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2. Let $A_{n}$ and $B_{n}$ be the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively. If $\frac{A_{n}}{B_{n}}=\frac{5 n-3}{n+9}$, then $\frac{a_{8}}{b_{8}}=$ $\qquad$
|
2.3.
Let $\left\{a_{n}\right\}, \left\{b_{n}\right\}$ have common differences $d_{1}, d_{2}$, respectively. Then $\frac{A_{n}}{B_{n}}=\frac{a_{1}+\frac{1}{2}(n-1) d_{1}}{b_{1}+\frac{1}{2}(n-1) d_{2}}=\frac{5 n-3}{n+9}$.
Let $d_{2}=d$. Then $d_{1}=5 d$.
Thus, $a_{1}=d, b_{1}=5 d$.
Therefore, $\frac{a_{8}}{b_{8}}=\frac{d+7 \times 5 d}{5 d+7 d}=\frac{36 d}{12 d}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $O$ be the origin, $A$ be a moving point on the parabola $x=\frac{1}{4} y^{2}+1$, and $B$ be a moving point on the parabola $y=x^{2}+4$. Then the minimum value of the area of $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3.2.
Let $A\left(s^{2}+1,2 s\right), B\left(t, t^{2}+4\right)$. Then $l_{O B}:\left(t^{2}+4\right) x-t y=0$.
Let the distance from point $A$ to line $O B$ be $h$, we have
$$
h=\frac{\left|\left(t^{2}+4\right)\left(s^{2}+1\right)-t \cdot 2 s\right|}{\sqrt{\left(t^{2}+4\right)^{2}+t^{2}}} \text {. }
$$
Therefore, $S_{\triangle O A B}=\frac{1}{2} O B h$
$$
\begin{array}{l}
=\frac{1}{2}\left|\left(t^{2}+4\right)\left(s^{2}+1\right)-2 s t\right| \\
=\frac{1}{2}\left[\left(s^{2} t^{2}+3 s^{2}+(s-t)^{2}+4\right] \geqslant 2 .\right.
\end{array}
$$
When $s=t=0$, the equality holds.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the complex number $z=\cos \frac{4 \pi}{7}+\mathrm{i} \sin \frac{4 \pi}{7}$. Then
$$
\left|\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}\right|
$$
is equal to $\qquad$ (answer with a number).
|
5. 2 .
Given that $z$ satisfies the equation $z^{7}-1=0$.
From $z^{7}-1=(z-1) \sum_{i=0}^{6} z^{i}$, and $z \neq 1$, we get $z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$. After combining the fractions of $\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}$, the denominator is
$$
\begin{array}{l}
\left(1+z^{2}\right)\left(1+z^{4}\right)\left(1+z^{6}\right) \\
=1+z^{2}+z^{4}+z^{6}+z^{6}+z^{8}+z^{10}+z^{12} \\
=1+z^{2}+z^{4}+z^{6}+z^{6}+z+z^{3}+z^{5}=z^{6},
\end{array}
$$
The numerator is
$$
\begin{array}{l}
z\left(1+z^{4}\right)\left(1+z^{6}\right)+z^{2}\left(1+z^{2}\right)\left(1+z^{6}\right)+ \\
z^{3}\left(1+z^{2}\right)\left(1+z^{4}\right) \\
=\left(1+z^{4}\right)(z+1)+\left(1+z^{2}\right)\left(z^{2}+z\right)+ \\
\left(1+z^{2}\right)\left(z^{3}+1\right) \\
=1+z+z^{4}+z^{5}+z+z^{2}+z^{3}+z^{4}+1+z^{2}+z^{3}+z^{5} \\
=2\left(1+z+z^{2}+z^{3}+z^{4}+z^{5}\right)=-2 z^{6} \text {. } \\
\end{array}
$$
Therefore, $\left|\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}\right|=\left|\frac{-2 z^{6}}{z^{6}}\right|=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $a, b, c, d$ be real numbers, satisfying
$$
a+2 b+3 c+4 d=\sqrt{10} \text {. }
$$
Then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}$ is $\qquad$
|
6. 1 .
From the given equation, we have
$$
\begin{array}{l}
(1-t) a+(2-t) b+(3-t) c+(4-t) d+ \\
t(a+b+c+d)=\sqrt{10} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
{\left[(1-t)^{2}+(2-t)^{2}+(3-t)^{2}+(4-t)^{2}+t^{2}\right] .} \\
{\left[a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\right] \geqslant 10} \\
\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2} \\
\quad \geqslant \frac{10}{5 t^{2}-20 t+30}=\frac{10}{5(t-2)^{2}+10}=1 .
\end{array}
$$
Equality holds if and only if
$$
t=2, a=-\frac{\sqrt{10}}{10}, b=0, c=\frac{\sqrt{10}}{10}, d=\frac{\sqrt{10}}{5}
$$
Therefore, the minimum value sought is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $\left(x^{2}-x-2\right)^{3}=a_{0}+a_{1} x+\cdots+a_{6} x^{6}$, then $a_{1}+a_{3}+a_{5}=$
|
3. -4 .
Let $x=0, x=1$, we get
$$
\begin{array}{l}
a_{0}=-8, \\
a_{0}+a_{1}+\cdots+a_{6}=\left(1^{2}-1-2\right)^{3}=-8 .
\end{array}
$$
Thus, $a_{1}+a_{2}+\cdots+a_{6}=0$.
Let $x=-1$, we get
$$
\begin{array}{l}
a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6} \\
=\left[(-1)^{2}-(-1)-2\right]^{3}=0 .
\end{array}
$$
Thus, $-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6}=8$.
Therefore, $a_{1}+a_{3}+a_{5}=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given an isosceles triangle with a vertex angle of $20^{\circ}$ and a base length of $a$, the length of the legs is $b$. Then the value of $\frac{a^{3}+b^{3}}{a b^{2}}$ is $\qquad$
|
4.3.
Given $a=2 b \sin 10^{\circ}$.
Thus $a^{3}+b^{3}=8 b^{3} \sin ^{3} 10^{\circ}+b^{3}$
$$
\begin{array}{l}
=8 b^{3} \cdot \frac{1}{4}\left(3 \sin 10^{\circ}-\sin 30^{\circ}\right)+b^{3}=6 b^{3} \sin 10^{\circ} \\
\Rightarrow \frac{a^{3}+b^{3}}{a b^{2}}=\frac{6 b^{3} \sin 10^{\circ}}{2 b \sin 10^{\circ} \cdot b^{2}}=3 .
\end{array}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given a regular tetrahedron $P-ABC$ with the side length of the base being 6 and the side length of the lateral edges being $\sqrt{21}$. Then the radius of the inscribed sphere of the tetrahedron is $\qquad$
|
7.1 .
Let $P O \perp$ plane $A B C$ at point $O$. Then $O$ is the center of the equilateral $\triangle A B C$. Connect $A O$ and extend it to intersect $B C$ at point $D$, and connect $P D$. Thus, $D$ is the midpoint of $B C$.
It is easy to find that $P D=2 \sqrt{3}, O D=\sqrt{3}, P O=3$.
Let the radius of the inscribed sphere of this tetrahedron be $r$. Then
$$
\begin{aligned}
r & =\frac{3 V_{\text {tetrahedron } P-A B C}}{S_{\triangle A B C}+3 S_{\triangle P A B}} \\
& =\frac{3 \times \frac{\sqrt{3}}{4} \times 6^{2}}{\frac{\sqrt{3}}{4} \times 6^{2}+3 \times \frac{1}{2} \times 6 \times 2 \sqrt{3}}=1 .
\end{aligned}
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The largest prime $p$ such that $\frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are both perfect squares is $\qquad$.
|
10.7.
Let $\frac{p+1}{2}=x^{2} ， \frac{p^{2}+1}{2}=y^{2}\left(x, y \in \mathbf{Z}_{+}\right)$.
Obviously, $p>y>x, p>2$.
From $p+1=2 x^{2}, p^{2}+1=2 y^{2}$, subtracting the two equations gives $p(p-1)=2(y-x)(y+x)$.
Since $p(p>2)$ is a prime number and $p>y-x$, then $p \mid(y+x)$.
Because $2 p>y+x$, so $p=y+x$.
Thus, $p-1=2(y-x) \Rightarrow p+1=4 x$
$$
\Rightarrow 4 x=2 x^{2} \Rightarrow x=2 \Rightarrow p=7, y=5 \text {. }
$$
Therefore, the only $p$ that satisfies the condition is $p=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. On a plane, there is an $8 \times 8$ grid colored in a black and white checkerboard pattern. Basil arbitrarily selects one of the cells. Each turn, Peter draws a polygon (which can be concave but not self-intersecting) on the grid, and Basil will honestly inform Peter whether the selected cell is inside or outside the polygon. To ensure that he can determine whether the cell Basil chose is white or black, what is the minimum number of turns Peter needs to ask?
|
2. If the polygon drawn by Peter includes only all the cells of a certain color, then this polygon must intersect itself. Therefore, Peter cannot determine the color of the cell chosen by Basil in just one round.
Next, two strategies are given that can determine the color of the selected cell in two rounds.
【Strategy 1】The two polygon drawing methods shown in Figure 1 can determine whether the selected cell is white or black in two rounds.
If the selected cell is white, the answers in both rounds will either be inside the drawn polygon or outside the drawn polygon. If the selected cell is black, then in these two rounds, it will appear once inside the polygon and once outside the polygon.
【Strategy 2】The two polygon drawing methods shown in Figure 2 can also determine the color of the selected cell in two rounds. In the first round, draw the polygon in Figure 2 (a). If the selected cell is outside the polygon, it must be in the 2nd, 4th, 6th, or 8th column from left to right. In the second round, draw the polygon in Figure 2 (b), and determine whether the selected cell is black or white based on whether it is inside or outside the polygon. If the answer in the first round is inside the polygon, then the cell must be in the 1st, 3rd, 5th, or 7th column from left to right, or in the last row from top to bottom. In the second round, use the polygon in Figure 2 (c).
If the answer is inside the polygon, the selected cell is white; if the answer is outside the polygon, the selected cell is black.
【Note】If the problem does not require drawing the polygon within the table (i.e., it can be understood as being able to draw outside the table), then Strategy 1 can be simplified as shown in Figure 3.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the complex numbers be
$$
\begin{array}{l}
z_{1}=(6-a)+(4-b) \mathrm{i}, \\
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$
where, $a, b \in \mathbf{R}$.
When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ reaches its minimum value, $3 a+4 b$ $=$
|
2. 12 .
Notice that,
$$
\begin{array}{l}
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right| \\
=|12+9 \mathrm{i}|=15 .
\end{array}
$$
Equality holds if and only if
$$
\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9} \text {, }
$$
i.e., when $a=\frac{7}{3}, b=\frac{5}{4}$, the minimum value 15 is achieved. Therefore, $3 a+4 b=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let the line $l$ passing through the origin intersect the graph of the function $y=|\sin x|$ $(x \geqslant 0)$ at exactly three points, with $\alpha$ being the largest of the x-coordinates of these intersection points. Then
$$
\frac{\left(1+\alpha^{2}\right) \sin 2 \alpha}{2 \alpha}=
$$
$\qquad$ .
|
3. 1 .
As shown in Figure 2, let the line $l$ be tangent to the function $y=|\sin x|(x \geqslant 0)$
at point $P(\alpha,-\sin \alpha)$, and $k_{l}=-\cos \alpha$.
Then the line $l: y+\sin \alpha=-(x-\alpha) \cos \alpha$.
Substituting $(0,0)$, we get $\alpha=\tan \alpha$.
Therefore, $\frac{\left(1+\alpha^{2}\right) \sin 2 \alpha}{2 \alpha}=\frac{\left(1+\tan ^{2} \alpha\right) \sin 2 \alpha}{2 \tan \alpha}$
$$
=\frac{\sin 2 \alpha}{2 \cos ^{2} \alpha \cdot \tan \alpha}=1 .
$$
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the number of solutions to the equation
$$
|| \cdots|||x|-1|-2| \cdots|-2011|=2011
$$
The number of solutions. ${ }^{[4]}$
|
【Analysis】Remove the absolute value symbols from outside to inside, step by step.
From the original equation, we get
$$
|| \cdots|||x|-1|-2| \cdots|-2010|=0
$$
or ||$\cdots|||x|-1|-2| \cdots|-2010|=4002$.
For equation (1), we have
$$
\begin{array}{l}
|| \cdots|||x|-1|-2| \cdots|-2009|=2010 \\
\Rightarrow|| \cdots|||x|-1|-2| \cdots|-2008| \\
\quad=2009+2010 \\
\Rightarrow|x|=1+2+\cdots+2010=2011 \times 1005 .
\end{array}
$$
Similarly, for equation (2), we have
$$
|x|=1+2+\cdots+2010+4022=2011 \times 1007 .
$$
Therefore, $x= \pm 2011 \times 1005$ or
$$
x= \pm 2011 \times 1007 \text {, }
$$
there are four solutions in total.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let real numbers $x, y$ satisfy
$$
x^{2}+\sqrt{3} y=4, y^{2}+\sqrt{3} x=4, x \neq y \text {. }
$$
Then the value of $\frac{y}{x}+\frac{x}{y}$ is $\qquad$
|
3. -5 .
From the conditions, we have
$$
\left\{\begin{array}{l}
x^{2}-y^{2}+\sqrt{3} y-\sqrt{3} x=0, \\
x^{2}+y^{2}+\sqrt{3}(x+y)=8 .
\end{array}\right.
$$
From equation (1) and $x \neq y$, we know $x+y=\sqrt{3}$.
Substituting into equation (2) gives $x^{2}+y^{2}=5$.
$$
\begin{array}{l}
\text { Also, } 2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=-2 \\
\Rightarrow x y=-1 . \\
\text { Therefore, } \frac{y}{x}+\frac{x}{y}=\frac{x^{2}+y^{2}}{x y}=-5 .
\end{array}
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups.
The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups.
|
7.6 .
Rewrite the given equation as
$$
\begin{array}{l}
x^{2}-x(2+y)+y^{2}-2 y=0 \\
\Rightarrow \Delta=(2+y)^{2}-4\left(y^{2}-2 y\right)=-3 y^{2}+12 y+4 \\
\quad=-3(y-2)^{2}+16 \geqslant 0 \\
\Rightarrow|y-2| \leqslant \frac{4}{\sqrt{3}}<3 .
\end{array}
$$
Since $y$ is an integer, thus, $y=0,1,2,3,4$.
Upon calculation, only when $y=0,2,4$, $x$ is an integer, and we obtain
$$
\begin{aligned}
(x, y)= & (0,0),(2,0),(0,2),(4,2), \\
& (2,4),(4,4) .
\end{aligned}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) As shown in Figure 4, the three sides of $\triangle ABC$ are all positive integers, and the perimeter is 35. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find the length of side $AB$.
|
11. Extend $G I$, intersecting $C B$ and $C A$ (or their extensions) at points $P$ and $Q$ respectively.
Since $C I$ is the angle bisector of $\angle C$ and $\angle G I C=90^{\circ}$, we know that $\triangle C P Q$ is an isosceles triangle.
Draw $G E \perp P C$ and $G F \perp C Q$ from point $G$, and draw $I R \perp Q C$ from point $I$.
Let the inradius of $\triangle A B C$ be $r$, then $I R=r$.
Since $G$ is a point on the base of the isosceles $\triangle C P Q$, and $I$ is the midpoint of the base, then
$$
G E+G F=2 I R=2 r .
$$
Since $G$ is the centroid of $\triangle A B C$, we have
$$
G E=\frac{1}{3} h_{a}, G F=\frac{1}{3} h_{b} \text {. }
$$
Thus, $S_{\triangle A B C}=\frac{1}{3}\left(h_{a}+h_{b}\right)=2 r$
$$
\begin{array}{l}
\Rightarrow \frac{1}{3}\left(\frac{S}{\frac{1}{2} a}+\frac{S}{\frac{1}{2} b}\right)=2 \times \frac{S}{\frac{1}{2}(a+b+c)} \\
\Rightarrow 6 a b=(a+b)(a+b+c) \\
\Rightarrow 6 a b=35(a+b) \Rightarrow 6 I(a+b) .
\end{array}
$$
Given $18 \leqslant a+b<35$, thus $a+b$ can take the values 18, 24, 30.
Upon verification, only when $(a+b, a b)=(24,140)$, $a$ and $b$ have positive integer solutions.
Therefore, $c=35-(a+b)=11$, i.e., $A B=11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the function $f: \mathbf{N} \rightarrow \mathbf{N}$ defined as follows:
$$
f(x)=\left\{\begin{array}{ll}
\frac{x}{2}, & x \text { is even; } \\
\frac{x+7}{2}, & x \text { is odd. }
\end{array}\right.
$$
Then the number of elements in the set $A=\{x \in \mathbf{N} \mid f(f(f(x)))=x\}$ is
|
- 1.8.
On one hand, when $x \in \{0,1, \cdots, 7\}$, it is calculated that $f(x) \in \{0,1, \cdots, 7\}$,
and it can be verified that $\{0,1, \cdots, 7\} \subseteq A$.
On the other hand, when $x \geqslant 8$,
$$
\frac{x}{2}<x \text{, and } \frac{x+7}{2}<\frac{2 x}{2}<x \text{. }
$$
Then $f(f(f(x)))<8$, which does not hold.
Therefore, $A=\{0,1, \cdots, 7\}$, with a total of 8 elements.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. (16 points) As shown in Figure 4, $A$ and $B$ are the common vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. $P$ and $Q$ are moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})(\lambda \in \mathbf{R},|\lambda|>1) \text {. }
$$
Let $k_{1} 、 k_{2} 、 k_{3} 、 k_{4}$, then $k_{1}+k_{2}+k_{3}+k_{4}$ is a constant.
|
Prove: (1) $O, P, Q$ are collinear;
(2) If the slopes of the lines $AP, BP, AQ, BQ$ are respectively
14. (1) Note that,
$\overrightarrow{A P}+\overrightarrow{B P}=2 \overrightarrow{O P}, \overrightarrow{A Q}+\overrightarrow{B Q}=2 \overrightarrow{O Q}$.
Also, $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})$, hence $\overrightarrow{O P}=\lambda \overrightarrow{O Q}$.
Therefore, $O, P, Q$ are collinear.
(2) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}^{2}-a^{2}=\frac{a^{2}}{b^{2}} y_{1}^{2} \text {. }
$$
Thus, $k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}$
$$
=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}} \text {. }
$$
Similarly, $k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}$.
From (1), we know that $O, P, Q$ are collinear, i.e., $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$.
Therefore, from equations (1) and (2), we get
$$
\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}} \\
=\frac{2 b^{2}}{a^{2}}\left(\frac{x_{1}}{y_{1}}-\frac{x_{2}}{y_{2}}\right)=0 .
\end{array}
$$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\sqrt{a+b+c+d}+\sqrt{a^{2}-2 a+3-b}- \\
\sqrt{b-c^{2}+4 c-8}=3 .
\end{array}
$$
Then the value of $a-b+c-d$ is ( ).
(A) $-7 \quad \square \quad$ -
(B) -8
(C) -4
(D) -6
|
5. A.
$$
\begin{array}{l}
\text { Given } a^{2}-2 a+3-b \geqslant 0, \\
b-c^{2}+4 c-8 \geqslant 0,
\end{array}
$$
we know
$$
\begin{array}{l}
b \leqslant-(a+1)^{2}+4 \leqslant 4, \\
b \geqslant(c-2)^{2}+4 \geqslant 4 .
\end{array}
$$
Thus, $b=4, a=-1, c=2$.
Substituting these into the known equations gives $d=4$.
Therefore, $a-b+c-d=-7$.
|
-7
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$
then $x+\frac{1}{x}=$
|
3. 4 .
Transform the given equation into
$$
\begin{aligned}
& \left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow & \left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{aligned}
$$
Notice that, $x^{3}+x^{-3}+53>0$.
Thus, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
Then $x^{3}+x^{-3}=\left(x+x^{-1}\right)\left[\left(x+x^{-1}\right)^{2}-3\right]$
$$
\begin{array}{l}
\Rightarrow b\left(b^{2}-3\right)=52 \\
\Rightarrow(b-4)\left(b^{2}+4 b+13\right)=0 \\
\Rightarrow b=4 .
\end{array}
$$
Therefore, $x+\frac{1}{x}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) There are 288 sets of cards, totaling 2016 cards, each set consisting of $1,2, \cdots, 7$ and stacked in the order $1,2, \cdots, 7$ from top to bottom. Now, these 288 sets of cards are stacked together from top to bottom. First, discard the top five cards, then place the top card at the bottom, and continue discarding the top five cards, then placing the top card at the bottom, and so on, until only one card remains.
(1) In the above process, when only 301 cards are left, how many cards have been discarded?
(2) What is the last remaining card, and which set does it belong to?
|
(1) For the first 42 cards (in 6 groups each), according to the operation rule, after discarding 5 cards, the cards placed at the bottom are $6,5, \cdots, 1,7$. Thus, each number discards 5 cards.
And $288 \div 6=48, 48 \times 5 \times 7=1680$ (cards), $48 \times 5=240$ (cards), $2016-1680=336$ (cards). That is, when 1680 cards are discarded, each number has discarded 240 cards, leaving 336 cards. The arrangement of these 336 cards is
$$
6,5, \cdots, 1,7,6, \cdots, 1,7,6, \cdots, 1,7,
$$
When 35 more cards are discarded, the number 7 discards 4 more cards, so the card 7 has discarded a total of 244 cards.
(2) Note that if there are only $6^{n}$ cards, the last card discarded is the card numbered $6^{n}$.
$$
\begin{array}{l}
\text { Also } 6^{4}=1296<2016<6^{5}, \\
2016-1296=720,
\end{array}
$$
After discarding 720 cards, the card placed at the bottom is the one we are looking for.
When 700 cards are discarded, each 6 groups discard 35 cards, at this point, it is already the $700 \div 35 \times 6=120$th group. When discarding up to the 720th card, it is already the 2nd card of the 124th group, and the 3rd card is placed at the bottom. Therefore, the last card is the 3rd card of the 124th group.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the set $A=\left\{x \left\lvert\, \frac{x+4}{x-3} \leqslant 0\right., x \in \mathbf{Z}\right\}$, and from set $A$ a random element $x$ is drawn, denoted by $\xi=x^{2}$. Then the mathematical expectation of the random variable $\xi$ is $\mathrm{E} \xi=$ $\qquad$
|
-1.5 .
From the conditions, we know that
$$
A=\{-4,-3,-2,-1,0,1,2\},
$$
The values of the random variable $\xi$ are $0, 1, 4, 9, 16$.
It is easy to obtain that the probability distribution of $\xi$ is shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline$\xi$ & 0 & 1 & 4 & 9 & 16 \\
\hline$P$ & $\frac{1}{7}$ & $\frac{2}{7}$ & $\frac{2}{7}$ & $\frac{1}{7}$ & $\frac{1}{7}$ \\
\hline
\end{tabular}
Therefore, $\mathrm{E} \xi$
$$
\begin{array}{l}
=0 \times \frac{1}{7}+1 \times \frac{2}{7}+4 \times \frac{2}{7}+9 \times \frac{1}{7}+16 \times \frac{1}{7} \\
=5 .
\end{array}
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $f(x)=x+g(x)$, where $g(x)$ is a function defined on $\mathbf{R}$ with the smallest positive period of 2. If the maximum value of $f(x)$ in the interval $[2,4)$ is 1, then the maximum value of $f(x)$ in the interval $[10,12)$ is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
2.9.
According to the problem, we have
$$
\begin{array}{l}
f(x+2)=(x+2)+g(x+2) \\
=x+g(x)+2=f(x)+2 .
\end{array}
$$
Given that the maximum value of $f(x)$ in the interval $[2,4)$ is 1, we know that the maximum value of $f(x)$ in the interval $[4,6)$ is 3, ... and the maximum value of $f(x)$ in the interval $[10,12)$ is $3+2 \times 3=9$.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $x, y, z$ satisfy $x^{2}+2 y^{2}+3 z^{2}=24$.
Then the minimum value of $x+2 y+3 z$ is $\qquad$ .
|
4. -12 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x+2 y+3 z)^{2} \\
=(1 \times x+\sqrt{2} \times \sqrt{2} y+\sqrt{3} \times \sqrt{3} z)^{2} \\
\leqslant\left[1^{2}+(\sqrt{2})^{2}+(\sqrt{3})^{2}\right]\left(x^{2}+2 y^{2}+3 z^{2}\right) \\
\quad=144 .
\end{array}
$$
Therefore, $x+2 y+3 z \geqslant-12$, with equality holding if and only if $x=y=z=-2$.
Hence, the minimum value of $x+2 y+3 z$ is -12.
|
-12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. If $\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\cdots+\sin \frac{n \pi}{9}=\frac{1}{2} \tan \frac{4 \pi}{9}$, then the smallest positive integer $n$ is $\qquad$.
|
10.4.
Notice,
$$
\begin{array}{l}
2 \sin \frac{\pi}{18}\left(\sum_{k=1}^{n} \sin \frac{k \pi}{9}\right) \\
=\sum_{k=1}^{n}\left(\cos \frac{(2 k-1) \pi}{18}-\cos \frac{(2 k+1) \pi}{18}\right) \\
=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \tan \frac{4 \pi}{9} \cdot \sin \frac{\pi}{18}=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \cos \frac{\pi}{18}=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \cos \frac{(2 n+1) \pi}{18}=0 \\
\Rightarrow n=9 k+4(k \in \mathbf{Z}) .
\end{array}
$$
Therefore, the smallest positive integer $n$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let the function $f(x)=\frac{(x+1)^{2}+\sin x}{x^{2}+1}$ have the maximum value and minimum value as $M$ and $N$, respectively. Then $M+N=$
|
3. 2 .
From the given information, we have
$$
f(x)=\frac{(x+1)^{2}+\sin x}{x^{2}+1}=1+\frac{2 x+\sin x}{x^{2}+1} \text {. }
$$
Notice that the function $g(x)=\frac{2 x+\sin x}{x^{2}+1}$ is an odd function. Therefore, the maximum value $M_{0}$ and the minimum value $N_{0}$ of $g(x)$ satisfy
$$
\begin{array}{l}
M_{0}+N_{0}=0 . \\
\text { Also, } M=M_{0}+1, N=N_{0}+1 \Rightarrow M+N=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Four numbers
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \cdot \sqrt{2-\sqrt{2-\sqrt{3}}} 、 \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}}
\end{array}
$$
The product of these is ( ).
(A) $2+\sqrt{3}$
(B) 2
(C) 1
(D) $2-\sqrt{3}$
|
$-1 . \mathrm{C}$.
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \times \\
\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \\
\quad \sqrt{2^{2}-\left(\sqrt{2-\sqrt{2-\sqrt{3}})^{2}}\right.} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \\
\quad \sqrt{2+\sqrt{2-\sqrt{3}}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2^{2}-(\sqrt{2-\sqrt{3}})^{2}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2+\sqrt{3}}=\sqrt{4-3}=1 .
\end{array}
$$
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Several different numbers are written on the blackboard, such that the sum of any three of them is a rational number, while the sum of any two is an irrational number. The maximum number of numbers that can be written on the blackboard is $\qquad$
|
4. 3 .
Assume that the numbers written on the blackboard are no less than four, denoted as $a, b, c, d$. Then, $a+b+c$ and $b+c+d$ are both rational numbers, which implies that their difference
$$
(b+c+d)-(a+b+c)=d-a
$$
is also a rational number.
Similarly, $b-a$ and $c-a$ are also rational numbers.
Therefore, $b=a+r_{1}, c=a+r_{2}, d=a+r_{3}$, where $r_{1}, r_{2}, r_{3}$ are rational numbers.
Furthermore, since $a+b+c=3a+r_{1}+r_{2}$ is a rational number, it follows that $a$ is also a rational number. This indicates that $a+b=2a+r_{1}$ is a rational number, which contradicts the condition that "the sum of any two is irrational."
Therefore, the numbers written on the blackboard do not exceed three, such as $\sqrt{2}$, $2\sqrt{2}$, and $-3\sqrt{2}$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$1.2014^{2015}$ 的个位数为
The unit digit of $1.2014^{2015}$ is
|
-1.4 .
Let $g(n)$ denote the unit digit of a natural number $n$.
$$
\begin{array}{l}
\text { Then } g\left(2014^{2015}\right) \\
=g\left((201 \times 10+4)^{2015}\right) \\
=g\left(4^{2015}\right)=g\left(\left(4^{2}\right)^{1007} \times 4\right) \\
=g\left((10+6)^{1007} \times 4\right)=g\left(6^{1007} \times 4\right) \\
=g\left(g\left(6^{1007}\right) g(4)\right)=g(6 \times 4)=4 .
\end{array}
$$
Therefore, the unit digit of $2014^{2015}$ is 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $P$ be any point on the graph of the function $f(x)=x+\frac{2}{x}(x>0)$, and draw perpendiculars from point $P$ to the $x$-axis and $y$-axis, with the feet of the perpendiculars being $A$ and $B$ respectively. Then the minimum value of $|P A|+|P B|$ is $\qquad$
|
2. 4 .
Let $P(x, y)$. According to the problem,
$$
|P A|=|y|=y,|P B|=|x|=x,
$$
where, $y=x+\frac{2}{x}(x>0)$.
$$
\begin{array}{l}
\text { Hence }|P A|+|P B|=y+x=2\left(x+\frac{1}{x}\right) \\
\geqslant 2 \times 2 \sqrt{x \cdot \frac{1}{x}}=4,
\end{array}
$$
The equality holds if and only if $x=\frac{1}{x}$.
Since $x>0$, thus, when $x=1$, the above inequality takes the equality.
Therefore, when and only when $x=1$, $|P A|+|P B|$ has the minimum value 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. As shown in Figure 3, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through point $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
then the value of $m+n$ is
$\qquad$
|
14. 2 .
Since $O$ is the midpoint of side $B C$, we have
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} \text {. }
$$
Since points $M, O, N$ are collinear, it follows that,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\frac{1}{2} \sin 2 y+a=0$. Then the value of $\cos (x+2 y)$ is $\qquad$
|
15. 1 .
Solving the system of equations and eliminating $a$ yields
$$
x=-2 y \Rightarrow \cos (x+2 y)=1
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. For any $\alpha, \beta \in\left(0, \frac{2 \pi}{3}\right)$, we have
$$
\begin{array}{l}
4 \cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+4 \cos ^{2} \beta- \\
3 \cos \alpha-3 \cos \beta-k<0 .
\end{array}
$$
Then the minimum value of $k$ is
|
5.6.
Substitution $\left(x_{1}, x_{2}\right)=(2 \cos \alpha, 2 \cos \beta)$.
Then the given inequality becomes
$$
2 x_{1}^{2}+x_{1} x_{2}+2 x_{2}^{2}-3 x_{1}-3 x_{2}-2 k<0 \text {. }
$$
Since $x_{1}, x_{2} \in(-1,2)$, we have
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{1}-2\right)+\left(x_{2}+1\right)\left(x_{2}-2\right)+ \\
\frac{1}{2}\left(x_{1}+1\right)\left(x_{2}-2\right)+\frac{1}{2}\left(x_{1}-2\right)\left(x_{2}+1\right)<0 \\
\Rightarrow 2 x_{1}^{2}+x_{1} x_{2}+2 x_{2}^{2}-3 x_{1}-3 x_{2}-12<0 .
\end{array}
$$
When $x_{1}=x_{2}=x \rightarrow 2$ (or -1), the left side of the above equation
$$
\rightarrow 0 \text {. }
$$
Therefore, $k_{\min }=6$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let $a_{1} \in \mathbf{Z}_{+}$, and $a_{1} \leqslant 18$, define the sequence $\left\{a_{n}\right\}:$
$$
a_{n+1}=\left\{\begin{array}{ll}
2 a_{n}, & a_{n} \leqslant 18 ; \\
2 a_{n}-36, & a_{n}>18
\end{array}(n=1,2, \cdots) .\right.
$$
Find the maximum number of elements in the set $M=\left\{a_{n} \mid n \in \mathbf{Z}_{+}\right\}$.
|
11. Given the positive integer $a_{1} \leqslant 18$ and the recursive formula for the sequence $\left\{a_{n}\right\}$, the following properties can be derived:
(1) $a_{n+1} \equiv 2 a_{n}(\bmod 36)$, i.e.,
$a_{n+1} \equiv 2 a_{n}(\bmod 4)$, and $a_{n+1} \equiv 2 a_{n}(\bmod 9)$;
(2) All terms are positive integers, and for any $n \in$ $\mathbf{Z}_{+}, a_{n} \leqslant 36$;
(3) From the third term onwards, all terms are multiples of 4, i.e., for any $n \geqslant 3,4 \mid a_{n}$;
(4) $3\left|a_{n} \Leftrightarrow 3\right| a_{1}$, i.e., if $a_{1}$ is a multiple of 3, then all terms are multiples of 3; otherwise, none of the terms are multiples of 3.
Consider the maximum value of $|M|$ in the following two scenarios.
【Scenario 1】If $3 \mid a_{1}$, then consider $a_{3}$.
(i) $\left\{\begin{array}{l}a_{3} \equiv 3(\bmod 9), \\ a_{3} \equiv 0(\bmod 4) .\end{array}\right.$
By the Chinese Remainder Theorem, we get
$$
\begin{array}{l}
a_{3} \equiv 12(\bmod 36) \Rightarrow a_{3} \equiv 12 \\
\Rightarrow\left(a_{3}, a_{4}, a_{5}, a_{6}, \cdots\right) \equiv(12,24,12,24, \cdots) .
\end{array}
$$
Considering the first two terms, we get $|M| \leqslant 4$.
(ii) $\left\{\begin{array}{l}a_{3} \equiv 6(\bmod 9) \\ a_{3} \equiv 0(\bmod 4)\end{array}\right.$.
By the Chinese Remainder Theorem, we get
$$
\begin{array}{l}
a_{3} \equiv 24(\bmod 36) \\
\Rightarrow\left(a_{3}, a_{4}, a_{5}, a_{6}, \cdots\right) \equiv(24,12,24,12, \cdots) .
\end{array}
$$
Considering the first two terms, we get $|M| \leqslant 4$.
(iii) $\left\{\begin{array}{l}a_{3} \equiv 0(\bmod 9), \\ a_{3} \equiv 0(\bmod 4)\end{array}\right.$.
Then $a_{3} \equiv 0(\bmod 36)$
$\Rightarrow\left(a_{3}, a_{4}, a_{5}, \cdots\right) \equiv(36,36,36, \cdots)$.
Considering the first two terms, we get $|M| \leqslant 3$.
【Scenario 2】If $a_{1}$ is not a multiple of 3, then none of the terms in the sequence $\left\{a_{n}\right\}$ are multiples of 3.
In this case, $a_{3} \equiv 1,2,4,5,7,8(\bmod 9)$.
Also, $a_{3} \equiv 0(\bmod 4)$, so by the Chinese Remainder Theorem, we get $a_{3} \equiv 28,20,4,32,16,8(\bmod 36)$.
Therefore, $a_{3} \in\{28,20,4,32,16,8\}$.
Hence, when $n \geqslant 3$, $a_{n} \in\{28,20,4,32,16,8\}$, and $\{28,20,4,32,16,8\} \subset M$.
Considering the first two terms, we get $|M| \leqslant 8$.
Taking $a_{1}=1$, we get the sequence $\left\{a_{n}\right\}$:
$1,2,4,8,16,32,28,20,4, \cdots$.
Thus, the maximum number of elements in the set $M$ is 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2011^{3}}$.
Then the integer part of $4 S$ is ( ). ${ }^{[2]}$
(A) 4
(B) 5
(C) 6
(D) 7
(2011, "Mathematics Weekly" Cup National Junior High School Mathematics Competition)
|
When $k=2,3, \cdots, 2011$,
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {. }
$$
Then $1<S=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots+\frac{1}{2011^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2011 \times 2012}\right)<\frac{5}{4} \text {. }
$$
Thus, $4<4 S<5$.
Hence, the integer part of $4 S$ is 4.
|
4
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. The smallest positive odd number that cannot be expressed as $7^{x}-3 \times 2^{y}\left(x 、 y \in \mathbf{Z}_{+}\right)$ is $\qquad$
|
-1.3 .
Since $x, y \in \mathbf{Z}_{+}$, therefore, $7^{x}-3 \times 2^{y}$ is always an odd number, and $7^{1}-3 \times 2^{1}=1$.
If $7^{x}-3 \times 2^{y}=3$, then $317^{x}$.
And $7^{x}=(1+6)^{x}=1(\bmod 3)$, thus, there do not exist positive integers $x, y$ such that
$$
7^{x}-3 \times 2^{y}=3 \text {. }
$$
Therefore, the smallest positive odd number is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $f(x)=\frac{\sin \pi x}{x^{2}}(x \in(0,1))$. Then
$$
g(x)=f(x)+f(1-x)
$$
the minimum value of $g(x)$ is . $\qquad$
|
7.8.
From the given, we have
$$
\begin{aligned}
f^{\prime}(x) & =\frac{\pi x \cos \pi x-2 \sin \pi x}{x^{3}}, \\
f^{\prime \prime}(x) & =\frac{\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x}{x^{3}} .
\end{aligned}
$$
Next, we need to prove
$$
\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x>0.
$$
Let $t=\pi x \in(0, \pi)$. Then
$\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x$
$$
=\left(6-t^{2}\right) \sin t-2 t \cos t=p(t).
$$
Thus, $p^{\prime}(t)=\left(4-t^{2}\right) \cos t$.
It is easy to see that $p^{\prime}(t)$ is increasing on the interval $\left(0, \frac{\pi}{2}\right)$, decreasing on the interval $\left(\frac{\pi}{2}, 2\right)$, and increasing on the interval $(2, \pi)$.
Also, $p(0)=0, p^{\prime}(0)>0$, so for $t \in(0, \pi)$, $p(t)>0$, which means that for $x \in(0,1)$, $f^{\prime \prime}(x)>0$.
Therefore, $f(x)$ is a convex function on the interval $(0,1)$. By Jensen's inequality, we know that for $x \in(0,1)$,
$$
\begin{array}{l}
\frac{f(x)+f(1-x)}{2} \\
\geqslant f\left(\frac{x+(1-x)}{2}\right)=f\left(\frac{1}{2}\right)=4.
\end{array}
$$
Thus, $g(x) \geqslant 8$, with equality holding if and only if $x=\frac{1}{2}$.
Therefore, the minimum value of $g(x)$ is 8.
|
8
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $2016+3^{n}$ is a perfect square, then the positive integer $n=$ . $\qquad$
|
4. 2 .
Obviously, $2016+3^{n}$ is an odd perfect square.
So $2016+3^{n} \equiv 1(\bmod 8)$
$$
\Rightarrow 3^{n} \equiv 1(\bmod 8)
$$
$\Rightarrow n$ must be even.
Let $n=2+2 k(k \in \mathbf{N})$. Then
$$
2016+3^{n}=9\left(224+3^{2 k}\right) \text {. }
$$
So $224+3^{2 k}=224+\left(3^{k}\right)^{2}$ is a perfect square.
Let $224+\left(3^{k}\right)^{2}=t^{2}\left(t \in \mathbf{Z}_{+}\right)$. Then
$$
\left(t+3^{k}\right)\left(t-3^{k}\right)=224=2^{5} \times 7 \text {, }
$$
and $t+3^{k}>t-3^{k}$, both are even.
So $\left(t+3^{k}, t-3^{k}\right)$
$$
=\left(2^{4} \times 7,2\right),\left(2^{3} \times 7,2^{2}\right),\left(2^{2} \times 7,2^{3}\right),\left(2^{4}, 2 \times 7\right) \text {. }
$$
Solving gives $k=0$, i.e., $n=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}} \neq x \text {, }
$$
find all positive integers $a$ that meet the requirement.
|
$$
\text { Three, let } \sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}}=t \text {. }
$$
Then $t^{2} x+x+t^{2}-a=0$,
$$
(t+1) x^{2}+t-a=0 \text {. }
$$
$t \times$ (2) $-x \times$ (1) gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\frac{a-x^{2}}{1+x^{2}}=\frac{a-x}{1-x}$
$$
\Rightarrow 2 x^{3}-(1+a) x^{2}+(1-a) x=0 \text {. }
$$
Given that $x$ is a positive integer,
$$
2 x^{2}-(1+a) x+(1-a)=0 \text {. }
$$
Therefore, the discriminant of equation (1) must be a perfect square.
Let $(1+a)^{2}-8(1-a)=k^{2}(k \in \mathbf{N})$, i.e.,
$$
(a+5+k)(a+5-k)=32 \text {. }
$$
Since $a+5+k$ and $a+5-k$ are both even, and $a+5+k > a+5-k$, we have $(a+5+k, a+5-k)=(16,2)$ or $(8,4)$.
Solving gives $a=4$ or 1.
Substituting into equation (1) yields
$(a, x)=(4,3)$ or $(1,1)$.
Upon verification, only $(a, x)=(1,1)$ satisfies the given conditions, i.e., only $a=1$ meets the requirement.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$ satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$
|
2. 2 .
According to the problem, for any $(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking $(x, y)=(1,0),(0,1)$, we get the fixed point $B(a, b)$ satisfying the necessary conditions $\left\{\begin{array}{l}a \leqslant 1, \\ b \leqslant 1,\end{array}\right.$, which implies $a+b \leqslant 2$.
Thus, $(a+b)_{\text {max }} \leqslant 2$.
For the fixed point $B(1,1)$, for any $A(x, y) \in D$, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=(1,1) \cdot(x, y)=x+y \leqslant 1 \text {, }
$$
with equality holding at point $A\left(\frac{1}{2}, \frac{1}{2}\right)$.
Therefore, the point $B(1,1)$ satisfies the problem's conditions, and in this case, $a+b=2$.
Hence, $(a+b)_{\text {max }}=2$.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given the incenter $I(-1,7)$ of a right-angled triangle $\triangle O A B$ with all three vertices as integer points, and the origin $O$ as the right-angle vertex. The number of such right-angled triangles $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
8.2.
As shown in Figure 4.
Let $\angle x O A=\alpha, \angle x O I=\beta$.
Then $\alpha=\beta-\frac{\pi}{4}$.
Also, $\tan \beta=-7$, so
$$
\begin{array}{l}
\tan \alpha=\tan \left(\beta-\frac{\pi}{4}\right) \\
=\frac{\tan \beta-1}{1+\tan \beta}=\frac{4}{3}, \\
\tan \angle x O B=\tan \left(\frac{\pi}{2}+\alpha\right) \\
=-\cot \alpha=-\frac{1}{\tan \alpha}=-\frac{3}{4} .
\end{array}
$$
Then the point $A(3 t, 4 t), B(-4 k, 3 k)\left(k \backslash t \in \mathbf{Z}_{+}\right)$.
Thus, $O A=5 t, O B=5 k$, and the inradius of the right triangle $\triangle O A B$ is
$$
\begin{array}{l}
r=O I \sin 45^{\circ}=\frac{\sqrt{2}}{2} O I=5 . \\
\text { Also } r=\frac{1}{2}(O A+O B-A B) \\
\Rightarrow \frac{1}{2}\left[5 t+5 k-\sqrt{(5 t)^{2}+(5 k)^{2}}\right]=5 \\
\Rightarrow t+k-2=\sqrt{t^{2}+k^{2}} \\
\Rightarrow(t-2)(k-2)=2 .
\end{array}
$$
Since $t$ and $k$ are both positive integers, we have $(t, k)=(3,4)$ or $(4,3)$.
Therefore, there are two right triangles $\triangle O A B$ that satisfy the conditions.
|
2
|
Other
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given real numbers $a, b$ satisfy
$$
a+\lg a=10, b+10^{b}=10 \text {. }
$$
Then $\lg (a+b)=$ $\qquad$ .
|
8.1.
Since $\lg a=10-a, 10^{b}=10-b$, therefore, $a$ is the x-coordinate of the intersection point of $y=\lg x$ and $y=10-x$, and $b$ is the y-coordinate of the intersection point of $y=10^{x}$ and $y=10-x$.
Also, $y=\lg x$ and $y=10^{x}$ are symmetric about the line $y=x$, and $y=10-x$ is symmetric about the line $y=x$, so the two intersection points are symmetric about the line $y=x$.
Thus $10-b=a \Rightarrow \lg (a+b)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $6^{11}+C_{11}^{1} 6^{10}+C_{11}^{2} 6^{9}+\cdots+C_{11}^{10} 6-1$ when divided by 8 yields a remainder of $\qquad$ .
|
3.5.
Notice that,
$$
\begin{array}{l}
6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\mathrm{C}_{11}^{2} 6^{9}+\cdots+\mathrm{C}_{11}^{10} 6-1 \\
=\mathrm{C}_{10}^{0} 6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\cdots+\mathrm{C}_{11}^{10} 6+\mathrm{C}_{11}^{11} 6^{0}-2 \\
=(6+1)^{11}-2=7^{11}-2 \\
\equiv(-1)^{11}-2 \equiv 5(\bmod 8) .
\end{array}
$$
Therefore, the required result is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 2, point $A$ is on the positive $y$-axis, point $B$ is on the positive $x$-axis, $S_{\triangle A O B}=9$, segment $A B$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at points $C$ and $D$. If $C D=$ $\frac{1}{3} A B$, and $A C=B D$, then $k=$ . $\qquad$
|
8. 4 .
Let point $A\left(0, y_{A}\right), B\left(x_{B}, 0\right)$.
Given $C D=\frac{1}{3} A B, A C=B D$, we know that $C$ and $D$ are the trisection points of segment $A B$.
Thus, $x_{C}=\frac{1}{3} x_{B}, y_{C}=\frac{2}{3} y_{A}$.
Therefore, $k=x_{C} y_{C}=\frac{2}{9} x_{B} y_{A}=\frac{4}{9} S_{\triangle A O B}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If $4^{a}=6^{b}=9^{c}$, then
$$
\frac{1}{a}-\frac{2}{b}+\frac{1}{c}=
$$
$\qquad$
|
8. 0 .
Let $4^{a}=6^{b}=9^{c}=k$.
Then $a=\log _{k} 4, b=\log _{k} 6, c=\log _{k} 9$
$$
\begin{array}{l}
\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=\log _{k} 4-2 \log _{k} 6+\log _{k} 9 \\
\quad=\log _{k} 1=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. In rectangle $A B C D$, $A B=3, A D=4, P$ is a point on the plane of rectangle $A B C D$, satisfying $P A=2$, $P C=\sqrt{21}$. Then $\overrightarrow{P B} \cdot \overrightarrow{P D}=$
|
11.0.
As shown in Figure 3, let $A C$ and $B D$ intersect at point $E$, and connect $P E$. Then $E$ is the midpoint of $A C$ and $B D$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{P B} \cdot \overrightarrow{P D}=\frac{1}{4}\left[(\overrightarrow{P B}+\overrightarrow{P D})^{2}-(\overrightarrow{P B}-\overrightarrow{P D})^{2}\right] \\
=\frac{1}{4}\left[(2 \overrightarrow{P E})^{2}-(\overrightarrow{D B})^{2}\right] \\
=|\overrightarrow{P E}|^{2}-\frac{1}{4}|\overrightarrow{D B}|^{2} .
\end{array}
$$
Similarly, $\overrightarrow{P A} \cdot \overrightarrow{P C}=|\overrightarrow{P E}|^{2}-\frac{1}{4}|\overrightarrow{C A}|^{2}$.
Since $|\overrightarrow{D B}|=|\overrightarrow{C A}|$, we have,
$$
\overrightarrow{P B} \cdot \overrightarrow{P D}=\overrightarrow{P A} \cdot \overrightarrow{P C}
$$
Given $A C=B D=\sqrt{3^{2}+4^{2}}=5=\sqrt{P A^{2}+P C^{2}}$
$$
\Rightarrow \angle A P C=90^{\circ} \text {. }
$$
Therefore, $\overrightarrow{P B} \cdot \overrightarrow{P D}=\overrightarrow{P A} \cdot \overrightarrow{P C}=0$.
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that all positive integers are in $n$ sets, satisfying that when $|i-j|$ is a prime number, $i$ and $j$ belong to two different sets. Then the minimum value of $n$ is $\qquad$
|
6. 4 .
It is easy to see that $n \geqslant 4$ (2, 4, 7, 9 must be in four different sets).
Also, when $n=4$, the sets
$$
A_{i}=\left\{m \in \mathbf{Z}_{+} \mid m \equiv i(\bmod 4)\right\}(i=0,1,2,3)
$$
satisfy the condition.
Therefore, the minimum value of $n$ is 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that the ellipse $C$ passes through the point $M(1,2)$, with two foci at $(0, \pm \sqrt{6})$, and $O$ is the origin, a line $l$ parallel to $OM$ intersects the ellipse $C$ at points $A$ and $B$. Then the maximum value of the area of $\triangle OAB$ is $\qquad$
|
7. 2 .
According to the problem, let $l_{A B}: y=2 x+m$.
The ellipse equation is $\frac{y^{2}}{8}+\frac{x^{2}}{2}=1$. By solving the system and eliminating $y$, we get $16 x^{2}+8 m x+2\left(m^{2}-8\right)=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{8 m}{16}, x_{1} x_{2}=\frac{2\left(m^{2}-8\right)}{16}$.
Then $|A B|=\sqrt{5\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]}$
$$
=\frac{\sqrt{5\left(16-m^{2}\right)}}{2} \text {. }
$$
The distance from point $O$ to line $A B$ is $d=\frac{|m|}{\sqrt{5}}$, hence
$$
S_{\triangle O A B}=\frac{1}{2}|A B| d=\frac{\sqrt{m^{2}\left(16-m^{2}\right)}}{4} \leqslant 2 .
$$
When and only when $m= \pm 2 \sqrt{2}$, the area of $\triangle O A B$ reaches its maximum value of 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given positive integers $a, b, c, x, y, z$ satisfying $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$,
and $\left\{\begin{array}{l}2 a+b+4 c=4 x y z, \\ 2 x+y+4 z=4 a b c .\end{array}\right.$
Then the number of six-tuples $(a, b, c, x, y, z)$ that satisfy the conditions is $\qquad$ groups.
|
8. 0 .
Assume $x \geqslant a$. Then
$$
\begin{array}{l}
4 x y z=2 a+b+4 c \leqslant 7 a \leqslant 7 x \\
\Rightarrow y z \leqslant \frac{7}{4} \Rightarrow y z \leqslant 1 \Rightarrow(y, z)=(1,1) \\
\Rightarrow\left\{\begin{array}{l}
a a+b+4 c=4 x, \\
2 x+5=4 a b c
\end{array}\right. \\
\Rightarrow 2 a+b+4 c+10=8 a b c .
\end{array}
$$
If $b \geqslant 2$, then
$$
2 a+b+4 c+10 \leqslant 12 a<8 a b c,
$$
a contradiction.
If $b=1$, then $\left\{\begin{array}{l}2 a+5=4 x, \\ 2 x+5=4 a\end{array}\right.$ has no positive integer solutions.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given $0<a<1$, and satisfies
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]=$ $\qquad$
|
Solve:
$$
\begin{array}{l}
0<a<1 \\
\Rightarrow 0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2 \\
\Rightarrow\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right] \text { is either }
\end{array}
$$
0 or 1.
By the problem, we know that 18 of them are equal to 1, and 11 are equal to 0.
$$
\begin{array}{l}
\text { Therefore, }\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\left[a+\frac{11}{30}\right]=0, \\
{\left[a+\frac{12}{30}\right]=\left[a+\frac{13}{30}\right]=\cdots=\left[a+\frac{29}{30}\right]=1 .}
\end{array}
$$
Thus, $0<a+\frac{11}{30}<1,1 \leqslant a+\frac{12}{30}<2$.
Solving this, we get $18 \leqslant 30 a<1 \Rightarrow 6 \leqslant 10 a<\frac{19}{3}$.
Therefore, $[10 a]=6$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $O$ is the circumcenter of acute $\triangle A B C$, $\angle B A C$ $=60^{\circ}$, extend $C O$ to intersect $A B$ at point $D$, extend $B O$ to intersect $A C$ at point $E$. Then $\frac{B D}{C E}=$ $\qquad$
|
$=, 7.1$.
Connect $O A, D E$.
$$
\begin{array}{l}
\text { Since } \angle B O C=2 \angle B A C=120^{\circ} \\
\Rightarrow \angle C O E=60^{\circ}=\angle D A E \\
\Rightarrow A, D, O, E \text { are concyclic } \\
\Rightarrow \angle D E B=\angle D A O=\angle D B E \\
\Rightarrow D B=D E .
\end{array}
$$
Similarly, $C E=D E$.
Thus, $\frac{B D}{C E}=\frac{D E}{D E}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+2015(x-1)=-1, \\
(y-1)^{3}+2015(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$
|
4. 2 .
Notice that the function $f(z)=z^{3}+2015 z$ is a monotonically increasing function on $(-\infty,+\infty)$.
From the given condition, we have $f(x-1)=f(1-y)$.
Therefore, $x-1=1-y \Rightarrow x+y=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-y+1 \geqslant 0 \\
y+1 \geqslant 0 \\
x+y+1 \leqslant 0 .
\end{array}\right.
$$
Then the maximum value of $2 x-y$ is $\qquad$
|
8. 1 .
When $x=0, y=-1$, $2 x-y$ takes the maximum value 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given that $a$, $b$, and $c$ are three distinct real numbers. If in the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
any two of these equations have exactly one common root, find the value of $a^{2}+$ $b^{2}+c^{2}$.
|
Let the equations (1) and (3) have only one common root \( x_{1} \), equations (1) and (2) have only one common root \( x_{2} \), and equations (2) and (3) have only one common root \( x_{3} \). Therefore, the roots of equation (1) are \( x_{1} \) and \( x_{2} \), the roots of equation (2) are \( x_{2} \) and \( x_{3} \), and the roots of equation (3) are \( x_{3} \) and \( x_{1} \).
Since \( x_{1} \) is a root of equations (1) and (3), we have
\[
\left\{\begin{array}{l}
x_{1}^{2} + a x_{1} + b = 0, \\
x_{1}^{2} + c x_{1} + a = 0
\end{array} \Rightarrow x_{1} = \frac{a - b}{a - c} .\right.
\]
Similarly, \( x_{2} = \frac{b - c}{b - a} \) and \( x_{3} = \frac{c - a}{c - b} \).
Thus, \( x_{1} x_{2} x_{3} = -1 \).
Clearly, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are distinct. Otherwise, assume \( x_{2} = x_{3} \).
\[
\begin{array}{l}
\text{Hence } \frac{b - c}{b - a} = \frac{c - a}{c - b} \\
\Rightarrow a^{2} + b^{2} + c^{2} - b c - c a - a b = 0 \\
\Rightarrow (b - c)^{2} + (c - a)^{2} + (a - b)^{2} = 0 \\
\Rightarrow a = b = c,
\end{array}
\]
which contradicts the given conditions.
Thus, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are distinct.
By Vieta's formulas, we have
\[
\begin{array}{l}
a = -x_{1} - x_{2} = x_{3} x_{1}, \\
b = -x_{2} - x_{3} = x_{1} x_{2}, \\
c = -x_{3} - x_{1} = x_{2} x_{3}. \\
\text{From the above three equations, we get} \\
-2(x_{1} + x_{2} + x_{3}) = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2}, \text{ (4) } \\
x_{1}(1 + x_{3}) x_{2}(1 + x_{1}) x_{3}(1 + x_{2}) \\
= (-x_{2})(-x_{3})(-x_{1}). \\
\text{Noting that } x_{1} x_{2} x_{3} = -1. \\
\text{Thus } (1 + x_{3})(1 + x_{1})(1 + x_{2}) = -1 \\
\Rightarrow x_{1} + x_{2} + x_{3} + x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -1. \text{ (5) } \\
\text{From equations (4) and (5), we get} \\
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2
\end{array}\right. \\
\Rightarrow \left\{\begin{array}{l}
a + b + c = -2, \\
b c + c a + a b = -1 \\
a b c = 1
\end{array}\right., \\
\Rightarrow a^{2} + b^{2} + c^{2} \\
= (a + b + c)^{2} - 2(b c + c a + a b) \\
= (-2)^{2} - 2(-1) = 6. \\
\end{array}
\]
Let's explore further.
From the above solution, we also obtained
\[
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2 \\
x_{1} x_{2} x_{3} = -1.
\end{array}\right.
\]
Therefore, \( a \), \( b \), and \( c \) are the roots of the equation \( x^{3} + 2 x^{2} - x - 1 = 0 \); simultaneously, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are the roots of the equation \( x^{3} - x^{2} - 2 x + 1 = 0 \). If we substitute \( x \rightarrow -\frac{1}{x} \) in the equation \( x^{3} + 2 x^{2} - x - 1 = 0 \), we get the equation \( x^{3} - x^{2} - 2 x + 1 = 0 \). This is an interesting conclusion.
If we use conclusion (6), we can derive many symmetric expressions involving \( a \), \( b \), and \( c \). However, if we need to find the values of \( a^{2} b + b^{2} c + c^{2} a \) and \( a b^{2} + b c^{2} + c a^{2} \), the above solution is not straightforward. We will use a different approach to find these values.
Clearly, the six roots of these three equations must be the roots of the equation \( \left(x^{2} + a x + b\right)\left(x^{2} + b x + c\right)\left(x^{2} + c x + a\right) = 0 \),
which is \( x^{6} + (a + b + c) x^{5} + (a + b + c + b c + c a + a b) x^{4} + (a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c) x^{3} + (a^{2} b + b^{2} c + c^{2} a + b c + c a + a b) x^{2} + (a b^{2} + b c^{2} + c a^{2}) x + a b c = 0 \).
Let \( s_{1} = x_{1} + x_{2} + x_{3} \),
\( s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} \), \( s_{3} = x_{1} x_{2} x_{3} \). By the relationship between the roots and coefficients, and according to conclusion (6), we have
\[
\begin{array}{l}
\left\{\begin{array}{l}
s_{1} = x_{1} + x_{2} + x_{3} = 1, \\
s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2, \\
s_{3} = x_{1} x_{2} x_{3} = -1.
\end{array}\right. \\
\left\{\begin{array}{l}
a + b + c = -2 s_{1}, \\
a + b + c + b c + c a + a b = s_{1}^{2} + 2 s_{2}, \\
a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c \\
= -2 s_{1} s_{2} - 2 s_{3}, \\
a^{2} b + b^{2} c + c^{2} a + b c + c a + a b \\
= s_{2}^{2} + 2 s_{1} s_{3}, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3}, \\
a b c = s_{3}^{2}.
\end{array}\right.
\end{array}
\]
From equations (7) and (8), we get
\[
\begin{array}{l}
a^{2} b + b^{2} c + c^{2} a = s_{2}^{2} + 2 s_{1} s_{3} - (b c + c a + a b) \\
= (-2)^{2} + 2 \times 1(-1) - (-1) = 3, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3} \\
= -2(-2)(-1) = -4.
\end{array}
\]
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: A rope of length 2009 is operated as follows: first, it is divided into two ropes of positive integer lengths, and the lengths of the two ropes are recorded, then the above operation is repeated on one of the ropes, ... until 2009 ropes of length 1 are obtained. If the lengths of the two ropes obtained in a certain operation are not equal, then this operation is called "good".
(1) Find the maximum and minimum values of the number of good operations;
(2) Prove: In all operation processes where the number of good operations reaches the minimum value, the number of different lengths of ropes recorded is the same.
|
【Analysis】(1) It is easy to know that a rope of length 2 can only be divided into two segments of length 1, meaning the last operation on the rope must not be a good operation. A rope of length 2009 can be divided into 2009 segments of length 1 after exactly 2008 operations, so the number of good operations is no more than
$2008-1=2007$.
Therefore, the maximum number of good operations is 2007.
The following is such an operation: each time, cut a segment of length 1, then the first 2007 operations are all good operations.
Generalize the problem: For a rope of length $n$, let $f(n)$ be the minimum number of good operations.
By conducting mathematical experiments with simple positive integers $n$ and summarizing the experience, the following two lemmas can be discovered.
Lemma 1: $f(n)=0 \Leftrightarrow n=2^{k}$.
Lemma 2: Suppose $n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{1}}\left(k_{1}>k_{2}\right.$ $\left.>\cdots>k_{l} \geqslant 0\right)$. Let $l=S_{2}(n) \triangleq$ the sum of the digits of $n$ in binary. Then $f(n)=S_{2}(n)-1$.
Proof of Lemma 2: Since after $l-1$ good operations, we can obtain $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{1}}$, then
$f(n) \leqslant l-1$.
We will prove by mathematical induction:
$f(n) \geqslant S_{2}(n)-1$.
When $n=1$, it is Lemma 1, and the conclusion is obviously true.
Assume that for $n \leqslant k$, $f(n) \geqslant S_{2}(n)-1$.
When $n=k+1$, by the definition of $f$, there exists an operation process containing $f(k+1)$ good operations and several non-good operations, which cuts a rope of length $k+1$ into $k+1$ segments of length 1. If the first operation in the above process is a good operation, cutting a rope of length $k+1$ into segments of lengths $a$ and $b$, i.e., $k+1=a+b, a \neq b$, then
$$
\begin{array}{l}
f(k+1)=f(a)+f(b)+1 \\
\geqslant\left(S_{2}(a)-1\right)+\left(S_{2}(b)-1\right)+1 \\
=S_{2}(a)+S_{2}(b)-1 \\
\geqslant S_{2}(a+b)-1=S_{2}(k+1)-1,
\end{array}
$$
The equality holds if and only if there is no carry in the binary addition of $a+b$.
If the first operation in the above process is not a good operation, i.e., $k+1=2a$, then
$$
f(k+1)=2 f(a) \geqslant 2\left(S_{2}(a)-1\right) .
$$
Since $S_{2}(2a) \leqslant S_{2}(a)+S_{2}(a)-1$, the equality holds if and only if $S_{2}(a)=1$, i.e.,
$$
2 S_{2}(a)-2 \geqslant S_{2}(2a)-1 .
$$
Using this result, we get
$$
\begin{array}{l}
f(k+1) \geqslant 2\left(S_{2}(a)-1\right) \\
\geqslant S_{2}(2a)-1=S_{2}(k+1)-1,
\end{array}
$$
The conclusion holds, and the equality holds if and only if $S_{2}(a)=1$.
In summary, Lemma 2 is proved.
Therefore, using the binary representation of 2009, we get
$$
f(2009)=S_{2}(2009)-1=7 \text {. }
$$
(2) If the binary representation of $n$ is
$$
n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{i}} \text {, }
$$
we can set the set $M=\left\{k_{1}, k_{2}, \cdots, k_{l}\right\}$.
If $l \geqslant 2$, then in any operation process where the number of good operations reaches the minimum, the first operation is a good operation, i.e., there exists a non-empty proper subset $L$ of $M$ such that the rope of length $n$ is cut into segments of lengths $a$ and $b$, where $a=\sum_{k \in L} 2^{k}, b=\sum_{k \in M-L} 2^{k}$. Performing $l-1$ good operations in succession produces $l$ different lengths $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{i}}$.
Then, the non-good operations produce lengths $2^{k_{1}-1}, 2^{k_{1}-2}, \cdots, 2,2^{0}$, and so on. Therefore, the number of different lengths recorded is $l-1+k_{1}$.
In solving this problem, it is found through mathematical experiments that the minimum number of good operations $f(n)$ is related to the binary representation of the positive integer $n$, and a basic result about the sum of digits is used:
$$
S_{p}(a+b)=S_{p}(a)+S_{p}(b)-k(p-1),
$$
where $k$ is the number of carries in the addition of $a+b$ in base $p$.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In a lottery with 100000000 tickets, each ticket number consists of eight digits. A ticket number is called "lucky" if and only if the sum of its first four digits equals the sum of its last four digits. Then the sum of all lucky ticket numbers, when divided by 101, leaves a remainder of $\qquad$
|
7.0.
If the eight-digit number $x=\overline{a b c d e f g h}$ is lucky, then $y=99999999-x$ is also lucky, and $x \neq y$.
$$
\begin{array}{l}
\text { and } x+y=99999999=9999 \times 10001 \\
=99 \times 101 \times 10001,
\end{array}
$$
Therefore, the sum of all lucky numbers must be a multiple of 101.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the sets
$$
A=\left\{n^{2}+1 \mid n \in \mathbf{Z}_{+}\right\}, B=\left\{n^{3}+1 \mid n \in \mathbf{Z}_{+}\right\} \text {. }
$$
Arrange all elements in $A \cap B$ in ascending order to form the sequence $a_{1}, a_{2}, \cdots$. Then the units digit of $a_{99}$ is
|
2. 2 .
From the given, we know that $A \cap B=\left\{n^{6}+1 \mid n \in \mathbf{Z}_{+}\right\}$.
Therefore, $a_{99}=99^{6}+1$.
Thus, its unit digit is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$, respectively. Point $P$ is on the right branch of the hyperbola $C$, and the excenter of $\triangle P F_{1} F_{2}$ opposite to $\angle P F_{1} F_{2}$ is $I$. The line $P I$ intersects the $x$-axis at point $Q$. Then
$$
\frac{|P Q|}{|P I|}+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}=
$$
$\qquad$
|
6. 4 .
Since $I F_{1}$ is the angle bisector of $\angle P F_{1} Q$, we have
$$
\frac{|P Q|}{|P I|}=1+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|} \text {. }
$$
Let $P\left(x_{0}, y_{0}\right)$. Then, $\left|P F_{1}\right|=\frac{3}{2} x_{0}+2$.
By the optical property of the hyperbola, the tangent line to the hyperbola at point $P$ is perpendicular to line $P Q$.
Thus, $l_{P Q}: y=\frac{-4 y_{0}}{5 x_{0}}\left(x-x_{0}\right)+y_{0}$.
Let $y=0$, we get $Q\left(\frac{9}{4} x_{0}, 0\right)$.
Therefore, $\frac{|P Q|}{|P I|}+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}=1+2 \frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|}$
$$
=1+2 \times \frac{\frac{9}{4} x_{0}+3}{\frac{3}{2} x_{0}+2}=4
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. Answer Questions (Total 56 points)
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=4, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}+1(n \geqslant 4) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2016}}<3$.
|
(1) For $n \geqslant 4$, summing up we get
$$
a_{n}-a_{n-2}=a_{3}-a_{1}+n-3=n \text {. }
$$
When $n=2 m\left(m \in \mathbf{Z}_{+}\right)$,
$$
\begin{aligned}
a_{n} & =a_{2}+\sum_{k=1}^{m-1}\left(a_{2 k+2}-a_{2 k}\right) \\
& =2+\sum_{k=1}^{m-1}(2 k+2)=\frac{1}{4} n(n+2) ;
\end{aligned}
$$
When $n=2 m+1\left(m \in \mathbf{Z}_{+}\right)$,
$$
a_{n}=a_{1}+\sum_{k=1}^{m}\left(a_{2 k+1}-a_{2 k-1}\right)=\frac{1}{4}(n+1)^{2} \text {. }
$$
Thus, $a_{n}=\left\{\begin{array}{ll}\frac{1}{4}(n+1)^{2}, & n \text { is odd; } \\ \frac{1}{4} n(n+2), & n \text { is even. }\end{array}\right.$
(2) When $n$ is odd,
$$
\begin{array}{l}
\frac{1}{a_{n}}=\frac{4}{(n+1)^{2}}<\frac{4}{(n-1)(n+1)} \\
=2\left(\frac{1}{n-1}-\frac{1}{n+1}\right) ;
\end{array}
$$
When $n$ is even,
$$
\begin{array}{l}
\frac{1}{a_{n}}=\frac{4}{n(n+2)}=2\left(\frac{1}{n}-\frac{1}{n+2}\right) . \\
\text { Therefore, } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{2016}} \\
=\left(\frac{1}{a_{1}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{2015}}\right)+ \\
\quad\left(\frac{1}{a_{2}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{2016}}\right) \\
<\left[1+2\left(\frac{1}{2}-\frac{1}{2016}\right)\right]+2\left(\frac{1}{2}-\frac{1}{2018}\right)<3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Roll a die six times, let the number obtained on the $i$-th roll be $a_{i}$. If there exists a positive integer $k$, such that $\sum_{i=1}^{k} a_{i}=6$ has a probability $p=\frac{n}{m}$, where $m$ and $n$ are coprime positive integers. Then
$$
\log _{6} m-\log _{7} n=
$$
|
2. 1.
When $k=1$, the probability is $\frac{1}{6}$;
When $k=2$,
$$
6=1+5=2+4=3+3 \text {, }
$$
the probability is $5\left(\frac{1}{6}\right)^{2}$;
When $k=3$,
$$
6=1+1+4=1+2+3=2+2+2 \text {, }
$$
the probability is $(3+6+1)\left(\frac{1}{6}\right)^{3}=10\left(\frac{1}{6}\right)^{3}$;
When $k=4$,
$$
6=1+1+1+3=1+1+2+2 \text {, }
$$
the probability is $(4+6)\left(\frac{1}{6}\right)^{4}=10\left(\frac{1}{6}\right)^{4}$;
When $k=5$,
$$
6=1+1+1+1+2 \text {, }
$$
the probability is $5\left(\frac{1}{6}\right)^{5}$;
When $k=6$, the probability is $\left(\frac{1}{6}\right)^{6}$.
$$
\begin{array}{l}
\text { Hence } p=\frac{1}{6}+5\left(\frac{1}{6}\right)^{2}+10\left(\frac{1}{6}\right)^{3}+10\left(\frac{1}{6}\right)^{4}+ \\
\quad 5\left(\frac{1}{6}\right)^{5}+\left(\frac{1}{6}\right)^{6} \\
=\frac{1}{6}\left(1+\frac{1}{6}\right)^{5}=\frac{7^{5}}{6^{6}} \\
\Rightarrow n=7^{5}, m=6^{6} \\
\Rightarrow \log _{6} m-\log _{7} n=1 .
\end{array}
$$
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given positive real numbers $x, y, z$ satisfying
$$
\begin{array}{l}
x y + y z + z x \neq 1, \\
\frac{\left(x^{2}-1\right)\left(y^{2}-1\right)}{x y} + \frac{\left(y^{2}-1\right)\left(z^{2}-1\right)}{y z} + \\
\frac{\left(z^{2}-1\right)\left(x^{2}-1\right)}{z x} = 4 .
\end{array}
$$
(1) Find the value of $\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z z}$;
(2) Prove:
$$
\begin{array}{l}
9(x+y)(y+z)(z+x) \\
\geqslant 8 x y z(x y + y z + z x) .
\end{array}
$$
|
(1) From the given equation, we have
$$
\begin{array}{l}
z\left(x^{2}-1\right)\left(y^{2}-1\right)+x\left(y^{2}-1\right)\left(z^{2}-1\right)+ \\
y\left(z^{2}-1\right)\left(x^{2}-1\right)=4 x y z \\
\Rightarrow \quad x y z(x y+y z+z x)-(x+y+z)(x y+y z+z x)+ \\
\quad(x+y+z)-x y z=0 \\
\Rightarrow \quad[x y z-(x+y+z)](x y+y z+z x-1)=0 .
\end{array}
$$
Since \( x y + y z + z x \neq 1 \), then
\( x y z - (x + y + z) = 0 \)
$$
\begin{array}{r}
\Rightarrow x y z = x + y + z \\
\Rightarrow \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} = 1 .
\end{array}
$$
(2) Since \( x, y, z \) are positive numbers, we have
$$
\begin{array}{l}
9(x+y)(y+z)(z+x)-8 x y z(x y+y z+z x) \\
= 9(x+y)(y+z)(z+x)- \\
8(x+y+z)(x y+y z+z x) \\
= x^{2} y + x y^{2} + z^{2} x + z x^{2} + y^{2} z + y z^{2} - 6 x y z \\
= x(y-z)^{2} + y(z-x)^{2} + z(x-y)^{2} \geqslant 0 \\
\Rightarrow 9(x+y)(y+z)(z+x) \\
\quad \geqslant 8 x y z(x y+y z+z x)
\end{array}
$$
|
1
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) As shown in Figure 3, in isosceles $\triangle ABC$, $AB = AC = \sqrt{5}$, $D$ is a point on side $BC$ other than the midpoint, the symmetric point of $C$ with respect to line $AD$ is $E$, and the extension of $EB$ intersects the extension of $AD$ at point $F$. Find the value of $AD \cdot AF$.
|
Three, connect $A E$, $E D$, $C F$. From the given conditions, we know
$$
\angle A B C=\angle A C B=\angle A E D \text {. }
$$
Then $A$, $E$, $B$, $D$ are concyclic
$$
\Rightarrow \angle B E D=\angle B A D \text {. }
$$
Since points $C$, $E$ are symmetric with respect to line $A D$
$$
\begin{array}{l}
\Rightarrow \angle B E D=\angle B C F \\
\Rightarrow \angle B A D=\angle B C F \\
\Rightarrow A, B, F, C \text { are concyclic. } \\
\text { Also } A B=A C \\
\Rightarrow \angle A B D=\angle A C B=\angle A F B \\
\Rightarrow \triangle A B D \backsim \triangle A F B \Rightarrow \frac{A B}{A F}=\frac{A D}{A B} \\
\Rightarrow A D \cdot A F=A B^{2}=5 .
\end{array}
$$
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $AB$ is a diameter of the smallest circle with center $C(0,1)$ that has common points with the graph of the function $y=\frac{1}{|x|-1}$, and $O$ is the origin. Then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ $=$
|
4. -2 .
For any point $P(x, y)$ on the function $y=\frac{1}{|x|-1}(x>1)$, we have
$$
R^{2}=x^{2}+(y-1)^{2}=x^{2}+\left(\frac{1}{x-1}-1\right)^{2} \text {. }
$$
Let $t=x-1$. Then $t>0$,
$$
\begin{array}{l}
R^{2}=(t+1)^{2}+\left(\frac{1}{t}-1\right)^{2} \\
=t^{2}+\frac{1}{t^{2}}+2\left(t-\frac{1}{t}\right)+2 .
\end{array}
$$
Let $u=t-\frac{1}{t}$. Then $R=\sqrt{u^{2}+2 u+4}$.
Thus, when $u=-1$, i.e., $t=\frac{-1 \pm \sqrt{5}}{2}$, or $x=\frac{1 \pm \sqrt{5}}{2}$, the minimum value of $R$ is $\sqrt{3}$.
$$
\begin{aligned}
\text { Hence } \overrightarrow{O A} \cdot \overrightarrow{O B}=(\overrightarrow{O C}+\overrightarrow{C A}) \cdot(\overrightarrow{O C}+\overrightarrow{C B}) \\
=(\overrightarrow{O C}+\overrightarrow{C A}) \cdot(\overrightarrow{O C}-\overrightarrow{C A}) \\
=\overrightarrow{O C}^{2}-\overrightarrow{C A}^{2}=1-R^{2}=-2 .
\end{aligned}
$$
|
-2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the function $f(x)$ is an odd function with a period of 3, and when $x \in[0,1)$, $f(x)=3^{x}-1$, then $f\left(\log _{\frac{1}{3}} 54\right)=$ $\qquad$ .
|
2. -1 .
Notice that,
$$
\log _{\frac{1}{3}} 54=\log _{\frac{1}{3}} 27+\log _{\frac{1}{3}} 2=-3-\log _{3} 2 \text {, }
$$
and $f(x)$ has a period of 3.
Therefore, $f\left(\log _{\frac{1}{3}} 54\right)=f\left(-\log _{3} 2\right)$.
Furthermore, since $f(x)$ is an odd function and $\log _{3} 2 \in[0,1)$, we have
$$
\begin{array}{l}
f\left(\log _{\frac{1}{3}} 54\right)=f\left(-\log _{3} 2\right)=-f\left(\log _{3} 2\right) \\
=-\left(3^{\log _{3} 2}-1\right)=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $x, y>0$, and $x+2 y=2$. Then the minimum value of $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ is . $\qquad$
|
8. 2 .
$$
\begin{array}{l}
\text { Let } \boldsymbol{a}=\left(\sqrt{\frac{x^{2}}{2 y}}, \sqrt{\frac{4 y^{2}}{x}}\right), \boldsymbol{b}=(\sqrt{2 y}, \sqrt{x}) . \\
\text { Then }|\boldsymbol{a}|=\sqrt{\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}},|\boldsymbol{b}|=\sqrt{2 y+x}=\sqrt{2} \\
\boldsymbol{a} \cdot \boldsymbol{b}=x+2 y=2 .
\end{array}
$$
From $|\boldsymbol{a}||\boldsymbol{b}| \geqslant \boldsymbol{a} \cdot \boldsymbol{b} \Rightarrow \frac{x^{2}}{2 y}+\frac{4 y^{2}}{x} \geqslant 2$, equality holds if and only if $\boldsymbol{a}$ is in the same direction as $\boldsymbol{b}$.
Therefore, when $x=2 y=1$, $\frac{x^{2}}{2 y}+\frac{4 y^{2}}{x}$ achieves its minimum value of 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the curve $C_{1}: y=\sqrt{-x^{2}+10 x-9}$ and point $A(1,0)$. If there exist two distinct points $B$ and $C$ on curve $C_{1}$, whose distances to the line $l: 3 x+1=0$ are $|A B|$ and $|A C|$, respectively, then $|A B|+|A C|=$
|
4.8.
Let points $B\left(x_{B}, y_{B}\right), C\left(x_{C}, y_{C}\right)$, the locus of points whose distance to point $A(1,0)$ is equal to the distance to the line $l: x=-\frac{1}{3}$ is given by the equation $y^{2}=\frac{8}{3} x-\frac{8}{9}$.
Solving the system of equations
$$
\begin{array}{l}
\left\{\begin{array}{l}
y^{2}=\frac{8}{3} x-\frac{8}{9}, \\
x^{2}+y^{2}-10 x+9=0
\end{array}\right. \\
\Rightarrow x^{2}-\frac{22}{3} x+\frac{73}{9}=0 \\
\Rightarrow|A B|+|A C|=x_{B}+x_{C}+\frac{2}{3}=8 \text {. } \\
\end{array}
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If the function $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a sum of its maximum and minimum values equal to 4, then $a+b=$ $\qquad$
|
5.3.
Given $f(x)=\frac{a+\sin x}{2+\cos x}+b \tan x$ has a maximum or minimum value, we know $b=0$.
Then $y=\frac{a+\sin x}{2+\cos x}$
$$
\begin{array}{l}
\Rightarrow \sin x-y \cos x=a-2 y \\
\Rightarrow \sin (x+\alpha)=\frac{a-2 y}{\sqrt{1+y^{2}}} \\
\Rightarrow|a-2 y| \leqslant \sqrt{1+y^{2}} .
\end{array}
$$
By Vieta's formulas, $\frac{4 a}{3}=4 \Leftrightarrow a=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Connecting the intersection points of $x^{2}+y^{2}=10$ and $y=\frac{4}{x}$ in sequence, a convex quadrilateral is formed. The area of this quadrilateral is $\qquad$
|
2. 12 .
Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$ be the intersection point of the two curves in the first quadrant.
Since the two curves are symmetric with respect to the origin and the line $y=x$, the coordinates of the other three intersection points are
$$
B\left(y_{0}, x_{0}\right), C\left(-x_{0},-y_{0}\right), D\left(-y_{0},-x_{0}\right) \text {. }
$$
Thus, quadrilateral $A B C D$ is a rectangle, and its area is
$$
\begin{array}{l}
S=|A B||A D| \\
= \sqrt{2\left(x_{0}-y_{0}\right)^{2}} \sqrt{2\left(x_{0}+y_{0}\right)^{2}} \\
= 2 \sqrt{x_{0}^{2}+y_{0}^{2}-2 x_{0} y_{0}} \sqrt{x_{0}^{2}+y_{0}^{2}+2 x_{0} y_{0}} \\
=2 \sqrt{10-8} \sqrt{10+8}=12 .
\end{array}
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the complex coefficient equation with respect to $x$
$$
(1+2 \mathrm{i}) x^{2}+m x+1-2 \mathrm{i}=0
$$
has real roots, then the minimum value of the modulus of the complex number $m$ is $\qquad$
|
4.2.
Let $\alpha$ be a real root of the original equation, $m=p+q$ i.
$$
\begin{array}{l}
\text { Then }(1+2 \mathrm{i}) \alpha^{2}+(p+q \mathrm{i}) \alpha+1-2 \mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
\alpha^{2}+p \alpha+1=0, \\
2 \alpha^{2}+q \alpha-2=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
p=-\left(\alpha+\frac{1}{\alpha}\right), \\
q=-2\left(\alpha-\frac{1}{\alpha}\right)
\end{array}\right. \\
\Rightarrow|m|^{2}=p^{2}+q^{2} \\
=\left(\alpha+\frac{1}{\alpha}\right)^{2}+4\left(\alpha-\frac{1}{\alpha}\right)^{2} \\
=5\left(\alpha^{2}+\frac{1}{\alpha^{2}}\right)-6 \geqslant 4 \\
\Rightarrow|m| \geqslant 2 .
\end{array}
$$
When and only when $\alpha^{4}=1$, i.e., $\alpha= \pm 1$, the equality holds, at this time, $m= \pm 2$.
Therefore, the minimum value of $|m|$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Title: A quadruple of positive integers $(p, a, b, c)$ that satisfies the following conditions is called a "Leyden quadruple":
(i) $p$ is an odd prime;
(ii) $a, b, c$ are distinct;
(iii) $p \mid (ab + 1), p \mid (bc + 1), p \mid (ca + 1)$.
(1) Prove that for each Leyden quadruple $(p, a, b, c)$, we have $p + 2 \leq \frac{a + b + c}{3}$;
(2) If the Leyden quadruple $(p, a, b, c)$ satisfies
$$
p + 2 = \frac{a + b + c}{3},
$$
find all possible values of $p$. ${ }^{[1]}$
|
From the form, the Leiden quadruples that satisfy the conditions have a certain rotational symmetry. To facilitate expression and discovery of its inherent laws, it is extended as follows:
Definition If the $(k+1)$-tuple of positive integers $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$ satisfies the following three properties:
(i) $p$ is an odd prime;
(ii) $a_{1}, a_{2}, \cdots, a_{k}$ are distinct;
(iii) $p \mid\left(a_{1} a_{2}+1\right), p \mid\left(a_{2} a_{3}+1\right), \cdots, p \mid\left(a_{k-1} a_{k}+1\right), p \mid\left(a_{k} a_{1}+1\right)$.
Then the $(k+1)$-tuple of positive integers $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$ is called a "Leiden $(k+1)$-tuple".
Thus, after investigating the original problem, Proposition 1 is obtained.
Proposition 1 For each Leiden quadruple $\left(p, a_{1}, a_{2}, a_{3}\right)$,
we have $\frac{a_{1}+a_{2}+a_{3}}{3} \geqslant p+2$; equality holds if and only if $p=5$.
Proof of Proposition 1 Without loss of generality, assume $a_{1}p+k-1$.
Proof of Proposition 2 Without loss of generality, assume
$a_{1}2+k-3+p=p+k-1$.
Through the above calculations, it is found that the inequality in Proposition 2
is not "tight" enough. Therefore, the inequality on the right is strengthened and generalized, leading to the following series of corollaries.
Corollary 1 For each Leiden $(k+1)$-tuple $\left(p, a_{1}, a_{2}, \cdots, a_{k}\right)$, we have
$\frac{a_{1}+a_{2}+\cdots+a_{k}}{k} \geqslant 2+\frac{k-1}{2} p$,
with equality holding if and only if $p=5$.
It is easy to see that the proof of Corollary 1 is the same as that of Proposition 2. The value of $p$ for which the inequality holds with equality can be derived similarly to the proof of Proposition 1, and readers can complete it themselves.
Corollary 2 For each Leiden $(k+1)$-tuple $(p, a_{1}, a_{2}, \cdots, a_{k})$, we have
$\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}}{k}>\frac{(4+(k-1) p)^{2}}{4}$. (1)
Proof of Corollary 2 By the Cauchy-Schwarz inequality,
$\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}}{k}$
$\geqslant\left(\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}\right)^{2}$
$\geqslant\left(2+\frac{k-1}{2} p\right)^{2}$
Since $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, the equality in the inequality cannot be achieved.
Therefore, equation (1) is proved.
Corollary 3 For each Leiden $(k+1)$-tuple $(p, a_{1}, a_{2}, \cdots, a_{k})$ and $n \in \mathbf{Z}_{+}$, we have
$\frac{a_{1}^{n}+a_{2}^{n}+\cdots+a_{k}^{n}}{k}>\left(\frac{4+(k-1) p}{2 k}\right)^{n}$.
Proof of Corollary 3 By the Hölder inequality,
$\left(\sum_{i=1}^{k} a_{i}^{n}\right)^{\frac{1}{n}}\left(\sum_{i=1}^{k} 1^{\frac{n}{n-1}}\right)^{\frac{n-1}{n}} \geqslant \sum_{i=1}^{k} a_{i}$
$\Rightarrow a_{1}^{n}+a_{2}^{n}+\cdots+a_{k}^{n}$
Since $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, the equality in the inequality cannot be achieved.
Therefore, equation (2) is proved.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Given that $a$, $b$, and $c$ are three distinct real numbers. If the quadratic equations:
$$
\begin{array}{l}
x^{2} + a x + b = 0, \\
x^{2} + b x + c = 0, \\
x^{2} + c x + a = 0
\end{array}
$$
each have exactly one common root with any other, find the value of $a^{2} + b^{2} + c^{2}$.
(The 4th Chen Shengshen Cup National High School Mathematics Olympiad)
Reference [1] provides a detailed solution to this problem. Through further research, it is found that a simpler solution can also be given to find the values of
$$
a^{2} + b^{2} + c^{2}, \quad a^{2} b + b^{2} c + c^{2} a, \quad a b^{2} + b c^{2} + c a^{2}
$$
|
From [1] we know
$$
x_{1}=\frac{a-b}{a-c}, x_{2}=\frac{b-c}{b-a}, x_{3}=\frac{c-a}{c-b},
$$
where $x_{1}$ is the common root of equations (1) and (3), $x_{2}$ is the common root of equations (1) and (2), and $x_{3}$ is the common root of equations (2) and (3).
Since $x_{1}$ and $x_{2}$ are the two roots of equation (1),
$\Rightarrow x_{1} x_{2}=b=-\frac{b-c}{a-c} \Rightarrow \frac{c-a}{c-b}=-\frac{1}{b}=x_{3}$.
Similarly, $x_{1}=-\frac{1}{c}, x_{2}=-\frac{1}{a}$, and
$x_{1} x_{2}=\frac{1}{a c}=b \Rightarrow a b c=1$.
Since $x_{2}=-\frac{1}{a}$ is a root of equation (1),
$\Rightarrow \frac{1}{a^{2}}-1+b=0 \Rightarrow \frac{1}{a^{2}}=1-b$.
Similarly, $\frac{1}{b^{2}}=1-c, \frac{1}{c^{2}}=1-a$.
Thus, $\frac{1}{a^{2}} \cdot \frac{1}{b^{2}} \cdot \frac{1}{c^{2}}=(1-b)(1-c)(1-a)$.
Combining with equation (4) we get
$1=1-a-b-c+a b+b c+c a-a b c$
$\Rightarrow a b+b c+c a=1+a+b+c$.
Since $x_{2}=-\frac{1}{a}$ is a root of equation (2), we have
$\frac{1}{a^{2}}-\frac{b}{a}+c=0 \Rightarrow 1-a b+a^{2} c=0$.
Combining with equation (4) we get
$a b c-a b+a^{2} c=0(a \neq 0)$
$\Rightarrow b c-b+c a=0$.
Similarly, $a b-a+b c=0, c a-c+a b=0$.
Adding these three equations, we get
$2(a b+b c+c a)=a+b+c$.
Combining equations (5) and (6) we get
$a+b+c=-2, a b+b c+c a=-1$.
Then $a^{2}+b^{2}+c^{2}$
$=(a+b+c)^{2}-2(a b+b c+c a)=6$.
Since $x_{1}=-\frac{1}{c}$ is a root of equation (3),
$\Rightarrow \frac{1}{c^{2}}-1+a=0 \Rightarrow 1-c^{2}+c^{2} a=0$.
Similarly, $1-b^{2}+b^{2} c=0,1-a^{2}+a^{2} b=0$.
Adding these three equations, we get
$a^{2} b+b^{2} c+c^{2} a=a^{2}+b^{2}+c^{2}-3=3$.
Since $x_{3}=-\frac{1}{b}$ is a root of equation (3), we have
$$
\frac{1}{b^{2}}-\frac{c}{b}+a=0 \Rightarrow 1-b c+b^{2} a=0 .
$$
Similarly, $1-a b+a^{2} c=0,1-c a+c^{2} b=0$.
Adding these three equations, we get
$b^{2} a+a^{2} c+c^{2} b=a b+b c+c a-3=-4$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $\sqrt{24-t^{2}}-\sqrt{8-t^{2}}=2$, then
$$
\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=
$$
$\qquad$
|
II. 1.8.
From the known equation and using the formula $a+b=\frac{a^{2}-b^{2}}{a-b}$, we get $\sqrt{24-t^{2}}+\sqrt{8-t^{2}}=\frac{24-8}{2}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the set $M=\{(a, b) \mid a \leqslant-1, b \leqslant m\}$. If for any $(a, b) \in M$, it always holds that $a \cdot 2^{b}-b-3 a \geqslant 0$, then the maximum value of the real number $m$ is $\qquad$.
|
3. 1 .
Notice that,
$$
a \cdot 2^{b}-b-3 a \geqslant 0 \Leftrightarrow\left(2^{b}-3\right) a-b \geqslant 0
$$
holds for any $a \leqslant-1$.
$$
\text { Then }\left\{\begin{array}{l}
2^{b}-3 \leqslant 0, \\
2^{b}+b \leqslant 3
\end{array} \Rightarrow b \leqslant 1\right. \text {. }
$$
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If $P$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{P A}+\overrightarrow{P B}+\lambda \overrightarrow{P C}=\mathbf{0}, \angle C=120^{\circ}$. Then the value of the real number $\lambda$ is $\qquad$.
|
5. -1 .
Let the circumradius of $\triangle ABC$ be $R$.
From the given information, we have
$$
\begin{array}{l}
|\overrightarrow{P A}+\overrightarrow{P B}|^{2}=\lambda^{2}|\overrightarrow{P C}|^{2} \\
=|\overrightarrow{P A}|^{2}+|\overrightarrow{P B}|^{2}-2|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{C}{2} \\
=R^{2}+R^{2}-2 R^{2} \cdot \frac{1}{2}=R^{2}=\lambda^{2} R^{2} \\
\Rightarrow \lambda= \pm 1 .
\end{array}
$$
From the problem, we get $\lambda=-1$.
|
-1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $\alpha, \beta$ satisfy the equations respectively
$$
\begin{array}{l}
\alpha^{3}-3 \alpha^{2}+5 \alpha-4=0, \\
\beta^{3}-3 \beta^{2}+5 \beta-2=0
\end{array}
$$
then $\alpha+\beta=$ $\qquad$
|
9. 2 .
We have
$$
\begin{array}{l}
(\alpha-1)^{3}+2(\alpha-1)-1=0, \\
(1-\beta)^{3}+2(1-\beta)-1=0,
\end{array}
$$
which means $\alpha-1$ and $1-\beta$ are solutions to the equation $x^{3}+2 x-1=0$.
Since $x^{3}+2 x-1=0$ has only one solution, then
$$
\alpha-1=1-\beta \Rightarrow \alpha+\beta=2
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $n$ be a positive integer such that $\sqrt{3}$ lies between $\frac{n+3}{n}$ and $\frac{n+4}{n+1}$. Then $n=$ $\qquad$
|
3. 4 .
Notice that,
$$
\frac{n+4}{n+1}=1+\frac{3}{n+1}<1+\frac{3}{n}=\frac{n+3}{n} \text {. }
$$
Since $\sqrt{3}$ is an irrational number, then
$$
\begin{array}{l}
1+\frac{3}{n+1}<\sqrt{3}<1+\frac{3}{n} \\
\Rightarrow \frac{3}{n+1}<\sqrt{3}-1<\frac{3}{n} \\
\Rightarrow n<\frac{3}{\sqrt{3}-1}<n+1 .
\end{array}
$$
Thus, $n=\left[\frac{3}{\sqrt{3}-1}\right]=4$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. In the sequence $\left\{a_{n}\right\}$, $a_{4}=1, a_{11}=9$, and the sum of any three consecutive terms is 15. Then $a_{2016}=$
|
Ni, 9.5.
According to the problem, for any $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}+a_{n+2}=a_{n+1}+a_{n+2}+a_{n+3}=15 \\
\Rightarrow a_{n+3}=a_{n} .
\end{array}
$$
Then $a_{1}=a_{4}=1, a_{2}=a_{11}=9$,
$$
a_{3}=15-a_{1}-a_{2}=5 \text {. }
$$
Therefore, $a_{2016}=a_{3 \times 672}=a_{3}=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Let $x \in \mathbf{R}$. Then the function
$$
f(x)=|2 x-1|+|3 x-2|+|4 x-3|+|5 x-4|
$$
has a minimum value of $\qquad$
|
12. 1 .
Notice,
$$
\begin{array}{l}
f(x)=2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+ \\
4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right| \\
=2\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{4}{5}\right|\right)+ \\
\quad 3\left(\left|x-\frac{2}{3}\right|+\left|x-\frac{4}{5}\right|\right)+4\left|x-\frac{3}{4}\right| \\
\geqslant 2\left|\left(x-\frac{1}{2}\right)-\left(x-\frac{4}{5}\right)\right|+3\left|\left(x-\frac{2}{3}\right)-\left(x-\frac{4}{5}\right)\right| \\
=2\left|\frac{4}{5}-\frac{1}{2}\right|+3\left|\frac{4}{5}-\frac{2}{3}\right|=1 .
\end{array}
$$
When and only when
$$
\begin{array}{l}
\left(x-\frac{1}{2}\right)\left(x-\frac{4}{5}\right) \leqslant 0, \\
\left(x-\frac{2}{3}\right)\left(x-\frac{4}{5}\right) \leqslant 0, x-\frac{3}{4}=0,
\end{array}
$$
i.e., $x=\frac{3}{4}$, the equality holds.
Thus, $f(x)_{\text {min }}=f\left(\frac{3}{4}\right)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of zeros of the function $f(x)=x^{2} \ln x+x^{2}-2$ is . $\qquad$
|
3.1 .
From the condition, we have
$$
f^{\prime}(x)=2 x \ln x+x+2 x=x(2 \ln x+3) \text {. }
$$
When $0\mathrm{e}^{-\frac{3}{2}}$, $f^{\prime}(x)>0$.
Thus, $f(x)$ is a decreasing function on the interval $\left(0, \mathrm{e}^{-\frac{3}{2}}\right)$ and an increasing function on the interval $\left(\mathrm{e}^{-\frac{3}{2}},+\infty\right)$.
Also, $00 .
\end{array}
$$
Therefore, the number of zeros of the function $f(x)$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $z \in \mathbf{C}$. If the equation $x^{2}-2 z x+\frac{3}{4}+\mathrm{i}=0$ (where $\mathrm{i}$ is the imaginary unit) has real roots, then the minimum value of $|z|$ is $\qquad$ .
|
7.1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be a real root of the original equation.
$$
\begin{array}{l}
\text { Then } x_{0}^{2}-2(a+b \mathrm{i}) x_{0}+\frac{3}{4}+\mathrm{i}=0 \\
\Rightarrow\left\{\begin{array}{l}
x_{0}^{2}-2 a x_{0}+\frac{3}{4}=0, \\
-2 b x_{0}+1=0
\end{array}\right. \\
\Rightarrow x_{0}=\frac{1}{2 b}, a=\frac{3 b^{2}+1}{4 b} \\
\Rightarrow|z|^{2}=a^{2}+b^{2}=\left(\frac{3 b^{2}+1}{4 b}\right)^{2}+b^{2} \\
\quad=\frac{25}{16} b^{2}+\frac{1}{16 b^{2}}+\frac{3}{8} \\
\quad \geqslant \frac{5}{8}+\frac{3}{8}=1,
\end{array}
$$
When and only when $b= \pm \frac{\sqrt{5}}{5}$, the equality holds.
Thus, the minimum value of $|z|$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $f(x)$ is a periodic function on $\mathbf{R}$ with the smallest positive period of 2, and when $0 \leqslant x<2$, $f(x)=x^{3}-x$. Then the number of intersections between the graph of the function $y=f(x)$ and the $x$-axis in the interval $[0,6]$ is $\qquad$ .
|
2.7.
When $0 \leqslant x<2$, let $f(x)=x^{3}-x=0$, we get $x=0$ or 1.
According to the properties of periodic functions, given that the smallest period of $f(x)$ is 2, we know that $y=f(x)$ has six zeros in the interval $[0,6)$.
Also, $f(6)=f(3 \times 2)=f(0)=0$, so $f(x)$ has 7 intersection points with the $x$-axis in the interval $[0,6]$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If $f(x)=\sum_{k=0}^{4034} a_{k} x^{k}$ is the expansion of $\left(x^{2}+x+2\right)^{2017}$, then $\sum_{k=0}^{1344}\left(2 a_{3 k}-a_{3 k+1}-a_{3 k+2}\right)=$ $\qquad$
|
8. 2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$. Then
$$
\begin{array}{l}
x^{2}+x+2=1 \\
\Rightarrow \sum_{k=0}^{1344}\left(a_{3 k}+a_{3 k+1} \omega+a_{3 k+2} \omega^{2}\right)=1
\end{array}
$$
Taking the conjugate of the above equation, we get
$$
\sum_{k=0}^{1344}\left(a_{3 k}+a_{3 k+1} \omega^{2}+a_{3 k+2} \omega\right)=1 \text {. }
$$
Adding the two equations, we get
$$
\sum_{k=0}^{1344}\left(2 a_{3 k}-a_{3 k+1}-a_{3 k+2}\right)=2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given that $M N$ is a moving chord of the circumcircle of equilateral $\triangle A B C$ with side length $2 \sqrt{6}$, $M N=4$, and $P$ is a moving point on the sides of $\triangle A B C$. Then the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is
|
9.4.
Let the circumcenter of $\triangle ABC$ be $O$.
It is easy to find that the radius of $\odot O$ is $r=2 \sqrt{2}$.
Also, $MN=4$, so $\triangle OMN$ is an isosceles right triangle, and
$$
\begin{aligned}
& \overrightarrow{O M} \cdot \overrightarrow{O N}=0,|\overrightarrow{O M}+\overrightarrow{O N}|=4 . \\
& \text { Let }|\overrightarrow{O P}|=x . \\
& \text { Then } \sqrt{2} \leqslant x \leqslant 2 \sqrt{2} \text {, and } \\
& \overrightarrow{M P} \cdot \overrightarrow{P N}=(\overrightarrow{O P}-\overrightarrow{O M}) \cdot(\overrightarrow{O N}-\overrightarrow{O P}) \\
& =-\overrightarrow{O P}{ }^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P}-\overrightarrow{O M} \cdot \overrightarrow{O N} \\
= & -x^{2}+(\overrightarrow{O M}+\overrightarrow{O N}) \cdot \overrightarrow{O P} \\
\leqslant & -x^{2}+|\overrightarrow{O M}+\overrightarrow{O N}||\overrightarrow{O P}| \\
= & -x^{2}+4 x=4-(x-2)^{2} \leqslant 4,
\end{aligned}
$$
The equality holds if and only if $x=2$ and $\overrightarrow{O P}$ is in the same direction as $\overrightarrow{O M}+\overrightarrow{O N}$.
Therefore, the maximum value of $\overrightarrow{M P} \cdot \overrightarrow{P N}$ is 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. As shown in Figure 1, in $\triangle A B C$,
$$
\begin{array}{l}
\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}, \\
\overrightarrow{A H} \cdot \overrightarrow{B C}=0, \\
\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 .
\end{array}
$$
Then the eccentricity of the hyperbola passing through point $C$ and having $A$ and $H$ as its foci is . $\qquad$
|
9. 2 .
Given $\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow(\overrightarrow{C B}-\overrightarrow{C A}) \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow A C=B C$.
From $\overrightarrow{A H} \cdot \overrightarrow{B C}=0 \Rightarrow A H \perp B C$.
Since $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}$, therefore,
$\sin \frac{C}{2}=\frac{\sqrt{5}}{5}, \tan \frac{C}{2}=\frac{1}{2}$.
Then $\tan C=\frac{2 \tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}}=\frac{4}{3}$.
In the right triangle $\triangle A C H$, let's assume $C H=3$.
Then $A H=4, A C=B C=\sqrt{A H^{2}+C H^{2}}=5$.
Therefore, the eccentricity of the hyperbola with foci at $A$ and $H$ is
$$
e=\frac{A H}{A C-C H}=2 .
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure $2, P$ is a point on the incircle of square $A B C D$, and let $\angle A P C=\alpha$, $\angle B P D=\beta$. Then $\tan ^{2} \alpha+\tan ^{2} \beta$ $=$
|
4. 8 .
As shown in Figure 5, establish a Cartesian coordinate system.
Let the equation of the circle be $x^{2}+y^{2}=r^{2}$.
Then the coordinates of the vertices of the square are
$$
A(-r,-r), B(r,-r), C(r, r), D(-r, r) \text {. }
$$
If $P(r \cos \theta, r \sin \theta)$, then the slopes of the lines $P A, P B, P C$, and $P D$ are respectively
$$
\begin{array}{l}
k_{P A}=\frac{1+\sin \theta}{1+\cos \theta}, k_{P B}=-\frac{1+\sin \theta}{1-\cos \theta}, \\
k_{P C}=\frac{1-\sin \theta}{1-\cos \theta}, k_{P D}=-\frac{1-\sin \theta}{1+\cos \theta} \text {. } \\
\text { Therefore, } \tan ^{2} \alpha=\left(\frac{k_{P C}-k_{P A}}{1+k_{P A} k_{P C}}\right)^{2}=4(\cos \theta-\sin \theta)^{2} \text {, } \\
\tan ^{2} \beta=\left(\frac{k_{P D}-k_{P B}}{1+k_{P B} k_{P D}}\right)^{2}=4(\cos \theta+\sin \theta)^{2} \text {. } \\
\end{array}
$$
Therefore, $\tan ^{2} \alpha+\tan ^{2} \beta=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { Find the largest positive integer } n, \text { such that for positive real } \\
\text { numbers } \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}, \text { we have } \\
\quad \sum_{i=1}^{n} \frac{\alpha_{i}^{2}-\alpha_{i} \alpha_{i+1}}{\alpha_{i}^{2}+\alpha_{i+1}^{2}} \geqslant 0\left(\alpha_{n+1}=\alpha_{1}\right) .
\end{array}
$$
|
Let $a_{i}=\frac{\alpha_{i+1}}{\alpha_{i}}(i=1,2, \cdots, n)$.
Then $\prod_{i=1}^{n} a_{i}=1$, and $a_{i}>0$.
The inequality to be proved becomes
$\sum_{i=1}^{n} \frac{1-a_{i}}{1+a_{i}^{2}} \geqslant 0$.
Let $x_{i}=\ln a_{i}(i=1,2, \cdots, n)$.
Then $\sum_{i=1}^{n} \frac{1-\mathrm{e}^{x_{i}}}{1+\left(\mathrm{e}^{x_{i}}\right)^{2}} \geqslant 0$, and $\sum_{i=1}^{n} x_{i}=0$.
Let $f(x)=\frac{1-\mathrm{e}^{x}}{1+\left(\mathrm{e}^{x}\right)^{2}}$.
Differentiation shows that $f(x)$ is decreasing on the interval $(-\infty, \ln (1+\sqrt{2}))$ and increasing on the interval $(\ln (1+\sqrt{2}),+\infty)$.
(1) If there exists $x_{i}x>y$ such that $f^{\prime \prime}(x)>0$.
(2) If there exist $x_{i} 、 x_{j}2$ such that $a_{2}^{8}>2 a_{2}^{7}$.
Hence, inequality (2) $>0$, holds.
(ii) When $0 \leqslant a_{2} \leqslant 2$, $2 a_{2}^{5} \geqslant a_{2}^{6}$.
Hence, inequality (2) $\geqslant a_{2}^{6}\left(a_{2}-1\right)^{2} \geqslant 0$, holds.
(iii) When $a_{2}<0$,
$$
\begin{array}{l}
\text { Inequality (2) }=a_{2}^{8}-2 a_{2}^{7}+2 a_{2}^{5}+a_{2}^{4}+a_{2}^{2}+ \\
2 a_{2}^{2}\left(a_{2}+1\right)^{2}+\left(a_{2}+2\right)^{2} \\
\geqslant a_{2}^{8}-2 a_{2}^{7}+2 a_{2}^{5}+a_{2}^{4}+a_{2}^{2} \\
=a_{2}^{2}\left(a_{2}^{3}+1\right)^{2}-2 a_{2}^{7}+a_{2}^{4} \geqslant 0, \\
\end{array}
$$
also holds.
Therefore, when $n=5$, the original inequality holds.
Thus, the maximum value of $n$ is 5.
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $f(x)=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$, then the maximum value of $f(x)$ is $\qquad$
|
$-、 1.11$.
From the fact that $f(x)$ is defined, we know
$$
0 \leqslant x \leqslant 13 \text {. }
$$
Then $\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$$
\begin{array}{l}
=\sqrt{\left(6 \sqrt{\frac{x+27}{36}}+2 \sqrt{\frac{13-x}{4}}+3 \sqrt{\frac{x}{9}}\right)^{2}} \\
\leqslant \sqrt{(6+2+3)\left(6 \times \frac{x+27}{36}+2 \times \frac{13-x}{4}+3 \times \frac{x}{9}\right)} \\
=11 .
\end{array}
$$
When $x=9$, the equality holds.
Therefore, the maximum value of $f(x)$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $a \in \mathbf{R}$, the equation ||$x-a|-a|=2$ has exactly three distinct roots. Then $a=$ $\qquad$
|
11. 2 .
The original equation can be transformed into $|x-a|=a \pm 2$.
For the equation to have exactly three distinct roots, then $a=2$. At this point, the equation has exactly three distinct roots:
$$
x_{1}=2, x_{2}=6, x_{3}=-2 \text {. }
$$
Thus, $a=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given that $a$, $b$, and $c$ are distinct integers. Then
$$
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2}
$$
the minimum value is $\qquad$
|
15.8.
Notice,
$$
\begin{array}{l}
4\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\
=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}+a^{2}+b^{2}+c^{2},
\end{array}
$$
its minimum value is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If three distinct real numbers $a$, $b$, and $c$ satisfy
$$
a^{3}+b^{3}+c^{3}=3 a b c \text {, }
$$
then $a+b+c=$ $\qquad$
|
2.0.
Notice,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\
=\frac{1}{2}(a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right) \\
=0 .
\end{array}
$$
Since $a, b, c$ are not all equal, we have
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \neq 0 \text {. }
$$
Therefore, $a+b+c=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a$ and $b$ are real numbers. If the quadratic function
$$
f(x)=x^{2}+a x+b
$$
satisfies $f(f(0))=f(f(1))=0$, and $f(0) \neq f(1)$, then the value of $f(2)$ is $\qquad$.
|
3. 3 .
It is known that $f(0)=b, f(1)=1+a+b$ are both roots of the equation $f(x)=0$.
$$
\begin{array}{l}
\text { Then } x^{2}+a x+b \equiv(x-b)(x-(1+a+b)) \\
\Rightarrow a=-1-a-2 b, b=b(1+a+b) \\
\Rightarrow a=-\frac{1}{2}, b=0 \\
\Rightarrow f(2)=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Prove: The function
$$
f(x)=x^{x+2}-(x+2)^{x}-2 x(x+1)(x+2)+2
$$
has only one integer zero in the interval $[0,+\infty)$
|
9. In fact, it is only necessary to prove that there is only one positive integer satisfying $f(x)=0$.
When $x \in \mathbf{Z}_{+}, x \geqslant 5$,
$$
\begin{array}{l}
(x+3)^{x}=\left(\frac{x+3}{2}\right)^{4}(x+3)^{x-4} 4^{2} \\
& (x+3)^{x}-(x-2)^{x}-2 x(x+1)(x+2)+2 \\
> & (x+3)(x+2)^{x-1}-(x+2)^{x}- \\
& 2 x(x+1)(x+2)+2 \\
= & (x+2)^{x-1}-2 x(x+1)(x+2)+2 \\
\geqslant & (x+2)^{4}-2 x(x+1)(x+2)+2 \\
= & x^{4}+6 x^{3}+18 x^{2}+28 x+18>0 .
\end{array}
$$
By taking $x=1,2,3,4$ in sequence, it is found that $f(x)=0$ if and only if $x=3$.
Therefore, this function has only one positive integer zero.
|
3
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
13. In $\triangle A B C$, $\angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. Let
$$
\begin{array}{l}
f(x)=\boldsymbol{m} \cdot \boldsymbol{n}, \boldsymbol{m}=(2 \cos x, 1), \\
\boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x), \\
f(A)=2, b=1, S_{\triangle A B C}=\frac{\sqrt{3}}{2} . \\
\text { Then } \frac{b+c}{\sin B+\sin C}=
\end{array}
$$
|
$=13.2$.
It is easy to know, $f(x)=1+2 \sin \left(2 x+\frac{\pi}{6}\right)$.
Combining the conditions, we get $\angle A=\frac{\pi}{3}, c=2, a=\sqrt{3}$. Therefore, $\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}+a_{n+1}=n(-1)^{\frac{a(a+1)}{2}} \text {, }
$$
the sum of the first $n$ terms is $S_{n}, m+S_{2015}=-1007, a_{1} m>0$. Then the minimum value of $\frac{1}{a_{1}}+\frac{4}{m}$ is $\qquad$ .
|
15.9.
$$
\begin{array}{l}
\text { Given } S_{2015}=a_{1}+\sum_{k=1}^{1007}\left(a_{2 k}+a_{2 k+1}\right) \\
=a_{1}-1008 \\
\Rightarrow m+a_{1}-1008=-1007 \\
\Rightarrow m+a_{1}=1 . \\
\text { Also, } m a_{1}>0 \text {, so } m>0, a_{1}>0 . \\
\text { Therefore, } \frac{1}{a_{1}}+\frac{4}{m}=\left(m+a_{1}\right)\left(\frac{1}{a_{1}}+\frac{4}{m}\right) \\
=5+\frac{m}{a_{1}}+\frac{4 a_{1}}{m} \geqslant 9 .
\end{array}
$$
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The equation
$$
\sqrt[3]{(x+7)(x+8)}-\sqrt[3]{(x+5)(x+10)}=2
$$
has $\qquad$ real solutions $x$ that are not equal.
|
2.4.
Let $a=\sqrt[3]{(x+7)(x+8)}$,
$$
b=\sqrt[3]{(x+5)(x+10)} \text {. }
$$
Then $a-b=2, a^{3}-b^{3}=6$.
Eliminating $a$ gives
$$
\begin{array}{l}
3 b^{2}+6 b+1=0 \\
\Rightarrow b_{1}=\frac{-3+\sqrt{6}}{3}, b_{2}=\frac{-3-\sqrt{6}}{3} .
\end{array}
$$
If $b_{1}=\frac{-3+\sqrt{6}}{3}$, then
$$
x^{2}+15 x+50+\left(\frac{3-\sqrt{6}}{3}\right)^{3}=0 \text {. }
$$
By $\Delta_{1}=15^{2}-4\left(50+\left(\frac{3-\sqrt{6}}{3}\right)^{3}\right)>0$, we know the original
equation has two distinct real roots $x_{1} 、 x_{2}$.
If $b_{2}=\frac{-3-\sqrt{6}}{3}$, then
$$
x^{2}+15 x+50+\left(\frac{3+\sqrt{6}}{3}\right)^{3}=0 \text {. }
$$
By $\Delta_{2}=15^{2}-4\left(50+\left(\frac{3+\sqrt{6}}{3}\right)^{3}\right)>0$, we know the original equation has two distinct real roots $x_{3} 、 x_{4}$.
Since $x_{1} 、 x_{2} 、 x_{3} 、 x_{4}$ are all distinct, the number of distinct real solutions $x$ of the original equation is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the complex number $z$ satisfies
$$
(a-2) z^{2018}+a z^{2017} \mathrm{i}+a z \mathrm{i}+2-a=0 \text {, }
$$
where, $a<1, \mathrm{i}=\sqrt{-1}$. Then $|z|=$ $\qquad$
|
6. 1 .
Notice,
$$
z^{2017}((a-2) z+a \mathrm{i})=a-2-a z \mathrm{i} \text {. }
$$
Thus, $|z|^{2017}|(a-2) z+a \mathrm{i}|=|a-2-a z \mathrm{i}|$.
Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
$$
\begin{array}{l}
\text { Then }|(a-2) z+a \mathrm{i}|^{2}-|a-2-a z \mathrm{i}|^{2} \\
=|(a-2) x+((a-2) y+a) \mathrm{i}|^{2}- \\
|a-2+y a-a x \mathrm{i}|^{2} \\
=((a-2) x)^{2}+((a-2) y+a)^{2}- \\
(a-2+y a)^{2}-(-a x)^{2} \\
= 4(1-a)\left(x^{2}+y^{2}-1\right) \\
= 4(1-a)\left(|z|^{2}-1\right) .
\end{array}
$$
If $|z|>1$, then
$$
\begin{array}{l}
|a-2-a z \mathrm{i}||a-2-a z \mathrm{i}|,
\end{array}
$$
which contradicts equation (1).
If $|z|<1$, it would also lead to a contradiction.
Therefore, $|z|=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 On the plane, there are $n(n \geqslant 5)$ distinct points, each point is exactly at a distance of 1 from four other points. Find the minimum value of such $n$. [2]
|
Keep points $A, B, D, E, G$ in Figure 1, construct $\square E A B C, \square D A G F, \square B A G I$ to get points $C, F, I$, and construct $\square C I F H$ to get point $H$.
$$
\begin{array}{l}
\text { From } B I=A G=A E=B C, \\
\angle I B C=\angle A B C-\angle A B I \\
=\left(180^{\circ}-\angle B A E\right)-\left(180^{\circ}-\angle B A G\right) \\
=\angle E A G=60^{\circ},
\end{array}
$$
we know that $\triangle B C I$ is an equilateral triangle.
Similarly, $\triangle F G I$ is an equilateral triangle.
Thus, $C E=A B=G I=F I=C H$.
And $\angle E C H=\angle G I F=60^{\circ}$, so $\triangle C E H$ is an equilateral triangle.
Similarly, $\triangle D F H$ is an equilateral triangle.
Therefore, the points that are 1 unit away from point $A$ are $B, E, D, G$; the points that are 1 unit away from point $B$ are $A, D, I, C$; the points that are 1 unit away from point $C$ are $B, E, I, H$; the points that are 1 unit away from point $D$ are $B, A, F, H$; the points that are 1 unit away from point $E$ are $A, G, H, C$; the points that are 1 unit away from point $F$ are $G, D, I, H$; the points that are 1 unit away from point $G$ are $A, E, I, F$; the points that are 1 unit away from point $H$ are $C, E, D, F$; the points that are 1 unit away from point $I$ are $B, C, F, G$. This is an example of $n=9$, as shown in Figure 2.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Find all positive integers $n$, such that all positive divisors of $n$ can be placed in the cells of a rectangular grid, satisfying the following constraints:
(1) Each cell contains a different divisor;
(2) The sum of the numbers in each row of cells is equal;
(3) The sum of the numbers in each column of cells is equal.
|
2. $n=1$.
Assume all positive divisors of $n$ can be placed in a $k \times l$ $(k \leqslant l)$ rectangular grid, and satisfy the conditions.
Let the sum of the numbers in each column of the grid be $s$.
Since $n$ is one of the numbers in a column of the grid, we have $s \geqslant n$, and the equality holds if and only if $n=1$.
For $j=1,2, \cdots, l$, let $d_{j}$ be the largest number in the $j$-th column of the grid. Without loss of generality, assume
$d_{1}>d_{2}>\cdots>d_{l}$.
Since $d_{1}, d_{2}, \cdots, d_{l}$ are all positive divisors of $n$, then
$d_{l} \leqslant \frac{n}{l}$.
And $d_{l}$ is the largest number in the $l$-th column of the grid, so
$d_{l} \geqslant \frac{s}{k} \geqslant \frac{n}{k}$.
From equations (1) and (2), we get $\frac{n}{l} \geqslant \frac{n}{k} \Rightarrow k \geqslant l$.
Thus, $k=l$.
Therefore, the equalities in equations (1) and (2) both hold.
In particular, we have $s=n$.
Hence, $n=1$, and $n=1$ satisfies the conditions.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are $n(n \geqslant 2)$ cards, each with a real number written on it, and these $n$ numbers are all distinct. Now, these cards are arbitrarily divided into two piles (each pile has at least one card). It is always possible to take one card from the first pile and place it in the second pile, and then take one card from the second pile and place it in the first pile (it can be the same card), such that the sum of the numbers on the cards in each pile is equal to 0. Find the maximum possible value of $n$.
|
6. The maximum possible value of $n$ is 7.
If given seven cards, each written with $0, \pm 1, \pm 2, \pm 3$, it is easy to verify that they meet the requirements.
Below is the proof that the number of cards cannot be more.
Take any one card as the first pile, and the remaining cards as the second pile. After the operation, the first pile will have only one card. Therefore, this card must be written with 0.
For a card written with a non-zero real number $a$, place it and the card written with 0 in the first pile, and the remaining cards in the second pile. After the operation, the first pile will have exactly two cards. Therefore, it is only possible to replace the card written with 0 with a card written with $-a$.
Thus, the real numbers on all cards must be 0 and several pairs of opposite numbers.
Assume the positive real numbers written on the cards are
$$
a_{1}<a_{2}<\cdots<a_{n} \text {. }
$$
If $n \geqslant 4$, first divide the cards written with $a_{1}, a_{2}, \cdots, a_{n}$ into the first pile, and the remaining cards into the second pile. Note that, after the operation, the first pile can only exchange one card. Therefore, it is only possible to replace the card written with $a_{n}$ with a card written with $-a_{n}$ (any other exchange will result in a final sum greater than 0). Thus,
$$
a_{n}=a_{1}+a_{2}+\cdots+a_{n-1} \text {. }
$$
Next, divide the cards written with $a_{2}, a_{3}, \cdots, a_{n}$ into the first pile, and the remaining cards into the second pile. Note that, if the first pile does not replace the card written with $a_{n}$ with a card written with $-a_{n}$, the sum will definitely be greater than 0. If the first pile replaces the card written with $a_{n}$ with a card written with $-a_{n}$, the sum will definitely be less than 0. Neither satisfies the requirement.
Therefore, $n \leqslant 3$, meaning the total number of cards does not exceed 7.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $k$ be a positive integer. Suppose that all positive integers can be colored using $k$ colors, and there exists a function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying:
(1) For any positive integers $m, n$ of the same color (allowing $m = n$), we have $f(m+n)=f(m)+f(n)$;
(2) There exist positive integers $m, n$ (allowing $m = n$) such that $f(m+n) \neq f(m)+f(n)$.
Find the minimum value of $k$.
|
2. The minimum value of $k$ is 3.
First, construct an example for $k=3$.
Let $f(n)=\left\{\begin{array}{ll}2 n, & n \equiv 0(\bmod 3) ; \\ n, & n \equiv 1,2(\bmod 3)\end{array}\right.$
Then $f(1)+f(2)=3 \neq f(3)$, satisfying condition (2).
At the same time, color the numbers that are congruent to $0, 1, 2 \pmod{3}$ with three different colors, respectively. Thus,
(i) For any $x \equiv y \equiv 0(\bmod 3)$, we have
$x+y \equiv 0(\bmod 3)$
$\Rightarrow f(x+y)=\frac{x+y}{3}=f(x)+f(y)$;
(ii) For any $x \equiv y \equiv 1(\bmod 3)$, we have
$x+y \equiv 2(\bmod 3)$
$\Rightarrow f(x+y)=x+y=f(x)+f(y)$;
(iii) For any $x \equiv y \equiv 2(\bmod 3)$, we have
$x+y \equiv 1(\bmod 3)$
$\Rightarrow f(x+y)=x+y=f(x)+f(y)$.
Thus, condition (1) is also satisfied.
Therefore, $k=3$ meets the requirements.
Next, prove that $k=2$ does not hold.
It suffices to prove that for any function $f$ and coloring scheme satisfying condition (1) when $k=2$, we have
$f(n)=n f(1)$ (for any $n \in \mathbf{Z}_{+}$),
which contradicts condition (2).
In condition (1), take $m=n$, then
$f(2 n)=2 f(n)$ (for any $n \in \mathbf{Z}_{+}$).
Next, prove:
$f(3 n)=3 f(n)$ (for any $n \in \mathbf{Z}_{+}$).
For any positive integer $n$, by equation (2) we know
$f(2 n)=2 f(n), f(4 n)=4 f(n)$,
$f(6 n)=2 f(3 n)$.
If $n$ and $2 n$ are the same color, then
$f(3 n)=f(2 n)+f(n)=3 f(n)$,
equation (3) holds;
If $2 n$ and $4 n$ are the same color, then
$$
\begin{array}{l}
f(3 n)=\frac{1}{2} f(6 n)=\frac{1}{2}(f(4 n)+f(2 n)) \\
=3 f(n),
\end{array}
$$
equation (3) also holds.
Otherwise, $2 n$ is a different color from both $n$ and $4 n$, so $n$ and $4 n$ are the same color. In this case, if $n$ and $3 n$ are the same color, then
$$
f(3 n)=f(4 n)-f(n)=3 f(n),
$$
equation (3) holds;
If $n$ and $3 n$ are different colors, then $2 n$ and $3 n$ are the same color,
$$
f(3 n)=f(4 n)+f(n)-f(2 n)=3 f(n) \text {, }
$$
equation (3) also holds.
Thus, equation (3) is proven.
Assume the proposition (1) does not hold. Then there exists a positive integer $m$ such that $f(m) \neq m f(1)$.
Without loss of generality, take the smallest $m$, then by equations (2) and (3), $m \geqslant 5$, and $m$ is odd. Otherwise, by the minimality of $m$,
$f\left(\frac{m}{2}\right)=\frac{m}{2} f(1)$.
Thus, $f(m)=2 f\left(\frac{m}{2}\right)=m f(1)$, a contradiction.
Consider $\frac{m-3}{2}<\frac{m+3}{2}<m$ these three numbers.
Similarly, by the minimality of $m$,
$f\left(\frac{m-3}{2}\right)=\frac{m-3}{2} f(1)$,
$f\left(\frac{m+3}{2}\right)=\frac{m+3}{2} f(1)$.
Thus, $\frac{m-3}{2}$ and $\frac{m+3}{2}$ are different colors. Otherwise,
$f(m)=f\left(\frac{m-3}{2}\right)+f\left(\frac{m+3}{2}\right)=m f(1)$,
a contradiction.
Therefore, $m$ is exactly the same color as one of $\frac{m-3}{2}, \frac{m+3}{2}$.
Let $m$ be the same color as $\frac{m+3 p}{2}(p \in\{-1,1\})$.
Notice that, $\frac{m+p}{2}<m$.
Then $f(m)+f\left(\frac{m+3 p}{2}\right)=f\left(3 \times \frac{m+p}{2}\right)$
$=3 f\left(\frac{m+p}{2}\right)=\frac{3(m+p)}{2} f(1)$
$\Rightarrow f(m)=m f(1)$,
a contradiction.
Thus, proposition (1) is proven, i.e., the minimum value of $k$ is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Several boxes are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each box does not exceed 1 ton. To ensure that these boxes can be transported away in one go, the question is: what is the minimum number of trucks with a carrying capacity of 3 tons needed?
|
II. First, notice that the weight of each box does not exceed 1 ton. Therefore, the weight of boxes that each vehicle can transport at once will not be less than 2 tons. Otherwise, another box can be added.
Let $n$ be the number of vehicles needed, and the weights of the boxes transported by each vehicle be $a_{1}, a_{2}, \cdots, a_{n}$.
Then $2 \leqslant a_{i} \leqslant 3(i=1,2, \cdots, n)$.
Let the total weight of all transported goods be $S$.
Thus, $2 n \leqslant S=a_{1}+a_{2}+\cdots+a_{n} \leqslant 3 n$
$\Rightarrow 2 n \leqslant 10 \leqslant 3 n \Rightarrow \frac{10}{3} \leqslant n \leqslant 5$
$\Rightarrow n=4$ or 5.
Furthermore, it is explained that 4 vehicles are not enough.
Suppose there are 13 boxes, each weighing 13 tons.
Since $\frac{10}{13} \times 4>3$, each vehicle can transport at most 3 boxes. Therefore, 4 vehicles cannot transport all the boxes. Thus, at least 5 vehicles are needed to transport all the boxes at once.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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