understanding the given conditions and drawing the auxiliary lines." }, { "problem_id": 1881, "question": "As shown in Figure (1), an upper limb traction device is installed in a park. Figure (2) is a simplified schematic diagram of the device in its stationary state (with $C E$ and $D F$ on the same horizontal line). The vertical post $A B$ is perpendicular to the horizontal ground $M N$, and the arms $A C = A E$, while the pull chains $C D = E F$ are always perpendicular to the ground. According to the information provided, the arms $A C = A E = 0.33 \\mathrm{~m}$, and the angle $\\angle C A E = 130^{\\circ}$. When the exerciser performs upper limb traction, the device is pulled from the stationary state to the state shown in Figure (3), where $\\angle C A B = 52^{\\circ}$. Calculate the height by which point $E$ rises. (The result should be accurate to $0.01 \\mathrm{~m}$. Reference data: $\\sin 65^{\\circ} \\approx 0.91, \\cos 65^{\\circ} \\approx 0.42, \\tan 65^{\\circ} \\approx 2.14$, $\\sin 78^{\\circ} \\approx 0.98, \\cos 78^{\\circ} \\approx 0.21, \\tan 78^{\\circ} \\approx 4.70$)\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0087_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0087_2.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0087_3.jpg" ], "is_multi_img": true, "answer": "$0.07 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Let the intersection point of $AB$ and $CE$ be $Q$, as shown in the figure:\n\n\n\nSince $CE \\parallel MN$ and $AB \\perp MN$,\n\nTherefore, $AQ \\perp CE$,\n\nGiven that $AC = AE$,\n\nThus, $\\angle CAQ = \\frac{1}{2} \\angle CAE = \\frac{1}{2} \\times 130^\\circ = 65^\\circ$,\n\nIn the right triangle $\\triangle ACQ$, $AQ = AC \\cos 65^\\circ = 0.33 \\times 0.42 = 0.1386 \\text{ m}$,\n\nDraw $EP \\perp AB$ through point $E$, with the foot of the perpendicular at $P$,\n\n\n\nGiven that $\\angle CAB = 52^\\circ$ and $\\angle CAE = 130^\\circ$,\n\nTherefore, $\\angle EAP = \\angle CAE - \\angle CAB = 130^\\circ - 52^\\circ = 78^\\circ$,\n\nIn the right triangle $\\triangle APE$, $AP = AE \\cos 78^\\circ = 0.33 \\times 0.21 = 0.0693 \\text{ m}$,\n\nThus, $AQ - AP = 0.1386 - 0.0693 \\approx 0.07 \\text{ m}$,\n\nTherefore, the height that point $E$ rises is $0.07 \\text{ m}$.\n【Insight】This problem examines the application of solving right triangles, with the key to solving it correctly being a proper grasp of the relationships in acute angle trigonometry. The challenge lies in how to add auxiliary lines to transform the problem into one of solving right triangles." }, { "problem_id": 1882, "question": "As shown in Figure 1 and Figure 2, these are the actual image and schematic diagram of a certain type of suitcase available online. According to the product description, the following information is obtained: the sliding rod \\( DE \\), the length of the suitcase \\( BC \\), and the handle \\( AB \\) are all equal, i.e., \\( DE = BC = AB \\). Points \\( B \\) and \\( F \\) are on segment \\( AC \\), point \\( C \\) is on \\( DE \\), the support rod \\( DF = 24 \\, \\text{cm} \\), \\( CE:CD = 1:3 \\), \\( \\angle DCF = 45^\\circ \\), and \\( \\angle CDF = 30^\\circ \\).\n\nBased on the above information, find the length of \\( AC \\). (Reference data: \\( \\sqrt{2} \\approx 1.41 \\), \\( \\sqrt{3} \\approx 1.73 \\), \\( \\sqrt{6} \\approx 2.45 \\).)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0098_1.jpg", "batch29-2024_06_14_2b988f01d3dfec88a9c4g_0098_2.jpg" ], "is_multi_img": true, "answer": "$87.36 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in the figure, draw $F H \\perp D E$ at point $H$.\n\n\n\n$\\therefore \\angle F H C = \\angle F H D = 90^{\\circ}$.\n\n$\\because \\angle F D C = 30^{\\circ}$, and $D F = 24 \\mathrm{~cm}$,\n\n$\\therefore F H = D F \\cdot \\sin 30^{\\circ} = 12 \\mathrm{~cm}$, and $D H = D F \\cdot \\cos 30^{\\circ} = 12 \\sqrt{3} \\mathrm{~cm}$.\n\n$\\because \\angle F C H = 45^{\\circ}$,\n\n$\\therefore C H = F H = 12 \\mathrm{~cm}$,\n\n$\\therefore C D = C H + D H = (12 + 12 \\sqrt{3}) \\mathrm{cm}$.\n\n$\\because C E : C D = 1 : 3$,\n\n$\\therefore D E = \\frac{4}{3} C D = (16 + 16 \\sqrt{3}) \\mathrm{cm}$.\n\n$\\because A B = B C = D E$,\n\n$\\therefore A C = (32 + 32 \\sqrt{3}) \\mathrm{cm} = 87.36 \\mathrm{~cm}$.\n\n【Insight】This problem examines the practical application of solving right triangles, and constructing a right triangle is the key to solving the problem." }, { "problem_id": 1883, "question": "The students of a school's mathematics club measured the height of \"Ding Bridge\" using a tape measure and a homemade angle measuring device. As shown in Figure 2, they set up the angle measuring device $CM$ on the ground $MB$. First, they measured the elevation angle $\\angle ACD = 22^\\circ$ to the highest point $A$ of \"Ding Bridge\" at point $M$. Then, they moved forward along $MB$ for $155 \\text{ m}$ to point $N$, where they measured the elevation angle $\\angle ADE = 45^\\circ$ to point $A$ (points $M, N, B$ are collinear). The height of the angle measuring device is $1.6 \\text{ m}$. Please use the students' measurement data to find the height $AB$ of the highest point $A$ of \"Ding Bridge\" above the ground. (The result should be accurate to $0.1 \\text{ m}$. Reference data: $\\sin 22^\\circ \\approx 0.37$, $\\cos 22^\\circ \\approx 0.93$, $\\tan 22^\\circ \\approx 0.40$, $\\sqrt{2} \\approx 1.41$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_2eb57e0827ca373bc3c8g_0050_1.jpg", "batch29-2024_06_14_2eb57e0827ca373bc3c8g_0050_2.jpg" ], "is_multi_img": true, "answer": "$104.9 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, extend $CD$ to intersect $AB$ at point $F$,\n\n\n\nFigure 2\n\nFrom the problem statement, we have $CD = MN = 155$, $DF = BN$, $\\angle AFD = 90^{\\circ}$, and $CM = DN = BF = 1.6$.\n\nLet $DF = x$, then $CF = x + 155$.\n\nIn the right triangle $\\triangle ADF$, $\\angle ADF = 45^{\\circ}$,\n\n$\\therefore AF = x$.\n\nIn the right triangle $\\triangle ACF$, $\\tan 22^{\\circ} = \\frac{AF}{CF} = \\frac{x}{x + 155} \\approx 0.40$,\n\n$\\therefore x \\approx 103.3$.\n\nAfter verification, $x \\approx 103.3$ is the solution to the original equation and fits the context of the problem,\n\n$\\therefore AB = AF + BF = 103.3 + 1.6 = 104.9 \\mathrm{~m}$,\n\n$\\therefore$ the height $AB$ of the highest point $\\mathrm{A}$ of the \"Ding Bridge\" above the ground is approximately $104.9 \\mathrm{~m}$.\n\nThus, the answer is: $104.9 \\mathrm{~m}$.\n\n【Insight】This problem tests the application of solving right triangles, measuring the height of an object using angles of elevation and depression. Mastering the definitions of trigonometric functions is key to solving this problem." }, { "problem_id": 1884, "question": "Entering the Party Emblem Square in Nanniwan, one is greeted by a massive sculpture of the Party emblem, which carries forward the red spirit. As shown in Figure (1), this is a photograph of the sculpture, and Figure (2) is its measurement diagram. First, at point $C$, an inclinometer is used to measure the elevation angle to the top of the sculpture, point $A$, which is $60^{\\circ}$. Then, on the other side at point $D$, the inclinometer measures the elevation angle to the top of the sculpture, point $A$, which is $37^{\\circ}$. It is known that points $A$, $B$, $C$, $D$, $E$, $F$, and $G$ are all on the same plane, with the distance between points $E$ and $F$ being $22.9 \\mathrm{~m}$, and the height of the inclinometer $CE = DF = 1.8 \\mathrm{~m}$. Calculate the height of the sculpture $AB$. (The result should be accurate to $0.1 \\mathrm{~m}$. Reference data: $\\sin 37^{\\circ} \\approx 0.6, \\cos 37^{\\circ} \\approx 0.8, \\tan 37^{\\circ} \\approx 0.75, \\sqrt{3} \\approx 1.73$)\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch29-2024_06_14_329c9e71248ba516a80bg_0042_1.jpg", "batch29-2024_06_14_329c9e71248ba516a80bg_0042_2.jpg" ], "is_multi_img": true, "answer": "$13.8 m$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in the figure, it is known that $EF = 22.9 \\mathrm{~m}$, $\\angle AEG = 60^\\circ$, $\\angle AFG = 37^\\circ$, $AG \\perp EF$, and $CE = DF = BG = 1.8 \\mathrm{~m}$.\n\nTherefore, in the right triangle $\\triangle AGE$, we have:\n\\[\nEG = \\frac{AG}{\\tan \\angle AEG} = \\frac{AG}{\\tan 60^\\circ} = \\frac{AG}{\\sqrt{3}}\n\\]\n\nSimilarly, in the right triangle $\\triangle AGF$, we have:\n\\[\nFG = \\frac{AG}{\\tan \\angle AFG} = \\frac{AG}{\\tan 37^\\circ} \\approx \\frac{AG}{0.75}\n\\]\n\nSince $EF = EG + GF = 22.9$, we can write:\n\\[\n\\frac{AG}{\\sqrt{3}} + \\frac{AG}{0.75} = 22.9\n\\]\nSolving this equation gives:\n\\[\nAG \\approx 12.0 \\mathrm{~m}\n\\]\n\nThus, the total height $AB$ is:\n\\[\nAB = AG + BG = 12.0 + 1.8 = 13.8 \\mathrm{~m}\n\\]\n\n**Key Insight:** This problem tests the application of solving right triangles, angles of elevation and depression, and the use of trigonometric functions. Correctly interpreting the diagram is crucial for solving the problem." }, { "problem_id": 1885, "question": "To further strengthen the epidemic prevention and control measures, a school has decided to install an infrared temperature detection device. This device can quickly measure the temperature of individuals entering the detection area (as shown in Figure 1). The infrared detection point \\( O \\) can be adjusted up and down on the vertical support rod \\( OP \\) (as shown in Figure 2). Given that the maximum detection angle (\\( \\angle OBC \\)) is \\( 58.0^\\circ \\) and the minimum detection angle (\\( \\angle OAC \\)) is \\( 26.6^\\circ \\), the school requires the width of the temperature detection area \\( AB \\) to be 2.53 meters. Please calculate the installation height \\( OC \\) of the device.\n\nReference data:\n\\[\n\\sin 58.0^\\circ \\approx 0.85, \\quad \\cos 58.0^\\circ \\approx 0.53, \\quad \\tan 58.0^\\circ \\approx 1.60,\n\\]\n\\[\n\\sin 26.6^\\circ \\approx 0.45, \\quad \\cos 26.6^\\circ \\approx 0.89, \\quad \\tan 26.6^\\circ \\approx 0.50.\n\\]\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_329c9e71248ba516a80bg_0068_1.jpg", "batch29-2024_06_14_329c9e71248ba516a80bg_0068_2.jpg" ], "is_multi_img": true, "answer": " $1.84 \\mathrm{~m}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "According to the problem, we have $OC \\perp AC$, $\\angle OBC = 58.0^\\circ$, $\\angle OAC = 26.6^\\circ$, and $AB = 2.53 \\, \\text{m}$. Therefore, $AC = AB + BC = 2.53 + BC$.\n\nIn the right triangle $\\triangle OBC$,\n\n\\[\n\\tan \\angle OBC = \\frac{OC}{BC},\n\\]\n\n\\[\n\\therefore OC = BC \\cdot \\tan \\angle OBC \\approx 1.60 \\times BC.\n\\]\n\nIn the right triangle $\\triangle OAC$,\n\n\\[\nOC = AC \\cdot \\tan \\angle OAC \\approx (2.53 + BC) \\times 0.50.\n\\]\n\n\\[\n\\therefore 1.60 \\, BC = (2.53 + BC) \\times 0.50.\n\\]\n\nSolving the equation, we find:\n\n\\[\nBC = 1.15 \\, \\text{m},\n\\]\n\n\\[\n\\therefore OC = 1.60 \\, BC = 1.84 \\, \\text{m}.\n\\]\n\nAnswer: The installation height $OC$ of the equipment is approximately $1.84 \\, \\text{m}$.\n\n【Key Insight】This problem tests the application of solving right triangles. Mastering the process of solving right triangles is crucial to solving this problem." }, { "problem_id": 1886, "question": "As shown in Figure 1, the image is a physical diagram of a concrete spreader. Figure 2 is a partial schematic diagram of its operation. $AC$ is the telescopic spreading arm, with its pivot point $A$ at a height $AH$ of 3.2 meters above the ground $BD$. When the length of the spreading arm $AC$ is 8 meters and the angle $\\angle HAC$ is $118^{\\circ}$, find the height of the spreader outlet $C$ above the ground. (The result should be rounded to one decimal place; reference data: $\\sin 28^{\\circ} \\approx 0.47, \\cos 28^{\\circ} \\approx 0.88, \\tan 28^{\\circ} \\approx 0.53$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_329c9e71248ba516a80bg_0094_1.jpg", "batch29-2024_06_14_329c9e71248ba516a80bg_0094_2.jpg" ], "is_multi_img": true, "answer": " 7.0 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $CE \\perp BD$ at point $E$, and draw $AF \\perp CE$ at point $F$.\n\n\n\nSince $\\angle AHE = \\angle HEF = \\angle AFE = 90^\\circ$,\n\nquadrilateral $AHEF$ is a rectangle.\n\nTherefore, $EF = AH = 3.2$, and $\\angle HAF = 90^\\circ$. Thus, $\\angle CAF = \\angle CAH - \\angle HAF = 118^\\circ - 90^\\circ = 28^\\circ$.\n\nIn right triangle $\\triangle ACF$, $\\sin \\angle CAF = \\frac{CF}{AC}$,\n\nso $CF = 8 \\times \\sin 28^\\circ \\approx 8 \\times 0.47 = 3.76$.\n\nTherefore, $CE = CF + EF = 3.76 + 3.2 \\approx 7.0$.\n\nAnswer: The height of the fabric opening $C$ from the ground is 7.0 meters.\n\n【Key Insight】The problem mainly tests the determination and properties of rectangles, and solving triangles using trigonometric functions of acute angles. Understanding the problem and drawing the appropriate auxiliary lines are key to solving it." }, { "problem_id": 1887, "question": "In a certain location, there is a large bridge (Figure 1). A middle school mathematics interest group wants to measure the height $CD$ of the highest point $D$ of the outer arch tower above the bridge deck. They selected a measurement point $A$ on the bridge deck and measured an elevation angle of $26.6^{\\circ}$ to point $D$. Then, they moved $40$ meters along the direction of $AC$ to measurement point $B$ (i.e., $AB = 40$ meters), where they measured an elevation angle of $37^{\\circ}$ to point $D$, as shown in Figure 2. Find the height $CD$ of the highest point $D$ of the outer arch tower above the bridge deck. [Reference data: $\\sin 37^{\\circ} \\approx 0.60$, $\\cos 37^{\\circ} \\approx 0.80$, $\\tan 37^{\\circ} \\approx 0.75$, $\\sin 26.6^{\\circ} \\approx 0.45$, $\\cos 26.6^{\\circ} \\approx 0.89$, $\\tan 26.6^{\\circ} \\approx 0.50$]\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_329c9e71248ba516a80bg_0096_1.jpg", "batch29-2024_06_14_329c9e71248ba516a80bg_0096_2.jpg" ], "is_multi_img": true, "answer": " $60 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( DC = x \\) meters.\n\nIn the right triangle \\( \\triangle ADC \\), \\( \\angle A = 26.6^\\circ \\),\n\n\\[\n\\tan 26.6^\\circ \\approx 0.50 = \\frac{CD}{AC}\n\\]\n\n\\[\n\\therefore AC = 2CD\n\\]\n\nIn the right triangle \\( \\triangle BDC \\), \\( \\angle DBC = 37^\\circ \\),\n\n\\[\n\\tan 37^\\circ \\approx 0.75 = \\frac{CD}{BC}\n\\]\n\n\\[\n\\therefore BC = \\frac{4}{3}CD\n\\]\n\nGiven that \\( AC - BC = 40 \\),\n\n\\[\n2CD - \\frac{4}{3}CD = 40\n\\]\n\nSolving for \\( CD \\), we find \\( CD = 60 \\).\n\nAnswer: The height \\( CD \\) of the highest point \\( D \\) of the outer arch tower from the bridge deck is \\( 60 \\) meters.\n\n[Highlight] This problem examines the application of solving right triangle problems. The general steps involve understanding the meanings of terms and concepts in the problem, such as angles of elevation, angles of depression, slope, and gradient. Then, based on the problem's context, draw a geometric figure and establish a mathematical model to relate the practical quantities to the right triangle problem. If the figure is not a right triangle, appropriate auxiliary lines can be added to divide it into right triangles or rectangles, identify the right triangle, and solve it." }, { "problem_id": 1888, "question": "Since the launch of the \"National Fitness Campaign,\" the number of people who enjoy outdoor walking for fitness has been increasing. To facilitate public walking fitness, a local government has decided to renovate a slope road as shown in Figure 1. As depicted in Figure 2, the original slope $A B$ is 200 meters long with a gradient of $1: \\sqrt{3}$; after reducing the height $A E$ of the slope $A B$ by $A C=20$ meters, the slope $A B$ is transformed into slope $C D$, which has a gradient of $1: 4$. Find the length of slope $C D$. (The result should be kept in radical form)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_4379ad893b2f21a86d58g_0029_1.jpg", "batch29-2024_06_14_4379ad893b2f21a86d58g_0029_2.jpg" ], "is_multi_img": true, "answer": " $80 \\sqrt{17}$ m.", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "Since \\(\\angle AEB = 90^\\circ\\) and \\(AB = 200\\), with a slope of \\(1: \\sqrt{3}\\),\n\n\\[\n\\tan \\angle ABE = \\frac{1}{\\sqrt{3}} = \\frac{\\sqrt{3}}{3},\n\\]\n\n\\[\n\\angle ABE = 30^\\circ,\n\\]\n\n\\[\nAE = \\frac{1}{2} AB = 100.\n\\]\n\nGiven \\(AC = 20\\),\n\n\\[\nCE = 80.\n\\]\n\nSince \\(\\angle CED = 90^\\circ\\) and the slope of \\(CD\\) is \\(1: 4\\),\n\n\\[\n\\frac{CE}{DE} = \\frac{1}{4},\n\\]\n\nwhich implies\n\n\\[\n\\frac{80}{ED} = \\frac{1}{4},\n\\]\n\nsolving for \\(ED\\), we get\n\n\\[\nED = 320.\n\\]\n\nTherefore,\n\n\\[\nCD = \\sqrt{80^2 + 320^2} = 80 \\sqrt{17} \\text{ meters}.\n\\]\n\nAnswer: The length of slope \\(CD\\) is \\(80 \\sqrt{17}\\) meters.\n\n**Highlight:** This problem tests the application of solving right triangle problems involving slope angles. The key to solving it lies in understanding the given conditions and using trigonometric functions along with geometric visualization." }, { "problem_id": 1889, "question": "The Henan Provincial Government plans to support the construction of a batch of new steel pipe prefabricated greenhouses to promote agricultural development and accelerate rural construction. As shown in Figure 1, line segments $\\mathrm{AB}$ and $\\mathrm{BD}$ represent the height of the greenhouse wall and the span respectively, and $\\mathrm{AC}$ represents the length of the insulation board. It is known that the wall height $\\mathrm{AB}$ is 3 meters, and the angle formed by the wall and the insulation board $\\angle \\mathrm{BAC} = 150^{\\circ}$. At point $\\mathrm{D}$, the elevation angles to points $\\mathrm{A}$ and $\\mathrm{C}$ are measured to be $9^{\\circ}$ and $15.6^{\\circ}$ respectively, as shown in Figure 2. What is the length of the insulation board $\\mathrm{AC}$ in meters? (Round to the nearest 0.1 meter) (Reference data: $\\sin 9^{\\circ} \\approx 0.16, \\cos 9^{\\circ} \\approx 0.99, \\tan 9^{\\circ} \\approx 0.16, \\sin 15.6^{\\circ} \\approx 0.27, \\cos 15.6^{\\circ} \\approx 0.96, \\tan 15.6^{\\circ} \\approx 0.28$, $\\sqrt{3} \\approx 1.73$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_4379ad893b2f21a86d58g_0078_1.jpg", "batch29-2024_06_14_4379ad893b2f21a86d58g_0078_2.jpg" ], "is_multi_img": true, "answer": " 2.2 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "As shown in the figure, a perpendicular line $\\mathrm{CE} \\perp \\mathrm{BD}$ is drawn from point $\\mathrm{C}$ to point $\\mathrm{E}$, and a perpendicular line $\\mathrm{AF} \\perp \\mathrm{CE}$ is drawn from point $\\mathrm{A}$ to point $\\mathrm{F}$. Consequently, quadrilateral $\\mathrm{ABEF}$ is a rectangle.\n\n$\\therefore A B=E F, A F=B E$\n\nLet $A F=x$, then $B E=x$\n\n$\\because \\angle B A C=150^{\\circ}, \\angle B A F=90^{\\circ}$\n\n$\\therefore \\angle C A F=60^{\\circ}$\n\nThus, $A C=\\frac{A F}{\\cos \\angle C A F}=2 x, C F=A F \\cdot \\tan \\angle C A F=\\sqrt{3} x$\n\nIn the right triangle $\\triangle A B D$, $A B=E F=3, \\angle A D B=9^{\\circ}$\n\n$\\therefore B D=\\frac{3}{\\tan 9^{\\circ}}$\n\nThen, $D E=B D-B E=\\frac{3}{\\tan 9^{\\circ}}-x, C E=E F+C F=3+\\sqrt{3} x$\n\nIn the right triangle $\\triangle C D E$, $\\tan \\angle C D E=\\frac{C E}{D E}$\n\n$\\therefore \\tan 15.6^{\\circ}=\\frac{3+\\sqrt{3} x}{\\frac{3}{\\tan 9^{\\circ}}-x}$\n\nSolving gives: $x \\approx 1.12$\n\nThus, $2 x \\approx 2.24$, meaning $A C \\approx 2.2$ meters.\n\nTherefore, the length of the insulation board $\\mathrm{AC}$ is approximately 2.2 meters.\n\n\n\n【Insight】This problem examines the application of solving right triangles. The key to solving it lies in constructing right triangles by drawing auxiliary lines." }, { "problem_id": 1890, "question": "Figure 1 shows an accessible ramp, and Figure 2 is a cross-sectional schematic. It is known that the slope angle $\\angle \\mathrm{BAC}=30^{\\circ}$, the ramp $\\mathrm{AB}=4 \\mathrm{~m}$, and $\\angle \\mathrm{ACB}=90^{\\circ}$. Now, the slope surface needs to be modified so that the modified slope angle $\\angle \\mathrm{BDC}=26.5^{\\circ}$. How much should the horizontal width $\\mathrm{AC}$ be increased (the result should be accurate to 0.1)? (Reference data: $\\sqrt{3} \\approx 1.73, \\sin 26.5^{\\circ} \\approx 0.45, \\cos 26.5^{\\circ} \\approx 0.90, \\tan 26.5^{\\circ} \\approx 0.50$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_4379ad893b2f21a86d58g_0094_1.jpg", "batch29-2024_06_14_4379ad893b2f21a86d58g_0094_2.jpg" ], "is_multi_img": true, "answer": " 0.5 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle \\mathrm{ABC}$, $\\angle \\mathrm{BAC}=30^{\\circ}$, and $\\mathrm{AB}=4$,\n\n$\\therefore \\mathrm{BC}=\\mathrm{AB} \\cdot \\sin 30^{\\circ}=2, \\quad \\mathrm{AC}=\\mathrm{AB} \\cdot \\cos 30^{\\circ}=2 \\sqrt{3}$.\n\nIn the right triangle $\\triangle \\mathrm{DBC}$, $\\angle \\mathrm{BDC}=26.5^{\\circ}$, and $\\tan \\angle \\mathrm{BDC}=\\frac{\\mathrm{BC}}{\\mathrm{DC}}$,\n\n$\\therefore \\mathrm{DC}=\\frac{\\mathrm{BC}}{\\tan \\angle \\mathrm{BDC}}=\\frac{2}{\\tan 26.5^{\\circ}}$,\n\n$\\therefore \\mathrm{DA}=\\frac{2}{\\tan 26.5^{\\circ}}-2 \\sqrt{3} \\approx 4-3.46 \\approx 0.5 \\quad(\\mathrm{~m})$.\n\nAnswer: The horizontal width needs to be increased by approximately 0.5 meters.\n\n【Key Insight】This problem tests the application of solving right triangles—specifically, determining the slope angle. Mastering the concept of slope and being familiar with the definitions of the trigonometric functions of acute angles are key to solving the problem." }, { "problem_id": 1891, "question": "With the continuous development of science and technology in our country, science fiction has become a reality. As shown in Figure 1, this is a model of a certain type of stealth fighter independently developed by our country, and the all-moving swept-back vertical tail is one of the highlights of this fighter. Figure 2 shows the axial section of the vertical tail model, and the following data is obtained by measuring the periphery of the vertical tail model: $BC = 8, CD = 2, \\angle D = 135^{\\circ}, \\angle C = 60^{\\circ}$, and $AB \\parallel CD$. Calculate the area of the vertical tail model $\\mathrm{ABCD}$. (The result should be rounded to the nearest whole number, with reference data: $\\sqrt{2} \\approx 1.414$, $\\sqrt{3} \\approx 1.732$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_43bfcc8f54d76cf4fc5fg_0098_1.jpg", "batch29-2024_06_14_43bfcc8f54d76cf4fc5fg_0098_2.jpg" ], "is_multi_img": true, "answer": "24", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\nDraw a perpendicular line \\( DE \\) from point \\( D \\) to the extension of line \\( AB \\), intersecting \\( BC \\) at point \\( F \\).\n\nSince \\( AB \\parallel CD \\), it follows that \\( DE \\perp CD \\). Therefore, \\( \\angle FEB = \\angle FDC = 90^\\circ \\).\n\nIn the right triangle \\( \\triangle CDF \\), given \\( CD = 2 \\) and \\( \\angle C = 60^\\circ \\), we find:\n- \\( \\angle CFD = 30^\\circ \\),\n- \\( CF = 4 \\),\n- \\( DF = 2\\sqrt{3} \\).\n\nGiven \\( BC = 8 \\), we deduce \\( BF = 4 \\), hence \\( BF = CF \\).\n\nIn triangles \\( \\triangle FEB \\) and \\( \\triangle FDC \\), the following conditions hold:\n- \\( \\angle FEB = \\angle FDC \\),\n- \\( \\angle CFD = \\angle BFE \\),\n- \\( BF = CF \\).\n\nThus, \\( \\triangle FEB \\cong \\triangle FDC \\) by the AAS (Angle-Angle-Side) congruence criterion. Consequently:\n- \\( BE = CD = 2 \\),\n- \\( DF = EF = 2\\sqrt{3} \\).\n\nGiven \\( \\angle ADC = 135^\\circ \\) and \\( \\angle FDC = 90^\\circ \\), it follows that \\( \\angle ADE = 45^\\circ \\). Therefore:\n- \\( AE = DE = 4\\sqrt{3} \\).\n\nThe area of quadrilateral \\( ABCD \\) is equal to the area of \\( \\triangle AED \\):\n\\[\nS_{ABCD} = S_{\\triangle AED} = \\frac{1}{2} \\times AE \\times DE = \\frac{1}{2} \\times 4\\sqrt{3} \\times 4\\sqrt{3} = 24.\n\\]\n\n**Key Insight:**\nThis problem primarily examines the properties of parallel lines, triangle congruence, and the Pythagorean theorem. Constructing a right triangle and applying the Pythagorean theorem to determine the lengths of the triangle's sides is crucial for solving the problem." }, { "problem_id": 1892, "question": "A middle school mathematics practice group decided to use their knowledge to measure the height of an ancient building (as shown in Figure 1). As shown in Figure 2, two points $E$ and $G$ were selected on the ground $BC$, where two 2-meter-high markers $EF$ and $GH$ were erected. The distance between the two markers $EG$ is 23 meters, and the ancient building $AB$, the markers $EF$ and $GH$ are all in the same vertical plane. Moving back 2 meters from the marker $EF$ to point $D$ (i.e., $ED = 2$ meters), observing point $A$ from $D$, points $A, F, D$ are collinear; moving back 4 meters from the marker $GH$ to point $C$ (i.e., $CG = 4$ meters), observing point $A$ from $C$, points $A, H, C$ are also collinear. It is known that $B, E, D, G, C$ are on the same straight line, $AB \\perp BC, EF \\perp BC, GH \\perp BC$. Please help the practice group calculate the height of the ancient building $AB$ based on the above measurement data.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_465a5273307553c57debg_0008_1.jpg", "batch29-2024_06_14_465a5273307553c57debg_0008_2.jpg" ], "is_multi_img": true, "answer": " $25 \\mathrm{~m}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( AB = x \\) and \\( BE = y \\).\n\nSince \\( AB \\perp BC \\) and \\( EF \\perp BC \\),\n\nit follows that \\( AB \\parallel EF \\).\n\nGiven that \\( \\angle ADB = \\angle FDE \\),\n\nwe have \\( \\triangle ABD \\sim \\triangle FED \\).\n\nTherefore, \\( \\frac{AB}{FE} = \\frac{BD}{DE} \\), which translates to \\( \\frac{x}{2} = \\frac{y+2}{2} \\).\n\nSimilarly, \\( \\triangle ABC \\sim \\triangle HGC \\),\n\nso \\( \\frac{AB}{HG} = \\frac{BC}{GC} \\).\n\nGiven that \\( BC = BE + EG + GC = y + 23 + 4 = y + 27 \\),\n\nwe have \\( \\frac{x}{2} = \\frac{y+27}{4} \\).\n\nThus, \\( \\frac{y+2}{2} = \\frac{y+27}{4} \\), solving which gives \\( y = 23 \\).\n\nSubstituting \\( y = 23 \\) into \\( \\frac{x}{2} = \\frac{y+2}{2} \\) yields \\( x = 25 \\).\n\nTherefore, the height of the ancient building \\( AB \\) is \\( 25 \\, \\text{m} \\).\n\n【Highlight】This problem tests the understanding of solving right-angled triangles, the criteria for similarity of triangles, and their properties. The key to solving the problem lies in using the properties of similar triangles to derive \\( \\frac{y+2}{2} = \\frac{y+27}{4} \\), solving for \\( y \\), and then further determining \\( x \\)." }, { "problem_id": 1893, "question": "In life, we often see some windows equipped with awnings, as shown in Figure 1. Now, we need to install a rectangular awning for a window facing due south. As shown in Figure 2, \\( AB \\) represents the height of the window, \\( CD \\) represents the awning, and \\( AB = 1.5 \\, \\text{m} \\). The angle between the awning and the plane of the window, \\( \\angle BCD \\), is \\( 75^\\circ \\). It is known that the angle between the sunlight and the horizontal line at the lowest point of the sun at noon in winter is \\( 30^\\circ \\); at the highest point of the sun at noon in summer, the angle between the sunlight and the horizontal line is \\( 60^\\circ \\). If we want the sunlight to enter the room to the maximum extent at the lowest point of the sun at noon in winter, and for the sunlight to just not enter the room at the highest point of the sun at noon in summer, please find the width \\( CD \\) of the awning.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0023_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0023_2.jpg" ], "is_multi_img": true, "answer": "$C D=\\frac{9 \\sqrt{2}-3 \\sqrt{6}}{4}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, $EB$ represents the winter sunlight, with $\\angle EDF = 30^\\circ$,\n\n$TA$ represents the summer sunlight, with $\\angle TAN = 60^\\circ$, and the horizontal line $DF \\parallel AN$,\n\n\n\n$\\therefore \\angle TDF = \\angle TAN = 60^\\circ$,\n\n$\\therefore \\angle TDE = \\angle BDA = 30^\\circ$,\n\n$\\because AC \\perp AN$,\n\n$\\therefore \\angle BAD = 30^\\circ = \\angle BDA$,\n\n$\\therefore \\angle CBD = \\angle BAD + \\angle BDA = 60^\\circ$, and $BA = BD = 1.5$,\n\n$\\because \\angle BCD = 75^\\circ$,\n\n$\\therefore \\angle CDB = 45^\\circ$,\n\nDraw $CH \\perp BD$ at $H$,\n\n$\\therefore \\angle HCD = \\angle CDH = 45^\\circ$,\n\n$\\therefore CH = DH$,\n\nFrom $\\tan \\angle CBH = \\frac{CH}{BH} = \\tan 60^\\circ = \\sqrt{3}$,\n\n$\\therefore BH = \\frac{\\sqrt{3}}{3} CH = \\frac{\\sqrt{3}}{3} DH$,\n$\\therefore \\frac{\\sqrt{3}}{3} DH + DH = 1.5$,\n\n$\\therefore DH = \\frac{9 - 3\\sqrt{3}}{4}$,\n\nFrom $\\cos \\angle CDH = \\frac{DH}{CD} = \\cos 45^\\circ = \\frac{\\sqrt{2}}{2}$,\n\n$\\therefore CD = \\sqrt{2} DH = \\sqrt{2} \\times \\frac{9 - 3\\sqrt{3}}{4} = \\frac{9\\sqrt{2} - 3\\sqrt{6}}{4}$.\n\nAfter verification: $CD = \\frac{9\\sqrt{2} - 3\\sqrt{6}}{4}$ satisfies the problem's conditions.\n\n【Key Insight】This problem examines the Triangle Angle Sum Theorem, the properties of exterior angles of triangles, the determination of isosceles triangles, operations with quadratic roots, and the application of solving right triangles. Mastering these concepts is crucial for solving the problem." }, { "problem_id": 1894, "question": "Xiao Hong needs to purchase a suitcase online for an upcoming celebration event. Figures 1 and 2 are the actual image and schematic diagram of a certain model of suitcase she found online, respectively. She has obtained the following information: The sliding rod \\( DE \\), the length of the suitcase \\( BC \\), and the length of the handle \\( AB \\) are all equal. Points \\( B \\) and \\( F \\) lie on \\( AC \\), and point \\( C \\) lies on \\( DE \\). The support rod \\( DF = 30 \\, \\text{cm} \\), \\( CE:CD = 1:3 \\), \\( \\sin \\angle DCF = \\frac{\\sqrt{2}}{2} \\), and \\( \\cos \\angle CDF = \\frac{\\sqrt{3}}{2} \\). Calculate the length of \\( AC \\) (the result should be expressed in surd form).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0032_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0032_2.jpg" ], "is_multi_img": true, "answer": " $(40+40 \\sqrt{3}) \\mathrm{cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from point \\( F \\) to \\( CD \\), meeting at point \\( G \\).\n\nIn the right triangle \\( \\triangle DFG \\), we have:\n\\[\n\\cos \\angle CDF = \\frac{DG}{DF} = \\frac{\\sqrt{3}}{2},\n\\]\nwhich implies:\n\\[\n\\angle FDG = 30^\\circ, \\quad DG = \\frac{\\sqrt{3}}{2} DF = 15\\sqrt{3} \\text{ cm}.\n\\]\nThus:\n\\[\nFG = \\frac{1}{2} DF = \\frac{1}{2} \\times 30 = 15 \\text{ cm}.\n\\]\n\nIn the right triangle \\( \\triangle CFG \\), we have:\n\\[\n\\sin \\angle DCF = \\frac{\\sqrt{2}}{2},\n\\]\nwhich implies:\n\\[\n\\angle FCG = 45^\\circ.\n\\]\nTherefore:\n\\[\nCG = FG = 15 \\text{ cm}.\n\\]\nHence:\n\\[\nCD = CG + DG = 15 + 15\\sqrt{3} \\text{ cm}.\n\\]\n\nGiven the ratio \\( CE : CD = 1 : 3 \\), we find:\n\\[\nEC = \\frac{1}{3} CD = 5 + 5\\sqrt{3} \\text{ cm}.\n\\]\nThus:\n\\[\nDE = CD + EC = 15 + 15\\sqrt{3} + 5 + 5\\sqrt{3} = 20 + 20\\sqrt{3} \\text{ cm}.\n\\]\nFinally:\n\\[\nAC = 2DE = 40 + 40\\sqrt{3} \\text{ cm}.\n\\]\n\nAnswer: The length of \\( AC \\) is \\( (40 + 40\\sqrt{3}) \\) cm.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n【Insight】This problem tests the application of solving right triangles, primarily involving the basic concepts and operations of trigonometric functions. The key is to use mathematical knowledge to solve practical problems." }, { "problem_id": 1895, "question": "Figure (1) shows a set of intelligent passage gates at a certain station. Figure (2) is a cross-sectional view of the two circular arc wings when they are extended. The sectors $ABC$ and $DEF$ are the \"circular arc wings\" of the gate, which are axisymmetric. Both $BC$ and $EF$ are perpendicular to the ground. The central angles of the sectors $\\angle ABC = \\angle DEF = 20^{\\circ}$, and the radii $BA = ED = 60 \\text{ cm}$. Points $A$ and $D$ are on the same horizontal line, with a distance of $10 \\text{ cm}$ between them. Find the width of the gate passage, i.e., the distance between $BC$ and $EF$ (reference data: $\\sin 20^{\\circ} \\approx 0.34$, $\\cos 20^{\\circ} \\approx 0.94$, $\\tan 20^{\\circ} \\approx 0.36$).\n\n\n\nFigure (1)\n\n", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0052_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0052_2.jpg" ], "is_multi_img": true, "answer": " $50.8 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: Connect $AD$ and extend it in both directions to intersect $BC$ and $EF$ at points $M$ and $N$, respectively. Since points $A$ and $D$ lie on the same horizontal line and both $BC$ and $EF$ are perpendicular to the ground, it follows that $MN \\perp BC$ and $MN \\perp EF$. Therefore, the length of $MN$ represents the distance between $BC$ and $EF$.\n\nAdditionally, due to the symmetrical formation of the two arc wings, $AM = DN$.\n\nIn the right triangle $\\triangle ABM$, $\\angle AMB = 90^\\circ$, $\\angle ABM = 20^\\circ$, and $AB = 60 \\text{ cm}$.\n\nSince $\\sin \\angle ABM = \\frac{AM}{AB}$,\n\nwe have $AM = AB \\cdot \\sin \\angle ABM = 60 \\cdot \\sin 20^\\circ \\approx 60 \\times 0.34 = 20.4$.\n\nThus, $MN = AM + DN + AD = 2AM + AD = 20.4 \\times 2 + 10 = 50.8$.\n\nTherefore, the distance between $BC$ and $EF$ is $50.8 \\text{ cm}$.\n\n\n\n[Key Insight] This problem tests the application of solving right triangles, the properties of symmetry, and the definition of trigonometric functions. Correctly understanding the problem is crucial for solving it." }, { "problem_id": 1896, "question": "In 2021, during the inspection for the creation of a national civilized city, it was discovered that some bus shelters in our city were damaged and in need of repair. New bus shelters have now been installed (Figure 1). Figure 2 shows a side view schematic, where \\( AB \\) is a horizontal line segment, \\( CD \\perp AB \\), point \\( E \\) is the foot of the perpendicular, \\( AB = 3.56 \\, \\text{m} \\), \\( AE = 2.78 \\, \\text{m} \\), point \\( C \\) is on the arc \\( AB \\), and point \\( O \\) is the center of the circle containing arc \\( AB \\), \\( \\angle OAB = 27^\\circ \\). What is the approximate length of \\( CE \\) in meters? (Reference data: \\( \\sin 27^\\circ \\approx 0.45 \\), \\( \\cos 27^\\circ \\approx 0.89 \\), \\( \\tan 27^\\circ \\approx 0.51 \\), \\( \\sqrt{2} \\approx 1.414 \\), \\( \\sqrt{3} \\approx 1.732 \\), result accurate to 0.01)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0062_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0062_2.jpg" ], "is_multi_img": true, "answer": "0.83 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Draw $O M \\perp A B$ at point $M$, and draw $O N \\perp C D$ at point $N$. Since $C D \\perp A B$,\n\nthe quadrilateral MONE is a rectangle.\n\nThus, $A M=\\frac{1}{2} A B=1.78(\\mathrm{~m})$, and $M E=O N=A E-A M=2.78-1.78=1(\\mathrm{~m})$.\n\nIn the right triangle OAM, $\\angle O A B=27^{\\circ}$, and $\\cos \\angle O A B=\\frac{A M}{O A}$.\n\nTherefore, $O A=\\frac{A M}{\\cos \\angle O A B}=2(\\mathrm{~m})$.\n\nSince $\\sin \\angle O A B=\\frac{O M}{O A}$,\n\n$O M=O A \\cdot \\sin \\angle O A B=0.9(\\mathrm{~m})=N E$.\n\n\n\nConnect $O C$, then in the right triangle $\\triangle O C N$,\n\n$C N=\\sqrt{O C^{2}-O N^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3} \\approx 1.732(\\mathrm{~m})$.\n\nTherefore, $C E=C N-N E \\approx 1.732-0.9 \\approx 0.83(\\mathrm{~m})$.\n\nHence, the length of $C E$ is approximately 0.83 meters.\n\n【Highlight】This problem tests the application of solving right triangles. Mastering the definitions of trigonometric functions and the Pythagorean theorem is crucial for solving it." }, { "problem_id": 1897, "question": "To build a new countryside and fully achieve the goal of \"Village Brightness,\" a city has installed solar streetlights, as shown in Figure 1, in every village within its jurisdiction. Figure 2 is a schematic diagram of the streetlight. $MN$ represents a part of the vertical pole, the arm $AC$, and the bracket $BC$ intersect with the vertical pole $MN$ at points $A$ and $B$ respectively, and the arm $AC$ intersects with the bracket $BC$ at point $C$. Given that $\\angle MAC = 75^\\circ$, $\\angle ACB = 15^\\circ$, $AB = 20 \\, \\text{cm}$, and $BN = 280 \\, \\text{cm}$, find the distance from point $C$ to the ground. (The result should be accurate to $1 \\, \\text{cm}$. Reference data: $\\tan 75^\\circ \\approx 3.732$, $\\tan 15^\\circ \\approx 0.268$, $\\sqrt{3} \\approx 1.732$)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0063_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0063_2.jpg" ], "is_multi_img": true, "answer": "$317 \\mathrm{~cm}$.", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a perpendicular line from point $C$ to $MN$ at point $D$,\n\nthen $\\angle CDB = 90^{\\circ}$.\n\nGiven that $\\angle MAC = 75^{\\circ}$ and $\\angle ACB = 15^{\\circ}$,\n\nit follows that $\\angle ABC = \\angle MAC - \\angle ACB = 60^{\\circ}$.\n\nIn the right triangle $\\triangle CDA$, $\\tan \\angle MAC = \\frac{CD}{AD}$,\n\nthus $CD = AD \\cdot \\tan 75^{\\circ}$.\n\nIn the right triangle $\\triangle CDB$, $\\tan \\angle ABC = \\frac{CD}{BD}$,\n\nhence $CD = BD \\cdot \\tan 60^{\\circ}$.\n\nTherefore, $AD \\cdot \\tan 75^{\\circ} = \\tan 60^{\\circ} \\cdot (AD + AB)$,\n\nwhich gives $AD \\approx 17.32$.\n\nThus, $DN = AD + AB + BN = 17.32 + 20 + 280 \\approx 317$ cm.\n\nAnswer: The distance from point $C$ to the ground is approximately $317$ cm.\n\n\n\n[Key Insight] This problem tests the application of solving right triangles, involving the tangent function, and is an important topic. The difficulty is relatively easy, and mastering the relevant knowledge is key to solving the problem." }, { "problem_id": 1898, "question": "As shown in Figure 1, there is a yogurt cup with a straw inserted. Figure 2 is a cross-sectional view of the cup (the cross-section passes through the center of the cup's mouth and bottom). The length of the cup wall is \\( AB = 10 \\, \\text{cm} \\), and the angle between \\( AB \\) and the table \\( EF \\) is \\( \\angle ABF = 83^\\circ \\). The straw \\( NC \\) passes through point \\( A \\) and forms an angle of \\( \\angle NCF = 45^\\circ \\) with the table \\( EF \\). Find the height of the cup \\( AM \\) and the diameter of the cup's bottom \\( BC \\). (Results should be accurate to \\( 0.1 \\, \\text{cm} \\). Reference data: \\( \\sin 83^\\circ \\approx 0.993 \\), \\( \\cos 83^\\circ \\approx 0.122 \\), \\( \\tan 83^\\circ \\approx 8.144 \\))\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0081_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0081_2.jpg" ], "is_multi_img": true, "answer": " $8.7 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: In the right triangle $\\triangle ABM$, since $\\sin \\angle ABM = \\frac{AM}{AB}$, with $\\angle ABM = 83^\\circ$ and $AB = 10 \\mathrm{~cm}$,\n\nwe have $\\sin 83^\\circ = \\frac{AM}{10}$,\n\nthus $AM \\approx 0.993 \\times 10 = 9.93 \\approx 9.9 \\mathrm{~cm}$.\n\nSince $\\cos \\angle ABM = \\frac{BM}{AB}$,\n\nwe have $\\cos 83^\\circ = \\frac{BM}{10}$,\n\nthus $BM \\approx 0.122 \\times 10 = 1.22$.\n\nIn the right triangle $\\triangle ACM$, since $\\angle ACM = 45^\\circ$,\n\nwe have $CM = AM \\approx 9.93$.\n\nTherefore, $BC = CM - BM \\approx 9.93 - 1.22 \\approx 8.7$.\n\nAnswer: The height $AM$ of the cup is approximately $9.9 \\mathrm{~cm}$, and the diameter $BC$ of the base of the cup is approximately $8.7 \\mathrm{~cm}$.\n\n[Highlight] This problem examines the application of solving right triangles and the properties of isosceles right triangles, emphasizing the importance of mastering the definitions of trigonometric functions for acute angles." }, { "problem_id": 1899, "question": "Figure 1 is a physical diagram of a floor-standing microphone stand placed on a horizontal ground, and Figure 2 is its schematic diagram. The support rod \\( AB \\) is perpendicular to the ground \\( l \\), and the movable rod \\( CD \\) is fixed at point \\( E \\) on the support rod. If \\( \\angle AED = 48^\\circ \\), \\( BE = 110 \\, \\text{cm} \\), \\( DE = 80 \\, \\text{cm} \\), find the height \\( DF \\) of the movable rod's end \\( D \\) from the ground. (The result should be accurate to \\( 1 \\, \\text{cm} \\), with reference data: \\( \\sin 48^\\circ \\approx 0.74 \\), \\( \\cos 48^\\circ \\approx 0.67 \\), \\( \\tan 48^\\circ \\approx 1.11 \\))\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0089_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0089_2.jpg" ], "is_multi_img": true, "answer": "$164 \\mathrm{~cm}$", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Draw a perpendicular line from point $E$ to $DF$, denoted as $EM \\perp DF$.\n\n\n\nSince $EM \\perp DF$, $AB \\perp BF$, and $DF \\perp BF$,\n$\\angle EMF = \\angle EBF = \\angle MFB = 90^{\\circ}$,\n\nTherefore, quadrilateral $EBFM$ is a rectangle,\n\nHence, $MF = BE = 110 \\mathrm{~cm}$,\n\nGiven that $\\angle AED = 48^{\\circ}$,\n\nIt follows that $\\angle EDM = \\angle AED = 48^{\\circ}$,\n\nThus, $DM = DE \\cdot \\cos \\angle EDM \\approx 80 \\times 0.67 = 53.6 \\mathrm{~cm}$,\n\nTherefore, $DF = DM + MF \\approx 164 \\mathrm{~cm}$.\n\n【Insight】This problem examines the practical application of solving right-angled triangles. The key to solving the problem lies in constructing the appropriate auxiliary lines to form a right-angled triangle." }, { "problem_id": 1900, "question": "As shown in Figure (1) and Figure (2), these are the physical diagram and schematic diagram of a certain model of treadmill. It is known that the handles of the treadmill $A B$ are parallel to the ground $D E$, the length of the pedal $C D$ is $1.5 \\mathrm{~m}$, the angle between $C D$ and the ground $D E$ is $\\angle C D E=15^{\\circ}$, the length of the support $A C$ is $1 \\mathrm{~m}$, and $\\angle A C D=75^{\\circ}$. Find the distance between the line where the treadmill handles $A B$ are located and the ground $D E$. (The result should be accurate to $0.1 \\mathrm{~m}$.)\n\nReference data: $\\sin 15^{\\circ} \\approx 0.26, \\cos 15^{\\circ} \\approx 0.97, \\tan 15^{\\circ} \\approx 0.27, \\sqrt{3} \\approx 1.73$ )\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0092_1.jpg", "batch29-2024_06_14_46bc9ff1ccf05ea88589g_0092_2.jpg" ], "is_multi_img": true, "answer": "$1.3 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw a line $FG \\perp AB$ through point $C$, intersecting $DE$ at point $G$.\n\nSince the angle between $CD$ and the ground $DE$, $\\angle CDE$, is $15^{\\circ}$, and $\\angle ACD$ is $75^{\\circ}$,\n\nwe have $\\angle ACF = \\angle FCD - \\angle ACD = \\angle CGD + \\angle CDE - \\angle ACD = 90^{\\circ} + 15^{\\circ} - 75^{\\circ} = 30^{\\circ}$.\n\nTherefore, $\\angle CAF = 60^{\\circ}$.\n\nIn the right triangle $\\triangle ACF$, $CF = AC \\cdot \\sin \\angle CAF = \\frac{\\sqrt{3}}{2} \\mathrm{~m}$.\n\nIn the right triangle $\\triangle CDG$, $CG = CD \\cdot \\sin \\angle CDE = 1.5 \\cdot \\sin 15^{\\circ}$.\n\nThus, $FG = FC + CG = \\frac{\\sqrt{3}}{2} + 1.5 \\cdot \\sin 15^{\\circ} \\approx 1.3 \\mathrm{~m}$.\n\nTherefore, the distance between the line of the treadmill handle $AB$ and the ground $DE$ is approximately $1.3 \\mathrm{~m}$.\n\n\n\nFigure (2)\n\n[Insight] This problem mainly tests the application of solving right triangles, fully demonstrating the close connection between mathematics and real life. The key to solving the problem lies in correctly constructing the right triangle." }, { "problem_id": 1901, "question": "The image in Figure 1 shows the \"Guan Yu Sacred Statue\" at a scenic spot. Construction began in January 2007, using 500 tons of copper and 2,000 tons of iron, resulting in a grand and majestic structure. A side view of the statue is shown in Figure 2. From point $B$, the angle of elevation to the top of the statue, point $\\mathrm{A}$, is $52.8^{\\circ}$, and from point $E$, the angle of elevation to the top of the statue is $63.4^{\\circ}$. It is known that $A C \\perp B C$ at point $C$, $E G \\perp B C$ at point $G$, $E F / / B C$, $B G=30$ meters, and $F C=19$ meters. Calculate the height of the statue, $A F$. (Round the result to the nearest whole number. Reference data: $\\sin 52.8^{\\circ} \\approx 0.80, \\cos 52.8^{\\circ} \\approx 0.60, \\tan 52.8^{\\circ} \\approx 1.32, \\sin 63.4^{\\circ} \\approx 0.89$,\n\n$\\left.\\cos 63.4^{\\circ} \\approx 0.45, \\tan 63.4^{\\circ} \\approx 2.00\\right)$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_4927d4ee12e9d51216c7g_0028_1.jpg", "batch29-2024_06_14_4927d4ee12e9d51216c7g_0028_2.jpg" ], "is_multi_img": true, "answer": " 61 m", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let \\( AF = x \\) meters,\n\nSince \\( AC \\perp BC \\), \\( EG \\perp BC \\), and \\( EF \\parallel BC \\),\n\nTherefore, quadrilateral \\( FCGE \\) is a rectangle,\n\nHence, \\( EF = GC \\),\n\nIn right triangle \\( \\triangle AEF \\), \\( \\tan \\angle AEF = \\frac{AF}{EF} \\),\n\nThus, \\( EF = \\frac{AF}{\\tan \\angle AEF} = \\frac{x}{\\tan 63.4^\\circ} \\approx \\frac{x}{2} \\),\n\nTherefore, \\( GC = \\frac{x}{2} \\),\n\nSince \\( BG = 30 \\) meters,\n\nThus, \\( BC = \\left(30 + \\frac{x}{2}\\right) \\) meters,\n\nIn right triangle \\( \\triangle ACB \\), \\( \\tan \\angle ABC = \\frac{AC}{BC} \\),\n\n\\( \\tan 52.8^\\circ = \\frac{x + 19}{30 + \\frac{x}{2}} \\),\n\nTherefore, \\( \\frac{x + 19}{30 + \\frac{x}{2}} \\approx 1.32 \\),\n\nSolving for \\( x \\), we get \\( x \\approx 61 \\),\n\nAnswer: The height \\( AF \\) of the statue is approximately 61 meters.\n\n[Key Insight] This problem primarily tests trigonometric functions. The key to solving it lies in using the definition of trigonometric functions in right triangles, combining the given conditions to set up an equation for \\( x \\), and solving the equation to find the solution." }, { "problem_id": 1902, "question": "The Chengnan Bridge is the first large-scale cable-stayed bridge in Yongzhou City, spanning the Xiangjiang River to connect the Lengshuitan District and the Economic Development Zone. The main bridge structure is a low-tower cable-stayed bridge, with a main span of 380 meters. Figure 2 is a plan view derived from Figure 1. Suppose you are standing on the bridge and measure that the angle between the cable $AB$ and the horizontal bridge deck is $30^{\\circ}$, and the angle between the cable $CD$ and the horizontal bridge deck is $60^{\\circ}$. The distance between the tops of the two cables $BC$ is 2 meters, and the distance between the bottoms of the two cables $AD$ is 20 meters. Calculate the length of the pillar $BH$. (The result should be accurate to 0.1 meters, $\\sqrt{3}=1.732$).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_4ecac8fbaa93a129c835g_0003_1.jpg", "batch29-2024_06_14_4ecac8fbaa93a129c835g_0003_2.jpg" ], "is_multi_img": true, "answer": "16.3 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the length of $DH$ be $x$ meters. According to the problem, $\\angle AHB = 90^\\circ$.\n\nSince $\\angle CDH = 60^\\circ$ and $\\angle AHB = 90^\\circ$,\n\nwe have $CH = DH \\cdot \\tan \\angle CDH = DH \\cdot \\tan 60^\\circ = \\sqrt{3}x$ meters.\n\nTherefore, $BH = CH + BC = (2 + \\sqrt{3}x)$ meters.\n\nGiven that $\\angle A = 30^\\circ$,\n\nit follows that $\\angle B = 60^\\circ$.\n\nThus, $AH = BH \\cdot \\tan B = BH \\cdot \\tan 60^\\circ = (2\\sqrt{3} + 3x)$ meters.\n\nSince $AH = AD + DH$,\n\nwe have $2\\sqrt{3} + 3x = x + 20$.\n\nSolving for $x$, we get $x = 10 - \\sqrt{3}$.\n\nTherefore, $BH = 2 + \\sqrt{3}(10 - \\sqrt{3}) = 10\\sqrt{3} - 1 \\approx 16.3$ meters.\n\nAnswer: The length of the column $BH$ is approximately 16.3 meters.\n\n\n\nFigure 2\n\n[Key Insight] This problem primarily tests the understanding of solving right triangles. The key to solving it lies in mastering the definitions of trigonometric functions and the methods for solving right triangles." }, { "problem_id": 1903, "question": "As shown in Figure 1, an escalator at the exit of a train station in Wuwei is depicted, and Figure 2 is its simplified schematic diagram. The escalator $A B$ has an inclination angle of $31^{\\circ}$. From point $D$ on the ground below the escalator, the angle of elevation to the top of the escalator $A$ is $62^{\\circ}$. The distance from point $B$ to point $D$ is $6 \\mathrm{~m}$. Calculate the height of the escalator from the ground $A C$ (accurate to $0.1 \\mathrm{~m}$). **[Reference data: $\\sin 31^{\\circ} \\approx 0.52, \\cos 31^{\\circ} \\approx 0.86, \\tan 31^{\\circ} \\approx 0.60, \\sin 62^{\\circ} \\approx 0.88, \\cos 62^{\\circ} \\approx 0.47, \\tan 62^{\\circ} \\approx 1.88 \\rrbracket$\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_4ecac8fbaa93a129c835g_0020_1.jpg", "batch29-2024_06_14_4ecac8fbaa93a129c835g_0020_2.jpg" ], "is_multi_img": true, "answer": " $5.3 \\mathrm{~m}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: Since $\\angle ADC$ is an exterior angle of $\\triangle ADB$,\n\n$\\therefore \\angle ADC = \\angle ABC + \\angle DAB$,\n\nGiven that $\\angle ADC = 62^\\circ$ and $\\angle ABC = 31^\\circ$,\n\n$\\therefore \\angle DAB = 31^\\circ$,\n\n$\\therefore AD = DB = 6 \\mathrm{~m}$,\n\nIn right triangle $\\triangle ACD$, $\\angle ADC = 62^\\circ$,\n\n$\\therefore \\sin \\angle ADC = \\sin 62^\\circ = \\frac{AC}{AD} = \\frac{AC}{6}$,\n\n$\\therefore AC = 6 \\times \\sin 62^\\circ \\approx 6 \\times 0.88 = 5.28 \\approx 5.3$,\n\nAnswer: The vertical height $AC$ of the escalator is approximately $5.3 \\mathrm{~m}$.\n\n[Highlight] This problem examines the use of trigonometric functions to determine the length of a line segment, involving solving right triangles, angles of elevation and depression, and slope angles. The key to solving this problem lies in understanding the definitions of angles of elevation and depression." }, { "problem_id": 1904, "question": "The Rangcheng Bridge, spanning both banks of the Tuan River in Dengzhou City, has been completed and opened to traffic, transforming the transportation landscape across the Tuan River and significantly enhancing the connectivity and convenience between the north and south banks of the Tuan River in Dengzhou City, as well as between the industrial agglomeration zone and the central urban area. The design of the Rangcheng Bridge tower draws inspiration from the character \"邓\" in oracle bone script and early bronze script, symbolizing hands holding a pottery bean filled with grain, with the top rising and the bottom covering, representing the acceptance of \"sweet dew\" from above and \"earthly energy\" from below (see Figure 1). To measure the height from the top of the bridge tower to the water surface (see Figure 2), Xiao Ming controlled a drone to measure the elevation angle of the top point A as 45° from point C, which is 1 meter above the water surface. He then moved forward 130 meters (EF = 130 m) to point D, which is 2 meters above the water surface, where he measured the elevation angle of the top point A as 53°. Calculate the height from the top of the bridge tower to the water surface AB (accurate to 1 meter, sin 53° ≈ 4/5, cos 53° ≈ 3/5, tan 53° ≈ 4/3).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_4ecac8fbaa93a129c835g_0079_1.jpg", "batch29-2024_06_14_4ecac8fbaa93a129c835g_0079_2.jpg" ], "is_multi_img": true, "answer": " $76 m$", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Draw $CG \\perp AB$ at point $G$ from point $C$, and draw $DH \\perp AB$ at point $H$ from point $D$.\n\n\n\nGiven that $CE \\perp EF$, $AB \\perp EF$, and $DF \\perp EF$,\n\n$\\therefore$ Quadrilaterals $BGCE$ and $BFDH$ are both rectangles,\n\n$\\therefore CG = BE$, $HD = BF$.\n\nLet $AB = x$ meters, according to the problem, $AG = (x - 1)$ meters, $AH = (x - 2)$ meters.\n\nIn right triangle $ACG$,\n\n$\\because \\angle ACG = 45^\\circ$,\n\n$\\therefore CG = AG = x - 1$.\n\nIn right triangle $AHD$,\n\n$\\because \\tan \\angle ADH = \\tan 53^\\circ = \\frac{AH}{HD} = \\frac{x - 2}{HD}$,\n\n$\\therefore HD \\approx \\frac{3}{4}(x - 2)$.\n\n$\\because CG + HD = BE + BF = EF$,\n\n$\\therefore x - 1 + \\frac{3}{4}(x - 2) = 130$.\n\nSolving gives $x \\approx 76$.\n\nAnswer: The height from the top of the tower to the water surface $AB$ is approximately $76$ meters.\n\n【Highlight】This problem tests the application of solving right triangle problems involving angles of elevation and depression. Understanding the concept of angles of elevation and being familiar with the definitions of trigonometric functions for acute angles are key to solving the problem." }, { "problem_id": 1905, "question": "The Baoben Temple Pagoda is located in Wugong Town, Shaanxi Province, east of the Qishui River and west of Xiangshan Mountain. The pagoda is built adjacent to the temple, creating a picturesque scene, making it one of the famous pagodas in Shaanxi. Xiao Hong wants to use her knowledge to measure the height of the Baoben Temple Pagoda (Figure 1). Her measurement plan is as follows: As shown in Figure 2, Xiao Hong adjusts the position of the theodolite and measures the elevation angle of the top of the pagoda, point A, at point C around the pagoda as 53° (the height of the theodolite is ignored). Then, in the sunlight, Xiao Hong walks forward 30 meters along the direction of BC (i.e., CD = 30 meters) to reach point D, the end of the pagoda's shadow in the sunlight. At point D, she erects a 2-meter-long pole DE, and the shadow length DF of Pole DE in the sunlight is 3 meters. It is known that points B, C, D, and F are on the same straight line, AB ⊥ BF, and DE ⊥ BF. Please use the above data to find the height of the pagoda AB. (Reference data: sin 53° ≈ 4/5, cos 53° ≈ 3/5, tan 53° ≈ 4/3)\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_51d595e10fe1a51588ebg_0012_1.jpg", "batch29-2024_06_14_51d595e10fe1a51588ebg_0012_2.jpg" ], "is_multi_img": true, "answer": " 40 m", "answer_type": "single-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: Let the height of the tower \\( AB \\) be \\( x \\) meters.\n\nSince \\( AB \\perp BC \\) and \\( \\angle ACB = 53^\\circ \\),\n\n\\[ BC = \\frac{x}{\\tan 53^\\circ} = \\frac{3x}{4}. \\]\n\nSince \\( AB \\perp BF \\) and \\( DE \\perp BF \\),\n\n\\[ \\angle ABD = \\angle EDF = 90^\\circ. \\]\n\nBecause the sun's rays are parallel,\n\n\\[ \\angle EFD = \\angle ADB, \\]\n\nthus, triangles \\( EDF \\) and \\( ABD \\) are similar.\n\nTherefore,\n\n\\[ \\frac{DE}{DF} = \\frac{AB}{BD}, \\]\n\n\\[ \\frac{2}{3} = \\frac{x}{\\frac{3x}{4} + 30}. \\]\n\nSolving the equation yields:\n\n\\[ x = 40. \\]\n\nAnswer: The height of the tower \\( AB \\) is 40 meters.\n\n[Key Insight] This problem tests the understanding of trigonometric functions of acute angles and the properties of similar triangles. The key to solving it lies in proving the similarity of the triangles to establish proportional relationships between corresponding sides and using algebraic methods to find the solution." }, { "problem_id": 1906, "question": "Place two squares of different sizes as shown in Figure 1 and Figure 2. If the area of the shaded part in Figure 1 is 36, and the area of the shaded part in Figure 2 is 27, then the side length of the larger square is $\\qquad$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch29-2024_06_14_56706cac8b3f2dc0544dg_0003_1.jpg", "batch29-2024_06_14_56706cac8b3f2dc0544dg_0003_2.jpg" ], "is_multi_img": true, "answer": "9", "answer_type": "single-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: Let the side length of the small square be \\( a \\), and the side length of the large square be \\( b \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nThen, we have the equations:\n\\[\n\\frac{1}{2} a \\times (b - a) + \\frac{1}{2} a b = 36, \\quad \\frac{1}{2} a b = 27\n\\]\nSimplifying, we get:\n\\[\na b - \\frac{1}{2} a^{2} = 36, \\quad a b = 54\n\\]\n\\[\n\\therefore \\frac{1}{2} a^{2} = 18\n\\]\n\\[\n\\therefore a = 6\n\\]\n\\[\n\\therefore b = 9\n\\]\nThus, the answer is: 9.\n\n**Key Insight:** This problem tests the understanding of polynomial multiplication and the area of geometric figures. Setting up the equations based on the given conditions is crucial for solving the problem." }, { "problem_id": 1907, "question": "Definition: A point on a side of a triangle divides the side into two line segments, and the product of these two line segments is equal to the square of the length of the line segment connecting this point and the vertex opposite to the side, then this point is called the 'singular point' of the side of the triangle. As shown in Figure (1), in $\\triangle A B C$, point $D$ is a point on the side $BC$, connect $AD$, if $AD^{2}=BD \\cdotCD$, then point $D$ is called the 'singular point' on the side $BC$ in $\\triangle A B C$. Problem solution: As shown in Figure (2), in $\\triangle A B C$, $BC=11, \\tan B=\\frac{3}{5}, \\tan C=\\frac{1}{2}$, point $D$ is the 'singular point' on the side $BC$, find the line segment $BD$ Length.\n\n\n\n(1)\n\n\n\n(2)", "input_image": [ "batch29-2024_06_14_a2209ef2c9ad0d2b4985g_0029_1.jpg", "batch29-2024_06_14_a2209ef2c9ad0d2b4985g_0029_2.jpg" ], "is_multi_img": true, "answer": "2 or $\\frac{17}{2}$", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: As shown in the figure, draw $AH \\perp BC$ at point $H$.\n\nGiven that $\\tan B = \\frac{AH}{BH} = \\frac{3}{5}$ and $\\tan C = \\frac{AH}{CH} = \\frac{1}{2}$,\n\nwe can set $AH = 3x$, then $BH = 5x$ and $CH = 6x$.\n\nThus, $BC = 11x = 11$,\n\nso $x = 1$.\n\nTherefore, $BH = 5$, $CH = 6$, and $AH = 3$.\n\nLet $DH = a$.\n\nAs shown in the figure, when point $D$ is on the left side of point $H$,\n\n\n\nsince point $D$ is a \"singular point\" on side $BC$,\n\nwe have $AD^2 = BD \\cdot CD$.\n\nIn right triangle $\\triangle ADH$, $AD^2 = AH^2 + DH^2$,\n\nso $AH^2 + DH^2 = BD \\cdot CD$,\n\nwhich gives $a^2 + 3^2 = (5 - a)(6 + a)$,\n\ni.e., $2a^2 + a - 21 = 0$.\n\nSolving this equation, we get $a = 3$ or $a = -\\frac{7}{2}$ (discarded).\n\nThus, $BD = 5 - a = 2$.\n\nAs shown in the figure, when point $D$ is on the right side of point $H$,\n\n\n\nsince $AD^2 = BD \\cdot CD$,\n\nwe have $a^2 + 3^2 = (5 + a)(6 - a)$,\n\ni.e., $2a^2 - a - 21 = 0$.\n\nSolving this equation, we get $a = \\frac{7}{2}$ or $a = -3$ (discarded).\n\nThus, $BD = 5 + a = \\frac{17}{2}$.\n\nTherefore, the length of $BD$ is either $2$ or $\\frac{17}{2}$.\n\n【Key Insight】This problem mainly tests the Pythagorean theorem, the application of quadratic equations, and solving right triangles. Understanding the new definition and using the method of classification and discussion are key to solving the problem." }, { "problem_id": 1908, "question": "As shown in the figure, a cyclist rides along the straight road \\(B C\\) due east of the Suya Lake. At \\(B\\), he measures that a building \\(A\\) on the island in the lake is \\(45^{\\circ}\\) in the north-east direction. After driving \\(12 \\mathrm{~min}\\), he arrives at \\(C\\) and measures that the building \\(A\\) is \\(60^{\\circ}\\) in the north-west direction. If the speed of this cyclist is \\(60 \\mathrm{~km} / \\mathrm{h}\\), calculate the distance from the building \\(A\\) to the road \\(B C\\). (The result retains the square root) [Rationalize the denominator: \\(\\frac{1}{\\sqrt{3}+1}=\\frac{\\sqrt{3}-1}{(\\sqrt{3}+1)(\\sqrt{3}-1)}=\\frac{\\sqrt{3}-1}{2}\\) ]\n\n\n\n Figure 1\n\nFigure 2", "input_image": [ "batch29-2024_06_14_b7b499f508104dcd2ceag_0083_1.jpg", "batch29-2024_06_14_b7b499f508104dcd2ceag_0083_2.jpg" ], "is_multi_img": true, "answer": " \\((6 \\sqrt{3}-6) \\mathrm{km}\\).", "answer_type": "single-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: Draw a perpendicular from point \\(A\\) to \\(BC\\), with the foot of the perpendicular at point \\(D\\). As shown in the figure,\n\n\n\nSouth\n\n\n\nAccording to the problem, \\(\\angle ABC = 45^\\circ\\) and \\(\\angle ACB = 30^\\circ\\), so \\(\\angle BAD = 45^\\circ\\).\n\n\\(\\therefore AD = BD\\),\n\nIn the right triangle \\(\\triangle ACD\\), since \\(\\angle ACB = 30^\\circ\\),\n\n\\(\\therefore AC = 2AD\\), and \\(CD = \\sqrt{3}AD\\),\n\nLet \\(AD = x \\mathrm{~km}\\), then \\(BD = AD = x\\), \\(AC = 2x\\), and \\(CD = \\sqrt{3}x\\),\n\nAlso, since \\(BC = 60 \\times \\frac{12}{60} = 12(\\mathrm{~km})\\),\n\\(\\therefore x + \\sqrt{3}x = 12\\), solving gives: \\(x = \\frac{12}{\\sqrt{3} + 1} = 6\\sqrt{3} - 6\\).\n\nTherefore, the distance from building \\(A\\) to the road \\(BC\\) is \\((6\\sqrt{3} - 6) \\mathrm{km}\\).\n\n[Highlight] This problem tests the application of solving right triangle problems involving directional angles: In solving problems related to directional angles, it is generally necessary to clarify the relationships between the angles in the figure based on the given information. Sometimes, the given directional angles may not be in a right triangle, and knowledge such as alternate interior angles of parallel lines or the complement of an angle may be needed to convert them into the required angles." }, { "problem_id": 1909, "question": "The 'two-section railing' at the exit of an underground garage is shown in Figure 1. Point $\\mathrm{A}$ is the fulcrum of the railing rotation, and point $\\mathrm{E}$ is the connection point of the two sections of the railing. When a vehicle passes by, the position of the railing $\\mathrm{AEF}$ after it is raised is shown in Figure 2, and its schematic diagram is shown in Figure 3, where $\\mathrm{AB} \\perp \\mathrm{BC}, \\mathrm{EF} / / \\mathrm{BC}, \\angle \\mathrm{EAB}=143^{\\circ}, \\mathrm{AB}=\\mathrm{AE}=1.2$ meters. Calculate the height of the railing $\\mathrm{EF}$ section from the ground when the vehicle passes by (that is, the distance from any point on the straight line $\\mathrm{EF}$ to the straight line $\\mathrm{BC}$). (The result is accurate to 0.1 meter, railing width is negligible Reference data: $\\sin 37^{\\circ} \\approx 0.60, \\cos 37^{\\circ} \\approx 0.80, \\tan 37^{\\circ} \\approx 0.75$. )\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch29-2024_06_14_bf26b8bd00513e46386cg_0056_1.jpg", "batch29-2024_06_14_bf26b8bd00513e46386cg_0056_2.jpg", "batch29-2024_06_14_bf26b8bd00513e46386cg_0056_3.jpg" ], "is_multi_img": true, "answer": "2.2 m", "answer_type": "single-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Problem Analysis: Extend the line $\\mathrm{BA}$ to intersect the extension of $\\mathrm{FE}$ at point $\\mathrm{D}$, forming the right triangle $\\triangle \\mathrm{ADE}$. By applying the cosine function, calculate the length of $\\mathrm{AD}$ and add it to $\\mathrm{AB}$ to obtain the desired result." }, { "problem_id": 1910, "question": "As shown in the figure, in the rectangle $\\mathrm{OABC}$, $\\mathrm{O}$ is the origin of the plane rectangular coordinate system, and the coordinates of points $\\mathrm{A}$ and $\\mathrm{C}$ are $(3, 0)$ and $(0, 5)$ respectively. Point $\\mathrm{B}$ is in the first quadrant.\n\n\n\n(Figure 1)\n\n\n\n(Figure 2)\n\n\n\n(Figure 3)\n\n(1) In Figure 1, write down the coordinates of point $\\mathrm{B}$;\n\n(2) In Figure 2, if a line $\\mathrm{CD}$ passing through point $\\mathrm{C}$ intersects $\\mathrm{AB}$ at point $\\mathrm{D}$, and divides the perimeter of rectangle $\\mathrm{OABC}$ into two parts in the ratio of $3: 1$, find the coordinates of point $\\mathrm{D}$;\n\n(3) In Figure 3, if the line segment $\\mathrm{CD}$ in (2) is translated downward by 2 units to form $\\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}$, calculate the area of quadrilateral $\\mathrm{OAD}^{\\prime} \\mathrm{C}^{\\prime}$.", "input_image": [ "batch16-2024_06_15_362fd3a118810e3be438g_0029_1.jpg", "batch16-2024_06_15_362fd3a118810e3be438g_0029_2.jpg", "batch16-2024_06_15_362fd3a118810e3be438g_0029_3.jpg" ], "is_multi_img": true, "answer": "(1) B (3, 5); (2) (3, 4);（3）7.5.", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: (1) According to the problem, point B has coordinates (3, 5).\n\n(2) From the problem statement, we know that $\\mathrm{OC} = \\mathrm{AB} = 3$.\n\nTherefore, $\\frac{DB + CB}{CO + OA + AD} = \\frac{1}{3}$.\n\nThis implies $(5 - \\mathrm{AD} + 3) : (5 + 3 + \\mathrm{AD}) = 1 : 3$, leading to $8 + \\mathrm{AD} = 3(8 - \\mathrm{AD})$.\n\nSolving this, we find $\\mathrm{AD} = 4$.\n\nThus, the coordinates of point D are $(3, 4)$.\n\n(3) According to the problem: $\\mathrm{C}^{\\prime}(0, 3)$, $\\mathrm{D}^{\\prime}(3, 2)$.\n\nFrom the diagram, we have: $\\mathrm{OA} = 3$, $\\mathrm{AD}^{\\prime} = 2$, $\\mathrm{OC}^{\\prime} = 3$.\n\nTherefore, the area of quadrilateral $OAD^{\\prime}C^{\\prime}$ is calculated as:\n\\[\n\\mathrm{S} = \\frac{1}{2} \\left(\\mathrm{OC}^{\\prime} + \\mathrm{AD}^{\\prime}\\right) \\cdot \\mathrm{OA} = \\frac{1}{2} \\times (3 + 2) \\times 3 = 7.5.\n\\]" }, { "problem_id": 1911, "question": "In the plane rectangular coordinate system, the point $A(m, n)$ satisfies $n=\\sqrt{m-4}-\\sqrt{4-m}+\\sqrt{m}$.\n\n(1) Directly write down the coordinates of point $\\mathrm{A}$;\n\n(2) As shown in Figure 1, the segment $OA$ is translated downward along the $y$-axis by $a$ units to obtain segment $BC$ (with point $O$ corresponding to point $B$). Point $C$ is drawn such that $CD \\perp y$-axis at point $D$. If $4OD=3BD$, find the value of $a$;\n\n\n\nFigure 1\n\n(3) As shown in Figure 2, point $E(0,5)$ lies on the $y$-axis, and $AE$ is connected. The segment $OA$ is translated upward along the $y$-axis by 3 units to obtain segment $FG$ (with point $O$ corresponding to point $F$), and $FG$ intersects $AE$ at point $P$. On the $y$-axis, does there exist a point $Q$ such that $S_{\\triangle APQ}=6$? If so, request the coordinates of point $Q$; if not, explain why.\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_3ce168a9eb8541c46c9dg_0035_1.jpg", "batch16-2024_06_15_3ce168a9eb8541c46c9dg_0035_2.jpg" ], "is_multi_img": true, "answer": "(1) $(4,2)$ ；（2） $a=\\frac{1}{2}$ or $\\frac{7}{2}$ ；（3）There exists a point $Q$ with coordinates $(0,0)$ or $(0,10)$", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: (1) Since point $A(m, n)$ satisfies $n=\\sqrt{m-4}-\\sqrt{4-m}+\\sqrt{m}$.\n\nTherefore, $m-4$ and $4-m$ must be non-negative,\n\nHence, $m=4$,\n\nThus, $n=\\sqrt{4}=2$,\n\nTherefore, $A(4,2)$.\n\n(2) Since the line segment $O A$ is translated downward along the $y$-axis by $a$ units to obtain the line segment $B C$, and $A(4,2)$,\n\nTherefore, $B(0,-a)$, $C(4,2-a)$, $D(0,2-a)$,\n\nThus, $O D=|2-a|$, $B D=2$,\n\n(1) When point $D$ is above the $x$-axis,\n\n\n\nFigure 1\n\nSince $4 O D=3 B D$,\n\nTherefore, $4(2-a)=3 \\times 2$,\n\nSolving gives $a=\\frac{1}{2}$;\n\n(2) When point $D$ is below the $x$-axis,\n\n\n\nFigure 2\n\nSince $4 O D=3 B D$,\n\nTherefore, $4(a-2)=3 \\times 2$,\n\nSolving gives $a=\\frac{7}{2}$.\n\nCombining the above results, $a=\\frac{1}{2}$ or $\\frac{7}{2}$;\n\n(3) Connect $A G$, draw a line parallel to the $x$-axis through point $P$, intersecting $A G$ at point $M$, and intersecting the $y$-axis at point $N$,\n\n\n\nFigure 3\n\nGiven that $A G=3$, $E F=2$, $M N=4$, $E O=5$,\n\nTherefore, $S_{\\triangle E P F}=\\frac{1}{2} E F \\cdot P N=P N$, $S_{\\triangle A P G}=\\frac{1}{2} A G \\cdot P M=\\frac{3}{2}(4-P N)$,\n\nThus, $S_{\\text{quadrilateral } A G F O}=3 \\times 4=12$, $S_{\\triangle A E O}=\\frac{1}{2} \\times 5 \\times 4=10$,\n\nTherefore, $S_{\\text{quadrilateral } A G F O}-S_{\\triangle A E O}=S_{\\triangle A P G}-S_{\\triangle P E F}=2$,\nThat is, $\\frac{3}{2}(4-P N)-P N=2$,\n\nSolving gives $P N=\\frac{8}{5}$,\n\nLet $Q(0, n)$, $E Q=|5-n|$,\n\nTherefore, $S_{\\triangle A P Q}=S_{\\triangle A E Q}-S_{\\triangle A E Q}=\\frac{1}{2} E Q \\cdot M N-\\frac{1}{2} E Q \\cdot P N=6$,\n\nThat is, $\\frac{1}{2} E Q \\cdot(M N-P N)=\\frac{1}{2} \\times \\frac{12}{5} \\times E Q=6$,\n\nSolving gives $E Q=5$,\n\nThat is, $|5-n|=5$,\n\nSolving gives $n=0$ or $n=10$,\n\nCombining the above results, the coordinates of point $Q$ are $(0,0)$ or $(0,10)$.\n\n【Insight】This problem belongs to a comprehensive geometric transformation question, examining the properties of square roots, translation transformations, and the area of quadrilaterals. The key to solving the problem lies in understanding the question and using parameters to construct equations, which is a common type of question in middle school exams." }, { "problem_id": 1912, "question": "As shown in the figure, in the plane rectangular coordinate system, the parabola \\( y = ax^2 + bx + c \\) (where \\( a < 0 \\)) intersects the \\( x \\)-axis at points \\( A(-2, 0) \\) and \\( B(4, 0) \\), and intersects the \\( y \\)-axis at point \\( C \\), with \\( OC = 2OA \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the analytical expression of the parabola;\n\n(2) The line \\( y = kx + 1 \\) (where \\( k > 0 \\)) intersects the \\( y \\)-axis at point \\( D \\), intersects the parabola in the first quadrant at point \\( P \\), and intersects the line \\( BC \\) at point \\( M \\). Let \\( m = \\frac{S_{\\triangle CPM}}{S_{\\triangle CDM}} \\). Find the maximum value of \\( m \\) and the coordinates of point \\( P \\) at this time;\n\n(3) Under the conditions of (2), when \\( m \\) takes its maximum value, does there exist a point \\( Q \\) on the \\( x \\)-axis and a point \\( N \\) in the coordinate plane such that the quadrilateral formed by points \\( P, D, Q, N \\) is a rectangle? If such points exist, directly write down the coordinates of all such points \\( Q \\) and \\( N \\); if not, explain the reason.", "input_image": [ "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0013_1.jpg", "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0013_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=-\\frac{1}{2} x^{2}+x+4$\n\n(2) $m$ has a maximum value of $\\frac{2}{3}$, and the coordinates of point $P$ are $(2,4)$\n\n(3) It exists, $N_{1}\\left(\\frac{7}{2}, 3\\right), Q_{1}\\left(\\frac{3}{2}, 0\\right) ; \\quad N_{2}(6,-3), Q_{2}(8,0)$\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "(1) $\\because A(-2,0)$,\n\n$$\n\\begin{aligned}\n& \\therefore O A=2 \\\\\n& \\because O C=2 O A \\\\\n& \\therefore O C=4 \\\\\n& \\therefore C(0,4)\n\\end{aligned}\n$$\n\n$\\because$ The parabola $y=a x^{2}+b x+c$ passes through points $A(-2,0), B(4,0), C(0,4)$,\n$\\therefore\\left\\{\\begin{array}{l}4 a-2 b+c=0 \\\\ 16 a+4 b+c=0 \\\\ c=4\\end{array}\\right.$, solving gives: $\\left\\{\\begin{array}{l}a=-\\frac{1}{2} \\\\ b=1 \\\\ c=4\\end{array}, \\therefore\\right.$ the equation of the parabola is $y=-\\frac{1}{2} x^{2}+x+4$;\n\n(2) As shown in Figure 1, draw a line $P E$ parallel to the $y$-axis intersecting line $B C$ at $E$, connect $C P$,\n\nLet the equation of line $B C$ be $y=k x+d$,\n\n$\\because B(4,0), \\quad C(0,4)$,\n\n$\\therefore\\left\\{\\begin{array}{l}4 k+d=0 \\\\ d=4\\end{array}\\right.$, solving gives: $\\left\\{\\begin{array}{l}k=-1 \\\\ d=4\\end{array}\\right.$,\n\n$\\therefore$ The equation of line $B C$ is $y=-x+4$,\n\nLet $P\\left(t,-\\frac{1}{2} t^{2}+t+4\\right)$, then $E(t,-t+4)$,\n\n$\\therefore P E=-\\frac{1}{2} t^{2}+t+4-(-t+4)=-\\frac{1}{2} t^{2}+2 t$,\n\n$\\because$ The line $y=k x+1(k>0)$ intersects the $y$-axis at point $D$,\n\n$\\therefore D(0,1)$,\n\n$\\therefore C D=4-1=3$,\n\n$\\because P E$ is parallel to the $y$-axis, i.e., $P E / / C D$,\n\n$\\therefore \\triangle E M P \\sim \\triangle C M D$,\n\n$\\therefore \\frac{P M}{D M}=\\frac{P E}{C D}=\\frac{-\\frac{1}{2} t^{2}+2 t}{3}=-\\frac{1}{6} t^{2}+\\frac{2}{3} t$,\n\n$\\because m=\\frac{S_{\\triangle C P M}}{S_{\\triangle C D M}}=\\frac{P M}{D M}$,\n\n$\\therefore m=-\\frac{1}{6} t^{2}+\\frac{2}{3} t=-\\frac{1}{6}(t-2)^{2}+\\frac{2}{3}$,\n\n$\\because-\\frac{1}{6}<0$,\n\n$\\therefore$ When $t=2$, $m$ reaches its maximum value $\\frac{2}{3}$, at which point the coordinates of $P$ are $(2,4)$;\n\n\n\nFigure 1\n\n(3) There exist points $Q$ and $N$ such that the quadrilateral formed by points $P, D, Q, N$ is a rectangle.\n\n(1) When $D P$ is a side of the rectangle, there are two scenarios,\n$a$、As shown in Figure 2-1, when quadrilateral $D Q N P$ is a rectangle,\n\n\n\nFigure 2-1\n\nFrom (2), we know $P(2,4)$, substituting into $y=k x+1$ gives $k=\\frac{3}{2}$,\n\n$\\therefore$ The equation of line $D P$ is $y=\\frac{3}{2} x+1$, giving $D(0,1), E\\left(-\\frac{2}{3}, 0\\right)$,\n\nFrom $D O E \\sim Q O D$ we get $\\frac{O D}{O Q}=\\frac{O E}{O D}$,\n\n$\\therefore \\mathrm{OD}^{2}=O E \\cdot O Q$,\n\n$\\therefore 1=\\frac{2}{3} \\cdot O Q$,\n\n$\\therefore O Q=\\frac{3}{2}$,\n\n$\\therefore Q\\left(\\frac{3}{2}, 0\\right)$.\n\nAccording to the properties of a rectangle, translating point $P$ right by $\\frac{3}{2}$ units and down by 1 unit gives point $N$,\n\n$\\therefore N\\left(2+\\frac{3}{2}, 4-1\\right)$, i.e., $N\\left(\\frac{7}{2}, 3\\right)$\n\n$b$、As shown in Figure 2-2, when quadrilateral $P D N Q$ is a rectangle,\n\n\n\n$\\because$ The equation of line $P D$ is $y=\\frac{3}{2} x+1, P Q \\perp P D$,\n\n$\\therefore$ The equation of line $P Q$ is $y=-\\frac{2}{3} x+\\frac{16}{3}$,\n\n$\\therefore Q(8,0)$,\n\nAccording to the properties of a rectangle, translating point $D$ right by 6 units and down by 4 units gives point $N$,\n\n$\\therefore N(0+6,1-4)$, i.e., $N(6,-3)$.\n\n(2) When $D P$ is the diagonal, let $Q(x, 0)$,\n\nthen $Q D^{2}=x^{2}+1, Q P^{2}=(x-2)^{2}+4^{2}, P D^{2}=13$,\n\n$\\because Q$ is the right angle vertex,\n\n$\\therefore Q D^{2}+Q P^{2}=P D^{2}$,\n\n$\\therefore x^{2}+1+(x-2)^{2}+4^{2}=13$,\n\nSimplifying gives $x^{2}-2 x+4=0$, which has no real solutions, hence this scenario does not exist,\n\nIn summary, the points satisfying the conditions are $N_{1}\\left(\\frac{7}{2}, 3\\right), Q_{1}\\left(\\frac{3}{2}, 0\\right) ; N_{2}(6,-3), Q_{2}(8,0)$.\n\n【Highlight】This problem is a challenging quadratic function problem, comprehensively testing knowledge points such as quadratic functions, the method of undetermined coefficients, maximum value problems, similar triangles, and rectangles. The third question involves an existence problem, which is somewhat difficult. In solving the problem, attention should be paid to the application of the integration of numbers and shapes, categorical discussion, and equation thinking." }, { "problem_id": 1913, "question": "In $\\triangle \\mathrm{ABC}$, $\\mathrm{AB}=5, \\mathrm{AC}=4, \\mathrm{BC}=6$.\n\n(1) As shown in Figure 1, if $\\mathrm{AD}$ is the angle bisector of $\\angle \\mathrm{BAC}$, and $\\mathrm{DE} / / \\mathrm{AB}$, find the length of $\\mathrm{CE}$ and the ratio $\\frac{C D}{B D}$;\n\n(2) As shown in Figure 2, fold side $\\mathrm{AC}$ such that $\\mathrm{AC}$ lies on side $\\mathrm{AB}$, with the crease being $\\mathrm{AM}$. Then fold side $\\mathrm{MB}$ such that $\\mathrm{MB}^{\\prime}$ coincides with $\\mathrm{MC}^{\\prime}$, with the crease being $\\mathrm{MN}$. Find the length of $\\mathrm{AN}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0026_1.jpg", "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0026_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\frac{C D}{B D}=\\frac{4}{5}$ (2) $\\frac{40}{9}$", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Problem Analysis:**\n\n1. **First, determine that triangle \\( ADE \\) is an isosceles triangle. Then, using the theorem that parallel lines divide segments proportionally, find the length of \\( CE \\).**\n\n2. **First, based on the equality of two corresponding angles, determine that \\( \\triangle ABC \\sim \\triangle NB'C' \\). Then, using the proportionality of corresponding sides in similar triangles, find the relationship between \\( NC' \\) and \\( B'N \\). Finally, given that the length of \\( BC' \\) is 1, find the length of \\( NC' \\), and subsequently determine the length of \\( AN \\).**\n\n**Problem Solution:**\n\n1. **As shown in Figure 1, since \\( AD \\) is the angle bisector of \\( \\angle BAC \\) and \\( DE \\parallel AB \\),**\n \n \\[\n \\angle EAD = \\angle BAD = \\angle EDA,\n \\]\n \n **therefore \\( ED = EA \\), meaning triangle \\( ADE \\) is isosceles.**\n \n **Let \\( CE = x \\), then \\( AE = 4 - x = DE \\).**\n \n **Since \\( DE \\parallel AB \\), we have**\n \n \\[\n \\frac{DE}{BA} = \\frac{CE}{CA}, \\quad \\text{that is,} \\quad \\frac{4 - x}{5} = \\frac{x}{4},\n \\]\n \n **solving for \\( x \\) gives \\( x = \\frac{16}{9} \\).**\n \n **Thus, \\( CE = \\frac{16}{9} \\).**\n \n **Since \\( DE \\parallel AB \\), we have**\n \n \\[\n \\frac{CD}{BD} = \\frac{CE}{EA} = \\frac{\\frac{16}{9}}{4 - \\frac{16}{9}},\n \\]\n \n **therefore, \\( \\frac{CD}{BD} = \\frac{4}{5} \\).**\n\n2. **From the folding, we have \\( \\angle B = \\angle B' \\), \\( \\angle C = \\angle MC'A = \\angle B'C'N \\), and \\( AC = AC' = 4 \\).**\n \n **Therefore, \\( \\triangle ABC \\sim \\triangle NB'C' \\).**\n \n **Thus,**\n \n \\[\n \\frac{NC'}{NB'} = \\frac{AC}{AB} = \\frac{4}{5}.\n \\]\n \n **Let \\( NC' = 4a \\), then \\( BN = B'N = 5a \\).**\n \n **Since \\( BC = AB - AC' = 5 - 4 = 1 \\),**\n \n \\[\n NC' + BN = 1, \\quad \\text{that is,} \\quad 4a + 5a = 1,\n \\]\n \n **solving for \\( a \\) gives \\( a = \\frac{1}{9} \\).**\n \n **Therefore, \\( NC' = \\frac{4}{9} \\).**\n \n **Thus, \\( AN = \\frac{4}{9} + 4 = \\frac{40}{9} \\).**\n\n**Key Insight:** This problem, set against the backdrop of a folding scenario, primarily examines the theorem of proportional division by parallel lines and the properties of similar triangles. It presents a certain level of difficulty. Folding is a type of symmetry transformation, belonging to axial symmetry, where the shape and size of the figure remain unchanged before and after folding. When solving such problems, it is crucial to focus on the equality of corresponding sides and angles." }, { "problem_id": 1914, "question": "Points $C$ and $D$ are any two points on segment $A B$.\n\n(1) As shown in Figure 1, if point $D$ is the midpoint of segment $B C$, $A D=18$, and $A C=6$, find the length of segment $B D$;\n\n(2) As shown in Figure 2, if point $C$ divides segment $A B$ into two segments in the ratio $2: 3$ ($A C\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0088_1.jpg", "batch16-2024_06_15_401aff4a5a3b9e0145e1g_0088_2.jpg" ], "is_multi_img": true, "answer": "（1）12；（2）30.\n\n", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: (1) Since \\( \\mathrm{AD} = 18 \\) and \\( \\mathrm{AC} = 6 \\),\n\n\\[\n\\mathrm{CD} = \\mathrm{AD} - \\mathrm{AC} = 12.\n\\]\n\nMoreover, since point \\( \\mathrm{D} \\) is the midpoint of segment \\( \\mathrm{BC} \\),\n\n\\[\n\\mathrm{BD} = \\mathrm{CD} = 12.\n\\]\n\n(2) According to the problem, let \\( \\mathrm{AD} = x \\), then \\( \\mathrm{BD} = 5x \\),\n\n\\[\n\\mathrm{AB} = 6x.\n\\]\n\nSince point \\( \\mathrm{C} \\) divides segment \\( \\mathrm{AB} \\) in the ratio \\( 2:3 \\),\n\n\\[\n\\mathrm{AC} = \\frac{2}{5} \\mathrm{AB} = \\frac{12}{5}x,\n\\]\n\n\\[\n\\mathrm{CD} = \\mathrm{AC} - \\mathrm{AD} = \\frac{12}{5}x - x = \\frac{7}{5}x.\n\\]\n\nGiven that \\( \\mathrm{CD} = 7 \\),\n\n\\[\n\\frac{7}{5}x = 7,\n\\]\n\nsolving for \\( x \\) gives \\( x = 5 \\).\n\nTherefore,\n\n\\[\n\\mathrm{AB} = 6x = 30.\n\\]\n\n**Key Insight:** This problem primarily examines the concept of proportional segments and the distance between two points. It is essential to utilize the meaning of a midpoint and the operations of segment addition and subtraction during the solution process." }, { "problem_id": 1915, "question": "Operation and Exploration:\n\n(1) Perform the following operation on point \\( P \\) on the number line: First, multiply the number represented by point \\( P \\) by \\(\\frac{1}{3}\\), then shift the resulting point 1 unit to the right to obtain the corresponding point \\( P' \\). Points \\( A \\) and \\( B \\) are on the number line. After performing the above operation on each point on segment \\( AB \\), we obtain segment \\( A'B' \\), where the corresponding points of \\( A \\) and \\( B \\) are \\( A' \\) and \\( B' \\), respectively. As shown in Figure 1, if the number represented by point \\( A \\) is \\(-3\\), then the number represented by point \\( A' \\) is _; if the number represented by point \\( B' \\) is \\( 2 \\), then the number represented by point \\( B \\) is _; it is known that a point \\( E \\) on segment \\( AB \\) after the above operation results in its corresponding point \\( E' \\) coinciding with point \\( E \\). The number represented by point \\( E \\) is _; _\n\n\n\nFigure 1\n\n(2) As shown in Figure 2, in the plane rectangular coordinate system \\( xoy \\), perform the following operation on each point inside the square \\( ABCD \\): Multiply both the abscissa and ordinate of each point by the same real number \\( a \\), then shift the resulting point \\( m \\) units to the right and \\( n \\) units upward (\\( m > 0, n > 0 \\)) to obtain the square \\( A'B'C'D' \\) and its internal points, where the corresponding points of \\( A \\) and \\( B \\) are \\( A' \\) and \\( B' \\), respectively. It is known that a point \\( F \\) inside square \\( ABCD \\) after the above operation results in its corresponding point \\( F' \\) coinciding with point \\( F \\). Find the coordinates of point \\( F \\).\n\n", "input_image": [ "batch16-2024_06_15_42442d52fac5a3a476e0g_0057_1.jpg", "batch16-2024_06_15_42442d52fac5a3a476e0g_0057_2.jpg" ], "is_multi_img": true, "answer": "(1) $0,3, \\frac{3}{2} ;$ （2）The coordinates of the point $F$ are $(1,4)$.", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "Solution: (1) $-3 \\times \\frac{1}{3}+1=0$, so the number corresponding to $A^{\\prime}$ is 0.\n\nLet the number corresponding to $B$ be $x$,\n\nthen $\\frac{1}{3} x+1=2$,\n\n$\\therefore x=3$, so the number corresponding to $B$ is 3.\n\nLet the number represented by point $E$ be $b$, then $\\frac{1}{3} a+1=b$, solving gives $b=\\frac{3}{2}$.\n\nTherefore, the number corresponding to $E$ is $\\frac{3}{2}$.\n\n(2) According to the problem, we have the system of equations:\n\\[\n\\left\\{\n\\begin{array}{l}\n-3 a + m = -1 \\\\\n3 a + m = 2 \\\\\n0 \\cdot a + n = 2\n\\end{array}\n\\right.\n\\]\nSolving this system gives:\n\\[\n\\left\\{\n\\begin{array}{l}\na = \\frac{1}{2} \\\\\nm = \\frac{1}{2} \\\\\nn = 2\n\\end{array}\n\\right.\n\\]\n\nLet the coordinates of point $F$ be $(x, y)$.\n\nSince the corresponding point $F^{\\prime}$ coincides with point $F$,\n\nwe have:\n\\[\n\\left\\{\n\\begin{array}{l}\n\\frac{1}{2} x + \\frac{1}{2} = x \\\\\n\\frac{1}{2} y + 2 = y\n\\end{array}\n\\right.\n\\]\nSolving this system gives:\n\\[\n\\left\\{\n\\begin{array}{l}\nx = 1 \\\\\ny = 4\n\\end{array}\n\\right.\n\\]\n\nTherefore, the coordinates of point $F$ are $(1,4)$.\n\n【Insight】This problem tests the understanding of new definitions, setting up systems of equations, and solving linear equations. Mastering the reasoning behind new definitions is key to solving the problem." }, { "problem_id": 1916, "question": "As shown in Figure 1. Translate the line segment $AB$ to $CD$ such that $A$ corresponds to $D$ and $B$ corresponds to $C$, and connect $AD$ and $BC$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Fill in the blanks: The relationship between $AB$ and $CD$ is $\\qquad$ The relationship between the sizes of $\\angle B$ and $\\angle D$ is $\\qquad$ ;\n\n(2) As shown in Figure 2, if $\\angle B = 60^\\circ$, and $F$ and $E$ are points on the extension of $BC$, $\\angle EFD = \\angle EDF$, and $DG$ bisects $\\angle CDE$ intersecting $BE$ at $G$, find $\\angle FDG$.\n\n(3) In (2), if $\\angle FDG = \\alpha$ and the other conditions remain unchanged, then $\\angle B =$ $\\qquad$", "input_image": [ "batch16-2024_06_15_442a221c0368ccf995c2g_0045_1.jpg", "batch16-2024_06_15_442a221c0368ccf995c2g_0045_2.jpg" ], "is_multi_img": true, "answer": "(1) $A B / / C D$, and $A B=C D ; \\angle B=\\angle D$\n\n(2) $\\angle F D G=30^{\\circ}$\n\n(3) $2 \\alpha$", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "(1) According to the properties of translation, we have: $AB // CD$, and $AB = CD$, $AD // BC$,\n\n$\\therefore \\angle B + \\angle C = 180^{\\circ}$, $\\angle D + \\angle C = 180^{\\circ}$,\n\n$\\therefore \\angle B = \\angle D$,\n\nHence, the answer is: $AB // CD$, and $AB = CD$, $\\angle B = \\angle D$;\n\n(2) $\\because AB // CD$,\n\n$\\therefore \\angle DCE = \\angle B$,\n\nBy the exterior angle property of a triangle, $\\angle CDF = \\angle DFE - \\angle DCE$,\n\n$\\therefore \\angle CDG = \\angle CDF + \\angle FDG = \\angle DFE - \\angle DCE + \\angle FDG$,\n\nIn triangle $DEF$, $\\angle DEF = 180^{\\circ} - 2 \\angle DFE$,\n\nIn triangle $DFG$, $\\angle DGF = 180^{\\circ} - \\angle FDG - \\angle DFE$,\n\n$\\therefore \\angle EDG = \\angle DGF - \\angle DEF$\n\n$= 180^{\\circ} - \\angle FDG - \\angle DFE - (180^{\\circ} - 2 \\angle DFE)$\n\n$= 2 \\angle DFE - \\angle FDG - \\angle DFE$,\n\n$\\because DG$ bisects $\\angle CDE$,\n\n$\\therefore \\angle CDG = \\angle EDG$,\n\n$\\therefore \\angle DFE - \\angle DCE + \\angle FDG = 2 \\angle DFE - \\angle FDG - \\angle DFE$,\n\n$\\therefore \\angle FDG = \\frac{1}{2} \\angle DCE$,\n\nThat is, $\\angle FDG = \\frac{1}{2} \\angle B$,\n\n$\\because \\angle B = 60^{\\circ}$,\n\n$\\therefore \\angle FDG = \\frac{1}{2} \\times 60^{\\circ} = 30^{\\circ}$;\n\n(3) According to (2), we know that $\\angle FDG = \\frac{1}{2} \\angle B$,\n\n$\\because \\angle FDG = \\alpha$,\n\n$\\therefore \\angle B = 2 \\alpha$,\n\nHence, the answer is: $2 \\alpha$.\n\n【Insight】This question examines the properties of translation, the exterior angle property of a triangle, the properties of angle bisectors, and related knowledge points. Solving the problem based on these properties is key." }, { "problem_id": 1917, "question": "As shown in Figure 1, given that $\\angle ABC = 50^\\circ$, there is a triangle $BDE$ sharing the vertex $B$ with $\\angle ABC$, where $\\angle EBD = 45^\\circ$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) If $BD$ bisects $\\angle ABC$, find the measure of $\\angle EBC$;\n\n(2) As shown in Figure 2, rotate the triangle clockwise around point $B$ by $\\alpha$ degrees $\\left(0^\\circ < \\alpha < 90^\\circ\\right)$, such that $\\angle ABC$ and $\\angle CBD$ are complementary,\nfind the measure of $\\angle EBC$.", "input_image": [ "batch16-2024_06_15_442a221c0368ccf995c2g_0067_1.jpg", "batch16-2024_06_15_442a221c0368ccf995c2g_0067_2.jpg" ], "is_multi_img": true, "answer": "(1)70\n\n(2) $5^{\\circ}$\n\n", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "(1) Solution: Since \\( BD \\) bisects \\( \\angle ABC \\),\n\n\\[\n\\angle DBC = \\frac{1}{2} \\angle ABC = \\frac{1}{2} \\times 50^\\circ = 25^\\circ,\n\\]\n\n\\[\n\\angle EBC = \\angle EBD + \\angle DBC = 45^\\circ + 25^\\circ = 70^\\circ.\n\\]\n\n(2) Solution: Since \\( AB \\perp BD \\),\n\n\\[\n\\angle ABD = 90^\\circ,\n\\]\n\n\\[\n\\angle ABE = 90^\\circ - 45^\\circ = 45^\\circ,\n\\]\n\n\\[\n\\angle EBC = \\angle ABC - \\angle ABE = 50^\\circ - 45^\\circ = 5^\\circ.\n\\]\n\n**Key Insight:** This problem examines the concept of an angle bisector, which is a ray that divides an angle into two equal parts from the vertex, the properties of rotation, and the meaning of complementary angles (angles that sum to 90 degrees). Mastering these properties and definitions is crucial for solving the problem." }, { "problem_id": 1918, "question": "As shown in Figure 1, place a pair of right-angled triangles on the same straight line $AB$, where $\\angle ONM = 30^\\circ$ and $\\angle OCD = 45^\\circ$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) Observation and conjecture: When the triangle ruler $OCD$ in Figure 1 is translated along the direction of $AB$ to the position in Figure 2, such that point $O$ coincides with point $N$, and $CD$ intersects $MN$ at point $E$, then $\\angle CEN =$ $\\qquad$ ;\n\n(2) Operational investigation: Rotate the triangle ruler $OCD$ in Figure 1 around point $O$ in a clockwise direction, so that one side $OD$ is inside $\\angle MON$, as shown in Figure 3, and $OD$ exactly bisects $\\angle MON$, $CD$ intersects $NM$ at point $E$, find the degree measure of $\\angle CEN$;\n\n(3) Deepening and expanding: Rotate the triangle ruler $OCD$ in Figure 1 around point $O$ in a clockwise direction for one full cycle. During the rotation, when the side $OC$ rotates by how many degrees, does the side $CD$ exactly become parallel to the side $MN$?", "input_image": [ "batch16-2024_06_15_442a221c0368ccf995c2g_0079_1.jpg", "batch16-2024_06_15_442a221c0368ccf995c2g_0079_2.jpg", "batch16-2024_06_15_442a221c0368ccf995c2g_0079_3.jpg" ], "is_multi_img": true, "answer": "(1) $105^{\\circ}$\n\n(2) $150^{\\circ}$\n\n(3) $75^{\\circ}$ or $255^{\\circ}$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "(1) Solution: Since $\\angle ECN = 45^\\circ$ and $\\angle ENC = 30^\\circ$,\n\nTherefore, in triangle $CEN$, $\\angle CEN = 180^\\circ - 45^\\circ - 30^\\circ = 105^\\circ$.\n\nHence, the answer is: $105^\\circ$.\n\n(2) Since $OD$ bisects $\\angle MON$,\n\n$$\n\\begin{aligned}\n& \\therefore \\angle DON = \\frac{1}{2} \\angle MON = \\frac{1}{2} \\times 90^\\circ = 45^\\circ, \\\\\n& \\therefore \\angle DON = \\angle D = 45^\\circ, \\\\\n& \\therefore CD \\parallel AB, \\\\\n& \\therefore \\angle CEN = 180^\\circ - \\angle MNO = 180^\\circ - 30^\\circ = 150^\\circ;\n\\end{aligned}\n$$\n\n(3) As shown in Figure 1, when $CD$ is above $AB$, let $OM$ intersect $CD$ at $F$,\n\n$$\n\\begin{aligned}\n& \\because CD \\parallel MN, \\\\\n& \\therefore \\angle OFD = \\angle M = 60^\\circ, \\\\\n& \\text{In triangle } ODF, \\angle MOD = \\\\\n& = 180^\\circ - 45^\\circ - 60^\\circ, \\\\\n& = 75^\\circ,\n\\end{aligned}\n$$\n\n$$\n\\text{In triangle } ODF, \\angle MOD = 180^\\circ - \\angle D - \\angle OFD,\n$$\n\nWhen $CD$ is below $AB$, let the line $OM$ intersect $CD$ at $F$,\n\n$\\because CD \\parallel MN$,\n\n$\\therefore \\angle DFO = \\angle M = 60^\\circ$,\n\nIn triangle $DOF$, $\\angle DOF = 180^\\circ - \\angle D - \\angle DFO = 180^\\circ - 45^\\circ - 60^\\circ = 75^\\circ$,\n\n$\\therefore$ The rotation angle is $75^\\circ + 180^\\circ = 255^\\circ$,\n\nIn summary, when side $OC$ rotates $75^\\circ$ or $255^\\circ$, side $CD$ is exactly parallel to side $MN$.\n\n\n\n(3) Figure 1\n\n\n\n【Insight】This question examines the properties of rotation, the triangle angle sum theorem, the property that an exterior angle of a triangle equals the sum of the two non-adjacent interior angles, and the property that the two acute angles of a right triangle are complementary. Familiarity with these properties and the angle characteristics of the triangle is key to solving the problem." }, { "problem_id": 1919, "question": "The parasol shown in Figure 1 has its handle perpendicular to the horizontal ground, and its schematic diagram is shown in Figure 2. When the umbrella is closed, point $\\mathrm{P}$ coincides with point $\\mathrm{A}$; as the umbrella slowly opens, the moving point $\\mathrm{P}$ moves from $\\mathrm{A}$ to $\\mathrm{B}$; when point $\\mathrm{P}$ reaches point $\\mathrm{B}$, the umbrella is half-open to its widest extent. It is known that during the opening process of the umbrella, there is always $\\mathrm{PM}=\\mathrm{PN}=\\mathrm{CM}=\\mathrm{CN}=6.0$ decimeters, $\\mathrm{CE}=\\mathrm{CF}=18.0$ decimeters, and $\\mathrm{BC}=2.0$ decimeters.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Determine the range of values for the length of $\\mathrm{AP}$;\n\n(2) Under vertical sunlight, when the umbrella is half-open to its widest extent, find the area $S$ of the shadow under the umbrella (assuming it is a circular surface, with the result retained in terms of $\\pi$).", "input_image": [ "batch16-2024_06_15_482298a4fedf523ec3dbg_0031_1.jpg", "batch16-2024_06_15_482298a4fedf523ec3dbg_0031_2.jpg" ], "is_multi_img": true, "answer": "（1） The value range of $\\mathrm{AP}$ is: $0 \\leq x \\leq 10$ ；（2） $\\mathrm{S}_{\\text {maximum}}=315 \\pi$ (square decimeter).\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "(1) Since BC = 2 decimeters and AC = CN + PN = 12 decimeters,\n\nTherefore, AB = AC - BC = 10 decimeters.\n\nThus, if AP = x, then the range of AP is: 0 ≤ x ≤ 10;\n\n(2) Connect MN and EF, intersecting AC at G and H respectively.\n\n\n\nFigure 2\n\nLet AP = x decimeters,\n\nSince PM = PN = CM = CN,\n\nTherefore, quadrilateral PNCM is a rhombus.\n\nThus, MN and PC bisect each other perpendicularly, and AC is the angle bisector of ∠ECF,\n\nPG = PC / 2\n\nIn right triangle MGP, PM = 6 decimeters,\n\nTherefore, MG² = PM² - PG² = 6² - (6 - 1/2 x)² = 6x - 1/4 x².\n\nSince CE = CF, and AC is the angle bisector of ∠ECF,\n\nTherefore, EH = HF, and EF is perpendicular to AC.\n\nSince ∠ECH = ∠MCG, and ∠EHC = ∠MGC = 90°,\n\nTherefore, triangle CMG is similar to triangle CEH.\n\nThus, MG / EH = CM / CE.\n\nTherefore, MG² / EH² = (6 / 18)² = 1/9\n\nThus, EH² = 9 * MG² = 9 * (6x - 1/4 x²).\n\nTherefore, S = π * EH² = 9π(6x - 1/4 x²),\n\nThat is, S = -9/4 πx² + 54πx,\n\nSince x = -b / 24 = 12, and 0 ≤ x ≤ 10,\n\nTherefore, when x = 10, S_max = -9/4 π * 100 + 54π * 10 = 315π (square decimeters).\n\n[Highlight] This problem mainly examines the application of similar triangles, the properties of a rhombus, and the application of quadratic functions. Mastering the properties of a rhombus and the properties of similar triangles to derive the functional relationship between S and x is key to solving the problem." }, { "problem_id": 1920, "question": "As shown in the figure, $O C$ is the median of side $A B$ in $\\triangle A B C$, $\\angle A B C=36^{\\circ}$, point $D$ is a point on $O C$, if $O D$ $=k \\cdot O C$, through $D$ draw $D E / / C A$ intersecting $B A$ at point $E$, point $M$ is the midpoint of $D E$, and the $\\triangle O D E$ is rotated clockwise around point $O$ by $\\alpha$ degrees (where $0^{\\circ}<\\alpha<180^{\\circ}$) after which the ray $O M$ intersects the line $B C$ at point $N$.\n\n(1) If the area of $\\triangle A B C$ is 26, find the area of $\\triangle O D E$ (expressed in algebraic form of $k$);\n\n(2) When $N$ and $B$ do not coincide, explore the functional relationship between the degree $y$ of $\\angle O N B$ and the degree of the rotation angle $\\alpha$;\n\n(3) Write down the degree of the rotation angle $\\alpha$ when $\\triangle O N B$ is an isosceles triangle.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch16-2024_06_15_482298a4fedf523ec3dbg_0088_1.jpg", "batch16-2024_06_15_482298a4fedf523ec3dbg_0088_2.jpg" ], "is_multi_img": true, "answer": "(1) $S_{\\triangle} O D E=13 k^{2}$; （2） $y=\\alpha\\left(0<\\alpha<144^{\\circ}\\right) ; y=180^{\\circ}-\\alpha\\left(144^{\\circ}<\\alpha<180^{\\circ}\\right)$; (3) $\\alpha=162^{\\circ}$.\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) Since \\( OC \\) is the median of side \\( AB \\) in triangle \\( ABC \\), and the area of triangle \\( ABC \\) is 26,\n\nTherefore, the area of triangle \\( OAC \\) is 13.\n\nSince \\( DE \\parallel AC \\),\n\nTherefore, triangles \\( ODE \\) and \\( OCA \\) are similar, and \\( \\angle OEM = \\angle OAC \\).\n\nThus, the ratio of the areas \\( \\frac{S_{\\triangle DBO}}{S_{\\triangle OAC}} = \\left(\\frac{OD}{OC}\\right)^2 \\), and \\( OD = k \\cdot OC \\),\n\nTherefore, the area of triangle \\( ODE \\) is \\( 13k^2 \\).\n\n(2) Since triangles \\( ODE \\) and \\( OCA \\) are similar,\n\nTherefore, \\( \\frac{OE}{OA} = \\frac{OD}{OC} = \\frac{DE}{AC} = k \\).\n\nSince \\( OC \\) is the median of side \\( AB \\) in triangle \\( ABC \\), and point \\( M \\) is the midpoint of \\( DE \\),\n\nTherefore, \\( AB = 2AO \\), and \\( EM = \\frac{1}{2}DE \\).\n\nThus, \\( \\frac{OE}{AB} = \\frac{k}{2} = \\frac{EM}{AC} \\), and \\( \\angle OEM = \\angle OAC \\),\n\nTherefore, triangles \\( OEM \\) and \\( BAC \\) are similar,\nThus, \\( \\angle EOM = \\angle ABC = 36^\\circ \\).\n\nAs shown in Figure 2, when \\( 0 < \\alpha < 144^\\circ \\),\n\nSince \\( \\angle AON = \\angle B + \\angle ONB \\),\n\nTherefore, \\( \\angle AOE + \\angle EOM = \\angle B + \\angle ONB \\)\n\nThus, \\( y = \\alpha \\).\n\nAs shown in Figure 3, when \\( 144^\\circ < \\alpha < 180^\\circ \\),\n\n\n\nFigure 3\n\nSince \\( \\angle BON = \\angle EOM - \\angle BOE = 36^\\circ - (180^\\circ - \\alpha) \\)\n\nTherefore, \\( \\angle NOB = \\alpha - 144^\\circ \\),\n\nSince \\( \\angle BNO = \\angle ABC - \\angle NOB = 36^\\circ - (\\alpha - 144^\\circ) = 180^\\circ - \\alpha \\);\n\n(3) When \\( 0 < \\alpha < 144^\\circ \\), if \\( OB = ON \\), then \\( \\angle ABC = \\angle BNO = 36^\\circ = \\alpha \\),\n\nIf \\( OB = BN \\), then \\( \\angle ONB = \\frac{180^\\circ - 36^\\circ}{2} = 72^\\circ = \\alpha \\),\n\nIf \\( ON = BN \\), then \\( \\angle ABC = \\angle BON = 36^\\circ \\),\n\nTherefore, \\( \\angle ONB = 180^\\circ - 2 \\times 36^\\circ = 108^\\circ = \\alpha \\),\n\nWhen \\( 144^\\circ < \\alpha < 180^\\circ \\),\n\nIf \\( OB = BN \\), then \\( \\angle N = \\angle NOB = 18^\\circ = 180^\\circ - \\alpha \\),\n\nTherefore, \\( \\alpha = 162^\\circ \\).\n\n【Insight】This problem mainly examines the determination and properties of similar triangles, and the use of categorical discussion to think about problems. This question is of high difficulty." }, { "problem_id": 1921, "question": "As shown in Figure 1, in right $\\triangle ABC$, $\\angle ACB = 90^\\circ, \\angle CAB = 30^\\circ$, point $D$ is on side $AB$ and an isosceles right $\\triangle CDP$ is constructed with $CD$ as the base (point $P$ and $A$ are on opposite sides of line $CD$), the ray $CP$ intersects line $AB$ at point $E$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) If point $D$ is the midpoint of $AB$ and $BC = 2$, find the length of $DP$;\n\n(2) When $\\triangle CDE$ is an isosceles triangle, find the degree measure of $\\angle BCE$;\n\n(3) As shown in Figure 2, let $AP = a$, find the minimum area of quadrilateral $ADPC$. (Express in terms of $a$)", "input_image": [ "batch28-2024_06_17_a4abb72b0c977f12d19ag_0025_1.jpg", "batch28-2024_06_17_a4abb72b0c977f12d19ag_0025_2.jpg", "batch28-2024_06_17_a4abb72b0c977f12d19ag_0025_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\sqrt{2}$ ；（2） $7.5^{\\circ}$ or $30^{\\circ}$ or $15^{\\circ}$\n(3) $\\frac{3-\\sqrt{3}}{6} a^{2}$\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since $\\triangle ABC$ is a right-angled triangle with $\\angle CAB = 30^\\circ$,\n\n$\\therefore AB = 2BC = 4$,\n\nSince point $D$ is the midpoint of $AB$,\n\n$\\therefore CD = \\frac{1}{2} AB = 2$,\n\nSince $\\triangle CDP$ is an isosceles right-angled triangle,\n\n$\\therefore DP = CD \\cdot \\cos 45^\\circ = \\frac{\\sqrt{2}}{2} CD = \\sqrt{2}$;\n\n(2) Since $\\triangle ABC$ is a right-angled triangle with $\\angle CAB = 30^\\circ$,\n\n$\\therefore \\angle ABC = 90^\\circ - \\angle CAB = 60^\\circ$,\n\nSince $\\triangle CDP$ is an isosceles right-angled triangle,\n\n$\\therefore \\angle DCP = \\angle CDP = 45^\\circ$,\n\nWe discuss three cases:\n\n(1) When $CD = CE$, as shown in Figure 1,\n\n\n\nThen $\\angle DEC = \\angle CDE = \\frac{1}{2}(180^\\circ - \\angle DCP) = 67.5^\\circ$,\n\n$\\therefore \\angle BCE = \\angle DEC - \\angle ABC = 67.5^\\circ - 60^\\circ = 7.5^\\circ$;\n\n(2) When $ED = EC$, as shown in Figure 2,\n\n\n\nThen $\\angle EDC = \\angle ECD = 45^\\circ$, $\\angle DEC = \\angle DPC = 90^\\circ$,\n\n$\\therefore$ Point $P$ coincides with point $E$,\n\n$\\therefore \\angle BCE = \\angle DEC - \\angle ABC = 30^\\circ$;\n\n(3) When $DC = DE$, as shown in Figure 3,\n\n\n\nFigure 3\n\nThen $\\angle DEC = \\angle DCE = 45^\\circ$,\n\n$\\therefore \\angle BCE = \\angle ABC - \\angle DEC = 15^\\circ$;\n\nIn summary: If $\\triangle CDE$ is an isosceles triangle, then the measure of $\\angle BCE$ is $7.5^\\circ$ or $30^\\circ$ or $15^\\circ$;\n\n(3) Draw $PF \\perp AP$ from point $P$, cut off $PF = AP$, and connect $CF, AF$,\n\n\n\nSince $\\angle DPC = \\angle APF = 90^\\circ$,\n\n$\\therefore \\angle FPC = \\angle APD$,\n\nAlso, since $PD = PC$,\n\n$\\therefore \\triangle FPC \\cong \\triangle APD$ (SAS),\n\n$\\therefore S_{\\triangle FPC} = S_{\\triangle APD}$, $\\angle PAD = \\angle PFC$, $CF = AD$,\n\n$\\therefore S_{\\text{quadrilateral } ADPC} = S_{\\triangle APD} + S_{\\triangle APC} = S_{\\triangle FPC} + S_{\\triangle APC} = S_{\\triangle APF} - S_{\\triangle ACF}$,\n\nSince $S_{\\triangle APF} = \\frac{1}{2} AP \\cdot PF = \\frac{1}{2} a^2$, $AF = \\sqrt{2} AP = \\sqrt{2} a$ are constants,\n\n$\\therefore$ When the area of $\\triangle AFC$ reaches its maximum, the area of quadrilateral $ADPC$ reaches its minimum,\n\nSince $\\angle ACF = 180^\\circ - \\angle CAF - \\angle AFC$,\n\n$= \\angle APF + \\angle CAP + \\angle CFP$,\n\n$= \\angle APF + \\angle CAP + \\angle PAD$,\n\n$= \\angle APF + \\angle CAB = 90^\\circ + 30^\\circ = 120^\\circ$,\n\n$\\therefore$ Only when $CA = CF$, the distance from point $C$ to $AF$ is maximized, and the area of $\\triangle AFC$ is maximized,\n\nThen $S_{\\triangle AFC} = \\frac{1}{2} AF \\cdot h = \\frac{1}{2} \\times \\sqrt{2} a \\cdot \\frac{\\sqrt{6}}{6} a = \\frac{\\sqrt{3}}{6} a^2$,\n\n$\\therefore$ The minimum area of quadrilateral $ADPC$ is $\\frac{1}{2} a^2 - \\frac{\\sqrt{3}}{6} a^2 = \\frac{3 - \\sqrt{3}}{6} a^2$.\n\n【Highlight】This problem examines the properties of a $30^\\circ$ right-angled triangle, the properties of an isosceles right-angled triangle, trigonometric functions of acute angles, and the determination and properties of congruent triangles. It is crucial to use case discussions: (1) when $CD = CE$, (2) when $ED = EC$, and (3) when $DC = DE$ to determine the measure of $\\angle BCE$; and when the area of $\\triangle AFC$ reaches its maximum, the area of quadrilateral $ADPC$ reaches its minimum, determining that $\\angle ACF = 120^\\circ$, and only when $CA = CF$, the distance from point $C$ to $AF$ is maximized, and the area of $\\triangle AFC$ is maximized." }, { "problem_id": 1922, "question": "In the rectangular coordinate system, a rectangle \\( OABC \\) is constructed passing through the origin \\( O \\) and points \\( A(8,0) \\), \\( C(0,6) \\). The line segment \\( OB \\) is connected, with point \\( D \\) being the midpoint of \\( OB \\). Point \\( E \\) is a moving point on segment \\( AB \\), and line segment \\( DE \\) is connected. A line \\( DF \\) is drawn perpendicular to \\( DE \\) and intersects \\( OA \\) at point \\( F \\), and \\( EF \\) is connected. It is known that point \\( E \\) starts from point \\( A \\) and moves along segment \\( AB \\) at a speed of 1 unit per second, with the moving time denoted as \\( t \\) seconds.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) As shown in Figure 1, when \\( t = 3 \\), find the length of \\( DF \\).\n\n(2) As shown in Figure 2, during the movement of point \\( E \\) on segment \\( AB \\), does the size of \\( \\angle DEF \\) change? If it changes, explain why; if it remains constant, find the value of \\( \\tan \\angle DEF \\).\n\n(3) Connect \\( AD \\). When \\( AD \\) divides \\( \\triangle DEF \\) into two parts with an area ratio of 1:2, find the corresponding value of \\( t \\).", "input_image": [ "batch28-2024_06_17_a4abb72b0c977f12d19ag_0078_1.jpg", "batch28-2024_06_17_a4abb72b0c977f12d19ag_0078_2.jpg" ], "is_multi_img": true, "answer": "(1) 3 ; （2） The size of $\\angle \\mathrm{DEF}$ does not change, $\\tan \\angle \\mathrm{DEF}=\\frac{3}{4}$; （3） $\\frac{75}{41}$ or $\\frac{75}{17}$.", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) When \\( t = 3 \\), point \\( E \\) is the midpoint of \\( AB \\).\n\nGiven \\( A(8, 0) \\) and \\( C(0, 6) \\), we have \\( OA = 8 \\) and \\( OC = 6 \\).\n\nSince point \\( D \\) is the midpoint of \\( OB \\), it follows that \\( DE \\parallel OA \\) and \\( DE = \\frac{1}{2} OA = 4 \\).\n\nGiven that quadrilateral \\( OABC \\) is a rectangle, \\( OA \\perp AB \\), hence \\( DE \\perp AB \\).\n\nTherefore, \\( \\angle OAB = \\angle DEA = 90^\\circ \\).\n\nAdditionally, since \\( DF \\perp DE \\), \\( \\angle EDF = 90^\\circ \\).\n\nThus, quadrilateral \\( DFAE \\) is a rectangle, and \\( DF = AE = 3 \\).\n\n(2) The size of \\( \\angle DEF \\) remains unchanged. The reasoning is as follows:\n\nDraw \\( DM \\perp OA \\) at \\( M \\) and \\( DN \\perp AB \\) at \\( N \\), as shown in Figure 2.\n\nSince quadrilateral \\( OABC \\) is a rectangle, \\( OA \\perp AB \\), making quadrilateral \\( DMAN \\) a rectangle.\n\nThus, \\( \\angle MDN = 90^\\circ \\), \\( DM \\parallel AB \\), and \\( DN \\parallel OA \\).\n\nTherefore, \\( \\frac{BD}{DO} = \\frac{BN}{NA} \\) and \\( \\frac{BD}{DO} = \\frac{AM}{OM} \\).\n\nSince point \\( D \\) is the midpoint of \\( OB \\), \\( M \\) and \\( N \\) are the midpoints of \\( OA \\) and \\( AB \\), respectively.\n\nHence, \\( DM = \\frac{1}{2} AB = 3 \\) and \\( DN = \\frac{1}{2} OA = 4 \\).\n\nGiven \\( \\angle EDF = 90^\\circ \\), \\( \\angle FDM = \\angle EDN \\).\n\nAlso, \\( \\angle DMF = \\angle DNE = 90^\\circ \\), so \\( \\triangle DMF \\sim \\triangle DNE \\).\n\nThus, \\( \\frac{DF}{DE} = \\frac{DM}{DN} = \\frac{3}{4} \\).\n\nSince \\( \\angle EDF = 90^\\circ \\), \\( \\tan \\angle DEF = \\frac{DF}{DE} = \\frac{3}{4} \\).\n\n(3) Draw \\( DM \\perp OA \\) at \\( M \\) and \\( DN \\perp AB \\) at \\( N \\).\n\nIf \\( AD \\) divides the area of \\( \\triangle DEF \\) into a ratio of \\( 1:2 \\), let \\( AD \\) intersect \\( EF \\) at point \\( G \\), which is a trisection point of \\( EF \\).\n\n(1) Before point \\( E \\) reaches the midpoint, as shown in Figure 3, \\( NE = 3 - t \\).\n\nFrom \\( \\triangle DMF \\sim \\triangle DNE \\), we get \\( MF = \\frac{3}{4}(3 - t) \\).\n\nThus, \\( AF = 4 + MF = -\\frac{3}{4}t + \\frac{25}{4} \\).\n\nSince point \\( G \\) is a trisection point of \\( EF \\), \\( G\\left(\\frac{3t + 71}{12}, \\frac{2}{3}t\\right) \\).\n\nLet the equation of line \\( AD \\) be \\( y = kx + b \\).\n\nSubstituting \\( A(8, 0) \\) and \\( D(4, 3) \\) gives:\n\\[\n\\begin{cases}\n8k + b = 0 \\\\\n4k + b = 3\n\\end{cases}\n\\]\nSolving, we find \\( k = -\\frac{3}{4} \\) and \\( b = 6 \\).\n\nThus, the equation of line \\( AD \\) is \\( y = -\\frac{3}{4}x + 6 \\).\n\nSubstituting \\( G\\left(\\frac{3t + 71}{12}, \\frac{2}{3}t\\right) \\) into the equation gives \\( t = \\frac{75}{41} \\).\n\n(2) After point \\( E \\) passes the midpoint, as shown in Figure 4, \\( NE = t - 3 \\).\n\nFrom \\( \\triangle DMF \\sim \\triangle DNE \\), we get \\( MF = \\frac{3}{4}(t - 3) \\).\n\nThus, \\( AF = 4 - MF = -\\frac{3}{4}t + \\frac{25}{4} \\).\n\nSince point \\( G \\) is a trisection point of \\( EF \\), \\( G\\left(\\frac{3t + 23}{6}, \\frac{1}{3}t\\right) \\).\n\nSubstituting into the equation of line \\( AD \\), \\( y = -\\frac{3}{4}x + 6 \\), gives \\( t = \\frac{75}{17} \\).\n\nIn summary, when \\( AD \\) divides \\( \\triangle DEF \\) into two parts with an area ratio of \\( 1:2 \\), the value of \\( t \\) is \\( \\frac{75}{41} \\) or \\( \\frac{75}{17} \\).\n\nSource: Comprehensive quadrilateral problem." }, { "problem_id": 1923, "question": "Given, as shown in Figure (1), in rectangle \\( A B C D \\), \\( A B = \\sqrt{3}, A D = 3 \\), point \\( E \\) is a moving point on side \\( B C \\), and point \\( E \\) is rotated counterclockwise by \\( 60^\\circ \\) around point \\( A \\) to obtain point \\( F \\). Lines \\( A E \\), \\( A F \\), \\( E F \\), and \\( D F \\) are connected.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n(1) When points \\( A \\), \\( F \\), and \\( C \\) are collinear, find the length of \\( D F \\).\n\n(2) As shown in Figure (2), point \\( M \\) is on the extension of \\( C B \\) with \\( B M = 1 \\). Connect \\( A M \\). As point \\( E \\) moves on \\( B C \\), does the area of \\( \\triangle A M F \\) change? If it remains constant, find this constant value; if it changes, explain why.\n\n(3) During the process where point \\( E \\) moves from \\( B \\) to \\( C \\), find the range of values for \\( D F \\).", "input_image": [ "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0024_1.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0024_2.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0024_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\sqrt{3}$\n\n(2) remains unchanged, $\\sqrt{3}$\n\n(3) $\\frac{\\sqrt{3}}{2} \\leq D F \\leq \\sqrt{3}$", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "(1) According to the definition of trigonometric functions, we find that $\\angle CAB = 60^\\circ$, and point $E$ coincides with point $B$. Based on this, we can solve the problem;\n\n(2) Prove that $\\triangle APF \\cong \\triangle EBA$, and then solve the problem;\n\n(3) Construct an equilateral triangle $\\triangle ADQ$ to the right of side $AD$, prove that $\\triangle ADF \\cong \\triangle AQE$, obtaining $DF = QE$. When $E$ is at $B$ (or $C$), $QE$ is maximum, and when $E$ is at the midpoint of $BC$, $QE$ is minimum. Based on this, solve the problem.\n\n(1)\n\nSolution: Connect $AC$,\n\n\n\nIn right triangle $\\triangle ABC$, $AB = \\sqrt{3}$, $AD = BC = 3$,\n\n$\\therefore \\tan \\angle CAB = \\frac{BC}{AB} = \\frac{3}{\\sqrt{3}} = \\sqrt{3}$,\n\n$\\therefore \\angle CAB = 60^\\circ$,\n\n$\\therefore$ Point $E$ coincides with point $B$,\n\n$\\therefore AF = AB = \\sqrt{3} = \\frac{1}{2} AC$,\n\n$\\therefore F$ is the midpoint of $AC$,\n\n$\\therefore DF = \\frac{1}{2} AC = \\sqrt{3}$;\n\nSolution: It remains unchanged,\n\nReason as follows:\n\n\n\nDraw a perpendicular from $F$ to $MA$, with the foot at $P$,\n\nIn right triangle $\\triangle ABM$, $AB = \\sqrt{3}$, $BM = 1$,\n\n$\\therefore AM = \\sqrt{(\\sqrt{3})^2 + 1^2} = 2$, $\\tan \\angle MAB = \\frac{BM}{AB} = \\frac{1}{\\sqrt{3}} = \\frac{\\sqrt{3}}{3}$,\n\n$\\therefore \\angle MAB = 30^\\circ$,\n\n$\\because \\angle MAF = \\angle P + \\angle PFA = 90^\\circ + \\angle PFA$,\n\n$\\angle MAF = \\angle FAE + \\angle BAE + \\angle MAB = 60^\\circ + \\angle BAE + 30^\\circ = 90^\\circ + \\angle BAE$,\n\n$\\therefore \\angle PFA = \\angle BAE$,\n\n$\\because AE = AF$, $\\angle P = \\angle ABE = 90^\\circ$,\n\n$\\therefore \\triangle APF \\cong \\triangle EBA$,\n\n$\\therefore FP = AB = \\sqrt{3}$,\n\n$S_{\\triangle AMF} = \\frac{1}{2} AM \\cdot FP = \\sqrt{3}$;\n\n(3)\n\nSolution: Construct an equilateral triangle $\\triangle ADQ$ to the right of side $AD$, connect $QE$,\n\n\n\n$\\because \\triangle ADQ$ and $\\triangle AEF$ are equilateral triangles,\n\n$\\therefore \\angle DAQ = \\angle FAE = 60^\\circ$, $AF = AE$, $AD = AQ$,\n\n$\\therefore \\angle DAF = \\angle QAE$,\n\n$\\therefore \\triangle ADF \\cong \\triangle AQE$,\n\n$\\therefore DF = QE$,\n\nWhen $E$ is at $B$ (or $C$), $QE$ is maximum,\n\n\n\nAs in (1), $\\angle ADB = 30^\\circ$, then $\\angle QDB = 30^\\circ$, and $AD = AQ$,\n\n$\\therefore DB$ is the perpendicular bisector of segment $AQ$,\n\n$\\therefore QE_{\\text{max}} = AB = \\sqrt{3}$;\n\nDraw $QG \\perp AD$ at point $G$, intersecting $BC$ at point $H$,\n\nWhen $E$ coincides with point $H$, which is when $E$ is at the midpoint of $BC$, $QE$ is minimum,\n\n\n\n$QG = OD \\sin 60^\\circ = \\frac{3\\sqrt{3}}{2}$,\n\n$QE_{\\text{min}} = QG - EG = \\frac{\\sqrt{3}}{2}$;\n\n$\\therefore \\frac{\\sqrt{3}}{2} \\leq DF \\leq \\sqrt{3}$.\n\n【Insight】This problem examines the properties of rectangles, the determination and properties of equilateral triangles, solving right triangles, and the determination and properties of congruent triangles. The key to solving the problem is to master the fundamental knowledge." }, { "problem_id": 1924, "question": "The side length of square \\( A B C D \\) is 4.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n(1) Fold square \\( A B C D \\) along the crease \\( E F \\), as shown in Figure (1). After unfolding the square, fold point \\( C \\) to point \\( N \\) on the crease \\( E F \\), with the crease being \\( B M \\). Find the length of \\( P F \\);\n\n(2) As shown in Figure (2), when \\( A E = C F \\), determine the range of the area of \\( \\triangle A D G \\) as point \\( E \\) moves from point \\( A \\) to the midpoint of \\( A D \\).", "input_image": [ "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0033_1.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0033_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\frac{2 \\sqrt{3}}{3}$\n\n(2) $8 \\leq S_{\\triangle A D G} \\leq 4+4 \\sqrt{2}$", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "(1) Connect $CN$. According to the properties of axial symmetry, the properties of the perpendicular bisector of a line segment, the determination theorem and properties of an equilateral triangle, find $\\angle CBM$. Based on the properties of a square and the relationship between the sides and angles of a right triangle, determine $PF$.\n\n(2) Connect $AC$ intersecting $EF$ at point $O$, connect $OB$, $OD$, $OG$, then draw a circle with $O$ as the center and $OA$ as the radius. Take the midpoint of $AD$ as $K$, connect $KO$ and extend it to intersect the circle $\\odot O$ at $J$. According to the properties of a square, the determination theorem and properties of congruent triangles, confirm that $G$, $A$, $B$, $C$, $D$ are concyclic. Then determine that point $G$ moves along the arc $\\overparen{CB}$, and further determine that when point $G$ coincides with point $C$ or point $B$, the area of $\\triangle ADG$ reaches its minimum. When point $G$ coincides with point $J$, the area of $\\triangle ADG$ reaches its maximum. Finally, based on the properties of a square, the Pythagorean theorem, the mid-segment theorem of a triangle, and the area formula of a triangle, solve the problem.\n\n(1)\n\nSolution: As shown in the figure below, connect $CN$.\n\n\n\n$\\because$ After folding square $ABCD$, the crease is $EF$, and the side length of square $ABCD$ is 4,\n$\\therefore EF$ is the perpendicular bisector of $BC$, $BC=4$.\n\n$\\therefore NB=NC$, $BF=2$.\n\n$\\because$ After folding, point $C$ moves to point $N$,\n\n$\\therefore NB=BC$, $\\angle NBM=\\angle CBM=\\frac{1}{2} \\angle NBC$.\n\n$\\therefore NB=NC=BC$.\n\n$\\therefore \\triangle NBC$ is an equilateral triangle.\n\n$\\therefore \\angle NBC=60^{\\circ}$.\n\n$\\therefore \\angle CBM=30^{\\circ}$.\n\n$\\therefore PF=BF \\times \\tan \\angle CBM=\\frac{2 \\sqrt{3}}{3}$.\n\n(2)\n\nSolution: As shown in the figure below, connect $AC$ intersecting $EF$ at point $O$, connect $OB$, $OD$, $OG$, then draw a circle with $O$ as the center and $OA$ as the radius. Take the midpoint of $AD$ as $K$, connect $KO$ and extend it to intersect the circle $\\odot O$ at $J$.\n\n\n\n$\\because$ Quadrilateral $ABCD$ is a square,\n\n$\\therefore AD / / BC$.\n\n$\\therefore \\angle OAE=\\angle OCF$, $\\angle OEA=\\angle OFC$.\n\n$\\because AE=CF$,\n\n$\\therefore \\triangle OAE \\cong \\triangle OCF$ (ASA).\n\n$\\therefore OA=OC$.\n\n$\\therefore O$ is the midpoint of $AC$.\n\n$\\therefore OA=OB=OC=OD$.\n\n$\\because$ After folding square $ABCD$, the corresponding point of $C$ is $G$,\n\n$\\therefore OG=OC$.\n\n$\\therefore B$, $C$, $D$, $G$ are all on the circle $\\odot O$.\n\n$\\therefore OA=OB=OC=OD=OG$.\n\n$\\therefore$ As point $E$ moves from point $A$ to the midpoint $K$ of $AD$, point $G$ moves along the arc $\\overparen{CB}$.\n\n$\\therefore$ When point $G$ coincides with point $C$ or point $B$, the area of $\\triangle ADG$ reaches its minimum.\n\n$\\because$ The side length of square $ABCD$ is 4,\n\n$\\therefore AD=CD=4$, $\\angle ADC=90^{\\circ}$.\n$\\therefore S_{\\triangle ADC}=\\frac{1}{2} AD \\cdot CD=8$, $\\quad AC=\\sqrt{AD^{2}+CD^{2}}=4 \\sqrt{2}$.\n\n$\\therefore$ The minimum area of $\\triangle ADG$ is $8$, $OA=\\frac{1}{2} AC=2 \\sqrt{2}$.\n\n$\\therefore OJ=OA=2 \\sqrt{2}$.\n\n$\\because K$ is the midpoint of $AD$, $O$ is the midpoint of $AC$,\n\n$\\therefore KO$ is the mid-segment of $\\triangle ACD$.\n\n$\\therefore KO / / CD$, $KO=\\frac{1}{2} CD=2$.\n\n$\\therefore \\angle AKO=\\angle ADC=90^{\\circ}$, $KO$ is a fixed value.\n\n$\\therefore OK \\perp AD$, i.e., $OJ \\perp AD$.\n\n$\\therefore$ When $OG \\perp AD$, i.e., when point $G$ coincides with point $J$, the area of $\\triangle ADG$ reaches its maximum.\n\n$\\therefore KJ=KO+OJ=2+2 \\sqrt{2}$.\n\n$\\therefore S_{\\triangle ADJ}=\\frac{1}{2} AD \\cdot KJ=4+4 \\sqrt{2}$.\n\n$\\therefore$ The maximum area of $\\triangle ADG$ is $4+4 \\sqrt{2}$.\n\n$\\therefore$ The range of the area of $\\triangle ADG$ is $8 \\leq S_{\\triangle ADG} \\leq 4+4 \\sqrt{2}$.\n\n【Insight】This problem examines the properties of axial symmetry, the properties of the perpendicular bisector of a line segment, the determination theorem and properties of an equilateral triangle, the properties of a square, solving right triangles, the mid-segment theorem of a triangle, the Pythagorean theorem, the area formula of a triangle. Comprehensive application of these knowledge points is key to solving the problem." }, { "problem_id": 1925, "question": "In an isosceles triangle \\(ABC\\) with \\(AC = BC\\), point \\(P\\) is a point on side \\(BC\\) (not coinciding with \\(B\\) or \\(C\\)). Connect \\(PA\\), and rotate segment \\(PA\\) clockwise around point \\(P\\) by an angle equal to \\(\\angle C\\), resulting in segment \\(PD\\). Connect \\(DB\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) Special Case Perception\n\nAs shown in Figure 1, when \\(\\angle C = 90^\\circ\\), find the measure of \\(\\angle DBA\\).\n\n(2) Extension and Exploration\n\nAs shown in Figure 2, if \\(\\angle C = \\alpha\\), express the measure of \\(\\angle DBA\\) in terms of \\(\\alpha\\).\n\n(3) Problem Solving\n\nAs shown in Figure 3, connect \\(AD\\). If \\(\\angle C = \\alpha\\) and \\(\\tan \\alpha = \\frac{5}{12}\\), \\(AC = 13\\), and \\(\\angle APC = 135^\\circ\\), find the length of \\(AD\\).", "input_image": [ "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0053_1.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0053_2.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0053_3.jpg" ], "is_multi_img": true, "answer": "(1) $90^{\\circ}$\n\n(2) $\\alpha$\n\n(3) $\\frac{10 \\sqrt{13}}{13}$", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "(1) Draw a line from point \\( D \\) perpendicular to the extension of \\( CB \\) at point \\( H \\), and connect \\( AD \\). First, prove that right triangle \\( \\triangle APC \\) is congruent to right triangle \\( \\triangle PDH \\). Then, prove that \\( \\triangle BHD \\) is an isosceles right triangle, which allows us to determine the measure of angle \\( \\angle DBA \\).\n\n(2) On the extension of \\( CB \\), mark a point \\( H \\) such that \\( BH = CP \\), and connect \\( DH \\). First, prove that \\( \\triangle HPD \\) is congruent to \\( \\triangle CAP \\), which gives \\( HD = HB \\). Then, use the triangle angle sum theorem to determine the measure of angle \\( \\angle DBA \\).\n\n(3) Draw \\( AH \\) perpendicular to \\( BC \\), and let \\( AH = 5k \\) and \\( CH = 12k \\). Using the Pythagorean theorem, set up an equation to find \\( k = 1 \\), which gives \\( AH = 5 \\) and \\( CH = 12 \\). Next, find the length of \\( AB \\), and prove that \\( \\triangle PAD \\) is similar to \\( \\triangle CAB \\). Finally, set up a proportion to find the length of \\( AD \\).\n\n---\n\n(1) As shown in Figure 1, draw a line from point \\( D \\) perpendicular to the extension of \\( CB \\) at point \\( H \\).\n\n\n\nFigure 1\n\nSince \\( AC = BC \\), rotate line segment \\( PA \\) clockwise by \\( 90^\\circ \\) to obtain line segment \\( PD = PA \\), with \\( \\angle APD = 90^\\circ \\).\n\nThus, \\( \\angle HPD + \\angle APC = 180^\\circ - 90^\\circ = 90^\\circ \\).\n\nSince \\( \\angle APC + \\angle CAP = 90^\\circ \\), it follows that \\( \\angle CAP = \\angle HPD \\).\n\nAlso, since \\( PD = PA \\), right triangle \\( \\triangle APC \\) is congruent to right triangle \\( \\triangle PDH \\) (by AAS).\n\nTherefore, \\( CA = PH \\) and \\( CP = DH \\).\n\nThus, \\( PH = PB + BH = CA = CB = CP + PB \\), so \\( BH = CP = DH \\).\n\nThis means \\( \\triangle BHD \\) is an isosceles right triangle, so \\( \\angle HBD = 45^\\circ \\).\n\nSince \\( \\angle CBA = 45^\\circ \\), it follows that \\( \\angle ABD = 180^\\circ - \\angle ABC - \\angle HBD = 90^\\circ \\).\n\n---\n\n(2) As shown in Figure 2, on the extension of \\( CB \\), mark a point \\( H \\) such that \\( BH = CP \\), and connect \\( DH \\).\n\n\n\nThus, \\( BP + PC = BP + BH \\), which means \\( BC = PH \\).\n\nSince \\( AC = BC \\), it follows that \\( AC = PH \\).\n\nSince \\( \\angle HPD + \\angle DPA = \\angle C + \\angle CAP \\), it follows that \\( \\angle HPD = \\angle CAP \\), and \\( PD = PA \\).\n\nTherefore, \\( \\triangle HPD \\) is congruent to \\( \\triangle CAP \\) (by SAS).\n\nThus, \\( \\angle H = \\angle C = \\alpha \\), and \\( HD = CP \\), so \\( HD = HB \\).\n\nThis means \\( \\angle HBD = \\frac{180^\\circ - \\alpha}{2} \\).\n\nSince \\( \\angle ABC = \\frac{180^\\circ - \\alpha}{2} \\), it follows that \\( \\angle ABD = 180^\\circ - \\angle HBD - \\angle ABC = 180^\\circ - \\frac{180^\\circ - \\alpha}{2} - \\frac{180^\\circ - \\alpha}{2} = \\alpha \\).\n\n---\n\n(3) As shown in Figure 3, draw \\( AH \\) perpendicular to \\( BC \\).\n\n\n\nFigure 3\n\nSince \\( AC = 13 \\) and \\( \\tan \\alpha = \\frac{5}{12} \\), let \\( AH = 5k \\) and \\( CH = 12k \\).\n\nUsing the Pythagorean theorem, we have \\( (5k)^2 + (12k)^2 = 13^2 \\).\n\nSolving for \\( k \\), we find \\( k = 1 \\), so \\( AH = 5 \\) and \\( CH = 12 \\).\n\nThus, \\( BH = BC - CH = 13 - 12 = 1 \\).\n\nIn right triangle \\( \\triangle AHB \\), using the Pythagorean theorem, we find:\n\n\\[\nAB = \\sqrt{AH^2 + BH^2} = \\sqrt{5^2 + 1^2} = \\sqrt{26}.\n\\]\n\nSince \\( \\angle APC = 135^\\circ \\), it follows that \\( \\angle APH = 45^\\circ \\), so \\( AP = \\sqrt{2} \\cdot AH = 5\\sqrt{2} \\).\n\nSince \\( \\angle APD = \\angle ACB = \\alpha \\), \\( AC = BC \\), and \\( AP = DP \\), it follows that \\( \\triangle PAD \\) is similar to \\( \\triangle CAB \\).\n\nThus, \\( \\frac{AD}{AB} = \\frac{AP}{AC} = \\frac{5\\sqrt{2}}{13} \\), so:\n\n\\[\nAD = \\frac{5\\sqrt{2}}{13} \\cdot AB = \\frac{5\\sqrt{2}}{13} \\cdot \\sqrt{26} = \\frac{10\\sqrt{3}}{13}.\n\\]\n\n---\n\n**Key Insight:** This problem is a comprehensive geometric transformation question, primarily testing the properties and proofs of congruent triangles, similar triangles, the Pythagorean theorem, isosceles triangles, and trigonometric functions. Mastering the properties and proofs of similar triangles is crucial for solving this problem and represents the main challenge." }, { "problem_id": 1926, "question": "As shown in Figure 1, in the plane rectangular coordinate system, point $O$ is the origin, the line $y=-\\sqrt{3} x+6 \\sqrt{3}$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. The coordinates of the vertex $F$ of the rectangle $CDEF$ are $(-2, 4\\sqrt{3})$, and point $D$ coincides with the origin. The rectangle $CDEF$ is translated along the positive direction of the $x$-axis at a speed of 2 units per second. The movement stops when point $D$ reaches point $A$. Let the time of movement be $t$ seconds, and the area of the overlapping part of the rectangle $CDEF$ and triangle $ABO$ be $S$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) Fill in the blank: When $t=$ $\\qquad$ seconds, point $E$ falls on the line $AB$;\n\n(2) As shown in Figure 2, when $0\n\nFigure 1\n\nWhen \\( x = 0 \\), \\( y = 6 \\sqrt{3} \\).\n\nWhen \\( y = 0 \\), \\( -\\sqrt{3} x + 6 \\sqrt{3} = 0 \\),\n\n\\(\\therefore x = 6\\),\n\n\\(\\therefore O B = 6 \\sqrt{3}\\), \\( O A = 6 \\),\n\n\\(\\therefore \\tan \\angle A B O = \\frac{O A}{O B} = \\frac{\\sqrt{3}}{3}\\),\n\nLet the line \\( E F \\) intersect \\( O B \\) at \\( F^{\\prime} \\),\n\n\\(\\therefore B F^{\\prime} = 6 \\sqrt{3} - 4 \\sqrt{3} = 2 \\sqrt{3}\\),\n\n\\(\\therefore E F^{\\prime} = B F^{\\prime} \\cdot \\tan \\angle A B O = 2 \\sqrt{3} \\times \\frac{\\sqrt{3}}{3} = 2\\),\n\n\\(\\therefore t = \\frac{2}{2} = 1\\).\n\nThus, the answer is: 1;\n\n(2)\n\nWhen \\( 0 < t < 1 \\),\n\n\\(\\because O D = 2 t\\), \\( D E = 4 \\sqrt{3} \\),\n\n\\(\\therefore S = 2 t \\cdot 4 \\sqrt{3} = 8 \\sqrt{3} t\\);\n\n(3)\n\n(1) When \\( 0 < t < 1 \\),\n\n\\( 8 \\sqrt{3} t = 4 \\sqrt{3} \\),\n\\(\\therefore t = \\frac{1}{2}\\),\n\nAs shown in the figure,\n\n\n\n## Figure 2\n\n(2) When \\( 2 < t \\leq 3 \\),\n\n\\(\\because \\tan \\angle B A O = \\frac{O B}{O A} = \\frac{6 \\sqrt{3}}{6} = \\sqrt{3}\\),\n\n\\(\\therefore D G = A D \\cdot \\tan \\angle B A O = \\sqrt{3}(6 - 2 t) = 6 \\sqrt{3} - 2 \\sqrt{3} t\\), \\( C H = \\sqrt{3} A C = \\sqrt{3}(8 - 2 t) = 8 \\sqrt{3} - 2 \\sqrt{3} t\\),\n\n\\(\\because S = \\frac{D G + C H}{2} \\cdot C D = 4 \\sqrt{3}\\),\n\n\\(\\therefore 14 \\sqrt{3} - 4 \\sqrt{3} t = 4 \\sqrt{3}\\),\n\n\\(\\therefore t = \\frac{5}{2}\\),\n\nIn summary, \\( t = \\frac{1}{2} \\) or \\( \\frac{5}{2} \\).\n\n【Insight】This problem examines the evaluation of linear functions, solving right triangles, and solving linear equations. The key to solving the problem is to understand the motion process, draw the graph, and represent the relevant quantities." }, { "problem_id": 1927, "question": "In right $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = 6$, $BC = 8$. Point $D$ is a moving point on side $BC$ ( $D$ does not coincide with points $B$ and $C$), and $DE \\perp AB$, with $E$ as the foot of the perpendicular. Connect $AD$, and let $CD = x$, $DE = y$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\n(1) When point $E$ is the midpoint of $AB$, find the length of $CD$;\n\n(2) Find the functional relationship of $y$ with respect to $x$, and specify the domain;\n\n(3) Draw a line through $B$ parallel to $DE$ intersecting the extension of $AD$ at $F$. When $\\triangle BDF$ is an isosceles triangle with $BD$ as one of its equal sides, directly write down the length of $CD$.", "input_image": [ "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0095_1.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0095_2.jpg", "batch28-2024_06_17_a83d792bb4e395c3c2cfg_0095_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\frac{7}{4}$;\n\n(2) $y=-\\frac{3}{5} x+\\frac{24}{5} \\quad(0\n\nWhen \\( B D = B F \\), \\( \\angle 1 = \\angle 2 = \\angle 3 \\),\n\nIn the right triangle \\( \\triangle A B F \\), \\( \\tan \\angle 1 = \\frac{A B}{B F} = \\frac{10}{8 - x} \\),\nIn the right triangle \\( \\triangle A C D \\), \\( \\tan \\angle 3 = \\frac{A C}{C D} = \\frac{6}{x} \\),\n\nTherefore, \\( \\frac{10}{8 - x} = \\frac{6}{x} \\), solving gives \\( x = 3 \\);\n\nWhen \\( B D = D F \\), \\( \\angle 1 = \\angle 4 \\), and since \\( \\angle A B F = 90^\\circ \\),\n\nTherefore, \\( \\angle 4 + \\angle 5 = 90^\\circ \\), \\( \\angle 1 + \\angle 6 = 90^\\circ \\),\n\nThus, \\( \\angle 5 = \\angle 6 \\), hence \\( B D = A D \\),\n\nTherefore, \\( (8 - x)^2 = x^2 + 36 \\), solving gives \\( x = \\frac{7}{4} \\);\n\nIn summary: When \\( \\triangle B D F \\) is an isosceles triangle with \\( B D \\) as the leg, the length of \\( C D \\) is either 3 or \\( \\frac{7}{4} \\).\n\n【Insight】This problem examines the Pythagorean theorem, properties of isosceles triangles, properties of perpendicular bisectors, and practical applications of linear functions. The key to solving is a thorough understanding of the Pythagorean theorem, properties of isosceles triangles, properties of perpendicular bisectors, and practical applications of linear functions. In part (3), attention must be paid to discussing different cases." }, { "problem_id": 1928, "question": "There are 5 opaque cards, identical in all aspects except for the patterns on their fronts. These 5 cards are shuffled with their backs facing up and placed on the table.\n\n\nA\n\n\n$B$\n\n\nC\n\n\n$D$\n\n\nE\n\n(1) The probability of randomly drawing 1 card and it having a pattern that is a central symmetry figure is $\\qquad$.\n\n(2) If 1 card is randomly drawn and not put back, and then another card is randomly drawn, please use a tree diagram or a list to find the probability that both cards drawn are axially symmetric figures.", "input_image": [ "batch28-2024_06_17_ab42dd41464cd1add41fg_0047_1.jpg", "batch28-2024_06_17_ab42dd41464cd1add41fg_0047_2.jpg", "batch28-2024_06_17_ab42dd41464cd1add41fg_0047_3.jpg", "batch28-2024_06_17_ab42dd41464cd1add41fg_0047_4.jpg", "batch28-2024_06_17_ab42dd41464cd1add41fg_0047_5.jpg" ], "is_multi_img": true, "answer": "(1) $\\frac{2}{5}$; (2) The probability that the cards drawn twice are both axially symmetrical is $\\frac{3}{10}$.\n", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Combinatorics", "image_relavance": "0", "analysis": "Solution: (1) The cards with centrally symmetric patterns are A and D. Therefore, the probability of randomly drawing one card with a centrally symmetric pattern is $\\frac{2}{5}$. Hence, the answer is $\\frac{2}{5}$.\n\n(2) The cards with axially symmetric patterns are B, C, and E.\n\nThe tree diagram is as follows:\n\n\nB C D E\n\n\nA C D E\n\n\nA B D E\n\nA B C E\n\nFrom the tree diagram, there are a total of 20 equally possible outcomes. Among these, there are 6 outcomes where both cards drawn are axially symmetric, specifically $(B, C)$, $(B, E)$, $(C, B)$, $(C, E)$, $(E, B)$, and $(E, C)$.\n\nThus, the probability that both cards drawn are axially symmetric is $\\frac{6}{20} = \\frac{3}{10}$.\n\n[Key Insight] This problem tests the knowledge of using tree diagrams or lists to calculate the probability of two events, as well as the definitions of centrally symmetric and axially symmetric patterns. Understanding these definitions and the methods for calculating probabilities using tree diagrams or lists is crucial for solving the problem." }, { "problem_id": 1929, "question": "As shown in Figure (1), fold the corner of a notebook paper over so that the vertex $A$ coincides with $A'$, with $BC$ as the crease;\n\n(1) In Figure (1), if $\\angle 1 = 30^\\circ$, then $\\angle A'BD =$ $\\qquad$;\n\n(2) If the size of $\\angle 1$ is changed in Figure (2), the position of $BA'$ will also change. Then, fold another corner of the paper diagonally so that the edge $BD$ coincides with $BA'$, with $BE$ as the crease. Will the degree of $\\angle CBE$ change? Please explain the reason.\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch28-2024_06_17_b71aea1a7f4cafcbcc17g_0006_1.jpg", "batch28-2024_06_17_b71aea1a7f4cafcbcc17g_0006_2.jpg" ], "is_multi_img": true, "answer": "(1) $120^{\\circ}$;（2）remain unchanged, $\\angle C B E=90^{\\circ}$.\n\n", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since $\\angle 1 = 30^{\\circ}$,\n\nTherefore, $\\angle 1 = \\angle ABC = 30^{\\circ}$,\n\nHence, $\\angle A'BD = 180^{\\circ} - 30^{\\circ} - 30^{\\circ} = 120^{\\circ}$.\n\n(2) Conclusion: $\\angle CBE$ remains unchanged.\n\nSince $\\angle 1 = \\frac{1}{2} \\angle ABA'$, and $\\angle 2 = \\frac{1}{2} \\angle A'BD$, with $\\angle ABA' + \\angle A'BD = 180^{\\circ}$,\n\nTherefore, $\\angle 1 + \\angle 2 = \\frac{1}{2} \\angle ABA' + \\frac{1}{2} \\angle A'BD$\n\n$= \\frac{1}{2}(\\angle ABA' + \\angle A'BD)$\n\n$= \\frac{1}{2} \\times 180^{\\circ}$\n\n$= 90^{\\circ}$.\n\nThat is, $\\angle CBE = 90^{\\circ}$.\n\n[Highlight] This problem mainly examines the calculation of angles, with the key to solving it being the properties of folding." }, { "problem_id": 1930, "question": "As shown in Figure 1, this is a physical image of a certain basketball hoop, and Figure 2 is a side view of it. It is known that the length of the base $\\mathrm{BC}$ is 0.6 meters, the angle between the base $\\mathrm{BC}$ and the support $\\mathrm{AC}$ is $\\angle \\mathrm{ACB}=75^{\\circ}$, the points $\\mathrm{A}, \\mathrm{H}, \\mathrm{~F}$ are on the same straight line, the length of the support segment $\\mathrm{AH}$ is 0.5 meters, the length of the segment $\\mathrm{HF}$ is 1.5 meters, and the length of the bottom support of the backboard $\\mathrm{HE}$ is 0.75 meters.\n\n(1) Find the degree measure of the angle between the bottom support of the backboard $\\mathrm{HE}$ and the support $\\mathrm{AF}$;\n\n(2) Find the distance from the top of the backboard $\\mathrm{F}$ to the ground (the result should be accurate to 0.1 meters, reference data: $\\sin 75^{\\circ} \\approx 1.0, \\cos 75^{\\circ} \\approx 0.3$,\n\n$\\left.\\tan 75^{\\circ} \\approx 3.7, \\quad \\sqrt{3} \\approx 1.7, \\sqrt{2} \\approx 1.4\\right)$\n\n\n\nFigure 1\n\n", "input_image": [ "batch29-2024_06_14_68a74991cf4df7ca3078g_0048_1.jpg", "batch29-2024_06_14_68a74991cf4df7ca3078g_0048_2.jpg" ], "is_multi_img": true, "answer": "(1) $60^{\\circ}$; (2) 3.9 m.\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: (1) From the problem statement, in the right triangle $\\triangle \\mathrm{HEF}$, $\\cos \\angle \\mathrm{HEF} = \\frac{HE}{HF} = \\frac{1}{2}$,\n\n$\\therefore \\angle \\mathrm{FHE} = 60^{\\circ}$,\n\n$\\therefore$ the angle between the bottom bracket $\\mathrm{HE}$ of the basketball backboard and the bracket $\\mathrm{AF}$ is $60^{\\circ}$;\n\n(2) Extend $\\mathrm{FE}$ to meet the extension of $\\mathrm{CB}$ at point $\\mathrm{M}$, and draw $\\mathrm{AG} \\perp \\mathrm{FM}$ at point $\\mathrm{G}$.\n\n\n\nIn the right triangle $\\triangle \\mathrm{ABC}$, $\\mathrm{AB} = \\mathrm{BC} \\cdot \\tan 75^{\\circ} \\approx 0.6 \\times 3.7 = 2.22$ meters\n\n$\\therefore \\mathrm{GM} = \\mathrm{AB} \\approx 2.22$ meters\n\nIn the right triangle $\\triangle \\mathrm{AGF}$, since $\\angle \\mathrm{FAG} = \\angle \\mathrm{FHE} = 60^{\\circ}$ and $\\mathrm{AF} = \\mathrm{AH} + \\mathrm{HF} = 2$ meters\n\n$\\therefore \\mathrm{FG} = \\mathrm{AF} \\cdot \\sin 60^{\\circ} = \\sqrt{3} \\approx 1.7$ meters\n\n$\\therefore \\mathrm{FM} = \\mathrm{FG} + \\mathrm{GM} \\approx 1.7 + 2.22 = 3.92 \\approx 3.9$ meters\n\n$\\therefore$ The distance from the top $\\mathrm{F}$ of the backboard to the ground is approximately 3.9 meters.\n\n[Insight] This problem mainly examines the comprehensive application of trigonometric functions, while also utilizing the properties of rectangles. The key to solving the problem lies in identifying the relationships between the angles and sides, and using trigonometric functions to find the solution." }, { "problem_id": 1931, "question": "Figure 1 shows a real object projector, and Figure 2 is its schematic diagram. The polyline $\\mathrm{B}-\\mathrm{A}-\\mathrm{O}$ represents the fixed support, with $\\mathrm{AO}$ perpendicular to the horizontal desktop $\\mathrm{OE}$ at point $\\mathrm{O}$. Point $\\mathrm{B}$ is the rotation point, and $\\mathrm{BC}$ can rotate. When $\\mathrm{BC}$ rotates clockwise around point $\\mathrm{B}$, the projection probe $\\mathrm{CD}$ remains perpendicular to the horizontal desktop $\\mathrm{OE}$. Measurements show: $\\mathrm{AO}=6.8 \\mathrm{~cm}, \\mathrm{CD}=8 \\mathrm{~cm}, \\mathrm{AB}=30 \\mathrm{~cm}$, $\\mathrm{BC}=35 \\mathrm{~cm}$. As shown in Figure 2, $\\angle \\mathrm{ABC}=70^{\\circ}, \\mathrm{BC} / / \\mathrm{OE}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Fill in the blank: $\\angle \\mathrm{BAO}=$ $\\qquad$ degrees;\n\n(2) Find the distance from the end point $\\mathrm{D}$ of the projection probe to the desktop $\\mathrm{OE}$ (result accurate to 0.1, reference data: $\\sin 70^{\\circ} \\approx 0.94$, $\\cos 20^{\\circ} \\approx 0.94$).", "input_image": [ "batch29-2024_06_14_68a74991cf4df7ca3078g_0052_1.jpg", "batch29-2024_06_14_68a74991cf4df7ca3078g_0052_2.jpg" ], "is_multi_img": true, "answer": "(1) 160 ;（2）27\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: (1) Draw a line $\\mathrm{AG}$ parallel to $\\mathrm{BC}$ through point $\\mathrm{A}$, as shown in the figure. Then, $\\angle \\mathrm{BAG} = \\angle \\mathrm{ABC} = 70^{\\circ}$.\n\n\n\nSince $\\mathrm{BC}$ is parallel to $\\mathrm{OE}$,\n\n$\\therefore \\mathrm{AG}$ is parallel to $\\mathrm{OE}$,\n$\\therefore \\angle \\mathrm{GAO} = \\angle \\mathrm{AOE} = 90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{BAO} = 90^{\\circ} + 70^{\\circ} = 160^{\\circ}$,\n\nThus, the answer is: 160;\n\n(2) Draw a perpendicular $\\mathrm{AF}$ from point $\\mathrm{A}$ to $\\mathrm{BC}$, intersecting at point $\\mathrm{F}$, as shown in the figure.\n\n\n\nThen, $\\mathrm{AF} = \\mathrm{AB} \\cdot \\sin \\angle \\mathrm{ABF} = 30 \\sin 70^{\\circ} \\approx 28.2(\\mathrm{~cm})$,\n\nThus, the distance from the endpoint $\\mathrm{D}$ of the projection probe to the desktop $\\mathrm{OE}$ is: $\\mathrm{AF} + \\mathrm{OA} - \\mathrm{CD} \\approx 28.2 + 6.8 - 8 = 27(\\mathrm{~cm})$.\n\n【Highlight】This problem mainly examines the application of solving right triangles, fully demonstrating the close connection between mathematics and real-life situations. The key to solving the problem lies in constructing a right triangle." }, { "problem_id": 1932, "question": "As shown in the figure, Figure (1) is a physical image of a popular online slingshot, with rubber bands tied at both ends. Pulling the rubber bands can form planar schematic diagrams as shown in Figures (2) and (3). The two sides of the slingshot can be considered parallel, i.e., $A B / / C D$. Various activity groups explored the quantitative relationship between $\\angle A P D$ and $\\angle A, \\angle C$, and made the following discoveries:\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n\n\nFigure (4)\n\n(1) In the figure shown in Figure (2), if $\\angle A = 30^{\\circ}, \\angle D = 35^{\\circ}$, then $\\angle A P D =$ $\\qquad$\n(2) In Figure (8), if $\\angle A = 150^{\\circ}, \\angle A P D = 60^{\\circ}$, then $\\angle D =$ $\\qquad$ ?\n\n(3) Some students, based on Figures (2) and (3), drew the figure shown in Figure (4), where $A B / / C D$. Please determine the relationship between $\\angle \\alpha$, $\\angle \\beta$, and $\\angle \\gamma$, and explain the reason.", "input_image": [ "batch29-2024_06_14_6c52953b6fab6d80779dg_0018_1.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0018_2.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0018_3.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0018_4.jpg" ], "is_multi_img": true, "answer": "(1) $65^{\\circ}$\n\n(2) $150^{\\circ}$\n\n(3) $\\angle \\beta=180^{\\circ}-\\angle \\alpha+\\angle \\gamma$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Solution: As shown in the figure, draw line $PQ$ through point $P$ such that $PQ / / AB$,\n\n$$\n\\begin{aligned}\n& \\because AB / / CD \\\\\n& \\therefore AB / / PQ / / CD\n\\end{aligned}\n$$\n\n$\\therefore \\angle APQ=\\angle A=30^{\\circ}, \\angle DPQ=\\angle D=35^{\\circ}$,\n\n$\\therefore \\angle APD=\\angle APQ+\\angle DPQ=65^{\\circ}$,\n\nTherefore, the answer is: $65^{\\circ}$;\n\n\n\nFigure (2)\n\n(2) Solution: As shown in the figure, draw line $PQ$ through point $P$ such that $PQ / / AB$,\n\n$\\because AB / / CD$,\n\n$\\therefore AB / / PQ / / CD$,\n\n$\\therefore \\angle APQ=180^{\\circ}-\\angle A=30^{\\circ}, \\angle D=180^{\\circ}-\\angle DPQ$,\n\n$\\because \\angle APD=60^{\\circ}$,\n\n$\\therefore \\angle DPQ=\\angle APD-\\angle APQ=30^{\\circ}$\n\n$\\therefore \\angle D=150^{\\circ}$,\n\nTherefore, the answer is: $150^{\\circ}$;\n\n\n\nFigure (3)\n\n(3) Solution: $\\angle \\beta=180^{\\circ}-\\angle \\alpha+\\angle \\gamma$, the reasoning is as follows:\n\nAs shown in the figure, draw line $PQ$ through point $P$ such that $PQ / / AB$,\n\n$\\because AB / / CD$,\n\n$\\therefore AB / / PQ / / CD$,\n\n$\\therefore \\angle BPQ=180^{\\circ}-\\angle B, \\angle DPQ=\\angle D$,\n\n$\\therefore \\angle BPD=\\angle BPQ+\\angle DPQ=180^{\\circ}-\\angle B+\\angle D$,\n\n$\\therefore \\angle \\beta=180^{\\circ}-\\angle \\alpha+\\angle \\gamma$.\n\n\n\nFigure (4)\n\n【Insight】This problem mainly examines the properties of parallel lines, and correctly drawing auxiliary lines is the key to solving the problem." }, { "problem_id": 1933, "question": "Given \\( A D / / B C, \\angle B = \\angle D \\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n(1) As shown in Figure (1), are \\( A B \\) and \\( C D \\) parallel? Why?\n\n(2) As shown in Figure (2), if points \\( E \\) and \\( F \\) are on segment \\( C D \\), and \\( A C \\) bisects \\( \\angle B A E \\), \\( A F \\) bisects \\( \\angle D A E \\), and \\( \\angle D = 100^\\circ \\). Find: the measure of \\( \\angle C A F \\);\n\n(3) If under the conditions of (2), the measure of \\( \\angle D \\) is changed to \\( \\angle D = \\alpha \\), with other conditions unchanged. Find: the sum of the measures of \\( \\angle D F A \\) and \\( \\angle A C B \\). (Express in terms of \\( \\alpha \\)) Just write down the answer directly.", "input_image": [ "batch29-2024_06_14_6c52953b6fab6d80779dg_0021_1.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0021_2.jpg" ], "is_multi_img": true, "answer": "(1) Parallel \n\n(2) $40^{\\circ}$\n\n(3) $\\angle D A F+\\angle A C E=270^{\\circ}-\\frac{3}{2} \\alpha$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "(1) Solution: \\( AB \\parallel BD \\),\n\nReason: Since \\( AD \\parallel BC \\),\n\nTherefore, \\( \\angle A + \\angle B = 180^\\circ \\) (When two lines are parallel, the consecutive interior angles are supplementary),\n\nSince \\( \\angle B = \\angle D \\),\n\nTherefore, \\( \\angle A + \\angle D = 180^\\circ \\),\n\nHence, \\( AB \\parallel BD \\) (If consecutive interior angles are supplementary, then the two lines are parallel);\n\n(2) Solution: Since \\( AC \\) bisects \\( \\angle BAE \\),\n\nTherefore, \\( \\angle CAE = \\frac{1}{2} \\angle BAE \\),\n\nSimilarly: \\( \\angle EAF = \\frac{1}{2} \\angle DAE \\),\n\nTherefore, \\( \\angle CAF = \\angle CAE + \\angle EAF = \\frac{1}{2}(\\angle BAE + \\angle DAE) \\)\n\n\\( = \\frac{1}{2} \\angle DAB \\),\n\nSince \\( \\angle D = 100^\\circ \\), and \\( \\angle DAB + \\angle D = 180^\\circ \\),\n\nTherefore, \\( \\angle DAB = 80^\\circ \\).\n\nThus, \\( \\angle CAF = \\frac{1}{2} \\angle DAB = 40^\\circ \\);\n\n(3) Solution: From (2), we have \\( \\angle CAF = \\frac{1}{2} \\angle DAB \\),\n\nSince \\( \\angle D = \\alpha \\), and \\( \\angle DAB + \\angle D = 180^\\circ \\),\n\nTherefore, \\( \\angle DAB = 180^\\circ - \\alpha \\),\n\nThus, \\( \\angle CAF = \\frac{1}{2}(180^\\circ - \\alpha) = 90^\\circ - \\frac{1}{2} \\alpha \\),\n\nSince \\( AD \\parallel BC \\),\n\nTherefore, \\( \\angle ACB = \\angle DAC \\),\n\nFrom (1), we know: \\( AB \\parallel CD \\),\n\nTherefore, \\( \\angle DFA = \\angle BAF \\),\n\nThus, \\( \\angle DFA + \\angle ACB = \\angle BAF + \\angle CAD = \\angle BAF + \\angle DAF + \\angle CAF = \\angle DAB + \\angle CAF = 180^\\circ - \\alpha + 90^\\circ - \\frac{1}{2} \\alpha = 270^\\circ - \\frac{3}{2} \\alpha \\).\n\n【Insight】This question tests the properties and determination of parallel lines, the definition of angle bisectors, and the importance of mastering the determination and properties of parallel lines for solving the problem." }, { "problem_id": 1934, "question": "Given \\( AB \\parallel CD \\), point \\( M \\) is a point in the plane, and \\(\\angle AMD = 90^\\circ\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) When point \\( M \\) is in the position shown in Figure 1, determine the numerical relationship between \\(\\angle MAB\\) and \\(\\angle D\\) (provide the reasoning process);\n\n(2) When point \\( M \\) is in the position shown in Figure 2, the numerical relationship between \\(\\angle MAB\\) and \\(\\angle D\\) is \\(\\qquad\\) (directly provide the answer);\n\n(3) Under the condition of (2), as shown in Figure 3, draw \\( ME \\perp AB \\) with the foot at \\( E \\), and the angle bisectors of \\(\\angle EMA\\) and \\(\\angle EMD\\) intersect the ray \\( EB \\) at points \\( F \\) and \\( G \\) respectively. Answer the following questions (directly provide the answers): The angle in the figure that is equal to \\(\\angle MAB\\) is \\(\\qquad\\), and \\(\\angle FMG = \\qquad\\) degrees.", "input_image": [ "batch29-2024_06_14_6c52953b6fab6d80779dg_0062_1.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0062_2.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0062_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle M A B+\\angle D=90^{\\circ}$; \n\n(2) $\\angle M A B-\\angle D=90^{\\circ}$\n\n(3) $\\angle M A B=\\angle E M D ; 45$\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Solution: As shown in Figure (1), draw line \\( MN \\) parallel to \\( AB \\) through point \\( M \\).\n\n\n\nFigure (1)\n\nSince \\( AB \\parallel CD \\),\n\n\\( MN \\parallel AB \\parallel CD \\) (if a line is parallel to one of two parallel lines, it is also parallel to the other).\n\nThus, \\( \\angle D = \\angle NMD \\).\n\nSince \\( MN \\parallel AB \\),\n\n\\( \\angle MAB + \\angle NMA = 180^\\circ \\).\n\nTherefore, \\( \\angle MAB + \\angle AMD + \\angle DMN = 180^\\circ \\).\n\nSince \\( \\angle AMD = 90^\\circ \\),\n\n\\( \\angle MAB + \\angle DMN = 90^\\circ \\).\n\nHence, \\( \\angle MAB + \\angle D = 90^\\circ \\);\n\n(2) Solution: As shown in Figure (2), draw line \\( MN \\) parallel to \\( AB \\) through point \\( M \\).\n\n\n\nFigure (2)\n\nSince \\( MN \\parallel AB \\),\n\n\\( \\angle MAB + \\angle AMN = 180^\\circ \\).\n\nSince \\( AB \\parallel CD \\),\n\n\\( MN \\parallel AB \\parallel CD \\).\n\nThus, \\( \\angle D = \\angle NMD \\).\n\nSince \\( \\angle AMD = 90^\\circ \\),\n\n\\( \\angle AMN = 90^\\circ - \\angle NMD \\).\n\nTherefore, \\( \\angle AMN = 90^\\circ - \\angle D \\).\n\nHence, \\( 90^\\circ - \\angle D + \\angle MAB = 180^\\circ \\).\n\nThus, \\( \\angle MAB - \\angle D = 90^\\circ \\).\n\nThat is, the quantitative relationship between \\( \\angle MAB \\) and \\( \\angle D \\) is: \\( \\angle MAB - \\angle D = 90^\\circ \\).\n\nTherefore, the answer is: \\( \\angle MAB - \\angle D = 90^\\circ \\).\n\n(3) Solution: As shown in Figure (3),\n\n\n\nFigure (3)\n\nSince \\( ME \\perp AB \\),\n\n\\( \\angle E = 90^\\circ \\).\n\nThus, \\( \\angle MAE + \\angle AME = 90^\\circ \\).\n\nSince \\( \\angle MAB + \\angle MAE = 180^\\circ \\),\n\n\\( \\angle MAB - \\angle AME = 90^\\circ \\).\n\nThat is, \\( \\angle MAB = 90^\\circ + \\angle AME \\).\n\nSince \\( \\angle AMD = 90^\\circ \\),\n\n\\( \\angle MAB = \\angle AMD + \\angle AME = \\angle EMD \\).\n\nSince \\( MF \\) bisects \\( \\angle EMA \\),\n\n\\( \\angle FME = \\angle FMA = \\frac{1}{2} \\angle EMA \\).\n\nSince \\( MG \\) bisects \\( \\angle EMD \\),\n\n\\( \\angle EMG = \\angle GMD = \\frac{1}{2} \\angle EMD \\).\n\nSince \\( \\angle FMG = \\angle EMG - \\angle EMF \\),\n\n\\( \\angle FMG = \\frac{1}{2} \\angle EMD - \\frac{1}{2} \\angle EMA = \\frac{1}{2} (\\angle EMD - \\angle EMA) \\).\n\nSince \\( \\angle EMD - \\angle EMA = 90^\\circ \\),\n\n\\( \\angle FMG = 45^\\circ \\).\n\nTherefore, the answer is: \\( \\angle MAB = \\angle EMD; 45 \\).\n\n【Key Insight】This problem mainly examines the properties and determination of parallel lines. Drawing \\( MN \\parallel AB \\) through point \\( M \\) is the key to solving the problem." }, { "problem_id": 1935, "question": "Investigation:\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n(1) As shown in Figure (1), given that \\( AB \\parallel CD \\), what is the relationship between \\(\\angle 1\\), \\(\\angle 2\\), and \\(\\angle 3\\) in the figure?\n\n(2) As shown in Figure (2), given that \\( AB \\parallel CD \\), what is the relationship between \\(\\angle 1\\), \\(\\angle 2\\), \\(\\angle 3\\), and \\(\\angle 4\\) in the figure?\n\n(3) As shown in Figure (3), given that \\( AB \\parallel CD \\), please directly state the relationship between \\(\\angle 1\\), \\(\\angle 2\\), \\(\\angle 3\\), \\(\\angle 4\\), and \\(\\angle 5\\) in the figure.", "input_image": [ "batch29-2024_06_14_6c52953b6fab6d80779dg_0081_1.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0081_2.jpg", "batch29-2024_06_14_6c52953b6fab6d80779dg_0081_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle 1+\\angle 3=\\angle 2$;\n\n(2) $\\angle 1+\\angle 3=\\angle 2+\\angle 4$;\n\n(3) $\\angle 1+\\angle 3+\\angle 5=\\angle 2+\\angle 4$.\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Descriptive Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\n(1) As shown in Figure (1), draw a line $EM$ through point $E$ such that $EM \\parallel AB$.\n\n\n\n(1)\n\nSince $AB \\parallel CD$,\n\n$\\therefore AB \\parallel CD \\parallel EM$,\n\n$\\therefore \\angle 1 = \\angle NEM$, $\\angle 3 = \\angle MEF$,\n\n$\\therefore \\angle 1 + \\angle 3 = \\angle NEM + \\angle MEF$,\n\nThat is, $\\angle 1 + \\angle 3 = \\angle 2$;\n\n(2) As shown in Figure (2), draw a line $NF$ through point $F$ such that $NF \\parallel AB$.\n\n\n\n(2)\n\nSince $AB \\parallel CD$,\n\n$\\therefore AB \\parallel CD \\parallel FN$,\n\n$\\therefore \\angle 4 = \\angle NFH$,\n\nFrom (1), we know that $\\angle 1 + \\angle EFN = \\angle 2$,\n$\\therefore \\angle 1 + \\angle EFN + \\angle NFH = \\angle 2 + \\angle 4$,\n\nThat is, $\\angle 1 + \\angle 3 = \\angle 2 + \\angle 4$;\n\n(3) As shown in Figure (3), draw a line $GM$ through point $G$ such that $GM \\parallel AB$.\n\n\n\n(3)\n\nSince $AB \\parallel CD$,\n\n$\\therefore AB \\parallel CD \\parallel GM$,\n\n$\\therefore \\angle 5 = \\angle MGN$,\n\nFrom (2), we have $\\angle 1 + \\angle 3 = \\angle 2 + \\angle FGM$,\n\n$\\therefore \\angle 1 + \\angle 3 + \\angle 5 = \\angle 2 + \\angle FGM + \\angle MGN$,\n\nThat is, $\\angle 1 + \\angle 3 + \\angle 5 = \\angle 2 + \\angle 4$.\n\n**Key Insight:** This problem tests the properties of parallel lines. Remembering that when two lines are parallel, the alternate interior angles are equal is crucial for solving the problem." }, { "problem_id": 1936, "question": "Exploring the Pattern: We have a conclusion that can be directly applied: If two lines are parallel, then for any point taken on one of the lines, regardless of its position on the line, the distance from this point to the other line is always equal. For example: As shown in Figure 1, two lines $m / / n$, points $H$ and $T$ are on line $m$, $H E \\perp n$ at $E$, $T F \\perp n$ at $F$, then $H E = T F$.\n\nAs shown in Figure 2, given lines $m / / n$, points $A$ and $B$ are on line $n$, and points $C$ and $D$ are on line $m$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Please list the pairs of triangles with equal areas in the figure: $\\qquad$.\n\n(2) If $A$, $B$, and $C$ are fixed points, and point $D$ moves on line $m$, then no matter where point $D$ moves, there will always be: $\\qquad$ with the same area as $\\triangle ABC$; the reason is: $\\qquad$.", "input_image": [ "batch29-2024_06_14_77c8754e3487a98bf2b3g_0031_1.jpg", "batch29-2024_06_14_77c8754e3487a98bf2b3g_0031_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\triangle A B C$ and $\\triangle A B D, \\triangle D C A$ and $\\triangle D C B, \\triangle A C O$ and $\\triangle D B O$\n\n(2) $\\triangle A B D$ Two triangles with the same base and height have the same area\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) There are three pairs respectively: $\\triangle ABC$ and $\\triangle ABD$, $\\triangle DCA$ and $\\triangle DCB$, $\\triangle ACO$ and $\\triangle DBO$, analyzed as follows:\n\n$\\triangle ABC$ and $\\triangle ABD$, both triangles share the common base $AB$, and from the conclusion of Figure 1, we know their heights are equal. According to the area formula of a triangle, the areas of the two triangles are equal;\n\n$\\triangle DCA$ and $\\triangle DCB$, both triangles have $CD$ as the base, and their heights are equal, meaning their areas are equal;\n\n$\\triangle ACO$ and $\\triangle DBO$, based on the equal areas of $\\triangle DCA$ and $\\triangle DCB$, both triangles subtract $\\triangle CDO$ simultaneously, resulting in equal areas for $\\triangle ACO$ and $\\triangle DBO$.\n\nTherefore, the answer is: $\\triangle ABC$ and $\\triangle ABD$, $\\triangle DCA$ and $\\triangle DCB$, $\\triangle ACO$ and $\\triangle DBO$;\n\n(2) If points $A$, $B$, and $C$ are three fixed points, and point $D$ moves on line $m$, then no matter where point $D$ moves, the areas of $\\triangle ABD$ and $\\triangle ABC$ are always equal. The reason is: two triangles with the same base and equal heights have equal areas; analyzed as follows:\n\n$\\because \\triangle ABD$ and $\\triangle ABC$ share the same base, and point $D$ moves on line $m$, no matter where point $D$ moves, the distance from point $D$ to the other line remains equal, making these two triangles: two triangles with the same base and equal heights, thus their areas are equal. Therefore, the answer is: $\\triangle ABD$, two triangles with the same base and equal heights, have equal areas.\n\n[Highlight] This question examines the concept that the distance between parallel lines is equal everywhere. The key to solving the problem is to understand the question thoroughly and provide a comprehensive answer." }, { "problem_id": 1937, "question": "Given: As shown in the figure, the parabola \\( y = ax^2 + bx + c \\) intersects the \\( x \\)-axis at points \\( A(-2, 0) \\) and \\( B(3, 0) \\), and intersects the \\( y \\)-axis at point \\( C(0, 6) \\).\n\n(1) Write the values of \\( a, b, \\) and \\( c \\);\n\n(2) Connect \\( BC \\), and let \\( P \\) be a point on the parabola in the first quadrant. Draw \\( AD \\) perpendicular to the \\( x \\)-axis through point \\( A \\), and draw \\( PD \\) perpendicular to \\( BC \\) intersecting the line \\( AD \\) at point \\( D \\). Let the abscissa of point \\( P \\) be \\( t \\) and the length of \\( AD \\) be \\( h \\).\n\n(1) Find the functional relationship between \\( h \\) and \\( t \\) and the maximum value of \\( h \\) (specify the range of the independent variable \\( t \\));\n\n(2) Through point \\( D \\) in the second quadrant, draw \\( DE \\parallel AB \\) intersecting \\( BC \\) at point \\( E \\). If \\( DP = CE \\), find the coordinates of point \\( P \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch29-2024_06_14_7dd1ea928c3a5b25e3b4g_0027_1.jpg", "batch29-2024_06_14_7dd1ea928c3a5b25e3b4g_0027_2.jpg", "batch29-2024_06_14_7dd1ea928c3a5b25e3b4g_0027_3.jpg" ], "is_multi_img": true, "answer": "(1) $a=-1, b=1, c=6$; （2）(1) $h=\\left\\{\\begin{array}{l}-t^{2}+\\frac{1}{2} t+5\\left(0\n\nTherefore, \\( \\tan \\angle BCO = \\tan \\angle PDK = \\frac{1}{2} \\), \\( DK = t + 2 \\), \\( PK = \\frac{1}{2} DK = \\frac{1}{2}(t + 2) \\).\n\nSince \\( DK \\parallel AB \\) and \\( AD \\perp AB \\), quadrilateral \\( ADKG \\) is a rectangle.\n\nThus, \\( AD = KG \\),\n\n\\( h = AD = KG = |PG - PK| = \\left| -t^2 + t + 6 - \\frac{1}{2}(t + 2) \\right| \\).\n\nSetting \\( h = 0 \\), \\( -t^2 + \\frac{1}{2}t + 5 = 0 \\), solutions are \\( t_1 = \\frac{5}{2} \\), \\( t_2 = -2 \\) (discarded as it does not fit the context).\n\nTherefore, \\( h = \\left\\{\\begin{array}{l} -t^2 + \\frac{1}{2}t + 5 \\left(0 < t \\leq \\frac{5}{2}\\right) \\\\ t^2 - \\frac{1}{2}t - 5 \\left(\\frac{5}{2} < t < 3\\right) \\end{array}\\right. \\).\n\nWhen \\( 0 < t \\leq \\frac{5}{2} \\), \\( h = -t^2 + \\frac{1}{2}t + 5 = -\\left(t - \\frac{1}{4}\\right)^2 + \\frac{81}{16} \\).\n\nThus, when \\( t = \\frac{1}{4} \\), \\( h \\) reaches its maximum value of \\( \\frac{81}{16} \\).\n\nWhen \\( \\frac{5}{2} < t < 3 \\), \\( h = t^2 - \\frac{1}{2}t - 5 \\) has no maximum.\n\n(2) As shown in the figure, draw \\( PH \\perp AD \\) extending \\( AD \\) to point \\( H \\).\n\nSince \\( PD \\perp BC \\), \\( \\angle PHD = \\angle ECE = 90^\\circ - \\angle CMH \\).\n\n\n\nIn triangles \\( PHD \\) and \\( CNE \\),\n\n\\( \\left\\{\\begin{array}{c} \\angle CNE = \\angle PHD \\\\ \\angle HPD = \\angle NCE \\\\ DP = CE \\end{array}\\right. \\).\n\nThus, \\( \\triangle PHD \\cong \\triangle CNE \\) (AAS),\n\nTherefore, \\( PH = CN = OC - ON \\),\nSince quadrilateral \\( ADNO \\) is a rectangle,\n\n\\( CN = 6 - \\left(-t^2 + \\frac{1}{2}t + 5\\right) = t^2 - \\frac{1}{2}t + 1 \\), \\( PH = t + 2 \\),\n\nThus, \\( t + 2 = t^2 - \\frac{1}{2}t + 1 \\),\n\nSolving gives \\( t_1 = 2 \\), \\( t_2 = -\\frac{1}{2} \\) (discarded),\n\nSubstituting \\( t = 2 \\) into the parabola \\( y = -x^2 + x + 6 = 4 \\), thus point \\( P(2, 4) \\).\n\nWhen point \\( D \\) is in the third quadrant, there is no point \\( P \\) satisfying \\( DP = CE \\).\n\nTherefore, the coordinates of point \\( P \\) that meet the conditions are \\( (2, 4) \\).\n\n[Highlight] This problem tests knowledge of trigonometric functions, properties and determination of rectangles, and comprehensive properties of quadratic functions. Drawing auxiliary lines is key." }, { "problem_id": 1938, "question": "Question Presentation:\n\nAs shown in Figure 1, in a square grid with side length 1, connect the grid points $\\mathrm{A}, \\mathrm{B}$ and $\\mathrm{C}, \\mathrm{D}$. The lines $\\mathrm{AB}$ and $\\mathrm{CD}$ intersect at point $\\mathrm{P}$. Find the value of $\\tan \\angle \\mathrm{CPB}$.\n\nMethod Summary: To find the trigonometric value of an acute angle, we often need to identify (or construct) a right triangle. Observing that $\\angle \\mathrm{CPB}$ is not in a right triangle in the problem, we typically use methods such as drawing parallel lines on the grid to solve such problems. For example, connecting grid points $\\mathrm{B}$ and $\\mathrm{E}$ gives $\\mathrm{BE} \\parallel \\mathrm{CD}$, so $\\angle \\mathrm{ABE} = \\angle \\mathrm{CPB}$. By connecting $\\mathrm{AE}$, $\\angle \\mathrm{CPB}$ is transformed into the right triangle $\\triangle \\mathrm{ABE}$.\n\nProblem Solving:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Directly write the value of $\\tan \\angle CPB$ in Figure 1 as $\\qquad$;\n\n(2) As shown in Figure 2, in a square grid with side length 1, $\\mathrm{AB}$ and $\\mathrm{CD}$ intersect at point $\\mathrm{P}$. Find the value of $\\cos \\angle CPB$.", "input_image": [ "batch29-2024_06_14_7dd1ea928c3a5b25e3b4g_0056_1.jpg", "batch29-2024_06_14_7dd1ea928c3a5b25e3b4g_0056_2.jpg" ], "is_multi_img": true, "answer": "(1) 2 ;（2） $\\frac{\\sqrt{2}}{2}$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: (1) Connect grid points B and E,\n\nSince BC // DE and BC = DE,\n\nTherefore, quadrilateral BCDE is a parallelogram,\n\nThus, DC // BE,\n\nHence, ∠CPB = ∠ABE,\n\nGiven AE = √(2² + 2²) = 2√2, BE = √(1² + 1²) = √2, AB = √(1² + 3²) = √10,\n\nAE² + BE² = AB²,\n\nTherefore, triangle ABE is a right-angled triangle with ∠AEB = 90°,\n\nThus, tan ∠CPB = tan ∠ABE = AE / BE = 2√2 / √2 = 2,\n\nThe answer is: 2;\n\n(2) As shown in Figure 2, take grid point M, connect CM and DM,\n\nSince CB // AM and CB = AM,\n\nTherefore, quadrilateral ABCM is a parallelogram,\n\nThus, CM // AB,\n\nHence, ∠CPB = ∠MCD,\n\nGiven CM = √(1² + 3²) = √10, CD = √(1² + 2²) = √5, MD = √(1² + 2²) = √5,\n\nCD² + MD² = CM²,\n\nTherefore, triangle CDM is a right-angled triangle with ∠CDM = 90°,\n\nThus, cos ∠CPB = cos ∠MCD = CD / CM = √5 / √10 = √2 / 2.\n\n\n\nFigure 2\n\n[Insight] This problem examines comprehensive triangle problems, properties of parallel lines, the Pythagorean theorem and its converse, the determination and properties of right-angled triangles, and other knowledge. The key to solving the problem is to learn to use the idea of combining numbers and shapes to solve problems and to think about problems with the idea of transformation." }, { "problem_id": 1939, "question": "Every year on November 9th is our country's \"National Fire Safety Education and Publicity Day.\" To enhance the public's awareness of disaster prevention and reduction, a fire brigade conducted a fire drill. As shown in Figure 1, the ladder $AB$ mounted on the fire truck is retractable (with a maximum length of $20 \\mathrm{~m}$) and can rotate around point $B$. The distance from the bottom $B$ to the ground, $BC$, is $2 \\mathrm{~m}$. When the top of the ladder $A$ is on the straight line where the building $EF$ is located, the distance from the bottom $B$ to $EF$, $BD$, is $9 \\mathrm{~m}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) If $\\angle ABD = 53^{\\circ}$, find the length of the ladder $AB$ at this time.\n\n(2) As shown in Figure 2, if an emergency occurs $19 \\mathrm{~m}$ directly above the bottom of the building $E$, can the ladder reach the emergency without moving the fire truck? Please explain your reasoning.\n\n(Reference data: $\\sin 53^{\\circ} \\approx 0.8, \\cos 53^{\\circ} \\approx 0.6, \\tan 53^{\\circ} \\approx 1.3$)", "input_image": [ "batch29-2024_06_14_7dd1ea928c3a5b25e3b4g_0099_1.jpg", "batch29-2024_06_14_7dd1ea928c3a5b25e3b4g_0099_2.jpg" ], "is_multi_img": true, "answer": "(1) 15m\n\n(2) As long as the fire truck does not move, the ladder can reach the dangerous area", "answer_type": "multi-step", "difficulty": "Low", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "(1) Solution: In the right triangle $\\triangle ABD$, $\\angle ABD = 53^\\circ$, $BD = 9 \\mathrm{~m}$,\n\nTherefore, $AB = \\frac{BD}{\\cos 53^\\circ} \\approx \\frac{9}{0.6} = 15 \\mathrm{~m}$,\nThus, the length of the ladder $AB$ at this moment is $15 \\mathrm{~m}$;\n\n(2) Solution: Under the premise that the fire truck does not move, the ladder can reach the emergency site. The reasoning is as follows:\n\nGiven that $DE = BC = 2 \\mathrm{~m}$,\n\nSince $AE = 19 \\mathrm{~m}$,\n\nTherefore, $AD = AE - DE = 19 - 2 = 17 \\mathrm{~m}$,\n\nIn the right triangle $\\triangle ABD$, $BD = 9 \\mathrm{~m}$,\n\nThus, $AB = \\sqrt{AD^2 + BD^2} = \\sqrt{17^2 + 9^2} = \\sqrt{370} \\mathrm{~m}$,\n\nSince $\\sqrt{370} \\mathrm{~m} < 20 \\mathrm{~m}$,\n\nTherefore, under the premise that the fire truck does not move, the ladder can reach the emergency site.\n\n[Highlight] This problem examines the application of solving right triangles, and mastering the definitions of trigonometric functions for acute angles is key to solving the problem." }, { "problem_id": 1940, "question": "Definition: A triangle is called a \"quasi-right triangle\" if the difference between two of its internal angles is $90^{\\circ}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) If $\\triangle ABC$ is a \"quasi-right triangle\" and $\\angle C > 90^{\\circ}$, $\\angle A = 50^{\\circ}$, then $\\angle B =$ $\\qquad$ ${}^{\\circ}$;\n\n(2) As shown in Figure 1, in $\\triangle ABC$, $\\angle C = 90^{\\circ}$, $AB = 6$, $BC = 2$. If $D$ is a point on $AC$ and $CD = \\frac{\\sqrt{2}}{2}$, determine whether $\\triangle ABD$ is a quasi-right triangle and explain the reason;\n\n(3) As shown in Figure 2, in quadrilateral $ABCD$, $CD = CB$, $\\angle ABD = \\angle BCD$, $AB = 5$, $BD = 8$, and $\\triangle ABC$ is a \"quasi-right triangle\". Find the area of $\\triangle BCD$.", "input_image": [ "batch29-2024_06_14_8102b08e9c068dccb7d9g_0019_1.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0019_2.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0019_3.jpg" ], "is_multi_img": true, "answer": "(1) 20\n\n(2) $\\triangle A B D$ is a \"quasi-right triangle\" (3) 48 or 24\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "**Solution:**\n\n(1) Since triangle \\( ABC \\) is a \"quasi-right triangle\" with \\( \\angle C > 90^\\circ \\) and \\( \\angle A = 50^\\circ \\):\n\n- **Case 1:** When \\( \\angle C - \\angle A = 90^\\circ \\), then \\( \\angle C = 140^\\circ \\). \n Therefore, \\( \\angle B = 180^\\circ - 140^\\circ - 50^\\circ = -10^\\circ \\) (which is invalid and discarded).\n\n- **Case 2:** When \\( \\angle C - \\angle B = 90^\\circ \\), then \\( \\angle C = \\angle B + 90^\\circ \\). \n Since \\( \\angle A + \\angle C + \\angle B = 180^\\circ \\), \n \\( 50^\\circ + 90^\\circ + 2\\angle B = 180^\\circ \\), \n thus \\( \\angle B = 20^\\circ \\).\n\n**Answer:** \\( 20 \\).\n\n---\n\n(2) **Explanation:** Triangle \\( ABD \\) is a quasi-right triangle for the following reasons:\n\n- In triangle \\( ABC \\), \\( \\angle C = 90^\\circ \\), \\( AB = 6 \\), and \\( BC = 2 \\). \n Therefore, \\( AC = \\sqrt{6^2 - 2^2} = 4\\sqrt{2} \\), \\( \\sin A = \\frac{BC}{AB} = \\frac{1}{3} \\), and \\( \\tan A = \\frac{BC}{AC} = \\frac{2}{4\\sqrt{2}} = \\frac{\\sqrt{2}}{4} \\).\n\n- Since \\( CD = \\frac{\\sqrt{2}}{2} \\), \\( AD = AC - CD = \\frac{7\\sqrt{2}}{2} \\).\n\n- Draw \\( DE \\perp AB \\) at point \\( E \\), so \\( \\angle DEA = 90^\\circ \\), \\( \\sin A = \\frac{DE}{AD} = \\frac{1}{3} \\), and \\( \\tan A = \\frac{DE}{AE} = \\frac{\\sqrt{2}}{4} \\).\n\n- Thus, \\( DE = \\frac{7\\sqrt{2}}{6} \\) and \\( AE = \\frac{14}{3} \\).\n\n- Therefore, \\( \\frac{DE}{CD} = \\frac{AE}{BC} = \\frac{7}{3} \\).\n\n- Since \\( \\angle C = \\angle DEA = 90^\\circ \\), triangles \\( BCD \\) and \\( AED \\) are similar. \n Hence, \\( \\angle A = \\angle CBD \\).\n\n- Since \\( \\angle BDA = \\angle CBD + \\angle C = \\angle A + \\angle C = \\angle A + 90^\\circ \\), \n \\( \\angle BDA - \\angle A = 90^\\circ \\), \n so triangle \\( ABD \\) is a quasi-right triangle.\n\n---\n\n(3) **Solution:** As shown in the figure, draw \\( CF \\perp BD \\) at \\( F \\) and \\( CE \\perp AB \\), intersecting the extension of \\( AB \\) at \\( E \\).\n\n- Let \\( \\angle ABD = \\angle BCD = 2\\alpha \\).\n\n- Since \\( BC = CD \\) and \\( CF \\perp BD \\), \n \\( \\angle CBD = \\angle CDB = 90^\\circ - \\alpha \\), and \\( BF = DF = \\frac{1}{2}BD = 4 \\).\n\n- Therefore, \\( \\angle CBE = 180^\\circ - 2\\alpha - (90^\\circ - \\alpha) = 90^\\circ - \\alpha = \\angle CBD \\).\n\n- Since \\( CF \\perp BD \\) and \\( CE \\perp AB \\), \\( CE = CF \\).\n\n- Since \\( BC = BC \\), right triangles \\( BCE \\) and \\( BCF \\) are congruent (HL). \n Thus, \\( BE = BF = 4 \\), and \\( AE = AB + BE = 9 \\).\n\n- **Case 1:** When \\( \\angle ABC - \\angle ACB = 90^\\circ \\), \n since \\( \\angle ABC - \\angle AEC = \\angle BCE \\), \n \\( \\angle BCA = \\angle BCE \\).\n\n- Draw \\( BG \\perp AC \\) at \\( G \\), so \\( BG = BE \\).\n\n- Therefore, \\( \\sin \\angle CAB = \\frac{CE}{AC} = \\frac{BG}{AB} = \\frac{BE}{AB} = \\frac{4}{5} \\).\n\n- Let \\( AC = 5a \\) and \\( CE = 4a \\), so \\( AE = 3a = 9 \\). \n Thus, \\( a = 3 \\), \\( CE = 12 = CF \\), \n and the area of triangle \\( BCD \\) is \\( \\frac{1}{2} \\times 12 \\times 8 = 48 \\).\n\n- **Case 2:** When \\( \\angle ABC - \\angle BAC = 90^\\circ \\), \n since \\( \\angle ABC - \\angle AEC = \\angle BCE \\), \n \\( \\angle BAC = \\angle BCE \\).\n\n- Since \\( \\angle E = \\angle E = 90^\\circ \\), triangles \\( BCE \\) and \\( CAE \\) are similar. \n Thus, \\( \\frac{CE}{AE} = \\frac{BE}{CE} \\), so \\( CE = 6 \\), \n and the area of triangle \\( BCD \\) is \\( \\frac{1}{2} \\times 6 \\times 8 = 24 \\).\n\n**Conclusion:** The area of triangle \\( BCD \\) is either 48 or 24.\n\n---\n\n**Key Insight:** This problem tests the understanding of the triangle angle sum theorem, exterior angle properties, congruence and similarity of triangles, and angle bisector properties. A strong grasp of the definition of a quasi-right triangle is crucial for solving the problem." }, { "problem_id": 1941, "question": "As shown in Figure 1, \\( A M \\parallel B C, A B \\perp B C, B C = 6, A B = 8 \\). \\( D \\) is a moving point on the ray \\( A M \\), connecting \\( C D \\), \\( E \\) is the midpoint of segment \\( C D \\), connecting \\( B E \\), and through point \\( E \\), draw \\( E F \\perp B E \\), intersecting the extension of \\( B C \\) at point \\( F \\). Let \\( A D = x \\) and \\( C F = y \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the function expression of \\( y \\) with respect to \\( x \\);\n\n(2) As shown in Figure 2, \\( N \\) is the midpoint of \\( A D \\), connecting \\( N E \\). If \\( \\triangle D E N \\) is similar to \\( \\triangle F E C \\), find the length of \\( A D \\);\n\n(3) If \\( \\triangle B C E \\) is an isosceles triangle, directly state the tangent value of \\( \\angle F \\).", "input_image": [ "batch29-2024_06_14_8102b08e9c068dccb7d9g_0026_1.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0026_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=\\frac{x^{2}+28}{12+2 x}$\n\n(2) $A D=5+5 \\sqrt{5}$ or $\\frac{14}{3}$\n\n(3) $\\tan F=\\frac{\\sqrt{5}}{2}$ or $\\frac{3}{4}$\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "**Solution:**\n\n1. **Extend** \\( FE \\) to intersect \\( AD \\) at \\( H \\), and connect \\( BH \\).\n\n \n\n Since \\( AM \\parallel BC \\) and \\( E \\) is the midpoint of \\( CD \\), in triangles \\( \\triangle FCE \\) and \\( \\triangle HDE \\):\n\n \\[\n \\begin{cases}\n \\angle FCE = \\angle HDE \\\\\n CE = DE \\\\\n \\angle FEC = \\angle HED\n \\end{cases}\n \\]\n\n Therefore, \\( \\triangle FCE \\cong \\triangle HDE \\) (ASA).\n\n Hence, \\( DH = CF = y \\), and \\( EF = EH \\).\n\n Since \\( AD = x \\), then \\( AH = x - y \\).\n\n Given \\( EF \\perp BE \\) and \\( EF = EH \\), it follows that \\( BH = BF = 6 + y \\).\n\n In right triangle \\( \\triangle ABH \\), by the Pythagorean theorem:\n\n \\[\n BH^2 = AH^2 + AB^2\n \\]\n\n Substituting the known values:\n\n \\[\n (6 + y)^2 = (x - y)^2 + 64\n \\]\n\n Solving for \\( y \\):\n\n \\[\n y = \\frac{x^2 + 28}{12 + 2x}\n \\]\n\n2. **Draw** \\( DG \\perp BC \\) at point \\( G \\), making quadrilateral \\( ADGB \\) a rectangle.\n\n \n\n Therefore, \\( AD = BG = x \\), and \\( DG = AB = 8 \\).\n\n Hence, \\( CG = BC - BG = 6 - x \\).\n\n By the Pythagorean theorem in \\( \\triangle CDG \\):\n\n \\[\n CD^2 = CG^2 + DG^2 = (6 - x)^2 + 8^2 = x^2 - 12x + 100\n \\]\n\n **Case 1:** When \\( \\triangle DEN \\sim \\triangle CEF \\), and \\( E \\) is the midpoint of \\( CD \\):\n\n \n\n Therefore, \\( \\angle NDE = \\angle FCE \\), and \\( CE = DE \\).\n\n \\[\n \\begin{cases}\n \\angle NDE = \\angle FCE \\\\\n CE = DE \\\\\n \\angle DEN = \\angle CEF\n \\end{cases}\n \\]\n\n Hence, \\( \\triangle DEN \\cong \\triangle CEF \\) (ASA).\n\n Therefore, \\( DN = CF = y \\).\n\n Since \\( N \\) is the midpoint of \\( AD \\), \\( AN = DN = \\frac{1}{2}x \\), implying \\( y = \\frac{1}{2}x \\).\n\n Substituting into the earlier equation:\n\n \\[\n \\frac{x^2 + 28}{12 + 2x} = \\frac{1}{2}x\n \\]\n\n Solving for \\( x \\):\n\n \\[\n x = \\frac{14}{3}\n \\]\n\n Verification confirms \\( x = \\frac{14}{3} \\) is a valid solution.\n\n Therefore, \\( AD = \\frac{14}{3} \\).\n\n **Case 2:** When \\( \\triangle DEN \\sim \\triangle CFE \\), then:\n\n \\[\n \\frac{DE}{CF} = \\frac{DN}{CE}\n \\]\n\n \n\n Therefore, \\( DE \\cdot CE = CF \\cdot DN \\).\n\n Since \\( E \\) is the midpoint of \\( CD \\), \\( DE = CE = \\frac{1}{2}CD \\).\n\n Hence:\n\n \\[\n \\frac{1}{4}CD^2 = \\frac{1}{2}xy\n \\]\n\n Substituting \\( CD^2 = x^2 - 12x + 100 \\):\n\n \\[\n \\frac{1}{4}(x^2 - 12x + 100) = \\frac{1}{2}x \\cdot \\frac{x^2 + 28}{2x + 12}\n \\]\n\n Solving the equation yields:\n\n \\[\n x_1 = 5 - 5\\sqrt{5}, \\quad x_2 = 5 + 5\\sqrt{5}\n \\]\n\n Verification confirms both solutions are valid.\n\n Since \\( AD > 0 \\), \\( AD = 5 + 5\\sqrt{5} \\).\n\n **Conclusion:** \\( AD = 5 + 5\\sqrt{5} \\) or \\( \\frac{14}{3} \\).\n\n3. **Draw** \\( PQ \\parallel AB \\) intersecting \\( AM \\) and \\( BF \\) at \\( P \\) and \\( Q \\), respectively.\n\n \n\n In triangles \\( \\triangle DPE \\) and \\( \\triangle CQE \\):\n\n \\[\n \\begin{cases}\n \\angle DPE = \\angle CQE \\\\\n \\angle DEP = \\angle CEQ \\\\\n CE = DE\n \\end{cases}\n \\]\n\n Therefore, \\( \\triangle DPE \\cong \\triangle CQE \\) (AAS).\n\n Hence, \\( PE = QE = 4 \\), and \\( DP = CQ \\).\n\n **Case 1:** When \\( BC = BE \\):\n\n Therefore, \\( BQ = \\sqrt{BE^2 - QE^2} = 2\\sqrt{5} \\).\n\n Hence, \\( DP = BQ - AD = 2\\sqrt{5} - x \\), and \\( CQ = DP = 2\\sqrt{5} - x \\).\n\n Since \\( BC = BQ + CQ \\):\n\n \\[\n 6 = 2\\sqrt{5} + 2\\sqrt{5} - x\n \\]\n\n Solving for \\( x \\):\n\n \\[\n x = 4\\sqrt{5} - 6\n \\]\n\n Therefore, \\( y = \\frac{18\\sqrt{5}}{5} - 6 \\).\n\n Hence, \\( FQ = \\frac{8\\sqrt{5}}{5} \\).\n\n Therefore, \\( \\tan F = \\frac{QE}{FQ} = \\frac{\\sqrt{5}}{2} \\).\n\n **Case 2:** When \\( BC = CE \\):\n\n \\( CQ = \\sqrt{CE^2 - QE^2} = 2\\sqrt{5} \\).\n\n Hence, \\( DP = 2\\sqrt{5} \\).\n\n Therefore, \\( BQ = AP = x + 2\\sqrt{5} \\).\n\n Since \\( BC = BQ + CQ \\):\n\n \\[\n 6 = x + 2\\sqrt{5} + 2\\sqrt{5}\n \\]\n\n Solving for \\( x \\):\n\n \\[\n x = 6 - 4\\sqrt{5}\n \\]\n\n Since \\( 6 - 4\\sqrt{5} < 0 \\), this case is invalid.\n\n **Case 3:** When \\( BE = CE \\):\n\n Since \\( PQ \\perp BF \\), \\( Q \\) is the midpoint of \\( BC \\).\n\n Therefore, \\( BQ = CQ = 3 \\).\n\n Hence, \\( DP = CQ = 3 \\).\n\n Therefore, \\( AP = AD + DP = x + 3 \\).\n\n Since \\( BC = BQ + CQ \\):\n\n \\[\n 6 = x + 3 + 3\n \\]\n\n Solving for \\( x \\):\n\n \\[\n x = 0\n \\]\n\n Therefore, \\( y = \\frac{7}{3} \\).\n\n Hence, \\( FQ = CQ + CF = \\frac{16}{3} \\).\n\n Therefore, \\( \\tan F = \\frac{QE}{FQ} = \\frac{3}{4} \\).\n\n **Conclusion:** \\( \\tan F = \\frac{\\sqrt{5}}{2} \\) or \\( \\frac{3}{4} \\).\n\n**Key Insight:** This problem tests the understanding of congruent and similar triangles, the Pythagorean theorem, properties of isosceles triangles, trigonometric functions, and solving equations derived from geometric relationships. Mastery of these concepts is essential for solving complex geometric problems." }, { "problem_id": 1942, "question": "As shown in the figure, in the rectangular paper $ABCD$, $AB=8$. The paper is folded such that the vertex $B$ coincides with point $E$ on side $AD$, and one end of the crease, point $G$, is on side $BC$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) In Figure 1, when the other end of the crease, $F$, is on side $AB$ and $AF=\\frac{8}{3}$, then $\\angle BGE=$ $\\qquad$ ;\n\n(2) In Figure 2, when the other end of the crease, $F$, is on side $AD$ and point $E$ coincides with point $D$, determine whether $\\triangle FHD$ and $\\triangle DCG$ are congruent. Please explain the reason.\n\n(3) If $BG=10$, when the other end of the crease, $F$, is on side $AD$, point $E$ does not coincide with any point on side $AD$, and the distance from point $E$ to side $AD$ is 2, directly write the length of $AF$.", "input_image": [ "batch29-2024_06_14_8102b08e9c068dccb7d9g_0039_1.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0039_2.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0039_3.jpg" ], "is_multi_img": true, "answer": "(1)60\n\n(2)congruent \n\n(3) $\\frac{22}{3}$ or 2\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "0", "analysis": "(1) Solution: From the properties of folding, we have: \\( EF = BF \\), \\( \\angle FEG = \\angle B = 90^\\circ \\).\n\nSince \\( AF = \\frac{8}{3} \\) and \\( AB = 8 \\),\n\nTherefore, \\( EF = BF = AB - AF = \\frac{16}{3} \\).\n\nSince quadrilateral \\( ABCD \\) is a rectangle,\n\nThus, \\( \\angle A = \\angle B = 90^\\circ \\), \\( AD \\parallel BC \\),\n\nTherefore, \\( \\angle AEG + \\angle BGE = 180^\\circ \\).\n\nIn right triangle \\( \\triangle AEF \\), \\( \\sin \\angle AEF = \\frac{AF}{EF} = \\frac{1}{2} \\),\n\nThus, \\( \\angle AEF = 30^\\circ \\),\n\nTherefore, \\( \\angle AEG = 120^\\circ \\),\n\nHence, \\( \\angle BGE = 60^\\circ \\);\n\nThe answer is: \\( 60^\\circ \\).\n\n(2) Solution: The triangles are congruent.\n\nProof: Since quadrilateral \\( ABCD \\) is a rectangle,\n\nThus, \\( \\angle A = \\angle B = \\angle C = \\angle ADC = 90^\\circ \\), \\( AB = CD \\).\n\nFrom the problem statement: \\( \\angle A = \\angle H = \\angle B = \\angle HDG = 90^\\circ \\), \\( CD = AB = HD \\),\n\nTherefore, \\( \\angle H = \\angle C = 90^\\circ \\), \\( \\angle HDF = \\angle GDC \\).\n\nIn triangles \\( \\triangle FHD \\) and \\( \\triangle DCG \\),\n\n\\[\n\\left\\{\n\\begin{aligned}\n\\angle H &= \\angle C \\\\\nHD &= CD \\\\\n\\angle HDF &= \\angle CDG\n\\end{aligned}\n\\right.\n\\]\n\nThus, \\( \\triangle FDH \\cong \\triangle DGC \\) (ASA).\n\n(3) Solution: As shown in the figure, when point \\( E \\) is below \\( AD \\), let \\( EH \\) intersect \\( AD \\) at point \\( K \\). Draw \\( MN \\parallel CD \\) through point \\( E \\), intersecting \\( AD \\) and \\( BC \\) at \\( M \\) and \\( N \\) respectively.\n\nSince the distance from \\( E \\) to \\( AD \\) is \\( 2 \\) cm,\n\nThus, \\( EM = 2 \\), \\( EN = 8 - 2 = 6 \\).\n\nIn right triangle \\( \\triangle ENG \\), \\( GN = \\sqrt{EG^2 - EN^2} = \\sqrt{10^2 - 6^2} = 8 \\).\n\nSince \\( \\angle GEN + \\angle KEM = 180^\\circ - \\angle GEH = 180^\\circ - 90^\\circ = 90^\\circ \\),\n\nAnd \\( \\angle GEN + \\angle NGE = 180^\\circ - 90^\\circ = 90^\\circ \\),\n\nTherefore, \\( \\angle KEM = \\angle NGE \\).\n\nAlso, since \\( \\angle ENG = \\angle KME = 90^\\circ \\),\n\nThus, \\( \\triangle GEN \\sim \\triangle EKM \\),\n\nTherefore, \\( \\frac{EK}{EG} = \\frac{KM}{EN} = \\frac{EM}{GN} \\),\n\nThat is, \\( \\frac{EK}{10} = \\frac{KM}{6} = \\frac{2}{8} \\),\n\nSolving gives \\( EK = \\frac{5}{2} \\), \\( KM = \\frac{3}{2} \\),\n\nThus, \\( KH = EH - EK = 8 - \\frac{5}{2} = \\frac{11}{2} \\).\n\nSince \\( \\angle FKH = \\angle EKM \\), \\( \\angle H = \\angle EMK = 90^\\circ \\),\n\nThus, \\( \\triangle FKH \\sim \\triangle EKM \\),\n\nTherefore, \\( \\frac{FH}{EM} = \\frac{KH}{KM} \\),\n\nThat is, \\( \\frac{FH}{2} = \\frac{\\frac{11}{2}}{\\frac{3}{2}} \\),\n\nSolving gives \\( FH = \\frac{22}{3} \\),\n\nThus, \\( AF = FH = \\frac{22}{3} \\).\n\nAs shown in the figure, when \\( EG \\parallel CD \\), point \\( E \\) is above \\( AD \\), then \\( MG = CD = AB = 8 \\), \\( EM \\perp AD \\).\n\nSince \\( EG = BG = 10 \\),\n\nThus, \\( EM = 2 \\),\n\nAt this time, the distance from \\( E \\) to \\( AD \\) is 2, which fits the problem statement.\n\nAccording to the problem statement: \\( \\angle H = \\angle E = \\angle EMF = 90^\\circ \\),\n\nThus, quadrilateral \\( EHFM \\) is a rectangle,\n\nTherefore, \\( FH = EM = 2 \\),\n\nThus, \\( AF = HF = 2 \\);\n\nIn summary, \\( AF = \\frac{22}{3} \\) or 2.\n\n[Key Insight] This problem examines the properties of folding transformations, the application of the Pythagorean theorem, and the determination and properties of similar triangles. The key to solving the problem lies in remembering that the segments and angles that coincide before and after folding are equal. The difficulty of this problem lies in part (3), where auxiliary lines are drawn to construct similar triangles." }, { "problem_id": 1943, "question": "A rectangular paper piece $OABC$ is placed in the plane rectangular coordinate system, with points $O(0,0)$, $A(3,0)$, and $C(0,6)$.\n\nPoint $P$ is on the side $OC$ of the rectangle. The paper is folded such that the crease line passes through point $P$ and intersects the positive $x$-axis at point $Q$, with $\\angle OPQ = 30^\\circ$. The corresponding point $O'$ of $O$ falls in the first quadrant. Let $O'Q = t$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nSpare Figure\n\n(1) In Figure (1), when $t = 1$, find the size of $\\angle O'QA$ and the coordinates of point $O'$.\n\n(2) In Figure (2), if the overlapping part after folding is a quadrilateral, with $O'Q$ and $O'P$ intersecting side $AB$ at points $E$ and $F$ respectively, express the area $S$ of the overlapping part in terms of $t$, and state the range of values for $t$.\n\n(3) ① When the crease $PQ$ exactly passes through point $A$, find the area of the overlapping part after folding.\n\n(2) When point $P$ coincides with point $C$, find the area of the overlapping part after folding. (Just write down the answer directly)", "input_image": [ "batch29-2024_06_14_8102b08e9c068dccb7d9g_0074_1.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0074_2.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0074_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle O^{\\prime} Q A=60^{\\circ}, O^{\\prime}\\left(\\frac{3}{2}, \\frac{\\sqrt{3}}{2}\\right)$;\n\n(2) $S=-\\sqrt{3} t^{2}+6 \\sqrt{3} t-6 \\sqrt{3}(2\n\n(2) As shown in the figure, from the problem statement,\n\n$OA = 3$, $QO = QO' = t$,\n\n$\\therefore QA = OA - OQ = 3 - t$,\n\nFrom (1), we know that $\\angle O'QA = 60^\\circ$, $\\angle O' = \\angle QAE = 90^\\circ$,\n\n$\\therefore \\angle QEA = \\angle O'EF = 30^\\circ$,\n\n$\\therefore QE = 2QA = 2(3 - t) = 6 - 2t$,\n\n$\\therefore O'E = O'Q - QE = t - 2(3 - t) = 3(t - 2)$,\n\n$\\because \\frac{O'F}{O'E} = \\tan 30^\\circ$,\n\n$\\therefore O'F = \\tan 30^\\circ O'E = \\frac{\\sqrt{3}}{3}(3t - 6) = \\sqrt{3}(t - 2)$,\n\nSimilarly, we can get $O'P = \\frac{O'Q}{\\tan 30^\\circ} = \\sqrt{3}t$,\n\nWhen $O'$ is on $AB$, $O'E = 3(t - 2) = 0$, i.e., $t = 2$,\n\nWhen $Q$ is on $AB$, $OQ = OA$, i.e., $t = 3$,\n\nWhen $2 < t < 3$,\n\n$S = S_{\\triangle PQO'} - S_{\\triangle EFO'} = \\frac{1}{2} QO' \\cdot PO' - \\frac{1}{2} EO' \\cdot FO'$\n\n$= \\frac{1}{2} \\times t \\times \\sqrt{3}t - \\frac{1}{2} \\times 3(t - 2) \\times \\sqrt{3}(t - 2)$\n\n$= -\\sqrt{3}t^2 + 6\\sqrt{3}t - 6\\sqrt{3}$\n\n$\\therefore S = -\\sqrt{3}t^2 + 6\\sqrt{3}t - 6\\sqrt{3} (2 < t < 3);$\n\n\n\n(3) From (2), when the crease $PQ$ just passes point $A$, $OQ = OA$, i.e., $t = 3$,\n\n$S = -\\sqrt{3} \\times 9 + 6\\sqrt{3} \\times 3 - 6\\sqrt{3} = 3\\sqrt{3}$,\n\nTherefore, the answer is: $3\\sqrt{3}$;\n\n(2) As shown in the figure, from the problem statement,\n\n$$\n\\angle OPQ = \\angle O'PQ = \\angle PED = 30^\\circ, \\quad PB = 3,\n$$\n\n$\\therefore \\angle BPD = 30^\\circ$, $PD = ED$,\n\n$\\because \\cos \\angle BPD = \\frac{PB}{PD}$,\n\n$\\therefore PD = \\frac{PB}{\\cos \\angle BPD} = \\frac{3}{\\frac{\\sqrt{3}}{2}} = 2\\sqrt{3}$,\n\n$PD = ED = 2\\sqrt{3}$,\n\n$S = \\frac{1}{2} DE \\times PB = \\frac{1}{2} \\times 2\\sqrt{3} \\times 3 = 3\\sqrt{3}$,\n\nTherefore, the answer is: $3\\sqrt{3}$.\n\n\n\n【Highlight】This problem examines the folding problem related to rectangles, the Pythagorean theorem, solving right triangles, and the properties of right triangles with a $30^\\circ$ angle; the key to solving the problem is to master the properties of folding and the use of trigonometric functions to solve right triangles." }, { "problem_id": 1944, "question": "## Problem Statement\n\n(1) As shown in Figure (1), $O P$ is the angle bisector of $\\angle A O B$, $P C \\perp O A$ at point $C$, and $P D \\perp O B$ at point $D$. If $S_{\\triangle O P C}=3$, then $S_{\\triangle O P D}=$ $\\qquad$.\n\n## Problem Exploration\n\n(2) As shown in Figure (2), $a$ and $b$ are two parallel lines, with a distance of 12 between them. Point $A$ lies on line $a$, and points $B$ and $C$ lie on line $b$, with $B$ to the left of $C$. If $\\angle B A C=45^{\\circ}$, find the minimum value of $B C$.\n\n## Problem Solving\n\n(3) As shown in Figure (3), quadrilateral $A B C D$ is a schematic representation of a parallelogram garden that the landscaping bureau plans to build. A pedestrian path is to be constructed along diagonal $B D$, and a garden irrigation channel is to be built along the angle bisector $A P$ of $\\angle B A D$ (with point $P$ on $B D$). According to the planning requirements, $\\angle A B C=120^{\\circ}$, $A P=120$ meters, and the area of the parallelogram $A B C D$ should be as small as possible. Does the area of the parallelogram $A B C D$ have a minimum value? If so, find this minimum value; if not, explain why.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch29-2024_06_14_8102b08e9c068dccb7d9g_0094_1.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0094_2.jpg", "batch29-2024_06_14_8102b08e9c068dccb7d9g_0094_3.jpg" ], "is_multi_img": true, "answer": "(1) 3 ；（2） $24 \\sqrt{2}-24$ ；（3）exists, $9600 \\sqrt{3}$ square meters\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "Solution: (1) Since \\( OP \\) is the angle bisector of \\( \\angle AOB \\), and \\( PC \\perp OA \\), \\( PD \\perp OB \\),\n\n\\[\n\\therefore PC = PD,\n\\]\n\n\\[\n\\because OP = OP,\n\\]\n\n\\[\n\\therefore \\text{Rt} \\triangle OPD \\cong \\text{Rt} \\triangle OPC (\\text{HL}),\n\\]\n\n\\[\n\\therefore S_{\\triangle OPD} = S_{\\triangle OPC} = 3,\n\\]\n\nHence, the answer is: 3;\n\n(2) Construct the circumcircle \\( O \\) of \\( \\triangle ABC \\), connect \\( OA \\), \\( OB \\), \\( OC \\), and draw \\( OG \\perp BC \\) at point \\( D \\),\n\n\\[\n\\because \\angle BAC = 45^\\circ, \\text{ then } \\angle BOC = 90^\\circ, \\text{ making } \\triangle BOC \\text{ an isosceles right triangle},\n\\]\n\nLet \\( OD = m \\), then \\( BC = 2m \\), \\( CO = \\sqrt{2}m = OA \\),\n\n\\[\n\\text{Then } AO + OD = (\\sqrt{2} + 1)m,\n\\]\n\nWhen \\( (\\sqrt{2} + 1)m \\) is minimized, \\( m \\) is minimized, and \\( BC \\) is minimized,\n\nWhen points \\( A \\), \\( O \\), \\( D \\) are collinear, \\( (\\sqrt{2} + 1)m \\) is minimized to 12,\n\n\\[\n\\text{Then the minimum value of } m \\text{ is: } \\frac{12}{\\sqrt{2} + 1} = 12(\\sqrt{2} - 1),\n\\]\n\n\\[\n\\text{Thus, the minimum value of } BC \\text{ is } 2m = 24\\sqrt{2} - 24;\n\\]\n\n(3) Existence, reasoning:\n\n\\[\n\\because BD \\text{ is the diagonal of parallelogram } ABCD, \\text{ thus to minimize the area of parallelogram } ABCD, \\text{ it suffices to minimize the area of } \\triangle ABD,\n\\]\n\n\\[\n\\because \\angle ABC = 120^\\circ, \\text{ then } \\angle BAD = 180^\\circ - \\angle ABC = 60^\\circ, \\text{ draw } PE \\perp AB \\text{ at } E, \\text{ and } PF \\perp AD \\text{ at } F,\n\\]\n\n\\[\n\\because AP \\text{ is the angle bisector of } \\angle BAD, \\text{ then } PE = PF,\n\\]\n\n\\[\n\\text{Also, } AP = AP, \\quad \\angle AEP = \\angle AFP = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\text{Rt} \\triangle PAE \\cong \\text{Rt} \\triangle PQF (\\text{HL}),\n\\]\n\n\\[\n\\because AP = 120, \\angle PAE = \\frac{1}{2} \\angle BAD = 30^\\circ, \\angle AEP = 90^\\circ,\n\\]\n\n\\[\n\\therefore PE = \\frac{1}{2} AE = 60, AE = 60\\sqrt{3},\n\\]\n\n\\[\n\\text{Then } S_{\\triangle APE} = \\frac{1}{2} \\times AE \\times PE = 1800\\sqrt{3},\n\\]\n\nOn \\( AE \\), cut off \\( EG = DF \\), connect \\( PG \\), as shown,\n\n\\[\n\\because DF = GE, \\quad \\angle PFD = \\angle PEG = 90^\\circ, \\quad PF = PE,\n\\]\n\n\\[\n\\therefore \\triangle PDF \\cong \\triangle PGE (\\text{SAS}),\n\\]\n\n\\[\n\\therefore S_{\\triangle BAD} = S_{\\triangle APF} + S_{\\triangle APE} + S_{\\triangle PDF} + S_{\\triangle PBE} = 2S_{\\triangle APE} + S_{\\triangle PBG} = 3600\\sqrt{3} + S_{\\triangle PBG},\n\\]\n\n\\[\n\\therefore \\text{To minimize the area of } \\triangle ABD, \\text{ it suffices to minimize the area of } \\triangle PBG,\n\\]\n\n\\[\n\\because \\angle EPF = 360^\\circ - 90^\\circ - 90^\\circ - 60^\\circ = 120^\\circ,\n\\]\n\n\\[\n\\therefore \\angle BPG = 180^\\circ - \\angle EPF = 60^\\circ,\n\\]\n\nConstruct the circumcircle \\( O \\) of \\( \\triangle BPG \\), connect \\( OP \\), \\( OB \\), \\( OG \\), and draw \\( OH \\perp BG \\) at \\( H \\),\n\n\\[\n\\text{Then } \\angle BOH = \\angle GOH = 60^\\circ,\n\\]\n\nLet \\( OP = OB = OG = R \\), then \\( OH = \\frac{1}{2}R \\),\n\n\\[\n\\because OP + HO \\geq PE \\text{ gives: } R + \\frac{1}{2}R \\geq 60, \\text{ i.e., } R \\geq 40,\n\\]\n\n\\[\n\\therefore GH = \\frac{\\sqrt{3}}{2}R \\geq 20\\sqrt{3},\n\\]\n\n\\[\n\\text{Then } BG \\geq 40\\sqrt{3},\n\\]\n\n\\[\n\\text{Then } S_{\\triangle BPG} = \\frac{1}{2} \\times BG \\times PE \\geq \\frac{1}{2} \\times 40\\sqrt{3} \\times 60 = 1200\\sqrt{3},\n\\]\n\n\\[\n\\therefore S_{\\text{quadrilateral } ABCD} = 2S_{\\triangle ABD} \\geq 7200\\sqrt{3} + 2400\\sqrt{3} = 9600\\sqrt{3},\n\\]\n\nHence, the minimum area of quadrilateral \\( ABCD \\) exists and is: \\( 9600\\sqrt{3} \\) square meters.\n\n【Insight】This problem examines comprehensive quadrilateral topics, involving solving right triangles, triangle congruence, triangles and their circumcircles, and properties of parallelograms. Mastering and applying these concepts comprehensively is key to solving the problem." }, { "problem_id": 1945, "question": "For a segment \\( AB \\) and a point \\( M \\) in the Cartesian coordinate system \\( xOy \\), a definition is given: If \\( M \\) satisfies \\( MA = MB \\), then \\( M \\) is called the \"Prosperity Point\" of segment \\( AB \\). Among them, when \\( 0^\\circ < \\angle AMB < 60^\\circ \\), \\( M \\) is called the \"Democracy Point\" of segment \\( AB \\); when \\( 60^\\circ \\leq \\angle AMB \\leq 180^\\circ \\), \\( M \\) is called the \"Civilization Point\".\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) As shown in Figure 1, the coordinates of points \\( A \\) and \\( B \\) are \\( (0, 2) \\) and \\( (2, 0) \\) respectively. Among the coordinates \\( M_1(0, 0) \\), \\( M_2(2, 3) \\), \\( M_3(4, 4) \\), the \"Prosperity Point\" of segment \\( AB \\) is: \\(\\qquad\\); the \"Civilization Point\" of segment \\( AB \\) is \\(\\qquad\\).\n\n(2) As shown in Figure 2, the coordinates of point \\( A \\) are \\( (-3, 0) \\), \\( AB = 2\\sqrt{3} \\), and \\( \\angle OAB = 30^\\circ \\). If \\( M \\) is the \"Democracy Point\" of segment \\( AB \\), directly write the range of the abscissa \\( m \\) of \\( M \\);\n\n(3) Under the conditions of (2), point \\( P \\) is a moving point on the \\( y \\)-axis (not coinciding with \\( B \\) and \\( BP \\neq AB \\)). If \\( T \\) is the \"Prosperity Point\" of \\( AB \\), when the sum of segments \\( TB \\) and \\( TP \\) is minimized, find the coordinates of \\( T' \\) and the coordinates of the symmetric point \\( S \\) of \\( T' \\) with respect to the line \\( AB \\).", "input_image": [ "batch29-2024_06_14_81fc1871d77e7918f5f8g_0016_1.jpg", "batch29-2024_06_14_81fc1871d77e7918f5f8g_0016_2.jpg", "batch29-2024_06_14_81fc1871d77e7918f5f8g_0016_3.jpg" ], "is_multi_img": true, "answer": "（1） $M_{1}, M_{3} ; M_{1}$ ；（2） The range of the horizontal coordinate $m$ of $M$ is: $m>0$ or $m<-3$ ；（3） $T^{\\prime}$ ( $\\left.-1,0\\right)$, the coordinates of the symmetric point $S$ of $T^{\\prime}$ about the line $A B$ are $(-2, \\sqrt{3})$.\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) As shown in the figure,\n\n\n\nFigure 1\n\nAccording to the definition, the \"Prosperity Points\" of segment $\\mathrm{AB}$ are $\\mathrm{M}_{1}, \\mathrm{M}_{3}$, and the \"Civilization Point\" of segment $A B$ is $M_{1}$.\n\nTherefore, the answer is: $M_{1}, M_{3} ; M_{1}$.\n\n(2) Draw the perpendicular bisector 1 of segment $\\mathrm{AB}$ through its midpoint $\\mathrm{C}$, intersecting the $\\mathrm{y}$-axis at point $\\mathrm{F}$. From $\\mathrm{A}$, draw $\\mathrm{AE} \\perp \\mathrm{x}$-axis intersecting line 1 at point $\\mathrm{E}$, then connect $\\mathrm{BE}$ and $\\mathrm{AF}$.\n\n\n\nFigure 2\n\n$\\because \\angle \\mathrm{OAB}=30^{\\circ}, \\angle \\mathrm{AOB}=90^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{ABO}=60^{\\circ}$,\n\nAlso, $\\because \\mathrm{EA}=\\mathrm{EB}$,\n\n$\\therefore \\triangle \\mathrm{ABE}$ is an equilateral triangle,\n\nSimilarly, $\\triangle \\mathrm{ABF}$ is also an equilateral triangle,\n\n$\\therefore \\angle \\mathrm{AEB}=\\angle \\mathrm{AFB}=60^{\\circ}$,\n\nFrom the figure, the x-coordinate of $\\mathrm{E}$ is $-3$, and the x-coordinate of $\\mathrm{F}$ is 0.\n\nWhen $M$ is above point $E$, or $M$ is below point $F$, it satisfies: $0^{\\circ}<\\angle A M B<60^{\\circ}$,\n\n$\\therefore$ The range of the x-coordinate $\\mathrm{m}$ of $\\mathrm{M}$ is: $\\mathrm{m}>0$ or $\\mathrm{m}<-3$.\n\n(3) As shown in the figure, draw the perpendicular bisector 1 of segment $\\mathrm{AB}$, then $\\mathrm{T}$ moves on line 1.\n\n$\\because \\mathrm{T}$ is the \"Prosperity Point\" of segment $\\mathrm{AB}$,\n\n$\\therefore \\mathrm{TA}=\\mathrm{TB}$,\n$\\therefore \\mathrm{TB}+\\mathrm{TP}=\\mathrm{TA}+\\mathrm{TP} \\geqslant \\mathrm{AP}^{\\prime}$, (Among all lines from a point to a line, the perpendicular is the shortest), at this time, the intersection point $\\mathrm{T}^{\\prime}$ of line 1 with the $\\mathrm{x}$-axis is the desired coordinate.\n\nIn Rt $\\triangle \\mathrm{ACT}^{\\prime}$, $\\angle \\mathrm{CAT}^{\\prime}=30^{\\circ}, \\mathrm{AC}=\\sqrt{3}$,\n\n$\\therefore \\mathrm{AT}^{\\prime}=\\frac{A C}{\\cos 30^{\\circ}}=2$,\n\n$\\therefore \\mathrm{OT}^{\\prime}=\\mathrm{OA}-\\mathrm{AT}^{\\prime}=1$,\n\n$\\therefore \\mathrm{T}^{\\prime}(-1,0)$,\n\nIn Rt $\\triangle \\mathrm{ABO}$, $\\angle \\mathrm{OAB}=30^{\\circ}$,\n\n$\\therefore \\mathrm{OB}=\\frac{1}{2} \\mathrm{AB}=\\sqrt{3}$,\n\nDraw the symmetric point $\\mathrm{S}$ of $\\mathrm{T}^{\\prime}$ with respect to line $\\mathrm{AB}$, and from point $\\mathrm{S}$, draw $\\mathrm{SM} \\perp \\mathrm{OA}$ at $\\mathrm{M}$. By symmetry, $\\angle \\mathrm{SAB}=$ $\\angle \\mathrm{OAB}=30^{\\circ}$,\n\n$\\therefore \\angle \\mathrm{SAT}^{\\prime}=60^{\\circ}$,\n\n$\\because \\angle \\mathrm{AT}^{\\prime} \\mathrm{S}=60^{\\circ}$,\n\n$\\therefore \\triangle \\mathrm{SAT}^{\\prime}$ is an equilateral triangle,\n\n$\\because \\mathrm{SM} \\perp \\mathrm{AT}^{\\prime}$,\n\n$\\therefore \\mathrm{AM}=\\mathrm{T}^{\\prime} \\mathrm{M}=1$,\n\n$\\therefore \\mathrm{SM}=\\sqrt{S A^{2}-A M^{2}}=\\sqrt{3}$,\n\n$\\therefore$ The coordinates of the symmetric point $\\mathrm{S}$ of $\\mathrm{T}^{\\prime}$ with respect to line $\\mathrm{AB}$ are: $(-2, \\sqrt{3})$.\n\n\n\nFigure 3\n\n【Insight】This problem belongs to the geometric transformation type, examining the definitions of \"Prosperity Point\", \"Democracy Point\", and \"Civilization Point\", the properties of the perpendicular bisector of a segment, the determination and properties of equilateral triangles, and the solution of right triangles. The key to solving the problem is understanding the meaning and flexibly applying the knowledge points to solve the problem." }, { "problem_id": 1946, "question": "As shown in Figure 1, in the plane rectangular coordinate system, quadrilateral $A O B C$ is a rectangle, with point $A$ having coordinates $(0,3)$ and point $B$ having coordinates $(4,0)$. Points $E$ and $F$ are moving points on sides $B C$ and $A C$ respectively, and neither coincides with the endpoints. Connecting $E F$, the triangle $\\triangle C E F$ is folded along the moving line $E F$ to form $\\triangle D E F$.\n\n(1) As shown in Figure 1, when the corresponding point $D$ of point $C$ falls on $A B$ and $E F / / A B$, then $C E=$ $\\qquad$ ;\n\n(2) As shown in Figure 2, point $G(0,2)$ is connected to $F G$, which intersects $A B$ at point $H$. The line $E D$ intersects $A B$ at point $I$. When quadrilateral $F H I E$ is a parallelogram, find the length of $C E$;\n\n(3) When points $E$ and $F$ are in the position described in problem (1), the triangle $\\triangle E D F$ is rotated counterclockwise around point $E$ by $\\alpha$ degrees $\\left(0^{\\circ}\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch29-2024_06_14_81fc1871d77e7918f5f8g_0056_1.jpg", "batch29-2024_06_14_81fc1871d77e7918f5f8g_0056_2.jpg", "batch29-2024_06_14_81fc1871d77e7918f5f8g_0056_3.jpg" ], "is_multi_img": true, "answer": "（1） $\\frac{3}{2}$ ；（2） $C E=\\frac{3}{7}$ ；（3） $A M=\\frac{13}{2}-\\frac{3}{2} \\sqrt{5}$ or $A M=\\frac{7}{2}+\\frac{3}{4} \\sqrt{5}$.\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since \\( EF \\parallel AB \\),\n\n\\[\n\\angle CEF = \\angle EBA, \\quad \\angle FED = \\angle EDB.\n\\]\n\nFrom the folding, we know:\n\n\\[\n\\angle CEF = \\angle FED, \\quad CE = DE.\n\\]\n\nThus,\n\n\\[\n\\angle EBA = \\angle EDB,\n\\]\n\nwhich implies\n\n\\[\nDE = BE,\n\\]\n\nand consequently,\n\n\\[\nCE = BE.\n\\]\n\nGiven that point \\( A \\) has coordinates \\( (0, 3) \\) and quadrilateral \\( AOBC \\) is a rectangle,\n\n\\[\nBC = OA = 3,\n\\]\n\nso\n\n\\[\nCE = BE = \\frac{3}{2}.\n\\]\n\nTherefore, the length of \\( CE \\) is \\( \\frac{3}{2} \\).\n\n(2) Since quadrilateral \\( FHIE \\) is a parallelogram,\n\n\\[\nEF \\parallel HI, \\quad EF = HI, \\quad FH \\parallel EI, \\quad FH = EI.\n\\]\n\nThus,\n\n\\[\n\\frac{CE}{CB} = \\frac{CF}{AC}, \\quad \\angle CEF = \\angle EBA, \\quad \\angle FED = \\angle EIB, \\quad \\angle FHI = \\angle EIB.\n\\]\n\nGiven that\n\n\\[\n\\angle FHI = \\angle AHG,\n\\]\n\nwe have\n\n\\[\n\\angle AHG = \\angle EIB. \\quad \\text{(1)}\n\\]\n\nFrom the folding, we know:\n\n\\[\n\\angle CEF = \\angle FED,\n\\]\n\nwhich implies\n\n\\[\n\\angle EBI = \\angle EIB. \\quad \\text{(2)}\n\\]\n\nThus,\n\n\\[\nEI = EB.\n\\]\n\nGiven that point \\( A \\) has coordinates \\( (0, 3) \\) and point \\( B \\) has coordinates \\( (4, 0) \\), and quadrilateral \\( AOBC \\) is a rectangle,\n\n\\[\nBC = OA = 3, \\quad AC = OB = 4, \\quad OA \\parallel BC.\n\\]\n\nThus,\n\n\\[\n\\frac{CE}{3} = \\frac{CF}{4}, \\quad \\angle GAH = \\angle EBI.\n\\]\n\nFrom (1), (2), and (3), we have:\n\n\\[\n\\angle OAB = \\angle AHG,\n\\]\n\nwhich implies\n\n\\[\nGA = GH.\n\\]\n\nGiven that \\( G(0, 2) \\),\n\n\\[\nGA = GH = 3 - 2 = 1.\n\\]\n\nLet \\( CE = x \\), then\n\n\\[\nFH = EI = BE = 3 - x, \\quad CF = \\frac{4}{3}x.\n\\]\n\nThus,\n\n\\[\nGF = GH + FH = 1 + 3 - x = 4 - x, \\quad AF = 4 - \\frac{4}{3}x.\n\\]\n\nTherefore,\n\n\\[\n\\left(4 - \\frac{4}{3}x\\right)^2 = (4 - x)^2 - 1,\n\\]\n\nwhich yields\n\n\\[\nx_1 = \\frac{3}{7}, \\quad x_2 = 3 \\quad (\\text{discarded as it does not fit the context}).\n\\]\n\nThus, the length of \\( CE \\) is \\( \\frac{3}{7} \\).\n\n(3) Given that point \\( A \\) has coordinates \\( (0, 3) \\) and point \\( B \\) has coordinates \\( (4, 0) \\), and quadrilateral \\( AOBC \\) is a rectangle,\n\n\\[\nBC = OA = 3, \\quad AC = OB = 4.\n\\]\n\nThus,\n\n\\[\nAB = \\sqrt{3^2 + 4^2} = 5.\n\\]\n\nWhen point \\( N \\) is above point \\( A \\), as shown in Figure (1), draw \\( AP \\parallel MN \\), intersecting the \\( x \\)-axis at point \\( P \\).\n\nThus,\n\n\\[\n\\angle OAP = \\angle ANM, \\quad \\angle PAB = \\angle AMN.\n\\]\n\nSince \\( AM = AN \\),\n\n\\[\n\\angle ANM = \\angle AMN,\n\\]\n\nwhich implies\n\n\\[\n\\angle OAP = \\angle PAB.\n\\]\n\nDraw \\( PQ \\perp AB \\), with \\( Q \\) as the foot of the perpendicular.\n\nThus,\n\n\\[\nPO = PQ, \\quad AO = AQ = 3.\n\\]\n\nTherefore,\n\n\\[\nBQ = AB - AQ = 5 - 3 = 2.\n\\]\n\nLet \\( PO = PQ = x \\), then\n\n\\[\nPB = 4 - x.\n\\]\n\nIn right triangle \\( \\triangle PBQ \\),\n\n\\[\nx^2 + 2^2 = (4 - x)^2,\n\\]\n\nwhich yields\n\n\\[\nx = \\frac{3}{2}.\n\\]\n\nThus,\n\n\\[\nPB = 4 - x = \\frac{5}{2}, \\quad AP = \\sqrt{OA^2 + OP^2} = \\frac{3\\sqrt{5}}{2}.\n\\]\n\n\\[\n\\sin \\angle ANM = \\sin \\angle OAP = \\frac{PO}{AP} = \\frac{\\sqrt{5}}{5}.\n\\]\n\nAs shown in Figure (2), connect \\( EA \\) and \\( EP \\).\n\nThus,\n\n\\[\nEA = \\sqrt{AC^2 + CE^2} = \\frac{\\sqrt{73}}{2}, \\quad EP = \\sqrt{PB^2 + BE^2} = \\frac{\\sqrt{34}}{2}.\n\\]\n\nDraw \\( ER \\perp AP \\), with \\( R \\) as the foot of the perpendicular. Let \\( AR = y \\), then\n\n\\[\nPR = AP - AR = \\frac{3\\sqrt{5}}{2} - y.\n\\]\n\nSince\n\n\\[\nEA^2 - AR^2 = EP^2 - PR^2,\n\\]\n\nwe have\n\n\\[\n\\left(\\frac{\\sqrt{73}}{2}\\right)^2 - y^2 = \\left(\\frac{\\sqrt{34}}{2}\\right)^2 - \\left(\\frac{3\\sqrt{5}}{2} - y\\right)^2,\n\\]\n\nwhich yields\n\n\\[\ny = \\frac{7\\sqrt{5}}{5}.\n\\]\n\nThus,\n\n\\[\nER = \\sqrt{EA^2 - AR^2} = \\frac{13\\sqrt{5}}{10}.\n\\]\n\nSince\n\n\\[\n\\angle ED'F' = \\angle EDF = \\angle C = 90^\\circ,\n\\]\n\nthe distance from point \\( E \\) to line \\( MN \\) is \\( ED' = ED = EC = \\frac{3}{2} \\).\n\nAs shown in Figure (1), draw \\( AT \\perp MN \\), with \\( T \\) as the foot of the perpendicular.\n\nThus,\n\n\\[\nAT = ER - ED' = \\frac{13\\sqrt{5}}{10} - \\frac{3}{2}.\n\\]\n\nTherefore,\n\n\\[\nAN = \\frac{\\frac{13\\sqrt{5}}{10} - \\frac{3}{2}}{\\sin \\angle ANM} = \\frac{13}{2} - \\frac{3}{2}\\sqrt{5},\n\\]\n\nand\n\n\\[\nAM = \\frac{13}{2} - \\frac{3}{2}\\sqrt{5}.\n\\]\n\nWhen point \\( N \\) is below point \\( A \\), as shown in Figure (3), draw \\( AP \\parallel MN \\), intersecting line \\( BC \\) at point \\( P \\). Draw \\( AT \\perp MN \\), \\( PQ \\perp AB \\), and \\( PQ' \\perp y \\)-axis.\n\nThus,\n\n\\[\nCP = AQ', \\quad CA = PQ' = 4.\n\\]\n\nSimilarly, we can find\n\n\\[\n\\sin \\angle AMT = \\sin \\angle PAQ' = \\frac{PQ'}{PA}, \\quad PQ' = PQ = 4, \\quad AQ' = AQ.\n\\]\n\nLet \\( AQ' = AQ = PC = m \\), then\n\n\\[\nPB = m + 3, \\quad QB = 5 - m.\n\\]\n\nIn right triangle \\( \\triangle PQB \\),\n\n\\[\nPQ^2 + QB^2 = PB^2,\n\\]\n\nwhich yields\n\n\\[\n4^2 + (5 - m)^2 = (m + 3)^2,\n\\]\n\nand solving gives\n\n\\[\nm = 2.\n\\]\n\nSimilarly, we can find\n\n\\[\nPA = 2\\sqrt{5}.\n\\]\n\nAs shown in Figure (4), connect \\( EA \\), and draw \\( ER \\perp AP \\), with \\( R \\) as the foot of the perpendicular. Let \\( AR = n \\).\n\nSimilarly, we can find\n\n\\[\nn = \\frac{13\\sqrt{5}}{10}.\n\\]\n\nThus,\n\n\\[\nER = \\sqrt{EA^2 - n^2} = \\frac{7}{5}\\sqrt{5}.\n\\]\n\nTherefore,\n\n\\[\nAT = ER + ED' = \\frac{7}{5}\\sqrt{5} + \\frac{3}{2},\n\\]\n\nand\n\n\\[\nAM = \\frac{AT}{\\sin \\angle AMT} = \\frac{\\frac{7\\sqrt{5}}{5} + \\frac{3}{2}}{\\frac{2\\sqrt{5}}{5}} = \\frac{7}{2} + \\frac{3}{4}\\sqrt{5}.\n\\]\n\nIn summary,\n\n\\[\nAM = \\frac{13}{2} - \\frac{3}{2}\\sqrt{5} \\quad \\text{or} \\quad AM = \\frac{7}{2} + \\frac{3}{4}\\sqrt{5}.\n\\]\n\n\n\n\n【Insight】This problem comprehensively examines the properties of rectangles, symmetry, angle bisectors, the Pythagorean theorem, trigonometric functions, and rotational properties. It requires students to understand and apply these concepts to solve complex problems. The problem is highly comprehensive, demanding strong analytical skills and incorporating the ideas of combining numbers with shapes and categorizing discussions." }, { "problem_id": 1947, "question": "(1) We know: As shown in Figure (1), point \\( B \\) divides the segment \\( AC \\) into two parts, and if \\(\\frac{BC}{AB} = \\frac{AB}{AC}\\), then point \\( B \\) is called the golden section point of segment \\( AC \\). The ratio is \\(\\qquad\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n(2) In Figure (1), if \\( AC = 20 \\, \\text{cm} \\), then the length of \\( AB \\) is \\(\\qquad\\) \\(\\text{cm}\\);\n\n(3) As shown in Figure (2), using a square paper with side length \\( 20 \\, \\text{cm} \\), perform the following operations: Fold the square \\( ABCD \\) to get the crease \\( EF \\), connect \\( CE \\), fold \\( CB \\) onto \\( CE \\) so that point \\( B \\) corresponds to point \\( H \\), and get the crease \\( CG \\). Explain why \\( G \\) is the golden section point of \\( AB \\);\n\n(4) As shown in Figure (3), Xiao Ming further explored: On the side \\( AD \\) of a square \\( ABCD \\) with side length \\( a \\), take any point \\( E \\) such that \\( AE > DE \\), connect \\( BE \\), draw \\( CF \\perp BE \\), intersecting \\( AB \\) at point \\( F \\), and extend \\( EF \\) and \\( CB \\) to intersect at point \\( P \\). He found that when \\( PB \\) and \\( BC \\) satisfy a certain relationship, \\( E \\) and \\( F \\) are precisely the golden section points of \\( AD \\) and \\( AB \\), respectively. Guess Xiao Ming's discovery and explain the reason.", "input_image": [ "batch29-2024_06_14_81fc1871d77e7918f5f8g_0057_1.jpg", "batch29-2024_06_14_81fc1871d77e7918f5f8g_0057_2.jpg", "batch29-2024_06_14_81fc1871d77e7918f5f8g_0057_3.jpg" ], "is_multi_img": true, "answer": "（1） $\\frac{\\sqrt{5}-1}{2}$; （2） $(10 \\sqrt{5}-10) \\mathrm{cm}$; (3) See analysis; (4) $B P=B C$, See analysis\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since point $B$ divides segment $A C$ into two parts such that $\\frac{B C}{A B}=\\frac{A B}{A C}$,\n\n$\\therefore$ point $B$ is the golden section point of segment $A C$, and $A B > B C$,\n\n$\\therefore$ their ratio is $\\frac{\\sqrt{5}-1}{2}$,\n\nHence, the answer is $\\frac{\\sqrt{5}-1}{2}$;\n\n(2) Since point $B$ is the golden section point of segment $A C$, and $A C = 20 \\mathrm{~cm}$,\n\n$\\therefore \\mathrm{AB} = \\frac{\\sqrt{5}-1}{2} \\times 20 = (10 \\sqrt{5}-10) \\mathrm{cm}$.\n\nHence, the answer is: $(10 \\sqrt{5}-10)$.\n\n(3) Extend $E A$ and $C G$ to meet at point $M$,\n\n\n\nSince quadrilateral $A B C D$ is a square,\n\n$\\therefore D M \\parallel B C$,\n\n$\\therefore \\angle E M C = \\angle B C G$,\n\nFrom the folding property, $\\angle E C M = \\angle B C G$,\n\n$\\therefore \\angle E M C = \\angle E C M$,\n\n$\\therefore E M = E C$,\n\nSince $D E = 10$ and $D C = 20$,\n\n$\\therefore E C = \\sqrt{D E^{2} + D C^{2}} = \\sqrt{10^{2} + 20^{2}} = 10 \\sqrt{5}$,\n\n$\\therefore E M = 10 \\sqrt{5}$,\n\n$\\therefore D M = 10 \\sqrt{5} + 10$,\n$\\therefore \\tan \\angle D M C = \\frac{D C}{D M} = \\frac{20}{10 \\sqrt{5} + 10} = \\frac{\\sqrt{5}-1}{2}$,\n\n$\\therefore \\tan \\angle B C G = \\frac{\\sqrt{5}-1}{2}$,\n\nThat is, $\\frac{B G}{B C} = \\frac{\\sqrt{5}-1}{2}$,\n\nSince $A B = B C$,\n\n$\\therefore \\frac{B G}{A B} = \\frac{\\sqrt{5}-1}{2}$,\n\n$\\therefore G$ is the golden section point of $A B$;\n\n(4) When $B P = B C$, the condition is satisfied.\n\nThe reasoning is as follows:\n\nSince quadrilateral $A B C D$ is a square,\n\n$\\therefore A B = B C$, $\\angle B A E = \\angle C B F = 90^{\\circ}$,\n\nSince $B E \\perp C F$,\n\n$\\therefore \\angle A B E + \\angle C F B = 90^{\\circ}$,\n\nAlso, since $\\angle B C F + \\angle B F C = 90^{\\circ}$,\n\n$\\therefore \\angle B C F = \\angle A B E$,\n\n$\\therefore \\triangle A B E \\cong \\triangle B C F$ (ASA),\n\n$\\therefore B F = A E$,\n\nSince $A D \\parallel C P$,\n\n$\\therefore \\triangle A E F \\sim \\triangle B P F$,\n\n$\\therefore \\frac{A E}{B P} = \\frac{A F}{B F}$,\n\nWhen $E$ and $F$ are exactly the golden section points of $A D$ and $A B$ respectively,\n\nSince $A E > D E$,\n\n$\\therefore \\frac{A F}{B F} = \\frac{B F}{A B}$\n\nSince $B F = A E$, $A B = B C$,\n\n$\\therefore \\frac{A F}{B F} = \\frac{B F}{A B} = \\frac{A E}{B C}$\n\n$\\therefore \\frac{A E}{B P} = \\frac{A E}{B C}$,\n\n$\\therefore B P = B C$.\n\n【Insight】This problem is a comprehensive question on similarity, examining the properties of folding transformations, properties of squares, the determination and properties of congruent triangles, the definition of the golden section point, trigonometric functions of acute angles, and the determination and properties of similar triangles. Mastering the determination and properties of similar triangles is key to solving this problem." }, { "problem_id": 1948, "question": "As shown in the figure, in quadrilateral \\(ABCD\\), \\(AD \\parallel BC\\), \\(AB = 10\\), \\(CD = 4\\sqrt{5}\\). A moving point \\(P\\) starts from point \\(A\\) and travels along the path \\(A-B-C\\), while another moving point \\(Q\\) starts from point \\(D\\) and travels along the path \\(D-A\\). They both reach their respective endpoints simultaneously. Let the distance traveled by point \\(P\\) be \\(x\\). The length of \\(AQ\\) is \\(y\\), and \\(y = -\\frac{2}{3}x + 16\\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) In Figure 1, find the lengths of \\(AD\\) and \\(BC\\), and \\(\\tan A\\).\n\n(2) In Figure 2, connect \\(PQ\\), and let the area of \\(\\triangle APQ\\) be \\(S\\). Find the relationship between \\(S\\) and \\(x\\).\n\n(3) In Figure 3, when \\(PQ\\) is perpendicular to one of the sides of quadrilateral \\(ABCD\\), find all values of \\(x\\) that satisfy this condition.", "input_image": [ "batch29-2024_06_14_a2209ef2c9ad0d2b4985g_0093_1.jpg", "batch29-2024_06_14_a2209ef2c9ad0d2b4985g_0093_2.jpg", "batch29-2024_06_14_a2209ef2c9ad0d2b4985g_0093_3.jpg" ], "is_multi_img": true, "answer": "(1) $A D=16, B C=14, \\tan A=\\frac{4}{3}$\n\n(2) $S=-\\frac{4}{15} x^{2}+\\frac{32}{5} x$\n\n(3) $x=\\frac{48}{7}$ or 12 or $\\frac{240}{43}$", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "(1) Since points \\( P \\) and \\( Q \\) reach the endpoint simultaneously, we have \\( AD = 16 \\) and \\( BC = 14 \\). Drawing \\( BM \\perp AD \\) from point \\( B \\) and \\( DN \\perp BC \\) from point \\( D \\), let \\( AM = m \\). Since \\( BM = DN \\), the Pythagorean theorem yields the equation. Using \\( \\tan A = \\frac{BM}{AM} \\), we can solve for \\( \\tan A \\).\n\n(2) When point \\( P \\) is on \\( AB \\), drawing \\( PH \\perp AD \\) at point \\( H \\), we get \\( PH = \\frac{4}{5} PA = \\frac{4}{5}x \\). This allows us to express \\( S \\) as a function of \\( x \\).\n\n(3) There are three scenarios: \\( QP \\perp AB \\), \\( QP \\perp BC \\), and \\( PQ \\perp CD \\). For each scenario, we can set up an equation in terms of \\( x \\) to solve the problem.\n\n**Solution:**\n\n(1) Since points \\( P \\) and \\( Q \\) reach the endpoint simultaneously, when \\( x = 0 \\), \\( y = 16 \\), so \\( AD = 16 \\). When \\( y = 0 \\), \\( x = 24 \\), so \\( BC = 14 \\). Since \\( AD \\parallel BC \\), drawing \\( BM \\perp AD \\) from point \\( B \\) and \\( DN \\perp BC \\) from point \\( D \\), let \\( AM = m \\). Then \\( MD = 16 - m \\), so \\( BN = 16 - m \\), and \\( CN = m - 2 \\). Since \\( BM = DN \\), we have:\n\\[\n10^2 - m^2 = (4\\sqrt{5})^2 - (m - 2)^2\n\\]\nSolving this, we get \\( m = 6 \\), so \\( BM = 8 \\). Therefore:\n\\[\n\\tan A = \\frac{BM}{AM} = \\frac{4}{3}\n\\]\n\n(2) When point \\( P \\) is on \\( AB \\), drawing \\( PH \\perp AD \\) at point \\( H \\), from part (1) we know \\( \\sin A = \\frac{4}{5} \\), so:\n\\[\nPH = \\frac{4}{5} PA = \\frac{4}{5}x\n\\]\nThus, the area \\( S \\) is:\n\\[\nS = \\frac{1}{2} \\times \\left(-\\frac{2}{3}x + 16\\right) \\times \\frac{4}{5}x = \\frac{4}{15}x^2 + \\frac{32}{5}x\n\\]\n\n(3) When \\( QP \\perp AB \\), we have:\n\\[\n\\cos A = \\frac{AP}{AQ} = \\frac{3}{5}\n\\]\nSo:\n\\[\n\\frac{x}{-\\frac{2}{3}x + 16} = \\frac{3}{5}\n\\]\nSolving this, we get \\( x = \\frac{48}{7} \\).\n\nWhen \\( QP \\perp BC \\), as shown in the figure, \\( BP = MQ \\), so:\n\\[\nx - 10 = -\\frac{2}{3}x + 16 - 6\n\\]\nSolving this, we get \\( x = 12 \\).\n\nWhen \\( PQ \\perp CD \\) at point \\( E \\), as shown in the figure, drawing \\( PG \\perp AQ \\) at point \\( G \\) and \\( DN \\perp BC \\) at point \\( N \\), we have:\n\\[\n\\angle CDN + \\angle QDE = 90^\\circ\n\\]\nSince \\( \\angle QDE + \\angle DQE = 90^\\circ \\), we have \\( \\angle CDN = \\angle DQE = \\angle PQG \\). Also, \\( \\angle CND = \\angle PGQ \\), so \\( \\triangle CND \\sim \\triangle PGQ \\). Therefore:\n\\[\n\\frac{CN}{ND} = \\frac{PG}{GQ}\n\\]\nSubstituting the known values:\n\\[\n\\frac{4}{8} = \\frac{\\frac{4}{5}x}{QG}\n\\]\nSo \\( QG = \\frac{8}{5}x \\). Thus:\n\\[\n\\frac{3}{5}x + \\frac{8}{5}x = -\\frac{2}{3}x + 16\n\\]\nSolving this, we get \\( x = \\frac{240}{43} \\).\n\nIn summary, \\( x = \\frac{48}{7} \\) or \\( 12 \\) or \\( \\frac{240}{43} \\).\n\n**Key Insight:** This problem is a comprehensive quadrilateral problem, primarily testing the properties and determination of rectangles, the Pythagorean theorem, the similarity of triangles, and the meaning of functional expressions. The key to solving the problem lies in using classification to draw different scenarios." }, { "problem_id": 1949, "question": "We often use different methods to measure an object. Please measure the length of the lamp post \\( AB \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) As shown in Figure 1, place an angle measuring instrument at point \\( D \\) which is \\( a \\) meters away from the base of the lamp post \\( AB \\). The height of the instrument is \\( b \\) meters. The angle of elevation from point \\( C \\) to point \\( A \\) is \\( \\alpha \\). Find the height of the lamp post \\( AB \\) (expressed in algebraic terms involving \\( a, b, \\alpha \\)).\n\n(2) The ancient Chinese mathematician Zhao Shuang's method of measuring objects using shadows is still relevant today. As shown in Figure 2, place a wooden rod \\( CG \\) with a height of 2 meters in front of the lamp post \\( AB \\). The length of its shadow \\( CH \\) is measured to be 1 meter. Then move the rod along the direction of \\( BC \\) by 1.8 meters to position \\( DE \\). At this position, the length of its shadow \\( DF \\) is measured to be 3 meters. Find the height of the lamp post \\( AB \\).", "input_image": [ "batch29-2024_06_14_b016b8373068379abdc0g_0085_1.jpg", "batch29-2024_06_14_b016b8373068379abdc0g_0085_2.jpg" ], "is_multi_img": true, "answer": "(1) $a \\tan \\alpha+b$ m\n\n(2)3.8 m\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "(1) Solution: As shown in the figure\n\n\n\nFigure 1\n\nGiven the conditions: $BD = a$, $CD = b$, $\\angle ACE = \\alpha$\n\n$\\angle B = \\angle D = \\angle CEB = 90^\\circ$\n\n$\\therefore$ Quadrilateral $CDBE$ is a rectangle,\n\nThus, $BE = CD = b$, $BD = CE = a$,\n\nIn right triangle $Rt \\triangle ACE$, $\\tan \\alpha = \\frac{AE}{CE}$,\n\nWe get $AE = CE \\times \\tan \\alpha = a \\tan \\alpha$\n\nAnd $AB = AE + BE$,\n\nTherefore, $AB = a \\tan \\alpha + b$\n\nAnswer: The height of the lamp post $AB$ is $a \\tan \\alpha + b$ meters.\n\n(2) Given the conditions: $AB \\parallel GC \\parallel ED$, $GC = ED = 2$, $CH = 1$, $DF = 3$, $CD = 1.8$\n\nSince $AB \\parallel ED$,\n\n$\\therefore \\triangle ABF \\sim \\triangle EDF$,\n\nAt this point, $\\frac{ED}{DF} = \\frac{AB}{BF}$\n\nThat is, $\\frac{2}{3} = \\frac{AB}{BC + 1.8 + 3}$ (1),\n\n$\\because AB \\parallel GC$\n\n$\\therefore \\triangle ABH \\sim \\triangle GCH$,\n\nAt this point, $\\frac{AB}{BH} = \\frac{GC}{CH}$,\n\n$\\frac{2}{1} = \\frac{AB}{BC + 1}$\n\nCombining (1) and (2) gives\n\n$$\n\\left\\{\\begin{array}{l}\n\\frac{AB}{BC + 4.8} = \\frac{2}{3} \\\\\n\\frac{AB}{BC + 1} = 2\n\\end{array}\\right.\n$$\n\nSolving gives: $\\left\\{\\begin{array}{l}AB = 3.8 \\\\ BC = 0.9\\end{array}\\right.$\n\nAnswer: The height of the lamp post $AB$ is 3.8 meters.\n\n【Highlight】This problem examines the application of similar triangles, the use of trigonometric functions of acute angles, and systems of linear equations. The key to solving the problem lies in understanding the given conditions and being familiar with the properties and criteria of similar triangles." }, { "problem_id": 1950, "question": "As shown in the figure, given that \\( AB \\parallel CD \\), points \\( E \\) and \\( F \\) are on \\( AB \\) and \\( CD \\) respectively, and point \\( G \\) is between \\( AB \\) and \\( CD \\). Lines \\( GE \\) and \\( GF \\) are connected.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nWhen \\(\\angle BEG = 40^\\circ\\), \\( EP \\) bisects \\(\\angle BEG \\) and \\( FP \\) bisects \\(\\angle DFG \\);\n\n(1) In Figure 1, when \\( EG \\perp FG \\), then \\(\\angle P = \\) \\(\\qquad\\) \\(\\circ\\)\n\n(2) In Figure 2, there is a point \\( Q \\) below \\( CD \\). If \\( EG \\) exactly bisects \\(\\angle BEQ \\) and \\( FD \\) exactly bisects \\(\\angle GFQ \\), find the measure of \\(\\angle Q + 2\\angle P \\).", "input_image": [ "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0028_1.jpg", "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0028_2.jpg" ], "is_multi_img": true, "answer": "(1)45\n\n(2) $120^{\\circ}$", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Draw $GN \\parallel AB$ and $PM \\parallel AB$ through points $G$ and $P$ respectively. According to the properties and determination of parallel lines, and the definition of angle bisectors, we can derive that $\\angle EPE = \\frac{1}{2}(\\angle BEG + \\angle GFD) = \\frac{1}{2} \\angle EGF$. Based on the definition of perpendicularity, $\\angle EGF = 90^\\circ$, which allows us to solve the problem; $\\square$\n\n(2) Draw $QK \\parallel CD$ through point $Q$, and let $\\angle GFD = \\angle QFD = \\alpha$. According to the properties of parallel lines and the definition of angle bisectors, we can derive that $\\angle FQE = 80^\\circ - \\alpha$. From (1), we know that $\\angle G = 2 \\angle P = \\angle BEG + \\angle GFD = 40^\\circ + \\alpha$, which allows us to solve the problem.\n\n(1)\n\nAs shown in the figure, draw $GN \\parallel AB$ and $PM \\parallel AB$ through points $G$ and $P$ respectively.\n\n\n\nFigure 1\n\n$$\n\\begin{aligned}\n& \\because AB \\parallel CD \\\\\n& \\therefore AB \\parallel CD \\parallel GN \\parallel PM \\\\\n& \\therefore \\angle BEG = \\angle EGN, \\angle BEP = \\angle EPM, \\\\\n& \\angle NGF = \\angle GFD, \\angle MPF = \\angle PFD, \\\\\n& \\angle EGF = \\angle EGN + \\angle NGF = \\angle BEG + \\angle GFD, \\\\\n& \\angle EPF = \\angle EPM + \\angle MPF = \\angle BEP + \\angle PFD, \\\\\n& \\because EG \\perp FG, \\\\\n& \\therefore \\angle EGF = 90^\\circ, \\\\\n& \\because EP \\text{ bisects } \\angle BEG, \\quad FP \\text{ bisects } \\angle DFG, \\\\\n& \\therefore \\angle BEP = \\frac{1}{2} \\angle BEG, \\angle PFD = \\frac{1}{2} \\angle DFG, \\\\\n& \\therefore \\angle EPE = \\frac{1}{2}(\\angle BEG + \\angle GFD) = \\frac{1}{2} \\angle EGF = 45^\\circ\n\\end{aligned}\n$$\n\nTherefore, the answer is: 45\n\nAs shown in the figure, draw $QK \\parallel CD$ through point $Q$.\n\n\n\nFigure 2\n\n$\\because \\angle BEG = 40^\\circ, EG$ bisects $\\angle BEQ, FD$ bisects $\\angle GFQ$,\n\n$\\therefore \\angle GEQ = \\angle BEG = 40^\\circ, \\angle GFD = \\angle QFD$,\n\nLet $\\angle GFD = \\angle QFD = \\alpha$,\n\n$\\because QK \\parallel CD, \\quad AB \\parallel CD$,\n\n$\\therefore QK \\parallel AB$,\n$\\therefore \\angle EQK = \\angle BEQ = 2 \\angle BEG = 80^\\circ$\n\n$\\angle FQK = \\angle QFD = \\alpha$,\n\n$\\therefore \\angle FQE = 80^\\circ - \\alpha$\n\nFrom (1), we know that,\n\n$\\angle G = 2 \\angle P = \\angle BEG + \\angle GFD = 40^\\circ + \\alpha$\n\n$\\therefore \\angle FQE + 2 \\angle P = 80^\\circ - \\alpha + 40^\\circ + \\alpha = 120^\\circ$\n\n【Insight】This problem examines the properties and determination of parallel lines, and the definition of angle bisectors. Mastering the properties and determination of parallel lines is key to solving the problem." }, { "problem_id": 1951, "question": "As shown in Figure 1, given that the lines \\( m \\parallel n \\), \\( AB \\) is a plane mirror. Light rays emanate from point \\( O \\) on line \\( m \\), reflect off point \\( P \\) on the plane mirror \\( AB \\), and reach point \\( Q \\) on line \\( n \\). We call \\( OP \\) the incident ray and \\( PQ \\) the reflected ray. It is known that plane mirror reflection has the following property: the angle formed between the incident ray and the plane mirror is equal to the angle formed between the reflected ray and the plane mirror, i.e., \\( \\angle OPA = \\angle QPB \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) In Figure 1, if \\( \\angle OPQ = 82^\\circ \\), find the measure of \\( \\angle OPA \\);\n\n(2) In Figure 2, if \\( \\angle AOP = 43^\\circ \\) and \\( \\angle BQP = 49^\\circ \\), find the measure of \\( \\angle OPA \\);\n\n(3) In Figure 3, three additional plane mirrors are placed, with two mirrors on lines \\( m \\) and \\( n \\) respectively, and one between the two lines, forming a quadrilateral \\( ABCD \\) with the four mirrors. Given that light rays emanate from point \\( O \\) at an appropriate angle, with a propagation path of \\( O \\rightarrow P \\rightarrow Q \\rightarrow R \\rightarrow O \\rightarrow P \\), directly write the relationship between \\( \\angle OPQ \\) and \\( \\angle ORQ \\).", "input_image": [ "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0036_1.jpg", "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0036_2.jpg", "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0036_3.jpg" ], "is_multi_img": true, "answer": "(1) $49^{\\circ}$\n\n(2) $44^{\\circ}$\n\n(3) $\\angle O P Q=\\angle O R Q$", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) Based on the fact that $\\angle OPA = \\angle QPB$, the measure of $\\angle OPA$ can be determined;\n\n(2) From $\\angle AOP = 43^\\circ$ and $\\angle BQP = 49^\\circ$, the measure of $\\angle OPQ$ can be determined, which can then be used to solve problem (1);\n\n(3) From the reasoning in (2), it can be deduced that: $\\angle OPQ = \\angle AOP + \\angle BQP$, and $\\angle ORQ = \\angle DOR + \\angle RQC$, thus $\\angle OPQ = \\angle ORQ$.\n\n(1)\n\nSolution: Since $\\angle OPA = \\angle QPB$ and $\\angle OPQ = 82^\\circ$,\n\nTherefore, $\\angle OPA = \\left(180^\\circ - \\angle OPQ\\right) \\times \\frac{1}{2} = \\left(180^\\circ - 82^\\circ\\right) \\times \\frac{1}{2} = 49^\\circ$;\n\n(2)\n\nDraw $PC // m$, where $m // n$,\n\nThus, $m // PC // n$,\n\nTherefore, $\\angle AOP = \\angle OPC = 43^\\circ$, and $\\angle BQP = \\angle QPC = 49^\\circ$,\n\nThus, $\\angle OPQ = \\angle OPC + \\angle QPC = 43^\\circ + 49^\\circ = 92^\\circ$,\n\nTherefore, $\\angle OPA = \\left(180^\\circ - \\angle OPQ\\right) \\times \\frac{1}{2} - \\left(180^\\circ - 92^\\circ\\right) \\times \\frac{1}{2} = 44^\\circ$;\n\n\n\n## Figure 2\n\n(3)\n\n$\\angle OPQ = \\angle ORQ$,\n\nThe reasoning is as follows: From (2), it is known that: $\\angle OPQ = \\angle AOP + \\angle BQP$, and $\\angle ORQ = \\angle DOR + \\angle RQC$,\n\nSince the angle of incidence equals the angle of reflection,\n\nTherefore, $\\angle AOP = \\angle DOR$, and $\\angle BQP = \\angle RQC$.\n\nThus, $\\angle OPQ = \\angle ORQ$.\n\n【Highlight】This question examines the properties of parallel lines, the inference of the parallel postulate, and interdisciplinary knowledge that the angle of incidence equals the angle of reflection. The key to solving the problem is understanding that the angle of incidence equals the angle of reflection." }, { "problem_id": 1952, "question": "As shown in the figure, points $C$ and $D$ are located on rays $OA$ and $OB$ respectively, not coinciding with point $O$, and $CE // DF$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) In Figure 1, while Xiao Ming was exploring the relationship between $\\angle ACE$, $\\angle AOB$, and $\\angle ODF$, he drew a line through $O$ parallel to $DF$. Please determine the relationship between these three angles based on his method.\n\n(2) In Figure 2, $DP$ bisects $\\angle ODF$. If $\\angle ACE = \\alpha$ and $\\angle AOB = \\beta$, express the measure of $\\angle PDO$ in terms of $\\alpha$ and $\\beta$. (Provide the answer directly)", "input_image": [ "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0040_1.jpg", "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0040_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle O D F+\\angle A O B+\\angle A C E=360^{\\circ}$;\n\n(2) $\\angle P D O=180^{\\circ}-\\frac{1}{2} \\alpha-\\frac{1}{2} \\beta$", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) Draw $OG // FD$ through point $O$. According to the properties of parallel lines, the consecutive interior angles are supplementary, the alternate interior angles are equal, and the definition of adjacent supplementary angles, we have $\\angle ODF + \\angle DOG = 180^\\circ$, $\\angle GOC = \\angle OCE$, and $\\angle ACE + \\angle OCE = 180^\\circ$. Adding these equations together, we obtain $\\angle ODF + \\angle AOB + \\angle ACE = 360^\\circ$.\n\n(2) Using the conclusion from part (1), express $\\angle ODF$ in terms of $\\alpha$ and $\\beta$. Then, according to the definition of an angle bisector, we can determine the measure of $\\angle PDO$.\n\n**Solution:**\n\n(1) Draw $OG // FD$ through point $O$, as shown in the figure.\n\n\n\nThen, $\\angle ODF + \\angle DOG = 180^\\circ$.\n\nSince $CE // FD$, it follows that $OG // CE$.\n\nTherefore, $\\angle GOC = \\angle OCE$.\n\nSince $\\angle ACE + \\angle OCE = 180^\\circ$, we have $\\angle ACE + \\angle GOC = 180^\\circ$.\n\nThus, $\\angle ODF + \\angle DOG + \\angle ACE + \\angle GOC = 360^\\circ$,\n\nwhich simplifies to $\\angle ODF + \\angle AOB + \\angle ACE = 360^\\circ$.\n\n(2) Since $DP$ bisects $\\angle ODF$, we have $\\angle PDO = \\frac{1}{2} \\angle ODF$.\n\nGiven that $\\angle ODF + \\angle AOB + \\angle ACE = 360^\\circ$, and $\\angle ACE = \\alpha$, $\\angle AOB = \\beta$,\n\nit follows that $\\angle ODF = 360^\\circ - \\angle AOB - \\angle ACE = 360^\\circ - \\alpha - \\beta$.\n\nTherefore, $\\angle PDO = \\frac{1}{2}(360^\\circ - \\alpha - \\beta) = 180^\\circ - \\frac{1}{2}\\alpha - \\frac{1}{2}\\beta$.\n\n**Key Insight:** This problem examines the properties of parallel lines and angle bisectors. A thorough understanding of these properties is crucial for solving the problem." }, { "problem_id": 1953, "question": "As shown in Figure 1, $AB \\parallel CD, \\angle PAB = 125^\\circ, \\angle PCD = 115^\\circ$, find the degree measure of $\\angle APC$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n\n\nSpare Figure\n\nXiaoming's approach is: Draw $PM \\parallel AB$ through point $P$, and use the properties of parallel lines to find $\\angle APC$.\n\n(1) Following Xiaoming's approach, it is easy to find that the degree measure of $\\angle APC$ is $\\qquad$ degrees;\n\n(2) As shown in Figure 2, $AB \\parallel CD$, point $P$ moves along line $a$, denoted $\\angle PAB = \\alpha, \\angle PCD = \\beta$. When point $P$ moves between points $B$ and $D$, what is the quantitative relationship between $\\angle APC$ and $\\alpha, \\beta$? Please explain the reason;\n\n(3) Under the conditions of (2), if point $P$ moves outside points $B$ and $D$ (point $P$ does not coincide with points $B$ and $D$), please directly write the quantitative relationship between $\\angle APC$ and $\\alpha, \\beta$.", "input_image": [ "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0071_1.jpg", "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0071_2.jpg", "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0071_3.jpg", "batch29-2024_06_14_b2cbc009474a7bbfb8c0g_0071_4.jpg" ], "is_multi_img": true, "answer": "(1) 120\n\n(2) $\\angle A P C=\\angle \\alpha+\\angle \\beta$\n\n(3)When $P$ is on the extension line of $B D$, $\\angle A P C=\\angle \\alpha-\\angle \\beta$; When $P$ is on the extension line of $D B$, $\\angle A P C=\\angle \\beta-\\angle \\alpha$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Algebra", "image_relavance": "0", "analysis": "(1) Solution: As shown in Figure 1, draw $P M / / A B$ through point $P$,\n\n$\\therefore \\angle A P M + \\angle P A B = 180^{\\circ}$,\n\n$\\therefore \\angle A P M = 180^{\\circ} - 125^{\\circ} = 55^{\\circ}$,\n\n$\\because A B / / C D$,\n\n$\\therefore P M / / C D$,\n\n$\\therefore \\angle C P M + \\angle P C D = 180^{\\circ}$,\n$\\therefore \\angle C P M = 180^{\\circ} - 115^{\\circ} = 65^{\\circ}$,\n\n$\\therefore \\angle A P C = 55^{\\circ} + 65^{\\circ} = 120^{\\circ}$;\n\nThus, the answer is: 120;\n\n(2) As shown in Figure 2, $\\angle A P C = \\angle \\alpha + \\angle \\beta$, the reasoning is as follows:\n\nDraw $P E / / A B$ through point $P$, intersecting $A C$ at point $E$,\n\n\n\nFigure 2\n\n$\\because A B / / C D$\n\n$\\therefore A B / / P E / / C D$,\n\n$\\therefore \\angle \\alpha = \\angle A P E, \\angle \\beta = \\angle C P E$,\n\n$\\therefore \\angle A P C = \\angle A P E + \\angle C P E = \\angle \\alpha + \\angle \\beta$;\n\n(3) As shown in Figure 3, when $P$ is on the extension of $B D$, $\\angle A P C = \\angle \\alpha - \\angle \\beta$; the reasoning is:\n\nDraw $P E / / A B$ through point $P$,\n\n$\\because A B / / C D$,\n\n$\\therefore A B / / P E / / C D$,\n\n$\\therefore \\angle \\alpha = \\angle A P E, \\angle \\beta = \\angle C P E$,\n\n$\\therefore \\angle A P C = \\angle A P E - \\angle C P E = \\angle \\alpha - \\angle \\beta$;\n\n\n\nFigure 3\n\nAs shown in Figure 4, when $P$ is on the extension of $D B$, $\\angle A P C = \\angle \\beta - \\angle \\alpha$; the reasoning is: Draw $P E / / A B$ through point $P$,\n\n\n\nFigure 4\n\n$$\n\\begin{aligned}\n& \\because A B / / C D, \\\\\n& \\therefore A B / / P E / / C D,\n\\end{aligned}\n$$\n\n$\\therefore \\angle \\alpha = \\angle A P E, \\angle \\beta = \\angle C P E$,\n\n$\\therefore \\angle A P C = \\angle C P E - \\angle A P E = \\angle \\beta - \\angle \\alpha$.\n\n【Highlight】This question examines the application of the properties of parallel lines, mainly testing students' reasoning ability. The key to solving the problem lies in constructing auxiliary lines to form alternate interior angles and consecutive interior angles." }, { "problem_id": 1954, "question": "The flood season of the Yangtze River is approaching, and the flood control headquarters has installed a searchlight on each side of a dangerous area to facilitate nighttime inspection of the river and the embankments on both sides. As shown in the figure, the beam of light from lamp $A$ rotates clockwise from $A M$ to $A N$ and then immediately reverses direction, while the beam of light from lamp $B$ rotates clockwise from $B P$ to $B Q$ and then immediately reverses direction. The two lights continuously cross-scan. If the rotational speed of lamp $A$ is $a^{\\circ}$ per second and the rotational speed of lamp $B$ is $b^{\\circ}$ per second, and $a$ and $b$ satisfy $|a-3b|+(a-3)^2=0$. Assuming that the embankments on both sides of the Yangtze River in this area are parallel, i.e., $P Q \\parallel M N$, and $\\angle B A N=45^{\\circ}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the values of $a$ and $b$;\n\n(2) If the beam of light from lamp $B$ starts rotating 20 seconds before the beam of light from lamp $A$, how many seconds does lamp $A$ rotate before the beams of the two lights are parallel before lamp $B$'s beam reaches $B Q$?\n\n(3) As shown in the figure, if both lights rotate simultaneously, before the beam of light from lamp $A$ reaches $A N$. If the beams of light intersect at point $C$, and a line $C D \\perp A C$ intersects $P Q$ at point $D$, does the numerical relationship between $\\angle B A C$ and $\\angle B C D$ change during the rotation? If it remains unchanged, please find the relationship; if it changes, please find the range of values.", "input_image": [ "batch29-2024_06_14_b5a462bda054f96852c5g_0040_1.jpg", "batch29-2024_06_14_b5a462bda054f96852c5g_0040_2.jpg" ], "is_multi_img": true, "answer": "(1) $a=3, \\quad b=1$\n\n(2) $t=10$ or $t=85$\n\n(3) $2 \\angle B A C=3 \\angle B C D$", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "(1) According to $|a-3 b|+(a-3)^{2}=0$, we get $a-3 b=0$ and $a-3=0$, from which the values of $a$ and $b$ can be determined;\n\n(2) Let $t$ seconds later, the two beams of light are parallel, (1) before the ray of light A turns to $A N$; (2) after the ray of light A turns to $A N$; based on the properties of parallel lines, the time when the two beams of light are parallel to each other can be determined;\n\n(3) Let the time of rotation of the ray of light A be $t$ seconds, then $\\angle C B D=t$, $\\angle C A N=180^{\\circ}-3 t$, according to $\\angle B A N=45^{\\circ}$, we get $\\angle B A C=\\angle B A N-\\angle C A N=45^{\\circ}-\\left(180^{\\circ}-3 t\\right)$; drawing $H G / / Q P$ through point $C$, based on the properties of parallel lines, we get $\\angle B C A=\\angle C A N+\\angle C B D$; according to $\\angle A C D=90^{\\circ}$, we get $\\angle B C D=\\angle A C D-\\angle B C A$, from which the quantitative relationship between $\\angle B A C$ and $\\angle B C D$ can be obtained.\n\n(1)\n\nSolution: $\\because|a-3 b|+(a-3)^{2}=0$\n\n$\\therefore|a-3 b| \\geq 0,(a-3)^{2} \\geq 0$\n\n$\\therefore a-3 b=0, a-3=0$\n\n$\\therefore a=3, b=1$\n\n(2)\n\nSolution: Let $t$ seconds later, the two beams of light are parallel\n\n(1) Before the ray of light A turns to $A N$\n\n$\\therefore 3 t=(20+t) \\times 1$\n\nSolving gives $t=10$\n\n(2) After the ray of light A turns to $A N$\n\n$\\therefore 3 t-3 \\times 60+(20+t) \\times 1=180^{\\circ}$\nSolving gives $t=85$\n\n$\\therefore$ When $t=10$ or $t=85$ seconds later, the two beams of light are parallel to each other.\n\n(3)\n\nSolution: Let the time of rotation of the ray of light A be $t$ seconds\n\n$\\therefore \\angle C B D=t, \\angle C A N=180^{\\circ}-3 t$\n\n$\\because \\angle B A N=45^{\\circ}$\n\n$\\therefore \\angle B A C=\\angle B A N-\\angle C A N=45^{\\circ}-\\left(180^{\\circ}-3 t\\right)$\n\n$\\therefore \\angle B A C=3 t-135^{\\circ}$\n\nDrawing $H G$ // $Q P$ through point $C$\n\n$\\therefore H G / / Q P / / M N$\n\n$\\therefore \\angle H C B=\\angle C B D, \\quad \\angle H C A=\\angle C A N$\n\n$\\therefore \\angle B C A=\\angle C B D+\\angle C A N$\n\n$\\therefore \\angle B C A=t+\\left(180^{\\circ}-3 t\\right)=180^{\\circ}-2 t$\n\nAlso $\\because \\angle A C D=90^{\\circ}$\n\n$\\therefore \\angle B C D=90^{\\circ}-\\angle B C A$\n\n$\\therefore \\angle B C D=90^{\\circ}-\\left(180^{\\circ}-2 t\\right)=2 t-90^{\\circ}$\n\n$\\therefore \\angle B A C: \\angle B C D=\\left(3 t-135^{\\circ}\\right): 2 t-90^{\\circ}$\n\n$\\therefore \\angle B A C: \\angle B C D=3\\left(t-45^{\\circ}\\right): 2\\left(t-45^{\\circ}\\right)$\n\n$\\angle B A C: \\angle B C D=3: 2$\n\n$\\therefore 2 \\angle B A C=3 \\angle B C D$.\n\n\n\n【Highlight】This question examines the properties of non-negative numbers, the properties of parallel lines, and other knowledge. The key to solving the problem is to grasp: the absolute value of any number is non-negative, $a^{2} \\geq 0$; the properties of parallel lines." }, { "problem_id": 1955, "question": "A car access barrier is installed at the entrance of a residential community. When the barrier is closed, as shown in Figure 1, quadrilateral $ABCD$ is a rectangle, with $AB$ measuring 3 meters and $AD$ measuring 1 meter. The point $D$ is 0.2 meters above the ground. During the opening process of the barrier, side $AD$ remains fixed, and the rods $AB$ and $CD$ rotate around points $A$ and $D$ respectively, with side $BC$ always remaining parallel to side $AD$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) As shown in Figure 2, when the barrier opens to $\\angle ADC = 45^\\circ$, the distance $PE$ from a point $P$ on side $CD$ to the ground is 1.2 meters. Find the length of $PF$, the distance from point $P$ to $MN$.\n\n(2) A car passes through the barrier. It is known that the car is 1.8 meters wide and 1.6 meters high. When the barrier opens to $\\angle ADC = 36^\\circ$, can the car enter the community? Please explain the reason. (Reference data: $\\sin 36^\\circ \\approx 0.59, \\cos 36^\\circ \\approx 0.81, \\tan 36^\\circ \\approx 0.73$)", "input_image": [ "batch29-2024_06_14_d86f38afd8743289227bg_0037_1.jpg", "batch29-2024_06_14_d86f38afd8743289227bg_0037_2.jpg" ], "is_multi_img": true, "answer": "(1) $P F=2$ m\n\n(2) The car can enter the community", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: \n\n(1) Draw a perpendicular line \\( DQ \\) from point \\( D \\) to \\( PE \\), with the foot of the perpendicular at \\( Q \\), as shown in the diagram:\n\n\n\nFrom the problem statement, we know that \\( \\angle ADC = 45^\\circ \\), \\( PE = 1.2 \\) meters, and \\( QE = 0.2 \\) meters.\n\nIn the right triangle \\( \\triangle PDQ \\), \\( \\angle PDQ = 45^\\circ \\), and \\( PQ = 1.2 - 0.2 = 1 \\) meter.\n\nTherefore, \\( \\angle DPQ = 90^\\circ - 45^\\circ = 45^\\circ \\).\n\nThus, \\( \\angle PDQ = \\angle DPQ = 45^\\circ \\).\n\nHence, \\( DQ = PQ = 1 \\) meter.\n\nTherefore, \\( PF = EN = AB - DQ = 3 - 1 = 2 \\) meters.\n\n(2) When \\( \\angle ADC = 36^\\circ \\) and \\( PE = 1.6 \\) meters, then \\( \\angle DPQ = 36^\\circ \\), and \\( PQ = 1.6 - 0.2 = 1.4 \\) meters.\n\nThus, \\( DQ = PQ \\cdot \\tan 36^\\circ \\approx 1.4 \\times 0.73 = 1.022 \\) meters.\n\nTherefore, \\( PF = 3 - 1.022 \\approx 1.98 \\) meters.\n\nSince \\( 1.98 > 1.8 \\), it is possible to pass.\n\n【Key Insight】This problem primarily examines the application of solving right triangles. Mastering the relationships between the sides and angles of right triangles is crucial for solving such problems." }, { "problem_id": 1956, "question": "A desk manufacturer has found that tabletops tilted between $12^{\\circ}$ and $24^{\\circ}$ are beneficial for students to maintain a natural body posture. Based on this research, the manufacturer decided to make the horizontal desktop adjustable in angle. The design of the new desktop is shown in Figure 1, where $AB$ can rotate around point $A$, and a fixed-length, rotatable support arm $CD$ is installed at point $C$, with $AD = 30 \\text{ cm}$.\n\n(1) In Figure 2, when $\\angle BAC = 24^{\\circ}$ and $CD \\perp AB$, find the length of the support arm $CD$.\n\n(2) In Figure 3, when $\\angle BAC = 12^{\\circ}$, find the length of $AD$. (The result should be kept in radical form)\n\n(Reference data: $\\sin 24^{\\circ} \\approx 0.40, \\cos 24^{\\circ} \\approx 0.91, \\tan 24^{\\circ} \\approx 0.46, \\sin 12^{\\circ} \\approx 0.20$)\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0013_1.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0013_2.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0013_3.jpg" ], "is_multi_img": true, "answer": "（1） $12 \\mathrm{~cm}$;（2） $12 \\sqrt{6}+6 \\sqrt{3}$ or $12 \\sqrt{6}-6 \\sqrt{3}$.\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since \\(\\angle \\mathrm{BAC} = 24^\\circ\\) and \\(CD \\perp AB\\),\n\n\\[\n\\sin 24^\\circ = \\frac{CD}{AC}\n\\]\n\n\\[\n\\therefore CD = AC \\sin 24^\\circ = 30 \\times 0.40 = 12 \\text{ cm}\n\\]\n\n\\[\n\\therefore \\text{The length of the support arm } CD \\text{ is } 12 \\text{ cm}\n\\]\n\n(2) As shown in the figure, draw \\(CE \\perp AB\\) at point \\(E\\),\n\nWhen \\(\\angle \\mathrm{BAC} = 12^\\circ\\),\n\n\\[\n\\sin 12^\\circ = \\frac{CE}{AC} = \\frac{CE}{30}\n\\]\n\n\\[\n\\therefore CE = 30 \\sin 12^\\circ = 30 \\times 0.20 = 6 \\text{ cm}\n\\]\n\n\\[\n\\text{Since } CD = 12 \\text{ cm},\n\\]\n\n\\[\n\\text{By the Pythagorean theorem: } DE = \\sqrt{CD^2 - CE^2} = 6\\sqrt{3}, \\quad AE = \\sqrt{AC^2 - CE^2} = \\sqrt{30^2 - 6^2} = 12\\sqrt{6}\n\\]\n\n\\[\n\\therefore \\text{The length of } AD \\text{ is } (12\\sqrt{6} + 6\\sqrt{3}) \\text{ cm or } (12\\sqrt{6} - 6\\sqrt{3}) \\text{ cm}\n\\]\n\n\n【Key Insight】This problem examines the application of solving right triangles, and proficiency in using trigonometric relationships is crucial for solving it." }, { "problem_id": 1957, "question": "(Figure 1) A physical diagram of an air conditioner bracket and its installation data. As shown in (Figure 2), the horizontal rod \\( AB = 42 \\, \\text{cm} \\), the vertical fixed rod \\( HP = AB \\), the length from the mounting hole \\( C \\) to \\( E \\) is \\( 36 \\, \\text{cm} \\), the length to the fixed screw \\( B \\) is twice the length of \\( AE \\), \\( BH = 24 \\, \\text{cm} \\), the length between the mounting holes \\( D \\) and \\( E \\) is \\( 16 \\, \\text{cm} \\), one endpoint \\( F \\) of the support rod \\( GF \\) is the midpoint of \\( AD \\), and \\( PG = AE \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the length of \\( AE \\).\n\n(2) Find the degree measure of the angle between the support rod \\( GF \\) and the vertical fixed rod \\( HP \\) (referring to the acute angle) (reference data: \\( \\tan 65^\\circ \\approx 2.14 \\), \\( \\tan 64^\\circ \\approx 2.05 \\), \\( \\tan 63^\\circ \\approx 1.96 \\), \\( \\sin 65^\\circ = 0.91 \\), \\( \\sin 64^\\circ = 0.90 \\), \\( \\sin 63^\\circ \\approx 0.89 \\)).", "input_image": [ "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0035_1.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0035_2.jpg" ], "is_multi_img": true, "answer": "(1) $A E=2 \\mathrm{~cm}$ (2) $\\angle B G F=64^{\\circ}$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since \\( CE = 36 \\, \\text{cm} \\) and \\( AB = 42 \\, \\text{cm} \\), we have:\n\\[ BC + AE = AB - CE = 42 - 36 = 6 \\, \\text{cm} \\]\nGiven that \\( BC = 2AE \\), it follows that:\n\\[ AE = 2 \\, \\text{cm} \\]\n\n(2) From part (1), we know \\( AE = 2 \\, \\text{cm} \\) and \\( DE = 16 \\, \\text{cm} \\), thus:\n\\[ AD = 18 \\, \\text{cm} \\]\nSince point \\( F \\) is the midpoint of \\( AD \\):\n\\[ DF = \\frac{1}{2} AD = 9 \\, \\text{cm} \\]\nTherefore:\n\\[ BF = BC + CD + DF = 33 \\, \\text{cm} \\]\nGiven \\( HP = 42 \\, \\text{cm} \\), \\( HB = 24 \\, \\text{cm} \\), and \\( GP = AE = 2 \\, \\text{cm} \\), we find:\n\\[ BG = HP - HB - GP = 16 \\, \\text{cm} \\]\nIn the right triangle \\( \\triangle BFG \\), the tangent of angle \\( \\angle BGF \\) is:\n\\[ \\tan \\angle BGF = \\frac{BF}{BG} = \\frac{33}{16} \\approx 2.05 \\]\nSince \\( \\tan 64^\\circ \\approx 2.05 \\), it follows that:\n\\[ \\angle BGF = 64^\\circ \\]\nThus, the angle between the support rod \\( GF \\) and the vertical fixed rod \\( HP \\) is \\( 64^\\circ \\).\n\nThe answers are: (1) \\( AE = 2 \\, \\text{cm} \\) (2) \\( \\angle BGF = 64^\\circ \\)\n\n【Highlight】This problem examines the application of segment addition and subtraction, as well as the use of trigonometric functions in right triangles. It demonstrates the ability to abstract mathematical problems from real-world scenarios and solve them using mathematical methods." }, { "problem_id": 1958, "question": "As shown in Figure (1), the fishing umbrella can be bent and rotated at any angle at point $O$ on the support pole $A N$ to block sunlight from different directions. Figure (2) is a schematic diagram of the fishing umbrella when it is upright. When the umbrella is fully opened, the semi-ribs $A B$ and $A C$ form angles of $\\angle A B C = \\angle A C B = 30^{\\circ}$ with the horizontal direction. The maximum horizontal distance between the ribs $A B$ and $A C$ is $B C = 2 \\text{ m}$, and $B C$ intersects $A N$ at point $M$. The length of the support pole $A N$ is $2.2 \\text{ m}$, and the distance from the fixed point $O$ to the ground is $O N = 1.6 \\text{ m}$.\n\n(1) As shown in Figure (2), when the umbrella is fully opened and upright, find the distance from point $B$ to the ground.\n\n(2) On a certain day at a certain time, to increase the area of shade from oblique sunlight, the fishing umbrella is tilted at an angle of $30^{\\circ}$ to the vertical line $H N$, as shown in Figure (3).\n\n(1) Find the distance from point $B$ to the ground at this time.\n\n(2) If the oblique sunlight is perpendicular to the line $B C$, find the length of the projection of $B C$ on the horizontal ground, approximately. (Note: $\\sqrt{3} \\approx 1.732$, the result should be accurate to $0.1 \\text{ m}$)\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)", "input_image": [ "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0058_1.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0058_2.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0058_3.jpg" ], "is_multi_img": true, "answer": "(1) The distance from point $B$ to the ground is about $1.6 \\mathrm{~m}$; (2) (1) At this time, the distance from point $B$ to the ground is about $1.1 \\mathrm{~m}$; (2) The length of the projection of $BC$ on the horizontal ground is about $2.3 \\mathrm{~m}$.\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) The distance from point $B$ to the ground is the length of $MN$,\n\n$MN = AN - AM = AN - BM \\tan 30^\\circ = 2.2 - \\frac{\\sqrt{3}}{3} \\approx 1.6 \\quad (\\mathrm{m})$.\n\nAnswer: The distance from point $B$ to the ground is approximately $1.6 \\mathrm{m}$.\n\n(2) As shown in Figure (1), draw perpendiculars from points $A$ and $B$ to the ground, with the feet of the perpendiculars being $Q$ and $T$ respectively,\n\n\n\n(1)\n\nSince $\\angle AOH = 30^\\circ$,\n\n$\\therefore \\angle OAQ = 30^\\circ$.\n\nSince $\\angle ABC = 30^\\circ$,\n\n$\\therefore \\angle BAO = 90^\\circ - \\angle ABC = 60^\\circ$,\n\n$\\therefore \\angle BAQ = \\angle BAO - \\angle OAQ = 30^\\circ$,\n\n$\\therefore \\angle ABS = 30^\\circ$,\n\n$\\therefore BS = BM = 1$.\n\n$\\therefore BT = OP + ON - SB = OA \\cos 30^\\circ + ON - SB = 0.6 \\times \\frac{\\sqrt{3}}{2} + 1.6 - 1 \\approx 1.1 \\quad (\\mathrm{m})$.\n\nAnswer: At this time, the distance from point $B$ to the ground is approximately $1.1 \\mathrm{m}$.\n\n(2) As shown in Figure (2), according to the problem, we know that $BC \\perp CD$, $\\angle CBD = 30^\\circ$.\n\n\n\n(2)\n\nSince $BC = 2$,\n\n$\\therefore BD = \\frac{4 \\sqrt{3}}{3} \\approx 2.3 (\\mathrm{m})$.\n\nAnswer: The length of the projection of $BC$ on the horizontal ground is approximately $2.3 \\mathrm{m}$.\n\n【Insight】This problem examines the practical application of solving right triangles, and mastering the method of solving right triangles is the key to solving the problem." }, { "problem_id": 1959, "question": "As shown in Figure 1, it is a physical diagram of a thermos and its abstract plane schematic. Points $\\mathrm{A}$ and $\\mathrm{B}$ are two fixed points on the thermos, connected to two movable rings. The handle $C D$ is connected to the two movable rings $A D$ and $B C$. It is measured that $A D=B C=2.6 \\mathrm{~cm}$, $A B=17 \\mathrm{~cm}$. As shown in Figure 2, when points $\\mathrm{A}$, $\\mathrm{D}$, and $\\mathrm{C}$ are collinear, $A C$ is perpendicular to $B C$.\n\n(1) Request the length of the handle $C D$;\n\n(2) As shown in Figure 3, when $C D$ is parallel to $A B$, find the degree of $\\angle A D C$.\n\n(Reference data: $\\sin 57.5^{\\circ}=0.843, \\cos 57.5^{\\circ}=0.538, \\tan 57.5^{\\circ}=1.570$)\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0079_1.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0079_2.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0079_3.jpg", "batch29-2024_06_14_dc9fce73f0ad95b0a72bg_0079_4.jpg" ], "is_multi_img": true, "answer": "(1) $C D=14.2 \\mathrm{~cm}$; （2） $\\angle A D C=122.5^{\\circ}$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: (1) As shown in Figure 2, in the right triangle $\\triangle ABC$, $AC = \\sqrt{AB^2 - BC^2} = \\sqrt{17^2 - 2.6^2} = 16.8 \\text{ cm}$.\n\nTherefore, $CD = AC - AD = 16.8 - 2.6 = 14.2 \\text{ cm}$.\n\n(2) As shown in Figure 3, draw perpendiculars from $C$ and $D$ to $AB$, meeting $AB$ at $E$ and $F$ respectively.\n\n\n\nFigure 3\n\nSince $CD \\parallel AB$,\n\n$\\angle CDF = 90^\\circ = \\angle DFE = \\angle CEF$,\n\nThus, quadrilateral $CDFE$ is a rectangle,\n\nTherefore, $DF = CE$ and $EF = CD$.\n\nSince $AD = BC$,\n\nRight triangles $\\triangle ADF$ and $\\triangle BCE$ are congruent,\n\nThus, $AF = BE = (AB - EF) \\div 2 = (17 - 14.2) \\div 2 = 1.4 \\text{ cm}$.\n\nTherefore, $\\cos \\angle DAF = \\frac{AF}{AD} = \\frac{1.4}{2.6} = \\frac{7}{13} \\approx 0.538$,\n\nThus, $\\angle DAF = 57.5^\\circ$.\n\nSince $CD \\parallel AB$,\n\n$\\angle ADC = 180^\\circ - 57.5^\\circ = 122.5^\\circ$.\n\n【Highlight】This problem examines the application of the Pythagorean theorem and the solution of right triangles. Mastering the Pythagorean theorem and trigonometric functions is key to solving such problems." }, { "problem_id": 1960, "question": "As shown in Figure (1), $\\angle 1$ and $\\angle 2$ are two non-adjacent exterior angles of quadrilateral $ABCD$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n(1) Conjecture and explain the quantitative relationship between $\\angle 1 + \\angle 2$ and $\\angle A, \\angle C$;\n\n(2) As shown in Figure (2), in quadrilateral $ABCD$, the bisectors of $\\angle ABC$ and $\\angle ADC$ intersect at point $O$. If $\\angle A = 58^\\circ, \\angle C = 152^\\circ$, find the degree measure of $\\angle BOD$;\n\n(3) As shown in Figure (3), $BO$ and $DO$ are the angle bisectors of the exterior angles $\\angle CBE$ and $\\angle CDF$ of quadrilateral $ABCD$. Please directly state the quantitative relationship between $\\angle A, \\angle C$ and $\\angle O$.", "input_image": [ "batch2-2024_06_13_1bb99e2296d49824b482g_0100_1.jpg", "batch2-2024_06_13_1bb99e2296d49824b482g_0100_2.jpg", "batch2-2024_06_13_1bb99e2296d49824b482g_0100_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle 1+\\angle 2=\\angle A+\\angle C$, \n\n(2) $133^{\\circ}$\n\n(3) $2 \\angle O=\\angle C-\\angle A$", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Based on the sum of interior and exterior angles of a polygon, the quantitative relationship between \\(\\angle 1 + \\angle 2\\) and \\(\\angle A\\), \\(\\angle C\\) can be explained;\n\n(2) First, according to the theorem of the sum of interior angles of a quadrilateral, find \\(\\angle ABC + \\angle ADC = 150^\\circ\\), then based on the definition of angle bisectors, find \\(\\angle OBC + \\angle ODC = 75^\\circ\\), and thus determine the measure of \\(\\angle BOD\\);\n\n(3) Combining the conclusion from (1), since \\(BO\\) and \\(DO\\) are the angle bisectors of the exterior angles \\(\\angle CBE\\) and \\(\\angle CDF\\) of quadrilateral \\(ABCD\\), the quantitative relationship between \\(\\angle A\\), \\(\\angle C\\), and \\(\\angle O\\) can be established.\n\n(1)\n\nSolution: Conjecture: \\(\\angle 1 + \\angle 2 = \\angle A + \\angle C\\), reasoning as follows:\n\nSince \\(\\angle 1 + \\angle ABC = \\angle 2 + \\angle ADC = 180^\\circ\\),\n\nTherefore, \\(\\angle 1 + \\angle ABC + \\angle 2 + \\angle ADC = 360^\\circ\\).\n\nAlso, since \\(\\angle A + \\angle ABC + \\angle C + \\angle ADC = 360^\\circ\\),\n\nTherefore, \\(\\angle 1 + \\angle 2 = \\angle A + \\angle C\\);\n\n(2)\n\nSolution: Since \\(\\angle A = 58^\\circ\\) and \\(\\angle C = 152^\\circ\\),\n\nTherefore, \\(\\angle ABC + \\angle ADC = 360^\\circ - \\angle A - \\angle C = 150^\\circ\\),\n\nAlso, since \\(BO\\) and \\(DO\\) bisect \\(\\angle ABC\\) and \\(\\angle ADC\\) respectively,\n\nTherefore, \\(\\angle OBC = \\frac{1}{2} \\angle ABC\\), \\(\\angle ODC = \\frac{1}{2} \\angle ADC\\),\n\nThus, \\(\\angle OBC + \\angle ODC = \\frac{1}{2}(\\angle ABC + \\angle ADC) = 75^\\circ\\),\n\nTherefore, \\(\\angle BOD = 360^\\circ - (\\angle OBC + \\angle ODC + \\angle C) = 133^\\circ\\);\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n(3)\n\nSolution: Since \\(BO\\) and \\(DO\\) are the angle bisectors of the exterior angles \\(\\angle CBE\\) and \\(\\angle CDF\\) of quadrilateral \\(ABCD\\),\n\nTherefore, \\(\\angle FDC = 2 \\angle FDO = 2 \\angle ODC\\), \\(\\angle EBC = 2 \\angle EBO = 2 \\angle CBO\\),\n\nFrom (1), it is known that: \\(\\angle FDO + \\angle EBO = \\angle A + \\angle O\\), \\(2 \\angle FDO + 2 \\angle EBO = \\angle A + \\angle C\\),\n\nTherefore, \\(2 \\angle A + 2 \\angle O = \\angle A + \\angle C\\),\n\nThus, \\(\\angle C - \\angle A = 2 \\angle O\\).\n\n【Key Point】This problem examines the theorem of the sum of interior angles of a polygon, the supplementary nature of adjacent angles, and the definition of angle bisectors. The key to solving the problem lies in proving the conclusion in (1) and applying it to find the solution." }, { "problem_id": 1961, "question": "Given that $\\mathrm{BC}$ is the diameter of $\\odot \\mathrm{O}$, $\\mathrm{BF}$ is a chord, $\\mathrm{AD}$ passes through the center $\\mathrm{O}$, $\\mathrm{AD} \\perp \\mathrm{BF}$, $\\mathrm{AE} \\perp \\mathrm{BC}$ at $\\mathrm{E}$, and FC is connected.\n\n(1) As shown in Figure 1, if $\\mathrm{OE}=2$, find $\\mathrm{CF}$;\n\n(2) As shown in Figure 2, connect $\\mathrm{DE}$ and extend it to intersect the extension of $\\mathrm{FC}$ at $\\mathrm{G}$, then connect $\\mathrm{AG}$. Please determine the positional relationship between the line $\\mathrm{AG}$ and $\\odot \\mathrm{O}$, and explain the reason.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch2-2024_06_13_298b9b5a7fbcbb4457b4g_0009_1.jpg", "batch2-2024_06_13_298b9b5a7fbcbb4457b4g_0009_2.jpg" ], "is_multi_img": true, "answer": "(1) 4; (2) The straight line $\\mathrm{AG}$ is tangent to $\\odot \\mathrm{O}$.\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since BC is the diameter of circle O, AD passes through the center O, AD is perpendicular to BF, and AE is perpendicular to BC at E,\n\nTherefore, ∠AEO = ∠BDO = 90°, and OA = OB.\n\nIn triangles AEO and BDO,\n\n\\[\n\\begin{cases}\n∠AEO = ∠BDO \\\\\n∠AOE = ∠BOD \\\\\nOA = OB\n\\end{cases}\n\\]\n\nThus, △AEO ≅ △BDO (AAS),\n\nSo, OE = OD = 2.\n\nSince BC is the diameter of circle O,\n\n∠CFB = 90°, which means CF is perpendicular to BF,\nTherefore, OD is parallel to CF.\n\nSince O is the midpoint of BC,\n\nOD is the median of triangle BFC,\n\nThus, CF = 2 × OD = 4.\n\n(2) The line AG is tangent to circle O, for the following reasons:\n\nConnect AB, as shown in the figure:\n\n\n\nSince OA = OB, and OE = OD,\n\nTriangles OAB and ODE are isosceles triangles.\n\nSince ∠AOB = ∠DOE,\n\n∠ADG = ∠OED = ∠BAD = ∠ABO.\n\nSince ∠GDF + ∠ADG = 90° = ∠BAD + ∠ABD,\n\n∠GDF = ∠ABD.\n\nSince OD is the median of triangle BFC,\n\nBD = DF.\n\nIn triangles ABD and GDF,\n\n\\[\n\\begin{cases}\n∠ABD = ∠GDF \\\\\nBD = DF \\\\\n∠ADB = ∠GFD = 90°\n\\end{cases}\n\\]\n\nThus, △ABD ≅ △GDF (ASA),\n\nSo, AD = GF.\n\nSince AD is perpendicular to BF, and GF is perpendicular to BF,\n\nAD is parallel to GF.\n\nTherefore, quadrilateral ADFG is a rectangle,\n\nSo, AG is perpendicular to OA,\n\nThus, the line AG is tangent to circle O.\n\n[Key Insight] This problem mainly examines the positional relationship between a line and a circle, the properties of tangents, and the congruence of triangles. The problem is comprehensive, requiring the integrated application of learned knowledge to solve." }, { "problem_id": 1962, "question": "Given point \\( C \\) on segment \\( AB \\), triangles \\( \\triangle ACD \\) and \\( \\triangle BCE \\) are constructed on the same side of segment \\( AB \\) with \\( AC \\) and \\( BC \\) as their respective sides, such that \\( CA = CD \\), \\( CB = CE \\), and \\( \\angle ACD = \\angle BCE \\). The lines \\( AE \\) and \\( BD \\) intersect at point \\( F \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) In Figure 1, if \\( \\angle ACD = 60^\\circ \\), then \\( \\angle AFD = \\) \\(\\qquad\\) -\n\n(2) In Figure 2, if \\( \\angle ACD = \\alpha \\), and \\( CF \\) is connected, then \\( \\angle AFC = \\) \\(\\qquad\\) (expressed in terms of \\( \\alpha \\))\n\n(3) In Figure 3, \\( \\triangle ACD \\) is rotated clockwise around point \\( C \\). After connecting \\( AE \\), \\( AB \\), and \\( BD \\), if \\( \\angle ABD = 80^\\circ \\), find the measure of \\( \\angle EAB \\).", "input_image": [ "batch2-2024_06_13_32362c34ea54cc68ed47g_0019_1.jpg", "batch2-2024_06_13_32362c34ea54cc68ed47g_0019_2.jpg", "batch2-2024_06_13_32362c34ea54cc68ed47g_0019_3.jpg" ], "is_multi_img": true, "answer": "（1） $60^{\\circ}$. (2) $90^{\\circ}-\\frac{1}{2} \\alpha$;（3） $\\angle \\mathrm{EAB}=140^{\\circ}$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since \\(\\angle \\mathrm{ACD} = \\angle \\mathrm{BCE}\\),\n\n\\(\\therefore \\angle \\mathrm{ACD} + \\angle \\mathrm{DCE} = \\angle \\mathrm{BCE} + \\angle \\mathrm{DCE}\\),\n\\(\\therefore \\angle \\mathrm{ACE} = \\angle \\mathrm{DCB}\\),\n\nIn triangles \\(\\triangle \\mathrm{ACE}\\) and \\(\\triangle \\mathrm{DCB}\\),\n\n\\[\n\\left\\{\n\\begin{array}{l}\nAC = DC \\\\\n\\angle ACE = \\angle DCB \\\\\nCE = CB\n\\end{array}\n\\right.\n\\]\n\n\\(\\therefore \\triangle \\mathrm{ACE} \\cong \\triangle \\mathrm{DCB}\\),\n\n\\(\\therefore \\angle \\mathrm{CAE} = \\angle \\mathrm{CDB}\\),\n\n\\(\\therefore \\angle \\mathrm{AFB} = \\angle \\mathrm{CDB} + \\angle \\mathrm{CDA} + \\angle \\mathrm{DAE}\\)\n\n\\(= \\angle \\mathrm{CDA} + \\angle \\mathrm{DAE} + \\angle \\mathrm{BAE}\\)\n\n\\(= \\angle \\mathrm{CDA} + \\angle \\mathrm{DAC}\\)\n\n\\(= 180^{\\circ} - 60^{\\circ}\\)\n\n\\(= 120^{\\circ}\\)\n\n\\(\\therefore \\angle \\mathrm{AFD} = 60^{\\circ}\\)\n\nThus, the answer is \\(60^{\\circ}\\).\n\n(2) Since \\(\\angle \\mathrm{ACD} = \\angle \\mathrm{BCE}\\),\n\n\\(\\therefore \\angle \\mathrm{ACD} + \\angle \\mathrm{DCE} = \\angle \\mathrm{BCE} + \\angle \\mathrm{DCE}\\),\n\n\\(\\therefore \\angle \\mathrm{ACE} = \\angle \\mathrm{DCB}\\),\n\nIn triangles \\(\\triangle \\mathrm{ACE}\\) and \\(\\triangle \\mathrm{DCB}\\),\n\n\\[\n\\left\\{\n\\begin{array}{l}\nAC = DC \\\\\n\\angle ACE = \\angle DCB \\\\\nCE = CB\n\\end{array}\n\\right.\n\\]\n\n\\(\\therefore \\triangle \\mathrm{ACE} \\cong \\triangle \\mathrm{DCB}\\),\n\n\\(\\therefore \\angle \\mathrm{CAE} = \\angle \\mathrm{CDB}\\),\n\n\\(\\therefore \\angle \\mathrm{AFB} = \\angle \\mathrm{CDB} + \\angle \\mathrm{CDA} + \\angle \\mathrm{DAE}\\)\n\n\\(= \\angle \\mathrm{CDA} + \\angle \\mathrm{DAE} + \\angle \\mathrm{BAE}\\)\n\n\\(= \\angle \\mathrm{CDA} + \\angle \\mathrm{DAC}\\)\n\n\\(= 180^{\\circ} - \\angle \\mathrm{ACD}\\)\n\n\\(= 180^{\\circ} - \\alpha\\),\n\nSince \\(\\angle \\mathrm{CAE} = \\angle \\mathrm{CDB}\\),\n\n\\(\\therefore\\) Points \\(A, C, F, D\\) are concyclic,\n\n\\(\\therefore \\angle \\mathrm{ADC} = \\angle \\mathrm{AFC}\\),\n\nSimilarly, \\(\\angle \\mathrm{CFB} = \\angle \\mathrm{BEC}\\),\n\nSince \\(AC = CD\\), \\(CB = CE\\), and \\(\\angle \\mathrm{ACD} = \\angle \\mathrm{BCE}\\),\n\n\\(\\therefore \\angle \\mathrm{ADC} = \\angle \\mathrm{BEC}\\),\n\n\\(\\therefore \\angle \\mathrm{AFC} = \\frac{1}{2} \\angle \\mathrm{AFB} = 90^{\\circ} - \\frac{1}{2} \\alpha\\);\n\nThus, the answer is \\(90^{\\circ} - \\frac{1}{2} \\alpha\\);\n\n(3) Since \\(\\triangle \\mathrm{ACD}\\) is an equilateral triangle,\n\n\\(\\therefore \\angle \\mathrm{ACD} = 60^{\\circ}\\),\n\n\\[\n\\begin{aligned}\n& \\because \\angle \\mathrm{ABD} = 80^{\\circ}, \\\\\n& \\therefore \\angle \\mathrm{CAB} + \\angle \\mathrm{CDB} = 360^{\\circ} - 60^{\\circ} - 80^{\\circ} = 220^{\\circ}, \\\\\n& \\because \\angle \\mathrm{ACD} = \\angle \\mathrm{BCE}, \\\\\n& \\therefore \\angle \\mathrm{ACE} = \\angle \\mathrm{BCD}, \\\\\n& \\text{In triangles } \\triangle \\mathrm{ACE} \\text{ and } \\triangle \\mathrm{BCD}, \\\\\n& \\left\\{\n\\begin{array}{l}\nCE = BC \\\\\n\\angle ACE = \\angle BCD \\\\\nAC = CD\n\\end{array}\n\\right. \\\\\n& \\therefore \\triangle \\mathrm{ACE} \\cong \\triangle \\mathrm{DCB}, \\\\\n& \\therefore \\angle \\mathrm{CAE} = \\angle \\mathrm{CDB}, \\\\\n& \\therefore \\angle \\mathrm{CAE} + \\angle \\mathrm{CAB} = 220^{\\circ}, \\\\\n& \\therefore \\angle \\mathrm{EAB} = 140^{\\circ}.\n\\end{aligned}\n\\]\n\n【Highlight】This problem examines the properties and determination of congruent triangles, the exterior angle property of triangles, and the application of the triangle angle sum theorem. The key is to deduce that \\(\\triangle \\mathrm{ACE} \\cong \\triangle \\mathrm{DCB}\\)." }, { "problem_id": 1963, "question": "Given: In $\\triangle A B C$, $B E$ is the angle bisector of $\\angle A B C$, $B D$ is the altitude on side $A C$, and through point $A$, $A F$ is drawn parallel to $B E$, intersecting line $B D$ at point $F$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) As shown in Figure 1, if $\\angle A B C=74^{\\circ}, \\angle C=32^{\\circ}$, then $\\angle A F B=$ $\\qquad$ ${ }^{\\circ} ;$\n\n(2) If $\\angle B A C=\\alpha, \\angle A C B=\\beta(\\alpha>\\beta)$ in (1), find $\\angle A F B$; (expressed in terms of $\\alpha, \\beta$);\n\n(3) As shown in Figure 2, does the conclusion in (2) still hold? If it holds, explain why; if it does not hold, find $\\angle A F B$. (expressed in terms of $\\alpha, \\beta$).", "input_image": [ "batch2-2024_06_13_869f5d82626b22845d16g_0011_1.jpg", "batch2-2024_06_13_869f5d82626b22845d16g_0011_2.jpg" ], "is_multi_img": true, "answer": "(1)21\n\n(2) $\\angle A F B=\\frac{1}{2}(\\alpha-\\beta)$;\n\n(3) The conclusion in (2) does not hold, $\\angle A F B=180^{\\circ}+\\frac{1}{2}(\\beta-\\alpha)$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Solution: Since \\(\\angle ABC = 74^\\circ\\) and \\(BE\\) bisects \\(\\angle ABC\\),\n\n\\[\n\\angle CBE = \\frac{1}{2} \\angle ABC = 37^\\circ.\n\\]\n\nIn \\(\\triangle CBE\\),\n\n\\[\n\\angle AEB = \\angle C + \\angle CBE = 32^\\circ + 37^\\circ = 69^\\circ.\n\\]\n\nSince \\(BF \\perp AC\\),\n\n\\[\n\\angle BDE = 90^\\circ.\n\\]\n\nThus,\n\n\\[\n\\angle EBD = 90^\\circ - 69^\\circ = 21^\\circ.\n\\]\n\nSince \\(AF \\parallel BE\\),\n\n\\[\n\\angle AFB = \\angle EBD = 21^\\circ.\n\\]\n\nTherefore, the answer is: 21.\n\n(2) Solution: Since \\(\\angle BAC = \\alpha\\) and \\(\\angle ACB = \\beta\\),\n\n\\[\n\\angle ABC = 180^\\circ - \\alpha - \\beta.\n\\]\n\nSince \\(BE\\) bisects \\(\\angle ABC\\),\n\n\\[\n\\angle CBE = \\frac{1}{2} \\angle ABC = 90^\\circ - \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta.\n\\]\n\nIn \\(\\triangle CBE\\),\n\n\\[\n\\angle AEB = \\angle C + \\angle CBE = \\beta + 90^\\circ - \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta = 90^\\circ - \\frac{1}{2} \\alpha + \\frac{1}{2} \\beta.\n\\]\n\nSince \\(\\angle BDE = 90^\\circ\\),\n\n\\[\n\\angle EBD = 90^\\circ - \\left(90^\\circ - \\frac{1}{2} \\alpha + \\frac{1}{2} \\beta\\right) = \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta.\n\\]\n\nSince \\(AF \\parallel BE\\),\n\n\\[\n\\angle AFB = \\angle EBD = \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta = \\frac{1}{2}(\\alpha - \\beta).\n\\]\n\n(3) Solution: As shown in Figure 2, the conclusion in (2) does not hold. The reasoning is as follows:\n\nSince \\(\\angle BAC = \\alpha\\) and \\(\\angle ACB = \\beta\\),\n\n\\[\n\\angle ABC = 180^\\circ - \\alpha - \\beta.\n\\]\n\nSince \\(BE\\) bisects \\(\\angle ABC\\),\n\n\\[\n\\angle ABE = \\frac{1}{2} \\angle ABC = 90^\\circ - \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta.\n\\]\n\nIn \\(\\triangle ABC\\),\n\n\\[\n\\angle DAB = 180^\\circ - \\angle BAC = 180^\\circ - \\alpha.\n\\]\n\nSince \\(AF \\parallel BE\\),\n\n\\[\n\\angle FAB = \\angle ABE.\n\\]\n\nSince \\(\\angle D = 90^\\circ\\),\n\n\\[\n\\angle ABD = 90^\\circ - \\angle DAB = 90^\\circ - (180^\\circ - \\alpha) = \\alpha - 90^\\circ.\n\\]\n\nThus,\n\n\\[\n\\angle AFB = 180^\\circ - \\angle FAB - \\angle ABD = 180^\\circ - \\left(90^\\circ - \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta\\right) - (\\alpha - 90^\\circ) = 180^\\circ + \\frac{1}{2} \\beta - \\frac{1}{2} \\alpha = 180^\\circ + \\frac{1}{2}(\\beta - \\alpha).\n\\]\n\n**Key Insight:** This problem examines the knowledge of the triangle angle sum theorem, properties of parallel lines, and the exterior angle theorem of triangles. The key to solving the problem lies in the flexible application of these concepts." }, { "problem_id": 1964, "question": "As shown in the figure, the line \\( H D \\) is parallel to \\( G E \\), point \\( A \\) is on line \\( H D \\), point \\( C \\) is on line \\( G E \\), and point \\( B \\) is between lines \\( D H \\) and \\( G E \\), with \\( \\angle D A B = 120^\\circ \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) In Figure 1, if \\( \\angle B C G = 40^\\circ \\), find the measure of \\( \\angle A B C \\);\n\n(2) In Figure 2, \\( A F \\) bisects \\( \\angle H A B \\), \\( B C \\) bisects \\( \\angle F C G \\), and \\( \\angle B C G = 20^\\circ \\). Compare the sizes of \\( \\angle B \\) and \\( \\angle F \\);\n\n(3) In Figure 3, point \\( P \\) is a point on segment \\( A B \\), \\( P N \\) bisects \\( \\angle A P C \\), and \\( C N \\) bisects \\( \\angle P C E \\). Directly state the relationship between the measures of \\( \\angle H A P \\) and \\( \\angle N \\).", "input_image": [ "batch2-2024_06_13_869f5d82626b22845d16g_0047_1.jpg", "batch2-2024_06_13_869f5d82626b22845d16g_0047_2.jpg", "batch2-2024_06_13_869f5d82626b22845d16g_0047_3.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle A B C=100^{\\circ}$;\n\n(2) $\\angle A B C>\\angle A F C$;\n\n(3) $\\angle N=90^{\\circ}-\\frac{1}{2} \\angle H A P$.", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) Draw a line \\( BM \\) through point \\( B \\) such that \\( BM \\parallel HD \\), then \\( HD \\parallel GE \\parallel BM \\). Using the properties of parallel lines, we can determine \\( \\angle ABM \\) and \\( \\angle CBM \\), and thus find the final result.\n\n(2) Draw a line \\( BP \\) through \\( B \\) such that \\( BP \\parallel HD \\), then \\( BP \\parallel HD \\parallel GE \\). Draw a line \\( FQ \\) through \\( F \\) such that \\( FQ \\parallel HD \\), then \\( FQ \\parallel HD \\parallel GE \\). By the properties of parallel lines, we have \\( \\angle ABC = \\angle HAB + \\angle BCG \\) and \\( \\angle AFC = \\angle HAF + \\angle FCG \\). Using the properties of angle bisectors and the given angle measures, we can find \\( \\angle HAF \\) and \\( \\angle FCG \\), and finally determine the result.\n\n(3) Draw a line \\( PK \\) through \\( P \\) such that \\( PK \\parallel HD \\), then \\( PK \\parallel HD \\parallel GE \\). First, using the properties of parallel lines, we prove that \\( \\angle ABC = \\angle HAB + \\angle BCG \\) and \\( \\angle AFC = \\angle HAF + \\angle FCG \\). Then, using the angle bisector, we find \\( \\angle NPC \\) and \\( \\angle PCN \\). Finally, using the triangle angle sum theorem, we can determine the result.\n\n(1)\n\nAs shown in Figure 1, draw a line \\( BM \\) through point \\( B \\) such that \\( BM \\parallel HD \\), then \\( HD \\parallel GE \\parallel BM \\).\n\n\n\nFigure 1\n\nTherefore, \\( \\angle ABM = 180^\\circ - \\angle DAB \\) and \\( \\angle CBM = \\angle BCG \\).\n\nGiven \\( \\angle DAB = 120^\\circ \\) and \\( \\angle BCG = 40^\\circ \\),\n\nwe have \\( \\angle ABM = 60^\\circ \\) and \\( \\angle CBM = 40^\\circ \\).\n\nThus, \\( \\angle ABC = \\angle ABM + \\angle CBM = 100^\\circ \\).\n\n(2)\n\nAs shown in Figure 2, draw a line \\( BP \\) through \\( B \\) such that \\( BP \\parallel HD \\), then \\( BP \\parallel HD \\parallel GE \\). Draw a line \\( FQ \\) through \\( F \\) such that \\( FQ \\parallel HD \\), then \\( FQ \\parallel HD \\parallel GE \\).\n\n\n\nFigure 2\n\nTherefore, \\( \\angle ABP = \\angle HAB \\), \\( \\angle CBP = \\angle BCG \\), \\( \\angle AFQ = \\angle HAF \\), and \\( \\angle CFQ = \\angle FCG \\).\n\nThus, \\( \\angle ABC = \\angle HAB + \\angle BCG \\) and \\( \\angle AFC = \\angle HAF + \\angle FCG \\).\n\nGiven \\( \\angle DAB = 120^\\circ \\),\n\nwe have \\( \\angle HAB = 180^\\circ - 120^\\circ = 60^\\circ \\).\n\nSince \\( AF \\) bisects \\( \\angle HAB \\) and \\( BC \\) bisects \\( \\angle FCG \\), and \\( \\angle BCG = 20^\\circ \\),\n\nwe have \\( \\angle HAF = 30^\\circ \\) and \\( \\angle FCG = 40^\\circ \\).\n\nThus, \\( \\angle ABC = 60^\\circ + 20^\\circ = 80^\\circ \\),\n\nand \\( \\angle AFC = 30^\\circ + 40^\\circ = 70^\\circ \\).\n\nTherefore, \\( \\angle ABC > \\angle AFC \\).\n\n(3)\n\nAs shown in Figure 3, draw a line \\( PK \\) through \\( P \\) such that \\( PK \\parallel HD \\), then \\( PK \\parallel HD \\parallel GE \\).\n\n\n\nFigure 3\n\nTherefore, \\( \\angle APK = \\angle HAP \\) and \\( \\angle CPK = \\angle PCG \\).\n\nThus, \\( \\angle APC = \\angle HAP + \\angle PCG \\).\n\nSince \\( PN \\) bisects \\( \\angle APC \\),\n\nwe have \\( \\angle NPC = \\frac{1}{2} \\angle HAP + \\frac{1}{2} \\angle PCG \\).\n\nGiven \\( \\angle PCE = 180^\\circ - \\angle PCG \\) and \\( CN \\) bisects \\( \\angle PCE \\),\n\nwe have \\( \\angle PCN = \\frac{1}{2} \\angle PCE = 90^\\circ - \\frac{1}{2} \\angle PCG \\).\n\nSince \\( \\angle N + \\angle NPC + \\angle PCN = 180^\\circ \\),\n\nwe have \\( \\angle N = 180^\\circ - \\frac{1}{2} \\angle HAP - \\frac{1}{2} \\angle PCG - 90^\\circ + \\frac{1}{2} \\angle PCG = 90^\\circ - \\frac{1}{2} \\angle HAP \\).\n\nThat is, \\( \\angle N = 90^\\circ - \\frac{1}{2} \\angle HAP \\).\n\n【Insight】This problem examines the definition of angle bisectors, the triangle angle sum theorem, and the properties and determination of parallel lines: if two lines are parallel, their corresponding angles are equal; if two lines are parallel, their consecutive interior angles are supplementary; if two lines are parallel, their alternate interior angles are equal. This problem is of moderate difficulty, and attention should be paid to the method of constructing auxiliary lines." }, { "problem_id": 1965, "question": "As shown in Figure 1, given that \\( AB \\parallel CD \\) and \\( AC \\parallel EF \\):\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Observation and conjecture: If \\( \\angle A = 45^\\circ \\) and \\( \\angle E = 65^\\circ \\), then the degree measure of \\( \\angle CDE \\) is\n\n(2) Investigation of the problem: Please explore in Figure 1 what quantitative relationship exists between \\( \\angle A \\), \\( \\angle CDE \\), and \\( \\angle E \\), and explain the reason.\n\n(3) Extension: If Figure 1 is changed to Figure 2 with the same conditions, what quantitative relationship exists between \\( \\angle CAB \\), \\( \\angle CDE \\), and \\( \\angle E \\) now? Please state the conclusion and explain the reason.", "input_image": [ "batch2-2024_06_13_8b5913f829bd624e35f3g_0009_1.jpg", "batch2-2024_06_13_8b5913f829bd624e35f3g_0009_2.jpg" ], "is_multi_img": true, "answer": "(1) $110^{\\circ}$\n\n(2) $\\angle C D E=\\angle A+\\angle E$;\n\n(3) $\\angle C A B=\\angle E+\\angle D$,\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) Solution: Extend $AB$ to intersect $DE$ at point $G$ and $EF$ at point $H$, as shown in the figure:\n\n\n\nSince $AC \\parallel EF$ and $\\angle A = 45^\\circ$,\n\n$\\therefore \\angle EHG = \\angle A = 45^\\circ$.\n\nGiven that $\\angle E = 65^\\circ$,\n\n$\\therefore \\angle DGH = \\angle E + \\angle EHG = 65^\\circ + 45^\\circ = 110^\\circ$.\n\nSince $AB \\parallel CD$,\n\n$\\therefore \\angle CDE = \\angle DGH = 110^\\circ$.\n\nThus, the answer is: $110^\\circ$.\n\n(2) Solution: $\\angle CDE = \\angle A + \\angle E$; the reasoning is as follows:\n\nExtend $AB$ to intersect $DE$ at point $G$ and $EF$ at point $H$, as shown in the figure:\n\n\n\nSince $AC \\parallel EF$,\n\n$\\therefore \\angle EHG = \\angle A$.\n\nThus, $\\angle DGH = \\angle E + \\angle EHG = \\angle E + \\angle A$.\n\nSince $AB \\parallel CD$,\n\n$\\therefore \\angle CDE = \\angle DGH = \\angle A + \\angle E$.\n\n(3) Solution: $\\angle CAB = \\angle E + \\angle D$, the reasoning is as follows:\n\nExtend $CA$ to intersect $DE$ at point $G$, and let $AB$ intersect $DE$ at point $H$, as shown in the figure:\n\n\n\nSince $AC \\parallel EF$,\n\n$\\therefore \\angle CGD = \\angle E$.\n\nSince $AB \\parallel CD$,\n\n$\\therefore \\angle AHG = \\angle D$.\n\nThus, $\\angle CAB = \\angle CGD + \\angle AHG = \\angle E + \\angle D$.\n\n【Key Insight】This problem primarily examines the properties of parallel lines and the exterior angle theorem of triangles. The key to solving the problem lies in a thorough understanding of the properties of parallel lines and the fact that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles." }, { "problem_id": 1966, "question": "As shown in the figure, in $\\triangle A B C$, $C D \\perp B A$, intersecting the extension of $B A$ at $D$, and $D E \\perp A C$ at $E$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) In Figure 1, if $\\angle B=35^{\\circ}$ and $\\angle C D E=60^{\\circ}$, find the measure of $\\angle A C B$;\n\n(2) In Figure 2, if $A C$ bisects $\\angle B C D$ and $B F \\perp A C$, intersecting the extension of $C A$ at $F$, directly write down the angles that are equal to $\\angle A C B$ (excluding $\\angle A C B$ itself).", "input_image": [ "batch2-2024_06_13_8b5913f829bd624e35f3g_0020_1.jpg", "batch2-2024_06_13_8b5913f829bd624e35f3g_0020_2.jpg" ], "is_multi_img": true, "answer": "(1) $25^{\\circ}$\n\n(2) $\\angle A C D, \\angle A B F, \\angle A D E$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "0", "analysis": "(1) Solution: Since \\( CD \\perp BA \\) and \\( \\angle B = 35^\\circ \\),\n\n\\[\n\\angle DCB = 90^\\circ - 35^\\circ = 55^\\circ.\n\\]\n\nSince \\( DE \\perp AC \\) and \\( \\angle CDE = 60^\\circ \\),\n\n\\[\n\\angle DCE = 90^\\circ - 60^\\circ = 30^\\circ.\n\\]\n\nTherefore,\n\n\\[\n\\angle ACB = \\angle BCD - \\angle DCE = 55^\\circ - 30^\\circ = 25^\\circ.\n\\]\n\n(2) Solution: Since \\( AC \\) bisects \\( \\angle BCD \\),\n\n\\[\n\\angle ACD = \\angle ACB.\n\\]\n\nSince \\( BF \\perp AC \\) and \\( CD \\perp BA \\),\n\n\\[\n\\angle F = \\angle ADC = 90^\\circ.\n\\]\n\nThus,\n\n\\[\n\\angle ABF + \\angle FAB = \\angle ACD + \\angle DAC.\n\\]\n\nSince \\( \\angle FAB = \\angle DAC \\),\n\n\\[\n\\angle ABF = \\angle ACD = \\angle ACB.\n\\]\n\nSince \\( DE \\perp AC \\),\n\nSimilarly, we can deduce that \\( \\angle ABF = \\angle ADE \\).\n\nTherefore,\n\n\\[\n\\angle ACB = \\angle ABF = \\angle ACD = \\angle ADE.\n\\]\n\n**Key Insight:** This problem examines the complementary angles in a right triangle, the properties of angle bisectors, and angle calculations. Mastery of these properties and the ability to perform angle transformations are crucial for solving the problem." }, { "problem_id": 1967, "question": "Figure 1 shows line segments \\( AB \\) and \\( CD \\) intersecting at point \\( O \\), with \\( AD \\) and \\( CB \\) connected. We refer to the shape in Figure 1 as the \"8-shape.\" In Figure 2, under the conditions of Figure 1, the angle bisectors \\( AP \\) and \\( CP \\) of \\( \\angle DAB \\) and \\( \\angle BCD \\) intersect at point \\( P \\), and they also intersect \\( CD \\) and \\( AB \\) at points \\( M \\) and \\( N \\), respectively. Answer the following questions:\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) In Figure 1, directly state the quantitative relationship between \\( \\angle A + \\angle D \\) and \\( \\angle B + \\angle C \\) as: _;\n\n(2) Carefully observe, in Figure 2, the number of \"8-shapes\": _;\n\n(3) In Figure 2, when \\( \\angle D \\) and \\( \\angle B \\) are arbitrary angles and the other conditions remain unchanged, what is the quantitative relationship between \\( \\angle P \\) and \\( \\angle D \\), \\( \\angle B \\)? Explain the reason.\n\n(4) Application: In Figure 2, when \\( \\angle D = 50^\\circ \\) and \\( \\angle B = 40^\\circ \\), directly state the degree measure of \\( \\angle P \\).", "input_image": [ "batch2-2024_06_13_8b5913f829bd624e35f3g_0024_1.jpg", "batch2-2024_06_13_8b5913f829bd624e35f3g_0024_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle A+\\angle D=\\angle B+\\angle C$\n\n(2)6\n\n(3) $\\angle D+\\angle B=2 \\angle P$\n\n(4) $45^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Solution: According to the triangle angle sum theorem,\n\n$\\angle A + \\angle D + \\angle AOD = 180^\\circ$ and $\\angle B + \\angle C + \\angle BOC = 180^\\circ$.\n\nSince $\\angle AOD = \\angle BOC$,\n\nTherefore, $\\angle A + \\angle D = \\angle B + \\angle C$.\n\nHence, the answer is: $\\angle A + \\angle D = \\angle B + \\angle C$.\n\n(2) The pairs of triangles are: $\\triangle AOD$ with $\\triangle BOC$, $\\triangle AMD$ with $\\triangle CMP$, $\\triangle AOM$ with $\\triangle CON$, $\\triangle ANP$ with $\\triangle CNB$, $\\triangle AOD$ with $\\triangle CON$, and $\\triangle AOM$ with $\\triangle BOC$, totaling six pairs.\n\nHence, the answer is: 6.\n\n(3) Since $\\angle D + \\angle DAM + \\angle AMD = 180^\\circ$ and $\\angle P + \\angle PCM + \\angle PMC = 180^\\circ$,\n\nTherefore, $\\angle D + \\angle DAM = \\angle P + \\angle PCM$.\n\nSimilarly, since $\\angle B + \\angle BCN + \\angle BNC = 180^\\circ$ and $\\angle P + \\angle PAN + \\angle ANP = 180^\\circ$,\n\nTherefore, $\\angle B + \\angle BCN = \\angle P + \\angle PAN$.\n\nGiven that $AP$ and $CP$ bisect $\\angle DAB$ and $\\angle BCD$ respectively,\n\nTherefore, $\\angle DAM = \\angle PAN$ and $\\angle BCN = \\angle PCM$.\n\nThus, $\\angle D + \\angle DAM + \\angle B + \\angle BCN = \\angle P + \\angle PCM + \\angle P + \\angle PAN$,\n\nWhich simplifies to $\\angle D + \\angle B = 2\\angle P$.\n\n(4) Given $\\angle D = 50^\\circ$ and $\\angle B = 40^\\circ$,\n\nFrom part (3), we know $\\angle D + \\angle B = 2\\angle P$,\n\nTherefore, $\\angle P = \\frac{1}{2}(\\angle D + \\angle B) = 45^\\circ$.\n\n[Key Insight] This problem primarily examines the triangle angle sum theorem and the definition of angle bisectors. Mastering the relevant knowledge and applying it flexibly is key to solving the problem." }, { "problem_id": 1968, "question": "As shown in the figure, given that the line \\( CP \\parallel OQ \\), points \\( B \\) and \\( A \\) are respectively on the rays \\( CP \\) and \\( OQ \\), and satisfy \\( AB \\parallel OC \\), \\( \\angle BCO = 100^\\circ \\). Point \\( F \\) is on the line \\( BC \\) and to the left of point \\( B \\), satisfying \\( \\angle FOB = \\angle FBO = \\alpha \\), and the angle bisector of \\( \\angle COF \\) intersects the line \\( CP \\) at point \\( E \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) In Figure 1, find the measure of \\( \\angle BOE \\);\n\n(2) In Figure 2, if \\( \\alpha = 45^\\circ \\), find the measure of \\( \\angle BOE \\);\n\n(3) If the segment \\( AB \\) is translated left and right, is there a possibility that \\( \\angle OEC = \\frac{3}{2} \\angle OBA \\)? If so, find the value of \\( \\alpha \\) at that time; if not, explain the reason.", "input_image": [ "batch2-2024_06_13_8b5913f829bd624e35f3g_0046_1.jpg", "batch2-2024_06_13_8b5913f829bd624e35f3g_0046_2.jpg", "batch2-2024_06_13_8b5913f829bd624e35f3g_0046_3.jpg" ], "is_multi_img": true, "answer": "(1) $40^{\\circ}$\n\n(2)$40^{\\circ}$\n\n(3)exist, $\\alpha=32^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\n(1) Since \\( CP \\parallel OQ \\), \\( \\angle BCO = 100^\\circ \\), and \\( \\angle FOB = \\angle FBO = \\alpha \\),\n\n\\[\n\\angle COA = 180^\\circ - \\angle BCO = 80^\\circ, \\quad \\angle BOA = \\angle FBO = \\angle FOB = \\alpha,\n\\]\n\n\\[\n\\angle COF = \\angle COA - \\angle FOB - \\angle BOA = 80^\\circ - 2\\alpha,\n\\]\n\nSince \\( OE \\) bisects \\( \\angle COF \\),\n\n\\[\n\\angle EOF = \\frac{1}{2} \\angle COF = 40^\\circ - \\alpha,\n\\]\n\n\\[\n\\angle BOE = \\angle EOF + \\angle FOB = 40^\\circ,\n\\]\n\nThus, the measure of \\( \\angle BOE \\) is \\( 40^\\circ \\).\n\n---\n\n(2) When \\( \\alpha = 45^\\circ \\), the supplementary diagram is as shown in Figure 1.\n\n\n\nFrom part (1), we know \\( \\angle COA = 80^\\circ \\), \\( \\angle BOA = 45^\\circ \\),\n\n\\[\n\\angle COB = \\angle COA - \\angle BOA = 35^\\circ,\n\\]\n\n\\[\n\\angle COF = \\angle FOB - \\angle COB = 10^\\circ,\n\\]\n\n\\[\n\\angle EOC = \\frac{1}{2} \\angle COF = 5^\\circ,\n\\]\n\n\\[\n\\angle BOE = \\angle EOC + \\angle COB = 40^\\circ,\n\\]\n\nThus, the measure of \\( \\angle BOE \\) is \\( 40^\\circ \\).\n\n---\n\n(3) A solution exists when \\( \\alpha = 32^\\circ \\).\n\nFrom the problem statement, we consider two cases: \\( F \\) is to the left of \\( C \\) and \\( F \\) is to the right of \\( C \\).\n\n**Case 1:** When \\( F \\) is to the left of \\( C \\), as shown in Figure 2.\n\n\n\nSince \\( AB \\parallel OC \\),\n\n\\[\n\\angle BAQ = \\angle COA = 80^\\circ,\n\\]\n\n\\[\n\\angle COB = \\angle OBA = \\angle BAQ - \\angle BOA = 80^\\circ - \\alpha,\n\\]\n\n\\[\n\\angle COF = \\angle FOB - \\angle COB = 2\\alpha - 80^\\circ,\n\\]\n\n\\[\n\\angle EOF = \\angle COE = \\frac{1}{2} \\angle COF = \\alpha - 40^\\circ,\n\\]\n\n\\[\n\\angle OFB = 180^\\circ - \\angle FOB - \\angle FBO = 180^\\circ - 2\\alpha,\n\\]\n\n\\[\n\\angle OEC = \\angle OFB + \\angle EOF = 140^\\circ - \\alpha,\n\\]\n\nSince \\( \\angle OEC = \\frac{3}{2} \\angle OBA \\),\n\n\\[\n140^\\circ - \\alpha = \\frac{3}{2}(80^\\circ - \\alpha),\n\\]\n\nSolving gives \\( \\alpha = -40^\\circ \\) (discarded).\n\n---\n\n**Case 2:** When \\( F \\) is to the right of \\( C \\), as shown in Figure 3.\n\n\n\nSimilarly,\n\n\\[\n\\angle COB = \\angle OBA = \\angle BAQ - \\angle BOA = 80^\\circ - \\alpha,\n\\]\n\n\\[\n\\angle COF = \\angle COB - \\angle FOB = 80^\\circ - 2\\alpha,\n\\]\n\n\\[\n\\angle EOF = \\angle COE = \\frac{1}{2} \\angle COF = 40^\\circ - \\alpha,\n\\]\n\n\\[\n\\angle OFC = \\angle FOB + \\angle FBO = 2\\alpha,\n\\]\n\n\\[\n\\angle OEC = \\angle OFC + \\angle EOF = 40^\\circ + \\alpha,\n\\]\n\nSince \\( \\angle OEC = \\frac{3}{2} \\angle OBA \\),\n\n\\[\n40^\\circ + \\alpha = \\frac{3}{2}(80^\\circ - \\alpha),\n\\]\n\nSolving gives \\( \\alpha = 32^\\circ \\).\n\n---\n\n**Conclusion:** When \\( \\alpha = 32^\\circ \\), it is possible for \\( \\angle OEC = \\frac{3}{2} \\angle OBA \\).\n\n**Key Insight:** This problem examines the properties of parallel lines, exterior angles of triangles, the triangle angle sum theorem, and angle bisectors. Clarifying the relationships between angles and solving for different cases is crucial to finding the solution." }, { "problem_id": 1969, "question": "As shown in Figure 1, in the plane rectangular coordinate system, $A O=A B, \\angle B A O=90^{\\circ}, B O=8 \\mathrm{~cm}$, a moving point $D$ starts from the origin $O$ and moves along the positive direction of the $x$-axis at a speed of $a \\mathrm{~cm} / \\mathrm{s}$, and a moving point $E$ also starts from the origin $O$ and moves along the $y$-axis at a speed of $b \\mathrm{~cm} / \\mathrm{s}$, where $a, b$ satisfy the relation $a^{2}+b^{2}-4 a-2 b+5=0$. Connect $O D, O E$, and let the time of motion be $t$ seconds.\n\n(1) Find the values of $a, b$;\n\n(2) For what value of $t$ is $\\triangle B A D \\cong \\triangle O A E$;\n\n(3) As shown in Figure 2, there exists a point $P$ in the first quadrant such that $\\angle A O P=30^{\\circ}, \\angle A P O=15^{\\circ}$, find $\\angle A B P$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch4-2024_06_14_0ee83ee022511a0918bag_0043_1.jpg", "batch4-2024_06_14_0ee83ee022511a0918bag_0043_2.jpg" ], "is_multi_img": true, "answer": "(1) $a=2, \\quad b=1$;\n\n（2） $\\mathrm{t}=\\frac{8}{3}$ or $\\mathrm{t}=8$;（3） $\\angle \\mathrm{ABP}=105^{\\circ}$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) Given the equation \\( a^{2} + b^{2} - 4a - 2b + 5 = 0 \\),\n\nWe can rewrite it as \\( (a-2)^{2} + (b-1)^{2} = 0 \\),\n\nThis implies \\( a - 2 = 0 \\) and \\( b - 1 = 0 \\),\n\nThus, \\( a = 2 \\) and \\( b = 1 \\);\n\n(2) From part (1), we know \\( a = 2 \\) and \\( b = 1 \\),\n\nFrom the motion, \\( OD = 2t \\) and \\( OE = t \\),\n\nGiven \\( OB = 8 \\),\n\nTherefore, \\( DB = |8 - 2t| \\)\n\nSince \\( \\triangle BAD \\cong \\triangle OAE \\),\n\nAnd \\( DB = OE \\),\n\nThus, \\( |8 - 2t| = t \\),\n\nSolving gives \\( t = \\frac{8}{3} \\) (as in Figure 1) or \\( t = 8 \\) (as in Figure 2);\n\n(3) As shown in Figure 3,\n\nDraw \\( AQ \\perp AP \\) such that \\( AQ = AP \\), and connect \\( OQ, BQ, PQ \\),\n\nThen \\( \\angle APQ = 45^{\\circ} \\) and \\( \\angle PAQ = 90^{\\circ} \\),\n\nSince \\( \\angle OAB = 90^{\\circ} \\),\n\nThus \\( \\angle PAQ = \\angle OAB \\),\n\nTherefore \\( \\angle OAB + \\angle BAP = \\angle PAQ + \\angle BAP \\),\n\nWhich means: \\( \\angle OAP = \\angle BAQ \\),\n\nSince \\( OA = AB \\) and \\( AD = AD \\),\n\nThus \\( \\triangle OAP \\cong \\triangle BAQ \\) (by SAS),\n\nHence \\( OP = BQ \\), \\( \\angle ABQ = \\angle AOP = 30^{\\circ} \\), and \\( \\angle AQB = \\angle APO = 15^{\\circ} \\),\n\nIn \\( \\triangle AOP \\), \\( \\angle AOP = 30^{\\circ} \\) and \\( \\angle APO = 15^{\\circ} \\),\n\nThus \\( \\angle OAP = 180^{\\circ} - \\angle AOP - \\angle APO = 135^{\\circ} \\),\n\nTherefore \\( \\angle OAQ = 360^{\\circ} - \\angle OAP - \\angle PAQ = 135^{\\circ} - 90^{\\circ} = 135^{\\circ} = \\angle OAP \\),\n\nSince \\( OA = AB \\) and \\( AD = AD \\),\n\nThus \\( \\triangle OAQ \\cong \\triangle BAQ \\) (by SAS),\n\nHence \\( \\angle OQA = \\angle BQA = 15^{\\circ} \\) and \\( OQ = BQ \\),\n\nSince \\( OP = BQ \\),\n\nThus \\( OQ = OP \\)\n\nSince \\( \\angle APQ = 45^{\\circ} \\) and \\( \\angle APO = 15^{\\circ} \\),\n\nThus \\( \\angle OPQ = \\angle APO + \\angle APQ = 60^{\\circ} \\),\n\nTherefore \\( \\triangle OPQ \\) is an equilateral triangle,\n\nThus \\( \\angle OQP = 60^{\\circ} \\),\n\nTherefore \\( \\angle BQP = \\angle OQP - \\angle OQA - \\angle BQA = 60^{\\circ} - 15^{\\circ} - 15^{\\circ} = 30^{\\circ} \\),\n\nSince \\( BQ = PQ \\),\n\nThus \\( \\angle PBQ = \\frac{1}{2}(180^{\\circ} - \\angle BQP) = 75^{\\circ} \\),\n\nTherefore \\( \\angle ABP = \\angle ABQ + \\angle PBQ = 30^{\\circ} + 75^{\\circ} = 105^{\\circ} \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n【Key Insight】This problem is a comprehensive triangle problem, mainly testing the method of completing the square, properties of non-negative numbers, the triangle angle sum theorem, properties and determination of equilateral triangles, and the properties and determination of congruent triangles. The key to solving the problem lies in constructing congruent triangles." }, { "problem_id": 1970, "question": "Concept Understanding: As shown in Figure (1), in $\\angle ABC$, if $\\angle ABD = \\angle DBE = \\angle EBC$, then $BD$ and $BE$ are called the \"trisectors\" of $\\angle ABC$. Among them, $BD$ is the \"adjacent to $AB$ trisector\", and $BE$ is the \"adjacent to $BC$ trisector\".\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n【Problem Solving】\n\n(1) As shown in Figure (1), $\\angle ABC = 60^\\circ$, $BD$ and $BE$ are the \"trisectors\" of $\\angle ABC$, then $\\angle ABE =$ $\\qquad$ ${}^\\circ$;\n\n(2) As shown in Figure (2), in $\\triangle ABC$, $\\angle A = 60^\\circ$, $\\angle B = 48^\\circ$, if the trisector $BD$ of $\\angle B$ intersects $AC$ at point $D$, then $\\angle BDC =$ $\\qquad$ ${}^\\circ$;\n\n(3) As shown in Figure (3), in $\\triangle ABC$, $BP$ and $CP$ are the trisectors of $\\angle ABC$ adjacent to $BC$ and $\\angle ACB$ adjacent to $BC$ respectively, and $\\angle BPC = 140^\\circ$, find the degree measure of $\\angle A$;\n\n(4) 【Extension and Generalization】\n\nIn $\\triangle ABC$, $\\angle ACD$ is the exterior angle of $\\triangle ABC$, the line containing the trisector of $\\angle B$ intersects the line containing the trisector of $\\angle ACD$ at point $P$. If $\\angle A = m^\\circ$, $\\angle B = n^\\circ$, directly write down the degree measure of $\\angle BPC$. (expressed in algebraic terms involving $m$ and $n$)", "input_image": [ "batch4-2024_06_14_0f505914d7b97d0e25dbg_0002_1.jpg", "batch4-2024_06_14_0f505914d7b97d0e25dbg_0002_2.jpg", "batch4-2024_06_14_0f505914d7b97d0e25dbg_0002_3.jpg" ], "is_multi_img": true, "answer": "(1) 40\n\n(2) 76 or 92\n\n(3) $60^{\\circ}$\n\n(4) The degree of $\\angle B P C$ is $\\frac{2}{3} m^{\\circ}$ or $\\frac{2 m^{\\circ}+n^{\\circ}}{3}$ or $\\frac{m^{\\circ}-n^{\\circ}}{3}$ or $\\frac{n^{\\circ}-m^{\\circ}}{3}$ or $\\frac{1}{3} m^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) Solution: Since \\(\\angle ABC = 60^\\circ\\), and \\(BD\\), \\(BE\\) are the \"trisectors\" of \\(\\angle ABC\\),\n\n\\[\n\\angle ABD = \\angle DBE = \\angle EBC = \\frac{1}{3} \\angle ABC = \\frac{1}{3} \\times 60^\\circ = 20^\\circ,\n\\]\n\n\\[\n\\angle ABE = \\angle ABD + \\angle DAE = 20^\\circ + 20^\\circ = 40^\\circ.\n\\]\n\nTherefore, the answer is: 40.\n\n(2) As shown in the figure,\n\n\n\nWhen \\(BD\\) is the \"adjacent \\(AB\\) trisector\",\n\n\\[\n\\angle A = 60^\\circ, \\angle ABC = 48^\\circ,\n\\]\n\n\\[\n\\angle BDC = \\angle A + \\angle ABD = 60^\\circ + \\frac{1}{3} \\times 48^\\circ = 76^\\circ;\n\\]\n\nWhen \\(BD'\\) is the \"adjacent \\(BC\\) trisector\",\n\n\\[\n\\angle A = 60^\\circ, \\angle B = 48^\\circ,\n\\]\n\n\\[\n\\angle BDC' = \\angle A + \\angle ABD' = 60^\\circ + \\frac{2}{3} \\times 48^\\circ = 92^\\circ;\n\\]\n\nIn summary, \\(\\angle BDC = 76^\\circ\\) or \\(92^\\circ\\),\n\nTherefore, the answer is: 76 or 92.\n\n(3) As shown in the figure,\n\n\n\nSince \\(BP \\perp CP\\),\n\n\\[\n\\angle BPC = 140^\\circ,\n\\]\n\n\\[\n\\angle PBC + \\angle PCB = 40^\\circ,\n\\]\n\nSince \\(BP\\) and \\(CP\\) are the \"adjacent \\(BC\\) trisector\" of \\(\\angle ABC\\) and the \"adjacent \\(AC\\) trisector\" of \\(\\angle ACB\\) respectively,\n\n\\[\n\\angle PBC = \\frac{1}{3} \\angle ABC, \\angle PCB = \\frac{1}{3} \\angle ACB,\n\\]\n\n\\[\n\\frac{1}{3} \\angle ABC + \\frac{1}{3} \\angle ACB = 40^\\circ,\n\\]\n\n\\[\n\\angle ABC + \\angle ACB = 120^\\circ,\n\\]\n\n\\[\n\\angle A = 180^\\circ - (\\angle ABC + \\angle ACB) = 180^\\circ - 120^\\circ = 60^\\circ;\n\\]\n\n(4) There are four cases:\n\nCase 1: As shown in Figure 1,\n\n\n\nWhen \\(BP\\) and \\(CP\\) are the \"adjacent \\(BC\\) trisector\" and \"adjacent \\(AC\\) trisector\" respectively,\n\nBy the exterior angle theorem: \\(\\angle PCD = \\frac{1}{3} \\angle ACD = \\frac{1}{3}(m^\\circ + n^\\circ)\\),\n\n\\[\n\\angle BPC = \\angle PCD - \\angle PBC = \\frac{1}{3}(m^\\circ + n^\\circ) - \\frac{1}{3} n^\\circ = \\frac{1}{3} m^\\circ;\n\\]\n\nCase 2: As shown in Figure 2,\n\n\n\nWhen \\(BP\\) and \\(CP\\) are the \"adjacent \\(BC\\) trisector\" and \"adjacent \\(AC\\) trisector\" respectively,\n\nBy the exterior angle theorem: \\(\\angle PCD = \\frac{2}{3} \\angle ACD = \\frac{2}{3}(m^\\circ + n^\\circ)\\),\n\n\\[\n\\angle BPC = \\angle PCD - \\angle PBC = \\frac{2}{3}(m^\\circ + n^\\circ) - \\frac{1}{3} n^\\circ = \\frac{2m^\\circ + n^\\circ}{3};\n\\]\n\nCase 3:\n\n\n\nWhen \\(BP\\) and \\(CP\\) are the \"adjacent \\(AB\\) trisector\" and \"adjacent \\(CD\\) trisector\" respectively,\n\nWhen \\(m^\\circ > n^\\circ\\), as shown in Figure 3,\n\nBy the exterior angle theorem: \\(\\angle PCD = \\frac{1}{3} \\angle ACD = \\frac{1}{3}(m^\\circ + n^\\circ)\\),\n\n\\[\n\\angle BPC = \\angle PCD - \\angle PBC = \\frac{1}{3}(m^\\circ + n^\\circ) - \\frac{2}{3} n^\\circ = \\frac{m^\\circ - n^\\circ}{3};\n\\]\n\nWhen \\(\\alpha < \\beta\\), as shown in Figure 4,\n\n\n\nBy the exterior angle and vertical angle theorem: \\(\\angle DCE = \\angle PCB = \\frac{1}{3} \\angle ACD = \\frac{1}{3}(m^\\circ + n^\\circ)\\),\n\n\\[\n\\angle BPC = \\angle FBC - \\angle PCB = \\frac{2}{3} n^\\circ - \\frac{1}{3}(m^\\circ + n^\\circ) = \\frac{n^\\circ - m^\\circ}{3};\n\\]\n\nCase 4: As shown in Figure 5,\n\n\n\nWhen \\(BP\\) and \\(CP\\) are the \"adjacent \\(BC\\) trisector\" and \"adjacent \\(CD\\) trisector\" respectively,\n\nBy the exterior angle theorem: \\(\\angle PCD = \\frac{1}{3} \\angle ACD = \\frac{1}{3}(m^\\circ + n^\\circ)\\),\n\n\\[\n\\angle BPC = \\angle PCD - \\angle PBC = \\frac{1}{3}(m^\\circ + n^\\circ) - \\frac{1}{3} n^\\circ = \\frac{1}{3} m^\\circ;\n\\]\n\nIn summary: The measure of \\(\\angle BPC\\) is \\(\\frac{2}{3} m^\\circ\\) or \\(\\frac{2m^\\circ + n^\\circ}{3}\\) or \\(\\frac{m^\\circ - n^\\circ}{3}\\) or \\(\\frac{n^\\circ - m^\\circ}{3}\\) or \\(\\frac{1}{3} m^\\circ\\).\n\n【Insight】This problem examines the exterior angle property and the triangle angle sum theorem. Combining geometry with algebra and accurately applying the classification discussion method is key to solving the problem." }, { "problem_id": 1971, "question": "As shown in the figure, the lines \\( AB \\parallel CD \\), and a triangular ruler intersects \\( AB \\) and \\( CD \\) at points \\( E \\) and \\( G \\) respectively, with \\(\\angle EFG = 90^\\circ\\),\n\n\\(\\angle AEF = \\alpha \\left(0^\\circ < \\alpha < 90^\\circ\\right)\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n(1) In Figure (1), if \\(\\alpha = 70^\\circ\\), then \\(\\angle CGF =\\);\n\n(2) In Figure (2), if \\(EH\\) and \\(GH\\) respectively bisect \\(\\angle AEF\\) and \\(\\angle CGF\\), find the measure of \\(\\angle EHG\\);\n\n(3) If \\(EH\\) bisects \\(\\angle AEF\\), \\(GK\\) bisects \\(\\angle FGD\\), and \\(EH \\parallel GL\\), does the measure of \\(\\angle KGL\\) change? If it remains unchanged, provide the answer directly; if it changes, explain the reason.", "input_image": [ "batch4-2024_06_14_0f505914d7b97d0e25dbg_0003_1.jpg", "batch4-2024_06_14_0f505914d7b97d0e25dbg_0003_2.jpg", "batch4-2024_06_14_0f505914d7b97d0e25dbg_0003_3.jpg" ], "is_multi_img": true, "answer": "(1) $20^{\\circ}$\n\n(2) $45^{\\circ}$\n\n(3) $\\angle K G L=45^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) Draw $F M / / A B$ through $F$, then $\\angle A E F = \\angle E F M$,\n\n\n\nSince $A B / / C D$,\n\nTherefore $F M / / C D$,\n\nThus $\\angle C G F = \\angle G F M$,\n\nSince $\\angle E F G = \\angle E F M + \\angle G F M = 90^{\\circ}$,\n\nTherefore $\\angle E F G = \\angle A E F + \\angle C G F = 90^{\\circ}$,\n\nSince $\\angle A E F = \\alpha = 70^{\\circ}$,\n\nTherefore $\\angle C G F = 20^{\\circ}$,\n\nHence the answer is: $20^{\\circ}$;\n\n(2) From the model in (1), we have $\\angle E H G = \\angle A E H + \\angle C G H$\n\n\n\nSince $E H$ and $G H$ bisect $\\angle A E F$ and $\\angle C G F$ respectively,\n\nTherefore $\\angle A E H = \\frac{1}{2} \\angle A E F$, $\\angle C G H = \\frac{1}{2} \\angle C G F$,\n\nThus $\\angle E H G = \\angle A E H + \\angle C G H = \\frac{1}{2} \\angle A E F + \\frac{1}{2} \\angle C G F = \\frac{1}{2} \\angle E F G = 45^{\\circ}$;\n\n(3) Draw $F N / / A B$ through $F$, extend $E F$ and $L G$ to intersect at $P$,\n\n\n\nSince $F N / / A B$\n\nTherefore $\\angle A E F = \\angle E F N$\nSince $A B / / C D$,\n\nTherefore $F M / / C D$,\n\nThus $\\angle C G D + \\angle G F N = 180^{\\circ}$,\n\nSince $\\angle E F G = \\angle E F N + \\angle G F N = 90^{\\circ}$,\n\nTherefore $\\angle E F G = \\angle A E F + 180^{\\circ} - \\angle F G D = 90^{\\circ}$,\n\nThus $\\angle F G D - \\angle A E F = 90^{\\circ}$,\n\nSince $E H$ bisects $\\angle A E F$, $G K$ bisects $\\angle F G D$,\n\nTherefore $\\angle A E F = \\angle H E F = \\frac{1}{2} \\angle A E F$, $\\angle K G D = \\frac{1}{2} \\angle F G D$\n\nThus $\\angle K G D - \\angle H E F = \\frac{1}{2} \\angle F G D - \\frac{1}{2} \\angle A E F = 45^{\\circ}$,\n\nSince $E H / / G L$\n\nTherefore $\\angle H E F = \\angle P = \\frac{1}{2} \\angle A E F$,\n\nSince $A B / / C D$\n\nTherefore $\\angle E Q D = \\angle A E F$,\n\nSince $\\angle E Q D = \\angle P + \\angle C G P$,\n\nTherefore $\\angle H E F = \\angle P = \\angle C G P$,\n\nThus $\\angle H E F = \\angle P = \\angle C G P = \\angle L G D$,\n\nTherefore $\\angle K G L = \\angle K G D - \\angle L G D = \\angle K G D - \\angle H E F = 45^{\\circ}$\n\nThat is, the measure of $\\angle K G L$ remains unchanged, $\\angle K G L = 45^{\\circ}$.\n\n【Highlight】This question examines the properties and determination of parallel lines, the definition of angle bisectors, the exterior angle properties of triangles, and the equality of opposite angles. The key to solving the problem is familiarity with several common models of parallel lines." }, { "problem_id": 1972, "question": "Exploration Question:\n\n\n\nFigure 1\n\n\n\n(1) **Basic Model**: As shown in Figure 1, $\\angle C B M$ and $\\angle B C N$ are exterior angles of $\\triangle A B C$, and the bisectors of $\\angle C B M$ and $\\angle B C N$ intersect at point $O$. Please write the quantitative relationship between $\\angle B O C$ and $\\angle A$, and explain the reason.\n\n(2) **Variation Application**: As shown in Figure 2, given that $A B$ is not parallel to $C D$, $A D$ and $B C$ are the angle bisectors of $\\angle B A P$ and $\\angle A B M$ respectively, and $D E$ and $C E$ are the angle bisectors of $\\angle A D C$ and $\\angle B C D$ respectively.\n\n(1) If $\\angle P O M = 80^{\\circ}$, will the size of $\\angle C E D$ change during the movement of points $A$ and $B$? If it changes, please explain the reason; if it does not change, try to find its value.\n\n(2) If $A P \\parallel D E$ and $B M \\parallel C E$, find the degree measure of $\\angle P O M$.", "input_image": [ "batch4-2024_06_14_100ba0c99e9dd746dc48g_0019_1.jpg", "batch4-2024_06_14_100ba0c99e9dd746dc48g_0019_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle O=90^{\\circ}-\\frac{1}{2} \\angle A$;\n\n(2)(1) will not change, $\\angle C E D=65^{\\circ}$; (2) $\\angle P O M=60^{\\circ}$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\n(1) From the problem statement, we know:\n\n$\\angle C B O=\\frac{1}{2} \\angle C B M, \\quad \\angle B C O=\\frac{1}{2} \\angle B C N$,\n\nSince $\\angle C B M=180^{\\circ}-\\angle A B C$ and $\\angle B C N=180^{\\circ}-\\angle A C B$,\n\nTherefore, $\\angle C B O=90^{\\circ}-\\frac{1}{2} \\angle A B C$, and $\\angle B C O=90^{\\circ}-\\frac{1}{2} \\angle A C B$,\n\nThus, $\\angle O=180^{\\circ}-\\angle C B O-\\angle B C O$,\n\n$=180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} \\angle A B C\\right)-\\left(90^{\\circ}-\\frac{1}{2} \\angle A C B\\right)$\n\n$=\\frac{1}{2}(\\angle A B C+\\angle A C B)$,\n\nSince $\\angle A+\\angle A B C+\\angle A C B=180^{\\circ}$,\n\nTherefore, $\\angle A B C+\\angle A C B=180^{\\circ}-\\angle A$,\n\nThus, $\\angle O=\\frac{1}{2} \\times\\left(180^{\\circ}-\\angle A\\right)=90^{\\circ}-\\frac{1}{2} \\angle A$;\n\n(2) As shown in the figure, combining the problem statement with part (1), we know:\n\n$\\therefore \\angle F=90^{\\circ}-\\frac{1}{2} \\angle P O M$,\n\n$\\angle C E D=90^{\\circ}-\\frac{1}{2} \\angle F$,\n\nTherefore, $\\angle C E D=90^{\\circ}-\\frac{1}{2} \\times\\left(90^{\\circ}-\\frac{1}{2} \\angle P O M\\right)$\n\n$=90^{\\circ}-45^{\\circ}+\\frac{1}{4} \\angle P O M$\n\n$=45^{\\circ}+\\frac{1}{4} \\angle P O M$,\n\nThus, $\\angle C E D=45^{\\circ}+\\frac{1}{4} \\angle P O M$,\n\nTherefore, the size of $\\angle C E D$ does not change.\n\nWhen $\\angle P O M=80^{\\circ}$,\n\n$\\angle C E D=45^{\\circ}+\\frac{1}{4} \\times 80^{\\circ}=65^{\\circ}$;\n\n\n\n(2) Let $\\angle P A D=x$, $\\angle N B C=y$,\n\nSince $A P \\| D E$ and $B M \\| C E$,\n\nTherefore, $\\angle C D E=\\angle E D A=\\angle P A D=x$, and $\\angle D C E=\\angle E C B=\\angle N B C=y$,\n\nThus, $\\angle O A B=180^{\\circ}-2x$, and $\\angle O B A=180^{\\circ}-2y$,\n\nTherefore, $\\angle E=180^{\\circ}-\\angle C D E-\\angle D C E$\n\n$=180^{\\circ}-(x+y)$,\n\nAnd $\\angle F=180^{\\circ}-\\angle F A B-\\angle F B A$\n\n$=180^{\\circ}-(x+y)$,\n\nFrom part (1), we know:\n\n$\\angle E=90^{\\circ}-\\frac{1}{2} \\angle F$,\n\nTherefore, $180^{\\circ}-(x+y)=90^{\\circ}-\\frac{1}{2} \\times\\left[180^{\\circ}-(x+y)\\right]$,\n\nThus, $x+y=120^{\\circ}$,\n\nTherefore, $\\angle P O M=180^{\\circ}-\\angle O B A-\\angle O A B$\n\n$=180^{\\circ}-\\left(180^{\\circ}-2x\\right)-\\left(180^{\\circ}-2y\\right)$\n\n$=2(x+y)-180^{\\circ}$\n\n$=2 \\times 120^{\\circ}-180^{\\circ}$\n\n$=60^{\\circ}$.\n\n**Key Insight:** This problem examines the properties of angle bisectors, the triangle angle sum theorem, and the properties of parallel lines. The key to solving the problem lies in flexibly using the angle bisector and triangle angle sum theorem to construct relationships between angles." }, { "problem_id": 1973, "question": "Given: As shown in the figure, in $\\triangle ABC$, $CE$ is the angle bisector of the exterior angle $\\angle ACD$ of $\\triangle ABC$, and $CE$ intersects the extension of $BA$ at point $E$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) In Figure 1, if $\\angle B = 30^\\circ$ and $\\angle E = 20^\\circ$, find the measure of $\\angle BAC$;\n\n(2) In Figure 2, construct $CF \\perp AB$ at $F$, and if $\\angle BAC - \\angle B = 20^\\circ$, find the measure of $\\angle FCE$.", "input_image": [ "batch4-2024_06_14_100ba0c99e9dd746dc48g_0059_1.jpg", "batch4-2024_06_14_100ba0c99e9dd746dc48g_0059_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle B A C=70^{\\circ}$\n\n(2) $\\angle F C E=80^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\n(1) Since \\(\\angle DCE\\) is an exterior angle of \\(\\triangle BCE\\),\n\n\\[\n\\angle DCE = \\angle B + \\angle E = 30^\\circ + 20^\\circ = 50^\\circ.\n\\]\n\nSince \\(CE\\) is the angle bisector of the exterior angle \\(\\angle ACD\\) of \\(\\triangle ABC\\),\n\n\\[\n\\angle ACD = 2 \\angle DCE = 100^\\circ.\n\\]\n\nTherefore,\n\n\\[\n\\angle BAC = \\angle ACD - \\angle B = 100^\\circ - 30^\\circ = 70^\\circ.\n\\]\n\n(2) Since \\(CE\\) is the angle bisector of the exterior angle \\(\\angle ACD\\) of \\(\\triangle ABC\\),\n\n\\[\n\\angle ACE = \\angle DCE.\n\\]\n\nLet \\(\\angle ACE = \\angle DCE = \\alpha\\).\n\nSince \\(\\angle DCE\\) is an exterior angle of \\(\\triangle BCE\\),\n\n\\[\n\\angle B = \\angle DCE - \\angle E = \\alpha - \\angle E.\n\\]\n\nSince \\(\\angle BAC\\) is an exterior angle of \\(\\triangle ACE\\),\n\n\\[\n\\angle BAC = \\angle ACE + \\angle E = \\alpha + \\angle E.\n\\]\n\nGiven that \\(\\angle BAC - \\angle B = 20^\\circ\\),\n\n\\[\n\\alpha + \\angle E - (\\alpha - \\angle E) = 20^\\circ,\n\\]\n\nwhich simplifies to\n\n\\[\n2 \\angle E = 20^\\circ.\n\\]\n\nThus,\n\n\\[\n\\angle E = 10^\\circ.\n\\]\n\nSince \\(CF \\perp AB\\),\n\n\\[\n\\angle CFE = 90^\\circ.\n\\]\n\nTherefore,\n\n\\[\n\\angle FCE = 90^\\circ - \\angle E = 80^\\circ.\n\\]\n\n**Key Insight:** This problem primarily examines the properties of exterior angles of triangles, the complementary nature of acute angles in right triangles, the definition of perpendicular lines, and the definition of angle bisectors. The key to solving the problem lies in understanding that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles." }, { "problem_id": 1974, "question": "In $\\triangle A B C$ and $\\triangle C D E$, $\\angle A C B=\\angle D C E=60^{\\circ}, A C=B C, D C=E C$, connect $A E, B D$.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n(1) As shown in Figure (1), when point $E$ is on the extension of $B C$, if $A E=6$, then $B D=$ $\\qquad$ ;\n\n(2) As shown in Figure (2), when point $D$ is on segment $B E$, find the measure of $\\angle A E D$.", "input_image": [ "batch4-2024_06_14_100ba0c99e9dd746dc48g_0062_1.jpg", "batch4-2024_06_14_100ba0c99e9dd746dc48g_0062_2.jpg" ], "is_multi_img": true, "answer": "(1)6\n\n(2) $60^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "Low", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Solution: Since \\(\\angle ACB = \\angle DCE = 60^\\circ\\),\n\n\\[\n\\angle ACB + \\angle ACD = \\angle DCE + \\angle ACD,\n\\]\n\n\\[\n\\angle BCD = \\angle ACE.\n\\]\n\nGiven that \\(AC = BC\\) and \\(CE = CD\\),\n\n\\[\n\\triangle ACE \\cong \\triangle BCD \\ (\\text{by SAS}),\n\\]\n\n\\[\nBD = AE = 6.\n\\]\n\nTherefore, the answer is: 6.\n\n(2) Solution: As shown in the figure,\n\n\\[\n\\angle ACB = \\angle DCE = 60^\\circ,\n\\]\n\n\\[\n\\angle ACB - \\angle ACD = \\angle DCE - \\angle ACD,\n\\]\n\n\\[\n\\angle BCD = \\angle ACE.\n\\]\n\nGiven that \\(AC = BC\\) and \\(CE = CD\\),\n\n\\[\n\\triangle ACE \\cong \\triangle BCD \\ (\\text{by SAS}),\n\\]\n\n\\[\n\\angle EAC = \\angle DBC.\n\\]\n\nSince \\(\\angle BFC = \\angle AFE\\),\n\n\\[\n\\angle AED = \\angle ACB = 60^\\circ.\n\\]\n\n**Key Insight:** This problem primarily examines the determination and properties of congruent triangles, as well as the triangle angle sum theorem. Mastering the determination and properties of congruent triangles is crucial for solving the problem." }, { "problem_id": 1975, "question": "As shown in the figure, it is given that \\( CD \\) is the perpendicular bisector of segment \\( AB \\), with the foot of the perpendicular at \\( D \\), and point \\( C \\) is above point \\( D \\). The angle \\( \\angle BAC = 30^\\circ \\), \\( P \\) is a moving point on the line \\( CD \\), \\( E \\) is a point on the ray \\( AC \\) other than \\( A \\), and \\( PB = PE \\). Connect \\( BE \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) In Figure 1, if point \\( P \\) coincides with point \\( C \\), find the measure of \\( \\angle ABE \\);\n\n(2) In Figure 2, if \\( P \\) is above point \\( C \\), conjecture the quantitative relationship among segments \\( PD \\), \\( AC \\), and \\( CE \\), and explain the reason;\n\n(3) If \\( AC = 6 \\) and \\( CE = 2 \\), then the value of \\( PD \\) is \\(\\qquad\\). (Write the result directly)", "input_image": [ "batch4-2024_06_14_107ffb15cafdb5e1e37ag_0062_1.jpg", "batch4-2024_06_14_107ffb15cafdb5e1e37ag_0062_2.jpg", "batch4-2024_06_14_107ffb15cafdb5e1e37ag_0062_3.jpg" ], "is_multi_img": true, "answer": "(1) $90^{\\circ}$\n\n(2) $P D+\\frac{1}{2} A C=C E$ \n\n(3) 1 or 5", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) Based on the properties of the perpendicular bisector of a line segment and the determination and properties of an equilateral triangle, we obtain: $\\triangle B P E$ is an equilateral triangle, hence $\\angle C B E=60^{\\circ}$, and therefore $\\angle A B E=90^{\\circ}$;\n\n(2) As shown in Figure 2, draw $P H \\perp A E$ at $H$, connect $B C$, and draw $P G \\perp B C$ extending to $G$ on the extension of $B C$, constructing a right triangle $\\triangle P C G$ with a 30-degree angle, a right triangle $\\triangle C P H$, and congruent triangles (Rt $\\triangle P G B \\cong \\mathrm{Rt} \\triangle P H E$). According to the properties of a right triangle with a 30-degree angle and the corresponding sides of congruent triangles being equal, the conclusion is proven;\n\n(3) Solve by considering two scenarios separately.\n\n(1)\n\nSolution: As shown in Figure 1, since point $P$ coincides with point $C$, and $C D$ is the perpendicular bisector of segment $A B$,\n\n\n\nFigure 1\n\n$\\therefore P A=P B$,\n\n$\\therefore \\angle P A B=\\angle P B A=30^{\\circ}$,\n\n$\\therefore \\angle B P E=\\angle P A B+\\angle P B A=60^{\\circ}$,\n\n$\\because P B=P E$,\n\n$\\therefore \\triangle B P E$ is an equilateral triangle,\n\n$\\therefore \\angle C B E=60^{\\circ}$,\n\n$\\therefore \\angle A B E=90^{\\circ}$;\n\n(2)\n\nSolution: Conclusion: $P D+\\frac{1}{2} A C=C E$.\n\nReason: As shown in Figure 2, draw $P H \\perp A E$ at $H$, connect $B C$, and draw $P G \\perp B C$ extending to $G$ on the extension of $B C$,\n\n\n\nFigure 2\n\n$\\because C D$ is the perpendicular bisector of $A B$,\n\n$\\therefore C A=C B$.\n\n$\\because \\angle B A C=30^{\\circ}$,\n\n$\\therefore \\angle A C D=\\angle B C D=60^{\\circ}$.\n\n$\\therefore \\angle G C P=\\angle H C P=\\angle B C E=\\angle A C D=\\angle B C D=60^{\\circ}$.\n\n$\\therefore P G=P H, C G=C H=\\frac{1}{2} C P, C D=\\frac{1}{2} A C$.\n\nIn Rt $\\triangle P G B$ and Rt $\\triangle P H E$,\n\n$\\left\\{\\begin{array}{c}P G=P H \\\\ P B=P E\\end{array}\\right.$,\n\n$\\therefore$ Rt $\\triangle P G B \\cong$ Rt $\\triangle P H E$ ( $H L$ ).\n\n$\\therefore B G=E H$, i.e., $C B+C G=C E-C H$.\n\n$\\therefore C B+\\frac{1}{2} C P=C E-\\frac{1}{2} C P$, i.e., $C B+C P=C E$.\n\nAlso, $\\because C B=A C$,\n\n$\\therefore C P=P D-C D=P D-\\frac{1}{2} A C$.\n$\\therefore P D+\\frac{1}{2} A C=C E$;\n\n(3)\n\nSolution: As shown in Figure 3, draw $P H \\perp A E$ at $H$, connect $B C$, and draw $P G \\perp B C$ at $G$,\n\n\n\nFigure 3\n\nHere, Rt $\\triangle P G B \\cong \\mathrm{Rt} \\triangle P H E （ H L ）$.\n\n$\\therefore B G=E H$, i.e., $C B-C G=C E+C H$.\n\n$\\therefore C B-\\frac{1}{2} C P=C E+\\frac{1}{2} C P$, i.e., $C P=C B-C E=6-2=4$.\n\nAlso, $\\because C B=A C$,\n\n$\\therefore P D=C P-C D=4-3=1$.\n\nAs shown in Figure 4,\n\n\n\nFigure 4\n\nSimilarly, $P C=E C+B C=8$,\n\n$P D=P C-C D=8-3=5$.\n\nTherefore, the answer is: 1 or 5.\n\n【Insight】This problem mainly examines the properties of equilateral triangles, the properties of right triangles with a 30-degree angle, the properties and determination of congruent triangles, and the properties of the perpendicular bisector of a line segment. Mastering these properties is key to solving the problem." }, { "problem_id": 1976, "question": "As shown in the figure, in the isosceles $\\triangle ABC$, $AB = AC = 3 \\text{ cm}$, $\\angle B = 30^\\circ$. Point $D$ moves uniformly along side $BC$ from $C$ towards $B$ (with $D$ not coinciding with $B$ or $C$), at a uniform speed of $1 \\text{ cm/s}$. Line segment $AD$ is connected, and $\\angle ADE = 30^\\circ$, with $DE$ intersecting line segment $AC$ at point $E$.\n\n(1) During this motion, $\\angle BDA$ gradually becomes $\\qquad$ (fill in \"larger\" or \"smaller\"); when point $D$ reaches the position shown in Figure 1, $\\angle BDA = 75^\\circ$, then $\\angle BAD = \\qquad$.\n\n(2) After point $D$ moves for $3$ seconds to reach the position shown in Figure 2, then $CD = \\qquad$. At this time, are $\\triangle ABD$ and $\\triangle DCE$ congruent? Please explain the reason;\n\n(3) During the motion of point $D$, the shape of $\\triangle ADE$ also changes. Determine the degree of $\\angle BDA$ when $\\triangle ADE$ is an isosceles triangle (please provide the result directly).\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch4-2024_06_14_109e722f3891d28b2e92g_0013_1.jpg", "batch4-2024_06_14_109e722f3891d28b2e92g_0013_2.jpg" ], "is_multi_img": true, "answer": "(1) larger; $75^{\\circ}$; (2) $3 \\mathrm{~cm} ; \\triangle A B D$ and $\\triangle D C E$ are equal, see explanation for reason; (3) $105^{\\circ}$ or $60^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: (1) During this motion, the angle $\\angle \\mathrm{BDA}$ gradually increases. When point $\\mathrm{D}$ moves to the position shown in Figure 1, $\\angle \\mathrm{BAD} = 180^\\circ - \\angle \\mathrm{B} - \\angle \\mathrm{BDA} = 75^\\circ$. Therefore, the answer is: larger; $75^\\circ$.\n\n(2) After point $\\mathrm{D}$ moves for $3 \\mathrm{~s}$, it reaches the position shown in Figure 2, where $CD = 3 \\mathrm{~cm}$. At this point, $\\triangle \\mathrm{ABD} \\cong \\triangle \\mathrm{DCE}$. The reasoning is as follows: Since $\\mathrm{AB} = \\mathrm{AC}$ and $\\angle \\mathrm{B} = 30^\\circ$, it follows that $\\angle \\mathrm{C} = 30^\\circ$. Given that $\\mathrm{CD} = \\mathrm{CA} = 3 \\mathrm{~cm}$, we have $\\angle \\mathrm{CAD} = \\angle \\mathrm{CDA} = \\frac{1}{2} \\times (180^\\circ - 30^\\circ) = 75^\\circ$. Consequently, $\\angle \\mathrm{ADB} = 105^\\circ$ and $\\angle \\mathrm{EDC} = 75^\\circ - 30^\\circ = 45^\\circ$. Therefore, $\\angle \\mathrm{DEC} = 180^\\circ - 45^\\circ - 30^\\circ = 105^\\circ$, and $\\angle \\mathrm{ADB} = \\angle \\mathrm{DEC}$. In triangles $\\triangle \\mathrm{ABD}$ and $\\triangle \\mathrm{DCE}$, the following conditions hold:\n\\[\n\\left\\{\n\\begin{array}{l}\n\\angle ADB = \\angle DEC \\\\\nAB = DC \\\\\n\\angle ABD = \\angle DCE\n\\end{array}\n\\right.\n\\]\nThus, $\\triangle \\mathrm{ABD} \\cong \\triangle \\mathrm{DCE}$ (ASA).\n\n(3) The triangle $\\triangle \\mathrm{ADE}$ can be an isosceles triangle in three scenarios:\n\n1. When $\\mathrm{AD} = \\mathrm{AE}$, $\\angle \\mathrm{ADE} = 30^\\circ$. Therefore, $\\angle \\mathrm{AED} = \\angle \\mathrm{ADE} = 30^\\circ$, and $\\angle \\mathrm{DAE} = 180^\\circ - \\angle \\mathrm{ADE} - \\angle \\mathrm{AED} = 120^\\circ$. Since $\\angle \\mathrm{BAC} = 180^\\circ - \\angle \\mathrm{B} - \\angle \\mathrm{C} = 120^\\circ$ and point $\\mathrm{D}$ does not coincide with points $\\mathrm{B}$ or $\\mathrm{C}$, it follows that $\\mathrm{AD} \\neq \\mathrm{AE}$.\n\n2. When $\\mathrm{DA} = \\mathrm{DE}$, $\\angle \\mathrm{ADE} = 30^\\circ$. Therefore, $\\angle \\mathrm{DAE} = \\angle \\mathrm{DEA} = \\frac{1}{2}(180^\\circ - \\angle \\mathrm{ADE}) = 75^\\circ$. Consequently, $\\angle \\mathrm{BDA} = \\angle \\mathrm{DEC} = 180^\\circ - \\angle \\mathrm{AED} = 105^\\circ$.\n\n3. When $\\mathrm{EA} = \\mathrm{ED}$, $\\angle \\mathrm{ADE} = 30^\\circ$. Therefore, $\\angle \\mathrm{EAD} = \\angle \\mathrm{EDA} = 30^\\circ$, and $\\angle \\mathrm{AED} = 180^\\circ - \\angle \\mathrm{EAD} - \\angle \\mathrm{EDA} = 120^\\circ$. Consequently, $\\angle \\mathrm{BDA} = \\angle \\mathrm{DEC} = 180^\\circ - \\angle \\mathrm{AED} = 60^\\circ$.\n\nIn summary, during the motion of point $\\mathrm{D}$, $\\triangle \\mathrm{ADE}$ can be an isosceles triangle, with $\\angle \\mathrm{BDA}$ measuring either $60^\\circ$ or $105^\\circ$.\n\n[Key Insight] This problem examines the properties of isosceles triangles, the triangle angle sum theorem, and the criteria for congruent triangles. Mastering the properties of isosceles triangles, the triangle angle sum theorem, and the ability to discuss different scenarios flexibly are key to solving this problem." }, { "problem_id": 1977, "question": "In right $\\triangle ABC$, $\\angle ACB = 90^\\circ$, $AC = 15$, $AB = 25$, point $D$ is a moving point on the hypotenuse $AB$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n(1) As shown in Figure 1, when $CD \\perp AB$, find the length of $CD$;\n\n(2) As shown in Figure 2, when $AD = AC$, draw $DE \\perp AB$ intersecting $BC$ at point $E$, find the length of $CE$;\n\n(3) As shown in Figure 3, during the movement of point $D$, connect $CD$, and when $\\triangle ACD$ is an isosceles triangle, directly write out the length of $AD$.", "input_image": [ "batch4-2024_06_14_109e722f3891d28b2e92g_0027_1.jpg", "batch4-2024_06_14_109e722f3891d28b2e92g_0027_2.jpg", "batch4-2024_06_14_109e722f3891d28b2e92g_0027_3.jpg" ], "is_multi_img": true, "answer": "（1）CD=12；（2）CE= $\\frac{15}{2}$ ；（3）When $\\triangle A C D$ is an isosceles triangle, the length of $\\mathrm{AD}$ is:\n\n15 or 18 or $\\frac{25}{2}$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "Solution: (1) As shown in the figure,\n\n\n\nIn right triangle $\\triangle \\mathrm{ABC}$, $\\angle \\mathrm{ACB}=90^{\\circ}$, $\\mathrm{AC}=15$, $\\mathrm{AB}=25$,\n\n$\\therefore \\mathrm{BC}=\\sqrt{A B^{2}-A C^{2}}=\\sqrt{25^{2}-15^{2}}=20$,\n\n$\\therefore S_{\\triangle A B C}=\\frac{1}{2} A B \\bullet C D=\\frac{1}{2} B C \\bullet A C$,\n\n$\\therefore \\frac{1}{2} \\times 25 \\bullet C D=\\frac{1}{2} \\times 20 \\times 15$,\n\nSolving gives: $C D=12$;\n\n(2) As shown in the figure, connect $\\mathrm{AE}$,\n\n\n\n$\\because \\mathrm{DE} \\perp \\mathrm{AB}$\n\n$\\therefore \\angle \\mathrm{ADE}=\\angle \\mathrm{C}=90^{\\circ}$,\n\nIn right triangles $\\triangle \\mathrm{ADE}$ and $\\triangle \\mathrm{ACE}$,\n\n$\\left\\{\\begin{array}{l}A D=A C \\\\ A E=A E\\end{array}\\right.$,\n\n$\\therefore \\mathrm{Rt} \\triangle \\mathrm{ADE} \\cong \\mathrm{Rt} \\triangle \\mathrm{ACE}$,\n\n$\\therefore \\mathrm{DE}=\\mathrm{CE}$;\n\nLet $D E=C E=x$, then $B E=20-x$, and $B D=25-15=10$,\n\nIn right triangle $\\triangle B D E$, by the Pythagorean theorem,\n\n$10^{2}+x^{2}=(20-x)^{2}$,\n\nSolving gives: $x=\\frac{15}{2}$,\n$\\therefore C E=\\frac{15}{2}$\n\n(3) In right triangle $\\triangle \\mathrm{ABC}$, with $\\mathrm{AB}=25$, $\\mathrm{AC}=15$, $\\mathrm{BC}=20$, the distance from point $\\mathrm{C}$ to $\\mathrm{AB}$ is 12; when $\\triangle A C D$ is an isosceles triangle, it can be divided into three cases:\n\n(1) When $\\mathrm{AD}=\\mathrm{AC}$, $\\mathrm{AD}=15$;\n\n(2) When $\\mathrm{AC}=\\mathrm{CD}$, as shown in the figure, draw $\\mathrm{CE} \\perp \\mathrm{AB}$ at point $\\mathrm{E}$, then $A D=2 A E$,\n\n\n\n$\\because \\mathrm{CE}=12$, by the Pythagorean theorem,\n\n$A E=\\sqrt{15^{2}-12^{2}}=9$,\n\n$\\therefore A D=2 A E=18$;\n\n(3) When $\\mathrm{AD}=\\mathrm{CD}$, as shown in the figure,\n\n\n\nIn right triangle $\\triangle \\mathrm{ABC}$, $\\angle \\mathrm{ACB}=90^{\\circ}$,\n\nWhen point $\\mathrm{D}$ is the midpoint of $\\mathrm{AB}$, $\\mathrm{AD}=\\mathrm{BD}=C D$,\n\n$\\therefore A D=\\frac{1}{2} A B=\\frac{1}{2} \\times 25=\\frac{25}{2}$;\n\nIn summary, when $\\triangle A C D$ is an isosceles triangle, the length of $\\mathrm{AD}$ is: 15 or 18 or $\\frac{25}{2}$.\n\n【Highlight】This problem examines the definition of an isosceles triangle, the determination and properties of congruent triangles, the properties of right triangles, and the Pythagorean theorem. The key to solving the problem is to master the properties and apply them, paying attention to the classification discussion when dealing with isosceles triangles." }, { "problem_id": 1978, "question": "$\\angle M O N=90^{\\circ}$, points $\\mathrm{A}, B$ move along $O M$ and $O N$ respectively (not coinciding with point $O$).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)\n\n(1) As shown in Figure (1), $A E$ and $B E$ are the angle bisectors of $\\angle B A O$ and $\\angle A B O$ respectively. As points $\\mathrm{A}$ and $B$ move, when $A O=B O$, $\\angle A E B=$ $\\qquad$ ${ }^{\\circ}$;\n\n(2) As shown in Figure (2), if $B C$ is the angle bisector of $\\angle A B N$, and the extension of $B C$ intersects the angle bisector of $\\angle O A B$ at point $D$, will the size of $\\angle D$ change as points $\\mathrm{A}$ and $B$ move? If not, find the degree measure of $\\angle D$; if so, explain why;\n\n(3) As shown in Figure (3), extend $M O$ to $Q$ and extend $B A$ to $G$. It is known that the angle bisectors of $\\angle B A O$ and $\\angle O A G$ intersect with the angle bisector of $\\angle B O Q$ and its extension at points $E$ and $F$ respectively. In $\\triangle A E F$, if one angle is three times another angle, find the degree measure of $\\angle A B O$.", "input_image": [ "batch4-2024_06_14_109e722f3891d28b2e92g_0059_1.jpg", "batch4-2024_06_14_109e722f3891d28b2e92g_0059_2.jpg", "batch4-2024_06_14_109e722f3891d28b2e92g_0059_3.jpg" ], "is_multi_img": true, "answer": "(1) $135^{\\circ}$\n\n(2) The degree of $\\angle D$ does not change with the movement of $A and B$, and its value is $45^{\\circ}$\n\n(3) $60^{\\circ}$ or $45^{\\circ}$\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "**Solution:**\n\n(1) **Given:** Since \\( AE \\) and \\( BE \\) are the angle bisectors of \\( \\angle BAO \\) and \\( \\angle ABO \\) respectively,\n\n\\[\n\\angle EBA = \\frac{1}{2} \\angle OBA, \\quad \\angle BAE = \\frac{1}{2} \\angle BAO.\n\\]\n\n**Given:** \\( \\angle MON = 90^\\circ \\),\n\n\\[\n\\angle EAB + \\angle EBA = 90^\\circ.\n\\]\n\n**Given:** \\( \\angle AEB + \\angle EAB + \\angle EBA = 180^\\circ \\),\n\n\\[\n\\angle AEB = 180^\\circ - \\angle EBA - \\angle BAE = 180^\\circ - \\frac{1}{2} (\\angle OBA + \\angle BAO) = 180^\\circ - \\frac{1}{2} \\times 90^\\circ = 180^\\circ - 45^\\circ = 135^\\circ.\n\\]\n\n(2) **Given:** The measure of \\( \\angle D \\) does not change as points \\( A \\) and \\( B \\) move. Let \\( \\angle BAD = \\alpha \\).\n\n**Given:** \\( AD \\) bisects \\( \\angle BAO \\),\n\n\\[\n\\angle BAO = 2\\alpha.\n\\]\n\n**Given:** \\( \\angle AOB = 90^\\circ \\),\n\n\\[\n\\angle ABN = 180^\\circ - \\angle ABO = \\angle AOB + \\angle BAO = 90^\\circ + 2\\alpha.\n\\]\n\n**Given:** \\( BC \\) bisects \\( \\angle ABN \\),\n\n\\[\n\\angle ABC = 45^\\circ + \\alpha.\n\\]\n\n**Given:** \\( \\angle ABC = 180^\\circ - \\angle ABD = \\angle D + \\angle BAD \\),\n\n\\[\n\\angle D = \\angle ABC - \\angle BAD = 45^\\circ + \\alpha - \\alpha = 45^\\circ.\n\\]\n\n(3) **Given:** The angle bisectors of \\( \\angle BAO \\) and \\( \\angle BOQ \\) intersect at point \\( E \\),\n\n\\[\n\\angle AOE = 135^\\circ.\n\\]\n\n\\[\n\\angle E = 180^\\circ - \\angle EAO - \\angle AOE = 45^\\circ - \\angle EAO = 45^\\circ - \\frac{1}{2} \\angle BAO = 45^\\circ - \\frac{1}{2} (180^\\circ - 90^\\circ - \\angle ABO) = \\frac{1}{2} \\angle ABO.\n\\]\n\n**Given:** \\( AE \\) and \\( AF \\) are the angle bisectors of \\( \\angle BAO \\) and \\( \\angle OAG \\) respectively,\n\n\\[\n\\angle EAF = \\frac{1}{2} \\angle BAO + \\frac{1}{2} \\angle GAO = \\frac{1}{2} \\times 180^\\circ = 90^\\circ.\n\\]\n\nIn \\( \\triangle AEF \\), if one angle is three times another, then:\n\n1. When \\( \\angle EAF = 3 \\angle E \\), we get \\( \\angle E = 30^\\circ \\), and thus \\( \\angle ABO = 60^\\circ \\).\n2. When \\( \\angle EAF = 3 \\angle F \\), we get \\( \\angle E = 60^\\circ \\), but \\( \\angle ABO = 120^\\circ > 90^\\circ \\), which is invalid.\n3. When \\( \\angle F = 3 \\angle E \\), we get \\( \\angle E = 22.5^\\circ \\), and thus \\( \\angle ABO = 45^\\circ \\).\n4. When \\( \\angle E = 3 \\angle F \\), we get \\( \\angle E = 67.5^\\circ \\), but \\( \\angle ABO = 135^\\circ > 90^\\circ \\), which is invalid.\n\n**Conclusion:** The measure of \\( \\angle ABO \\) is either \\( 60^\\circ \\) or \\( 45^\\circ \\).\n\n**Key Insight:** The first two parts require a good grasp of the triangle angle sum theorem, the complementary angles in a right triangle, the equality of opposite angles, and the properties of angle bisectors. The third part involves proving \\( \\angle EAF = 90^\\circ \\) and then discussing different cases, where the triangle angle sum theorem and angle bisector properties are crucial for solving the problem." }, { "problem_id": 1979, "question": "As shown in Figure 1, in the Cartesian coordinate system, point $\\mathrm{B}(8,0)$, point $\\mathrm{C}(0,6)$, point $\\mathrm{A}$ is on the negative half of the x-axis, and $\\mathrm{AB}=\\mathrm{BC}$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nSpare Figure\n\n(1) Find the coordinates of point $\\mathrm{A}$;\n\n(2) As shown in Figure 2, if point $\\mathrm{E}$ is the midpoint of $\\mathrm{BC}$, the moving point $\\mathrm{M}$ starts from point $\\mathrm{A}$ and moves uniformly along the segment $\\mathrm{AB}$ towards point $\\mathrm{B}$ at a speed of 1 unit length per second, and the time of movement of point $\\mathrm{M}$ is denoted as $\\mathrm{t}$ (seconds);\n\n(1) If the area of $\\triangle \\mathrm{OME}$ is 2, find the value of $\\mathrm{t}$;\n\n(2) As shown in Figure 3, during the movement of point $\\mathrm{M}$, can $\\triangle \\mathrm{OME}$ become a right triangle? If so, find the value of $\\mathrm{t}$ at that time and write the corresponding coordinates of point $\\mathrm{M}$; if not, explain why.", "input_image": [ "batch4-2024_06_14_dc27078b77f9a2d53186g_0010_1.jpg", "batch4-2024_06_14_dc27078b77f9a2d53186g_0010_2.jpg", "batch4-2024_06_14_dc27078b77f9a2d53186g_0010_3.jpg", "batch4-2024_06_14_dc27078b77f9a2d53186g_0010_4.jpg" ], "is_multi_img": true, "answer": "(1) A $(-2,0)$;\n\n(2) (1) $t=\\frac{2}{3}$ or $t=\\frac{10}{3}$;\n\n(2)When $t=6$, $M (4,0)$ or $\\mathrm{t}=\\frac{33}{4}$, M $\\left(\\frac{25}{4}, 0\\right)$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since points B(8,0) and C(0,6) are given,\n\nTherefore, OB = 8, OC = 6,\n\nThus, BC = √(8² + 6²) = 10.\n\nSince AB = BC = 10,\n\nTherefore, OA = 2, hence A(-2,0).\n\n(2) (1) Draw EH perpendicular to OB at H,\n\n\n\nFigure 1\n\nSince in right triangle △BOC, point E is the midpoint of side BC,\n\nTherefore, OE = BE.\n\nAlso, since EH is perpendicular to OB,\n\nTherefore, H is the midpoint of OB.\n\nThus, EH = 1/2 OC = 1/2 × 6 = 3.\n\nWhen point M is to the left of point O, OM = 2 - t,\n\nTherefore, 1/2 × (2 - t) × 3 = 2,\n\nThus, t = 2/3;\n\nWhen point M is to the right of point O,\n\n\n\nFigure 2\n\nOM = t - 2, 1/2 × (t - 2) × 3 = 2,\n\nThus, t = 10/3.\n\nIn summary, if the area of △OME is 2, t = 2/3 or t = 10/3.\n\n(2) When point M is on AO, i.e., 0 ≤ t < 2,\n\n△OME is an obtuse triangle and cannot be a right triangle;\n\nWhen t = 2, point M moves to point O, and △OME does not form a triangle.\n\nWhen point M is on OB, i.e., 2 < t ≤ 10,\n\nAs shown in Figure 3, when ∠OME = 90°,\n\n\n\nFigure 3\n\nSince OE = BE,\n\nTherefore, OM = 1/2 OB = 1/2 × 8 = 4,\n\nThus, t - 2 = 4,\n\nTherefore, t = 6, M(4,0);\n\nAs shown in Figure 4, when ∠OEM = 90°, draw EH perpendicular to OB at H,\n\n\n\nFigure 4\n\nSince OE² + EM² = OM²,\n\nTherefore, 5² + (t - 6)² + 3² = (t - 2)²,\n\nThus, t = 33/4, M(25/4, 0).\n\nIn summary, the required values are t = 6, M(4,0) or t = 33/4, M(25/4, 0).\n\n【Key Point】This problem is a comprehensive triangle problem, examining the Pythagorean theorem, the area of a triangle, properties of right triangles, and the relationship between coordinates and geometric figures. Correctly drawing the figures and classifying the discussions is key to solving the problem." }, { "problem_id": 1980, "question": "As shown in Figure 1, in the plane rectangular coordinate system, $A(a, 0), B(b, 0), C(-1,2)$, and $|a+2|+(b-4)^{2}=0$.\n\n(1) Find the values of $a$ and $b$.\n\n(2) Is there a point $M$ on the coordinate axes such that the area of $\\triangle C O M$ is equal to $\\frac{1}{2}$ of the area of $\\triangle A B C$? If so, find the coordinates of point $M$.\n\n(3) As shown in Figure 2, draw $C D \\perp y$-axis intersecting the $y$-axis at point $D$. Point $P$ is a moving point on the extension of segment $C D$. Connect $O P$, and $O E$ bisects $\\angle A O P$, with $O F \\perp O E$. As point $P$ moves, will the value of $\\frac{\\angle O P D}{\\angle D O E}$ change? If it remains constant, find its value; if it changes, explain why.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch4-2024_06_14_ddcbf2b8af7471867e91g_0028_1.jpg", "batch4-2024_06_14_ddcbf2b8af7471867e91g_0028_2.jpg" ], "is_multi_img": true, "answer": "(1) $-2,4$; (2) Yes, $(3,0),(-3,0),(0,6),(0,-6)$; (3) remain unchanged, 2\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) Since $|a+2| + (b-4)^2 = 0$,\n\n$\\therefore a + 2 = 0$, $b - 4 = 0$,\n\n$\\therefore a = -2$, $b = 4$.\n\n(2) From (1), we have: $A(-2, 0)$, $B(4, 0)$. And $C(-1, 2)$,\n\n$\\therefore S_{\\triangle ABC} = \\frac{1}{2} \\times (4 + 2) \\times 2 = 6$.\n\nWhen $M$ is on the x-axis, let $M(x, 0)$,\n\n$\\therefore S_{\\triangle COM} = \\frac{1}{2}|x| \\times 2 = 3$,\n\n$\\therefore x = \\pm 3$,\n\n$\\therefore M(3, 0)$ or $M(-3, 0)$.\n\nWhen $M$ is on the y-axis, let $M(0, y)$,\n\n$\\therefore S_{\\triangle COM} = \\frac{1}{2}|y| \\times 1 = 3$,\n\n$\\therefore y = \\pm 6$,\n\n$\\therefore M(0, 6)$ or $M(0, -6)$.\n\nIn summary: The coordinates of $M$ are $(3, 0)$, $(-3, 0)$, $(0, 6)$, $(0, -6)$.\n\n(3) $\\frac{\\angle OPD}{\\angle DOE} = 2$, the reasoning is as follows:\n\nLet $\\angle AOP = 2a$, $OE$ bisects $\\angle AOP$,\n\n$\\therefore \\angle AOE = \\angle POE = a$, $\\angle POB = 180^\\circ - 2a$.\n\nSince $CD \\perp y$-axis, $\\angle BOD = 90^\\circ$, then $CD \\parallel AB$,\n\n$\\therefore \\angle DPO = \\angle POB = 180^\\circ - 2a$, $\\angle ODP = 90^\\circ$,\n\n$\\therefore \\angle DOP = 90^\\circ - (180^\\circ - 2a) = 2a - 90^\\circ$,\n\n$\\therefore \\angle DOE = a - (2a - 90^\\circ) = 90^\\circ - a$,\n\n$\\therefore \\frac{\\angle OPD}{\\angle DOE} = \\frac{180^\\circ - 2a}{90^\\circ - a} = \\frac{2(90^\\circ - a)}{90^\\circ - a} = 2$.\n\n【Key Insight】This problem examines the properties of non-negative numbers, coordinates and shapes, properties of parallel lines, and the triangle angle sum theorem. Mastering these concepts is key to solving the problem." }, { "problem_id": 1981, "question": "In $\\triangle A B C$, $\\angle B>\\angle C, A D$ bisects $\\angle B A C$, intersecting $B C$ at point $D, P$ is a point on $A D$ (not coinciding with point $D), P E \\perp B C$ at point $E$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) If $\\angle B=2 \\angle C=60^{\\circ}$, as shown in Figure 1, when point $P$ coincides with point $A$, find the measure of $\\angle E P D$.\n\n(2) When $\\triangle A B C$ is an acute triangle, as shown in Figure 2, if $\\angle E P D=20^{\\circ}$, find the value of $\\angle B-\\angle C$.", "input_image": [ "batch4-2024_06_14_ddcbf2b8af7471867e91g_0082_1.jpg", "batch4-2024_06_14_ddcbf2b8af7471867e91g_0082_2.jpg" ], "is_multi_img": true, "answer": "(1) $15^{\\circ}$; (2) $40^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "Solution: (1) Since \\(\\angle B = 2 \\angle C = 60^\\circ\\),\n\n\\(\\therefore \\angle C = 30^\\circ\\), and \\(\\angle BAC = 180^\\circ - 60^\\circ - 30^\\circ = 90^\\circ\\).\n\nSince \\(AD\\) bisects \\(\\angle BAC\\),\n\n\\(\\therefore \\angle BAD = \\angle DAC = \\frac{1}{2} \\angle BAC = 45^\\circ\\).\n\n\\(\\therefore \\angle ADE = \\angle C + \\angle DAC = 30^\\circ + 45^\\circ = 75^\\circ\\).\n\nSince \\(PE \\perp BC\\),\n\n\\(\\therefore \\angle PED = 90^\\circ\\).\n\n\\(\\therefore \\angle EPD = 90^\\circ - 75^\\circ = 15^\\circ\\);\n\n(2) As shown in the figure, draw \\(AF \\perp BC\\) at \\(F\\),\n\n\n\n\\(\\therefore \\angle PED = \\angle AFD = 90^\\circ\\).\n\n\\(\\therefore PE \\parallel AF\\).\n\n\\(\\therefore \\angle DAF = \\angle DPE = 20^\\circ\\).\n\nSince \\(\\angle ADE = 90^\\circ - 20^\\circ = 70^\\circ\\),\n\n\\(\\therefore \\angle DAC = 70^\\circ - \\angle C\\).\n\nSince \\(AF \\perp BC\\),\n\n\\(\\therefore \\angle FAB = 90^\\circ - \\angle B\\).\n\nSince \\(AD\\) bisects \\(\\angle BAC\\),\n\n\\(\\therefore \\angle BAD = \\angle DAC\\).\n\n\\(\\therefore 90^\\circ - \\angle B + 20^\\circ = 70^\\circ - \\angle C\\).\n\n\\(\\therefore \\angle B - \\angle C = 40^\\circ\\).\n\n【Key Insight】This problem mainly examines the properties and determination of parallel lines, the definition of angle bisectors, and the fact that the two acute angles in a right triangle are complementary. Mastering the properties and determination of parallel lines, the definition of angle bisectors, and the complementary nature of acute angles in right triangles is crucial for solving the problem." }, { "problem_id": 1982, "question": "Given that in $\\triangle ABD$, $\\angle ABD = 45^\\circ$, $\\angle ADB = 90^\\circ$, the point $B$ is symmetric to point $E$ with respect to line $AD$, connect $AE$, point $C$ is on the ray $DE$, draw line $AC$, draw $EN \\perp AC$ at $N$, and $BM \\perp AC$ at $M$.\n\n\n\n(Figure 1)\n\n\n\n(Alternate Figure)\n\n(1) If point $C$ is to the right of point $E$, as shown in Figure 1, and if $EN = 1$, $BM = 3$, find the length of $MN$;\n\n(2) When point $C$ moves along segment $DE$, find the quantitative relationship between $EN$, $BM$, and $MN$.", "input_image": [ "batch4-2024_06_14_e1852eab246ef162002dg_0014_1.jpg", "batch4-2024_06_14_e1852eab246ef162002dg_0014_2.jpg" ], "is_multi_img": true, "answer": "(1) 4\n\n(2) $B M=M N+E N$", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) According to the properties of symmetry, $\\angle ABD = 45^\\circ$, $\\angle ADB = 90^\\circ$, we obtain: $AB = AE$, $\\angle BAE = 90^\\circ$. Since the complements of the same angle are equal, we get $\\angle MBA = \\angle EAN$. This proves: $\\triangle MBA = \\triangle NAE$. Using $MN = AM + AN$, the solution can be derived.\n\n(2) As shown in the figure, similarly to (1), we prove $\\triangle MBA = \\triangle NAE$, obtaining $AM = EN$, $AN = BM$, which allows us to find the solution.\n\n(1)\n\nSolution: Points $B$ and $E$ are symmetric with respect to line $AD$,\n\n$\\therefore AD$ is the perpendicular bisector of segment $BE$,\n\n$\\therefore AB = AE$,\n\n$\\therefore \\angle AED = \\angle ABD = 45^\\circ$,\n\n$\\therefore \\angle BAE = 180^\\circ - \\angle ABD - \\angle AED = 90^\\circ$,\n\n$\\because EN \\perp AC$, $BM \\perp AC$\n\n$\\therefore \\angle BMA = \\angle ANE = 90^\\circ$\n\n$\\therefore \\angle BAM + \\angle ABM = 90^\\circ$,\n\n$\\because \\angle BAM + \\angle EAN = 180^\\circ - \\angle BAE = 90^\\circ$,\n\n$\\therefore \\angle ABM = \\angle NAE$\n\n$\\therefore \\triangle MBA = \\triangle NAE$ (AAS),\n\n$\\therefore AM = EN = 1$, $AN = BM = 3$,\n\n$\\therefore MN = AM + AN = 1 + 3 = 4$;\n\n(2)\n\nSolution: Points $B$ and $E$ are symmetric with respect to line $AD$,\n\n$\\therefore AD$ is the perpendicular bisector of segment $BE$,\n\n$\\therefore AB = AE$,\n\n$\\therefore \\angle AED = \\angle ABD = 45^\\circ$,\n\n$\\therefore \\angle BAE = 180^\\circ - \\angle ABD - \\angle AED = 90^\\circ$,\n\n$\\because EN \\perp AC$, $BM \\perp AC$\n\n$\\therefore \\angle BMA = \\angle ANE = 90^\\circ$\n\n$\\therefore \\angle BAM + \\angle ABM = 90^\\circ$,\n\n$\\because \\angle BAM + \\angle EAN = 90^\\circ$,\n\n$\\therefore \\angle ABM = \\angle NAE$\n\n$\\therefore \\triangle MBA = \\triangle NAE$ (AAS),\n\n$\\therefore AM = EN$, $AN = BM$,\n\n$\\because AN = AM + MN$\n\n$\\therefore BM = MN + EN$.\n\n\n\n【Key Insight】This problem examines the determination and properties of congruent triangles. Mastering the properties of symmetry and the equality of complementary angles is crucial for solving the problem. When solving for the sum and difference relationships between line segments, congruent triangles are typically used for proof." }, { "problem_id": 1983, "question": "Given that the side length of a cube is \\( a \\).\n\n(1) What is the surface area of a cube? What is its volume?\n\n(2) When two cubes are stacked together as shown in Figure (2), what is its surface area? What is its volume?\n\n(3) When \\( n \\) cubes are stacked together in the same way as shown in Figure (2), what is its surface area? What is its volume?\n\n\n\nFigure (1)\n\n\n\nFigure (2)", "input_image": [ "batch5-2024_06_14_3a30e9069b53a92047dbg_0026_1.jpg", "batch5-2024_06_14_3a30e9069b53a92047dbg_0026_2.jpg" ], "is_multi_img": true, "answer": "（1）Surface area $=6 a^{2}$, volume $=a^{3}$;（2）Surface area $=10 a^{2}$, volume $=2 a^{3}$;（3）Surface area $=(4 n+2)$ $\\mathrm{a}^{2}$, volume $=\\mathrm{na}^{3}$.\n\n ", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "1", "analysis": "Solution: (1) According to the problem, the surface area of a cube $=6 \\times$ the area of a square $=6a^{2}$, and the volume $=a^{3}$;\n\n(2) When two cubes are stacked together, their combined surface area $=6a^{2} \\times 2 - 2a^{2} = 10a^{2}$, and the volume $=2a^{3}$;\n\n(3) When $n$ cubes are stacked together, their combined surface area $=n \\cdot 6a^{2} - (n - 1) \\cdot 2a^{2} = (4n + 2)a^{2}$, and the volume $=na^{3}$.\n\n[Key Insight] This problem primarily examines the properties of geometric solids, with the key to solving it being familiar with the formulas for the surface area and volume of geometric solids." }, { "problem_id": 1984, "question": "Given that the sides of one angle are perpendicular to the sides of another angle, based on the diagram, explore the quantitative relationship between these two angles.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) As shown in Figure 1, $AB \\perp DE, BC \\perp EF, \\angle 1$ and $\\angle 2$ have the following quantitative relationship: $\\qquad$.\n\n(2) As shown in Figure 2, $AB \\perp DE, BC \\perp EF, \\angle 1$ and $\\angle 2$ have the following quantitative relationship: $\\qquad$.\n\n(3) From (1) and (2), your conclusion is: If $\\qquad$, then $\\qquad$.\n\n(4) If the sides of two angles are mutually perpendicular, and one angle is 40 degrees less than three times the other angle, find the measures of these two angles.", "input_image": [ "batch5-2024_06_14_488c0aee42adf18c46b1g_0046_1.jpg", "batch5-2024_06_14_488c0aee42adf18c46b1g_0046_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\angle 1=\\angle 2$\n\n(2) $\\angle 1+\\angle 2=180^{\\circ}$\n\n(3) If the two sides of an angle are perpendicular to the two sides of another angle, the two angles are equal or complementary\n\n(4) $20^{\\circ} 、 20^{\\circ}$ or $55^{\\circ} 、 125^{\\circ}$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "1", "analysis": "(1) Solution: Since \\( AB \\perp DE \\) and \\( BC \\perp EF \\),\n\nTherefore, \\( \\angle EGB = 90^\\circ \\) and \\( \\angle EHB = 90^\\circ \\),\nThus, \\( \\angle 2 + \\angle 4 = 90^\\circ \\) and \\( \\angle 1 + \\angle 3 = 90^\\circ \\),\n\nSince \\( \\angle 3 = \\angle 4 \\)\n\nTherefore, \\( \\angle 1 = \\angle 2 \\),\n\nHence, the answer is \\( \\angle 1 = \\angle 2 \\);\n\n\n\n(2) Solution: Since \\( AB \\perp DE \\) and \\( BC \\perp EF \\),\n\nTherefore, \\( \\angle EGB = 90^\\circ \\) and \\( \\angle EHB = 90^\\circ \\),\n\nSince \\( \\angle 1 + \\angle 2 + \\angle EGB + \\angle EHB = 360^\\circ \\),\n\nTherefore, \\( \\angle 1 + \\angle 2 = 180^\\circ \\),\n\nHence, the answer is \\( \\angle 1 + \\angle 2 = 180^\\circ \\).\n\n\n\n(3) Solution: From (1) and (2), it can be concluded that if two sides of one angle are respectively perpendicular to two sides of another angle, then these two angles are either equal or supplementary,\n\nHence, the answer is: If two sides of one angle are respectively perpendicular to two sides of another angle, then these two angles are either equal or supplementary;\n\n(4) Solution: Let the measure of the other angle be \\( \\alpha \\), then the measure of this angle is \\( 3\\alpha - 40^\\circ \\),\n\nAccording to the problem, we have \\( \\alpha = 3\\alpha - 40^\\circ \\) or \\( \\alpha + 3\\alpha - 40^\\circ = 180^\\circ \\)\n\nSolving gives \\( \\alpha = 20^\\circ \\) or \\( 55^\\circ \\),\n\nWhen \\( \\alpha = 20^\\circ \\), \\( 3\\alpha - 40^\\circ = 20^\\circ \\),\n\nWhen \\( \\alpha = 55^\\circ \\), \\( 3\\alpha - 40^\\circ = 125^\\circ \\),\n\nTherefore, the measures of these two angles are \\( 20^\\circ, 20^\\circ \\) or \\( 55^\\circ, 125^\\circ \\).\n\n【Highlight】This problem mainly examines the definitions of supplementary angles and perpendicularity, the properties of complementary angles, the properties of right triangles, and the sum of interior angles of a quadrilateral. In the process of solving, attention should be paid to categorical discussion." }, { "problem_id": 1985, "question": "There is a rectangular strip of paper, where \\( E \\) and \\( F \\) are points on sides \\( AD \\) and \\( BC \\) respectively, and \\( \\angle DEF = \\alpha \\) degrees (\\( 0 < \\alpha < 90 \\)). The strip is folded along \\( EF \\) to form Figure 1, and then folded along \\( GF \\) to form Figure 2.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) In Figure 1, when \\( \\alpha = 30 \\) degrees, \\( \\angle GFC' = \\) _ degrees;\n\n(2) In Figure 2, if \\( \\angle GFN = 4 \\angle GFE \\), find the value of \\( \\alpha \\);\n\n(3) Construct \\( GP \\) to bisect \\( \\angle MGF \\) intersecting line \\( EF \\) at point \\( P \\). Directly state the relationship between \\( \\angle GEP \\) and \\( \\angle GPE \\).", "input_image": [ "batch5-2024_06_14_488c0aee42adf18c46b1g_0051_1.jpg", "batch5-2024_06_14_488c0aee42adf18c46b1g_0051_2.jpg", "batch5-2024_06_14_488c0aee42adf18c46b1g_0051_3.jpg" ], "is_multi_img": true, "answer": "(1) 120\n\n(2) $\\alpha=30$\n\n(3) $\\angle G P E=2 \\angle G E P$ or $2 \\angle G E P-\\angle G P E=180^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\n(1) By folding, we have $\\angle GEF = \\angle DEF$.\n\nSince the opposite sides of a rectangle are parallel,\n\n$\\angle BFE = \\angle DEF = 30^\\circ$,\n\n$\\therefore \\angle GEF = \\angle DEF = \\angle BFE = 30^\\circ$,\n\n$\\therefore \\angle EGB = \\angle BFE + \\angle GEF = 60^\\circ$,\n\n$\\therefore \\angle FGD' = \\angle EGB = 60^\\circ$,\n\n$\\therefore \\angle GFC' = 180^\\circ - \\angle FGD' = 120^\\circ$;\n\nThus, when $\\alpha = 30^\\circ$, the measure of $\\angle GFC'$ is $120^\\circ$.\n\n**Answer:** 120.\n\n---\n\n(2) By folding, we have $\\angle GEF = \\angle DEF = \\alpha^\\circ$, $\\angle GFC' = \\angle GFN$.\n\nSince the opposite sides of a rectangle are parallel,\n\n$\\angle GFE = \\angle DEF = \\alpha^\\circ$,\n\n$\\therefore \\angle EGB = \\angle GFE + \\angle GEF = 2\\alpha^\\circ$, $\\angle GFN = 4\\angle GFE = 4\\alpha^\\circ$,\n\n$\\therefore \\angle FGD' = \\angle EGB = 2\\alpha^\\circ$,\n\nSince $\\angle GFC' + \\angle FGD' = 180^\\circ$,\n\n$\\therefore 4\\alpha^\\circ + 2\\alpha^\\circ = 180^\\circ$,\n\n$\\therefore \\alpha^\\circ = 30^\\circ$,\n\n$\\therefore$ the value of $\\alpha$ is 30.\n\n---\n\n(3) **Solution:**\n\n(1) As shown in Figure 3, the conclusion is: $\\angle GPE = 2\\angle GEP$.\n\n**Reason:** Since $\\angle MGF = \\angle D'GF = 180^\\circ - \\angle EGF = 180^\\circ - (180^\\circ - \\angle DEG) = \\angle DEG = 2\\alpha$,\n\nand since $GP$ bisects $\\angle MGF$,\n\n$\\therefore \\angle PGF = \\alpha$,\n\n$\\therefore \\angle GPE = \\angle PGE + \\angle PFG = \\alpha + \\alpha = 2\\alpha = 2\\angle GEP$.\n\n(2) As shown in Figure 4, the conclusion is: $2\\angle GEP - \\angle GPE = 180^\\circ$.\n\n**Reason:** Since $2\\angle GEP = 2(180^\\circ - \\alpha)$,\n\n$\\angle GPE = 180^\\circ - \\angle PGF - \\angle PFG = 180^\\circ - 2\\alpha$,\n\n$\\therefore 2\\angle GEP - \\angle GPE = 2(180^\\circ - \\alpha) - (180^\\circ - 2\\alpha) = 180^\\circ$.\n\n**Answer:** $\\angle GPE = 2\\angle GEP$ or $2\\angle GEP - \\angle GPE = 180^\\circ$.\n\n---\n\n**Key Insight:** This problem primarily examines the properties of folding, parallel lines, the exterior angle property of triangles, the definition of angle bisectors, and the use of supplementary angles to determine angle measures. The key to solving the problem lies in drawing the figure and combining geometric intuition with algebraic reasoning, while paying attention to case analysis." }, { "problem_id": 1986, "question": "As shown in the figure, the geometric shape in Figure (1) is called a triangular prism, which has 6 vertices, 9 edges, and 5 faces. The geometric shapes in Figure (2) and Figure (3) are respectively a quadrangular prism and a pentagonal prism.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n(1) A quadrangular prism has $\\qquad$ vertices, $\\qquad$ edges, and $\\qquad$ faces;\n\n(2) A pentagonal prism has $\\qquad$ vertices, $\\qquad$ edges, and $\\qquad$ faces;\n\n(3) Then, an $n$-gonal prism has $\\qquad$ vertices, $\\qquad$ edges, and $\\qquad$ faces.", "input_image": [ "batch5-2024_06_14_62d4abfbd6f580c7356eg_0003_1.jpg", "batch5-2024_06_14_62d4abfbd6f580c7356eg_0003_2.jpg", "batch5-2024_06_14_62d4abfbd6f580c7356eg_0003_3.jpg" ], "is_multi_img": true, "answer": "(1)8, 12, 6\n\n(2) $10,15,7$\n\n(3) $2 n, 3 n,(n+2)$\n\n ", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Solid Geometry", "image_relavance": "0", "analysis": "(1) Solution: From the geometric characteristics of a prism, we know that a quadrilateral prism has 8 vertices, 12 edges, and 6 faces. Therefore, the answer is: $8, 12, 6$.\n\n(2) Solution: From the geometric characteristics of a prism, we know that a pentagonal prism has 10 vertices, 15 edges, and 7 faces. Therefore, the answer is: $10, 15, 7$.\n\n(3) Solution: From the geometric characteristics of a prism, we know that an $n$-gonal prism has $2n$ vertices, $3n$ edges, and $(n+2)$ faces. Therefore, the answer is: $2n, 3n, (n+2)$.\n\n[Key Insight] This question mainly examines the characteristics of prisms. Understanding the geometric features of prisms is key to solving the problem." }, { "problem_id": 1987, "question": "Observe the dot matrix below and the corresponding equations, and explore the patterns within:\n\n$\\bullet$\n\n\n\n$$\n\\text { : }\n$$\n\n\n\n(1) Write the corresponding equation on the line following (4):\n(1) $1=1^{2}$;\n(2) $1+3=2^{2}$;\n(3) $1+3+5=3^{2}$;\n(4)\n; (5) $1+3+5+7+9=5^{2} ; \\quad \\ldots$\n\n(2) Write the $n$th equation;\n\n(3) Using the equation from (2), calculate: $41+43+45+\\ldots+199$.", "input_image": [ "batch5-2024_06_14_62d4abfbd6f580c7356eg_0010_1.jpg", "batch5-2024_06_14_62d4abfbd6f580c7356eg_0010_2.jpg" ], "is_multi_img": true, "answer": "(1) $1+3+5+7=4^{2}$\n\n$(2) 1+3+\\ldots+(2 n-1)=n^{2}$\n\n(3) 9600\n\n ", "answer_type": "multi-step", "difficulty": "Low", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "(1) According to the problem, the fourth term is \\(1 + 3 + 5 + 7 = 16 = 4^{2}\\), so the answer is \\(4^{2}\\).\n\n(2) From the figure, we observe:\n\n\\[\n1 = 2 \\times 1 - 1 = 1^{2};\n\\]\n\\[\n1 + 3 = 1 + (2 \\times 2 - 1) = 2^{2}\n\\]\n\\[\n\\ldots\n\\]\nThe \\(n\\)th equation is:\n\\[\n1 + 3 + 5 + \\ldots + (2n - 1) = \\frac{1 + 2n - 1}{2} \\times n = n^{2}\n\\]\nThus, the answer is:\n\\[\n1 + 3 + 5 + \\ldots + (2n - 1) = n^{2}\n\\]\n\n(3) \n\\[\n41 + 43 + 45 + \\ldots + 199\n\\]\n\\[\n= (1 + 3 + 5 + \\ldots + 199) - (1 + 3 + 5 + \\ldots + 39)\n\\]\n\\[\n= [1 + 3 + 5 + \\ldots + (2 \\times 100 - 1)] - [1 + 3 + 5 + \\ldots + (2 \\times 20 - 1)]\n\\]\n\\[\n= 100^{2} - 20^{2}\n\\]\n\\[\n= 9600\n\\]\n\n**Key Insight:** This problem tests the understanding of numerical patterns. Carefully observing the figure and identifying the underlying pattern is crucial to solving this problem." }, { "problem_id": 1988, "question": "(1) In $\\triangle ABC$, if $\\angle A = 60^\\circ$\n\n(1) As shown in Figure 1, if the angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point $O$, then $\\angle BOC =$ $\\qquad$.\n\n(2) As shown in Figure 2, if the trisectors of $\\angle ABC$ and $\\angle ACB$ intersect at points $O_1$ and $O_2$, then $\\angle BO_1C =$ $\\qquad$.\n\n(2) In $\\triangle ABC$, if $\\angle A = x^\\circ$\n\n(1) As shown in Figure 1, if the angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point $O$, then express $\\angle BOC$ in terms of $x$ as $\\qquad$ degrees.\n\n(2) As shown in Figure 2, if the trisectors of $\\angle ABC$ and $\\angle ACB$ intersect at points $O_1$ and $O_2$, then express $\\angle BO_1C$ in terms of $x$ as $\\qquad$ degrees.\n\n(3) As shown in Figure 3, if the $n$-sectors of $\\angle ABC$ and $\\angle ACB$ intersect at points $O_1, O_2, \\ldots, O_{n-1}$, then express $\\angle BO_1C$ in terms of $x$ as $\\qquad$ degrees (the result does not need to be simplified).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch5-2024_06_14_f7dd8e06445035e43841g_0010_1.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0010_2.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0010_3.jpg" ], "is_multi_img": true, "answer": "(1) (1) $120^{\\circ}$; (2) $100^{\\circ}$; (2) (1) $90^{\\circ}+\\frac{1}{2} x^{\\circ}$; (2) $60^{\\circ}+\\frac{2}{3} x^{\\circ}$; (3) $\\frac{180^{\\circ}+(n-1) x^{\\circ}}{n}$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "Solution: \n\n(1) \n$\\because \\angle A=60^{\\circ}$,\n\n$\\therefore \\angle ABC + \\angle ACB = 180^{\\circ} - 60^{\\circ} = 120^{\\circ}$,\n\n$\\because$ the angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point $O$,\n\n$\\therefore \\angle OBC + \\angle OCB = \\frac{1}{2}(\\angle ABC + \\angle ACB) = 60^{\\circ}$,\n\n$\\therefore \\angle BOC = 180^{\\circ} - (\\angle OBC + \\angle OCB) = 120^{\\circ}$;\n\n(2) \n$\\because \\angle A = 60^{\\circ}$,\n\n$\\therefore \\angle ABC + \\angle ACB = 180^{\\circ} - 60^{\\circ} = 120^{\\circ}$,\n\n$\\because$ the trisectors of $\\angle ABC$ and $\\angle ACB$ intersect at points $O_1$ and $O_2$,\n\n$\\therefore \\angle O_1BC + \\angle O_1CB = \\frac{2}{3}(\\angle ABC + \\angle ACB) = 80^{\\circ}$,\n\n$\\therefore \\angle BO_1C = 180^{\\circ} - (\\angle O_1BC + \\angle O_1CB) = 100^{\\circ}$;\n\n(2) \n(1) \n$\\because \\angle A = x^{\\circ}$,\n\n$\\therefore \\angle ABC + \\angle ACB = 180^{\\circ} - x^{\\circ}$,\n\n$\\because$ the angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect at point $O$,\n\n$\\therefore \\angle OBC + \\angle OCB = \\frac{1}{2}(\\angle ABC + \\angle ACB) = 90^{\\circ} - \\frac{1}{2}x^{\\circ}$,\n\n$\\therefore \\angle BOC = 180^{\\circ} - (\\angle OBC + \\angle OCB) = 90^{\\circ} + \\frac{1}{2}x^{\\circ}$;\n\n(2) \n$\\because \\angle A = x^{\\circ}$,\n\n$\\therefore \\angle ABC + \\angle ACB = 180^{\\circ} - x^{\\circ}$,\n\n$\\because$ the trisectors of $\\angle ABC$ and $\\angle ACB$ intersect at points $O_1$ and $O_2$,\n\n$\\therefore \\angle O_1BC + \\angle O_1CB = \\frac{2}{3}(\\angle ABC + \\angle ACB) = 120^{\\circ} - \\frac{2}{3}x^{\\circ}$,\n\n$\\therefore \\angle BO_1C = 180^{\\circ} - (\\angle O_1BC + \\angle O_1CB) = 60^{\\circ} + \\frac{2}{3}x^{\\circ}$;\n\n(3) \n$\\because \\angle A = x^{\\circ}$,\n\n$\\therefore \\angle ABC + \\angle ACB = 180^{\\circ} - x^{\\circ}$,\n\n$\\because$ the $n$-sectors of $\\angle ABC$ and $\\angle ACB$ intersect at points $O_1, O_2, \\ldots, O_{n-1}$,\n\n$\\therefore \\angle O_1BC + \\angle O_1CB = \\frac{n-1}{n}(\\angle ABC + \\angle ACB) = \\frac{n-1}{n}(180^{\\circ} - x^{\\circ})$,\n\n$\\therefore \\angle BO_1C = 180^{\\circ} - (\\angle O_1BC + \\angle O_1CB) = \\frac{180^{\\circ} + (n-1)x^{\\circ}}{n}$.\n\n【Key Insight】This problem examines the concepts of angle bisectors, trisectors, and $n$-sectors in triangles, as well as the application of the triangle angle sum theorem. Familiarity with these concepts and their flexible application is key to solving this problem." }, { "problem_id": 1989, "question": "As shown in Figure 1, it is a rectangular piece of paper, which is folded along $E F$ to form the shape shown in Figure 1.\n\nZhang Ming, a student, noticed that after folding, the quadrilateral $C D E F$ is identical to the quadrilateral $C^{\\prime} D^{\\prime} E F$, thus the fold line is exactly the angle bisector of $\\angle D E D^{\\prime}$.\n\n(1) In Figure 1, if $\\angle D E F = 70^{\\circ}$, find the value of $\\angle E M B$;\n\n\n\nFigure 1\n\n(2) Fold the right side of the rectangular paper along $E F$ and the left side along $E G$, as shown in Figure 2. If the angle formed by the two fold lines $\\angle F E G = 70^{\\circ}$, find the degree measure of the angle $\\angle F N E$ formed by $F C$ and $E A$.\n\n\n\nFigure 2\n\n(3) Fold the right side of the rectangular paper along $E F$ and the left side along $G H$, as shown in Figure 3. Explore the quantitative relationship between the angle $\\angle P$ formed by the two fold lines and the angle $\\angle E O H$ formed by $D E$ and $B H$.\n\n\n\nFigure 3", "input_image": [ "batch5-2024_06_14_f7dd8e06445035e43841g_0019_1.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0019_2.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0019_3.jpg" ], "is_multi_img": true, "answer": "(1) $140^{\\circ}$\n\n(2) $40^{\\circ}$\n\n(3) $\\angle E O H+2 \\angle P=180^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Algebra", "image_relavance": "1", "analysis": "(1) Since \\( EF \\) bisects \\( \\angle DED' \\) and \\( \\angle FED = 70^\\circ \\),\n\n\\[\n\\angle DED' = 2 \\angle FED = 140^\\circ.\n\\]\n\nSince \\( AD \\parallel BC \\),\n\n\\[\n\\angle EMB = \\angle DED' = 140^\\circ,\n\\]\n\nwhich means the desired angle is \\( 140^\\circ \\).\n\n(2) As shown in the figure,\n\n\n\nFigure 2\n\nAccording to the problem, considering the fold, \\( \\angle GEA = \\angle GEA' \\) and \\( \\angle DEF = \\angle D'EF \\).\n\n\\[\n\\angle A'ED = 180^\\circ - \\angle DEF - \\angle D'EF = 180^\\circ - 2 \\angle DEF.\n\\]\n\nSince \\( \\angle FEG = 70^\\circ \\),\n\n\\[\n\\angle DEF - \\angle DEG = \\angle FEG = 70^\\circ,\n\\]\n\n\\[\n\\angle DEF = \\angle DEG + 70^\\circ.\n\\]\n\nSince \\( \\angle DEA = \\angle GEA + \\angle GEA' - \\angle A'ED \\),\n\n\\[\n\\angle DEA = 2 \\angle GEA - \\left(180^\\circ - 2 \\angle DEF\\right),\n\\]\n\n\\[\n\\angle DEA = 2 \\angle GEA - \\left[180^\\circ - 2\\left(\\angle DEG + 70^\\circ\\right)\\right],\n\\]\n\n\\[\n\\angle DEA = 2(\\angle GEA + \\angle DEG) - 40^\\circ,\n\\]\n\n\\[\n\\angle DEA = 2 \\angle DEA - 40^\\circ,\n\\]\n\n\\[\n\\angle DEA = 40^\\circ.\n\\]\n\nSince \\( DE \\parallel CF \\),\n\n\\[\n\\angle ENF = \\angle DEA = 40^\\circ,\n\\]\n\n\\[\n\\angle ENF = 40^\\circ.\n\\]\n\n(3) As shown in the figure,\n\n\n\nFigure 3\n\nAccording to the problem, considering the fold, \\( \\angle BHG = \\angle B'HG \\) and \\( \\angle DEF = \\angle D'EF \\).\n\nSince \\( A'D' \\parallel B'C' \\) and \\( DE \\parallel CF \\),\n\n\\[\n\\angle A'ET = \\angle ETH, \\quad \\angle D'EF = \\angle EFH.\n\\]\n\nSince \\( \\angle EFH = \\angle P + \\angle FHP \\),\n\n\\[\n\\angle EFH = \\angle P + \\angle B'HG,\n\\]\n\nwhich means \\( \\angle EFH - \\angle P = \\angle B'HG \\).\n\nSince \\( \\angle EOH = \\angle ETH + \\angle BHT \\),\n\n\\[\n\\angle EOH = \\angle ETH + 2 \\angle B'HG,\n\\]\n\n\\[\n\\angle EOH = \\angle ETH + 2(\\angle EFH - \\angle P),\n\\]\n\n\\[\n\\angle EOH + 2 \\angle P = \\angle ETH + 2 \\angle EFH.\n\\]\n\nSince \\( \\angle D'EF = \\angle EFH \\) and \\( \\angle DEF = \\angle D'EF \\),\n\n\\[\n\\angle DEF = \\angle EFH,\n\\]\n\n\\[\n\\angle EOH + 2 \\angle P = \\angle ETH + 2 \\angle EFH = 180^\\circ,\n\\]\n\nwhich satisfies the relationship \\( \\angle EOH + 2 \\angle P = 180^\\circ \\).\n\n【Key Insight】This problem examines the properties of rectangle folding, parallel lines, the triangle angle sum theorem, and the exterior angle theorem of triangles. The key to solving this problem lies in correctly identifying the equal angle relationships based on the fold." }, { "problem_id": 1990, "question": "Given $\\triangle A B C$, where $B E$ bisects $\\angle A B C$, and point $P$ lies on ray $B E$.\n\n\n\n(1)\n\n\n\n(2)\n\n\n\n(3)\n\n(1) As shown in Figure (1), if $\\angle A B C=46^{\\circ}$ and $C P / / A B$, find the measure of $\\angle B P C$;\n\n(2) As shown in Figure (2), if $\\angle B A C=110^{\\circ}$ and $\\angle P B C=\\angle P C A$, find the measure of $\\angle B P C$;\n\n(3) As shown in Figure (3), if $\\angle A B C=46^{\\circ}$ and $\\angle A C B=34^{\\circ}$, and line $C P$ is perpendicular to one side of $\\triangle A B C$, then the measure of $\\angle B P C$ is $\\qquad$. (Write the answer directly)", "input_image": [ "batch5-2024_06_14_f7dd8e06445035e43841g_0032_1.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0032_2.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0032_3.jpg" ], "is_multi_img": true, "answer": "(1) $23^{\\circ}$\n\n(2) $110^{\\circ}$\n\n(3) $113^{\\circ}, 33^{\\circ}, 67^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\n(1) **Given:** Since \\( BE \\) bisects \\( \\angle ABC \\) and \\( \\angle ABC = 46^\\circ \\),\n\n\\[\n\\angle ABP = \\frac{1}{2} \\angle ABC = \\frac{1}{2} \\times 46^\\circ = 23^\\circ.\n\\]\n\n**Given:** \\( CP \\parallel AB \\),\n\n\\[\n\\angle BPC = \\angle ABP = 23^\\circ.\n\\]\n\n---\n\n(2) **Given:** Since \\( BE \\) bisects \\( \\angle ABC \\), \\( \\angle PBC = \\angle PCA \\),\n\n\\[\n\\angle ABP = \\angle PBC = \\angle PCA.\n\\]\n\n**In triangle \\( ABO \\):**\n\n\\[\n\\angle A + \\angle ABP + \\angle AOB = 180^\\circ.\n\\]\n\n**In triangle \\( PCO \\):**\n\n\\[\n\\angle BPC + \\angle PCA + \\angle POC = 180^\\circ.\n\\]\n\n**Since \\( \\angle ABP = \\angle PCA \\) and \\( \\angle AOB = \\angle POC \\),**\n\n\\[\n\\angle A = \\angle BPC = 110^\\circ.\n\\]\n\nThus, \\( \\angle BPC = 110^\\circ \\).\n\n---\n\n(3) **Solution:**\n\n- **When \\( CP \\perp BC \\):**\n\n\\[\n\\angle BPC = 180^\\circ - 90^\\circ - 23^\\circ = 67^\\circ.\n\\]\n\n- **When \\( CP \\perp AC \\):**\n\n\\[\n\\angle BPC = 180^\\circ - 90^\\circ - 23^\\circ - 34^\\circ = 33^\\circ.\n\\]\n\n- **When \\( CP \\perp AB \\):**\n\n\\[\n\\angle BPC = 90^\\circ + 23^\\circ = 113^\\circ.\n\\]\n\n**Final Answer:** \\( 113^\\circ, 33^\\circ, 67^\\circ \\).\n\n---\n\n**Key Insight:** This problem primarily examines the **Triangle Angle Sum Theorem** and the definition of **angle bisectors**. A solid understanding of the fact that the sum of angles in a triangle equals \\( 180^\\circ \\) is crucial for solving such problems." }, { "problem_id": 1991, "question": "As shown in Figure 1, given that $\\angle M O N=76^{\\circ}$, $O E$ bisects $\\angle M O N$, and points $A, B, C$ are moving points on rays $O M, O E$, and $O N$ respectively, distinct from point $O$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\n\n\nSpare Figure\n\n(1) In Figure 1, connect $A B$. If $A B / / O C$, then the degree measure of $\\angle A B E$ is $\\qquad$\n(2) In Figure 2, connect $A C$. If ray $A B$ bisects $\\angle M A C$, then the relationship between the measures of $\\angle A B O$ and $\\angle A C O$ is ; $\\qquad$\n(3) In Figure 3, connect $A C$ intersecting ray $O E$ at point $D$ (distinct from point $B$). When $A B \\perp O M$ and $\\triangle A D B$ has two equal angles, find the degree measure of $\\angle O A C$.", "input_image": [ "batch5-2024_06_14_f7dd8e06445035e43841g_0039_1.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0039_2.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0039_3.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0039_4.jpg" ], "is_multi_img": true, "answer": "(1)142\n\n(2) $\\angle A B O=\\frac{1}{2} \\angle A C O$\n\n(3) The degree of $\\angle O A C$ is $14^{\\circ}$ or $38^{\\circ}$ or $26^{\\circ}$.\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) Solution: Since \\(\\angle MON = 76^\\circ\\) and \\(OE\\) bisects \\(\\angle MON\\),\n\n\\(\\therefore \\angle AOB = \\angle BOC = \\frac{1}{2} \\angle MON = \\frac{1}{2} \\times 76^\\circ = 38^\\circ\\),\n\nSince \\(AB \\parallel OC\\),\n\n\\(\\therefore \\angle ABO = \\angle BOC = 38^\\circ\\),\n\n\\(\\therefore \\angle ABE = 180^\\circ - \\angle ABO = 180^\\circ - 38^\\circ = 142^\\circ\\).\n\nTherefore, the answer is: 142;\n\n(2) Solution: \\(\\angle ABO = \\frac{1}{2} \\angle ACO\\). The reasoning is as follows:\n\nSince \\(OE\\) bisects \\(\\angle MON\\) and \\(AB\\) bisects \\(\\angle MAC\\),\n\n\\(\\therefore \\angle AOB = \\frac{1}{2} \\angle AOC\\), \\(\\angle MAB = \\frac{1}{2} \\angle MAC\\),\n\n\\(\\therefore \\angle ABO = \\angle MAB - \\angle AOB = \\frac{1}{2} \\angle MAC - \\frac{1}{2} \\angle AOC = \\frac{1}{2}(\\angle MAC - \\angle AOC)\\),\n\nSince \\(\\angle MAC - \\angle AOC = \\angle ACO\\),\n\n\\(\\therefore \\angle ABO = \\frac{1}{2} \\angle ACO\\),\n\nTherefore, the answer is: \\(\\angle ABO = \\frac{1}{2} \\angle ACO\\);\n\n(3) Solution: Since \\(AB \\perp OM\\),\n\n\\(\\therefore \\angle BAO = 90^\\circ\\),\n\nSince \\(\\angle AOB = \\frac{1}{2} \\angle AOC = 38^\\circ\\),\n\n\\(\\therefore \\angle ABO = 90^\\circ - 38^\\circ = 52^\\circ\\),\n\nWhen \\(\\angle ABD = \\angle ADB\\), \\(\\angle OAC = \\angle ADB - \\angle AOD = 52^\\circ - 38^\\circ = 14^\\circ\\);\n\nWhen \\(\\angle ABD = \\angle BAD\\), \\(\\angle OAC = 90^\\circ - \\angle BAD = 90^\\circ - 52^\\circ = 38^\\circ\\);\n\nWhen \\(\\angle ADB = \\angle BAD\\), \\(\\angle BAD = \\frac{180^\\circ - 52^\\circ}{2} = 64^\\circ\\),\n\nThen \\(\\angle OAC = 90^\\circ - \\angle BAD = 90^\\circ - 64^\\circ = 26^\\circ\\).\n\nTherefore, the measure of \\(\\angle OAC\\) is \\(14^\\circ\\) or \\(38^\\circ\\) or \\(26^\\circ\\).\n\n【Highlight】This problem examines the properties of parallel lines, the triangle angle sum theorem, and the application of the exterior angle property of triangles. Note: The sum of the interior angles of a triangle is 180 degrees, and an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles." }, { "problem_id": 1992, "question": "Given \\( M N / / P Q \\), point \\( D \\) is a fixed point on the line \\( P Q \\).\n\n\n\nFigure 1\n\n\n\nFigure 3\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) As shown in Figure 1, there is a right triangle with a \\( 30^\\circ \\) angle (\\( \\angle C A B = 30^\\circ, \\angle A C B = 60^\\circ, \\angle A B C = 90^\\circ \\)). The point \\( A \\) is fixed on the line \\( M N \\), and it is placed as shown in Figure 1, such that \\( \\angle M A C = 30^\\circ \\), and point \\( B \\) exactly lies on the ray \\( D E \\). At this moment, \\( \\angle P D E = 20^\\circ \\). Find the measure of \\( \\angle A B D \\).\n\n(2) Now, the ray \\( D E \\) starts rotating clockwise around point \\( D \\) from the position in Figure 1 at a speed of \\( 2^\\circ \\) per second, stopping when it coincides with \\( D Q \\). The triangle remains stationary as placed in Figure 1. Let the rotation time be \\( t \\) seconds. During the rotation, when \\( D E \\) is parallel to one side of the triangle, find the value of \\( t \\).\n\n(3) If the ray \\( D E \\) starts rotating clockwise around point \\( D \\) at a speed of \\( 2^\\circ \\) per second from the position in Figure 1, and simultaneously, the triangle \\( A B C \\) also starts rotating counterclockwise around point \\( A \\) at a speed of \\( 4^\\circ \\) per second from the position in Figure 1,\n\nThe angle bisector \\( A H \\) of \\( \\angle M A C \\) intersects the angle bisector \\( D F \\) of \\( \\angle P D E \\) at point \\( O \\).\n\n(1) As shown in Figure 2, when \\( D F / / B C \\), \\( \\angle A O D = \\) \\(\\qquad\\) degrees;\n\n(2) As shown in Figure 3, when \\( D F / / B A \\), \\( \\angle A O D = \\) \\(\\qquad\\) degrees.", "input_image": [ "batch5-2024_06_14_f7dd8e06445035e43841g_0046_1.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0046_2.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0046_3.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0046_4.jpg" ], "is_multi_img": true, "answer": "(1) $80^{\\circ}$\n\n(2)5 or 50 or 65\n\n(3)) 37 (2) 91\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Junior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "(1) As shown in Figure 1, draw line $BK$ through point $B$ such that $BK / / MN$,\n\n\n\nFigure 1\n\n$\\therefore \\angle ABK = \\angle MAB$.\n\n$\\because PQ / / MN$,\n\n$\\therefore BK / / PQ$,\n\n$\\therefore \\angle KBD = \\angle PDE$\n\n$\\because \\angle PDE = 20^{\\circ}$,\n\n$\\therefore \\angle KBD = 20^{\\circ}$,\n\n$\\because \\angle MAC = 30^{\\circ}, \\angle CAB = 30^{\\circ}$,\n\n$\\therefore \\angle MAB = \\angle MAC + \\angle CAB = 60^{\\circ}$,\n\n$\\therefore \\angle ABK = 60^{\\circ}$,\n\n$\\therefore \\angle ABD = \\angle ABK + \\angle KBD = 80^{\\circ}$.\n\n(2) According to the problem, $\\angle PDE = (20 + 2t)^{\\circ}$. We discuss the following three cases:\n\n(1) As shown in Figure 4, when $DE / / BC$, $DE$ intersects $AB$ at point $R$,\n\n$\\because DE / / BC, \\angle ABC = 90^{\\circ}$,\n\n$\\therefore \\angle BRD = \\angle ABC = 90^{\\circ}$,\n\n$\\therefore \\angle ARD = 180^{\\circ} - \\angle BRD = 90^{\\circ}$,\n\n$\\because \\angle MAB = 60^{\\circ}$,\n\n$\\therefore \\angle PDE = \\angle ARD - \\angle MAB = 30^{\\circ}$,\n\n$\\therefore 20 + 2t = 30$, solving gives $t = 5$.\n\n\n\nFigure 4\n\n(2) As shown in Figure 5, when $DE / / AB$, $DE$ intersects $MN$ at point $S$,\n\n$\\because DE / / AB, \\angle MAB = 60^{\\circ}$,\n\n$\\therefore \\angle DSM = \\angle MAB = 60^{\\circ}$,\n\n$\\because MN / / PQ$,\n\n$\\therefore \\angle PDE + \\angle DSM = 180^{\\circ}$,\n\n$\\therefore \\angle PDE = 180^{\\circ} - \\angle DSM = 120^{\\circ}$,\n\n$\\therefore 20 + 2t = 120$, solving gives $t = 50$.\n\n\n\nFigure 5\n\n(3) As shown in Figure 6, when $DE / / AC$, $DE$ intersects $MN$ at point $T$,\n\n$\\because DE / / AC, \\angle MAC = 30^{\\circ}$,\n\n$\\therefore \\angle MTD = \\angle MAC = 30^{\\circ}$,\n\n$\\because MN / / PQ$,\n\n$\\therefore \\angle MTD + \\angle PDE = 180^{\\circ}$,\n\n$\\therefore \\angle PDE = 180^{\\circ} - \\angle MTD = 150^{\\circ}$,\n\n$\\therefore 20 + 2t = 150$, solving gives $t = 65$.\n\n\n\nFigure 6\n\nIn summary, the value of $t$ is 5, 50, or 65.\n\n(3) According to the problem, $\\angle PDE = (20 + 2t)^{\\circ}, \\angle MAC = (30 + 4t)^{\\circ}$,\n\n$\\because$ The angle bisector $AH$ of $\\angle MAC$ and the angle bisector $DF$ of $\\angle PDE$ intersect at point $O$,\n\n$\\therefore \\angle PDO = \\frac{1}{2} \\angle PDE = 10 + t, \\angle MAO = \\angle OAC = \\frac{1}{2} \\angle MAC = 15 + 2t$,\n\nFrom the model in (1), we have $\\angle AOD = \\angle MAO + \\angle PDO = 25 + 3t$,\n\n(1) When $DF / / BC$, extend $AB$ to intersect $DF$ at $G$,\n\n\n\nFigure 2\n\n$\\therefore \\angle ABC = \\angle OGA = 90^{\\circ}, \\angle OAG = \\angle OAC + \\angle BAC = 45 + 2t$\n\n$\\because \\angle AOD + \\angle GAO + \\angle OGA = 180^{\\circ}$,\n\n$\\therefore 25 + 3t + 90 + 45 + 2t = 180$,\n\nSolving gives $t = 4$,\n\n$\\angle AOD = \\angle MAO + \\angle PDO = 25 + 3t = 37^{\\circ}$,\n\nTherefore, the answer is: 37;\n\n(2) When $DF / / BA$, extend $AC$ to intersect $DF$ at $G$,\n\n\n\nFigure 3\n\n$\\therefore \\angle BAC = \\angle OGA = 30^{\\circ}, \\angle OAG = \\angle OAC = 15 + 2t$\n\n$\\because \\angle AOD + \\angle GAO + \\angle OGA = 180^{\\circ}$,\n$\\therefore 25 + 3t + 30 + 15 + 2t = 180$,\n\nSolving gives $t = 22$,\n\n$\\angle AOD = \\angle MAO + \\angle PDO = 25 + 3t = 91^{\\circ}$,\n\nTherefore, the answer is: 91;\n\n【Insight】This problem examines the drawing of translation transformations, the determination and properties of parallel lines, and the interior angles of triangles. The key to solving the problem is understanding the problem and using a classification discussion approach to think about the problem, which is a common type of question in middle school exams." }, { "problem_id": 1993, "question": "(1) As shown in Figure 1, the angle bisector $AE$ of $\\angle BAD$ intersects the angle bisector $CE$ of $\\angle BCD$ at point $E$.\n\nGiven $\\angle ABC = 60^\\circ$ and $\\angle ADC = 140^\\circ$, the measure of $\\angle AEC$ is $\\qquad$;\n\n(2) As shown in Figure 2, the angle bisector $AE$ of $\\angle BAD$ intersects the angle bisector $CE$ of $\\angle BCD$ at point $E$.\n\nGiven $\\angle ABC = \\alpha$ and $\\angle ADC = \\beta (\\alpha > \\beta)$, find the measure of $\\angle AEC$; (express in terms of $\\alpha$ and $\\beta$)\n\n(3) As shown in Figure 3, in $\\triangle ABC$, $\\angle ACB = \\alpha$ and $\\angle ABC = \\beta (\\alpha > \\beta)$. $AD$ is the angle bisector of $\\triangle ABC$, and point $E$ is a point on the extension of $AD$. Construct $EF \\perp BC$ at point $F$. Is the value of $\\frac{\\angle AEF}{\\alpha - \\beta}$ constant? If so, find its value; if not, explain why.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch5-2024_06_14_f7dd8e06445035e43841g_0087_1.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0087_2.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0087_3.jpg" ], "is_multi_img": true, "answer": "（1） $100^{\\circ}$;（2） $\\frac{1}{2}(\\alpha-\\beta)$ ；（3）remain unchanged, $\\frac{1}{2}$\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "Solution: (1) As shown in the figure, extend $CD$ to intersect $AB$ at point $H$, and draw ray $BF$ passing through point $E$.\n\n\n\nSince $\\angle ADC = \\angle DAH + \\angle AHD$ and $\\angle ADC = 140^\\circ$,\n\n$\\therefore \\angle DAH + \\angle AHD = 140^\\circ$.\n\nSince $\\angle AHD = \\angle ABC + \\angle BCD$,\n\n$\\therefore \\angle ABC + \\angle BCD + \\angle DAH = 140^\\circ$.\n\nGiven that $\\angle ABC = 60^\\circ$,\n\n$\\therefore \\angle BCD + \\angle DAH = 80^\\circ$.\n\nSince the angle bisector $AE$ of $\\angle BAD$ and the angle bisector $CE$ of $\\angle BCD$ intersect at point $E$,\n\n$\\therefore \\angle BCE + \\angle BAE = 40^\\circ$.\n\nSince $\\angle CEF = \\angle CBE + \\angle BCE$ and $\\angle AEF = \\angle ABE + \\angle BAE$,\n\n$\\therefore \\angle AEC = \\angle CEF + \\angle AEF = \\angle BCE + \\angle CBE + \\angle ABE + \\angle BAE = \\angle ABC + \\angle BCE + \\angle BAE = 60^\\circ + 40^\\circ = 100^\\circ$.\n\nTherefore, the answer is: $100^\\circ$.\n\n(2) Draw ray $AG$ passing through point $C$, as shown in the figure.\n\n\n\n$\\therefore \\angle BCD = \\angle BCG + \\angle DCG = \\angle B + \\angle BAC + \\angle D + \\angle DAC = \\alpha + \\beta + \\angle BAD$.\n\nSince the angle bisector $AE$ of $\\angle BAD$ and the angle bisector $CE$ of $\\angle BCD$ intersect at point $E$,\n\n$\\therefore \\angle BAF = \\frac{1}{2} \\angle BAD$, $\\angle BCE = \\frac{1}{2} \\angle BCD = \\frac{1}{2} \\alpha + \\frac{1}{2} \\beta + \\frac{1}{2} \\angle BAD$.\n\nSince $\\angle BFE = \\angle B + \\angle BAF = \\alpha + \\frac{1}{2} \\angle BAD$,\n\n$\\therefore \\angle AEC = \\angle BFE - \\angle BCE = \\alpha + \\frac{1}{2} \\angle BAD - \\left(\\frac{1}{2} \\alpha + \\frac{1}{2} \\beta + \\frac{1}{2} \\angle BAD\\right) = \\frac{1}{2}(\\alpha - \\beta)$.\n\n(3) The value of $\\frac{\\angle AEF}{\\alpha - \\beta}$ remains constant and is always $\\frac{1}{2}$. The reasoning is as follows:\n\nSince $\\angle ACB = \\alpha$ and $\\angle ABC = \\beta$,\n\n$\\therefore \\angle BAC = 180^\\circ - \\alpha - \\beta$.\n\nSince $AD$ is the angle bisector of $\\triangle ABC$,\n\n$\\therefore \\angle BAD = \\angle CAD = \\frac{1}{2} \\angle BAC = 90^\\circ - \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta$.\n\n$\\therefore \\angle EDF = \\angle B + \\angle BAD = \\beta + 90^\\circ - \\frac{1}{2} \\alpha - \\frac{1}{2} \\beta = 90^\\circ - \\frac{1}{2} \\alpha + \\frac{1}{2} \\beta$.\n\nSince $EF \\perp BC$,\n\n$\\therefore \\angle AEF = 90^\\circ - \\angle EDF = \\frac{1}{2}(\\alpha - \\beta)$.\n\n$\\therefore \\frac{\\angle AEF}{\\alpha - \\beta} = \\frac{1}{2}$.\n\nTherefore, the value of $\\frac{\\angle AEF}{\\alpha - \\beta}$ remains constant and is always $\\frac{1}{2}$.\n\n【Key Insight】This problem examines the triangle angle sum theorem, the properties of exterior angles of a triangle, the properties of parallel lines, the properties of perpendicular lines, and the definition of angle bisectors. Mastering the properties of parallel lines, the properties of exterior angles of a triangle, and the properties of angle bisectors is crucial for solving the problem." }, { "problem_id": 1994, "question": "In $\\triangle A B C$, $A B>A C$, point $E$ is on side $B C$, connect $A E$, and fold $\\triangle A E C$ along $A E$ so that point $D$ falls on side $A B$ to form $\\triangle A E D$, connect $D C$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) As shown in Figure 1, $\\angle A B C=50^{\\circ}, \\angle A C B=90^{\\circ}$, find the measure of $\\angle B C D$.\n\n(2) As shown in Figure 2, if $A B=B C, B D=D E$, find the measure of $\\angle B C D$.", "input_image": [ "batch5-2024_06_14_f7dd8e06445035e43841g_0090_1.jpg", "batch5-2024_06_14_f7dd8e06445035e43841g_0090_2.jpg" ], "is_multi_img": true, "answer": "(1) $20^{\\circ}$\n\n(2) $18^{\\circ}$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Junior", "subject": "Transformation Geometry", "image_relavance": "1", "analysis": "(1) Solution: In triangle \\( ABC \\), since \\( \\angle ABC = 50^\\circ \\) and \\( \\angle ACB = 90^\\circ \\),\n\n\\[\n\\angle BAC = 180^\\circ - 90^\\circ - 50^\\circ = 40^\\circ.\n\\]\n\nWhen triangle \\( AEC \\) is folded along \\( AE \\) to form triangle \\( AED \\),\n\n\\[\nAD = AC,\n\\]\n\n\\[\n\\angle ADC = \\angle ACD = \\frac{1}{2} \\times (180^\\circ - \\angle DAC) = \\frac{180^\\circ - 40^\\circ}{2} = 70^\\circ,\n\\]\n\n\\[\n\\angle BCD = \\angle ACB - \\angle ACD = 90^\\circ - 70^\\circ = 20^\\circ.\n\\]\n\n(2) Solution: Since \\( AB = BC \\),\n\n\\[\n\\angle BAC = \\angle BCA.\n\\]\n\nWhen triangle \\( AEC \\) is folded along \\( AE \\) to form triangle \\( AED \\),\n\n\\[\n\\triangle ADE \\cong \\triangle ACE,\n\\]\n\n\\[\n\\angle ADE = \\angle ACE, \\quad DE = EC,\n\\]\n\n\\[\n\\angle ADE = \\angle ACE = \\angle BAC, \\quad \\angle EDC = \\angle ECD.\n\\]\n\nSince \\( BD = DE \\),\n\n\\[\n\\angle B = \\angle DEB,\n\\]\n\n\\[\n\\angle B = \\frac{1}{2} \\angle ADE = \\frac{1}{2} \\angle BAC = \\frac{1}{2} \\angle BCA.\n\\]\n\nGiven that \\( \\angle B + \\angle BAC + \\angle BCA = 180^\\circ \\),\n\n\\[\n\\frac{1}{2} \\angle BCA + \\angle BCA + \\angle BCA = 180^\\circ,\n\\]\n\n\\[\n\\angle BCA = 72^\\circ,\n\\]\n\n\\[\n\\angle B = \\angle BED = \\frac{1}{2} \\times 72^\\circ = 36^\\circ.\n\\]\n\nSince \\( \\angle EDC + \\angle DCE = \\angle DEB \\),\n\n\\[\n\\angle DCB = \\frac{1}{2} \\times 36^\\circ = 18^\\circ.\n\\]\n\n【Insight】This problem examines the properties of folding, the properties of isosceles triangles, the triangle angle sum theorem, and the properties of exterior angles of triangles. The key to solving the problem is understanding that corresponding sides and angles are equal after folding, and that in an isosceles triangle, equal sides correspond to equal angles." }, { "problem_id": 1995, "question": "As shown in Figure 1, in the plane rectangular coordinate system \\( xOy \\), the parabola \\( y = -x^2 + 2x + 3 \\) intersects the \\( x \\)-axis at points \\( A \\) and \\( B \\), and intersects the \\( y \\)-axis at point \\( C \\). Line segment \\( BC \\) is connected.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the coordinates of points \\( B \\) and \\( C \\);\n\n(2) As shown in Figure 2, point \\( P \\) is a moving point on the parabola, moving along the parabola from point \\( B \\) to point \\( A \\). Line segments \\( PO \\) and \\( PB \\) are connected, and a parallelogram \\( POQB \\) is formed with \\( PO \\) and \\( PB \\) as its sides.\n\n(1) When the area of the parallelogram \\( POQB \\) is 9, find the coordinates of point \\( P \\);\n\n(2) During the entire motion process, find the maximum distance between point \\( Q \\) and line segment \\( BC \\).", "input_image": [ "batch6-2024_06_14_51e39c963cc74becf9f8g_0018_1.jpg", "batch6-2024_06_14_51e39c963cc74becf9f8g_0018_2.jpg" ], "is_multi_img": true, "answer": "(1) $B(3,0), C(0,3)$\n\n(2)(1) $(0,3)$ or $(2,3)$; (2) $\\frac{21 \\sqrt{2}}{8}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "(1) Solution: Since \\( y = -x^{2} + 2x + 3 \\),\n\nLet \\( x = 0 \\), then \\( y = 3 \\),\n\nTherefore, \\( C(0, 3) \\),\n\nLet \\( y = 0 \\), then \\( -x^{2} + 2x + 3 = 0 \\),\n\nSolving gives: \\( x = 3 \\) or \\( x = -1 \\),\n\nTherefore, \\( A(-1, 0) \\), \\( B(3, 0) \\), \\( OB = 3 \\),\n\nThus, \\( B(3, 0) \\), \\( C(0, 3) \\).\n\n(2) (1) Since the area of parallelogram \\( POQB \\) is 9,\n\nTherefore, the area of \\( \\triangle POB \\) is 4.5,\n\nThus, \\( \\frac{1}{2} OB \\cdot y_{P} = 4.5 \\), i.e., \\( \\frac{1}{2} \\times 3 \\times y_{P} = 4.5 \\),\n\nTherefore, \\( y_{P} = 3 \\),\n\nThus, \\( -x^{2} + 2x + 3 = 3 \\),\n\nSolving gives: \\( x = 0 \\) or \\( x = 2 \\),\n\nTherefore, the coordinates of point \\( P \\) are \\( (0, 3) \\) or \\( (2, 3) \\);\n\n(2) As shown in the figure, connect \\( CQ \\), and draw \\( QH \\) parallel to the y-axis intersecting \\( BC \\) at \\( H \\),\n\nLet \\( P(t, -t^{2} + 2t + 3) \\), \\( Q(m, n) \\),\n\nSince quadrilateral \\( POQB \\) is a parallelogram,\n\nTherefore, \\( PQ \\) and \\( OB \\) bisect each other, i.e., the midpoints of \\( PQ \\) and \\( OB \\) coincide,\n\nThus, \n\\[\n\\begin{cases}\n0 + 3 = t + m \\\\\n0 + 0 = -t^{2} + 2t + 3 + n\n\\end{cases}\n\\]\n\nEliminating \\( t \\), we get: \\( n = m^{2} - 4m \\),\n\nTherefore, \\( Q(m, m^{2} - 4m) \\),\n\nSince \\( B(3, 0) \\), \\( C(0, 3) \\),\n\nTherefore, \\( BC = \\sqrt{OC^{2} + OB^{2}} = \\sqrt{3^{2} + 3^{2}} = 3\\sqrt{2} \\),\n\nLet the equation of line \\( BC \\) be \\( y = kx + b \\),\n\nThus, \n\\[\n\\begin{cases}\n3k + b = 0 \\\\\nb = 3\n\\end{cases}\n\\]\n\nSolving gives: \n\\[\n\\begin{cases}\nk = -1 \\\\\nb = 3\n\\end{cases}\n\\]\n\nTherefore, the equation of line \\( BC \\) is \\( y = -x + 3 \\),\n\nThus, let \\( H(m, -m + 3) \\),\n\nTherefore, \\( QH = (-m + 3) - (m^{2} - 4m) = -m^{2} + 3m + 3 = -\\left(m - \\frac{3}{2}\\right)^{2} + \\frac{21}{4} \\),\n\nThus, \\( S_{\\triangle BCQ} = \\frac{1}{2} QH \\cdot |x_{B} - x_{C}| = \\frac{1}{2} \\times \\left[-\\left(m - \\frac{3}{2}\\right)^{2} + \\frac{21}{4}\\right] \\times 3 = -\\frac{3}{2}\\left(m - \\frac{3}{2}\\right)^{2} + \\frac{63}{8} \\),\n\nLet the distance from point \\( Q \\) to line segment \\( BC \\) be \\( h \\),\n\nThus, \\( S_{\\triangle BCQ} = \\frac{1}{2} BC \\cdot h = -\\frac{3}{2}\\left(m - \\frac{3}{2}\\right)^{2} + \\frac{63}{8} \\),\n\nTherefore, \\( \\frac{1}{2} \\times 3\\sqrt{2} h = -\\frac{3}{2}\\left(m - \\frac{3}{2}\\right)^{2} + \\frac{63}{8} \\),\n\nThus, \\( h = -\\frac{\\sqrt{2}}{2}\\left(m - \\frac{3}{2}\\right)^{2} + \\frac{21\\sqrt{2}}{8} \\),\n\nSince \\( -\\frac{\\sqrt{2}}{2} < 0 \\),\n\nTherefore, when \\( m = \\frac{3}{2} \\), \\( h \\) reaches its maximum value of \\( \\frac{21\\sqrt{2}}{8} \\).\n\nThus, the maximum distance from point \\( Q \\) to line segment \\( BC \\) is \\( \\frac{21\\sqrt{2}}{8} \\).\n\n\n\n【Key Insight】This problem examines the comprehensive application of quadratic functions, involving the characteristics of point coordinates on the function graph, triangle area, and parallelogram. The key to solving the problem lies in expressing the coordinates of related points and the lengths of related line segments using algebraic expressions." }, { "problem_id": 1996, "question": "As shown in Figure 1, the parabola \\( y = ax^2 + bx - 3 \\) intersects the \\( x \\)-axis at points \\( A(4,0) \\) and \\( B(-1,0) \\), and intersects the \\( y \\)-axis at point \\( C \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the equation of this parabola;\n\n(2) Point \\( P \\) is a moving point on the parabola below the line \\( AC \\). Connect \\( PA \\) and \\( PC \\). Find the maximum area of \\( \\triangle ACP \\);\n\n(3) As shown in Figure 2, line \\( l \\) is the axis of symmetry of the parabola. Does there exist a point \\( M \\) on line \\( l \\) such that \\( \\triangle BCM \\) is a right triangle? If so, find the coordinates of point \\( M \\); if not, explain why.", "input_image": [ "batch6-2024_06_14_51e39c963cc74becf9f8g_0070_1.jpg", "batch6-2024_06_14_51e39c963cc74becf9f8g_0070_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=\\frac{3}{4} x^{2}-\\frac{9}{4} x-3$\n\n(2) $S_{\\triangle A C P}$ has a maximum value of 6\n\n(3) The coordinates of $M$ are $M\\left(\\frac{3}{2}, \\frac{5}{6}\\right)$ or $M\\left(\\frac{3}{2},-\\frac{5}{2}\\right)$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\n(1) Substituting the points \\( A(4,0) \\) and \\( B(-1,0) \\) into the parabolic equation \\( y = ax^2 + bx - 3 \\) yields:\n\n\\[\n\\begin{cases}\n0 = 16a + 4b - 3 \\\\\n0 = a - b - 3\n\\end{cases}\n\\]\n\nSolving the system of equations gives:\n\n\\[\n\\begin{cases}\na = \\frac{3}{4} \\\\\nb = -\\frac{9}{4}\n\\end{cases}\n\\]\n\nThus, the equation of the parabola is:\n\n\\[\ny = \\frac{3}{4}x^2 - \\frac{9}{4}x - 3\n\\]\n\n(2) Drawing a perpendicular line from point \\( P \\) to the \\( x \\)-axis, intersecting line \\( AC \\) at point \\( E \\):\n\nWhen \\( x = 0 \\), \\( y = -3 \\), so \\( C(0, -3) \\).\n\nLet point \\( P \\) be \\( \\left(m, \\frac{3}{4}m^2 - \\frac{9}{4}m - 3\\right) \\). The equation of line \\( AC \\) is \\( y = kx + b \\).\n\nSubstituting points \\( A(4,0) \\) and \\( C(0,-3) \\) into the equation gives:\n\n\\[\n\\begin{cases}\n4k + b = 0 \\\\\nb = -3\n\\end{cases}\n\\]\n\nSolving the system yields:\n\n\\[\n\\begin{cases}\nk = \\frac{3}{4} \\\\\nb = -3\n\\end{cases}\n\\]\n\nThus, the equation of line \\( AC \\) is:\n\n\\[\ny = \\frac{3}{4}x - 3\n\\]\n\nSince \\( PE \\) is perpendicular to the \\( x \\)-axis and intersects \\( AC \\) at \\( E \\), the coordinates of \\( E \\) are:\n\n\\[\nE\\left(m, \\frac{3}{4}m - 3\\right)\n\\]\n\nTherefore, the length \\( PE \\) is:\n\n\\[\nPE = \\frac{3}{4}m - 3 - \\left(\\frac{3}{4}m^2 - \\frac{9}{4}m - 3\\right) = -\\frac{3}{4}m^2 + 3m \\quad (0 < m < 4)\n\\]\n\nThe area of triangle \\( ACP \\) is:\n\n\\[\nS_{\\triangle ACP} = S_{\\triangle AEP} + S_{\\triangle CEP} = \\frac{1}{2} \\times PE \\times 4 = 2PE = -\\frac{3}{2}m^2 + 6m = -\\frac{3}{2}(m^2 - 4m)\n\\]\n\n\\[\n= -\\frac{3}{2}(m - 2)^2 + 6\n\\]\n\nThus, when \\( m = 2 \\), \\( S_{\\triangle ACP} \\) reaches its maximum value of 6.\n\n(3) There exists a point \\( M \\) such that triangle \\( BCM \\) is a right triangle. The reasoning is as follows:\n\nThe axis of symmetry of the parabola is the line \\( x = \\frac{3}{2} \\). Let \\( M \\) be \\( \\left(\\frac{3}{2}, t\\right) \\).\n\nGiven \\( B(-1,0) \\) and \\( C(0,-3) \\), we have:\n\n\\[\nBC^2 = 3^2 + 1^2 = 10, \\quad MC^2 = \\left(\\frac{3}{2}\\right)^2 + (t + 3)^2 = t^2 + 6t + \\frac{45}{4}, \\quad MB^2 = \\left(\\frac{5}{2}\\right)^2 + t^2 = \\frac{25}{4} + t^2\n\\]\n\n- **Case 1:** When \\( \\angle CBM = 90^\\circ \\):\n\n\\[\nMC^2 = BC^2 + BM^2\n\\]\n\n\\[\nt^2 + 6t + \\frac{45}{4} = 10 + t^2 + \\frac{25}{4}\n\\]\n\nSolving for \\( t \\) gives:\n\n\\[\nt = \\frac{5}{6}\n\\]\n\nThus, \\( M\\left(\\frac{3}{2}, \\frac{5}{6}\\right) \\).\n\n- **Case 2:** When \\( \\angle BCM = 90^\\circ \\):\n\n\\[\nBM^2 = BC^2 + MC^2\n\\]\n\n\\[\nt^2 + \\frac{25}{4} = 10 + t^2 + 6t + \\frac{45}{4}\n\\]\n\nSolving for \\( t \\) gives:\n\n\\[\nt = -\\frac{5}{2}\n\\]\n\nThus, \\( M\\left(\\frac{3}{2}, -\\frac{5}{2}\\right) \\).\n\n- **Case 3:** When \\( \\angle BMC = 90^\\circ \\):\n\n\\[\nBC^2 = BM^2 + MC^2\n\\]\n\n\\[\n10 = t^2 + \\frac{25}{4} + t^2 + 6t + \\frac{45}{4}\n\\]\n\nSimplifying:\n\n\\[\n2t^2 + 6t + 7.5 = 0\n\\]\n\nThe discriminant \\( \\Delta = b^2 - 4ac = 36 - 4 \\times 2 \\times 7.5 = -24 < 0 \\), so there is no real solution.\n\nIn conclusion, the coordinates of \\( M \\) are either \\( M\\left(\\frac{3}{2}, \\frac{5}{6}\\right) \\) or \\( M\\left(\\frac{3}{2}, -\\frac{5}{2}\\right) \\).\n\n**Key Insight:** This problem examines the properties of quadratic functions, the distance between two points, and the Pythagorean theorem. Mastery of these concepts is essential for solving such problems." }, { "problem_id": 1997, "question": "In the plane rectangular coordinate system, the parabola \\( y = a(x+1)(x-3) \\) (where \\( a \\neq 0 \\)) intersects the \\( x \\)-axis at points \\( A \\) and \\( B \\) (with point \\( A \\) to the left of point \\( B \\)), and intersects the \\( y \\)-axis at point \\( C(0,3) \\). Point \\( D \\) is the vertex of the parabola, and point \\( P \\) is a point on the axis of symmetry of the parabola.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nSpare Figure\n\n(1) Find the equation of the parabola and the coordinates of point \\( D \\);\n\n(2) As shown in Figure (1), connect \\( PB \\) and \\( PD \\), and find the minimum value of \\( PB + \\frac{\\sqrt{2}}{2} PD \\);\n\n(3) As shown in Figure (2), connect \\( CP \\), \\( PB \\), and \\( BC \\). If \\( \\angle CPB = 135^\\circ \\), find the coordinates of point \\( P \\).", "input_image": [ "batch6-2024_06_14_51e39c963cc74becf9f8g_0083_1.jpg", "batch6-2024_06_14_51e39c963cc74becf9f8g_0083_2.jpg", "batch6-2024_06_14_51e39c963cc74becf9f8g_0083_3.jpg" ], "is_multi_img": true, "answer": "(1) $y=-x^{2}+2 x+3, \\quad D(1,4)$\n\n(2) $3 \\sqrt{2}$\n\n(3) $P(1,2 \\sqrt{2})$ or $P(1,3-\\sqrt{5})$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) Let \\( y = 0 \\), then \\( a(x+1)(x-3) = 0 \\),\n\nSolving gives: \\( x = 3 \\) or \\( x = -1 \\),\n\n\\(\\therefore A(-1,0), \\quad B(3,0)\\),\n\n\\(\\because C(0,3)\\) lies on the parabola,\n\n\\(\\therefore -3a = 3\\),\n\n\\(\\therefore a = -1\\)\n\n\\(\\therefore y = -x^{2} + 2x + 3\\),\n\\(\\therefore y = -x^{2} + 2x + 3 = -(x-1)^{2} + 4\\)\n\n\\(\\therefore D(1,4)\\)\n\n(2) \\(\\because y = -x^{2} + 2x + 3 = -(x-1)^{2} + 4\\),\n\n\\(\\therefore\\) the axis of symmetry of the parabola is the line \\( x = 1 \\),\n\nConnect \\( C D \\), draw \\( P H \\perp C D \\) at point \\( H \\), connect \\( B C \\) intersecting \\( D P \\) at point \\( E \\), and draw \\( C M \\perp D E \\) from point \\( C \\),\n\n\n\nFigure (1)\n\n\\(\\because C(0,3), \\quad D(1,4)\\),\n\n\\(\\therefore D M = C M = 1\\),\n\n\\(\\therefore \\angle C D M = 45^{\\circ}\\),\n\n\\(\\therefore P H = \\frac{\\sqrt{2}}{2} P D\\),\n\n\\(\\therefore P B + \\frac{\\sqrt{2}}{2} P D = P B + P H\\),\n\n\\(\\therefore\\) when points \\( B, P, H \\) are collinear, \\( P B + \\frac{\\sqrt{2}}{2} P D \\) has a minimum value,\n\n\\(\\because \\angle B C O = 45^{\\circ}\\),\n\n\\(\\therefore \\angle C E D = 45^{\\circ}\\),\n\n\\(\\therefore \\angle D C B = 90^{\\circ}\\),\n\n\\(\\therefore B C \\perp C D\\),\n\n\\(\\therefore\\) the minimum value of \\( P B + \\frac{\\sqrt{2}}{2} P D \\) is the length of \\( B C \\),\n\n\\(\\because C(0,3), \\quad B(3,0)\\),\n\n\\(\\therefore B C = 3 \\sqrt{2}\\),\n\n\\(\\therefore\\) the minimum value of \\( P B + \\frac{\\sqrt{2}}{2} P D \\) is \\( 3 \\sqrt{2} \\)\n\n(3) When point \\( P \\) is above the line segment \\( B C \\), draw a circle with \\( O \\) as the center and \\( O C \\) as the radius, intersecting the axis of symmetry of the parabola above \\( B C \\) at point \\( P \\), at this time \\( \\angle C P B = 135^{\\circ} \\), connect \\( O P \\),\n\n\n\nWhen point \\( P \\) is below the line segment \\( B C \\), construct an isosceles right triangle \\( C M B \\) with \\( B C \\) as the hypotenuse above \\( B C \\), draw a circle with \\( M \\) as the center and \\( C M \\) as the radius, intersecting the axis of symmetry of the parabola below \\( B C \\) at point \\( P \\), at this time \\( \\angle C P B = 135^{\\circ} \\), draw \\( P H \\perp B M \\) from point \\( P \\),\n\n\n\n\\(\\because P M = C M = 3, \\quad P H = 2\\),\n\n\\(\\therefore M H = \\sqrt{5}\\),\n\n\\(\\because \\angle C B O = 45^{\\circ}, \\angle C B M = 45^{\\circ}\\),\n\n\\(\\therefore B M \\perp x\\)-axis,\n\nSimilarly, \\( C M \\perp y\\)-axis,\n\n\\(\\therefore M(3,3)\\),\n\n\\(\\therefore P(1,3-\\sqrt{5})\\)\n\nIn summary: the coordinates of point \\( P \\) are \\( P(1,2 \\sqrt{2}) \\) or \\( P(1,3-\\sqrt{5}) \\).\n\n【Highlight】This problem examines the image and properties of quadratic functions. Mastering the image and properties of quadratic functions, the method of finding the minimum value using the \"Hu Bugui\" approach, and constructing auxiliary circles are key to solving the problem." }, { "problem_id": 1998, "question": "As shown in the figure, in the plane rectangular coordinate system, the parabola \\( y = ax^2 + bx - 4 \\) intersects the \\( x \\)-axis at points \\( A(-2, 0) \\) and \\( B \\), and its axis of symmetry, the line \\( x = 2 \\), intersects the \\( x \\)-axis at point \\( D \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the function expression of the parabola as \\(\\qquad\\);\n\n(2) As shown in Figure 1, point \\( P \\) is a moving point on the parabola in the fourth quadrant. Connect \\( CD \\), \\( PB \\), and \\( PC \\). Find the maximum area of quadrilateral \\( BDCP \\) and the coordinates of point \\( P \\) at that time;\n\n(3) As shown in Figure 2, the parabola is translated to the left to form the parabola \\( y' \\). When the parabola \\( y' \\) passes through the origin, it intersects the axis of symmetry of the original parabola at point \\( E \\). Point \\( F \\) is a point on the axis of symmetry of the parabola \\( y' \\), and point \\( M \\) is a point in the plane. If the quadrilateral with vertices \\( A \\), \\( E \\), \\( F \\), and \\( M \\) is a rhombus with \\( AE \\) as one of its sides, directly write the coordinates of point \\( M \\) that satisfy the condition \\(\\qquad\\).", "input_image": [ "batch6-2024_06_14_51e39c963cc74becf9f8g_0088_1.jpg", "batch6-2024_06_14_51e39c963cc74becf9f8g_0088_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=\\frac{1}{3} x^{2}-\\frac{4}{3} x-4$\n\n(2) The maximum value of $S_{\\text {Quadrilateral } P B D C}$ is 17, and the coordinates of the point $P$ are $(3,-5)$\n\n(3) $\\left(0, \\frac{2 \\sqrt{127}+20}{3}\\right)$ or $\\left(0, \\frac{-2 \\sqrt{127}+20}{3}\\right)$ or $\\left(-8, \\frac{2 \\sqrt{55}}{3}\\right)$ or $\\left(-8,-\\frac{2 \\sqrt{55}}{3}\\right)$\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\n(1) From the problem statement, we have the following system of equations:\n\n$$\n\\left\\{\\begin{array}{l}\n4 a-2 b-4=0 \\\\\n-\\frac{b}{2 a}=2\n\\end{array}\\right.\n$$\n\nSolving this system, we find:\n\n$$\n\\left\\{\\begin{array}{l}a=\\frac{1}{3} \\\\ b=-\\frac{4}{3}\\end{array}\\right.\n$$\n\nThus, the equation of the parabola is:\n\n$$\ny=\\frac{1}{3} x^{2}-\\frac{4}{3} x-4\n$$\n\n---\n\n(2) Let \\( P\\left(m, \\frac{1}{3} m^{2}-\\frac{4}{3} m-4\\right) \\). From the problem statement:\n\n- When \\( y=0 \\), solving \\( \\frac{1}{3} x^{2}-\\frac{4}{3} x-4=0 \\) gives \\( x_{1}=-2 \\) and \\( x_{2}=6 \\), so \\( B(6,0) \\).\n- When \\( x=0 \\), \\( y=-4 \\), so \\( C(0,-4) \\).\n- The axis of symmetry \\( x=2 \\) intersects the \\( x \\)-axis at point \\( D \\), so \\( D(2,0) \\).\n\nThe area of quadrilateral \\( BDCP \\) is:\n\n\\[\nS_{\\text{四边形 } BDCP}=S_{\\triangle OCP}+S_{\\triangle OBP}-S_{\\triangle OCD}\n\\]\n\n\\[\n=\\frac{1}{2} \\times 4 \\times m+\\frac{1}{2} \\times 6 \\times\\left(-\\frac{1}{3} m^{2}+\\frac{4}{3} m+4\\right)-\\frac{1}{2} \\times 2 \\times 4\n\\]\n\n\\[\n=-m^{2}+6m+8\n\\]\n\n\\[\n=-(m-3)^{2}+17\n\\]\n\nSince \\( a=-1<0 \\), the maximum area occurs when \\( m=3 \\), with a maximum value of 17.\n\nWhen \\( m=3 \\), \\( y=\\frac{1}{3} \\times 3^{2}-\\frac{4}{3} \\times 3-4=-5 \\), so \\( P(3,-5) \\).\n\n---\n\n(3) From the problem statement, point \\( B \\) has moved to point \\( O \\), meaning the function has shifted left by 6 units. The new equation is:\n\n\\[\ny^{\\prime}=\\frac{1}{3}(x+6)^{2}-\\frac{4}{3}(x+6)-4=\\frac{1}{3} x^{2}+\\frac{8}{3} x\n\\]\n\nWhen \\( x=2 \\), \\( y^{\\prime}=\\frac{20}{3} \\), so the coordinates of \\( E \\) are \\( E\\left(2, \\frac{20}{3}\\right) \\).\n\nLet \\( F \\) be \\( F(-4, n) \\).\n\n- **Case 1:** When \\( AF=AE \\), \\( FM \\parallel AE \\), and \\( FM=AE \\), using the properties of translation:\n\n\\[\nM\\left(0, n+\\frac{20}{3}\\right)\n\\]\n\nFrom \\( AF=AE \\):\n\n\\[\n2^{2}+n^{2}=4^{2}+\\left(\\frac{20}{3}\\right)^{2}\n\\]\n\n\\[\nn= \\pm \\frac{2 \\sqrt{127}}{3}\n\\]\n\nThus, \\( M_{1}\\left(0, \\frac{20+2 \\sqrt{127}}{3}\\right) \\) or \\( M_{2}\\left(0, \\frac{20-2 \\sqrt{127}}{3}\\right) \\).\n\n- **Case 2:** When \\( FE=AE \\), \\( FM \\parallel AE \\), and \\( FM=AE \\), using the properties of translation:\n\n\\[\nM\\left(-8, n+\\frac{20}{3}\\right)\n\\]\n\nFrom \\( FE=AE \\):\n\n\\[\n6^{2}+\\left(\\frac{20}{3}-n\\right)^{2}=4^{2}+\\left(\\frac{20}{3}\\right)^{2}\n\\]\n\nSolving gives:\n\n\\[\nn=\\frac{-20 \\pm 2 \\sqrt{55}}{3}\n\\]\n\nThus, \\( M_{3}\\left(-8, \\frac{2 \\sqrt{55}}{3}\\right) \\) or \\( M\\left(-8,-\\frac{2 \\sqrt{55}}{3}\\right) \\).\n\nIn summary, the coordinates of \\( M \\) are:\n\n\\[\n\\left(0, \\frac{2 \\sqrt{127}+20}{3}\\right), \\left(0, \\frac{-2 \\sqrt{127}+20}{3}\\right), \\left(-8, \\frac{2 \\sqrt{55}}{3}\\right), \\text{ or } \\left(-8,-\\frac{2 \\sqrt{55}}{3}\\right)\n\\]\n\n---\n\n**Key Insight:** This problem involves finding the equation of a quadratic function, analyzing its properties, and solving for the coordinates of points forming a rhombus. The key steps are determining the quadratic equation, setting up the conditions for the rhombus, and solving the resulting equations." }, { "problem_id": 1999, "question": "Many streets in the city are either perpendicular or parallel to each other, so it is often impossible to walk in a straight line to reach the destination, and one can only walk by making right-angle turns. A Cartesian coordinate system \\(xOy\\) can be established according to the perpendicular and parallel directions of the streets. For two points \\(A\\left(x_{1}, y_{1}\\right)\\) and \\(B\\left(x_{2}, y_{2}\\right)\\), the distance between them is defined as: \\(d(A, B)=\\left|x_{1}-x_{2}\\right|+\\left|y_{1}-y_{2}\\right|\\).\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n(1) Given point \\(A(-2,1)\\), then \\(d(O, A)=\\) \\(\\qquad\\).\n\n(2) The graph of the function \\(y=-2x+4\\) for \\(0 \\leq x \\leq 2\\) is shown in Figure (1). Point \\(B\\) is a point on the graph, and \\(d(O, B)=3\\). Find the coordinates of point \\(B\\).\n\n(3) The graph of the function \\(y=x^{2}-5x+7\\) for \\(x \\geq 0\\) is shown in Figure (2). Point \\(D\\) is a point on the graph. Find the minimum value of \\(d(O, D)\\) and the corresponding coordinates of point \\(D\\).", "input_image": [ "batch6-2024_06_14_54c8b3c3e1f0939737abg_0056_1.jpg", "batch6-2024_06_14_54c8b3c3e1f0939737abg_0056_2.jpg" ], "is_multi_img": true, "answer": "(1)(1) 3 , (2) $(1,2)$\n\n(2) $3,(2,1)$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "(1) Given points \\( A(-2,1) \\) and \\( O(0,0) \\),\n\n\\[\nd(O, A) = |x_1 - x_2| + |y_1 - y_2| = |0 - (-2)| + |0 - 1| = 2 + 1 = 3,\n\\]\n\nThus, the answer is: 3.\n\n(2) Since point \\( B \\) lies on the graph of the function \\( y = -2x + 4 \\) for \\( 0 \\leq x \\leq 2 \\),\n\nand the coordinates of points on the graph of \\( y = -2x + 4 \\) are non-negative,\n\n\\[\nx_B \\geq 0, \\quad y_B \\geq 0, \\quad y_B = -2x_B + 4.\n\\]\n\nGiven \\( d(O, B) = 3 \\),\n\n\\[\n|0 - x_B| + |0 - y_B| = 3,\n\\]\n\n\\[\nx_B + y_B = 3.\n\\]\n\nSubstituting \\( y_B = -2x_B + 4 \\),\n\n\\[\n\\begin{cases}\ny_B = -2x_B + 4 \\\\\nx_B + y_B = 3\n\\end{cases}\n\\]\n\nSolving the system:\n\n\\[\n\\begin{cases}\nx_B = 1 \\\\\ny_B = 2\n\\end{cases}\n\\]\n\nThus, the coordinates of point \\( B \\) are: \\( (1,2) \\).\n\n(2) The function \\( y = x^2 - 5x + 7 \\) can be rewritten in vertex form as:\n\n\\[\ny = \\left(x - \\frac{5}{2}\\right)^2 + \\frac{3}{4},\n\\]\n\n\\[\ny = \\left(x - \\frac{5}{2}\\right)^2 + \\frac{3}{4} \\geq \\frac{3}{4}.\n\\]\n\nSince \\( x \\geq 0 \\) and point \\( D \\) lies on the graph,\n\n\\[\ny_D \\geq \\frac{3}{4}, \\quad x_D \\geq 0, \\quad y_D = x_D^2 - 5x_D + 7.\n\\]\n\n\\[\nd(O, D) = |0 - x_D| + |0 - y_D| = x_D + y_D,\n\\]\n\n\\[\nd(O, D) = x_D + x_D^2 - 5x_D + 7 = x_D^2 - 4x_D + 7,\n\\]\n\n\\[\nd(O, D) = (x_D - 2)^2 + 3.\n\\]\n\nTherefore, when \\( x_D = 2 \\), \\( d(O, D) \\) reaches its minimum value of 3.\n\n\\[\ny_D = x_D^2 - 5x_D + 7 = 2^2 - 5 \\times 2 + 7 = 1.\n\\]\n\nThus, the coordinates of point \\( D \\) are: \\( (2,1) \\).\n\nIn summary, the minimum distance is 3, and the coordinates of point \\( D \\) are \\( (2,1) \\).\n\n**Key Insight:** This problem primarily examines the graph and properties of quadratic functions. A thorough understanding of the defined distance between two points, \\( d(A, B) = |x_1 - x_2| + |y_1 - y_2| \\), is crucial for solving this problem." }, { "problem_id": 2000, "question": "As shown in Figure 1, in the Cartesian coordinate plane \\(xOy\\), the line \\(y = \\frac{1}{2}x - 4\\) intersects the \\(x\\)-axis at point \\(A\\) and the \\(y\\)-axis at point \\(B\\). The parabola \\(y = \\frac{1}{4}x^2 + bx + c\\) passes through points \\(A\\) and \\(B\\), and intersects the \\(x\\)-axis at another point \\(C\\).\n\n(1) Find the function expression for the parabola;\n\n(2) Point \\(M\\) is a moving point on the parabola below the line \\(AB\\).\n\n(1) As shown in Figure 2, the line \\(CM\\) intersects segment \\(AB\\) at point \\(N\\). Find the minimum value of \\(\\frac{CN}{NM}\\);\n\n(2) As shown in Figure 3, connect \\(BM\\) and draw \\(MD \\perp AB\\) at \\(D\\). Is there a point \\(M\\) such that one of the angles in \\(\\triangle BMD\\) is exactly twice the angle \\(\\angle CAB\\)? If such a point exists, find the coordinates of point \\(M\\); if not, explain why.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3", "input_image": [ "batch6-2024_06_14_54c8b3c3e1f0939737abg_0085_1.jpg", "batch6-2024_06_14_54c8b3c3e1f0939737abg_0085_2.jpg", "batch6-2024_06_14_54c8b3c3e1f0939737abg_0085_3.jpg" ], "is_multi_img": true, "answer": "(1) $y=\\frac{1}{4} x^{2}-\\frac{3}{2} x-4$; (2) (1) When $x_{0}=4$, the minimum value of $\\frac{C N}{M N}$ is $\\frac{5}{4}$; (2) There exists a point $\\mathrm{M}$ whose coordinates are $\\left(\\frac{58}{11},-\\frac{600}{121}\\right)$ or $(4,-6)$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\n**(1)** On the line \\( y = \\frac{1}{2}x - 4 \\), set \\( x = 0 \\) and \\( y = 0 \\) respectively. We obtain points \\( \\mathrm{A}(8, 0) \\) and \\( \\mathrm{B}(0, -4) \\).\n\nSubstituting \\( \\mathrm{A}(8, 0) \\) and \\( \\mathrm{B}(0, -4) \\) into the equation \\( y = \\frac{1}{4}x^2 + bx + c \\), we have:\n\\[\n\\begin{cases}\n\\frac{1}{4} \\times 8^2 + 8b + c = 0 \\\\\nc = -4\n\\end{cases}\n\\]\nSolving the system, we find:\n\\[\n\\begin{cases}\nb = -\\frac{3}{2} \\\\\nc = -4\n\\end{cases}\n\\]\nThus, the equation becomes:\n\\[\ny = \\frac{1}{4}x^2 - \\frac{3}{2}x - 4\n\\]\n\n**(2)** \n\n**(i)** As shown in Figure 1, draw a line \\( CE \\) parallel to the \\( y \\)-axis intersecting line \\( AB \\) at point \\( E \\), and draw a line \\( MF \\) parallel to the \\( y \\)-axis intersecting line \\( AB \\) at point \\( F \\). It follows that \\( CE \\parallel MF \\), so:\n\\[\n\\frac{CN}{MN} = \\frac{CE}{MF}\n\\]\nLet \\( M\\left(x_0, \\frac{1}{4}x_0^2 - \\frac{3}{2}x_0 - 4\\right) \\). Since \\( MF \\) is parallel to the \\( y \\)-axis and intersects line \\( AB \\) at \\( F \\), and line \\( AB \\) is given by \\( y = \\frac{1}{2}x - 4 \\), we have:\n\\[\nF\\left(x_0, \\frac{1}{2}x_0 - 4\\right)\n\\]\nThus:\n\\[\nMF = \\frac{1}{2}x_0 - 4 - \\left(\\frac{1}{4}x_0^2 - \\frac{3}{2}x_0 - 4\\right) = -\\frac{1}{4}x_0^2 + 2x_0\n\\]\nWe find point \\( C(-2, 0) \\), and drawing \\( CE \\) parallel to the \\( y \\)-axis intersecting line \\( AB \\) at \\( E \\), we get:\n\\[\nE(-2, -5), \\quad CE = 5\n\\]\nTherefore:\n\\[\n\\frac{CN}{MN} = \\frac{5}{-\\frac{1}{4}x_0^2 + 2x_0} = \\frac{5}{-\\frac{1}{4}(x_0 - 4)^2 + 4}\n\\]\nWhen \\( x_0 = 4 \\), \\( \\frac{CN}{MN} \\) reaches its minimum value of \\( \\frac{5}{4} \\).\n\n**(ii)** Such a point \\( M \\) exists. The reasoning is as follows:\n\nGiven points \\( C(-2, 0) \\), \\( B(0, -4) \\), and \\( A(8, 0) \\), we have:\n\\[\nOC = 2, \\quad OB = 4, \\quad OA = 8\n\\]\nIt can be shown that \\( \\triangle BOC \\sim \\triangle ABC \\), with \\( \\angle ABC = \\angle AOB = 90^\\circ \\). Since \\( MD \\perp AB \\) at \\( D \\), we have:\n\\[\n\\angle BDM = \\angle ABC = 90^\\circ, \\quad \\angle BAC < 45^\\circ\n\\]\nThus, in \\( \\triangle BMD \\), either \\( \\angle BMD = 2\\angle BAC \\) or \\( \\angle MBD = 2\\angle BAC \\). In Figure 2, taking the midpoint \\( H \\) of \\( AC \\) and connecting \\( BH \\), we find:\n\\[\n\\angle BHO = 2\\angle BAC, \\quad OH = OA - AH = 3, \\quad \\tan \\angle BHO = \\frac{OB}{OH} = \\frac{4}{3}\n\\]\nDrawing \\( DT \\perp y \\)-axis at \\( T \\) and \\( MG \\perp TD \\) intersecting its extension at \\( G \\), we can prove:\n\\[\n\\triangle TBD \\sim \\triangle GDM, \\quad \\frac{BT}{DG} = \\frac{TD}{GM} = \\frac{BD}{MD}\n\\]\nSince \\( DM \\perp AB \\), we have:\n\\[\n\\tan \\angle DMB = \\frac{BD}{MD}, \\quad \\tan \\angle DBM = \\frac{MD}{BD}\n\\]\nWhen \\( \\angle BMD = 2\\angle BAC \\), we have:\n\\[\n\\frac{BD}{MD} = \\frac{4}{3}\n\\]\nWhen \\( \\angle MBD = 2\\angle BAC \\), we have:\n\\[\n\\frac{MD}{BD} = \\frac{4}{3}\n\\]\nLet \\( D\\left(a, \\frac{1}{2}a - 4\\right) \\) and \\( M\\left(m, \\frac{1}{4}m^2 - \\frac{3}{2}m - 4\\right) \\) with \\( 8 > a > 0 \\) and \\( 8 > m > 0 \\). Then:\n\\[\nT\\left(0, \\frac{1}{2}a - 4\\right), \\quad G\\left(m, \\frac{1}{2}a - 4\\right)\n\\]\nThus:\n\\[\nDT = a, \\quad DG = m - a, \\quad BT = \\frac{1}{2}a - 4 - (-4) = \\frac{1}{2}a\n\\]\nWhen \\( \\angle BMD = 2\\angle BAC \\), we have:\n\\[\n\\frac{BD}{MD} = \\frac{4}{3}, \\quad \\frac{BT}{DG} = \\frac{TD}{GM} = \\frac{BD}{MD}\n\\]\nSolving:\n\\[\n\\frac{\\frac{1}{2}a}{m - a} = \\frac{a}{\\frac{1}{2}a - \\frac{1}{4}m^2 + \\frac{3}{2}m} = \\frac{4}{3}\n\\]\nWe find \\( m_1 = 0 \\) and \\( m_2 = \\frac{58}{11} \\). Since \\( 0 < m < 8 \\), we take \\( m = \\frac{58}{11} \\), and the coordinates of \\( M \\) are:\n\\[\n\\left(\\frac{58}{11}, -\\frac{600}{121}\\right)\n\\]\nWhen \\( \\angle MBD = 2\\angle BAC \\), we have:\n\\[\n\\frac{MD}{BD} = \\frac{4}{3}, \\quad \\frac{BT}{DG} = \\frac{TD}{GM} = \\frac{BD}{MD}\n\\]\nSolving:\n\\[\n\\frac{\\frac{1}{2}a}{m - a} = \\frac{a}{\\frac{1}{2}a - \\frac{1}{4}m^2 + \\frac{3}{2}m} = \\frac{3}{4}\n\\]\nWe find \\( m_1 = 0 \\) and \\( m_2 = 4 \\). Since \\( 0 < m < 8 \\), we take \\( m = 4 \\), and the coordinates of \\( M \\) are:\n\\[\n(4, -6)\n\\]\nIn conclusion, the coordinates of point \\( M \\) that satisfy the conditions are:\n\\[\n\\left(\\frac{58}{11}, -\\frac{600}{121}\\right) \\quad \\text{or} \\quad (4, -6)\n\\]\n\n**Key Insight:** This problem primarily examines the properties of quadratic functions and the application of the Pythagorean theorem in coordinate geometry." }, { "problem_id": 2001, "question": "As shown in the figure, the parabola \\( y = ax^2 + bx + 6 \\) (where \\( a \\neq 0 \\)) intersects the \\( x \\)-axis at points \\( A(-1, 0) \\) and \\( B(3, 0) \\), and intersects the \\( y \\)-axis at point \\( C \\).\n\n\n\nFigure 1\n\n\n\nSpare Figure\n\n(1) Find the equation of the parabola;\n\n(2) If there is a point \\( M \\) on segment \\( BC \\) such that \\( \\angle BMO = 45^\\circ \\), and a line through point \\( O \\) is drawn perpendicular to \\( OM \\) intersecting the extension of \\( CB \\) at point \\( H \\), find the coordinates of point \\( H \\);\n\n(3) Under the conditions of (2), let \\( P \\) be a moving point on the positive half of the \\( y \\)-axis, connect \\( PM \\), and through \\( M \\) draw \\( MQ \\perp PM \\) intersecting the \\( x \\)-axis at \\( Q \\). \\( N \\) is the midpoint of \\( PQ \\). Find the minimum value of \\( BN \\).", "input_image": [ "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0022_1.jpg", "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0022_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=-2 x^{2}+4 x+6$\n\n(2)When $\\angle O M B=45^{\\circ}$, the coordinates of point $H$ are $\\left(\\frac{18}{5},-\\frac{6}{5}\\right)$\n\n(3) The minimum value of $B N$ is $\\frac{3 \\sqrt{10}}{10}$\n\n", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\n**(1)** \nGiven that the parabola \\( y = ax^{2} + bx + 6 \\) passes through points \\( A(-1, 0) \\) and \\( B(3, 0) \\), we have the following system of equations:\n\n\\[\n\\begin{cases}\na - b + 6 = 0 \\\\\n9a + 3b + 6 = 0\n\\end{cases}\n\\]\n\nSolving this system yields:\n\n\\[\n\\begin{cases}\na = -2 \\\\\nb = 4\n\\end{cases}\n\\]\n\nThus, the equation of the parabola is:\n\n\\[\ny = -2x^{2} + 4x + 6\n\\]\n\n**(2)** \nFrom part (1), we know that point \\( C(0, 6) \\) lies on the parabola. Let the equation of line \\( BC \\) be \\( y = kx + c \\). Since line \\( BC \\) passes through points \\( B(3, 0) \\) and \\( C(0, 6) \\), we have:\n\n\\[\n\\begin{cases}\n3k + c = 0 \\\\\nc = 6\n\\end{cases}\n\\]\n\nSolving this system gives:\n\n\\[\n\\begin{cases}\nk = -2 \\\\\nc = 6\n\\end{cases}\n\\]\n\nTherefore, the equation of line \\( BC \\) is:\n\n\\[\ny = -2x + 6\n\\]\n\nLet point \\( H \\) have coordinates \\( (m, -2m + 6) \\). As shown in the figure, draw \\( HK \\perp y \\)-axis at point \\( K \\) and \\( MS \\perp y \\)-axis at point \\( S \\). Then, \\( \\angle MSO = \\angle OKH = 90^\\circ \\). Since \\( OH \\perp OM \\), \\( \\angle MOH = 90^\\circ \\). Given \\( \\angle OMB = 45^\\circ \\), triangle \\( MOH \\) is an isosceles right triangle, so \\( OM = OH \\).\n\nBecause \\( \\angle MOS + \\angle KOH = 90^\\circ \\) and \\( \\angle OHK + \\angle KOH = 90^\\circ \\), it follows that \\( \\angle MOS = \\angle OHK \\). Therefore, triangles \\( OMS \\) and \\( HOK \\) are congruent (AAS), which implies \\( MS = OK \\) and \\( OS = HK \\). Hence, point \\( M \\) has coordinates \\( (2m - 6, m) \\).\n\nSince point \\( M(2m - 6, m) \\) lies on the line \\( y = -2x + 6 \\), we have:\n\n\\[\n-2(2m - 6) + 6 = m\n\\]\n\nSolving for \\( m \\) gives:\n\n\\[\nm = \\frac{18}{5}\n\\]\n\nThus, \\( -2m + 6 = -\\frac{6}{5} \\), and the coordinates of point \\( H \\) when \\( \\angle OMB = 45^\\circ \\) are:\n\n\\[\n\\left( \\frac{18}{5}, -\\frac{6}{5} \\right)\n\\]\n\n**(3)** \nFrom part (2), the coordinates of point \\( M \\) are \\( \\left( \\frac{6}{5}, \\frac{18}{5} \\right) \\). Since \\( PM \\perp MQ \\), triangle \\( PMQ \\) is a right triangle. Point \\( N \\) is the midpoint of segment \\( PQ \\), so \\( MN = \\frac{1}{2} PQ \\). Additionally, since triangle \\( POQ \\) is a right triangle and \\( N \\) is the midpoint of \\( PQ \\), \\( ON = \\frac{1}{2} PQ \\). Therefore, \\( MN = ON \\), which means point \\( N \\) lies on the perpendicular bisector of segment \\( OM \\).\n\nDraw the perpendicular bisector \\( l \\) of \\( OM \\). As shown in the figure, when \\( BN \\perp l \\), \\( BN \\) is minimized (point \\( N \\) is at position \\( N_1 \\) in the figure).\n\nGiven the coordinates of point \\( M \\), the coordinates of point \\( H \\) are \\( \\left( \\frac{3}{5}, \\frac{9}{5} \\right) \\), and \\( OH = \\frac{3\\sqrt{10}}{5} \\). Since \\( \\tan \\angle COM = \\frac{1}{3} \\) and \\( \\angle COM = \\angle OEH \\), we have:\n\n\\[\n\\tan \\angle OEH = \\frac{OH}{HE} = \\frac{\\frac{3\\sqrt{10}}{5}}{HE} = \\frac{1}{3}\n\\]\n\nSolving for \\( HE \\) gives:\n\n\\[\nHE = \\frac{9\\sqrt{10}}{5}\n\\]\n\nThus, \\( OE = \\sqrt{OH^2 + HE^2} = 6 \\), and \\( \\tan \\angle OEF = \\frac{OF}{OE} = \\frac{1}{3} \\), so \\( OF = 2 \\). Therefore, the coordinates of points \\( E \\) and \\( F \\) are \\( (6, 0) \\) and \\( (0, 2) \\), respectively.\n\nThe equation of line \\( l \\) is:\n\n\\[\ny = -\\frac{1}{3}x + 2\n\\]\n\nGiven \\( BE = 3 \\) and \\( EF = \\sqrt{6^2 + 2^2} = 2\\sqrt{10} \\), and since \\( \\sin \\angle N_1EB = \\sin \\angle OEF \\), we have:\n\n\\[\n\\frac{BN_1}{BE} = \\frac{OF}{EF} \\quad \\Rightarrow \\quad \\frac{BN_1}{3} = \\frac{2}{2\\sqrt{10}}\n\\]\n\nSolving for \\( BN_1 \\) gives:\n\n\\[\nBN_1 = \\frac{3\\sqrt{10}}{10}\n\\]\n\nThus, the minimum value of \\( BN \\) is \\( \\frac{3\\sqrt{10}}{10} \\).\n\n**Key Insight:** This problem involves determining the equations of a parabola and a line using the method of undetermined coefficients, analyzing the properties of coordinates and geometric figures, and applying the properties of congruent triangles, isosceles right triangles, and the characteristics of linear functions. Mastery of these concepts is essential for solving such problems." }, { "problem_id": 2002, "question": "As shown in the figure, the parabola \\( y = ax^2 + bx + 4 \\) intersects the \\( x \\)-axis at points \\( A(-3, 0) \\) and \\( B(4, 0) \\), and intersects the \\( y \\)-axis at point \\( C \\). Point \\( P \\) is a moving point on the parabola. A line \\( PD \\) perpendicular to the \\( x \\)-axis is drawn through point \\( P \\), with the foot of the perpendicular at \\( D \\). The line \\( PD \\) intersects the line \\( BC \\) at point \\( E \\). Another line \\( PF \\) perpendicular to the \\( y \\)-axis is drawn through point \\( P \\), with the foot of the perpendicular at \\( F \\). The line \\( PF \\) intersects the line \\( BC \\) at point \\( G \\). Let the abscissa of point \\( P \\) be \\( m \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the equation of the parabola;\n\n(2) If \\( -3 < m < 4 \\) and \\( m \\neq 0 \\), when \\( PE = 2CF \\), find the coordinates of point \\( P \\);\n\n(3) If \\( 0 < m < 4 \\), draw the line \\( AC \\). There is a moving point \\( Q \\) on the line \\( AC \\). Connect \\( BQ \\) and \\( GQ \\). When \\( \\angle BQG = 45^\\circ \\), directly find the minimum value of \\( BG \\) and the coordinates of point \\( P \\) at that time.", "input_image": [ "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0025_1.jpg", "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0025_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=-\\frac{1}{3} x^{2}+\\frac{1}{3} x+4$\n\n(2) $P\\left(2, \\frac{10}{3}\\right)$ or $P(-2,2)$\n\n(3) $P\\left(\\frac{3+\\sqrt{105}}{6}, \\frac{28}{9}\\right)$, minimum value $B G=\\frac{28}{9} \\sqrt{2}$.\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "0", "analysis": "**Solution:**\n\n(1) Since the parabola \\( y = ax^2 + bx + 4 \\) intersects the x-axis at points \\( A(-3, 0) \\) and \\( B(4, 0) \\), we have:\n\\[\n\\begin{cases}\n9a - 3b + 4 = 0 \\\\\n16a + 4b + 4 = 0\n\\end{cases}\n\\]\nSolving the system of equations, we find:\n\\[\n\\begin{cases}\na = -\\frac{1}{3} \\\\\nb = \\frac{1}{3}\n\\end{cases}\n\\]\nThus, the equation of the parabola is:\n\\[\ny = -\\frac{1}{3}x^2 + \\frac{1}{3}x + 4\n\\]\n\n(2) Let \\( x = 0 \\), then:\n\\[\ny = -\\frac{1}{3}(0)^2 + \\frac{1}{3}(0) + 4 = 4\n\\]\nSo, point \\( C(0, 4) \\).\n\nLet the line \\( BC \\) be \\( y = kx + 4 \\). Since \\( B(4, 0) \\) lies on this line:\n\\[\n4k + 4 = 0 \\implies k = -1\n\\]\nThus, the equation of line \\( BC \\) is:\n\\[\ny = -x + 4\n\\]\n\nLet the x-coordinate of point \\( P \\) be \\( m \\), where \\( -3 < m < 4 \\) and \\( m \\neq 0 \\). Then:\n\\[\nP\\left(m, -\\frac{1}{3}m^2 + \\frac{1}{3}m + 4\\right)\n\\]\nPoints \\( E \\) and \\( F \\) are:\n\\[\nE(m, -m + 4), \\quad F\\left(0, -\\frac{1}{3}m^2 + \\frac{1}{3}m + 4\\right)\n\\]\nThe distances \\( EP \\) and \\( CF \\) are:\n\\[\nEP = \\left| -\\frac{1}{3}m^2 + \\frac{1}{3}m + 4 + m - 4 \\right| = \\left| -\\frac{1}{3}m^2 + \\frac{4}{3}m \\right|\n\\]\n\\[\nCF = \\left| 4 + \\frac{1}{3}m^2 - \\frac{1}{3}m - 4 \\right| = \\left| \\frac{1}{3}m^2 - \\frac{1}{3}m \\right|\n\\]\nGiven \\( PE = 2CF \\), we have:\n\\[\n\\left| -\\frac{1}{3}m^2 + \\frac{4}{3}m \\right| = 2\\left| \\frac{1}{3}m^2 - \\frac{1}{3}m \\right| = \\left| \\frac{2}{3}m^2 - \\frac{2}{3}m \\right|\n\\]\nSolving the equation:\n\\[\n-\\frac{1}{3}m^2 + \\frac{4}{3}m = \\frac{2}{3}m^2 - \\frac{2}{3}m \\implies m^2 - 2m = 0\n\\]\nSolutions are \\( m_1 = 2 \\) and \\( m_2 = 0 \\) (discarded as it does not satisfy the conditions). Thus:\n\\[\nP\\left(2, \\frac{10}{3}\\right)\n\\]\nAlternatively, solving:\n\\[\n-\\frac{1}{3}m^2 + \\frac{4}{3}m + \\frac{2}{3}m^2 - \\frac{2}{3}m = 0 \\implies m^2 + 2m = 0\n\\]\nSolutions are \\( m_1 = -2 \\) and \\( m_2 = 0 \\) (discarded). Thus:\n\\[\nP(-2, 2)\n\\]\nIn summary, the possible coordinates for \\( P \\) are \\( \\left(2, \\frac{10}{3}\\right) \\) or \\( (-2, 2) \\).\n\n(3) As shown in the figure, for \\( 0 < m < 4 \\), point \\( P \\) lies on the parabola in the first quadrant.\n\nGiven points \\( B(4, 0) \\), \\( C(0, 4) \\), and \\( A(-3, 0) \\), we have:\n\\[\nOB = OC = 4, \\quad AC = \\sqrt{3^2 + 4^2} = 5, \\quad \\angle OBC = \\angle OCB = 45^\\circ\n\\]\nLet \\( GK \\) be perpendicular to the axis at \\( K \\), with \\( KG = KB \\). Drawing a circle centered at \\( K \\) with radius \\( KG \\) intersecting line \\( AC \\) at \\( Q \\), we have:\n\\[\n\\angle BQG = \\frac{1}{2} \\angle BKG = 45^\\circ\n\\]\nThus, \\( PD = GK \\).\n\nWhen the circle is tangent to \\( AC \\), \\( BG \\) is minimized, and \\( Q \\) is the point of tangency with \\( KQ \\perp AC \\). Connecting \\( QK \\), let:\n\\[\nP\\left(m, -\\frac{1}{3}m^2 + \\frac{1}{3}m + 4\\right)\n\\]\nSince \\( G \\) lies on \\( AC \\):\n\\[\n-\\frac{1}{3}m^2 + \\frac{1}{3}m + 4 = -x + 4 \\implies x = \\frac{1}{3}m^2 - \\frac{1}{3}m\n\\]\nThus:\n\\[\nG\\left(\\frac{1}{3}m^2 - \\frac{1}{3}m, -\\frac{1}{3}m^2 + \\frac{1}{3}m + 4\\right), \\quad K\\left(\\frac{1}{3}m^2 - \\frac{1}{3}m, 0\\right)\n\\]\nThe distance \\( AK \\) is:\n\\[\nAK = \\frac{1}{3}m^2 - \\frac{1}{3}m + 3\n\\]\nAnd \\( KQ = GK = -\\frac{1}{3}m^2 + \\frac{1}{3}m + 4 \\).\n\nUsing the sine of angle \\( \\angle CAO \\):\n\\[\n\\sin \\angle CAO = \\frac{CO}{AC} = \\frac{4}{5} = \\frac{QK}{AK}\n\\]\nThus:\n\\[\n\\frac{-\\frac{1}{3}m^2 + \\frac{1}{3}m + 4}{\\frac{1}{3}m^2 - \\frac{1}{3}m + 3} = \\frac{4}{5} \\implies 3m^2 - 3m - 8 = 0\n\\]\nSolutions are:\n\\[\nm_1 = \\frac{3 + \\sqrt{105}}{6}, \\quad m_2 = \\frac{3 - \\sqrt{105}}{6} \\quad (\\text{discarded})\n\\]\nTherefore:\n\\[\n-\\frac{1}{3}m^2 + \\frac{1}{3}m + 4 = -\\frac{1}{3}\\left(m - \\frac{1}{2}\\right)^2 + \\frac{49}{12} = \\frac{28}{9}\n\\]\nThus:\n\\[\nP\\left(\\frac{3 + \\sqrt{105}}{6}, \\frac{28}{9}\\right)\n\\]\nAt this point, the length \\( BG \\) is:\n\\[\nBG = \\sqrt{2} \\cdot KG = \\sqrt{2} \\cdot PD = \\frac{28}{9}\\sqrt{2}\n\\]\n\n**Key Insight:** This problem involves determining the equation of a parabola using given points, solving quadratic equations, applying the properties of circles and tangents, and using trigonometric relationships to find the minimal length \\( BG \\). The critical step is identifying the position of \\( Q \\) when \\( BG \\) is minimized." }, { "problem_id": 2003, "question": "As shown in Figure 1, in the plane rectangular coordinate system, the parabola \\( y = -\\frac{1}{4} x^2 + bx + c \\) intersects the \\( x \\)-axis at points \\( A \\) and \\( B \\), and the \\( y \\)-axis at point \\( C \\), where \\( OC = 4 \\) and \\( 2b - c = -1 \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the analytical expression of the parabola;\n\n(2) As shown in Figure 2, if point \\( P \\) is a point on the parabola in the first quadrant, and connecting \\( PB \\) such that \\( \\angle PBC = \\angle CBO \\), find the coordinates of point \\( P \\).\n\n(3) As shown in Figure 3, point \\( D \\) is a moving point on the parabola in the first quadrant, connecting \\( OD \\) which intersects \\( BC \\) at point \\( E \\). When the value of \\( \\frac{S_{\\triangle DBE}}{S_{\\triangle OBE}} \\) is maximized, find the coordinates of point \\( D \\) and the maximum value of \\( \\frac{S_{\\triangle DBE}}{S_{\\triangle OBE}} \\).", "input_image": [ "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0029_1.jpg", "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0029_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=-\\frac{1}{4} x^{2}+\\frac{3}{2} x+4$\n\n(2) $P\\left(\\frac{10}{3}, \\frac{56}{9}\\right)$\n\n(3) $1, D(4,6)$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\n(1) Given that \\( OC = 4 \\), \n\\(\\therefore C(0,4)\\). \n\nSince the parabola \\( y = -\\frac{1}{4}x^2 + bx + c \\) intersects the \\( x \\)-axis at points \\( A \\) and \\( B \\), and the \\( y \\)-axis at point \\( C \\), and \\( 2b - c = -1 \\), \n\\(\\therefore \\begin{cases} c = 4 \\\\ 2b - c = -1 \\end{cases}\\), \nSolving gives: \\( \\begin{cases} b = \\frac{3}{2} \\\\ c = 4 \\end{cases} \\). \n\n\\(\\therefore\\) The equation of the parabola is \\( y = -\\frac{1}{4}x^2 + \\frac{3}{2}x + 4 \\).\n\n---\n\n(2) As shown in the figure, draw \\( CQ \\parallel BP \\) intersecting \\( OB \\) at point \\( Q \\). \n\\(\\therefore \\angle QCB = \\angle PBC\\). \nSince \\( \\angle PBC = \\angle CBO \\), \n\\(\\therefore \\angle QCB = \\angle QBC\\), \n\\(\\therefore CQ = BQ\\). \n\nSince the parabola \\( y = -\\frac{1}{4}x^2 + \\frac{3}{2}x + 4 \\) intersects the \\( x \\)-axis at points \\( A \\) and \\( B \\), \n\\(\\therefore\\) when \\( y = 0 \\), \\( -\\frac{1}{4}x^2 + \\frac{3}{2}x + 4 = 0 \\), \nSolving gives: \\( x_1 = -2 \\), \\( x_2 = 8 \\), \n\\(\\therefore B(8,0) \\), and \\( OB = 8 \\). \n\nLet \\( OQ = m \\), \n\\(\\therefore CQ = BQ = 8 - m \\). \n\nIn right triangle \\( \\triangle OCQ \\), \\( OC^2 + OQ^2 = CQ^2 \\), \n\\(\\therefore 4^2 + m^2 = (8 - m)^2 \\), \nSolving gives: \\( m = 3 \\), \n\\(\\therefore Q(3,0) \\). \n\nLet the equation of line \\( CQ \\) be \\( y = k_{CQ}x + b_{CQ} \\), passing through points \\( C(0,4) \\) and \\( Q(3,0) \\), \n\\(\\therefore \\begin{cases} b_{CQ} = 4 \\\\ 3k_{CQ} + b_{CQ} = 0 \\end{cases}\\), \n\\(\\therefore \\begin{cases} k_{CQ} = -\\frac{4}{3} \\\\ b_{CQ} = 4 \\end{cases}\\). \n\n\\(\\therefore\\) The equation of line \\( CQ \\) is \\( y = -\\frac{4}{3}x + 4 \\). \n\nSince \\( CQ \\parallel BP \\), let the equation of line \\( BP \\) be \\( y = -\\frac{4}{3}x + b_{BP} \\), passing through point \\( B(8,0) \\), \n\\(\\therefore -\\frac{4}{3} \\times 8 + b_{BP} = 0 \\), \nSolving gives: \\( b_{BP} = \\frac{32}{3} \\), \n\\(\\therefore\\) The equation of line \\( BP \\) is \\( y = -\\frac{4}{3}x + \\frac{32}{3} \\). \n\nSolving the system: \n\\[\n\\begin{cases}\ny = -\\frac{1}{4}x^2 + \\frac{3}{2}x + 4 \\\\\ny = -\\frac{4}{3}x + \\frac{32}{3}\n\\end{cases}\n\\]\ngives: \\( x_1 = 8 \\), \\( x_2 = \\frac{10}{3} \\). \n\nWhen \\( x = \\frac{10}{3} \\), \\( y = -\\frac{4}{3} \\times \\frac{10}{3} + \\frac{32}{3} = \\frac{56}{9} \\), \n\\(\\therefore P\\left(\\frac{10}{3}, \\frac{56}{9}\\right) \\). \n\n\\(\\therefore\\) The coordinates of point \\( P \\) are \\( \\left(\\frac{10}{3}, \\frac{56}{9}\\right) \\).\n\n---\n\n(3) As shown in the figure, draw \\( ON \\perp BC \\) intersecting \\( BC \\) at point \\( N \\), draw \\( DM \\perp BC \\) intersecting \\( BC \\) at point \\( M \\), and draw \\( DF \\perp x \\)-axis intersecting \\( BC \\) at point \\( F \\). \n\n\\(\\therefore \\frac{S_{\\triangle DBE}}{S_{\\triangle OBE}} = \\frac{\\frac{1}{2}BE \\cdot DM}{\\frac{1}{2}BE \\cdot ON} = \\frac{DM}{ON} \\). \n\nSince \\( ON \\perp BC \\) and \\( DM \\perp BC \\), \n\\(\\therefore \\angle DME = \\angle ONE = 90^\\circ \\), \nand since \\( \\angle DEM = \\angle OEN \\), \n\\(\\therefore \\triangle DME \\sim \\triangle ONE \\), \n\\(\\therefore \\frac{DM}{ON} = \\frac{DE}{OE} \\). \n\nSince \\( DF \\perp x \\)-axis, \n\\(\\therefore DF \\parallel y \\)-axis, \n\\(\\therefore \\frac{DE}{OE} = \\frac{DF}{OC} \\). \n\nLet the equation of line \\( BC \\) be \\( y = k_{BC}x + b_{BC} \\), passing through points \\( C(0,4) \\) and \\( B(8,0) \\), \n\\(\\therefore \\begin{cases} b_{BC} = 4 \\\\ 8k_{BC} + b_{BC} = 0 \\end{cases}\\), \n\\(\\therefore \\begin{cases} k_{BC} = -\\frac{1}{2} \\\\ b_{BC} = 4 \\end{cases}\\). \n\n\\(\\therefore\\) The equation of line \\( BC \\) is \\( y = -\\frac{1}{2}x + 4 \\). \n\nLet \\( D\\left(x, -\\frac{1}{4}x^2 + \\frac{3}{2}x + 4\\right) \\), then \\( F\\left(x, -\\frac{1}{2}x + 4\\right) \\), \n\\(\\therefore DF = -\\frac{1}{4}x^2 + \\frac{3}{2}x + 4 - \\left(-\\frac{1}{2}x + 4\\right) = -\\frac{1}{4}x^2 + 2x \\). \n\n\\(\\therefore \\frac{S_{\\triangle DBE}}{S_{\\triangle OBE}} = \\frac{DM}{ON} = \\frac{DE}{OE} = \\frac{DF}{OC} = \\frac{-\\frac{1}{4}x^2 + 2x}{4} = -\\frac{1}{16}x^2 + \\frac{1}{2}x = -\\frac{1}{16}(x - 4)^2 + 1 \\). \n\nSince \\( -\\frac{1}{16} < 0 \\), \n\\(\\therefore\\) when \\( x = 4 \\), \\( \\frac{S_{\\triangle DBE}}{S_{\\triangle OBE}} \\) reaches its maximum value of 1. \n\nAt this point, \\( y = -\\frac{1}{4} \\times 4^2 + \\frac{3}{2} \\times 4 + 4 = 6 \\), \n\\(\\therefore\\) The coordinates of point \\( D \\) are \\( (4,6) \\). \n\n\\(\\therefore\\) The maximum value of \\( \\frac{S_{\\triangle DBE}}{S_{\\triangle OBE}} \\) is 1, and the coordinates of point \\( D \\) are \\( (4,6) \\). \n\n---\n\n**Key Points:** \nThis problem is a comprehensive quadratic function question, testing properties of quadratic functions, solving for equations using the method of undetermined coefficients, coordinates of points on quadratic function graphs, intersection points of quadratic and linear functions, the Pythagorean theorem, and the properties of similar triangles. The key to solving the problem lies in flexibly applying relevant knowledge." }, { "problem_id": 2004, "question": "As shown in Figure 1, the parabola \\( y = -x^2 + bx + c \\) passes through points \\( A(-1, 0) \\) and \\( B(3, 0) \\), intersecting the \\( y \\)-axis at point \\( C \\). There is a moving point \\( E(m, 0) \\) on the \\( x \\)-axis where \\( 0 < m < 3 \\). A line \\( ME \\) perpendicular to the \\( x \\)-axis is drawn through point \\( E \\), intersecting the parabola at point \\( M \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) Find the equation of the parabola;\n\n(2) When \\( m = 1 \\), point \\( D \\) is a point on line \\( ME \\) in the first quadrant. If \\( \\triangle ACD \\) is a right triangle with \\( CA \\) as the hypotenuse, find the coordinates of point \\( D \\);\n\n(3) As shown in Figure 2, connect \\( BC \\), which intersects \\( ME \\) at point \\( F \\). Connect \\( AF \\), and the areas of \\( \\triangle ACF \\) and \\( \\triangle BFM \\) are \\( S_1 \\) and \\( S_2 \\) respectively. When \\( S_1 = 4S_2 \\), find the coordinates of point \\( E \\).", "input_image": [ "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0030_1.jpg", "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0030_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=-x^{2}+2 x+3$\n\n(2) The coordinates of point $D$ are $(1,1)$ or $(1,2)$\n\n(3) $E(2,0)$\n\n", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "**Solution:**\n\n**(1)** By substituting the points \\( A(-1, 0) \\) and \\( B(3, 0) \\) into the equation \\( y = -x^{2} + b x + c \\), we obtain the system of equations:\n\\[\n\\begin{cases}\n-1 - b + c = 0 \\\\\n-9 + 3b + c = 0\n\\end{cases}\n\\]\nSolving this system yields:\n\\[\n\\begin{cases}\nb = 2 \\\\\nc = 3\n\\end{cases}\n\\]\nThus, the equation of the parabola is:\n\\[\ny = -x^{2} + 2x + 3\n\\]\n\n**(2)** Let the coordinates of point \\( D \\) be \\( (1, a) \\). Connect points \\( A \\) and \\( D \\), and points \\( C \\) and \\( D \\). Draw a perpendicular from point \\( C \\) to \\( ME \\), intersecting at point \\( H \\).\n\nUsing the Pythagorean theorem:\n- In right triangle \\( \\triangle CHD \\):\n\\[\nCD^{2} = CH^{2} + HD^{2} = 1^{2} + (3 - a)^{2}\n\\]\n- In right triangle \\( \\triangle AED \\):\n\\[\nAD^{2} = AE^{2} + ED^{2} = 2^{2} + a^{2}\n\\]\n- In right triangle \\( \\triangle ACD \\):\n\\[\nAC^{2} = AD^{2} + CD^{2} = 1^{2} + 3^{2} = 10\n\\]\nTherefore:\n\\[\n1^{2} + (3 - a)^{2} + 2^{2} + a^{2} = 10\n\\]\nSolving this equation gives:\n\\[\na_{1} = 1, \\quad a_{2} = 2\n\\]\nHence, the coordinates of point \\( D \\) are \\( (1, 1) \\) or \\( (1, 2) \\).\n\n**(3)** Let the equation of line \\( CB \\) be \\( y = kx + 3 \\). Substituting point \\( B(3, 0) \\) into the equation gives:\n\\[\n3k + 3 = 0 \\quad \\Rightarrow \\quad k = -1\n\\]\nThus, the equation of line \\( CB \\) is:\n\\[\ny = -x + 3\n\\]\nGiven point \\( E(m, 0) \\), the coordinates of points \\( M \\) and \\( F \\) are:\n\\[\nM\\left(m, -m^{2} + 2m + 3\\right), \\quad F(m, -m + 3)\n\\]\nThe length \\( BE \\) is:\n\\[\nBE = 3 - m\n\\]\nThe length \\( MF \\) is:\n\\[\nMF = -m^{2} + 2m + 3 - (-m + 3) = -m^{2} + 3m\n\\]\nTherefore, the area \\( S_{2} \\) is:\n\\[\nS_{2} = \\frac{1}{2} MF \\cdot BE = \\frac{1}{2} (-m^{2} + 3m) \\cdot (3 - m)\n\\]\nThe area \\( S_{1} \\) is:\n\\[\nS_{1} = S_{\\triangle ABC} - S_{\\triangle ABF} = \\frac{1}{2} AB \\cdot OC - \\frac{1}{2} AB \\cdot FE = \\frac{1}{2} AB \\cdot (OC - FE)\n\\]\n\\[\n= \\frac{4[3 - (-m + 3)]}{2} = 2m\n\\]\nGiven \\( S_{1} = 4S_{2} \\), we have:\n\\[\n-2m = 4 \\cdot \\frac{1}{2} (-m^{2} + 3m) \\cdot (3 - m)\n\\]\nSimplifying this equation yields:\n\\[\n(m - 3)^{2} = 1\n\\]\nSolving for \\( m \\) gives:\n\\[\nm_{1} = 2, \\quad m_{2} = 4 \\quad (\\text{Discarded as it does not fit the context})\n\\]\nTherefore, the coordinates of point \\( E \\) are \\( (2, 0) \\).\n\n**Key Insight:** This problem involves the comprehensive application of quadratic functions, including the properties of linear functions and the Pythagorean theorem. The key to solving it lies in the use of graphical analysis and algebraic manipulation." }, { "problem_id": 2005, "question": "As shown in the figure, the parabola \\( y = -\\frac{1}{6} x^2 + \\frac{\\sqrt{3}}{6} x + 3 \\) intersects the \\( x \\)-axis at points \\( A \\) and \\( B \\) (with \\( A \\) to the left of \\( B \\)), and intersects the \\( y \\)-axis at point \\( C \\). The axis of symmetry of the parabola intersects the line \\( BC \\) at point \\( E \\).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) In Figure 1, find the expression for the line \\( BC \\);\n\n(2) In Figure 1, point \\( P \\) is a moving point on the parabola above the line \\( BC \\). A line parallel to the \\( y \\)-axis through point \\( P \\) intersects the line \\( BC \\) at point \\( Q \\), and a line parallel to the \\( x \\)-axis through point \\( P \\) intersects the line \\( BC \\) at point \\( H \\). Find the maximum perimeter of \\( \\triangle PQH \\) and the coordinates of point \\( P \\) at that time;\n\n(3) In Figure 2, the parabola is translated 4 units along the direction of the ray \\( BC \\) to form a new parabola \\( y' \\), which intersects the \\( y \\)-axis at point \\( M \\). Point \\( D \\) is symmetric to point \\( C \\) with respect to the \\( x \\)-axis. Connect \\( BD \\), and translate \\( \\triangle BCD \\) along the line \\( AC \\) to form \\( \\triangle B'C'D' \\). During the translation, does there exist a point \\( N \\) on the line \\( ME \\) such that the quadrilateral with vertices \\( N, B', C', D' \\) is a rhombus? If so, directly give the coordinates of point \\( N \\) and provide the process for solving the coordinates of one of the points \\( N \\).", "input_image": [ "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0047_1.jpg", "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0047_2.jpg" ], "is_multi_img": true, "answer": "(1) $y=-\\frac{\\sqrt{3}}{3} x+3$\n\n(2) $\\frac{9 \\sqrt{3}+27}{8}, P\\left(\\frac{3 \\sqrt{3}}{2}, \\frac{21}{8}\\right)$\n\n(3) $N\\left(-\\frac{\\sqrt{3}}{9}, \\frac{13}{3}\\right)$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Analytic Geometry", "image_relavance": "1", "analysis": "(1) Given the parabola \\( y = -\\frac{1}{6}x^2 + \\frac{\\sqrt{3}}{6}x + 3 \\) intersecting the \\( x \\)-axis at points \\( A \\) and \\( B \\) (with \\( A \\) to the left of \\( B \\)) and the \\( y \\)-axis at point \\( C \\).\n\nSetting \\( y = 0 \\):\n\n\\[\n-\\frac{1}{6}x^2 + \\frac{\\sqrt{3}}{6}x + 3 = 0,\n\\]\n\nsolving for \\( x \\):\n\n\\[\nx_1 = 3\\sqrt{3}, \\quad x_2 = -2\\sqrt{3};\n\\]\n\nSetting \\( x = 0 \\):\n\n\\[\ny = 3,\n\\]\n\nthus:\n\n\\[\nA(-2\\sqrt{3}, 0), \\quad B(3\\sqrt{3}, 0), \\quad C(0, 3).\n\\]\n\nLet the equation of line \\( BC \\) be \\( y = kx + 3 \\):\n\n\\[\n3\\sqrt{3}k + 3 = 0,\n\\]\n\nsolving for \\( k \\):\n\n\\[\nk = -\\frac{\\sqrt{3}}{3},\n\\]\n\nhence the equation of line \\( BC \\) is:\n\n\\[\ny = -\\frac{\\sqrt{3}}{3}x + 3.\n\\]\n\n(2) Given \\( B(3\\sqrt{3}, 0) \\) and \\( C(0, 3) \\):\n\n\\[\nBO = 3\\sqrt{3}, \\quad OC = 3,\n\\]\n\n\\[\n\\tan \\angle CBO = \\frac{OC}{OB} = \\frac{3}{3\\sqrt{3}} = \\frac{\\sqrt{3}}{3},\n\\]\n\n\\[\n\\angle CBO = 30^\\circ.\n\\]\n\nSince \\( PH \\parallel x \\)-axis and \\( PQ \\parallel y \\)-axis:\n\n\\[\n\\angle PHQ = 30^\\circ, \\quad \\angle QPH = 90^\\circ,\n\\]\n\n\\[\nHQ = 2PQ, \\quad PH = \\sqrt{3}PQ,\n\\]\n\nthus the perimeter of \\( \\triangle PQH \\) is:\n\n\\[\n(3 + \\sqrt{3})PQ.\n\\]\n\nGiven the parabola and line equations:\n\n\\[\ny = -\\frac{1}{6}x^2 + \\frac{\\sqrt{3}}{6}x + 3, \\quad y = -\\frac{\\sqrt{3}}{3}x + 3,\n\\]\n\nlet point \\( P \\) be \\( \\left(n, -\\frac{1}{6}n^2 + \\frac{\\sqrt{3}}{6}n + 3\\right) \\), then point \\( Q \\) is \\( \\left(n, -\\frac{\\sqrt{3}}{3}n + 3\\right) \\):\n\n\\[\nPQ = -\\frac{1}{6}n^2 + \\frac{\\sqrt{3}}{2}n = -\\frac{1}{6}\\left(n - \\frac{3\\sqrt{3}}{2}\\right)^2 + \\frac{9}{8}.\n\\]\n\nThus, when \\( n = \\frac{3\\sqrt{3}}{2} \\), \\( PQ \\) reaches its maximum value of \\( \\frac{9}{8} \\).\n\nAt this point:\n\n\\[\n-\\frac{1}{6}n^2 + \\frac{\\sqrt{3}}{6}n + 3 = \\frac{21}{8}, \\quad (3 + \\sqrt{3})PQ = \\frac{27 + 9\\sqrt{3}}{8},\n\\]\n\nhence the maximum perimeter of \\( \\triangle PQH \\) is:\n\n\\[\n\\frac{9\\sqrt{3} + 27}{8},\n\\]\n\nand the coordinates of point \\( P \\) are:\n\n\\[\nP\\left(\\frac{3\\sqrt{3}}{2}, \\frac{21}{8}\\right).\n\\]\n\n(3) Since the parabola is translated 4 units along the direction of ray \\( BC \\), and \\( \\angle CBO = 30^\\circ \\):\n\n\\[\n4 \\sin 30^\\circ = 2, \\quad 4 \\cos 30^\\circ = 2\\sqrt{3},\n\\]\n\nthus the parabola \\( y = -\\frac{1}{6}\\left(x - \\frac{\\sqrt{3}}{2}\\right)^2 + \\frac{25}{8} \\) is translated left by \\( 2\\sqrt{3} \\) units and up by 2 units, resulting in:\n\n\\[\ny = -\\frac{1}{6}\\left(x - \\frac{\\sqrt{3}}{2} + 2\\sqrt{3}\\right)^2 + \\frac{25}{8} + 2,\n\\]\n\nsimplifying to:\n\n\\[\ny = -\\frac{1}{6}x^2 - \\frac{\\sqrt{3}}{2}x + 4.\n\\]\n\nThus, the coordinates of \\( M \\) are \\( (0, 4) \\).\n\nGiven the line \\( BC \\) equation \\( y = -\\frac{\\sqrt{3}}{3}x + 3 \\) and the vertex of the parabola at \\( \\left(\\frac{\\sqrt{3}}{2}, \\frac{25}{8}\\right) \\):\n\n\\[\nE\\left(\\frac{\\sqrt{3}}{2}, \\frac{5}{2}\\right).\n\\]\n\nLet the equation of line \\( ME \\) be \\( y = px + 4 \\):\n\n\\[\n\\frac{\\sqrt{3}}{2}p + 4 = \\frac{5}{2},\n\\]\n\nsolving for \\( p \\):\n\n\\[\np = -\\sqrt{3},\n\\]\n\nhence the equation of line \\( ME \\) is:\n\n\\[\ny = -\\sqrt{3}x + 4.\n\\]\n\nGiven \\( A(-2\\sqrt{3}, 0) \\) and \\( C(0, 3) \\):\n\n\\[\n\\frac{OC}{OA} = \\frac{3}{2\\sqrt{3}} = \\frac{\\sqrt{3}}{2},\n\\]\n\nlet the translation be right by \\( 2t \\) units and up by \\( \\sqrt{3}t \\) units.\n\nGiven \\( B(3\\sqrt{3}, 0) \\), \\( C(0, 3) \\), and \\( D(0, -3) \\):\n\n\\[\nB'(3\\sqrt{3} + 2t, \\sqrt{3}t), \\quad C'(2t, \\sqrt{3}t + 3), \\quad D'(2t, \\sqrt{3}t - 3).\n\\]\n\nSince the diagonals of a rhombus bisect each other perpendicularly, \\( C'D' \\perp x \\)-axis and \\( C'D' \\perp B'N \\):\n\n\\[\nNB' \\parallel x \\text{-axis},\n\\]\n\nthus the intersection point of the diagonals is \\( (2t, \\sqrt{3}t) \\), and \\( N(x, \\sqrt{3}t) \\):\n\n\\[\n\\frac{3\\sqrt{3} + 2t + x}{2} = 2t,\n\\]\n\nsolving for \\( x \\):\n\n\\[\nx = 2t - 3\\sqrt{3},\n\\]\n\nhence:\n\n\\[\nN(2t - 3\\sqrt{3}, \\sqrt{3}t).\n\\]\n\nSubstituting into the line equation:\n\n\\[\n\\sqrt{3}t = -\\sqrt{3}(2t - 3\\sqrt{3}) + 4,\n\\]\n\nsolving for \\( t \\):\n\n\\[\nt = \\frac{13\\sqrt{3}}{9},\n\\]\n\nthus:\n\n\\[\nN\\left(-\\frac{\\sqrt{3}}{9}, \\frac{13}{3}\\right).\n\\]\n\n【Insight】This problem examines the intersection of parabolas with quadratic equations, applications of trigonometry, constructing quadratic functions to find maxima, translation rules, determining linear equations, properties of rhombuses, and the midpoint formula. Mastery of trigonometric applications, constructing quadratic functions for optimization, translation rules, properties of rhombuses, and the midpoint formula is essential." }, { "problem_id": 2006, "question": "From a vertex of a polygon, two rays are drawn to form an angle, the sides of which intersect with the sides of the polygon. The portion of the polygon within this angle, along with the sides of the angle, forms a shape called the \"projection figure\" of the angle on the polygon.\n\n(1) As shown in Figure 1, a square \\(ABCD\\) with sides of length 4, the angle \\(\\angle EAF\\) intersects the sides \\(BC\\) and \\(CD\\) of the square at points \\(E\\) and \\(F\\) respectively. The \"projection figure\" of \\(\\angle EAF\\) on the square \\(ABCD\\) is the quadrilateral \\(AECF\\). If \\(CE = 3\\) and \\(CF = 2\\), then the area of the quadrilateral \\(AECF\\) is \\(\\qquad\\).\n\n\n\nFigure 1\n\n(2) As shown in Figure 2, there is a rhombus-shaped lawn \\(ABCD\\). The planning department plans to plant four types of flowers within this empty space. They plan to take points \\(E\\) and \\(F\\) on the sides \\(BC\\) and \\(CD\\) respectively, and use the three paths \\(AE\\), \\(AF\\), and \\(EF\\) to divide the lawn into four planting areas. Given \\(AB = 200 \\text{ m}\\) and \\(\\angle D = 120^\\circ\\), please help the planning department solve the following problems according to the given requirements:\n\n\n\nFigure 2\n\n\n\nSpare Figure\n\n(1) If the area of the quadrilateral \\(AECF\\), the \"projection figure\" of \\(\\angle EAF\\) on the rhombus \\(ABCD\\), is half the area of the rhombus \\(ABCD\\), find the value of \\(CE + CF\\).\n\n(2) Under the conditions of (1), to make the central planting area (\\(\\triangle AEF\\)) as small as possible, find the minimum area of \\(\\triangle AEF\\).", "input_image": [ "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0048_1.jpg", "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0048_2.jpg", "batch6-2024_06_14_60aac5b7e7ea0a8e6ddcg_0048_3.jpg" ], "is_multi_img": true, "answer": "(1) 10\n\n(2)(1) $C F+C E=200$; (2) The minimum area of ​​$\\triangle A E F$ is $2500 \\sqrt{3}$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Combinatorial Geometry", "image_relavance": "0", "analysis": "(1) Connect $AC$, as shown in the figure:\n\n\n\nSince the side of square $ABCD$ is 4,\n\n$\\therefore AB = BC = CD = AD = 4$,\n\nSince $CE = 3$, and $CF = 2$,\n\n$\\therefore S_{\\text{quadrilateral } AECF} = S_{\\triangle ACE} + S_{\\triangle ACF} = \\frac{1}{2} \\times CE \\times AB + \\frac{1}{2} \\times CF \\times AD = \\frac{1}{2} \\times 3 \\times 4 + \\frac{1}{2} \\times 2 \\times 4 = 10$,\n\nTherefore, the answer is: 10.\n\n(2) (1) Connect $AC$, draw $AM \\perp BC$ intersecting the extension of $CB$ at point $M$, and draw $AN \\perp CD$ intersecting the extension of $CD$ at point $N$,\n\n\n\nSince $ABCD$ is a rhombus, $AB = 200 \\mathrm{~m}$, and $\\angle D = 120^{\\circ}$,\n\n$\\therefore AB = BC = CD = AD = 200$, and $\\angle ACD = \\angle BCD = 30^{\\circ}$,\n\n$\\therefore \\angle NAD = \\angle MAB = 30^{\\circ}$,\n\n$\\therefore BM = \\frac{1}{2} AB = 100$, $AM = 100 \\sqrt{3}$, and $ND = \\frac{1}{2} AD = 100$, $AN = 100 \\sqrt{3}$,\n\n$\\therefore S_{\\text{quadrilateral } AECF} = S_{\\triangle ACE} + S_{\\triangle ACF} = \\frac{1}{2} CE \\cdot AM + \\frac{1}{2} CF \\cdot AN$\n\n$= \\frac{1}{2} \\times 100 \\sqrt{3} \\times CF + \\frac{1}{2} \\times 100 \\sqrt{3} \\times CE = 50 \\sqrt{3} \\times (CF + CE)$,\n\n$S_{\\text{rhombus } ABCD} = CD \\cdot AN = 100 \\sqrt{3} \\times 200 = 20000 \\sqrt{3}$,\n\nSince the area of quadrilateral $AECF$ is half the area of rhombus $ABCD$,\n\n$\\therefore 50 \\sqrt{3} \\times (CF + CE) = \\frac{1}{2} \\times 20000 \\sqrt{3}$,\n\nSolving gives $CF + CE = 200$,\n\n(2) Let $CF = x$, from (1) $CF + CE = 200$ we get $CE = 200 - x$,\n\nDraw $FP \\perp BC$ at point $P$, then $CP = \\frac{1}{2} CF = \\frac{1}{2} x$, $FP = \\frac{\\sqrt{3}}{2} x$,\n\n\n\n$\\therefore S_{\\triangle ECF} = \\frac{1}{2} CE \\cdot FP = \\frac{1}{2}(200 - x) \\frac{\\sqrt{3}}{2} x = \\frac{\\sqrt{3}}{4}(x - 200 y) = \\frac{\\sqrt{3}}{4}(x - 10)^{2} + 2500 \\sqrt{3}$\n\n$\\therefore$ When $x = 100$, $S_{\\triangle ECF} = 2500 \\sqrt{3}$ is minimized,\n\nThat is, the minimum area of $\\triangle AEF$ is $2500 \\sqrt{3}$.\n\n【Highlight】This problem examines the properties of special quadrilaterals, the maximum and minimum values of quadratic functions, the properties of $30^{\\circ}$ right triangles, and the Pythagorean theorem. The key to solving the problem is to master the relevant basic properties and apply them flexibly." }, { "problem_id": 2007, "question": "(1) As shown in Figure (1), in the right triangle \\( \\triangle ABC \\), \\( \\angle ACB = 90^\\circ \\), \\( AC = BC \\). A line \\( l \\) is drawn through point \\( C \\), and perpendiculars \\( AM \\perp l \\) at \\( M \\) and \\( BN \\perp l \\) at \\( N \\) are drawn from points \\( A \\) and \\( B \\) respectively. The quantitative relationship between the segments \\( MN \\), \\( AM \\), and \\( BN \\) is \\(\\qquad\\);\n\n(2) As shown in Figure (2), in the right triangle \\( \\triangle ABC \\), \\( \\angle C = 90^\\circ \\), \\( AC = 30 \\), \\( BC = 40 \\). Point \\( P \\) is on \\( AB \\), and points \\( E \\) and \\( F \\) are on sides \\( AC \\) and \\( BC \\) respectively, such that \\( \\angle ABC = \\angle FPB \\), \\( PE \\perp PF \\). Let \\( BP = x \\). Find the functional relationship between the area \\( y \\) of quadrilateral \\( CEPF \\) and \\( x \\);\n\n(3) As shown in Figure (3), there is a circular plaza, where the quadrilateral \\( ACBD \\) is planned as a garden area (all four vertices are on the circle), and it is required that \\( \\angle ACB = 90^\\circ \\), \\( AC = 30 \\) meters, \\( BC = 40 \\) meters. Lines \\( AB \\) and \\( CD \\) intersect at point \\( P \\). To enhance the environment, points \\( E \\) and \\( F \\) need to be determined on sides \\( AC \\) and \\( BC \\) respectively, satisfying \\( \\angle ABC = \\angle FPB \\), \\( PE \\perp PF \\). For overall layout, flowers will be planted within quadrilateral \\( CEPF \\), and grass will be planted in the remaining area of quadrilateral \\( ACBD \\). The price per square meter for flowers is 60 yuan, and for grass is 90 yuan. From a practical perspective, it is desired to maximize the area of quadrilateral \\( CEPF \\). According to the design requirements, find the total cost of planting flowers and grass when the area of quadrilateral \\( CEPF \\) is maximized.\n\n\n\nFigure (1)\n\n\n\nFigure (2)\n\n\n\nFigure (3)", "input_image": [ "batch7-2024_06_14_3b9c127d95f826cc17d7g_0004_1.jpg", "batch7-2024_06_14_3b9c127d95f826cc17d7g_0004_2.jpg", "batch7-2024_06_14_3b9c127d95f826cc17d7g_0004_3.jpg" ], "is_multi_img": true, "answer": "（1） $M N=A M+B N$; （2） $y=-\\frac{25}{48} x^{2}+\\frac{100}{3} x-\\frac{700}{3}$ ；（3）99000\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "0", "analysis": "Solution: (1) Since triangle \\( ABC \\) is an isosceles right-angled triangle,\n\n\\[\n\\therefore AC = BC, \\quad \\angle ACB = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle ACM + \\angle BCN = 90^\\circ.\n\\]\n\nSince \\( AM \\perp l \\) at \\( M \\) and \\( BN \\perp l \\) at \\( N \\),\n\n\\[\n\\therefore \\angle ACM + \\angle CAM = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle BCN = \\angle CAM.\n\\]\n\nSince \\( \\angle AMC = \\angle CNB = 90^\\circ \\) and \\( AC = BC \\),\n\n\\[\n\\therefore \\triangle AMC \\cong \\triangle CNB,\n\\]\n\n\\[\n\\therefore AM = CN, \\quad CM = BN.\n\\]\n\nSince \\( MN = CM + CN \\),\n\n\\[\n\\therefore MN = AM + BN.\n\\]\n\nThus, the answer is: \\( MN = AM + BN \\).\n\n(2) Since \\( \\angle ACB = 90^\\circ \\),\n\n\\[\n\\therefore \\angle B + \\angle A = 90^\\circ.\n\\]\n\nSince \\( FP \\perp EP \\),\n\n\\[\n\\therefore \\angle FPE = 90^\\circ,\n\\]\n\n\\[\n\\therefore \\angle FPB + \\angle EPA = 90^\\circ.\n\\]\n\nSince \\( \\angle B = \\angle FPB \\),\n\n\\[\n\\therefore \\angle EPA = \\angle A,\n\\]\n\n\\[\n\\therefore EP = EA.\n\\]\n\nDraw \\( FM_1 \\perp BP \\) through point \\( F \\), \\( EN_1 \\perp PA \\) through point \\( E \\), and \\( CH \\perp PA \\), as shown in the figure.\n\nSince \\( \\angle C = 90^\\circ \\), \\( AC = 30 \\), and \\( BC = 40 \\),\n\n\\[\n\\therefore AB = \\sqrt{30^2 + 40^2} = 50.\n\\]\n\nSince \\( \\angle ABC = \\angle FPB \\),\n\n\\[\n\\therefore FB = FP,\n\\]\n\n\\[\n\\therefore BM_1 = PM_1 = \\frac{1}{2} BP = \\frac{x}{2}.\n\\]\n\nSimilarly, we can find,\n\n\\[\nAN_1 = PN_1 = \\frac{1}{2} AP = \\frac{1}{2}(50 - x) = \\frac{50 - x}{2}.\n\\]\n\nIt is easy to see that \\( \\triangle BFM_1 \\sim \\triangle BAC \\),\n\n\\[\n\\therefore \\frac{BM_1}{BC} = \\frac{FM_1}{AC}, \\quad \\text{that is}, \\quad \\frac{x}{20} = \\frac{FM_1}{30},\n\\]\n\n\\[\n\\therefore FM_1 = \\frac{3}{8}x.\n\\]\n\nIt is easy to see that \\( \\triangle AN_1E \\sim \\triangle ACB \\),\n\n\\[\n\\therefore \\frac{AN_1}{AC} = \\frac{EN_1}{BC}, \\quad \\text{that is}, \\quad \\frac{50 - x}{20} = \\frac{EN_1}{40},\n\\]\n\n\\[\n\\therefore EN_1 = \\frac{100 - 2x}{3}.\n\\]\n\nSince the area of quadrilateral \\( CEPF \\) is equal to the area of \\( \\triangle ABC \\) minus the areas of \\( \\triangle FBP \\) and \\( \\triangle EPA \\),\n\n\\[\n\\therefore y = \\frac{1}{2} \\times 40 \\times 30 - \\frac{1}{2} x \\cdot \\frac{3}{8}x - \\frac{1}{2} \\cdot (50 - x) \\cdot \\frac{100 - 2x}{3},\n\\]\n\nSimplifying, we get:\n\n\\[\ny = -\\frac{25}{48}x^2 + \\frac{100}{3}x - \\frac{700}{3}.\n\\]\n\n(3) From part (2), we have:\n\n\\[\ny = -\\frac{25}{48}x^2 + \\frac{100}{3}x - \\frac{700}{3}.\n\\]\n\nThe axis of symmetry of the quadratic function is \\( x = 32 \\), at which point \\( y \\) reaches its maximum.\n\nSubstituting \\( x = 32 \\) into the equation:\n\n\\[\ny = -\\frac{25}{48} \\times 32^2 + \\frac{100}{3} \\times 32 - \\frac{700}{3},\n\\]\n\nwe find:\n\n\\[\ny = 300.\n\\]\n\nThus, the maximum area of quadrilateral \\( CEPF \\) is 300.\n\nWhen \\( CH \\perp AB \\),\n\n\\[\n\\frac{1}{2} \\times 40 \\times 30 = \\frac{1}{2} \\times 50 \\times CH = 600,\n\\]\n\n\\[\n\\therefore CH = 24.\n\\]\n\n\\[\n\\therefore BH = \\sqrt{40^2 - 24^2} = 32, \\quad AH = 18.\n\\]\n\nAt this point, when point \\( P \\) coincides with point \\( H \\), the area of quadrilateral \\( CEPF \\) is maximized.\n\nThe remaining area is:\n\n\\[\n600 \\times 2 - 300 = 900.\n\\]\n\nThus, the total cost is:\n\n\\[\n60 \\times 300 + 90 \\times 900 = 99000 \\text{ yuan}.\n\\]\n\n**Key Points:** This problem examines the determination and properties of similar triangles, the Pythagorean theorem for solving triangles, the properties of quadratic functions, the determination and properties of congruent triangles, and the properties of isosceles triangles. The key to solving the problem lies in understanding the given conditions, correctly drawing auxiliary lines, identifying critical points, and proceeding with the solution accordingly." }, { "problem_id": 2008, "question": "As shown in Figure 1, the graph of the quadratic function $\\mathrm{y}=\\mathrm{ax}^{2}-3 \\mathrm{ax}-4 \\mathrm{a}$ intersects the $\\mathrm{x}$-axis at points $\\mathrm{A}$ and $\\mathrm{B}$ (with point $\\mathrm{A}$ to the left of point $\\mathrm{B}$), and intersects the $\\mathrm{y}$-axis at point $\\mathrm{C}(0,-3)$.\n\n(1) Determine the expression of the quadratic function and the coordinates of points $\\mathrm{A}$ and $\\mathrm{B}$;\n\n(2) If point $\\mathrm{D}$ lies on the graph of the quadratic function and $S_{\\triangle D B C}=\\frac{4}{5} S_{\\triangle A B C}$, find the abscissa of point $\\mathrm{D}$;\n\n(3) Translate the line $\\mathrm{BC}$ downward, intersecting the graph of the quadratic function at points $\\mathrm{M}$ and $\\mathrm{N}$ (with $\\mathrm{M}$ to the left of $\\mathrm{N}$), as shown in Figure 2. Draw $M E / / y$-axis, intersecting the line $B C$ at point $E$, and draw $N F / / y$-axis, intersecting the line $B C$ at point $F$. When the value of $\\mathrm{MN}+\\mathrm{ME}$ is maximized, find the coordinates of point $\\mathrm{M}$.\n\n\n\nFigure 1\n\n\n\nFigure 2", "input_image": [ "batch7-2024_06_14_3b9c127d95f826cc17d7g_0085_1.jpg", "batch7-2024_06_14_3b9c127d95f826cc17d7g_0085_2.jpg" ], "is_multi_img": true, "answer": "(1) $\\mathrm{y}=\\frac{3}{4} x^{2}-\\frac{9}{4} x-3, \\mathrm{~A}(-1,0), \\mathrm{B}(4,0)$ ；（2） $2+2 \\sqrt{2}$ or $2-2 \\sqrt{2}$ or 2 ；（3） $\\mathrm{M}\\left(\\frac{1}{3},-\\frac{11}{3}\\right)$\n\n ", "answer_type": "multi-step", "difficulty": "High", "grade": "Senior", "subject": "Algebra", "image_relavance": "1", "analysis": "(1) The equation $\\mathrm{y}=\\mathrm{ax}^{2}-3 \\mathrm{ax}-4 \\mathrm{a}$ intersects the y-axis at point C $(0,-3)$,\n\n$\\therefore \\mathrm{a}=\\frac{3}{4}$,\n\n$\\therefore \\mathrm{y}=\\frac{3}{4} x^{2}-\\frac{9}{4} x-3$,\n\nIt intersects the x-axis at points $\\mathrm{A}(-1,0)$ and $\\mathrm{B}(4,0)$;\n\n(2) Let the equation of line $BC$ be $\\mathrm{y}=\\mathrm{kx}+\\mathrm{b}$,\n\n$\\therefore\\left\\{\\begin{array}{l}4 k+b=0 \\\\ b=-3\\end{array}\\right.$,\n\n$\\therefore\\left\\{\\begin{array}{l}k=-\\frac{3}{4} \\\\ b=-3\\end{array}\\right.$,\n\n$\\therefore \\mathrm{y}=\\frac{3}{4} \\mathrm{x}-3$;\n\nDraw a line through point $\\mathrm{D}$ parallel to the y-axis, intersecting line $\\mathrm{BC}$ at point $\\mathrm{H}$,\n\nLet H $\\left(x, \\frac{3}{4} x-3\\right)$, D$\\left(x, \\frac{3}{4} x^{2}-\\frac{9}{4} x-3\\right)$,\n\n$\\therefore \\mathrm{DH}=\\left|\\frac{3}{4} \\mathrm{x}^{2}-3 \\mathrm{x}\\right|$,\n\n$\\because \\mathrm{S}_{\\triangle \\mathrm{ABC}}=\\frac{1}{2} \\times 5 \\times 3=\\frac{15}{3}$,\n\n$\\therefore \\mathrm{S}_{\\triangle \\mathrm{DBC}}=\\frac{4}{5} \\times \\frac{15}{2}=6$,\n\n$\\therefore \\mathrm{S}_{\\triangle \\mathrm{DBC}}=2 \\times\\left|\\frac{3}{4} \\mathrm{x}^{2}-3 \\mathrm{x}\\right|=6$,\n\n$\\therefore \\mathrm{x}=2+2 \\sqrt{2}, \\mathrm{x}=2-2 \\sqrt{2}, \\mathrm{x}=2$;\n\n$\\therefore$ The x-coordinate of point D is $2+2 \\sqrt{2}, 2-2 \\sqrt{2}, 2$;\n\n(3) Draw a line through point $M$ parallel to the x-axis, intersecting the extension of $F N$ at point $G$,\n\nLet $M\\left(m, \\frac{3}{4} m^{2}-\\frac{9}{4} m-3\\right)$, $N\\left(n, \\frac{3}{4} n^{2}-\\frac{9}{4} n-3\\right)$,\n\nThen $E\\left(m, \\frac{3}{4} m-3\\right)$, $F\\left(n, \\frac{3}{4} n-3\\right)$,\n\n$\\therefore \\mathrm{ME}=-\\frac{3}{4} \\mathrm{~m}^{2}+3 \\mathrm{~m}$, $\\mathrm{NF}=-\\frac{3}{4} \\mathrm{n}^{2}+3 \\mathrm{n}$,\n\n$\\because \\mathrm{EF} / / \\mathrm{MN}$, $\\mathrm{ME} / / \\mathrm{NF}$,\n\n$\\therefore$ Quadrilateral MNFE is a parallelogram,\n\n$\\therefore \\mathrm{ME}=\\mathrm{NF}$,\n\n$\\therefore-\\frac{3}{4} m^{2}+3 m=-\\frac{3}{4} n^{2}+3 n$,\n\n$\\therefore \\mathrm{m}+\\mathrm{n}=4$,\n\n$$\n\\begin{aligned}\n& \\therefore M G=n-m=4-2 m \\text {, } \\\\\n& \\therefore \\angle \\mathrm{NMG}=\\angle \\mathrm{OBC} \\text {, } \\\\\n& \\therefore \\cos \\angle \\mathrm{NMG}=\\cos \\angle \\mathrm{OBC}=\\frac{M G}{M N}=\\frac{O B}{B C} \\text {, } \\\\\n& \\because B(4,0), \\text { C }(0,-3), \\\\\n& \\therefore \\mathrm{OB}=4, \\quad \\mathrm{OC}=3 \\text {, }\n\\end{aligned}\n$$\n\n\n\n【Insight】This problem examines the properties of quadratic and linear functions, requiring proficiency in the method of undetermined coefficients for determining function equations and solving problems using the properties of triangles." }, { "problem_id": 2009, "question": "In right $\\triangle ABC$, points $D$ and $E$ are on $AC$ and $BC$ respectively, with\n\n$AC=CB=7, \\angle ACB=90^{\\circ}, CD=3$. Connect $DE$, and take points $M$ and $N$ on $DE$ and $AB$ respectively. Connect $CM$ and $MN$, always satisfying $CM=MN$, and let $\\frac{ME}{DM}=\\frac{BN}{AN}=m$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n(1) As shown in Figure 1, when $m=1$, connect $DN$ and $NE$, and draw $NG \\perp BC$ at $G$. The length of segment $EG$ is $\\qquad$;\n\n(2) As shown in Figure 2, when $m=2$, the length of segment $CE$ is $\\qquad$.", "input_image": [ "batch8-2024_06_14_7baacbfddb223e39e04bg_0002_1.jpg", "batch8-2024_06_14_7baacbfddb223e39e04bg_0002_2.jpg" ], "is_multi_img": true, "answer": "(1)$\\quad \\frac{1}{2} (2)\\quad \\frac{11}{2}$\n\n ", "answer_type": "multi-step", "difficulty": "Medium", "grade": "Senior", "subject": "Metric Geometry", "image_relavance": "1", "analysis": "Solution: (1) As shown in the figure, establish a plane rectangular coordinate system with point $C$ as the origin, and $C B$ and $C A$ as the $x$-axis and $y$-axis, respectively.\n\n\n\nWhen $m=1$, points $M$ and $N$ are the midpoints of $D E$ and $A B$, respectively.\n\nThus, $D(0,3)$, $A(0,7)$, and $N\\left(\\frac{7}{2}, \\frac{7}{2}\\right)$.\n\nLet $E(x, 0)$, then $M\\left(\\frac{1}{2} x, \\frac{3}{2}\\right)$.\n\n$\\therefore C M^{2}=\\frac{1}{4} x^{2}+\\frac{9}{4}$.\n\n$M N^{2}=\\left(\\frac{1}{2} x-\\frac{7}{2}\\right)^{2}+\\left(\\frac{3}{2}-\\frac{7}{2}\\right)^{2}=\\frac{1}{4} x^{2}-\\frac{7}{2} x+\\frac{49}{4}+4$.\n\n$\\because C M=M N$,\n\n$\\therefore C M^{2}=M N^{2}$.\n\n$\\therefore \\frac{1}{4} x^{2}+\\frac{9}{4}=\\frac{1}{4} x^{2}-\\frac{7}{2} x+\\frac{49}{4}+4$.\n\nSolving gives $x=4$,\n\n$\\therefore C E=4$.\n\n$\\because N G \\perp B C$,\n\n$\\therefore C G=\\frac{7}{2}$.\n\n$\\therefore E G=E C-C G=\\frac{1}{2}$.\n\nHence, the answer is: $\\frac{1}{2}$.\n\n(2) When $m=2$, points $M$ and $N$ are the trisection points of $D E$ and $A B$, respectively.\n\nLet $E(y, 0)$, then $M\\left(\\frac{1}{3} y, 2\\right)$, $N\\left(\\frac{7}{3}, \\frac{14}{3}\\right)$.\n\n$\\therefore C M^{2}=\\frac{1}{9} y^{2}+4$.\n\n$M N^{2}=\\left(\\frac{1}{3} y-\\frac{7}{3}\\right)^{2}+\\left(2-\\frac{14}{3}\\right)^{2}$.\n\n$\\because C M=M N$,\n\n$\\therefore \\frac{1}{9} y^{2}+4=\\left(\\frac{1}{3} y-\\frac{7}{3}\\right)^{2}+\\left(2-\\frac{14}{3}\\right)^{2}$.\n\nSolving gives $y=\\frac{11}{2}$.\n\n$\\therefore C E=\\frac{11}{2}$.\n\nHence, the answer is: $\\frac{11}{2}$.\n\n【Key Insight】This problem examines the plane rectangular coordinate system, the Pythagorean theorem, and solving linear equations. The key to solving the problem lies in establishing the coordinate system and using coordinates to construct equations." } ]