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Problem 2. Dasha wrote the numbers $9,10,11, \ldots, 22$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the rema... | Answer: 203.
Solution. The sum of numbers from 9 to 22 is 217. If at least one number is erased, the sum of the remaining numbers does not exceed 208. Let's sequentially consider the options:
- if the sum is 208, then Dasha could have erased only the number 9; then the remaining numbers can be divided into two groups... | 203 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,137 |
Problem 3. In the surgical department, there are 4 operating rooms: I, II, III, and IV. In the morning, they were all empty. At some point, an operation began in operating room I, then after some time in operating room II, then after some more time in operating room III, and finally in operating room IV.
All four oper... | Answer: Only the duration of the operation in Operating Room IV can be determined.
Solution. First, let's prove that the durations of the operations in Operating Rooms I, II, and III cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are 70, 39, 33, 10 or 56, 5... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,138 |
Problem 4. In triangle $A B C$, the lengths of the sides are 4, 5, and $\sqrt{17}$. Find the area of the figure consisting of those and only those points $X$ inside triangle $A B C$ for which the condition $X A^{2}+X B^{2}+X C^{2} \leqslant 21$ is satisfied. | Answer: $5 \pi / 9$.
Solution. Let $B C=a, A C=b, A B=c, \rho^{2}=21$.
Let $G$ be the point of intersection of the medians of triangle $A B C$. Represent
$$
\overrightarrow{X A}=\overrightarrow{G A}-\overrightarrow{G X}, \overrightarrow{X B}=\overrightarrow{G B}-\overrightarrow{G X}, \overrightarrow{X C}=\overrighta... | \frac{5\pi}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,139 |
Problem 5. Solve the equation: $4\left(x^{4}+3 x^{2}+3\right)\left(y^{4}-7 y^{2}+14\right)=21$. | Answer: $\left(0 ; \pm \sqrt{\frac{7}{2}}\right)$.
Solution. Note that $x^{4}+3 x^{2}+3 \geqslant 3$ for each $x$, and $y^{4}-7 y^{2}+14=\left(y^{2}-7 / 2\right)^{2}+7 / 4 \geqslant 7 / 4$ for each $y$. Therefore, the left side of the equation is greater than or equal to $4 \cdot 3 \cdot 7 / 4=21$, and equality is ach... | (0;\\sqrt{\frac{7}{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,140 |
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{43} \cdot \operatorname{tg} \frac{2 \pi}{43}+\operatorname{tg} \frac{2 \pi}{43} \cdot \operatorname{tg} \frac{3 \pi}{43}+\ldots+\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}+\ldots+\operatorname{tg} \frac{2019 \pi}{43} \cdot \operat... | Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alp... | -2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,141 |
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-2 x+a x^{2}\right)^{8}$ be equal to -1540? | Answer: -5.
Solution. Applying the polynomial formula, we get
$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_{... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,142 |
Problem 8. Given an isosceles triangle $K L M(K L=L M)$ with the angle at the vertex equal to $114^{\circ}$. Point $O$ is located inside triangle $K L M$ such that $\angle O M K=30^{\circ}$, and $\angle O K M=27^{\circ}$. Find the measure of angle $\angle L O M$. | Answer: $150^{\circ}$.
Solution. Let $L H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $M O$ and segment $L H$. Note that $K S=S M$. For example, since in triangle $K S M$ the median SH coincides with the height.
 \neq x$ for each integer $x$. We call a number $a$ beautiful if for any integer $x$, $g(x) = g(a-x)$. Can both numbers 739 and 741 be beautiful? | Answer: No, it cannot.
Solution. Suppose that both numbers 739 and 741 turned out to be beautiful. Then
$$
g(x+2)=g(741-(x+2))=g(739-x)=g(x) .
$$
Thus, there exist integers $a$ and $b$ such that the function $g$ takes the value $a$ at all even numbers and the value $b$ at all odd numbers. On the other hand, at point... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,144 |
Task 10. Vasya made a prism from glass rods. The prism has 171 lateral edges and the same number of edges in each of the bases. Vasya wondered: "Is it possible to parallelly translate each of the 513 edges of the prism so that they form a closed broken line in space?" Is Vasya's idea feasible? | Answer: No, it is not possible.
Solution. Suppose that the implementation of Vasya's idea is possible, and consider the closed broken line formed by 513 edges. Introduce a coordinate system such that the plane $Oxy$ is parallel to the bases of the prism, the axis $Oz$ is perpendicular to the bases of the prism, and th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,145 |
Problem 1. The sum of the first three terms of an arithmetic progression, as well as the sum of the first seven terms, are natural numbers. In addition, its first term $c_{1}$ satisfies the inequality $c_{1} \geqslant \frac{1}{3}$. What is the smallest value that $c_{1}$ can take? | Answer: $5 / 14$
Solution. Let $c_{n}$ be the $n$-th term of the progression, $d$ be the common difference of the progression, and $S_{n}=c_{1}+c_{2}+\ldots+c_{n}$ be the sum of the first $n$ terms of the progression.
Express $c_{1}$ in terms of $S_{3}$ and $S_{7}$. Note that $S_{3}=3 c_{1}+3 d, S_{7}=7 c_{1}+21 d$, ... | \frac{5}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,146 |
Problem 2. Sasha wrote the numbers $7, 8, 9, \ldots, 17$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the rema... | Answer: 121.
Solution. The sum of numbers from 7 to 17 is 132. If at least one number is erased, the sum of the remaining numbers does not exceed 125. Let's sequentially consider the options:
- if the sum is 125, then Sasha could have erased only the number 7; then the remaining numbers can be divided into five group... | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,147 |
Task 3. In the surgical department, there are 4 operating rooms: 1, 2, 3, and 4. In the morning, they were all empty. At some point, a surgery began in operating room 1, after some time - in operating room 2, then after some more time - in operating room 3, and finally in operating room 4.
All four surgeries ended sim... | Answer: Only the duration of the operation in operating room 4 can be determined.
Solution. First, let's prove that the durations of operations in operating rooms 1, 2, and 3 cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are 58, 29, 27, 13 minutes or 46, 4... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,148 |
Problem 4. In triangle $X Y Z$, the lengths of the sides are 2, 7, and $5 \sqrt{3}$. Find the area of the figure consisting of those and only those points $A$ inside triangle $X Y Z$ for which the condition $A X^{2}+A Y^{2}+A Z^{2} \leqslant 43$ is satisfied. | Answer: $\pi / 9$.
Solution. Let $Y Z=a, X Z=b, X Y=c, \rho^{2}=43$.
Let $G$ be the intersection point of the medians of triangle $X Y Z$. Represent
$$
\overrightarrow{A X}=\overrightarrow{G X}-\overrightarrow{G A}, \overrightarrow{A Y}=\overrightarrow{G Y}-\overrightarrow{G A}, \overrightarrow{A Z}=\overrightarrow{... | \frac{\pi}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,149 |
Problem 5. Solve the equation: $4\left(x^{4}+3 x^{2}+1\right)\left(y^{4}-5 y^{2}+19\right)=51$. | Answer: $\left(0 ; \pm \sqrt{\frac{5}{2}}\right)$.
Solution. Note that $x^{4}+3 x^{2}+1 \geqslant 1$ for each $x$, and $y^{4}-5 y^{2}+19=\left(y^{2}-5 / 2\right)^{2}+51 / 4 \geqslant 51 / 4$ for each $y$. Therefore, the left side of the equation is greater than or equal to $4 \cdot 1 \cdot 51 / 4=51$, and equality is ... | (0;\\sqrt{\frac{5}{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,150 |
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{47} \cdot \operatorname{tg} \frac{2 \pi}{47}+\operatorname{tg} \frac{2 \pi}{47} \cdot \operatorname{tg} \frac{3 \pi}{47}+\ldots+\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}+\ldots+\operatorname{tg} \frac{2021 \pi}{47} \cdot \operat... | Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alp... | -2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,151 |
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-3 x+a x^{2}\right)^{8}$ be equal to $70$? | Answer: -4.
Solution. Applying the polynomial formula, we get
$$
\left(1-3 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-3 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-3)^{n_{... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,152 |
Problem 9. The function $G$ is defined on integers and takes integer values, and for each integer $c$ there exists a number $x$ such that $G(x) \neq c$. We will call a number $a$ non-standard if for any integer $x$ it holds that $G(x) = G(a - x)$. Can each of the numbers 267 and 269 be non-standard? | Answer: No, it cannot.
Solution. Suppose that both numbers 267 and 269 turned out to be non-standard. Then
$$
G(x+2)=G(269-(x+2))=G(267-x)=G(x)
$$
Thus, there exist integers $a$ and $b$ such that the function $G$ takes the value $a$ at all even numbers and the value $b$ at all odd numbers. On the other hand, at poin... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,154 |
Task 10. Anya crafted a prism from glass rods. The prism has 373 lateral edges and the same number of edges in each of the bases. Anya wondered: "Is it possible to parallelly translate each of the 1119 edges of the prism so that they form a closed broken line in space?" Is the realization of Anya's idea possible? | Answer: No, it is not possible.
Solution. Suppose that the implementation of Anya's idea is possible, and consider the closed broken line formed by 1119 edges. Introduce a coordinate system such that the plane $Oxy$ is parallel to the bases of the prism, the axis $Oz$ is perpendicular to the bases of the prism, and th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,155 |
Problem 1. The sum of the first four terms of an arithmetic progression, as well as the sum of the first seven of its terms, are natural numbers. In addition, its first term $a_{1}$ satisfies the inequality $a_{1} \leqslant \frac{2}{3}$. What is the greatest value that $a_{1}$ can take? | Answer: $9 / 14$
Solution. Let $a_{n}$ be the $n$-th term of the progression, $d$ be the common difference of the progression, and $S_{n}=a_{1}+a_{2}+\ldots+a_{n}$ be the sum of the first $n$ terms of the progression.
Express $a_{1}$ in terms of $S_{4}$ and $S_{7}$. Note that $S_{4}=4 a_{1}+6 d, S_{7}=7 a_{1}+21 d$, ... | \frac{9}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,156 |
Problem 2. Masha wrote the numbers $4,5,6, \ldots, 16$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remain... | Answer: 121.
Solution. The sum of numbers from 4 to 16 is 130. If at least one number is erased, the sum of the remaining numbers does not exceed 126. Let's sequentially consider the options:
- if the sum is 126, then Masha could have erased only the number 4; then the remaining numbers can be divided into two groups... | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,157 |
Problem 3. In a surgical department, there are 4 operating rooms: A, B, V, and G. In the morning, they were all empty. At some point, an operation began in operating room A, after some time - in operating room B, then after some more time - in V, and then in $\Gamma$.
All four operations ended simultaneously, and the ... | Answer: Only the duration of the operation in operating room $\Gamma$ can be determined.
Solution. First, let's prove that the durations of operations in operating rooms A, B, and C cannot be determined uniquely. Indeed, it is easy to verify that if the durations of the operations are $65, 45, 44, 31$ or $56, 55, 43, ... | 31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,158 |
Problem 4. In triangle $P Q R$, the lengths of the sides are 4, 7, and 9. Find the area of the figure consisting of those and only those points $M$ inside triangle $P Q R$ for which the condition $M P^{2}+$ $M Q^{2}+M R^{2} \leqslant 50$ is satisfied. | Answer: $4 \pi / 9$.
Solution. Let $Q R=a, P R=b, P Q=c, \rho^{2}=50$.
Let $G$ be the point of intersection of the medians of triangle $P Q R$. Represent
$$
\overrightarrow{M P}=\overrightarrow{G P}-\overrightarrow{G M}, \overrightarrow{M Q}=\overrightarrow{G Q}-\overrightarrow{G M}, \overrightarrow{M R}=\overrighta... | \frac{4\pi}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,159 |
Problem 5. Solve the equation: $\left(x^{4}+9 x^{2}+4\right)\left(y^{4}-3 y^{2}+17\right)=59$. | Answer: $\left(0 ; \pm \sqrt{\frac{3}{2}}\right)$.
Solution. Note that $x^{4}+9 x^{2}+4 \geqslant 4$ for each $x$, and $y^{4}-3 y^{2}+17=\left(y^{2}-3 / 2\right)^{2}+59 / 4 \geqslant 59 / 4$ for each $y$. Therefore, the left side of the equation is greater than or equal to $4 \cdot 59 / 4=59$, and equality is achieved... | (0;\\sqrt{\frac{3}{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,160 |
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{43} \cdot \operatorname{tg} \frac{2 \pi}{43}+\operatorname{tg} \frac{2 \pi}{43} \cdot \operatorname{tg} \frac{3 \pi}{43}+\ldots+\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}+\ldots+\operatorname{tg} \frac{2021 \pi}{43} \cdot \operat... | Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alp... | -2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,161 |
Problem 7. For what least value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-2 x+a x^{2}\right)^{8}$ be equal to $-1540?$ | Answer: -19.
Solution. Applying the polynomial formula, we get
$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_... | -19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,162 |
Problem 8. Given an isosceles triangle $A B C(A B=B C)$ with the angle at the vertex equal to $102^{\circ}$. Point $O$ is located inside triangle $A B C$ such that $\angle O C A=30^{\circ}$, and $\angle O A C=21^{\circ}$. Find the measure of angle $\angle B O A$. | Answer: $81^{\circ}$.
Solution. Let $B H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $C O$ and segment $B H$. Note that $A S=S C$. For example, in triangle $A S C$, the median $S H$ coincides with the height.
 \neq c$. We call a number $a$ interesting if for any integer $x$ the following holds: $F(x) = F(a - x)$. Can each of the numbers 412, 414, and 451 be interesting? | Answer: No, it cannot.
Solution. Suppose that each of the numbers 412, 414, and 451 turned out to be interesting. Then
$$
F(x+2)=F(414-(x+2))=F(412-x)=F(x)
$$
This means there exist integers $a$ and $b$ such that the function $F$ takes the value $a$ at all even numbers and the value $b$ at all odd numbers. On the ot... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,164 |
Task 10. Petya made a pyramid out of glass rods. The pyramid has 373 lateral edges and the same number of edges in the base. Petya wondered: "Is it possible to parallelly translate each of the 746 edges of the pyramid so that they form a closed broken line in space?" Is Petya's idea feasible? | Answer: No, it is not possible.
Solution. Suppose that the implementation of Petya's idea is possible, and consider the closed broken line formed by 746 edges. Introduce a coordinate system such that the plane $Oxy$ is parallel to the base of the pyramid, the axis $Oz$ is perpendicular to the base of the pyramid, and ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,165 |
Problem 1. The sum of the first four terms of an arithmetic progression, as well as the sum of the first nine terms, are natural numbers. In addition, its first term $b_{1}$ satisfies the inequality $b_{1} \leqslant \frac{3}{4}$. What is the greatest value that $b_{1}$ can take? | Answer: $11 / 15$
Solution. Let $b_{n}$ be the $n$-th term of the progression, $d$ be the common difference of the progression, and $S_{n}=b_{1}+b_{2}+\ldots+b_{n}$ be the sum of the first $n$ terms of the progression.
Express $b_{1}$ in terms of $S_{4}$ and $S_{9}$. Note that $S_{4}=4 b_{1}+6 d, S_{9}=9 b_{1}+36 d$,... | \frac{11}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,166 |
Problem 2. Pasha wrote the numbers $4,5,6, \ldots, 14$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remain... | # Answer: 91.
Solution. The sum of the numbers from 4 to 14 is 99. If at least one number is erased, the sum of the remaining numbers does not exceed 95. Let's sequentially consider the options:
- if the sum is 95, then Pasha could have erased only the number 4; then the remaining numbers can be divided into five gro... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,167 |
Problem 3. In a surgical department, there are 4 operating rooms: A, B, C, and D. In the morning, they were all empty. At some point, a surgery began in operating room A, after some time - in operating room B, then after some more time - in C, and finally in D.
All four surgeries ended simultaneously, and the total du... | Answer: Only the duration of the operation in operating room D can be determined.
Solution. First, let's prove that the durations of the operations in operating rooms A, B, and C cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are $72, 35, 34, 17$ or $56, 55... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,168 |
Problem 4. In triangle $K L M$, the lengths of the sides are $8, 3 \sqrt{17}$, and 13. Find the area of the figure consisting of those and only those points $P$ inside triangle $K L M$ for which the condition $P K^{2}+P L^{2}+P M^{2} \leqslant 145$ is satisfied. | Answer: $49 \pi / 9$.
Solution. Let $L M=a, K M=b, K L=c, \rho^{2}=145$.
Let $G$ be the point of intersection of the medians of triangle $K L M$. Represent
$$
\overrightarrow{P K}=\overrightarrow{G K}-\overrightarrow{G P}, \overrightarrow{P L}=\overrightarrow{G L}-\overrightarrow{G P}, \overrightarrow{P M}=\overrigh... | \frac{49\pi}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,169 |
Problem 5. Solve the equation: $2\left(x^{4}+3 x^{2}+6\right)\left(y^{4}-5 y^{2}+12\right)=69$. | Answer: $\left(0 ; \pm \sqrt{\frac{5}{2}}\right)$.
Solution. Note that $x^{4}+3 x^{2}+6 \geqslant 6$ for each $x$, and $y^{4}-5 y^{2}+12=\left(y^{2}-5 / 2\right)^{2}+23 / 4 \geqslant 23 / 4$ for each $y$. Therefore, the left side of the equation is greater than or equal to $2 \cdot 6 \cdot 23 / 4=69$, and equality is ... | (0;\\sqrt{\frac{5}{2}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,170 |
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{47} \cdot \operatorname{tg} \frac{2 \pi}{47}+\operatorname{tg} \frac{2 \pi}{47} \cdot \operatorname{tg} \frac{3 \pi}{47}+\ldots+\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}+\ldots+\operatorname{tg} \frac{2019 \pi}{47} \cdot \operat... | Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alp... | -2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,171 |
Problem 7. For what least value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-3 x+a x^{2}\right)^{8}$ be equal to $70 ?$ | Answer: -50.
Solution. Applying the polynomial formula, we get
$$
\left(1-3 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-3 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-3)^{n_... | -50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,172 |
Problem 8. Given an isosceles triangle $X Y Z(X Y=Y Z)$ with the angle at the vertex equal to $96^{\circ}$. Point $O$ is located inside triangle $X Y Z$ such that $\angle O Z X=30^{\circ}$, and $\angle O X Z=18^{\circ}$. Find the measure of the angle $\angle Y O X$. | Answer: $78^{\circ}$.
Solution. Let $Y H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $Z O$ and segment $Y H$. Note that $X S=S Z$. For example, since in triangle $X S Z$ the median $S H$ coincides with the height.
 \neq x$ for each integer $x$. We call a number $a$ curious if for any integer $x$, $f(x) = f(a-x)$. Can each of the numbers 60, 62, and 823 be curious? | Answer: No, it cannot.
Solution. Suppose that each of the numbers 60, 62, and 823 turned out to be lovely. Then
$$
f(x+2)=f(62-(x+2))=f(60-x)=f(x)
$$
This means there exist integers $a$ and $b$ such that the function $f$ takes the value $a$ at all even numbers and the value $b$ at all odd numbers. On the other hand,... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,174 |
Task 10. Lena made a pyramid out of glass rods. The pyramid has 171 lateral edges and the same number of edges in the base. Lena wondered: "Is it possible to parallelly translate each of the 342 edges of the pyramid so that they form a closed broken line in space?" Is Lena's idea feasible? | Answer: No, it is not possible.
Solution. Suppose that the implementation of Lenin's idea is possible, and consider the closed broken line formed by 342 edges. Introduce a coordinate system such that the plane $Oxy$ is parallel to the base of the pyramid, the axis $Oz$ is perpendicular to the base of the pyramid, and ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,175 |
Problem 5. Since $\left|a+\frac{1}{a}\right| \geqslant 2$ and equality is achieved only when $|a|=1$, the first equation of the system is equivalent to $\operatorname{tg} x= \pm 1, \sin y= \pm 1$. Then from the second equation, $\cos z=0$. | Answer: $\left(\frac{\pi}{4}+\frac{\pi n}{2} ; \frac{\pi}{2}+\pi k ; \frac{\pi}{2}+\pi l\right)$. | (\frac{\pi}{4}+\frac{\pin}{2};\frac{\pi}{2}+\pik;\frac{\pi}{2}+\pi) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,177 |
Task 1. Vasya and Masha got married in 1994. Since then, they have had four children, and they welcomed the new year 2015 as a family of six. By a strange coincidence, all the children were born on February 6, and on February 7, 2016, Vasya noticed that the age of the oldest child is equal to the product of the ages of... | Solution. Note that from the condition, it follows that today all the children are at least two years old - they were all born no later than February 6, 2014. Suppose there are no twins, then the ages of all four children are different. Then the product of the ages of the three younger ones is no less than $2 \cdot 3 \... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 9,186 |
Problem 2. The number 27 is written on the board. Every minute, the number is erased from the board and replaced with the product of its digits, increased by 12. For example, after one minute, the number on the board will be $2 \cdot 7+12=26$. What will be on the board after an hour? | # Answer: 14
Solution. Let's find the next few numbers that will appear on the board. After 26, it will be 24, then $20, 12, 14, 16, 18$ and again 20. Notice that the sequence has looped and each subsequent number will coincide with the one that is 5 positions earlier. Therefore, after an hour, that is, 60 minutes, th... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,187 |
Problem 3. For real numbers $x, y, z$, it is known that
$$
x y+z=y z+x=z x+y
$$
Prove that at least two of the numbers $x, y, z$ are equal. | Solution. Suppose that all three numbers are distinct. Consider the equation $x y+z=y z+x$. Move all terms to one side and factorize, we get $(x-z)(y-1)=0$. Since $x \neq z$, it must be that $y=1$. Now consider another equation, for example $y z+x=z x+y$. Similarly, we get that $z=1$. But then $y=z$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 9,188 |
Problem 4. Trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $AC = BC + AD$, and one of the angles between the lines $AC$ and $BD$ is $60^{\circ}$. Prove that $ABCD$ is an isosceles trapezoid. | Solution. Construct point $E$ on line $A D$ such that $C E$ is parallel to $B D$. Then $B C E D$ is a parallelogram. Consider triangle $A C E$. $A C = A D + B C = A D + D E = A E$, and angle $A C E$ is either $60^{\circ}$ or $120^{\circ}$. However, the angle at the base of the isosceles triangle $A C E$ cannot be obtus... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,189 |
Problem 5. Petya has 50 balls of three colors: red, blue, and green. It is known that among any 34 balls, there is at least one red; among any 35, there is at least one blue; among any 36, there is at least one green. How many green balls can Petya have? | Answer: Petya can have 15, 16, or 17 green balls.
Solution. Consider the first condition. From the fact that among any 34 balls there is at least one red ball, it follows that there are no more than 33 non-red balls. From this, we get that there are no fewer than $50-33=17$ red balls. Similarly, there are no fewer tha... | 15,16,or17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,190 |
Problem 6. Find all real numbers $x$ such that both numbers $x+\sqrt{3}$ and $x^{2}+\sqrt{3}$ are rational. | Answer: $\frac{1}{2}-\sqrt{3}$.
Solution. Let the number $x+\sqrt{3}$ be denoted by $a$, and the number $x^{2}+\sqrt{3}$ by $b$, then $x=a-\sqrt{3}, b=$ $(a-\sqrt{3})^{2}+\sqrt{3}=a^{2}-2 a \sqrt{3}+3+\sqrt{3}$. From this we get $\left(a^{2}+3-b\right)+(1-2 a) \sqrt{3}=0$. On the right side, we have the rational numbe... | \frac{1}{2}-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,191 |
Problem 7. On the sides $AD$ and $BC$ of a convex quadrilateral $ABCD$, points $F$ and $E$ are chosen respectively such that $AF / FD = BE / EC = AB / CD$. The extension of segment $EF$ beyond point $F$ intersects line $AB$ at point $P$, and line $CD$ at point $Q$. Prove that $\angle BPE = \angle CQE$. | Solution. Construct points $G$ and $H$ such that $A B G F$ and $C D F H$ are parallelograms. Since $B G, A D$ and $C H$ are parallel, $\angle G B E = \angle H C E$. Also, $B G / C H = A F / D F = A B / C D = B E / C E$. Therefore, triangles $B G E$ and $C H E$ are similar. From this, points $G, E$ and $H$ lie on the sa... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,192 |
Problem 8. Karlson wrote the fraction 5/8. Little Man can:
- add the same natural number to both the numerator and the denominator simultaneously,
- multiply the numerator and the denominator by the same natural number.
Can Little Man use these actions to get a fraction equal to $3 / 5 ?$ | Answer: No, he will not be able to.
Solution. Note that with both allowed actions, the fraction does not decrease and always remains less than one. Indeed, the value of the fraction does not change with the second action; it remains to check the first action. Suppose at some step we have the fraction $a / b$, where $a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,193 |
Task 2. Angelica wants to choose a three-digit code for her suitcase. To make it easier to remember, Angelica wants all the digits in her code to be in non-decreasing order. How many different options does Angelica have for choosing the code? | Problem 2. Angelica wants to choose a three-digit code for her suitcase. To make it easier to remember, Angelica wants all the digits in her code to be in non-decreasing order. How many different ways can Angelica choose her code? Answer. 220. | 220 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,195 |
Problem 5. A semicircle is inscribed inside a quarter circle as shown in the figure. Find the ratio of the area of the semicircle to the area of the quarter circle. If necessary, round the answer to 0.01.
+\log _{4}\left(\log _{8} x\right)+\log _{8}\left(\log _{2} x\right)=1$. Find the value of the expression $\log _{4}\left(\log _{2} x\right)+\log _{8}\left(\log _{4} x\right)+\log _{2}\left(\log _{8} x\right)$. If necessary, round the answer to 0.... | Problem 6. The number $x$ is such that $\log _{2}\left(\log _{4} x\right)+\log _{4}\left(\log _{8} x\right)+\log _{8}\left(\log _{2} x\right)=1$. Find the value of the expression $\log _{4}\left(\log _{2} x\right)+\log _{8}\left(\log _{4} x\right)+\log _{2}\left(\log _{8} x\right)$. If necessary, round the answer to 0.... | \frac{5}{3}-\log_{4}3\approx0.87 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,198 |
1. (Option 1) The decimal representation of the natural number $n$ contains sixty-three digits. Among these digits, there are twos, threes, and fours. No other digits are present. The number of twos is 22 more than the number of fours. Find the remainder when $n$ is divided by 9. | # Answer. 5.
(Option 2) The decimal representation of a natural number $n$ contains sixty-one digits. Among these digits, there are threes, fours, and fives. No other digits are present. The number of threes is 11 more than the number of fives. Find the remainder when $n$ is divided by 9.
Answer. 8.
Criteria. "干" Th... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,199 |
3. Before the trial run of one of the units of the hydropower station under construction, it was found that a fishing net is located $S$ km upstream from the dam. The river current speed is $v$ km/h. The hydropower station workers decided to go there by boat. Removing the net will take 5 minutes. What should be the boa... | Answer. $x \geqslant \frac{3 S+\sqrt{9 S^{2}+4 v^{2}}}{2}$.
Criteria. «+.» Answer $x=\ldots$ instead of $x \geqslant \ldots ;$ «士» answer found in different units of measurement; «干» in the correct solution, the comparison of the roots of the quadratic equation with zero was not performed, leading to an additional piec... | x\geqslant\frac{3S+\sqrt{9S^{2}+4v^{2}}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,200 |
6. Find the sum $\frac{\sin \frac{\pi}{3}}{2}+\frac{\sin \frac{2 \pi}{3}}{2^{2}}+\cdots+\frac{\sin \frac{2010 \pi}{3}}{2^{2010}}$. | Answer: $\frac{\sqrt{3}}{3}\left(1-\frac{1}{2^{2010}}\right)$.
Criteria. "士" In the correct solution, the sums of finite geometric progressions are approximated by the sums of the corresponding infinite geometric progressions (and the answer $\approx \frac{\sqrt{3}}{3}$ is obtained). | \frac{\sqrt{3}}{3}(1-\frac{1}{2^{2010}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,202 |
8. (Option 1) Find all solutions of the system $\left\{\begin{aligned} x y-t^{2} & =9 ; \\ x^{2}+y^{2}+z^{2} & =18\end{aligned}\right.$ | Answer. $(3 ; 3 ; 0 ; 0),(-3 ;-3 ; 0 ; 0)$.
(Variant 2) Find all $x, y$ and $z$ satisfying the equations
$\int \log _{2}\left(x^{2}+1\right)+\log _{2}\left(y^{2}+1\right)=4$
$x^{2}+y^{2}=2 \cos ^{2} z+4$.
Answer. $( \pm \sqrt{3} ; \pm \sqrt{3} ; \pi n), n \in \mathbb{Z}$. | (3;3;0;0),(-3;-3;0;0),(\\sqrt{3};\\sqrt{3};\pin),n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,203 |
9. One farmer made cheese in the form of an irregular pentagonal prism, while another made it in the form of a regular quadrilateral pyramid, the height of which is half the length of the side of the base. At night, mice ate all the cheese particles from the vertices of these polyhedra that were no more than 1 cm away ... | Answer. The damage to the first farmer is 4.5 times greater.
Criteria. «Shi» Correctly calculated the volume of at least one of the two eaten parts. | 4.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,204 |
10. On the coordinate plane, depict the set of points $(a, b)$ such that the system of equations $\left\{\begin{array}{c}x^{2}+y^{2}=a^{2}, \\ x+y=b\end{array}\right.$ has at least one solution. | Answer. $|b| \leq \sqrt{2}|a|$.
Criteria. «士» the set of parameters $a$ and $b$ is correctly found, but it is not plotted on the coordinate plane; «Ғ» the answer is correct for $a>0$, but not for $a<0$. | |b|\leq\sqrt{2}|| | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,205 |
Task 1. Prove the inequality
$$
\log _{2015} 2017>\frac{\log _{2015} 1+\log _{2015} 2+\ldots+\log _{2015} 2016}{2016}
$$ | Solution. After multiplying both sides by 2016 and some transformations, we get that it is sufficient to prove the inequality
$$
\log _{2015} 2017^{2016}>\log _{2015}(1 \cdot 2 \cdot \ldots \cdot 2016)
$$
The stated inequality follows from the fact that $2017^{2016}>1 \cdot 2 \cdot \ldots \cdot 2016$, and the latter ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 9,210 |
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 200 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't... | Answer: 21.
Solution. First, let's prove that when $n \leqslant 20$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+18=171=200-29
$$
the last mushroom picker collected at least ... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,211 |
Problem 3. Vasya wants to find all integers $a$ such that the expression $10 n^{3}-3 n^{5}+7$ is divisible by 15 for all integers $n$. What remainders can the number $a$ give when divided by 15? Provide all possible answers or prove that there are no such integers $a$. | Answer: 14.
Solution. First, let's show that $n^{3} \equiv n(\bmod 3)$ and $n^{5} \equiv n(\bmod 5)$ for any natural $n$. This can be done in several ways. We will present only three of them.
First method. We will divide all natural numbers into groups based on their remainders when divided by 3 and 5:
| $n$ | $n^{3... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,212 |
Problem 4. A circle is inscribed in trapezoid $ABCD$, touching the lateral side $AD$ at point $K$. Find the area of the trapezoid if $AK=16, DK=4$ and $CD=6$. | Answer: 432.
Solution. Let $L, M, N$ be the points of tangency of the inscribed circle with the sides $BC, AB, CD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $DN = DK = 4$ (the first equality follows from the equality of the segments of ta... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,213 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}=(y-z)^{2}-3 \\
y^{2}=(z-x)^{2}-7 \\
z^{2}=(x-y)^{2}+21
\end{array}\right.
$$ | Answer: $(-1,-3,-5),(1,3,5)$.
Solution. Move the square of the difference in each equation to the left side and apply the formula for the difference of squares:
$$
\left\{\begin{array}{l}
(x-y+z)(x+y-z)=-3 \\
(y-z+x)(y+z-x)=-7 \\
(z-x+y)(z+x-y)=21
\end{array}\right.
$$
Let $X=-x+y+z, Y=x-y+z, Z=x+y-z$. Then
$$
\lef... | (-1,-3,-5),(1,3,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,214 |
Task 6. Illustrate (with justification) on the coordinate plane $O x y$ the set of solutions to the inequality
$$
\left(y^{2}-\arcsin ^{2}(\sin x)\right) \cdot\left(y^{2}-\arcsin ^{2}(\sin (x+\pi / 3))\right) \cdot\left(y^{2}-\arcsin ^{2}(\sin (x-\pi / 3))\right)<0
$$ | Solution. Note that when $x$ is increased or decreased by $2 \pi$, the value of the expression on the left does not change, so it is sufficient to construct a solution for some interval of length $2 \pi$.
First, let's look at the set of points satisfying the equation
$$
\left(y^{2}-\arcsin ^{2}(\sin x)\right) \cdot\l... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,215 |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $40^{\circ}$. Fin... | Answer: $60^{\circ}$ and $80^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,216 |
Problem 8. For what values of the parameter $a$ does the equation
$$
3^{x^{2}-2 a x+a^{2}}=a x^{2}-2 a^{2} x+a^{3}+a^{2}-4 a+4
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x-a$ by $t$. Note that the number of solutions of the equation does not change with such a substitution. Then the original equation will take the form
$$
3^{t^{2}}=a t^{2}+a^{2}-4 a+4.
$$
Notice that the expressions on both sides do not change when $t$ is replaced by ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,217 |
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits can repeat. It is known that the sum of the digits of the number $2M$ is 35, and the sum of the digits of the number $M / 2$ is 29. What values can the sum of the digits of the number $M$ take? Li... | # Answer: 31.
Solution. Let's denote the sum of the digits of a natural number $n$ by $S(n)$. Notice the following facts, each of which is easy to verify if you add numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,218 |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=1: 2, B N: B B_{1}=1: 3, C K: C C_{1}=1: 4$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | Answer: 4.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, r... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,219 |
Task 1. Prove the inequality
$$
\log _{2017} 2019>\frac{\log _{2017} 1+\log _{2017} 2+\ldots+\log _{2017} 2018}{2018}
$$ | Solution. After multiplying both sides by 2018 and some transformations, we get that it is sufficient to prove the inequality
$$
\log _{2017} 2019^{2018}>\log _{2017}(1 \cdot 2 \cdot \ldots \cdot 2018)
$$
The stated inequality follows from the fact that $2019^{2018}>1 \cdot 2 \cdot \ldots \cdot 2018$, and the latter ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 9,220 |
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 338 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't... | Answer: 27.
Solution. First, let's prove that when $n \leqslant 26$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+24=300=338-38
$$
the last mushroom picker collected at least ... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,221 |
Problem 3. Vasya wants to find all integers $a$ such that the expression $5 n^{3}+6 n^{5}+4 a n$ is divisible by 15 for all integers $n$. What remainders can the number $a$ give when divided by 15? Provide all possible answers or prove that there are no such integers $a$. | # Answer: 1.
Solution. First, let's show that $n^{3} \equiv n(\bmod 3)$ and $n^{5} \equiv n(\bmod 5)$ for any natural $n$. This can be done in several ways. We will present only three of them.
First method. We will divide all natural numbers into groups based on their remainders when divided by 3 and 5:
| $n$ | $n^{... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,222 |
Problem 4. A circle with radius 4 is inscribed in trapezoid $ABCD$, touching the base $AB$ at point $M$. Find the area of the trapezoid if $BM=16$ and $CD=3$. | Answer: 108.
Solution. Let $K, L, N$ be the points of tangency of the inscribed circle with the sides $AD, BC, CD$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $BL = BM = 16$.
^{2}-3 \\
y^{2}=(z-x)^{2}-9 \\
z^{2}=(x-y)^{2}+27
\end{array}\right.
$$ | Answer: $(-1,-4,-6),(1,4,6)$.
Solution. Move the square of the difference in each equation to the left side and apply the formula for the difference of squares:
$$
\left\{\begin{array}{l}
(x-y+z)(x+y-z)=-3 \\
(y-z+x)(y+z-x)=-9 \\
(z-x+y)(z+x-y)=27
\end{array}\right.
$$
Let $X=-x+y+z, Y=x-y+z, Z=x+y-z$. Then
$$
\lef... | (-1,-4,-6),(1,4,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,224 |
Task 6. Illustrate (with justification) on the coordinate plane $O x y$ the set of solutions to the inequality
$$
\left(y^{2}-\arccos ^{2}(\cos x)\right) \cdot\left(y^{2}-\arccos ^{2}(\cos (x+\pi / 3))\right) \cdot\left(y^{2}-\arccos ^{2}(\cos (x-\pi / 3))\right)<0 .
$$ | Solution. Note that when $x$ is increased or decreased by $2 \pi$, the value of the expression on the left does not change, so it is sufficient to construct the solution for some interval of length $2 \pi$.
First, let's look at the set of points satisfying the equation
$$
\left(y^{2}-\arccos ^{2}(\cos x)\right) \cdot... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,225 |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $50^{\circ}$. Fin... | Answer: $60^{\circ}$ and $70^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,226 |
Problem 8. For what values of the parameter $a$ does the equation
$$
3^{x^{2}+6 a x+9 a^{2}}=a x^{2}+6 a^{2} x+9 a^{3}+a^{2}-4 a+4
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x+3a$ by $t$. Note that the number of solutions to the equation does not change with this substitution. Then the original equation will take the form
$$
3^{t^{2}}=a t^{2}+a^{2}-4 a+4
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,227 |
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits may repeat. It is known that the sum of the digits of the number $2M$ is 31, and the sum of the digits of the number $M / 2$ is 28. What values can the sum of the digits of the number $M$ take? Li... | Answer: 29.
Solution. Let $S(n)$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,228 |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=2: 3, B N: B B_{1}=3: 5, C K: C C_{1}=4: 7$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | # Answer: 6.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$,... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,229 |
Task 1. Prove the inequality
$$
\log _{2016} 2018>\frac{\log _{2016} 1+\log _{2016} 2+\ldots+\log _{2016} 2017}{2017}
$$ | Solution. After multiplying both sides by 2017 and some transformations, we get that it is sufficient to prove the inequality
$$
\log _{2016} 2018^{2017}>\log _{2016}(1 \cdot 2 \cdot \ldots \cdot 2017)
$$
The stated inequality follows from the fact that $2018^{2017}>1 \cdot 2 \cdot \ldots \cdot 2017$, and the latter ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 9,230 |
Task 2. $n$ mushroom pickers went to the forest and brought a total of 450 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answe... | Answer: 30.
Solution. First, let's prove that when $n \leqslant 29$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+28=406=450-44
$$
the last mushroom picker collected at least 44 mus... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,231 |
Problem 3. Vasya wants to find all integers $a$ such that the expression $10 n^{3}+3 n^{5}+7 a n$ is divisible by 15 for all integers $n$. What remainders can the number $a$ give when divided by 15? Provide all possible answers or prove that there are no such integers $a$. | Answer: 11.
Solution. First, let's show that $n^{3} \equiv n(\bmod 3)$ and $n^{5} \equiv n(\bmod 5)$ for any natural $n$. This can be done in several ways. We will present only three of them.
First method. We will divide all natural numbers into groups based on their remainders when divided by 3 and 5:
| $n$ | $n^{3... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,232 |
Problem 4. A circle is inscribed in trapezoid $ABCD$, touching the lateral side $BC$ at point $L$. Find the area of the trapezoid if $BL=4, CL=\frac{1}{4}$, and $AB=6$. | Answer: $6.75$.
Solution. Let $M, N, K$ be the points of tangency of the inscribed circle with the sides $AB, CD, AD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $BM = BL = 4$ (the first equality follows from the equality of tangent segment... | 6.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,233 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}=(y-z)^{2}-8 \\
y^{2}=(z-x)^{2}-16 \\
z^{2}=(x-y)^{2}+32
\end{array}\right.
$$ | Answer: $(-1,-3,-6),(1,3,6)$.
Solution. Move the square of the difference in each equation to the left side and apply the formula for the difference of squares:
$$
\begin{cases}(x-y+z)(x+y-z) & =-8 \\ (y-z+x)(y+z-x) & =-16 \\ (z-x+y)(z+x-y) & =32\end{cases}
$$
Let $X=-x+y+z, Y=x-y+z, Z=x+y-z$. Then
$$
\begin{cases}... | (-1,-3,-6),(1,3,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,234 |
Task 6. Illustrate (with justification) on the coordinate plane $O x y$ the set of solutions to the inequality
$$
\left(y^{2}-\arcsin ^{2}(\sin x)\right) \cdot\left(y^{2}-\arcsin ^{2}(\sin (x+\pi / 6))\right) \cdot\left(y^{2}-\arcsin ^{2}(\sin (x-\pi / 6))\right)<0
$$ | Solution. Note that when $x$ is increased or decreased by $2 \pi$, the value of the expression on the left does not change, so it is sufficient to construct a solution for some interval of length $2 \pi$.
First, let's look at the set of points satisfying the equation
$$
\left(y^{2}-\arcsin ^{2}(\sin x)\right) \cdot\l... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,235 |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $70^{\circ}$. Fin... | Answer: $60^{\circ}$ and $50^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,236 |
Task 8. For what values of the parameter $a$ does the equation
$$
5^{x^{2}-6 a x+9 a^{2}}=a x^{2}-6 a^{2} x+9 a^{3}+a^{2}-6 a+6
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x-3a$ by $t$. Note that the number of solutions to the equation does not change with this substitution. Then the original equation will take the form
$$
5^{t^{2}}=a t^{2}+a^{2}-6a+6
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-t... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,237 |
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits can repeat. It is known that the sum of the digits of the number $2M$ is 43, and the sum of the digits of the number $M / 2$ is 31. What values can the sum of the digits of the number $M$ take? Li... | # Answer: 35.
Solution. Let $S(n)$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ a... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,238 |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=3: 7, B N: B B_{1}=2: 5, C K: C C_{1}=4: 9$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | Answer: 8.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, r... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,239 |
Task 1. Prove the inequality
$$
\log _{2018} 2020>\frac{\log _{2018} 1+\log _{2018} 2+\ldots+\log _{2018} 2019}{2019}
$$ | Solution. After multiplying both sides by 2019 and some transformations, we get that it is sufficient to prove the inequality
$$
\log _{2018} 2020^{2019}>\log _{2018}(1 \cdot 2 \cdot \ldots \cdot 2019)
$$
The stated inequality follows from the fact that $2020^{2019}>1 \cdot 2 \cdot \ldots \cdot 2019$, and the latter ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 9,240 |
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 162 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, declared: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify you... | Answer: 18.
Solution. First, let's prove that when $n \leqslant 17$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+16=136=162-26
$$
the last mushroom picker collected at least 26 mus... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,241 |
Problem 3. Vasya wants to find all integers $a$ such that the expression $5 n^{3}+9 n^{5}+8+a$ is divisible by 15 for all integers $n$. What remainders can the number $a$ give when divided by 15? Provide all possible answers or prove that there are no such integers $a$. | Answer: 2.
Solution. First, let's show that $n^{3} \equiv n(\bmod 3)$ and $n^{5} \equiv n(\bmod 5)$ for any natural $n$. This can be done in several ways. We will present only three of them.
First method. We will divide all natural numbers into groups based on their remainders when divided by 3 and 5:
| $n$ | $n^{3}... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,242 |
Problem 4. A circle with radius 2 is inscribed in trapezoid $ABCD$, touching the base $CD$ at point $N$. Find the area of the trapezoid if $DN=1$ and $AB=12$. | Answer: 27.
Solution. Let $K, L, M$ be the points of tangency of the inscribed circle with the sides $AD, BC, AB$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $DK = DN = 1$.
^{2}-8 \\
y^{2}=(z-x)^{2}-20 \\
z^{2}=(x-y)^{2}+40
\end{array}\right.
$$ | Answer: $(-1,-4,-7),(1,4,7)$.
Solution. Move the square of the difference in each equation to the left side and apply the formula for the difference of squares:
$$
\begin{cases}(x-y+z)(x+y-z) & =-8 \\ (y-z+x)(y+z-x) & =-20 \\ (z-x+y)(z+x-y) & =40\end{cases}
$$
Let $X=-x+y+z, Y=x-y+z, Z=x+y-z$. Then
$$
\begin{cases}... | (-1,-4,-7),(1,4,7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,244 |
Task 6. Illustrate (with justification) on the coordinate plane $O x y$ the set of solutions to the inequality
$$
\left(y^{2}-\arccos ^{2}(\cos x)\right) \cdot\left(y^{2}-\arccos ^{2}(\cos (x+\pi / 6))\right) \cdot\left(y^{2}-\arccos ^{2}(\cos (x-\pi / 6))\right)<0 .
$$ | Solution. Note that when $x$ is increased or decreased by $2 \pi$, the value of the expression on the left does not change, so it is sufficient to construct the solution for some interval of length $2 \pi$.
First, let's look at the set of points satisfying the equation
$$
\left(y^{2}-\arccos ^{2}(\cos x)\right) \cdot... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,245 |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $80^{\circ}$. Fin... | Answer: $60^{\circ}$ and $40^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,246 |
Problem 8. For what values of the parameter $a$ does the equation
$$
5^{x^{2}+2 a x+a^{2}}=a x^{2}+2 a^{2} x+a^{3}+a^{2}-6 a+6
$$
have exactly one solution? | Answer: Only when $a=1$.
Solution. Let's denote $x+a$ by $t$. Note that the number of solutions of the equation does not change with such a substitution. Then the original equation will take the form
$$
5^{t^{2}}=a t^{2}+a^{2}-6 a+6
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,247 |
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits may repeat. It is known that the sum of the digits of the number $2M$ is 39, and the sum of the digits of the number $M / 2$ is 30. What values can the sum of the digits of the number $M$ take? Li... | Answer: 33.
Solution. Let $S(n)$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,248 |
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=5: 6, B N: B B_{1}=6: 7, C K: C C_{1}=2: 3$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K... | Answer: 10.
Solution. Suppose we have found the position of point $P$ at which the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, r... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,249 |
Problem 5. Multiply both sides of the equation by $\sin \frac{x}{2}$ (this transformation is equivalent, as this expression does not turn to zero on the specified interval). Rewrite $2 \sin \frac{x}{2} \sin k x$ as $\cos \frac{2 k-1}{2} x-\cos \frac{2 k+1}{2} x$. After this, all terms will cancel out except the first a... | Answer: $\frac{11 \pi}{5}$.
Remark. $\arccos \frac{-1+\sqrt{5}}{4}=\frac{2 \pi}{5}$ (in particular, the answer $\pi+\arccos \left(\frac{-1-\sqrt{5}}{4}\right)+\arccos \left(\frac{-1+\sqrt{5}}{4}\right)$ is also correct). | \frac{11\pi}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,250 |
Problem 6. We will "flip" each of the equations in the system:
$$
\left\{\begin{array}{l}
\frac{x+y}{x y}=1 \\
\frac{y+z}{y z}=\frac{1}{2} \\
\frac{x+z}{x z}=\frac{1}{3}
\end{array}\right.
$$
(The transformation is equivalent since none of the right-hand sides can become zero).
Notice that $\frac{x+y}{x y}=\frac{1}{... | Answer. $x=\frac{12}{5}, y=\frac{12}{7}, z=-12$ | \frac{12}{5},\frac{12}{7},-12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,251 |
Problem 5. Multiply both sides of the equation by $\sin \frac{x}{2}$ (this transformation is equivalent, as this expression does not turn to zero on the specified segment). Rewrite $2 \sin \frac{x}{2} \sin k x$ as $\cos \frac{2 k-1}{2} x-\cos \frac{2 k+1}{2} x$. After this, all terms will cancel out except the first an... | Answer: $\frac{29 \pi}{5}$.
Remark. $\arccos \frac{-1+\sqrt{5}}{4}=\frac{2 \pi}{5}$ (in particular, the answer $7 \pi-\arccos \left(\frac{-1-\sqrt{5}}{4}\right)-\arccos \left(\frac{-1+\sqrt{5}}{4}\right)$ is also correct). | \frac{29\pi}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,253 |
Problem 6. "Flip" each of the equations in the system:
$$
\left\{\begin{array}{l}
\frac{a+b}{a b}=1 \\
\frac{b+c}{b c}=\frac{1}{2} \\
\frac{c+a}{c a}=\frac{1}{4}
\end{array}\right.
$$
(The transformation is equivalent since none of the right-hand sides can turn into zero).
Notice that $\frac{a+b}{a b}=\frac{1}{a}+\f... | Answer. $a=\frac{8}{3}, b=\frac{8}{5}, c=-8$. | =\frac{8}{3},b=\frac{8}{5},=-8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,254 |
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