problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
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:
$$
x_{... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The equation $x^{2}+a x+2=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,
$$
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}... | 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:
$$
x_{1... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. param 1 people participated in a survey. They were given a list of $N$ movies. Each person was asked to name their favorite movies from this list. It turned out that everyone named at least two movies. Moreover, any pair of respondents had no more than one movie in common among those they named. Find the smallest po... | 8. param 1 people participated in a survey. They were given a list of $N$ movies. Each person was asked to name their favorite movies from this list. It turned out that everyone named at least two movies. Moreover, any pair of respondents had no more than one movie in common among the ones they named. Find the smallest... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] Given the numbers $\log _{\sqrt{5 x-1}}(4 x+1), \log _{4 x+1}\left(\frac{x}{2}+2\right)^{2}, \log _{\frac{x}{2}+2}(5 x-1)$. For which $x$ are two of these numbers equal, and the third one less than them by 1? | Answer: $x=2$.
Solution. From the condition, it follows that the functions $4 x+1, \frac{x}{2}+2,5 x-1$ are positive and do not take the value 1 for all $x$ from the domain of admissible values. Let $a=\log _{\sqrt{5 x-1}}(4 x+1), b=\log _{4 x+1}\left(\frac{x}{2}+2\right)^{2}, c=$ $\log _{\frac{x}{2}+2}(5 x-1)$. Then
... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] Given the numbers $\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), \log _{6 x-14}(x-1)^{2}, \log _{x-1}\left(\frac{x}{3}+3\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1? | Answer: $x=3$.
Solution. From the condition, it follows that the functions $6 x-14, x-1, \frac{x}{3}+3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), b=\log _{6 x-14}(x-1)^{2}, c=$ $\log _{x-1}\left(\frac{x}{3}+3\right)$. Then
$$
... | 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}}\... | 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
$$
\begin{aligned}
& a b c=\log _{\sqrt{... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. [5 points] Given the numbers $\log _{\sqrt{2 x-3}}(x+1), \log _{2 x^{2}-3 x+5}(2 x-3)^{2}, \log _{x+1}\left(2 x^{2}-3 x+5\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1? | Answer: $x=4$.
Solution. From the condition, it follows that the functions $x+1, 2x-3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{2 x-3}}(x+1), b=\log _{2 x^{2}-3 x+5}(2 x-3)^{2}, c=\log _{x+1}\left(2 x^{2}-3 x+5\right)$. Then
$$
\begin{aligned}
a b c=... | 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 ... | 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 | 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 _... | -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)$. The... | -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,... | 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 t... | 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, ... | 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 f... | 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 ^{... | -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
$$
\begin{aligned}
\cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x & =\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& =\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x
\end{aligned}
$$
$$
\begin{aligned}
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos 2 y}=\frac{1}{6}+\sin ^{2}(x+2 y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin 2 y}=\frac{5}{6}+\cos ^{2}(x+2 y)$. Find all possible values of the expression $\cos (x+6 y)$, given that there are at least two. Answer: -1 or $-\frac{2}{3}$. | Solution. Note that
$$
\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+... | -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 ... | 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... | 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 remainin... | 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... | 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-coordinat... | -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 $a=b \neq c$... | 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.
| ... | 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.
| ... | 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,... | 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 form... | 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 \Leftrightarro... | 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,... | 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 form... | 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 \Left... | 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)... | 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... | 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:
$$
\frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \gamma}{1-\operatorname{tg} \alpha \opera... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that $\operatorname{tg}(2 \alpha-\beta)-4 \operatorname{tg} 2 \alpha+4 \operatorname{tg} \beta=0, \operatorname{tg} \alpha=-3$. Find $\operatorname{ctg} \beta$. | Answer: -1 or $\frac{4}{3}$.
Solution. $\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{-6}{1-9}=\frac{3}{4}$. Then the given equality can be transformed as follows:
$$
\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \operatorna... | -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 additi... | 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... | 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... | 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
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - 3 y ^ { 2 } + 4 x + 4 \leqslant 4 x ^ { 2 } + 4 x + 1 , } \\
{ ( x + 2 ) ^ { 2 } - 3 y ^ { 2 } \geqslant 0 , } \\
{ 2 x + 1 \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7. Plot the figure $\Phi$ on the plane, consisting of points $(x ; y)$ of the coordinate plane such that the system of inequalities is satisfied
$$
\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^{... | 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... | 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... | -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 $f_{0}(4)=4, f_{1}(4)=-2, f_{2}(4)=1, f_{3}(4)=4 \ldots$
... | -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)=... | -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.... | -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: decrease... | # 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}=\f... | 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... | # 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 ... | 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}... | 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} / \mat... | 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 he... | 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 tota... | # 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... | 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... | # 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 m... | 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... | # 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 m... | 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 ... | 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... | 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_... | 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 inextensib... | 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... | 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... | 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... | 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)
,
where each of the resistors has a resistance of $R_{0}=10$ Ohms. As a result, the total resistance is: $R=\frac... | 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... | 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 o... | 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}... | 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 eve... | 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. F... | -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{align... | 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 collis... | # 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 occu... | 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,... | -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, obt... | -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 $... | -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 $\s... | 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 ... | 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 cosi... | 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 th... | 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 l... | 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?
 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 th... | 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 a... | 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 th... | 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_{1... | 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}}=\f... | 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 ... | 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 peopl... | 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?
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... | 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?
 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)}
$$
for any integer $n \geq 0$. Find $f_{2023}\left(\frac{\pi}{6}\right)$. | Answer: $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Solution. It is easy to compute: $f_{3}(x)=f_{0}(x)$, therefore $f_{2023}(x)=f_{1}(x)=$ $\frac{9}{3-3 \sin x}$. Consequently, $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Remark. One can immediately compute the values of the functions at the given point. This results in a cy... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (12 points) The sequence of functions is defined by the formulas
$$
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 cy... | 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 tha... | 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 part... | 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 minut... | 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 ... | 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:
 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 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.