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3. The equation $x^{2}+a x+4=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{2}-\frac{20}{3 x_{2}^{3}}=x_{2}^{2}-\frac{20}{3 x_{1}^{3}}
$$
Find all possible values of $a$.
|
Answer: $a=-10$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-16>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=4$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-4$.
Transform the given equation:
$$
x_{1}^{2}-x_{2}^{2}-\frac{20}{3} \cdot \frac{x_{1}^{3}-x_{2}^{3}}{x_{1}^{3} x_{2}^{3}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)-\frac{20\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)}{3\left(x_{1} x_{2}\right)^{3}}=0
$$
Since the roots are distinct, $x_{1}-x_{2} \neq 0$. Dividing both sides by $x_{1}-x_{2}$ and substituting the values given above, we get $-a-\frac{20}{3} \cdot \frac{a^{2}-4}{64}=0 \Leftrightarrow 5 a^{2}+48 a-20=0$, from which $a=-10$ or $a=0.4$. Only $a=-10$ satisfies the inequality $a^{2}-16>0$.
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The equation $x^{2}+a x+2=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{3}+\frac{14}{x_{2}^{2}}=x_{2}^{3}+\frac{14}{x_{1}^{2}}
$$
Find all possible values of $a$.
|
Answer: $a=4$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-8>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=2$. Then $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-2$.
Transform the given equation:
$$
x_{1}^{3}-x_{2}^{3}+14 \cdot \frac{x_{1}^{2}-x_{2}^{2}}{x_{1}^{2} x_{2}^{2}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)+\frac{14\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)}{\left(x_{1} x_{2}\right)^{2}}=0
$$
Since the roots are distinct, $x_{1}-x_{2} \neq 0$. Dividing both sides by $x_{1}-x_{2}$ and substituting the values given above, we get $a^{2}-2-\frac{14 a}{4}=0 \Leftrightarrow 2 a^{2}-7 a-4=0$, from which $a=-0.5$ or $a=4$. Only $a=4$ satisfies the inequality $a^{2}-8>0$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The equation $x^{2}+a x+3=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
x_{1}^{3}-\frac{99}{2 x_{2}^{2}}=x_{2}^{3}-\frac{99}{2 x_{1}^{2}}
$$
Find all possible values of $a$.
|
Answer: $a=-6$.
Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-12>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=3$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-3$.
Transform the given equation:
$$
x_{1}^{3}-x_{2}^{3}-\frac{99}{2} \cdot \frac{x_{1}^{2}-x_{2}^{2}}{x_{1}^{2} x_{2}^{2}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)-\frac{99\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)}{2\left(x_{1} x_{2}\right)^{2}}=0
$$
Since the roots are distinct, $x_{1}-x_{2} \neq 0$. Dividing both sides by $x_{1}-x_{2}$ and substituting the values given above, we get $a^{2}-3+\frac{99 a}{18}=0 \Leftrightarrow 2 a^{2}+11 a-6=0$, from which $a=0.5$ or $a=-6$. Only $a=-6$ satisfies the inequality $a^{2}-12>0$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. param 1 people participated in a survey. They were given a list of $N$ movies. Each person was asked to name their favorite movies from this list. It turned out that everyone named at least two movies. Moreover, any pair of respondents had no more than one movie in common among those they named. Find the smallest possible value of $N$.
| param1 | |
| :---: | :---: |
| 30 | |
| 39 | |
| 49 | |
| 57 | |
| 60 | |
|
8. param 1 people participated in a survey. They were given a list of $N$ movies. Each person was asked to name their favorite movies from this list. It turned out that everyone named at least two movies. Moreover, any pair of respondents had no more than one movie in common among the ones they named. Find the smallest possible value of $N$.
| param1 | Answer |
| :---: | :---: |
| 30 | 9 |
| 39 | 10 |
| 49 | 11 |
| 57 | 12 |
| 60 | 12 |
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\sqrt{5 x-1}}(4 x+1), \log _{4 x+1}\left(\frac{x}{2}+2\right)^{2}, \log _{\frac{x}{2}+2}(5 x-1)$. For which $x$ are two of these numbers equal, and the third one less than them by 1?
|
Answer: $x=2$.
Solution. From the condition, it follows that the functions $4 x+1, \frac{x}{2}+2,5 x-1$ are positive and do not take the value 1 for all $x$ from the domain of admissible values. Let $a=\log _{\sqrt{5 x-1}}(4 x+1), b=\log _{4 x+1}\left(\frac{x}{2}+2\right)^{2}, c=$ $\log _{\frac{x}{2}+2}(5 x-1)$. Then
$$
\begin{aligned}
& a b c=\log _{\sqrt{5 x-1}}(4 x+1) \cdot \log _{4 x+1}\left(\frac{x}{2}+2\right)^{2} \cdot \log _{\frac{x}{2}+2}(5 x-1)= \\
& \quad=2 \log _{5 x-1}(4 x+1) \cdot 2 \log _{4 x+1}\left(\frac{x}{2}+2\right) \cdot \frac{\log _{4 x+1}(5 x-1)}{\log _{4 x+1}\left(\frac{x}{2}+2\right)}=4
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c+1 \\
& b=c \text { and } c=a+1 \\
& c=a \text { and } \quad a=b+1
\end{aligned}
$$
Consider case (3). Substituting $b=a$ and $c=a-1$ into the equation $a b c=4$, we have $a \cdot a \cdot(a-1)=4$, from which $a^{3}-a^{2}-4=0,(a-2)\left(a^{2}+a+2\right)=0$. Since the polynomial $a^{2}+a+2$ has no roots, the only solution to the equation is $a=2$, so the system is satisfied by the triplet of numbers $a=2, b=2, c=1$. Cases (4) and (5) are considered similarly; from them we get that either $a=1, b=2, c=2$, or $a=2, b=1, c=2$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $c=1$, then $5 x-1=\frac{x}{2}+2$, that is, $x=\frac{2}{3}$. Therefore, $a=2 \log _{\frac{7}{3}} \frac{11}{3} \neq 2$, that is, there are no values of $x$ for which $a=b=2, c=1$.
If $a=1$, then $4 x+1=\sqrt{5 x-1}$. Squaring both sides of the last equation, we get the equation $16 x^{2}+3 x+2=0$, which has no roots, so the case $a=1, b=c=2$ also does not fit.
If $b=1$, then $\left(\frac{x}{2}+2\right)^{2}=4 x+1$. This equation is equivalent to the equation $x^{2}-8 x+12=0$, the roots of which are $x=2$ and $x=6$, but $x=6$ does not fit, because in this case $a=\log _{\sqrt{29}} 25 \neq 2$. The value $x=2$ fits: $a=\log _{\sqrt{9}} 9=2, c=\log _{3} 9=2$.
Thus, $x=2$ is the only solution to the problem.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), \log _{6 x-14}(x-1)^{2}, \log _{x-1}\left(\frac{x}{3}+3\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1?
|
Answer: $x=3$.
Solution. From the condition, it follows that the functions $6 x-14, x-1, \frac{x}{3}+3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), b=\log _{6 x-14}(x-1)^{2}, c=$ $\log _{x-1}\left(\frac{x}{3}+3\right)$. Then
$$
\begin{aligned}
& a b c=\log _{\sqrt{\frac{x}{3}+3}}(6 x-14) \cdot \log _{6 x-14}(x-1)^{2} \cdot \log _{x-1}\left(\frac{x}{3}+3\right)= \\
& \quad=2 \log _{\frac{x}{3}+3}(6 x-14) \cdot 2 \log _{6 x-14}(x-1) \cdot \frac{\log _{6 x-14}\left(\frac{x}{3}+3\right)}{\log _{6 x-14}(x-1)}=4
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c+1 \\
& b=c \text { and } c=a+1 \\
& c=a \text { and } \quad a=b+1
\end{aligned}
$$
Consider case (8). Substituting $b=a$ and $c=a-1$ into the equation $a b c=4$ obtained above, we have $a \cdot a \cdot(a-1)=4$, from which $a^{3}-a^{2}-4=0,(a-2)\left(a^{2}+a+2\right)=0$. Since the polynomial $a^{2}+a+2$ has no roots, the only solution to the equation is $a=2$, so the system is satisfied by the triplet of numbers $a=2, b=2, c=1$. Cases (9) and (10) are considered similarly; from them we get that either $a=1, b=2, c=2$, or $a=2, b=1, c=2$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $c=1$, then $x-1=\frac{x}{3}+3$, that is, $x=6$. Therefore, $a=2 \log _{\sqrt{5}} 22 \neq 2$, that is, there are no values of $x$ for which $a=b=2, c=1$.
If $a=1, b=c=2$, then $c=2 \Leftrightarrow(x-1)^{2}=\frac{x}{3}+3$. This equation is equivalent to the equation $3 x^{2}-7 x-6=0$, the roots of which are $x=3$ and $x=-\frac{2}{3}$, but $x=-\frac{2}{3}$ does not satisfy the domain of admissible values. For $x=3 a=2$, i.e., this root also does not fit in this case.
If $b=1$, then $(x-1)^{2}=(6 x-14)$. This equation is equivalent to the equation $x^{2}-8 x+15=0$, the roots of which are $x=3$ and $x=5$, but $x=5$ does not fit, as in this case $a=\log \sqrt{\frac{14}{3}} 16 \neq 2$. The value $x=3$ fits: $a=\log _{\sqrt{4}} 4=2, c=\log _{2} 4=2$.
Thus, $x=3$ is the only solution to the problem.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\left(\frac{x}{2}-1\right)^{2}}\left(\frac{x}{2}-\frac{1}{4}\right), \log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right), \log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right)^{2}$. For which $x$ are two of these numbers equal, and the third one greater than them by 1?
|
Answer: $x=5$.
Solution. From the condition, it follows that the functions $\frac{x}{2}-1, \frac{x}{2}-\frac{1}{4}, x-\frac{11}{4}$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right), b=\log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right)^{2}, c=$ $\log _{\left(\frac{x}{2}-1\right)^{2}}\left(\frac{x}{2}-\frac{1}{4}\right)$. Then
$$
\begin{aligned}
a b c=\log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right) \cdot \log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right)^{2} & \cdot \log _{\left(\frac{x}{2}-1\right)^{2}}\left(\frac{x}{2}-\frac{1}{4}\right)= \\
& =2 \log _{x-\frac{11}{4}}\left(\frac{x}{2}-1\right) \cdot 2 \log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right) \cdot \frac{\log _{x-\frac{11}{4}}\left(\frac{x}{2}-\frac{1}{4}\right)}{2 \log _{x-\frac{11}{4}}\left(\frac{x}{2}-1\right)}=2
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c-1 \\
& b=c \text { and } \quad c=a-1 \\
& c=a \text { and } \quad a=b-1
\end{aligned}
$$
Consider case (13). Substituting $b=a$ and $c=a+1$ into the equation $a b c=2$ obtained above, we have $a \cdot a \cdot(a+1)=2$, from which $a^{3}+a^{2}-2=0,(a-1)\left(a^{2}+2 a+2\right)=0$. Since the polynomial $a^{2}+2 a+2$ has no roots, the only solution to the equation is $a=1$, so the system is satisfied by the triplet of numbers $a=1, b=1, c=2$. Cases (14) and (15) are considered similarly; from them we get that either $a=2, b=1, c=1$, or $a=1, b=2, c=1$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $b=2$, then $\frac{x}{2}-\frac{1}{4}=x-\frac{11}{4}$, that is, $x=5$. This value fits: $c=\log _{\frac{9}{4}} \frac{9}{4}=1, a=\log _{\frac{3}{2}} \frac{3}{2}=1$.
If $b=1$, then $\frac{x}{2}-\frac{1}{4}=\left(x-\frac{11}{4}\right)^{2}$. This equation is equivalent to the equation $x^{2}-6 x+\frac{125}{16}=0$, which has roots $x=3 \pm \frac{\sqrt{19}}{4}$. The value $x=3-\frac{\sqrt{19}}{4}$ does not satisfy the domain of admissible values (since $x<\frac{11}{4}$). If $x=3+\frac{\sqrt{19}}{4}$, then $a=\log _{\sqrt{\frac{1+\sqrt{19}}{4}}}\left(\frac{4+\sqrt{19}}{8}\right)$, which is different from 1 and 2. Therefore, the cases $a=b=1, c=2$ and $b=c=1, a=2$ do not fit.
Thus, $x=5$ is the only solution to the problem.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\sqrt{2 x-8}}(x-4), \log _{(x-4)^{2}}(5 x-26), \log _{\sqrt{5 x-26}}(2 x-8)$. For which $x$ are two of these numbers equal, and the third one greater than them by 1?
|
Answer: $x=6$.
Solution. From the condition, it follows that the functions $x-4, 5x-26$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{2 x-8}}(x-4), b=\log _{\sqrt{5 x-26}}(2 x-8), c=\log _{(x-4)^{2}}(5 x-26)$. Then
$$
\begin{aligned}
& a b c=\log _{\sqrt{2 x-8}}(x-4) \cdot \log _{\sqrt{5 x-26}}(2 x-8) \cdot \log _{(x-4)^{2}}(5 x-26)= \\
& \quad=2 \log _{2 x-8}(x-4) \cdot 2 \log _{5 x-26}(2 x-8) \cdot \frac{\log _{2 x-8}(5 x-26)}{\log _{2 x-8}(x-4)^{2}}=2
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c-1 \\
& b=c \text { and } c=a-1 \\
& c=a \text { and } \quad a=b-1
\end{aligned}
$$
Consider case (18). Substituting $b=a$ and $c=a+1$ into the equation $a b c=2$, we have $a \cdot a \cdot(a+1)=2$, from which $a^{3}+a^{2}-2=0,(a-1)\left(a^{2}+2 a+2\right)=0$. Since the polynomial $a^{2}+2 a+2$ has no roots, the only solution to the equation is $a=1$, so the system is satisfied by the triplet of numbers $a=1, b=1, c=2$. Cases (19) and (20) are considered similarly; from them we get that either $a=2, b=1, c=1$, or $a=1, b=2, c=1$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $b=2$, then $5 x-26=2 x-8$, that is, $x=6$. This value fits: $a=\log _{\sqrt{4}} 2=1, c=\log _{2^{2}} 4=1$.
If $b=1$, then $\sqrt{5 x-26}=2 x-8$. This equation is equivalent to the equation $4 x^{2}-37 x+90=0$, which has no roots, so the cases $a=b=1, c=2$ and $b=c=1, a=2$ do not fit.
Thus, $x=6$ is the only solution to the problem.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\sqrt{2 x-3}}(x+1), \log _{2 x^{2}-3 x+5}(2 x-3)^{2}, \log _{x+1}\left(2 x^{2}-3 x+5\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1?
|
Answer: $x=4$.
Solution. From the condition, it follows that the functions $x+1, 2x-3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{2 x-3}}(x+1), b=\log _{2 x^{2}-3 x+5}(2 x-3)^{2}, c=\log _{x+1}\left(2 x^{2}-3 x+5\right)$. Then
$$
\begin{aligned}
a b c=\log _{\sqrt{2 x-3}}(x+1) \cdot & \log _{2 x^{2}-3 x+5}(2 x-3)^{2} \cdot \log _{x+1}\left(2 x^{2}-3 x+5\right)= \\
& =2 \log _{2 x-3}(x+1) \cdot 2 \log _{2 x^{2}-3 x+5}(2 x-3) \cdot \frac{\log _{2 x-3}\left(2 x^{2}-3 x+5\right)}{\log _{2 x-3}(x+1)}=4
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c+1 \\
& b=c \text { and } c=a+1 \\
& c=a \quad \text { and } \quad a=b+1
\end{aligned}
$$
Consider case (23). Substituting $b=a$ and $c=a-1$ into the equation $a b c=4$, we have $a \cdot a \cdot(a-1)=4$, from which $a^{3}-a^{2}-4=0,(a-2)\left(a^{2}+a+2\right)=0$. Since the polynomial $a^{2}+a+2$ has no roots, the only solution to the equation is $a=2$, so the system is satisfied by the triplet of numbers $a=2, b=2, c=1$. Cases (24) and (25) are considered similarly; from them we get that either $a=1, b=2, c=2$, or $a=2, b=1, c=2$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $c=1$, then $x+1=2 x^{2}-3 x+5$. This equation has no roots, so there are no values of $x$ for which $a=b=2, c=1$.
If $b=1$, then $2 x^{2}-3 x+5=(2 x-3)^{2}$. This equation is equivalent to the equation $2 x^{2}-9 x+4=0$, the roots of which are $x=\frac{1}{2}$, which does not satisfy the domain of admissible values, and $x=4$. The value $x=4$ is valid: $a=\log _{\sqrt{5}} 5=2$, $c=\log _{5} 25=2$.
If $a=1$, then $\sqrt{2 x-3}=x+1$. Squaring both sides of the equation, we get the equation $x^{2}=-4$, which has no roots. Therefore, there are no values of $x$ for which $b=c=2, a=1$.
Thus, $x=4$ is the only solution to the problem.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{7 x}{2}-\frac{17}{4}\right), \log _{\sqrt{\frac{7 x}{2}-\frac{17}{4}}}\left(\frac{3 x}{2}-6\right)^{2}, \log _{\sqrt{\frac{3 x}{2}-6}}\left(\frac{x}{2}+1\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1?
|
Answer: $x=7$.
Solution. From the condition, it follows that the functions $\left(\frac{x}{2}+1\right),\left(\frac{7 x}{2}-\frac{17}{4}\right),\left(\frac{3 x}{2}-6\right)$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{7 x}{2}-\frac{17}{4}\right), b=$ $\log _{\sqrt{\frac{7 x}{2}-\frac{17}{4}}}\left(\frac{3 x}{2}-6\right)^{2}, c=\log _{\sqrt{\frac{3 x}{2}-6}}\left(\frac{x}{2}+1\right)$. Then
$$
\begin{aligned}
a b c=\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{7 x}{2}-\frac{17}{4}\right) \cdot & \log _{\sqrt{\frac{7 x}{2}-\frac{17}{4}}}\left(\frac{3 x}{2}-6\right)^{2} \cdot \log _{\sqrt{\frac{3 x}{2}-6}}\left(\frac{x}{2}+1\right)= \\
& =\frac{1}{2} \log _{\left(\frac{x}{2}+1\right)}\left(\frac{7 x}{2}-\frac{17}{4}\right) \cdot 4 \log _{\frac{7 x}{2}-\frac{17}{4}}\left(\frac{3 x}{2}-6\right) \cdot 2 \frac{\log _{\frac{7 x}{2}-\frac{17}{4}}\left(\frac{x}{2}+1\right)}{\log _{\frac{7 x}{2}-\frac{17}{4}}\left(\frac{3 x}{2}-6\right)}=4
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c+1 \\
& b=c \text { and } c=a+1 \\
& c=a \quad \text { and } \quad a=b+1
\end{aligned}
$$
Consider case (28). Substituting $b=a$ and $c=a-1$ into the equation $a b c=4$, we have $a \cdot a \cdot(a-1)=4$, from which $a^{3}-a^{2}-4=0,(a-2)\left(a^{2}+a+2\right)=0$. Since the polynomial $a^{2}+a+2$ has no roots, the only solution to the equation is $a=2$, so the system is satisfied by the triplet of numbers $a=2, b=2, c=1$. Cases (29) and (30) are considered similarly; from them we get that either $a=1, b=2, c=2$, or $a=2, b=1, c=2$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $c=2$, then $\frac{3 x}{2}-6=\frac{x}{2}+1$, that is, $x=7$. This value fits: $a=\log _{\left(\frac{9}{2}\right)^{2}} \frac{81}{4}=1, b=\log _{\sqrt{\frac{81}{4}}}\left(\frac{9}{2}\right)=2$. If $c=1$, then $\sqrt{\frac{3 x}{2}-6}=\frac{x}{2}+1$. From this equation, after squaring, we get that $x^{2}-2 x+28=0$, that is, there are no roots.
Thus, $x=7$ is the only solution to the problem.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\sqrt{x+34}}(2 x+23), \log _{(x+4)^{2}}(x+34), \log _{\sqrt{2 x+23}}(-x-4)$. For which $x$ are two of these numbers equal, and the third one greater than them by 1?
|
Answer: $x=-9$.
Solution. From the condition, it follows that the functions $2 x+23, x+34,-x-4$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{x+34}}(2 x+23), b=\log _{(x+4)^{2}}(x+34), c=\log _{\sqrt{2 x+23}}(-x-4)$. Then
$$
\begin{aligned}
& a b c=\log _{\sqrt{x+34}}(2 x+23) \cdot \log _{(x+4)^{2}}(x+34) \cdot \log _{\sqrt{2 x+23}}(-x-4)= \\
& \quad=2 \log _{x+34}(2 x+23) \cdot \frac{1}{2} \log _{(-x-4)}(x+34) \cdot 2 \frac{\log _{(x+34)}(5 x-1)}{\log _{(x+34)}(-x-4)}=2
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c-1 \\
& b=c \text { and } c=a-1 \\
& c=a \text { and } \quad a=b-1
\end{aligned}
$$
Consider case (33). Substituting $b=a$ and $c=a+1$ into the equation $a b c=2$, we have $a \cdot a \cdot(a+1)=2$, from which $a^{3}+a^{2}-2=0,(a-1)\left(a^{2}+2 a+2\right)=0$. Since the polynomial $a^{2}+2 a+2$ has no roots, the only solution to the equation is $a=1$, so the system is satisfied by the triplet of numbers $a=1, b=1, c=2$. Cases (34) and (35) are considered similarly; from them we get that either $a=2, b=1, c=1$, or $a=1, b=2, c=1$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $c=1$, then $2 x+23=(x+4)^{2}$. This equation is equivalent to the equation $x^{2}+6 x-7=0$, the roots of which are $x=1$ and $x=-7$. $x=1$ does not satisfy the domain of admissible values. For $x=-7$, $a=\log _{\sqrt{27}} 9 \neq 2$. Moreover, $a \neq 1$. Therefore, this case does not fit.
If $c=2$, then $2 x+23=-x-4$, that is, $x=-9$. The value $x=-9$ fits: $a=\log _{\sqrt{25}} 5=1$, $b=\log _{5^{2}} 25=1$.
Thus, $x=-9$ is the only solution to the problem.
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5 points] Given the numbers $\log _{\sqrt{29-x}}\left(\frac{x}{7}+7\right), \log _{(x+1)^{2}}(29-x), \log _{\sqrt{\frac{x}{7}+7}}(-x-1)$. For which $x$ are two of these numbers equal, and the third one greater than them by 1?
|
Answer: $x=-7$.
Solution. From the condition, it follows that the functions $\frac{x}{7}+7, 29-x, -x-1$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{29-x}}\left(\frac{x}{7}+7\right), b=\log _{(x+1)^{2}}(29-x), c=$ $\log _{\sqrt{\frac{x}{7}+7}}(-x-1)$. Then
$$
\begin{aligned}
& a b c=\log _{\sqrt{29-x}}\left(\frac{x}{7}+7\right) \cdot \log _{(x+1)^{2}}(29-x) \cdot \log \sqrt{\frac{x}{7}+7}(-x-1)= \\
& \quad=2 \log _{29-x}\left(\frac{x}{7}+7\right) \cdot \frac{1}{2} \log _{(-x-1)}(29-x) \cdot 2 \frac{\log _{29-x}(-x-1)}{\log _{29-x}\left(\frac{x}{7}+7\right)}=2
\end{aligned}
$$
By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:
$$
\begin{aligned}
& a=b \text { and } \quad a=c-1 \\
& b=c \text { and } c=a-1 \\
& c=a \text { and } \quad a=b-1
\end{aligned}
$$
Consider case (38). Substituting $b=a$ and $c=a+1$ into the equation $a b c=2$, we have $a \cdot a \cdot(a+1)=2$, from which $a^{3}+a^{2}-2=0,(a-1)\left(a^{2}+2 a+2\right)=0$. Since the polynomial $a^{2}+2 a+2$ has no roots, the only solution to the equation is $a=1$, so the system is satisfied by the triplet of numbers $a=1, b=1, c=2$. Cases (39) and (40) are considered similarly; from them we get that either $a=2, b=1, c=1$, or $a=1, b=2, c=1$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.
If $c=1$, then $\frac{x}{7}+7=x^{2}+2 x+1$. This equation is equivalent to the equation $7 x^{2}+13 x-42=0$, the roots of which are $x=-\frac{-13 \pm \sqrt{1345}}{14}$. The larger root does not satisfy the domain of admissible values, and for the smaller root $a \neq 1$ and $a \neq 2$, so it also does not fit.
If $c=2$, then $\frac{x}{7}+7=-x-1$, that is, $x=-7$. The value $x=-7$ fits: $a=\log _{\sqrt{36}} 6=1, b=\log _{6^{2}} 36=1$. Thus, $x=-7$ is the only solution to the problem.
|
-7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given quadratic trinomials $f_{1}(x)=x^{2}+a x+3, f_{2}(x)=x^{2}+2 x-b, f_{3}(x)=x^{2}+2(a-1) x+b+6$ and $f_{4}(x)=x^{2}+(4-a) x-2 b-3$. Let the differences of their roots be $A, B, C$ and $D$, respectively, and given that $|A| \neq|B|$. Find the ratio $\frac{C^{2}-D^{2}}{A^{2}-B^{2}}$. The values of $A, B, C, D, a, b$ are not specified.
|
Answer: 3.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|} \cdot$ Applying this formula four times, we get
$$
A=\sqrt{a^{2}-12}, B=\sqrt{4+4 b}, C=\sqrt{(2 a-2)^{2}-4(6+b)}, D=\sqrt{(4-a)^{2}+4(2 b+3)}
$$
From this, it follows that $C^{2}-D^{2}=\left(\left(4 a^{2}-8 a-4 b-20\right)-\left(a^{2}-8 a+8 b+28\right)\right)=3\left(a^{2}-4 b-16\right)$, $A^{2}-B^{2}=a^{2}-4 b-16$. Therefore, the desired ratio is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given quadratic trinomials $f_{1}(x)=x^{2}+2 x+a, f_{2}(x)=x^{2}+b x-1, f_{3}(x)=2 x^{2}+(6-b) x+3 a+1$ and $f_{4}(x)=2 x^{2}+(3 b-2) x-a-3$. Let the differences of their roots be $A, B, C$ and $D$, respectively, and given that $|A| \neq|B|$. Find the ratio $\frac{C^{2}-D^{2}}{A^{2}-B^{2}}$. The values of $A, B, C, D, a, b$ are not specified.
|
Answer: 2.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|}$. Applying this formula four times, we get
$$
A=\sqrt{4-4 a}, B=\sqrt{b^{2}+4}, C=\frac{1}{2} \sqrt{(6-b)^{2}-8(1+3 a)}, D=\frac{1}{2} \sqrt{(3 b-2)^{2}+8(a+3)}
$$
From this, it follows that $C^{2}-D^{2}=\frac{1}{4}\left(\left(b^{2}-12 b-24 a+28\right)-\left(9 b^{2}-12 b+8 a+28\right)\right)=-2\left(b^{2}+4 a\right)$, $A^{2}-B^{2}=-\left(b^{2}+4 a\right)$. Therefore, the desired ratio is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos y}=\frac{2}{5}+\cos ^{2}(x+y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin y}=\frac{3}{5}+\sin ^{2}(x+y)$. Find all possible values of the expression $\cos (x+3 y)$, given that there are at least two.
|
Answer: -1 or $\frac{1}{5}$.
Solution. Note that
$$
\begin{aligned}
& \cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x=\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& \quad=\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x \\
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{2} x+\sin x \cos 2 x=\sin x(1+\cos 2 x)+\sin x \cos 2 x=(2 \cos 2 x+1) \sin x
\end{aligned}
$$
Taking this into account, the given equalities take the form
$$
\left\{\begin{array}{l}
\frac{\cos x}{\cos y}=\frac{2}{5}+\cos ^{2}(x+y) \\
\frac{\sin x}{\sin y}=\frac{3}{5}+\sin ^{2}(x+y)
\end{array}\right.
$$
(it is also necessary to consider that $\cos 2 x \neq \pm \frac{1}{2}$). Adding the equations (??) we get $\frac{\cos x}{\cos y}+$ $\frac{\sin x}{\sin y}=2$, from which $\cos y \neq 0, \sin y \neq 0$ and $\cos x \sin y+\sin x \cos y=2 \sin y \cos y$. The last equality is equivalent to the following:
$$
\sin (x+y)=\sin 2 y \Leftrightarrow\left[\begin{array}{l}
x + y = 2 y + 2 \pi k \\
x + y = \pi - 2 y + 2 \pi k , k \in \mathbb { Z }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=y+2 \pi k \\
x=\pi-3 y+2 \pi k, k \in \mathbb{Z}
\end{array}\right.\right.
$$
Let's consider these two possibilities separately
If $x=y+2 \pi k, k \in \mathbb{Z}$, then the system (??) takes the form $\left\{\begin{array}{l}1=\frac{2}{5}+\cos ^{2} 2 y, \\ 1=\frac{3}{5}+\sin ^{2} 2 y ;\end{array} \quad\right.$ from which $\sin ^{2} 2 y=\frac{2}{5}$, $\cos ^{2} 2 y=\frac{3}{5}$. It is easy to see that this system has solutions. In this case, $\cos (x+3 y)=\cos 4 y=$ $2 \cos ^{2} 2 y-1=2 \cdot \frac{3}{5}-1=-\frac{1}{5}$.
If $x=\pi-3 y+2 \pi k, k \in \mathbb{Z}$, then (??) takes the form
$$
\left\{\begin{array}{l}
- \frac { \cos 3 y } { \cos y } = \frac { 2 } { 5 } + \cos ^ { 2 } 2 y , \\
\frac { \sin 3 y } { \sin y } = \frac { 3 } { 5 } + \sin ^ { 2 } 2 y
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3-4 \cos ^{2} y=\frac{2}{5}+\cos ^{2} 2 y \\
3-4 \sin ^{2} y=\frac{3}{5}+\sin ^{2} 2 y
\end{array}\right.\right.
$$
Each of the equations in the last system reduces to the equation $\cos ^{2} 2 y+2 \cos 2 y-\frac{3}{5}=0$, which has solutions, since $\cos 2 y=-1+\sqrt{\frac{8}{5}}$. Therefore, this case is also possible, and we get that $\cos (x+3 y)=\cos (\pi+2 \pi k)=-1$.
Thus, $\cos (x+3 y)$ can be equal to $\frac{1}{5}$ or -1.
Note that in both cases all the restrictions that appeared in connection with the domain of definition $\left(\cos 2 x \neq \pm \frac{1}{2}\right.$, $\cos y \neq 0, \sin y \neq 0$ ) are satisfied.
Remark. Since the problem states that the expression $\cos (x+6 y)$ has at least two different values, checking the possibility of the cases is not mandatory.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos y}=\frac{2}{3}+\cos ^{2}(x-y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin y}=-\frac{1}{3}-\sin ^{2}(x-y)$. Find all possible values of the expression $\cos (x-3 y)$, given that there are at least two. Answer: -1 or $-\frac{1}{3}$.
|
Solution. Note that
$$
\begin{aligned}
\cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x & =\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& =\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x
\end{aligned}
$$
$$
\begin{aligned}
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{2} x+\sin x \cos 2 x=\sin x(1+\cos 2 x)+\sin x \cos 2 x=(2 \cos 2 x+1) \sin x
\end{aligned}
$$
Taking this into account, the given equalities take the form
$$
\left\{\begin{array}{l}
\frac{\cos x}{\cos y}=\frac{2}{3}+\cos ^{2}(x-y) \\
\frac{\sin x}{\sin y}=-\frac{1}{3}-\sin ^{2}(x-y)
\end{array}\right.
$$
(it is also necessary to take into account that $\cos 2 x \neq \pm \frac{1}{2}$). Subtract the second equation from the first and we get $\frac{\cos x}{\cos y}-\frac{\sin x}{\sin y}=2$, from which $\cos y \neq 0, \sin y \neq 0$ and $\cos x \sin y-\sin x \cos y=2 \sin y \cos y$. The last equality is equivalent to the following:
$$
\sin (y-x)=\sin 2 y \Leftrightarrow\left[\begin{array} { l }
{ y - x = 2 y + 2 \pi k } \\
{ y - x = \pi - 2 y + 2 \pi k , k \in \mathbb { Z } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=-y-2 \pi k \\
x=3 y-\pi-2 \pi k, k \in \mathbb{Z}
\end{array}\right.\right.
$$
Consider these two possibilities separately.
If $x=-y-2 \pi k, k \in \mathbb{Z}$, then the system (??) takes the form $\left\{\begin{array}{l}1=\frac{2}{3}+\cos ^{2} 2 y, \\ -1=-\frac{1}{3}-\sin ^{2} 2 y ;\end{array} \quad\right.$ from which $\sin ^{2} 2 y=$ $\frac{2}{3}, \cos ^{2} 2 y=\frac{1}{3}$. It is easy to see that this system has solutions. In this case, $\cos (x-3 y)=\cos 4 y=$ $2 \cos ^{2} 2 y-1=2 \cdot \frac{1}{3}-1=-\frac{1}{3}$.
If $x=3 y-\pi-2 \pi k, k \in \mathbb{Z}$, then (??) takes the form
$$
\left\{\begin{array} { l }
{ - \frac { \operatorname { c o s } 3 y } { \operatorname { c o s } y } = \frac { 2 } { 3 } + \operatorname { c o s } ^ { 2 } 2 y , } \\
{ - \frac { \operatorname { s i n } 3 y } { \operatorname { s i n } y } = - \frac { 1 } { 3 } - \operatorname { s i n } ^ { 2 } 2 y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3-4 \cos ^{2} y=\frac{2}{3}+\cos ^{2} 2 y \\
4 \sin ^{2} y-3=-\frac{1}{3}-\sin ^{2} 2 y
\end{array}\right.\right.
$$
Each of the equations in the last system reduces to the equation $\cos ^{2} 2 y+2 \cos 2 y-\frac{1}{3}=0$, which has solutions, since $\cos 2 y=-1+\frac{2}{\sqrt{3}}$. Therefore, this case is also possible, and we get that $\cos (x-3 y)=\cos (-\pi-2 \pi k)=-1$.
Thus, $\cos (x-3 y)$ can be equal to $-\frac{2}{3}$ or -1.
Note that in both cases all the restrictions that appeared in connection with the domain of definition $\left(\cos 2 x \neq \pm \frac{1}{2}\right.$, $\cos y \neq 0, \sin y \neq 0$) are satisfied.
Remark. Since the problem states that the expression $\cos (x-3 y)$ has at least two different values, checking the possibility of cases is not mandatory.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos 2 y}=\frac{1}{6}+\sin ^{2}(x+2 y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin 2 y}=\frac{5}{6}+\cos ^{2}(x+2 y)$. Find all possible values of the expression $\cos (x+6 y)$, given that there are at least two. Answer: -1 or $-\frac{2}{3}$.
|
Solution. Note that
$$
\begin{aligned}
& \cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x=\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& \quad=\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x \\
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{2} x+\sin x \cos 2 x=\sin x(1+\cos 2 x)+\sin x \cos 2 x=(2 \cos 2 x+1) \sin x
\end{aligned}
$$
Taking this into account, the given equalities take the form
$$
\left\{\begin{array}{l}
\frac{\cos x}{\cos 2 y}=\frac{1}{6}+\sin ^{2}(x+2 y) \\
\frac{\sin x}{\sin 2 y}=\frac{5}{6}+\cos ^{2}(x+2 y)
\end{array}\right.
$$
(it is also necessary to take into account that $\cos 2 x \neq \pm \frac{1}{2}$). Adding the equations (??) we get $\frac{\cos x}{\cos 2 y}+$ $\frac{\sin x}{\sin 2 y}=2$, from which $\cos 2 y \neq 0, \sin 2 y \neq 0$ and $\cos x \sin 2 y+\sin x \cos 2 y=2 \sin 2 y \cos 2 y$. The last equality is equivalent to the following:
$$
\sin (x+2 y)=\sin 4 y \Leftrightarrow\left[\begin{array} { l }
{ x + 2 y = 4 y + 2 \pi k } \\
{ x + 2 y = \pi - 4 y + 2 \pi k , k \in \mathbb { Z } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=2 y+2 \pi k \\
x=\pi-6 y+2 \pi k, k \in \mathbb{Z}
\end{array}\right.\right.
$$
Let's consider these two possibilities separately.
If $x=2 y+2 \pi k, k \in \mathbb{Z}$, then the system (??) takes the form $\left\{\begin{array}{l}1=\frac{1}{6}+\sin ^{2} 4 y, \\ 1=\frac{5}{6}+\cos ^{2} 4 y ;\end{array} \quad\right.$ from which $\sin ^{2} 4 y=\frac{5}{6}$, $\cos ^{2} 4 y=\frac{1}{6}$. It is easy to see that this system has solutions. In this case, $\cos (x+6 y)=\cos 8 y=$ $2 \cos ^{2} 4 y-1=2 \cdot \frac{1}{6}-1=-\frac{2}{3}$.
If $x=\pi-6 y+2 \pi k, k \in \mathbb{Z}$, then (??) takes the form
$$
\left\{\begin{array} { l }
{ - \frac { \operatorname { c o s } 6 y } { \operatorname { c o s } 2 y } = \frac { 1 } { 6 } + \operatorname { s i n } ^ { 2 } 4 y , } \\
{ \frac { \operatorname { s i n } 6 y } { \operatorname { s i n } 2 y } = \frac { 5 } { 6 } + \operatorname { c o s } ^ { 2 } 4 y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3-4 \cos ^{2} 2 y=\frac{1}{6}+\sin ^{2} 4 y \\
3-4 \sin ^{2} 2 y=\frac{5}{6}+\cos ^{2} 4 y
\end{array}\right.\right.
$$
Each of the equations in the last system reduces to the equation $\cos ^{2} 4 y-2 \cos 4 y-\frac{1}{6}=0$, which has solutions, since $\cos 4 y=1-\sqrt{\frac{7}{6}}$. Therefore, this case is also possible, and we get that $\cos (x+6 y)=\cos (\pi+2 \pi k)=-1$.
Thus, $\cos (x+6 y)$ can be equal to $-\frac{2}{3}$ or -1.
Note that in both cases all the restrictions that appeared in connection with the domain of definition $\left(\cos 2 x \neq \pm \frac{1}{2}\right.$, $\cos 2 y \neq 0, \sin 2 y \neq 0$) are satisfied.
Remark. Since the problem states that the expression $\cos (x+6 y)$ has at least two different values, checking the possibility of the cases is not mandatory.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-3 x+2\right)$, $\log _{x^{2}} \frac{x^{2}}{x-2}$, and $\log _{x^{2}} \frac{x^{2}}{x-1}$ is equal to the sum of the other two.
|
Answer: $x=3$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-2} \cdot \frac{x^{2}}{x-1}\left(x^{2}-3 x+2\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining numbers by $a$ and $b$. Then $c=a+b$ and $a+b+c=2$, from which it follows that $c=1$, i.e., one of the three given logarithms is equal to 1.
The converse is also true, namely, if one of the three given logarithms is equal to 1, then since the sum of all three logarithms is 2, the sum of the remaining two is 1, i.e., their sum is equal to the first logarithm.
Thus, the condition of the problem is satisfied if and only if one of the logarithms is equal to 1 (and all logarithms exist). A logarithm is equal to 1 when its base is equal to the logarithmic expression. We obtain the system
$$
\left[\begin{array}{l}
x^{2}=\frac{x^{2}}{x-2}, \\
x^{2}=\frac{x^{2}}{x-1}, \\
x^{2}=x^{2}-3 x+2
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=0 \\
x=3 \\
x=2 \\
x=\frac{2}{3}
\end{array}\right.\right.
$$
Of the found values of the variable, only $x=3$ satisfies the domain of definition.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-7 x+12\right)$, $\log _{x^{2}} \frac{x^{2}}{x-3}$, and $\log _{x^{2}} \frac{x^{2}}{x-4}$ is equal to the sum of the other two.
|
Answer: $x=5$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-3} \cdot \frac{x^{2}}{x-4}\left(x^{2}-7 x+12\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining numbers by $a$ and $b$. Then $c=a+b$ and $a+b+c=2$, from which it follows that $c=1$, i.e., one of the three given logarithms is equal to 1.
The converse is also true, namely, if one of the three given logarithms is equal to 1, then since the sum of all three logarithms is 2, the sum of the remaining two is 1, i.e., their sum is equal to the first logarithm.
Thus, the requirement of the problem is satisfied if and only if one of the logarithms is equal to 1 (and all logarithms exist). A logarithm is equal to 1 when its base is equal to the logarithmic expression. We obtain the system
$$
\left[\begin{array}{l}
x^{2}=\frac{x^{2}}{x-3}, \\
x^{2}=\frac{x^{2}}{x-4}, \\
x^{2}=x^{2}-7 x+12
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=0 \\
x=4 \\
x=5 \\
x=\frac{12}{7}
\end{array}\right.\right.
$$
Of the found values of the variable, only $x=5$ satisfies the domain of definition.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-10 x+21\right)$, $\log _{x^{2}} \frac{x^{2}}{x-7}$, and $\log _{x^{2}} \frac{x^{2}}{x-3}$ is equal to the sum of the other two.
|
Answer: $x=8$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-3} \cdot \frac{x^{2}}{x-7}\left(x^{2}-10 x+21\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining numbers by $a$ and $b$. Then $c=a+b$ and $a+b+c=2$, from which it follows that $c=1$, i.e., one of the three given logarithms is equal to 1.
The converse is also true, namely, if one of the three given logarithms is equal to 1, then since the sum of all three logarithms is 2, the sum of the remaining two is 1, i.e., their sum is equal to the first logarithm.
Thus, the condition of the problem is satisfied if and only if one of the logarithms is equal to 1 (and all logarithms exist). A logarithm is equal to 1 when its base is equal to the logarithmic expression. We obtain the system
$$
\left[\begin{array}{l}
x^{2}=\frac{x^{2}}{x-7}, \\
x^{2}=\frac{x^{2}}{x-3}, \\
x^{2}=x^{2}-10 x+21
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=0 \\
x=8 \\
x=4 \\
x=\frac{21}{10}
\end{array}\right.\right.
$$
Of the found values of the variable, only $x=8$ satisfies the domain of definition.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-7 x+10\right)$, $\log _{x^{2}} \frac{x^{2}}{x-2}$, and $\log _{x^{2}} \frac{x^{2}}{x-5}$ is equal to the sum of the other two.
|
Answer: $x=6$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-2} \cdot \frac{x^{2}}{x-5}\left(x^{2}-7 x+10\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining numbers by $a$ and $b$. Then $c=a+b$ and $a+b+c=2$, from which it follows that $c=1$, i.e., one of the three given logarithms is equal to 1.
The converse is also true, namely, if one of the three given logarithms is equal to 1, then since the sum of all three logarithms is 2, the sum of the remaining two is 1, i.e., their sum is equal to the first logarithm.
Thus, the requirement of the problem is satisfied if and only if one of the logarithms is equal to 1 (and all logarithms exist). A logarithm is equal to 1 when its base is equal to the logarithmic expression. We obtain the system
$$
\left[\begin{array}{l}
x^{2}=\frac{x^{2}}{x-2}, \\
x^{2}=\frac{x^{2}}{x-5}, \\
x^{2}=x^{2}-7 x+10
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=0 \\
x=3 \\
x=6 \\
x=\frac{10}{7}
\end{array}\right.\right.
$$
Of the found values of the variable, only $x=6$ satisfies the domain of definition.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}-$ $3 f(x)$, if the minimum value of the function $(f(x))^{2}-3 g(x)$ is $\frac{11}{2}$.
|
Answer: -10.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}-3 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}-3(a x+c)=a^{2} x^{2}+a(2 b-3) x+b^{2}-3 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coordinate of the vertex is $x_{\mathrm{B}}=-\frac{2 b-3}{2 a}$; the y-coordinate of the vertex is $h\left(x_{\mathrm{B}}\right)=3 b-\frac{9}{4}-3 c$.
Similarly, we find that the minimum value of the expression $(g(x))^{2}-3 f(x)$ is $3 c-\frac{9}{4}-3 b$. Note that the sum of these two minimum values is $-\frac{9}{2}$, therefore, if one of these minimum values is $\frac{11}{2}$, then the other is $-\frac{9}{2}-\frac{11}{2}=-10$.
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Non-zero numbers $a, b$, and $c$ are such that the equalities $a^{2}(b+c-a)=b^{2}(a+c-b)=c^{2}(b+a-c)$ hold. What is the greatest value that the expression $\frac{2 b+3 c}{a}$ can take?
|
Answer: 5.
Solution: By equating the first and second expressions, after transformation, we get: $(a-b)\left(a^{2}+b^{2}-a c-b c\right)=0$. Similarly, we obtain the equalities $(b-c)\left(b^{2}+c^{2}-a b-a c\right)=0$ and $(a-c)\left(a^{2}+c^{2}-a b-c b\right)=0$.
We will prove that $a=b=c$.
Assume that $a=b \neq c$. Then from the second equality, we get: $0=b^{2}+c^{2}-a b-a c=b^{2}+c^{2}-b^{2}-b c=c^{2}-b c$, from which $b=c$ (since $c \neq 0$). Contradiction.
Thus, either the numbers are pairwise distinct, or they are all equal to each other.
Assume that the numbers are pairwise distinct, then $a^{2}+b^{2}-a c-b c=0$, $b^{2}+c^{2}-a b-a c=0$ and $a^{2}+c^{2}-a b-c b=0$. Adding these equalities, we get $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$, from which $a=b=c$. Contradiction.
Thus, the only possible case is $a=b=c$. But then $\frac{2 b+3 c}{a}=\frac{2 a+3 a}{a}=5$.
Comment: The correct answer is obtained by considering a special case - 1 point.
It is claimed that $a=b=c$, but this statement is not proven - no points are added.
9.4-1. It is known that the quadratic equation $a x^{2}+b x+c=0$ has no solutions. Is it true that the quadratic equation $(2 a+c) x^{2}+3 b x+(a+2 c)=0$ also has no solutions?
Answer: True.
Solution: Note that if the quadratic equation $a x^{2}+b x+c=0$ has no solutions, then the quadratic equation $c x^{2}+b x+a=0$ also has no solutions, as these equations have the same discriminant. Moreover, if the quadratic equation $a x^{2}+b x+c=0$ has no solutions, then the signs of the coefficients $a$ and $c$ will be the same (otherwise, the discriminant would be non-negative). This means that the graphs of the quadratic functions $f(x)=a x^{2}+b x+c$ and $g(x)=c x^{2}+b x+a$ either both lie above the $O x$ axis or both lie below the $O x$ axis. But then the graph of the sum $h(x)=2 f(x)+g(x)=(2 a+c) x^{2}+3 b x+(a+2 c)$ also does not intersect the $O x$ axis.
Comment: The correct answer without justification - 0 points.
9.4-2. It is known that the quadratic equation $a x^{2}+b x+c=0$ has no solutions. Is it true that the quadratic equation $(a+3 c) x^{2}+4 b x+(3 a+c)=0$ also has no solutions?
Answer: True.
Solution: Note that if the quadratic equation $a x^{2}+b x+c=0$ has no solutions, then the quadratic equation $c x^{2}+b x+a=0$ also has no solutions, as these equations have the same discriminant. Moreover, if the quadratic equation $a x^{2}+b x+c=0$ has no solutions, then the signs of the coefficients $a$ and $c$ will be the same (otherwise, the discriminant would be non-negative). This means that the graphs of the quadratic functions $f(x)=a x^{2}+b x+c$ and $g(x)=c x^{2}+b x+a$ either both lie above the $O x$ axis or both lie below the $O x$ axis. But then the graph of the sum $h(x)=f(x)+3 g(x)=(a+3 c) x^{2}+4 b x+(3 a+c)$ also does not intersect the $O x$ axis.
Comment: The correct answer without justification - 0 points.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
28. Circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii intersect at points $B$ and $C$. A point $A$ is chosen on circle $\Omega_{1}$. Ray $A B$ intersects circle $\Omega_{2}$ at point $D$ (point $B$ lies between points $A$ and $D$). On ray $D C$, a point $E$ is chosen such that $D C = C E$. Find $A E$, if param1.
| param1 | Answer |
| :---: | :---: |
| $A C=5, A D=6$ | |
| $A C=5, A D=8$ | |
| $A C=13, A D=10$ | |
| $A C=13, A D=24$ | |
|
28. Circles $\Omega_{1}$ and $\Omega_{2}$ of equal radii intersect at points $B$ and $C$. A point $A$ is chosen on circle $\Omega_{1}$. Ray $A B$ intersects circle $\Omega_{2}$ at point $D$ (point $B$ lies between points $A$ and $D$). On ray $D C$, a point $E$ is chosen such that $D C = C E$. Find $A E$, if param1.
| param1 | Answer |
| :---: | :---: |
| $A C=5, A D=6$ | 8 |
| $A C=5, A D=8$ | 6 |
| $A C=13, A D=10$ | 24 |
| $A C=13, A D=24$ | 10 |
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given quadratic trinomials $f_{1}(x)=x^{2}-a x-3, f_{2}(x)=x^{2}+2 x-b, f_{3}(x)=3 x^{2}+(2-2 a) x-6-b$ and $f_{4}(x)=3 x^{2}+(4-a) x-3-2 b$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C, D, a, b$ are not given.
|
Answer: 3.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=\frac{\sqrt{T}}{|a|}$. Applying this formula four times, we get
$$
A=\sqrt{a^{2}+12}, B=\sqrt{4+4 b}, C=\frac{1}{3} \sqrt{(2-2 a)^{2}+12(6+b)}, D=\frac{1}{3} \sqrt{(4-a)^{2}+12(3+2 b)}
$$
From this, it follows that $C^{2}-D^{2}=\frac{1}{9}\left(\left(4 a^{2}-8 a+12 b+76\right)-\left(a^{2}-8 a+24 b+52\right)\right)=\frac{1}{3}\left(a^{2}-4 b+8\right), A^{2}-$ $B^{2}=a^{2}-4 b+8$. Therefore, the desired ratio is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|y|+|4+y| \leqslant 4 \\
\frac{x-y^{2}-4 y-3}{2 y-x+3} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area.
|
Answer: 8.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $y<0$. Then $y+4-y \leqslant 4 \Leftrightarrow 4 \leqslant 4$, i.e., the solution is $y \in \mathbb{R}$, but since $y<0$, we have $y \in (-\infty; 0)$.
2) $y=0$. Then $y+4-y \leqslant 4 \Leftrightarrow 4 \leqslant 4$, i.e., the solution is $y=0$.
3) $y>0$. Then $y+4+y \leqslant 4 \Leftrightarrow y \leqslant 0$, i.e., there are no solutions.
Combining the results, we get that $y \in[-4 ; 0]$.
Now, let's move to the second inequality. The denominator of the fraction in its left part is zero at points belonging to the line $x=2 y+3$ (let's call it $\ell$; in this case, the inequality is not satisfied, as the fraction is undefined). The numerator of the fraction is zero when $x-y^{2}-4 y-3=0 \Leftrightarrow x=(y+2)^{2}-1$. This set of points is a parabola with branches to the right and the vertex at point $C(-1 ;-2)$. The points of intersection of the line and the parabola can be determined from the system of equations $\left\{\begin{array}{l}x=2 y+3, \\ x=y^{2}+4 y+3\end{array} \Leftrightarrow\left\{\begin{array}{l}x=2 y+3, \\ y^{2}+2 y=0 .\end{array}\right.\right.$ From this, we get two points - $A(3 ; 0)$ and $C(-1 ;-2)$.
The second inequality is satisfied:
- at points on the parabola (except points $A$ and $C$);
- at points to the right of the parabola and above the line (in this case, both the numerator and the denominator of the fraction are positive);
- at points to the left of the parabola and below the line (both the numerator and the denominator of the fraction are negative).
Considering also the restriction $y \in[-4 ; 0]$ from the first inequality, we get that the set $M$ is the union of two sets $M_{1}$ and $M_{2}$; the first of them is a curvilinear triangle $B C D$, where $B(3 ;-4)$ and $D(-5 ;-4)$ are the points of intersection of the parabola and the line $\ell$ with the line $y=-4$ (its sides are segments $C D, B D$ and the arc of the parabola $B C$), and the second is the region bounded by the segment $A C$ and the arc of the parabola $A C$ (in this case, all points of the line $A C$ do not belong to the set, and all other boundary points do belong).
From the symmetry of the parabola relative to its axis (i.e., the line $y=-2$), it follows that the area of the figure $M_{3}$, bounded by the segment $B C$ and the arc of the parabola $B C$, is equal to the area of $M_{2}$. But $M_{1} \cup M_{3}=\triangle B C D$, and the area of this triangle is easy to find: $S_{\triangle B C D}=\frac{1}{2} \cdot 8 \cdot 2=8$.
## Ticket 2
|
8
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given quadratic trinomials $f_{1}(x)=x^{2}-2 x+a, f_{2}(x)=x^{2}+b x-2, f_{3}(x)=4 x^{2}+(b-6) x+3 a-2$ and $f_{4}(x)=4 x^{2}+(3 b-2) x-6+a$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C, D, a, b$ are not given.
|
Answer: 2.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=\frac{\sqrt{T}}{|a|}$. Applying this formula four times, we get
$$
A=\sqrt{4-4 a}, B=\sqrt{b^{2}+8}, C=\frac{1}{4} \sqrt{(b-6)^{2}-16(3 a-2)}, D=\frac{1}{4} \sqrt{(3 b-2)^{2}-16(a-6)}
$$
From this, it follows that $C^{2}-D^{2}=\frac{1}{16}\left(\left(b^{2}-48 a-12 b+68\right)-\left(9 b^{2}-16 a-12 b+100\right)\right)=\frac{1}{2}\left(-b^{2}-4 a-4\right)$, $A^{2}-B^{2}=-b^{2}-4 a-4$. Therefore, the desired ratio is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|x|+|4-x| \leqslant 4 \\
\frac{x^{2}-4 x-2 y+2}{y-x+3} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area.
|
Answer: 4.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $x<0$. Then $-x-x+4 \leqslant 4 \Leftrightarrow -2x \leqslant 0 \Leftrightarrow x \geqslant 0$, i.e., there are no solutions in this case.
2) $0 \leqslant x \leqslant 4$. Then $x-x+4 \leqslant 4 \Leftrightarrow 4 \leqslant 4$, which is always true, so all $x \in [0; 4]$ are solutions.
3) $x>4$. Then $x+x-4 \leqslant 4 \Leftrightarrow x \leqslant 4$, i.e., there are no solutions in this case either.
Combining the results, we get that $x \in [0; 4]$.
Now let's move to the second inequality. The denominator of the fraction in its left part is zero at points belonging to the line $y=x-3$ (in this case, the inequality is not satisfied, as the fraction is undefined). The numerator of the fraction is zero when $x^{2}-4 x-2 y+2=0 \Leftrightarrow y=\frac{1}{2}(x-2)^{2}-1$. This set of points is a parabola with branches upwards and vertex at point $C(2; -1)$. Note also that the parabola intersects the y-axis at point $B(0; 1)$, and the line intersects the y-axis at point $D(0; -3)$. The points of intersection of the line and the parabola can be determined from the system of equations $\left\{\begin{array}{l}y=x-3, \\ x^{2}-4 x-2 y+2=0\end{array} \Leftrightarrow\left\{\begin{array}{l}y=x-3, \\ x^{2}-6 x+8=0 .\end{array}\right.\right.$
From this, we get two points - $A(4; 1)$ and $C(2; -1)$.
The second inequality is satisfied:
- at points on the parabola (except points $A$ and $C$);
- at points below the parabola and above the line (in this case, both the numerator and the denominator of the fraction are positive);
- at points above the parabola and below the line (both the numerator and the denominator of the fraction are negative).
Considering also the restriction $x \in [0; 4]$ from the first inequality, we get that the set $M$ is the union of two sets $M_{1}$ and $M_{2}$; the first of which is a curvilinear triangle $B C D$ (its sides are segments $C D, B D$ and the arc of the parabola $B C$), and the second is the region bounded by segment $A C$ and the arc of the parabola $A C$ (in this case, all points of the line $A C$ do not belong to the set, while all other boundary points do belong).
From the symmetry of the parabola relative to its axis (i.e., the line $x=2$), it follows that the area of the figure $M_{3}$, bounded by segment $B C$ and the arc of the parabola $B C$, is equal to the area of $M_{2}$. But $M_{1} \cup M_{3}=\triangle B C D$, and the area of this triangle is easy to find: $S_{\triangle B C D}=\frac{1}{2} \cdot 4 \cdot 2=4$.
The problem is considered fully solved (and the maximum number of points is awarded), only if all necessary transformations are provided in the solution text and all logical steps are fully explained; the obtained answers are also reduced to a simplified form.
The presence of a correct answer does not guarantee a positive score for the problem. A correct answer without justification does not add points.
A full and justified solution to the problem is awarded the full number of points (indicated in parentheses after the problem number). Some partial progress is evaluated according to the instructions. In other cases, the score is set at the discretion of the checker.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x-y \geqslant|x+y| \\
\frac{x^{2}-6 x+y^{2}-8 y}{3 y-x+6} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area.
|
Answer: 3.
Solution. The first inequality is equivalent to the system ${ }^{1}$ $\left\{\begin{array}{l}x+y \leqslant x-y, \\ x+y \geqslant y-x\end{array} \Leftrightarrow\left\{\begin{array}{l}y \leqslant 0, \\ x \geqslant 0 .\end{array}\right.\right.$
Consider the second inequality. It can be written as $\frac{(x-3)^{2}+(y-4)^{2}-25}{3 y-x+6} \geqslant 0$. The numerator of the fraction on the left side of the inequality is zero on the circle of radius 5 centered at point $Q(3 ; 4)$ (let's call it $\omega$). The denominator of the fraction is zero on the line $y=-2+\frac{x}{3}$ (let's call it $\ell$). The points of intersection of the circle and the line are determined from the system of equations
$$
\left\{\begin{array}{l}
x = 3 y + 6, \\
x ^ { 2 } - 6 x + y ^ { 2 } - 8 y = 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x = 3 y + 6 \\
( 3 y + 6 ) ^ { 2 } - 6 ( 3 y + 6 ) + y ^ { 2 } - 8 y = 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=3 y+6 \\
y^{2}+y=0
\end{array}\right.\right.\right.
$$
from which we obtain two points $A(3 ;-1)$ and $B(6 ; 0)$. Also, denote the origin by $O$, and the point of intersection of the line $\ell$ with the $O y$ axis by $C$ (it is not difficult to determine that the coordinates of point $C$ are $(0 ;-2)$).
The inequality is satisfied:
- at all points of the circle $\omega$ except points $A$ and $B$ (then the numerator of the fraction is zero);
- inside the circle $\omega$ at points located below the line $\ell$ (the numerator and denominator are negative)
- outside the circle $\omega$ at points located above the line $\ell$ (the numerator and denominator are positive).
Describe the set of points $M$ that satisfy the original system of inequalities. It consists of the segment of the circle bounded by the chord $A B$ and located below this chord, as well as the curvilinear triangle $A O C$, the boundaries of which are the arc $A O$ of the circle $\omega$ and the segments $A C$ and $C O$ (while the points of the line $\ell$ do not belong to the set, and the other boundary points do belong).
Note that due to symmetry, the segment of the circle located below the chord $A B$ is equal to the segment of the circle located below the chord $A O$. Therefore, the area of the figure $M$ is equal to the area of the triangle $A C O$, i.e., $\frac{1}{2} \cdot 3 \cdot 2=3$.[^0]
## Ticket 10
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y+x \geqslant|x-y| \\
\frac{x^{2}-8 x+y^{2}+6 y}{x+2 y-8} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area.
|
Answer: 8.
Solution. The first inequality is equivalent to the system $\left\{\begin{array}{l}x-y \leqslant x+y, \\ x-y \geqslant-x-y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geqslant 0, \\ y \geqslant 0 .\end{array}\right.\right.$.
Consider the second inequality. It can be written as $\frac{(x-4)^{2}+(y+3)^{2}-25}{x+2 y-8} \leqslant 0$. The numerator of the fraction on the left side of the inequality is zero on the circle of radius 5 centered at the point $Q(4 ;-3)$ (let's call it $\omega$). The denominator of the fraction is zero on the line $y=4-\frac{x}{2}$ (let's call it $\ell$). The points of intersection of the circle and the line are determined from the system of equations
$$
\left\{\begin{array}{l}
x=8-2 y, \\
x^{2}-8 x+y^{2}+6 y=0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=8-2 y \\
(8-2 y)^{2}-8(8-2 y)+y^{2}+6 y=0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=8-2 y \\
y^{2}-2 y=0
\end{array}\right.\right.\right.
$$
from which we obtain two points $A(4 ; 2)$ and $B(8 ; 0)$. Also, denote the origin by $O$, and the intersection of the line $\ell$ with the $O y$ axis by $C$ (it is not difficult to determine that the coordinates of point $C$ are $(0 ; 4)$). The inequality is satisfied:
- at all points of the circle $\omega$ except points $A$ and $B$ (then the numerator of the fraction is zero);
- inside the circle $\omega$ at points located above the line $\ell$ (the numerator is negative, and the denominator is positive)
- outside the circle $\omega$ at points located below the line $\ell$ (the numerator is positive, and the denominator is negative).
Describe the set of points $M$ that satisfy the original system of inequalities. It consists of the segment of the circle bounded by the chord $A B$ and located above this chord, as well as the curvilinear triangle $A O C$, the boundaries of which are the arc $A O$ of the circle $\omega$ and the segments $A C$ and $C O$ (while the points of the line $\ell$ do not belong to the set, and the other boundary points do belong).
Note that due to symmetry, the segment of the circle located above the chord $A B$ is equal to the segment of the circle located above the chord $A O$. Therefore, the area of the figure $M$ is equal to the area of the triangle $A C O$, i.e., $\frac{1}{2} \cdot 4 \cdot 4=8$.[^1]
The problem is considered fully solved (and the maximum number of points is awarded), only if all necessary transformations are provided in the solution text and all logical steps are fully explained; the answers obtained are reduced to a simplified form.
The presence of a correct answer does not guarantee a positive score for the problem. A correct answer without justification does not add points.
A full and justified solution to the problem is awarded the full number of points (indicated in parentheses after the problem number). Some partial progress is evaluated according to the instructions. In other cases, the score is set at the discretion of the checker.
|
8
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $p$, for each of which the numbers $p-2$, $2 \cdot \sqrt{p}$, and $-3-p$ are respectively the first, second, and third terms of some geometric progression.
|
Answer: $p=1$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(p-2)(-p-3)$, from which
$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+5 p-6=0
\end{array} \Leftrightarrow p=1\right.
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $p$, for each of which the numbers $-p-12, 2 \cdot \sqrt{p}$, and $p-5$ are respectively the first, second, and third terms of some geometric progression.
|
Answer: $p=4$.
Solution. For the specified numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(2 \sqrt{p})^{2}=(-p-12)(p-5)$, from which
$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+11 p-60=0
\end{array} \Leftrightarrow p=4\right.
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $p$, for each of which the numbers $p-2$, $3 \cdot \sqrt{p}$, and $-8-p$ are respectively the first, second, and third terms of some geometric progression.
|
Answer: $p=1$.
Solution. For the given numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 \sqrt{p})^{2}=(p-2)(-8-p)$, from which
$$
\left\{\begin{array}{l}
p>0 \\
p^{2}+15 p-16=0
\end{array} \Leftrightarrow p=1\right.
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all values of $p$, for each of which the numbers $-p-8$, $3 \cdot \sqrt{p}$, and $p-7$ are the first, second, and third terms, respectively, of some geometric progression.
|
Answer: $p=4$.
Solution. For the specified numbers to be consecutive terms of a geometric progression, it is necessary and sufficient that they are non-zero and $(3 \sqrt{p})^{2}=(-p-8)(p-7)$, from which
$$
\left\{\begin{array}{l}
p>0, \\
p^{2}+10 p-56=0
\end{array} \Leftrightarrow p=4\right.
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. It is known that $\operatorname{tg}(\alpha+2 \gamma)+2 \operatorname{tg} \alpha-4 \operatorname{tg}(2 \gamma)=0, \operatorname{tg} \gamma=\frac{1}{3}$. Find $\operatorname{ctg} \alpha$.
|
Answer: 2 or $\frac{1}{3}$.
Solution. $\operatorname{tg} 2 \gamma=\frac{2 \operatorname{tg} \gamma}{1-\operatorname{tg}^{2} \gamma}=\frac{2 / 3}{1-1 / 9}=\frac{3}{4}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \gamma}{1-\operatorname{tg} \alpha \operatorname{tg} 2 \gamma}+2 \operatorname{tg} \alpha-3=0 \Leftrightarrow \frac{\operatorname{tg} \alpha+\frac{3}{4}}{1-\frac{3}{4} \operatorname{tg} \alpha}+2 \operatorname{tg} \alpha-3=0
$$
By bringing to a common denominator and simplifying, we obtain the equivalent equation on the domain of definition $2 \operatorname{tg}^{2} \alpha-7 \operatorname{tg} \alpha+3=0$, from which $\operatorname{tg} \alpha=3$ or $\operatorname{tg} \alpha=\frac{1}{2}$. Both options are valid (since the denominators do not become zero). Therefore, $\operatorname{ctg} \alpha=\frac{1}{3}$ or $\operatorname{ctg} \alpha=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. It is known that $\operatorname{tg}(2 \alpha-\beta)-4 \operatorname{tg} 2 \alpha+4 \operatorname{tg} \beta=0, \operatorname{tg} \alpha=-3$. Find $\operatorname{ctg} \beta$.
|
Answer: -1 or $\frac{4}{3}$.
Solution. $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{-6}{1-9}=\frac{3}{4}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \operatorname{tg} \beta}-4 \operatorname{tg} 2 \alpha+4 \operatorname{tg} \beta=0 \Leftrightarrow \frac{\frac{3}{4}-\operatorname{tg} \beta}{1+\frac{3}{4} \operatorname{tg} \beta}-3+4 \operatorname{tg} \beta=0
$$
By bringing to a common denominator and simplifying, we obtain the equivalent equation on the domain of definition $4 \operatorname{tg}^{2} \beta+\operatorname{tg} \beta-3=0$, from which $\operatorname{tg} \beta=\frac{3}{4}$ or $\operatorname{tg} \beta=-1$. Both options are valid (since the denominators do not become zero). Therefore, $\operatorname{ctg} \beta=\frac{4}{3}$ or $\operatorname{ctg} \beta=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. [5 points] Point $D$ lies on side $A C$ of triangle $A B C$. The circle with diameter $B D$ intersects sides $A B$ and $B C$ at points $P$ and $T$ respectively. Points $M$ and $N$ are the midpoints of segments $A D$ and $C D$ respectively. It is known that $P M \| T N$.
a) Find the angle $A B C$.
b) Suppose additionally that $M P=1, N T=\frac{3}{2}, B D=\sqrt{5}$. Find the area of triangle $A B C$.
|
Answer: (a) $90^{\circ} ;$ ( $\mathbf{\text { ( ) }} 5$.
Solution. (a) Points $P$ and $T$ lie on the circle with diameter $B D$, therefore $\angle B P D=\angle B T D=90^{\circ}$. Consequently, triangles $A D P$ and $D C T$ are right-angled; $P M$ and $T N$ are their medians. Since the median of a right-angled triangle, drawn to the hypotenuse, is equal to half of it, $T N=C N=D N, P M=$ $A M=D M$. Let $\angle T C D=\gamma$. Since triangle $C T N$ is isosceles, and $\angle C T N=\gamma, \angle T N D=$ $2 \gamma$ (as the external angle of $\triangle C T N$ ). Angles $P M A$ and $T N D$ are equal due to the parallelism of lines $P M$ and $T N$. And since triangle $A M P$ is also isosceles, $\angle P A M=90^{\circ}-\frac{1}{2} \angle P M A=90^{\circ}-\frac{1}{2} \angle T N D=$ $90^{\circ}-\gamma$. Therefore, the sum of angles $A$ and $C$ of triangle $A B C$ is $90^{\circ}$, and its third angle $\angle A B C$ is also $90^{\circ}$.
(b) As stated above, $C D=2 N T=3, A D=2 M P=2$. Let $\angle A D B=\psi$. Then $\angle B D C=$ $180^{\circ}-\psi$. By the cosine theorem for triangles $A B D$ and $A C D$, we get that $A B^{2}=4+5-4 \sqrt{5} \cos \psi$, $B C^{2}=9+5-6 \sqrt{5} \cos \left(180^{\circ}-\psi\right)$. But by the Pythagorean theorem $A B^{2}+B C^{2}=A C^{2}=25$, from which it follows that $9-4 \sqrt{5} \cos \psi+14+6 \sqrt{5} \cos \psi=25, \cos \psi=\frac{1}{\sqrt{5}}$. Further, we find: $\sin \psi=\sqrt{1-\cos ^{2} \psi}=\frac{2}{\sqrt{5}}, S_{\triangle A B C}=$ $S_{\triangle A B D}+S_{\triangle B C D}=\frac{1}{2} D A \cdot D B \sin \psi+\frac{1}{2} D C \cdot D B \sin \left(180^{\circ}-\psi\right)=\frac{1}{2} A C \cdot B D \sin \psi=\frac{1}{2} \cdot 5 \cdot \sqrt{5} \cdot \frac{2}{\sqrt{5}}=5$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. [6 points] Solve the equation $\sqrt{x+1}-\sqrt{4-x}+3=2 \sqrt{4+3 x-x^{2}}$.
|
Answer: $3, \frac{3-2 \sqrt{6}}{2}$.
Solution. Let $\sqrt{x+1}-\sqrt{4-x}=t$. Squaring both sides of this equation, we get $(x+1)-2 \sqrt{(x+1)(4-x)}+(4-x)=t^{2}$, from which $2 \sqrt{4+3 x-x^{2}}=5-t^{2}$. The equation becomes $t+3=5-t^{2}$; hence $t^{2}+t-2=0$, i.e., $t=1$ or $t=-2$. We consider each case separately.
$$
\begin{aligned}
& \sqrt{x+1}-\sqrt{4-x}=1 \Leftrightarrow \sqrt{x+1}=1+\sqrt{4-x} \Leftrightarrow\left\{\begin{array}{l}
x+1=1+4-x+2 \sqrt{4-x} \\
-1 \leqslant x \leqslant 4
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l }
{ \sqrt { 4 - x } = x - 2 , } \\
{ - 1 \leqslant x \leqslant 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
4-x=x^{2}-4 x+4 \\
2 \leqslant x \leqslant 4
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l }
{ x ^ { 2 } - 3 x = 0 , } \\
{ 2 \leqslant x \leqslant 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=0 \text { or } x=3, \\
2 \leqslant x \leqslant 4
\end{array} \quad \Leftrightarrow x=3 ;\right.\right. \\
& \sqrt{x+1}-\sqrt{4-x}=-2 \Leftrightarrow \sqrt{x+1}+2=\sqrt{4-x} \Leftrightarrow\left\{\begin{array}{l}
x+1+4 \sqrt{x+1}+4=4-x \\
-1 \leqslant x \leqslant 4
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array} { l }
{ 4 \sqrt { x + 1 } = - 1 - 2 x , } \\
{ - 1 \leqslant x \leqslant 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
16 x+16=1+4 x+4 x^{2}, \\
-1 \leqslant x \leqslant-0.5
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l }
{ 4 x ^ { 2 } - 1 2 x - 1 5 = 0 , } \\
{ - 1 \leqslant x \leqslant - 0 . 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{3 \pm 2 \sqrt{6}}{2}, \\
-1 \leqslant x \leqslant-0.5
\end{array} \quad \Leftrightarrow x=\frac{3-2 \sqrt{6}}{2} .\right.\right.
\end{aligned}
$$
Thus, the equation has two roots: $x=\frac{3-2 \sqrt{6}}{2}$ and $x=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
\sqrt{x^{2}-3 y^{2}+4 x+4} \leqslant 2 x+1 \\
x^{2}+y^{2} \leqslant 4
\end{array}\right.
$$
Determine how many parts the figure $\Phi$ consists of.
|
Solution. The first inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - 3 y ^ { 2 } + 4 x + 4 \leqslant 4 x ^ { 2 } + 4 x + 1 , } \\
{ ( x + 2 ) ^ { 2 } - 3 y ^ { 2 } \geqslant 0 , } \\
{ 2 x + 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2} \geqslant 1 \\
(x+2-y \sqrt{3})(x+2+y \sqrt{3}) \geqslant 0 \\
x \geqslant-0.5
\end{array}\right.\right.
$$
The first of these inequalities, together with the second inequality of the original system, defines the set of points located between two concentric circles centered at ( $0 ; 0$ ) with radii 1 and 2. The third inequality defines the half-plane to the right of the line $x=-0.5$. The second inequality defines two vertical angles, the boundaries of which are the lines $\ell_{1}$ and $\ell_{2}$ with equations $y= \pm \frac{x+2}{\sqrt{3}}$ (such that the point $(0 ; 0)$ lies inside one of these angles). The lines $\ell_{1}$ and $\ell_{2}$ both pass through the point ( $-2 ; 0$ ), which lies on the larger circle, and are tangent to the smaller circle at the points $\left(-\frac{1}{2} ; \pm \frac{\sqrt{3}}{2}\right)$ (the coordinates of the points of tangency can be found by solving the corresponding systems of equations).
By intersecting all the specified sets, we obtain the figure $\Phi$, which, as can be easily seen, consists of one part.
## TICKET 14
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied
$$
\left\{\begin{array}{l}
\sqrt{y^{2}-8 x^{2}-6 y+9} \leqslant 3 y-1 \\
x^{2}+y^{2} \leqslant 9
\end{array}\right.
$$
Determine how many parts the figure $\Phi$ consists of.
|
Solution. The first inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ y ^ { 2 } - 8 x ^ { 2 } - 6 y + 9 \leqslant 9 y ^ { 2 } - 6 y + 1 , } \\
{ ( y - 3 ) ^ { 2 } - 8 x ^ { 2 } \geqslant 0 , } \\
{ 3 y - 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2} \geqslant 1 \\
(y-3-2 x \sqrt{2})(y-3+2 x \sqrt{2}) \geqslant 0 \\
y \geqslant \frac{1}{3}
\end{array}\right.\right.
$$
The first of these inequalities, together with the second inequality of the original system, defines the set of points lying between two concentric circles centered at ( $0 ; 0$ ) with radii 1 and 2. The third inequality defines the half-plane above the line $y=\frac{1}{3}$. The second inequality defines two vertical angles, the boundaries of which are the lines $\ell_{1}$ and $\ell_{2}$ with equations $y=3 \pm 2 x \sqrt{2}$ (such that the point $(0 ; 0)$ lies inside one of these angles). The lines $\ell_{1}$ and $\ell_{2}$ both pass through the point $(3 ; 0)$ and are tangent to the smaller circle at the points $\left( \pm \frac{2 \sqrt{2}}{3} ; \frac{1}{3}\right)$ (the coordinates of the points of tangency can be found by solving the corresponding systems of equations).
By intersecting all the specified sets, we obtain the figure $\Phi$, which, as can be easily seen, consists of one part.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
30. It is known that for pairwise distinct numbers $a, b, c$, the equality param1 holds. What is the smallest value that the expression $a+b+c$ can take?
| param1 | Answer |
| :---: | :---: |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+2\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -2 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+4\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -4 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+6\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -6 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+8\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -8 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+10\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -10 |
|
30. It is known that for pairwise distinct numbers $a, b, c$, the equality param1 holds. What is the smallest value that the expression $a+b+c$ can take?
| param1 | Answer |
| :---: | :---: |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+2\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -2 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+4\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -4 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+6\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -6 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+8\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -8 |
| $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)+10\left(a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)\right)=0$ | -10 |
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (12 points) The sequence of functions is defined by the formulas:
$$
f_{0}(x)=2 \sqrt{x}, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}, n=0,1,2 \ldots, x \in[4 ; 9]
$$
Find $f_{2023}(4)$.
|
Answer: -2.
Solution. It is easy to calculate that $f_{3}(x)=f_{0}(x)$, therefore
$$
f_{2023}(x)=f_{1}(x)=\frac{2}{1-\sqrt{x}}
$$
Then $f_{2023}(4)=-2$.
Remark. One can immediately compute the values of the functions at the given point. The sequence will be $f_{0}(4)=4, f_{1}(4)=-2, f_{2}(4)=1, f_{3}(4)=4 \ldots$
Evaluation criteria. A fully justified solution - 12 points. Calculation errors - minus 1 point. The relation $f_{3}(x)=f_{0}(x)$ found - 7 points, the equality $f_{2023}(x)=f_{1}(x)$ found - another 4 points.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (12 points) The sequence of functions is defined by the formulas:
$$
f_{0}(x)=2 \sqrt{x}, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}, n=0,1,2 \ldots, x \in[4 ; 9]
$$
Find $f_{2023}(9)$.
|
Answer: -1.
Solution. It is easy to calculate that $f_{3}(x)=f_{0}(x)$, therefore,
$$
f_{2023}(x)=f_{1}(x)=\frac{2}{1-\sqrt{x}}
$$
Then $f_{2023}(9)=-1$.
Remark. One can immediately compute the values of the functions at the given point. The sequence will be $f_{0}(9)=6, f_{1}(9)=-1, f_{2}(9)=\frac{4}{3}, f_{3}(9)=6 \ldots$
Evaluation Criteria. A fully justified solution - 12 points. Calculation errors - minus 1 point. The relation $f_{3}(x)=f_{0}(x)$ found - 7 points, the equality $f_{2023}(x)=f_{1}(x)$ found - another 4 points.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Solve the equation
$$
2 x+2+\operatorname{arctg} x \cdot \sqrt{x^{2}+1}+\operatorname{arctg}(x+2) \cdot \sqrt{x^{2}+4 x+5}=0
$$
|
Answer: -1.
Solution. Let $f(x)=x+\operatorname{arctg} x \cdot \sqrt{x^{2}+1}$. The original equation can be rewritten as $f(x)+f(x+2)=0$. Note that the function $f(x)$ is odd. It is increasing on the positive half-axis (as the sum of increasing functions). Due to its oddness, it is increasing on the entire real line. Further, we have
$$
f(x)=-f(x+2) ; \quad f(x)=f(-x-2)
$$
Since an increasing function takes each of its values exactly once, it follows that $x=-x-2$, from which $x=-1$.
Remark. The monotonicity of the left side of the original equation could also have been established using the derivative.
Evaluation. 12 points for a correct solution. If the answer is guessed but not proven that there are no other solutions, 3 points.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem No. 6 (10 points)
The density of a body is defined as the ratio of its mass to the volume it occupies. A homogeneous cube with a volume of \( V = 8 \, \text{m}^3 \) is given. As a result of heating, each of its edges increased by 4 mm. By what percentage did the density of this cube change?
Answer: decreased by \( 6 \% \)
|
# Solution and evaluation criteria:
Volume of the cube: $v=a^{3}$, where $a$ is the length of the edge, therefore:
$a=2 \partial m=200 mm$.
Final edge length: $a_{\kappa}=204$ mm.
Thus, the final volume: $V_{\kappa}=a_{K}^{3}=2.04^{3}=8.489664 \partial \mu^{3} \approx 1.06 V$.
Therefore, the density:
$\rho_{K}=\frac{m}{V_{K}}=\frac{m}{1.06 V}=0.94 \frac{m}{V} \approx 0.94 \rho$. The density has decreased by approximately $6 \%$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem № 7 (10 points)
A person and his faithful dog started moving along the perimeter of a block from point A at the same time $t_{0}=0$ min. The person moved with a constant speed clockwise, while the dog ran with a constant speed counterclockwise (see fig.). It is known that they met for the first time after $t_{1}=1$ min from the start of the movement. This meeting occurred at point $B$. Given that they continued to move in the same direction and at the same speed after this, determine at what moment in time they will next be at point $B$ simultaneously. Note that $A B=C D=100$ m, $B C=A D=300$ m.
|
# Answer: in 9 min
## Solution and evaluation criteria:
In $t_{1}=1$ min, the person and the dog together covered a distance equal to the perimeter of the block, with the person moving 100 meters from the starting point of the journey.
That is, during each subsequent meeting, the person will be 100 meters away from the location of the previous meeting.
Perimeter of the block: $A B+B C+C D+D A=100+300+100+300=800$ meters.
So, it will take another 8 minutes.
And the next time the person and the dog will be at the point at the moment: $t_{\kappa}=9$ min.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem No. 6 (10 points)
The density of a body is defined as the ratio of its mass to the volume it occupies. A homogeneous cube with a volume of \( V = 27 \partial \mu^{3} \) is given. As a result of heating, each of its edges increased by 9 mm. By what percentage did the density of this cube change?
|
# Answer: decreased by $8 \%$
## Solution and evaluation criteria:
Volume of a cube: $v=a^{3}$, where $a$ is the length of the edge, therefore:
$a=3 dm=300 mm$.
Final edge length: $a_{K}=309 mm$.
Thus, the final volume: $V_{K}=a_{K}^{3}=3.09^{3}=29.503629 dm^{3} \approx 1.09 V$.
Therefore, the density:
$\rho_{K}=\frac{m}{V_{K}}=\frac{m}{1.09 V}=0.92 \frac{m}{V} \approx 0.92 \rho$. The density has decreased by approximately $8 \%$. (4 points)
#
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (16 points) Fresh mushrooms contain $80 \%$ water by mass, while dried mushrooms contain $20 \%$ water. How many kg of dried mushrooms can be obtained from 20 kg of fresh mushrooms?
|
# Answer: 5 kg
Solution. The dry matter in fresh mushrooms is 4 kg, which constitutes $80 \%$ of the weight in dried mushrooms. By setting up the corresponding proportion, we will find the weight of the dried mushrooms.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (20 points) A ball was thrown from the surface of the Earth at an angle of $30^{\circ}$ with a speed of $v_{0}=20 \mathrm{M} / \mathrm{c}$. How long will it take for the velocity vector of the ball to turn by an angle of $60^{\circ}$? Neglect air resistance. The acceleration due to gravity is $g=10 \mathrm{M} / \mathrm{c}^{2}$.
|
Answer: $t=2 s$
Solution. In fact, the problem requires finding the total flight time of the ball. The $y$-coordinate of the ball changes according to the law: $y=v_{0 y} t-\frac{g t^{2}}{2}=v_{0} \sin 30^{\circ} t-\frac{g t^{2}}{2}=10 \cdot t-5 t^{2}=0$
From this, we obtain that: $t=2 s$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Let
$$
\sqrt{49-a^{2}}-\sqrt{25-a^{2}}=3
$$
Calculate the value of the expression
$$
\sqrt{49-a^{2}}+\sqrt{25-a^{2}} .
$$
|
Answer: 8.
Solution. Let
$$
\sqrt{49-a^{2}}+\sqrt{25-a^{2}}=x
$$
Multiplying this equality by the original one, we get $24=3x$.
Evaluation. Full solution: 11 points.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The infantry column stretched out over 1 km. Sergeant Kim, riding out on a gyro-scooter from the end of the column, reached its beginning and returned to the end. The infantrymen walked 2 km 400 m during this time. How far did the sergeant travel during this time?
|
Answer: 3 km $600 \mathrm{~m}$
Solution. Let the speed of the column be $x$ km/h, and the sergeant travels $k$ times faster, i.e., at a speed of $k x$ km/h. To reach the end of the column, Kim traveled $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction, $t_{2}=\frac{1}{k x+x}$ hours (meeting head-on). During this time, the column covered 2.4 km, i.e., $x\left(t_{1}+t_{2}\right)=2.4$. Substituting the expressions for $t_{1}$ and $t_{2}$, we get
$$
\frac{x}{k x-x}+\frac{x}{k x+x}=2.4 ; \quad \frac{1}{k-1}+\frac{1}{k+1}=2.4 ; \quad 2 k=2.4\left(k^{2}-1\right) .
$$
The obtained quadratic equation has a single positive root $k=\frac{3}{2}$. The sergeant travels 1.5 times faster than the column. Therefore, the distance he covers will also be 1.5 times greater.
Evaluation. Full solution: 12 points.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) A tourist travels from point $A$ to point $B$ in 1 hour 56 minutes. The route from $A$ to $B$ goes uphill first, then on flat ground, and finally downhill. What is the length of the road on flat ground if the tourist's speed downhill is 6 km/h, uphill is 4 km/h, and on flat ground is 5 km/h, and the total distance between $A$ and $B$ is $x$ km? Additionally, the distances uphill and on flat ground are whole numbers of kilometers.
|
# Answer: 3
Solution. Let $x$ km be the distance the tourist walks uphill, $y$ km - on flat ground, then $9-x-y$ km - downhill. We get $\frac{x}{4}+\frac{y}{5}+\frac{9-x-y}{6}=\frac{29}{15}$. After transformations, $5 x+2 y=26$. It is obvious that $x$ must be even and $x+y \leq 9$. The only solution is $x=4, y=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (16 points) A chess player played 40 chess games and scored 25 points (1 point for each win, -0.5 points for a draw, 0 points for a loss). Find the difference between the number of his wins and the number of his losses.
#
|
# Answer: 10
Solution. Let the chess player have $n$ wins and $m$ losses. Then we get $n+0.5 \cdot(40-n-m)=25$. In the end, $n-m=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) A tourist travels from point $A$ to point $B$ in 2 hours and 14 minutes. The route from $A$ to $B$ goes uphill first, then on flat terrain, and finally downhill. What is the length of the uphill road if the tourist's speed downhill is 6 km/h, uphill is 4 km/h, and on flat terrain is 5 km/h, and the total distance between $A$ and $B$ is 10 km? Additionally, the distances uphill and on flat terrain are whole numbers of kilometers.
|
Answer: 6
Solution. Let $x$ km be the distance the tourist walks uphill, $y$ km - on flat ground, then $10-x-y$ km - downhill. We get $\frac{x}{4}+\frac{y}{5}+\frac{10-x-y}{6}=\frac{67}{30}$. After transformations, $5 x+2 y=34$. It is obvious that $x$ must be even and $x+y \leq 10$. The only solution is $x=6, y=2$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) Find the smallest root of the equation
$$
\sqrt{x+2}+2 \sqrt{x-1}+3 \sqrt{3 x-2}=10
$$
|
Answer: 2
Solution. It is clear that 2 is a root of the equation. The function on the left side of the equation is increasing (as the sum of increasing functions). Therefore, there are no other roots.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (16 points) Kolya, after walking a quarter of the way from home to school, realized he had forgotten his workbook. If he does not go back for it, he will arrive at school 5 minutes before the bell, but if he does go back, he will be one minute late. How much time (in minutes) does the journey to school take?
|
# Answer: 12 min
Solution. The extra $2 / 4$ of the journey takes 6 min. Therefore, the entire journey to school will take 12 min.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) Find the largest root of the equation
$$
3 \sqrt{x-2}+2 \sqrt{2 x+3}+\sqrt{x+1}=11
$$
|
Answer: 3
Solution. It is clear that 3 is a root of the equation. The function on the left side of the equation is increasing (as the sum of increasing functions). Therefore, there are no other roots.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (16 points) A truck left the village of Mirny at a speed of 40 km/h. At the same time, a car left the city of Tikhaya in the same direction as the truck. In the first hour of the journey, the car traveled 50 km, and in each subsequent hour, it traveled 5 km more than in the previous hour. How many hours will it take for the car to catch up with the truck if the distance between the village and the city is 135 km?
|
# Answer: 6.
Solution. In the first hour of travel, the pursuit speed of the car relative to the truck was $50-40=10$ km/h. In each subsequent hour, the pursuit speed increases by 5 km/h. Thus, the pursuit speeds form an arithmetic progression $10,15,20 \ldots$ km/h. Let's find the number of hours $n$ until the cars meet. The sum of this progression is equal to the total distance the car will travel until it meets the truck $\frac{20+5(n-1)}{2} \cdot n=135$. After transformation, we get the equation $n^{2}+3 n-54=0$, the roots of which are $6,-9$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (16 points) A truck left the village of Mirny at a speed of 40 km/h. At the same time, a car left the city of Tikhaya in the same direction as the truck. In the first hour of the journey, the car traveled 50 km, and in each subsequent hour, it traveled 5 km more than in the previous hour. How many hours will it take for the car to catch up with the truck if the distance between the village and the city is 175 km?
|
# Answer: 7.
Solution. In the first hour of travel, the pursuit speed of the car relative to the truck was $50-40=10$ km/h. In each subsequent hour, the pursuit speed increases by 5 km/h. Thus, the pursuit speeds form an arithmetic progression $10,15,20 \ldots$ km/h. Let's find the number of hours $n$ until the cars meet. The sum of this progression is equal to the total distance the car will travel until it meets the truck $\frac{20+5(n-1)}{2} \cdot n=175$. After transformation, we get the equation $n^{2}+3 n-70=0$, the roots of which are $7,-10$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (15 points) Two heaters are connected sequentially to the same DC power source. The water in the pot boiled after $t_{1}=3$ minutes from the first heater. The same water, taken at the same initial temperature, boiled after $t_{2}=6$ minutes from the second heater. How long would it take for the water to boil if the heaters were connected in parallel? Neglect heat dissipation to the surroundings.
|
Answer: 2 min.
Solution. The amount of heat required for heating in the first case $Q=I^{2} R_{1} t_{1}$. The amount of heat required for heating in the second case $Q=$ $I^{2} R_{2} t_{2}$. The amount of heat required for heating in the case of parallel connection $Q=I^{2} \frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}} t$. As a result, we get:
$$
t=\frac{t_{1} t_{2}}{t_{1}+t_{2}}=\frac{3 \cdot 6}{3+6}=2 \text { min. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Two identical resistors with resistance $R$ each are connected in series and connected to a source of constant voltage $U$. An ideal voltmeter is connected in parallel to one of the resistors. Its readings were $U_{v}=10 B$. After that, the voltmeter was replaced with an ideal ammeter. The ammeter readings were $-I_{A}=10 \mathrm{~A}$. Determine the value of $R$. (10 points)
|
Answer: 2 Ohms
Solution. Voltage of the source: $U=U_{v}+U_{v}=20 V$ (4 points). The resistance of an ideal ammeter: $r_{A}=0$ Ohms (3 points). Therefore, the resistance of the resistor: $R=\frac{U}{I_{A}}=\frac{20}{10}=2$ Ohms (3 points).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The structure shown in the figure is in equilibrium. It is known that the length of the homogeneous rod $l=50 \mathrm{~cm}$, and its mass $m_{2}=2$ kg. The distance between the attachment points of the left thread to the rod $S=10$ cm. Determine the mass $m_{1}$ of the load. All threads are weightless and inextensible. The pulleys are weightless. (15 points)
|
Answer: 10 kg
Solution. From the equilibrium condition for the large block, it follows that the tension force in the left thread: $T_{n}=\frac{m g}{2}$ (5 points). The equilibrium condition for the rod relative to the attachment point of the right thread: $T_{n} \cdot l=T_{n} \cdot(l-S)+m_{2} g \cdot \frac{1}{2} l$ (5 points).
As a result, we get: $m_{1}=\frac{m_{2} l}{S}=\frac{2 \cdot 0.5}{0.1}=10$ kg (5 points).
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the cells of a $3 \times 3$ square, the numbers $1,2,3, \ldots, 9$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$
#
|
# Answer: 7.
Solution. Let's color the cells in a checkerboard pattern: let the corner and central cells be black, and the rest white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white ones, we get that the odd numbers are in the black cells. Their total sum is $1+3+5+7+9=$ 25. Therefore, the number in the central cell is $7=25-18$.
Evaluation. 13 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences
## Final Stage
$2017-2018$ academic year
## Problems, answers, and evaluation criteria
## 6th Grade Variant 1
## physics
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the cells of a $3 \times 3$ square, the numbers $0,1,2, \ldots, 8$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is 18?
|
Answer: 2.
Solution. Let's color the cells in a checkerboard pattern: let the corner and central cells be black, and the rest white. From the condition, it follows that cells of different colors contain numbers of different parity. Since there are five black cells and four white cells, we get that the even numbers are in the black cells. Their total sum is \(0+2+4+6+8=20\). Therefore, the number in the central cell is \(2=20-18\).
Evaluation. 13 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences
## Final Stage
2017-2018 academic year
## Problems, Answers, and Evaluation Criteria
## 6th Grade Variant 2
physics
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Thirteen identical metal rods are connected as follows (see fig.). It is known that the resistance of one rod \( R_{0}=10 \) Ohms. Determine the resistance of the entire structure if it is connected to a current source at points \( A \) and \( B \). (10 points)
|
Answer: $4 O m$
Solution. The equivalent circuit looks as follows (5 points),
where each of the resistors has a resistance of $R_{0}=10$ Ohms. As a result, the total resistance is: $R=\frac{4}{10} R_{0}=4$ Ohms (5 points).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the cells of a $3 \times 3$ square, the numbers $1,2,3, \ldots, 9$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$
|
Answer: 7.
Solution. Let's color the cells in a checkerboard pattern: suppose the corner and central cells are black, and the rest are white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white cells, we get that the odd numbers are in the black cells. Their total sum is $1+3+5+7+9=$ 25. Therefore, the number in the central cell is $7=25-18$.
Evaluation. 12 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A person is walking parallel to a railway track at a constant speed. A train also passes by him at a constant speed. The person noticed that depending on the direction of the train, it passes by him either in $t_{1}=1$ minute or in $t_{2}=2$ minutes. Determine how long it would take the person to walk from one end of the train to the other.
## $(15$ points)
|
Answer: 4 min
Solution. When the train and the person are moving towards each other:
$l=\left(v_{n}+v_{u}\right) \cdot t_{1}$ (3 points), where $l$ - the length of the train, $v_{n}$ - its speed, $v_{u}$ - the speed of the person. If the directions of the train and the person are the same, then:
$l=\left(v_{n}-v_{u}\right) \cdot t_{2} \quad\left(3\right.$ points). In the situation where the person is walking on the train: $l=v_{u} \cdot t_{3}$. (3 points). Solving this system of equations, we get the final answer:
$t_{3}=\frac{2 t_{1} t_{2}}{t_{2}-t_{1}}=\frac{2 \cdot 1 \cdot 2}{2-1}=4$ minutes (6 points).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the cells of a $3 \times 3$ square, the numbers $0,1,2, \ldots, 8$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$
|
Answer: 2.
Solution. Let's color the cells in a checkerboard pattern: suppose the corner and central cells are black, and the rest are white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white cells, we get that the even numbers are in the black cells. Their total sum is $0+2+4+6+8=20$. Therefore, the number in the central cell is $2=20-18$.
Evaluation. 12 points for a correct solution. If an example of the arrangement of numbers satisfying the problem's condition is provided but the uniqueness of the answer is not proven, 6 points.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The cold water tap fills the bathtub in 17 minutes, and the hot water tap in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it?
|
Answer: in 3 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 8.5 minutes. Therefore, the hot water tap should be open for 3 minutes longer.
Evaluation. 12 points for the correct solution.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The cold water tap fills the bathtub in 19 minutes, and the hot water tap in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water?
|
Answer: in 2 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 9.5 minutes. Therefore, the hot water tap should be open for 2 minutes longer.
Evaluation. 12 points for the correct solution.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Solve the equation
$$
2 x+2+x \sqrt{x^{2}+1}+(x+2) \sqrt{x^{2}+4 x+5}=0 .
$$
|
Answer: -1.
Solution. Let $f(x)=x\left(1+\sqrt{x^{2}+1}\right)$. The original equation can be rewritten as $f(x)+f(x+2)=0$. Note that the function $f(x)$ is odd. It is increasing on the positive half-axis (as the product of positive increasing functions). Due to its oddness, it is increasing on the entire real line. Further, we have
$$
f(x)=-f(x+2) ; \quad f(x)=f(-x-2)
$$
Since an increasing function takes each of its values exactly once, it follows that $x=-x-2$, from which $x=-1$.
Remark. The monotonicity of the left side of the original equation could also have been established using the derivative.
Evaluation. 13 points for a correct solution. If the answer is guessed but not proven that there are no other solutions, 3 points.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (17 points) Point $M$ lies inside segment $A B$, the length of which is 40 cm. Points are chosen: $N$ at the midpoint of $A M, P$ at the midpoint of $M B, C$ at the midpoint of $N M, D$ at the midpoint of $M P$. Find the length of segment $C D$ in cm.
|
Answer: 10.
## Solution.
$$
\begin{aligned}
& N P=N M+M P=\frac{1}{2} A M+\frac{1}{2} M B=\frac{1}{2} A B=20, \\
& C D=C M+M D=\frac{1}{2} N M+\frac{1}{2} M P=\frac{1}{2} N P=10 .
\end{aligned}
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem № 6 (10 points)
Five identical balls are rolling towards each other on a smooth horizontal surface. The speeds of the first and second are \( v_{1} = v_{2} = 0.5 \) m/s, while the others are \( v_{3} = v_{4} = v_{5} = 0.1 \) m/s. The initial distances between the balls are the same, \( l = 2 \) m. All collisions are perfectly elastic. How much time will pass between the first and last collisions in this system?
1
2
3
4
5
## Time: 10 min
#
|
# Solution and Evaluation Criteria:
In the case of a perfectly elastic collision, identical balls "exchange" velocities.
Therefore, the situation can be considered as if the balls pass through each other with unchanged speeds. The first collision occurs between the second and third balls. The last collision will occur at the moment when the first ball "passes" by the fifth.
(4 points)
The corresponding time: \( t = \frac{3 l}{v_{1} + v_{5}} = \frac{3 \cdot 2}{0.6} = 10 \text{s} \)
#
|
10
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (17 points) Solve the equation ||$|x-1|+2|-3|=-2 x-4$.
|
Answer: -4.
Solution. The set of possible solutions for the inequality $-2 x-4 \geq 0$, that is, $x \leq-2$. Under this condition, the inner absolute value is uniquely determined. Using the property of the absolute value $|-a|=|a|$, we get the equation $||x-3|-3|=-2 x-4$. Similarly, we expand the inner absolute value, obtaining the equation $|x|=-2 x-4$. Under the obtained restriction, the last absolute value is uniquely determined, resulting in the equation $-x=-2 x-4$, that is, $x=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (17 points) Solve the equation ||$|x-2|+3|-4|=-3 x-9$.
|
Answer: -5.
Solution. The set of possible solutions for the inequality $-3 x-9 \geq 0$, that is, $x \leq-3$. Under this condition, the inner absolute value is uniquely determined. Using the property of absolute value $|-a|=|a|$, we get the equation $||x-5|-4|=-3 x-9$. Similarly, we expand the inner absolute value, obtaining the equation $|x-1|=-3 x-9$. Under the obtained restriction, the last absolute value is uniquely determined, resulting in the equation $-x+1=-3 x-9$, that is, $x=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (12 points) Solve the equation
$$
\sqrt[3]{(7-x)^{2}}-\sqrt[3]{(7-x)(9+x)}+\sqrt[3]{(9+x)^{2}}=4
$$
|
Answer: -1
Solution. Let $a=7-x, b=9+x$. Notice that $a+b=16$. We obtain the system of equations
$$
\left\{\begin{array}{c}
\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}}=4 \\
a+b=16
\end{array}\right.
$$
Multiply the first equation by $\sqrt[3]{a}+\sqrt[3]{b}$, using the formula for the difference of cubes, we get $\sqrt[3]{a}+\sqrt[3]{b}=4$. Square this equality, and subtract the first equation of the system from the result, we get $\sqrt[3]{a b}=4$.
The system $\left\{\begin{array}{c}\sqrt[3]{a b}=4, \\ a+b=16,\end{array}\right.$ reduces to a quadratic equation. Solving it, we get $a=8 ; x=-1$. Let's check the solution.
Evaluation criteria. Full justified solution - 12 points. Correct answer obtained without justification (guessed) - 3 points.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (12 points) Solve the equation
$$
\sqrt[3]{(9-x)^{2}}-\sqrt[3]{(9-x)(7+x)}+\sqrt[3]{(7+x)^{2}}=4
$$
|
Answer: 1.
Solution. Let $a=9-x, b=7+x$. Note that $a+b=16$. We obtain the system of equations
$$
\left\{\begin{array}{c}
\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}}=4 \\
a+b=16
\end{array}\right.
$$
Multiply the first equation by $\sqrt[3]{a}+\sqrt[3]{b}$, using the formula for the difference of cubes, we get $\sqrt[3]{a}+\sqrt[3]{b}=4$. Square this equality, and subtract the first equation of the system from the result, we get $\sqrt[3]{a b}=4$.
The system $\left\{\begin{array}{c}\sqrt[3]{a b}=4, \\ a+b=16\end{array}\right.$ reduces to a quadratic equation. Solving it, we get $a=8 ; x=1$. Let's check the solution.
Evaluation criteria. Full justified solution - 12 points. Correct answer obtained without justification (guessed) - 3 points.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (15 points) A mosquito was moving over the water in a straight line at a constant speed of \( v = 0.5 \) m/s and at the end of its movement, it landed on the water surface. 20 seconds before landing, it was at a height of \( h = 6 \) m above the water surface. The cosine of the angle of incidence of the sunlight on the water surface is 0.6. The incident sunlight, which creates the shadow of the mosquito, and its trajectory lie in the same vertical plane. Determine the speed at which the shadow of the mosquito moved along the bottom of the water body.
|
Answer: 0 m/s or $0.8 \mathrm{~m} / \mathrm{s}$.
Solution. The mosquito flew a distance of $s=v t=0.5 \cdot 20=10$ m before landing.
That is, it moved along the trajectory $A B$.
The cosine of the angle between the trajectory and the water surface is 0.8.
Obviously, the speed of the shadow moving along the bottom coincides with the speed of the shadow moving on the surface of the pond.
Since the angle of incidence of the sunlight coincides with the angle between the trajectory and the water surface, there are two possible scenarios.
The first - the mosquito is moving along the sunbeam.
The speed of the shadow on the bottom of the pond $v_{\text {shadow }}=0 \mathrm{~m} / \mathrm{s}$.
The second - the mosquito is flying towards the sunbeam. 2
The speed of the shadow $v_{\text {shadow }}=2 v \cos \beta=2 \cdot 0.5 \cdot 0.8=0.8 \mathrm{~m} / \mathrm{s}$.
|
0
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (13 points) What is the minimum number of participants that could have been in the school drama club if the number of fifth graders was more than $25 \%$ but less than $35 \%$, the number of sixth graders was more than $30 \%$ but less than $40 \%$, and the number of seventh graders was more than $35 \%$ but less than $45 \%$ (there were no participants from other grades).
|
Answer: 11. Fifth-graders -3, sixth-graders -4, seventh-graders - 4.
Solution. Let the required number of participants be
$$
n=a+b+c
$$
where $a$ is the number of fifth-graders, $b$ is the number of sixth-graders, $c$ is the number of seventh-graders. Note: $a<c$ and all three numbers constitute more than 0.25 and less than 0.5 of the total number of participants according to the condition. Consider all remaining cases of the sum $(*)$ taking into account these observations.
$5=1+2+2$. The condition for sixth-graders is not met.
$7=2+2+3$. The condition for sixth-graders is not met.
$8=2+3+3$. The condition for fifth-graders is not met.
$9=3+3+3$. The condition $a<c$ is not met.
$10=3+3+4$. The condition for sixth-graders is not met.
$11=3+4+4$. All conditions are met.
Evaluation criteria. Correct answer based on a complete enumeration without losing cases - 13 points. Correct answer, but not all possible cases are considered - 7 points. For the correct answer without explanation 2 points.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) An industrial robot travels from point $A$ to point $B$ according to a pre-determined algorithm. The diagram shows a part of its repeating trajectory. Determine how many times faster it would reach from point $A$ to point $B$ if it moved in a straight line at three times the speed?
|
Answer: 9 times.
Solution. The time spent by the robot moving along the given trajectory:
$t_{1}=N \frac{2 a+3 a+a+3 a}{v}=9 \frac{N a}{v}$.
The time spent by the robot moving in a straight line:
$t_{2}=N \frac{2 a+a}{3 v}=\frac{N a}{v}$.
We obtain that the robot will arrive $\frac{t_{1}}{t_{2}}=9$ times faster.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (13 points) What is the minimum number of participants that could have been in the school drama club if the number of fifth graders was more than $22 \%$ but less than $27 \%$, the number of sixth graders was more than $25 \%$ but less than $35 \%$, and the number of seventh graders was more than $35 \%$ but less than $45 \%$ (there were no participants from other grades).
|
Answer: 9. Fifth-graders - 2, sixth-graders - 3, seventh-graders - 4.
Solution. Let the required number of participants be
$$
n=a+b+c
$$
where $a$ is the number of fifth-graders, $b$ is the number of sixth-graders, and $c$ is the number of seventh-graders. Note: $a<c$ and all three numbers constitute more than 0.2 and less than 0.5 of the total number of participants according to the condition. Consider all remaining sum cases ( * with these notes in mind.
$5=1+2+2$. The condition for sixth-graders is not met.
$7=2+2+3$. The condition for fifth-graders is not met.
$8=2+3+3$. The condition for sixth-graders is not met.
$9=2+3+4$. All conditions are met.
Evaluation criteria. Correct answer based on a complete enumeration without losing cases - 13 points. Correct answer, but not all possible cases are considered - 7 points.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (10 points) A body moves along the Ox axis. The dependence of velocity on time is shown in the figure. Determine the distance traveled by the body in 6 seconds.
|
Answer: 5 meters.
Solution. In the first second, the body traveled 2 meters.
In the second - 1 meter.
In the fifth - 2 meters.
In total, the entire distance traveled is 5 meters.
|
5
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Fifteen numbers are arranged in a circle. The sum of any six consecutive numbers is 50. Petya covered one of the numbers with a card. The two numbers adjacent to the card are 7 and 10. What number is under the card?
|
Answer: 8.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 15$.) Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{4}=a_{10}=a_{1} .
$$
Now it is clear that for any $i$, $a_{i}=a_{i+3}$, i.e., the numbers go in the following order:
$$
a, b, c, a, b, c, \ldots, a, b, c
$$
From the condition, it follows that
$$
2(a+b+c)=50
$$
Thus, the sum of any three consecutive numbers is 25. Hence the answer.
Scoring. 12 points for a complete solution.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Twenty numbers are arranged in a circle. It is known that the sum of any six consecutive numbers is 24. What is the number in the 12th position if the number in the 1st position is 1?
|
Answer: 7.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 20)$. Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{19}=a_{5}=a_{11}=a_{17}=a_{3}=a_{9}=a_{15}=a_{1}
$$
Now it is clear that all numbers at odd positions are equal to each other. The same is true for numbers at even positions. Therefore, the numbers go like this:
$$
x, y, x, y, \ldots, x, y
$$
From the condition, it follows that
$$
3(x+y)=24, \quad x=1
$$
Hence, $y=7$. Therefore, ones are at odd positions, and sevens are at even positions.
Evaluation. 12 points for a complete solution.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (15 points) The mass of a vessel that is completely filled with kerosene is 20 kg. If this vessel is completely filled with water, its mass will be 24 kg. Determine the mass of the empty vessel. The density of water $\rho_{W}=1000 \kappa g / \mu^{3}$, the density of kerosene $\rho_{K}=800 \kappa g / \mathrm{M}^{3}$.
|
Answer: 4 kg
Solution. The mass of the vessel filled with kerosene: $m_{1}=m_{c}+\rho_{K} V$, where $m_{c}$ - the mass of the empty vessel, $V$ - the volume occupied by the kerosene.
The mass of the vessel filled with water $m_{2}=m_{c}+\rho_{B} V$.
(4 points)
We get that $V=\frac{m_{2}-m_{1}}{\rho_{B}-\rho_{K}}=\frac{24-20}{1000-800}=0.02 \mu^{3}$.
The mass of the empty vessel: $m_{c}=m_{1}-\rho_{\kappa} V=20-800 \cdot 0.02=4$ kg.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (13 points) Sixteen people are standing in a circle: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle?
|
Answer: 10.
Solution. A truthful person can only be next to liars. Three liars in a row cannot stand, so between any two nearest truthful persons, there is one or two liars. Then, if there are 5 or fewer truthful persons, in the intervals between them, there can be no more than 10 liars in total, so there are no more than 15 people in total. This leads to a contradiction. Therefore, there are no fewer than 6 truthful persons, and no more than 10 liars. An example with 10 liars is easily provided.
Evaluation Criteria. Full solution - 13 points. Example constructed - 3 points. Estimate proven - 9 points. Correct estimate named but not justified by considering examples - 2 points.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (13 points) In a circle, there are 17 people: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle?
|
Answer: 11.
Solution. A truth-teller can only be surrounded by liars. Three liars in a row cannot stand, so between any two nearest truth-tellers, there is one or two liars. Then, if there are 5 or fewer truth-tellers, there can be no more than 10 liars in the gaps between them, making a total of no more than 15 people. This leads to a contradiction. Therefore, there must be at least 6 truth-tellers, and no more than 12 liars, but there are 17 people in total, so there must be 11 liars. An example with 11 liars is easily provided.
Grading Criteria. Full solution - 13 points. Example constructed - 3 points. Estimate proven - 9 points. Correct estimate named but not justified by considering examples - 2 points.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) Two wheels rotate, meshed with each other, around fixed axes passing through the centers of wheels $A$ and $B$. The radii of the wheels differ by a factor of three. The smaller wheel makes 30 revolutions per minute. Determine how many seconds the larger wheel spends on one revolution?
|
Answer: $6 s$
Solution. The speeds of points lying on the edge of the wheels are the same.
(2 points)
The distances traveled by these points differ by three times.
The small wheel spends on one revolution: $t_{\text {small }}=\frac{1 \text { min }}{30 \text { revolutions }}=\frac{60 s}{30 \text { revolutions }}=2 s$. (3 points)
Therefore, the large wheel spends on one revolution:
$t_{\text {large }}=3 t_{\text {small }}=3 \cdot 2=6 s$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) Two wheels rotate, meshed with each other, around fixed axes passing through the centers of wheels $A$ and $B$. The radii of the wheels differ by a factor of three. The larger wheel makes 10 revolutions per minute. Determine how many seconds the smaller wheel spends on one revolution?
|
Answer: $2 c$
Solution. The speeds of points lying on the edge of the wheels are the same.
The distances traveled by these points differ by three times.
The large wheel spends on one revolution
$$
t_{\text {large }}=\frac{1 \text { min }}{10 \text { revolutions }}=\frac{60 \text { s }}{10 \text { revolutions }}=6 \text { s }
$$
Therefore, the small wheel spends on one revolution $t_{\text {small }}=\frac{1}{3} t_{\text {large }}=\frac{6}{3}=2 c$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (12 points) The sequence of functions is defined by the formulas
$$
f_{0}(x)=3 \sin x, f_{n+1}(x)=\frac{9}{3-f_{n}(x)}
$$
for any integer $n \geq 0$. Find $f_{2023}\left(\frac{\pi}{6}\right)$.
|
Answer: $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Solution. It is easy to compute: $f_{3}(x)=f_{0}(x)$, therefore $f_{2023}(x)=f_{1}(x)=$ $\frac{9}{3-3 \sin x}$. Consequently, $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Remark. One can immediately compute the values of the functions at the given point. This results in a cyclic sequence
$$
f_{0}\left(\frac{\pi}{6}\right)=\frac{3}{2} ; f_{1}\left(\frac{\pi}{6}\right)=6 ; f_{2}\left(\frac{\pi}{6}\right)=-3 ; f_{3}\left(\frac{\pi}{6}\right)=\frac{3}{2} ; \ldots
$$
Grading criteria. Full solution - 12 points. The relation $f_{3}(x)=f_{0}(x)$ is found - 7 points, the equality $f_{2023}(x)=f_{1}(x)$ is found - another 4 points. Calculation errors - minus 1 point.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (12 points) The sequence of functions is defined by the formulas
$$
f_{0}(x)=2 \cos x, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}
$$
for any integer $n \geq 0$. Find $f_{2023}\left(\frac{\pi}{3}\right)$.
|
Answer: $f_{2023}\left(\frac{\pi}{3}\right)=4$.
Solution. It is easy to compute: $f_{3}(x)=f_{0}(x)$, therefore $f_{2023}(x)=f_{1}(x)=$ $\frac{4}{2-2 \cos x}$. Consequently, $f_{2023}\left(\frac{\pi}{3}\right)=4$.
Remark. One can immediately compute the values of the functions at the given point. This results in a cyclic sequence
$$
f_{0}\left(\frac{\pi}{3}\right)=1 ; f_{1}\left(\frac{\pi}{3}\right)=4 ; f_{2}\left(\frac{\pi}{3}\right)=-2 ; f_{3}\left(\frac{\pi}{3}\right)=1 ; \ldots
$$
Grading criteria. Full solution - 12 points. The relation $f_{3}(x)=f_{0}(x)$ is found - 7 points, the equality $f_{2023}(x)=f_{1}(x)$ is found - another 4 points. Calculation errors - minus 1 point.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (16 points) Solve the equation $x-5=\frac{3 \cdot|x-2|}{x-2}$. If the equation has multiple roots, write their sum in the answer.
#
|
# Answer: 8.
Solution. The equation has a restriction on the variable $x \neq 2$. We open the modulus: for $x>2, x-5=3, x=8$. For $x<2, \quad x-5=-3, x=2$ - an extraneous root.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (16 points) Solve the equation $x-7=\frac{4 \cdot|x-3|}{x-3}$. If the equation has multiple roots, write their sum in the answer.
|
Answer: 11.
Solution. The equation has a restriction on the variable $x \neq 3$. We open the modulus: for $x>3, x-7=4, x=11$. For $x<3, \quad x-7=-4, x=3-$ extraneous root.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The cold water tap fills the bathtub in 17 minutes, and the hot water tap in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it?
|
Answer: in 3 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 8.5 minutes. Therefore, the hot water tap should be open for 3 minutes longer.
Evaluation. 12 points for the correct solution.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A certain mechanism consists of 30 parts, some of which are large, and some are small. It is known that among any 12 parts taken, there will be at least one small part, and among any 20 parts - at least one large part. How many of each type of part does the mechanism contain?
|
Answer: 11 large parts and 19 small parts.
Solution. Since among any 12 parts there is a small one, there are no more than 11 large parts. Since among any 20 parts there is a large one, there are no more than 19 small parts. If there were fewer than 11 large parts or fewer than 19 small parts, there would be fewer than 30 parts in total, but according to the condition, there are 30. Therefore, there are 11 large parts and 19 small parts.
Grading. 13 points for a correct solution. 2 points for a correct answer (without justification). If the correctness of the answer is shown but its uniqueness is not justified, 6 points.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The cold water tap fills the bathtub in 19 minutes, while the hot water tap fills it in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it?
|
Answer: in 2 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 9.5 minutes. Therefore, the hot water tap should be open for 2 minutes longer.
Evaluation. 12 points for the correct solution.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A certain mechanism consists of 25 parts, some of which are large, and some are small. It is known that among any 12 parts taken, there will be at least one small part, and among any 15 parts, there will be at least one large part. How many of each type of part does the mechanism contain?
|
Answer: 11 large parts and 14 small parts.
Solution. Since among any 12 parts there is a small one, there are no more than 11 large parts. Since among any 15 parts there is a large one, there are no more than 14 small parts. If there were fewer than 11 large parts or fewer than 14 small parts, the total number of parts would be less than 25, but according to the condition, there are 25. Therefore, there are 11 large parts and 14 small parts.
Grading. 13 points for a correct solution. 2 points for a correct answer (without justification). If the correctness of the answer is shown but its uniqueness is not justified, 6 points.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (15 points) A pedestrian is moving towards a crosswalk along a straight path at a constant speed of 3.6 km/h. At the initial moment, the distance from the pedestrian to the crosswalk is 20 m. The length of the crosswalk is $5 \mathrm{~m}$. At what distance from the crosswalk will the pedestrian be after half a minute?
|
Answer: 5 m.
Solution. The pedestrian's speed is 3.6 km/h = 1 m/s. In half a minute, he walked $s = v t = 1 \cdot 30 s = 30$ m. From the crossing, he is at a distance of $l = s - 20 - 5 = 5$ m.
## Tasks, answers, and evaluation criteria
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem No. 5 (10 points)
When constructing this structure, a homogeneous wire of constant cross-section was used. It is known that points $B, D, F$ and $H$ are located equally at the midpoints of the corresponding sides of the square ACEG. The resistance of segment $A B$ is $R_{0}=1 \Omega$. Determine the resistance of the entire structure if it is connected to the electrical circuit at points C and $G$.
|
Answer: 2 Ohms
## Solution and grading criteria:
The resistance of a resistor is proportional to its length.
Taking this into account, the proposed circuit can be replaced by an equivalent one:
Its resistance: $\quad R=2 R_{0} \approx 2$ Ohms
#
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (17 points) When walking uphill, the tourist walks 2 km/h slower, and downhill 2 km/h faster, than when walking on flat ground. Climbing the mountain takes the tourist 10 hours, while descending the mountain takes 6 hours. What is the tourist's speed on flat ground?
|
# Answer: 8
Solution. Let $x$ km/h be the tourist's speed on flat terrain. According to the problem, we get the equation $10(x-2)=6(x+2)$. From this, we find $x=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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