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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 |
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 |
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 |
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
$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_... | -19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. Given an isosceles triangle $A B C(A B=B C)$ with the angle at the vertex equal to $102^{\circ}$. Point $O$ is located inside triangle $A B C$ such that $\angle O C A=30^{\circ}$, and $\angle O A C=21^{\circ}$. Find the measure of angle $\angle B O A$. | Answer: $81^{\circ}$.
Solution. Let $B H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $C O$ and segment $B H$. Note that $A S=S C$. For example, in triangle $A S C$, the median $S H$ coincides with the height.
 \pi}{47}+\ldots+\operatorname{tg} \frac{2019 \pi}{47} \cdot \operat... | Answer: -2021.
Solution. From the formula
$$
\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 |
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 |
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.
 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 |
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
$$
the last mushroom picker collected at least ... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A circle is inscribed in trapezoid $ABCD$, touching the lateral side $AD$ at point $K$. Find the area of the trapezoid if $AK=16, DK=4$ and $CD=6$. | Answer: 432.
Solution. Let $L, M, N$ be the points of tangency of the inscribed circle with the sides $BC, AB, CD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $DN = DK = 4$ (the first equality follows from the equality of the segments of ta... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $40^{\circ}$. Fin... | Answer: $60^{\circ}$ and $80^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. For what values of the parameter $a$ does the equation
$$
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 |
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 |
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 |
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 |
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$.
$ 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 |
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 |
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
$$
the last mushroom picker collected at least 44 mus... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $70^{\circ}$. Fin... | Answer: $60^{\circ}$ and $50^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 8. For what values of the parameter $a$ does the equation
$$
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 |
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 |
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 |
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 |
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$.
$ 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 |
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 |
Problem 9. In a football championship, 20 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | Answer: 90 games.
Solution. We will consider the games that have not been played. The condition means that the unplayed games do not form triangles. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the estimate is obvious).
Inductive... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other?
# | # Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played against each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played with each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the ... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 30? | Answer: 32.
Solution. First, let's provide an example of 32 numbers whose sum is 0 and the sum of their squares is 30. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{16}=\sqrt{\frac{15}{16}}, x_{17}=x_{18}=\ldots=x_{32}=-\sqrt{\frac{15}{16}}$ will work.
Now, we will prove that fewer than 32 numbers will not suffice... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $ABC$, a circle is constructed on the leg $AC$ as its diameter, which intersects the hypotenuse $AB$ at point $E$. A tangent to the circle is drawn through point $E$, which intersects the leg $CB$ at point $D$. Find the length of $DB$, if $AE=6$, and $BE=2$. | Answer: 2.
Solution. The solution is based on two simple observations. First, $\angle A E C=90^{\circ}$, since it subtends the diameter. Second, $D E$ and $D C$ are tangents to the circle from the condition, so $D E=D C$. Therefore, in the right triangle $C E B$, a point $D$ is marked on the hypotenuse $B C$ such that... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-11 x^{2}+a x-8=0$ have three distinct real roots that form a geometric progression? | Answer: only 22.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-11 x^{2}+a x-8=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=11 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 42? | Answer: 44.
Solution. First, let's provide an example of 44 numbers whose sum is 0 and the sum of their squares is 42. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{22}=\sqrt{\frac{21}{22}}, x_{23}=x_{24}=\ldots=x_{44}=-\sqrt{\frac{21}{22}}$ will work.
Now, we will prove that it is impossible to achieve this with ... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $P Q R$, a circle is constructed on the leg $P R$ as its diameter, which intersects the hypotenuse $P Q$ at point $T$. A tangent to the circle is drawn through point $T$, which intersects the leg $R Q$ at point $S$. Find the length of $S Q$, if $P T=15$, and $Q T=5$. | Answer: 5.
Solution. The solution is based on two simple observations. First, $\angle P T R=90^{\circ}$, since it subtends the diameter. Second, $S T$ and $S R$ are tangents to the circle from the given conditions, so $S T=S R$. Therefore, in the right triangle $R T Q$, a point $S$ is marked on the hypotenuse $Q R$ su... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-14 x^{2}+a x-27=0$ have three distinct real roots that form a geometric progression? | Answer: only 42.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-14 x^{2}+a x-27=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=14 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=27
\e... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 36?
# | # Answer: 38.
Solution. First, let's provide an example of 38 numbers whose sum is 0 and the sum of their squares is 36. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{19}=\sqrt{\frac{18}{19}}, x_{20}=x_{21}=\ldots=x_{38}=-\sqrt{\frac{18}{19}}$ will work.
Now, we will prove that fewer than 38 numbers will not suffi... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $X Y Z$, a circle is constructed on the leg $X Z$ as its diameter, which intersects the hypotenuse $X Y$ at point $W$. A tangent to the circle is drawn through point $W$, which intersects the leg $Z Y$ at point $V$. Find the length of $V Y$, if $X W=12$, and $Y W=4$. | Answer: 4.
Solution. The solution is based on two simple observations. First, $\angle X W Z=90^{\circ}$, since it subtends the diameter. Second, $V W$ and $V Z$ are tangents to the circle from the condition, so $V W=V Z$. Therefore, in the right triangle $Z W Y$, a point $V$ is marked on the hypotenuse $Y Z$ such that... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-15 x^{2}+a x-64=0$ have three distinct real roots that form a geometric progression? | Answer: only 60.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-15 x^{2}+a x-64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=15 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 40? | Answer: 42.
Solution. First, let's provide an example of 42 numbers whose sum is 0 and the sum of their squares is 40. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{21}=\sqrt{\frac{20}{21}}, x_{22}=x_{23}=\ldots=x_{42}=-\sqrt{\frac{20}{21}}$ will work.
Now, we will prove that fewer than 42 numbers will not suffice... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In a right triangle $KLM$, a circle is constructed on the leg $KM$ as its diameter, which intersects the hypotenuse $KL$ at point $G$. A tangent to the circle is drawn through point $G$, intersecting the leg $ML$ at point $F$. Find the length of $FL$, if $KG=5$ and $LG=4$. | Answer: 3.
Solution. The solution is based on two simple observations. First, $\angle K G M=90^{\circ}$, since it subtends the diameter. Second, $F G$ and $F M$ are tangents to the circle from the condition, so $F G=F M$. Therefore, in the right triangle $M G L$, a point $F$ is marked on the hypotenuse $L M$ such that... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation $x^{3}+16 x^{2}+a x+64=0$ have three distinct real roots that form a geometric progression? | Answer: only 64.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}+16 x^{2}+a x+64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=-16 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=-64
... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. Each knight gives one affirmative answer to four questions, while a liar gives three. In total, there were $105+45+85+65=300$ affirmative answers. If all the residents of the city were knights, the total number of affirmative answers would be 200. The 100 extra "yes" answers come from the lies of the liars. ... | Answer: in Block B, on 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10. The sum of the surface areas of the polyhedra into which a parallelepiped is divided by sections is equal to the sum of the surface area of the parallelepiped and the areas of the internal surfaces. The sum of the areas of the internal surfaces is equal to twice the sum of the areas of the sections.
Let's ... | Answer: 194.
[^0]: ${ }^{1}$ This statement is a particular case of Cauchy's inequality. | 194 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Each of the two workers was assigned to process the same number of parts. The first completed the work in 8 hours. The second spent more than 2 hours on setting up the equipment and with its help finished the work 3 hours earlier than the first. It is known that the second worker processed as many parts in 1... | Answer: 4 times.
Solution: Let $x$ be the time spent on equipment setup. Then the second worker worked (on the equipment) $8-3-x=5-x$ hours, producing as much per hour as the first worker in $x+1$ hours. Therefore, $\frac{8}{5-x}=\frac{x+1}{1}$. We get $x^{2}-4 x+3=0$. But by the condition $x>2$, so $x=3$, and the req... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$. | Answer: 4.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$. | Answer: 3.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$. | Answer: 4.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$. | Answer: 3.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 13 adventurers have rubies; exactly 9 have emeralds; exactly 15 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has sapphires, then they have either emeralds or diamonds (but not both at the ... | Answer: 22.
Solution. Note that the number of adventurers who have sapphires is equal to the total number of adventurers who have emeralds or diamonds. Then, from the first condition, it follows that 9 adventurers have sapphires and emeralds, and 6 have sapphires and diamonds. That is, every adventurer who has emerald... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 3. A team of workers was working on pouring the rink on the large and small fields, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 4 more workers than in the part that worked on the small field. When the pouring of the la... | Answer: 10.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+4$, and the total number of people in the team is $2n+4$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=8$ and $AC=4$. Find the length of side $AB$ if the length of the vector $4 \overrightarrow{OA} - \overrightarrow{OB} - 3 \overrightarrow{OC}$ is 10. | Answer: 5.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what values of the parameter $a$ does the equation
$$
\log _{2}^{2} x+(a-6) \log _{2} x+9-3 a=0
$$
have exactly two roots, one of which is four times the other? | Answer: $-2.2$
Solution. Let $t=\log _{2} x$, then the equation becomes $t^{2}+(a-6) t+(9-3 a)=0$. Notice that $3 \cdot(3-a)=9-3 a, 3+(3-a)=6-a$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -3 and $3-a$. We make the reverse substitution: $\log _{2} x=3$ or $\log _{2} x=3-a$, ... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. In triangle $A B C$, side $A C=42$. The bisector $C L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $2: 1$, counting from the vertex. Find the length of side $A B$, if the radius of the circle inscribed in triangle $A B C$ is 14. | Answer: 56.
Answer: Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $AI$ is the angle bisector in triangle $ALC$, by the angle bisector theorem, we have: $AC: AL = CI: IL = 2: 1$, from which $AL = AC / 2 = 21$.
Next, $AC \cdot AL \c... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $F$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $F(n a, n b, n c)=n \cdot F(a, b, c)$, $F(a+n, b+n, c+n)=F(a, b, c)+n$, $F(a, b, c)=F(c, b, a)$ hold. Find $F(58,59,60)$. | Answer: 59.
Solution. Note that $F(-1,0,1)=F(1,0,-1)=(-1) \cdot F(-1,0,1)$, from which $F(-1,0,1)=0$. Then $F(58,59,60)=F(-1,0,1)+59=59$.
Comment. The function $F$ cannot be uniquely determined. For example, the functions $F(a, b, c)=(a+b+c) / 3$, $F(a, b, c)=b$, and $F(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 5 adventurers have rubies; exactly 11 have emeralds; exactly 10 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has diamonds, then they have either emeralds or sapphires (but not both at the ... | Answer: 16.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or diamonds. Then, from the second condition, it follows that 5 adventurers have rubies and emeralds, and 6 have emeralds and diamonds. That is, every adventurer who has diamonds must... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 3. A landscaping team worked on a large and a small football field, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 6 more workers than in the part that worked on the small field. When the landscaping of the large field wa... | Answer: 16.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+6$, and the total number of people in the team is $2n+6$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=8$ and $AC=5$. Find the length of side $BC$ if the length of the vector $\overrightarrow{OA}+3 \overrightarrow{OB}-4 \overrightarrow{OC}$ is 10. | Answer: 4.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $f$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $f(n a, n b, n c)=n \cdot f(a, b, c)$, $f(a+n, b+n, c+n)=f(a, b, c)+n$, $f(a, b, c)=f(c, b, a)$ hold. Find $f(24,25,26)$. | Answer: 25.
Solution. Note that $f(-1,0,1)=f(1,0,-1)=(-1) \cdot f(-1,0,1)$, from which $f(-1,0,1)=0$. Then $f(24,25,26)=f(-1,0,1)+25=25$.
Comment. The function $f$ cannot be uniquely determined. For example, the functions $f(a, b, c)=(a+b+c) / 3$, $f(a, b, c)=b$, and $f(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 4 adventurers have rubies; exactly 10 have emeralds; exactly 6 have sapphires; exactly 14 have diamonds. Moreover, it is known that
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same t... | Answer: 18.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or sapphires. Then, from the second condition, it follows that 4 adventurers have rubies and emeralds, and 6 have emeralds and sapphires. That is, every adventurer who has rubies must... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A team of workers was laying linoleum in a store's warehouse and in the cash hall, with the warehouse area being 3 times larger than the cash hall area. In the part of the team working in the warehouse, there were 5 more workers than in the part working in the cash hall. When the work in the warehouse was co... | Answer: 9.
Solution. Let the number of workers in the cash hall be denoted as $n$, then the number of workers in the warehouse is $n+5$, and the total number of people in the team is $2n+5$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivity o... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=5$ and $AB=4$. Find the length of side $AC$ if the length of the vector $3 \overrightarrow{OA}-4 \overrightarrow{OB}+\overrightarrow{OC}$ is 10. | Answer: 8.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $G$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$, the equalities $G(n a, n b, n c)=n \cdot G(a, b, c)$, $G(a+n, b+n, c+n)=G(a, b, c)+n$, $G(a, b, c)=G(c, b, a)$ hold. Find $G(89,90,91)$. | Answer: 90.
Solution. Note that $G(-1,0,1)=G(1,0,-1)=(-1) \cdot G(-1,0,1)$, from which $G(-1,0,1)=0$. Then $G(89,90,91)=G(-1,0,1)+90=90$.
Comment. The function $G$ cannot be uniquely determined. For example, the functions $G(a, b, c)=(a+b+c) / 3, G(a, b, c)=b$ and $G(a, b, c)=$ median of the numbers $\{a, b, c\}$ are... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 9 adventurers have rubies; exactly 8 have emeralds; exactly 2 have sapphires; exactly 11 have diamonds. Moreover, it is known that
- if an adventurer has diamonds, then they have either rubies or sapphires (but not both at the same t... | Answer: 17.
Solution. Note that the number of adventurers who have diamonds is equal to the total number of adventurers who have rubies or sapphires. Then, from the first condition, it follows that 9 adventurers have rubies and diamonds, and 2 have sapphires and diamonds. That is, every adventurer who has rubies must ... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A team of lumberjacks was cutting trees on a large and a small plot, with the area of the small plot being 3 times less than that of the large plot. In the part of the team that worked on the large plot, there were 8 more lumberjacks than in the part that worked on the small plot. When the tree harvesting on... | Answer: 14.
Solution. Let the number of workers on the smaller plot be denoted as $n$, then the number of workers on the larger plot is $n+8$, and the total number of workers in the team is $2n+8$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the product... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=5, AC=8$, and $BC=4$. Find the length of the vector $\overrightarrow{O A}-4 \overrightarrow{O B}+3 \overrightarrow{O C}$. | Answer: 10.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \c... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. In triangle $A B C$, side $B C=28$. The bisector $B L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $4: 3$, counting from the vertex. Find the radius of the circumscribed circle around triangle $A B C$, if the radius of the inscribed circle in it is 12. | Answer: 50.
Solution. Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $CI$ is the angle bisector in triangle $BLC$, by the angle bisector theorem, we have: $BC: CL = BI: IL = 4: 3$, from which $CL = 3BC / 4 = 21$.
Next, $BC \cdot CL... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. The function $g$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $g(n a, n b, n c)=$ $n \cdot g(a, b, c), g(a+n, b+n, c+n)=g(a, b, c)+n, g(a, b, c)=g(c, b, a)$ hold. Find $g(14,15,16)$. | Answer: 15.
Solution. Note that $g(-1,0,1)=g(1,0,-1)=(-1) \cdot g(-1,0,1)$, from which $g(-1,0,1)=0$. Then $g(14,15,16)=g(-1,0,1)+15=15$.
Comment. The function $g$ cannot be uniquely determined. For example, the functions $g(a, b, c)=(a+b+c) / 3$, $g(a, b, c)=b$, and $g(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In the $1^{\text{st}}$ grade class, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the ans... | Answer: 15 boys and 12 girls.
Solution. First solution. Let's denote the children who gave the answers (13,11), (17,11), (14,14) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of the girls has the same parity as $m$, and in the answers of the boys - the... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 4$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $4 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016. | Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 4$, from which $\overrightarrow{B L}=5 / 9 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{5}{9} \cdot \overrightarrow{B C}=\overrigh... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 256 wrestlers is held on an Olympic system: at ... | Answer: 16
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 256 participants, there are 8 rounds, so the number of the tournament winner does not exceed $1+2 \cdot... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In $1^{st}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers received wer... | Answer: 16 boys and 14 girls.
Solution. First solution. Let's denote the children who gave the answers (15,18), (15,10), (12,13) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of girls has the same parity as $m$, while in the answers of boys, it has the... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=4: 3$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $3 \cdot \overrightarrow{A B}+4 \cdot \overrightarrow{A C}$ is 2016. | Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=4: 3$, from which $\overrightarrow{B L}=4 / 7 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrigh... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 512 wrestlers is held on an Olympic system: at ... | Answer: 18.
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 512 participants, there are 9 rounds, so the number of the tournament winner does not exceed $1+2 \cdo... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In $1^{\text {st }}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the answers rec... | Answer: 13 boys and 16 girls.
Solution. First solution. Let's denote the children who gave the answers (12,18), (15,15), (11,15) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. T... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016. | Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 2$, from which $\overrightarrow{B L}=5 / 7 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrig... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. In the $1^{\text{st}}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers r... | Answer: 14 boys and 15 girls.
Solution. First solution. Let's denote the children who gave the answers $(10,14),(13,11),(13,19)$ as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. T... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. How many natural numbers $n>1$ exist, for which there are $n$ consecutive natural numbers, the sum of which is equal to 2016? | Answer: 5.
Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a+n-1$ have different parity. Therefore,... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=7: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+7 \cdot \overrightarrow{A C}$ is 2016. | Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=7: 2$, from which $\overrightarrow{B L}=7 / 9 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{7}{9} \cdot \overrightarrow{B C}=\overrig... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. Egor borrowed 28 rubles from Nikita, and then returned them in four payments. It turned out that Egor always returned a whole number of rubles, and the amount of each payment increased and was divisible by the previous one. What was the last amount Egor paid? | Answer: 18 rubles
Solution:
1) If Egor paid $a$ rubles the first time, then the second time - not less than $2a$, the third - not less than $4a$, the fourth - not less than $8a$, and in total - not less than $15a$. Since $15a \leq 28$, we get that $a=1$.
2) The second time he paid 2 or 3 rubles (because if 4, then he... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. On the surface of a pentagonal pyramid (see fig.), several gnomes live in pairwise distinct points, and they can live both inside the faces and on the edges or at the vertices. It turned out that on each face (including the vertices and edges that bound it) a different number of gnomes live. What is the minimum nu... | Answer: 6
Solution: There are 6 faces in total, so at least 5 gnomes live on the most "populated" one. If there are exactly 5 gnomes in total, then all of them live on one face (let's call it face $A$), so the faces where 4 and 3 gnomes live (let's call them $B$ and $C$ respectively) are adjacent to it. Then, on the e... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. Ten chess players over nine days played a full round-robin tournament, during which each of them played exactly one game with each other. Each day, exactly five games were played, with each chess player involved in exactly one of them. For what maximum $n \leq 9$ can it be claimed that, regardless of the schedule... | Answer. $n=5$.
Solution. By the end of the eighth day, each chess player has played 8 games, meaning they have not played one. The unplayed games divide the chess players into 5 non-intersecting pairs. By the Pigeonhole Principle, among any six chess players, there will always be two belonging to the same pair, that i... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. At a round table, 410 deputies sat down, each of whom was either a knight, who always tells the truth, or a liar, who always lies. Each of the deputies said: "Among my twenty neighbors to the left and twenty neighbors to the right, there are exactly 20 liars in total." It is known that at least half of the people ... | Answer: None.
Solution: Let's divide all the people sitting at the table into ten groups of 41 people each. Then, at least one group will have at least 21 liars. Otherwise, in each group, there would be a maximum of 20, i.e., no more than \(20 \times 10 = 200\) in total, which is less than half of the total number. Co... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. On a circular route, two buses operate with the same speed and a movement interval of 21 minutes. What will be the movement interval if 3 buses operate on this route at the same constant speed? | Answer: 14.
Solution: Since the interval of movement with two buses on the route is 21 minutes, the length of the route in "minutes" is 42 minutes. Therefore, the interval of movement with three buses on the route is $42: 3=14$ minutes.
Criteria: only answer, answer with verification - 3 points. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4. Natasha and Inna bought the same box of tea bags. It is known that one tea bag is enough for two or three cups of tea. Natasha's box was enough for only 41 cups of tea, while Inna's was enough for 58 cups. How many tea bags were in the box? | Answer: 20.
Solution: Let there be $n$ tea bags in the box. Then the number of brewings can vary from $2n$ to $3n$. Therefore, 58 is not greater than $3n$, which means $19<n$. Additionally, 41 is not less than $2n$, so $n<21$. Since the number of tea bags must be a natural number that is less than 21 but greater than ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. On a line, one hundred points are marked: green, blue, and red. It is known that between any two red points there is a blue one, and between any two blue points there is a green one. In addition, there are no fewer red points than blue ones, and no fewer blue points than green ones. How many points are painted blu... | Answer: 33.
Solution: Let the number of red points be $n$, then the number of blue points is not less than $n-1$ (the number of intervals between "adjacent" red points), and since by condition the number of red points is not less than the number of blue points, the number of blue points is either $n$ or $n-1$. Similar... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. On a circular route, 12 trams run in one direction at the same speed and at equal intervals. How many trams need to be added so that at the same speed, the intervals between trams decrease by one fifth?
# | # Answer: 3.
Solution: Let's take the entire distance as 60 arbitrary units, which means the trams are currently 5 arbitrary units apart. We want this distance to be reduced by $1 / 5$, making it equal to 4 arbitrary units. For this, we need $60: 4=15$ trams, which is 3 more than the current number.
Criteria: only th... | 3 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.5. Each digit of the natural number $N$ is strictly greater than the one to its left. What is the sum of the digits of the number $9 N$? | Answer: 9.
Solution: Note that $9 N=10 N-N$. Let's perform this subtraction in a column. In the units place, there will be the difference between 10 and the last digit of the number $N$, in the tens place - the last and the second-to-last digit, decreased by 1. In all subsequent places, there will be the difference be... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. In the morning, a dandelion blooms, it flowers yellow for three days, on the fourth morning it turns white, and by the evening of the fifth day, it withers. On Monday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and on Wednesday - 15 yellow and 11 white. How many white dandelions will the... | Answer. Six dandelions.
Solution. A blooming dandelion is white on the fourth and fifth day. This means that on Saturday, the dandelions that bloomed on Tuesday or Wednesday will be white. Let's determine how many there are. The dandelions that were white on Monday had already flown away by Wednesday, and 20 yellow on... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3. Inside a semicircle of radius 12, there are a circle of radius 6 and a small semicircle, each touching the others pairwise, as shown in the figure. Find the radius of the small semicircle. | Answer: 4.
Solution. Let the radius of the small semicircle be $x$, the center of the large semicircle be $A$, the center of the circle be $B$, and the center of the small semicircle be $C$. The centers of the tangent circles and semicircle, and the corresponding points of tangency, lie on the same straight line, so $... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.4. In a semicircle with a radius of 18 cm, a semicircle with a radius of 9 cm is constructed on one of the halves of the diameter, and a circle is inscribed, touching the larger semicircle from the inside, the smaller semicircle from the outside, and the second half of the diameter. Find the radius of this circle.
... | # Answer: 8 cm.
Solution. Let O, O_1, O_2 be the centers of the large semicircle, the small semicircle, and the inscribed circle, respectively, and let P, Q, R be the points of tangency of the inscribed circle with the diameter of the large semicircle, the small semicircle, and the large semicircle, respectively. Then... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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