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3. The polynomial $P(x)$ of the third degree, as well as its first, second, and third derivatives, take the value 1 at $x=-3$. Find $\mathrm{P}(0)$. | Answer: 13
## Examples of how to write answers:
$1 / 4$
0.25
$-10$
# | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,733 |
3. In space, there are 37 different vectors with integer non-negative coordinates, starting from the point $(0 ; 0 ; 0)$. What is the smallest value that the sum of all their coordinates can take? | Answer: 115
## Examples of answer notation:
239
Answer: $[-1 ; 7]$ | 115 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,734 |
3. Given the equalities $\log _{a} b+\log _{b} c+\log _{c} a=\log _{b} a+\log _{c} b+\log _{a} c=-1$. Find $\log _{a} b$.
If there are multiple answers, list them in ascending order separated by a semicolon. | Answer: $-1 ; 1$
## Examples of answer notation:
$1 / 4$
0,25 ; 0,5$
# | -1;1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,737 |
3. Four spheres of radius $r$ touch each other externally. A sphere of radius $\sqrt{6}+2$ touches all of them internally. Find $r$. | Answer: 2
## Examples of how to write answers:
$1 / 4$
0.25
10
# | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,739 |
3. For what values of parameters $a$ and $b$ ( $a$ not equal to $b$ ) does the set of values of the function $a(\sqrt{(a-x)(x-b)}-b)$ coincide with its domain? In your answer, list all possible values of the number $b$ in any order, separated by a semicolon. | Answer: $0 ;-1 \mid-1 ; 0$
## Examples of answer notation:
$1 / 4$
0, $25 ; 0,5$ | 0;-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,742 |
3. For what values of the parameter $a$ does the inequality $\left(a^{2}-6 a-7\right)|x|+(-a-19) \sqrt{x^{2}-9}>0$ have no solutions? Write the answer as an interval. If the set of answers consists of several intervals, list them separated by a semicolon. | Answer: $[-1 ; 7]$
## Examples of answer notation:
$[1 ; 2)$
$[0 ; 1] ;(2 ; 3)$
# | [-1;7] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 20,743 |
3. In the city of Gentle-city, there live 19 gentlemen, any two of whom are either friends or enemies. At some point, each gentleman asked each of his friends to send a hate card to each of his enemies (gentleman A asks gentleman B to send a card to all enemies of gentleman B). Each gentleman fulfilled all the requests... | Answer: 1538.
## Examples of answer recording:
100
# | 1538 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,744 |
3. $\mathrm{ABCD}$ is a trapezoid with bases $A D=15$ and $\mathrm{BC}=10$. $O$ is one of the intersection points of the circles constructed on the lateral sides of the trapezoid as diameters, and this point lies inside the trapezoid. Triangle $B C M$ is constructed on side $B C$ on the external side relative to the tr... | Answer: 4
## Examples of answer notations:
$1 / 4$
0.25
10 | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,745 |
# Problem 4. (3 points)
Given the function $f(x)=x^{2}+3$. Solve the equation $\underbrace{f(f(\ldots(7 x-3) \ldots))}_{10 \text { times }}=\underbrace{f(f(\ldots(x) \ldots))}_{11 \text { times }}$.
If there are multiple solutions, list them all in ascending order, separated by commas or semicolons. | Answer: $-7 ; 0 ; 1 ; 6$
# | -7,0,1,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,746 |
Problem 6. (3 points)
Let $f(x)=\sin (\pi x)$. How many roots does the function $\underbrace{f(f(f(\ldots f(x) \ldots)))}_{20 \text { times }}$ have on the interval $[0 ; 1]$? | Answer: $524289=2^{19}+1$
# | 524289=2^{19}+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 20,747 |
# Problem 8. (4 points)
Solve the equation $p^{2}-58 p=2 q^{2}-26 q-697$ in prime numbers. In your answer, list all possible values of $p$ in ascending order, separated by commas or semicolons. | Answer: $17 ; 19 ; 37 ; 41$
# | 17,19,37,41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,748 |
1. (2 points) In a row without spaces, all natural numbers are written in ascending order: $1234567891011 . .$. What digit stands at the 2017th place in the resulting long number? | Answer: 7.
Solution: The first 9 digits are contained in single-digit numbers, the next 180 - in two-digit numbers. $2017-180-9=1828$. Next, $1828: 3=609 \frac{1}{3}$. This means that the 2017-th digit is the first
 In a class, 10 people gathered, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each of them was asked to first name the number of knights in the room, and then the number of liars. It turned out that each number from 0 to 9 was named exactly twice. How many knight... | Answer: from 0 to 2.
Solution: Since knights give the same answer to all questions, there cannot be more of them than the maximum number of identical answers, i.e., 2. It is easy to construct examples of all cases. If there are two knights, they tell the truth, and each liar chooses a number other than 2 and 8, and me... | from0to2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,750 |
3. (3 points) Petya and Vasya are playing a game on an initially white $101 \times 101$ grid. Petya goes first and can paint one cell black with his first move. Each subsequent move allows a player to paint black any vertical or horizontal white rectangular strip $1 \times n$ on the grid, where $n$ is a natural number,... | Answer: The first player, Petya, wins.
Solution: The winning strategy for the first player is as follows: on the first move, he repaints the central cell, and then he moves symmetrically to the opponent relative to the center of the field. Thus, after each of his moves, the position on the field is symmetric relative ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,751 |
4. (3 points) Does there exist a four-digit natural number with the sum of its digits being 23, which is divisible by 23? | Answer: Yes, for example $7682=23 \cdot 334$.
Solution: Generally speaking, the solution is contained in the answer, but let's explain how to find such a number. The sum of the digits of the product gives the same remainder when divided by 9 as the sum of the original numbers. Represent 23 as the difference between a ... | 7682 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,752 |
6. (3 points) Given the cryptarithm: ЖАЛО + ЛОЖА = ОСЕНЬ. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter А. | Answer: 8
Solution: The rebus can be rewritten as ОСЕНЬ $=($ ЖА + ЛО $) \cdot 101$. First, this means that the last digit of ЖА + ЛО is Ь. Second, if ЖА + ЛО $<100$, the result will be a four-digit number. Let $Ж А+Л О=1 Х Ь$, where $X$ is some digit.
Then ОСЕНЬ $=1 Х Ь 00+1 Х Ь$. If $\mathrm{b}<9$, then the second a... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,753 |
8. (4 points) A triangle is divided into 1000 triangles. What is the maximum number of different points at which the vertices of these triangles can be located? | Answer: 1002
Solution: The sum of the angles of a triangle is $180^{\circ}$, thousands of triangles $-180000^{\circ}$. Where did the extra $179820^{\circ}$ come from? Each internal vertex, where only the angles of triangles meet, adds $360^{\circ}$. Each vertex on the side of the original triangle or on the side of on... | 1002 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,754 |
1. (2 points) In a row without spaces, all natural numbers from 999 to 1 are written in descending order: 999998 ...321. What digit is in the 2710th position of the resulting long number? | Answer: 9.
Solution: The first $3 \cdot 900=2700$ digits are contained in three-digit numbers. Therefore, we need to count another 10 digits in two-digit numbers. That is, we need the first digit of the fourth largest two-digit number. This number is 96, so we need the digit 9. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,755 |
2. (2 points) In a class, 12 people gathered, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each of them was asked to first name the number of knights in the room, and then the number of liars. It turned out that each number from 1 to 12 was named exactly twice. How many knigh... | Answer: from 0 to 2.
Solution: Since knights give the same answer to all questions, there cannot be more of them than the maximum number of identical answers, i.e., 2. It is easy to construct examples of all cases. If there are two knights, they tell the truth, and each liar chooses a number other than 2 and 10, and m... | from0to2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,756 |
3. (3 points) Petya and Vasya are playing a game on an initially white $100 \times 100$ grid. Petya goes first and can paint one cell black with his first move. Each subsequent move allows a player to paint black any vertical or horizontal white rectangular strip $1 \times n$ on the grid, where $n$ is a natural number,... | Answer: The second player, Vasya, wins.
Solution: He moves symmetrically to his opponent relative to the center of the board. Thus, after each of his moves, the position on the board is symmetric relative to the center, which means that for any move of the opponent, he again has a symmetric response. Therefore, the on... | Vasya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,757 |
4. (3 points) Does there exist a five-digit natural number with the sum of its digits equal to 31, which is divisible by 31? Answer: Yes, for example $93775=31 \cdot 3025$. | In general, the solution is contained in the answer, but let's explain how to find such a number. The sum of the digits of the product gives the same remainder when divided by 9 as the sum of the original numbers. Let's represent 23 as the difference between a number divisible by 5 and a number divisible by 9: $31=5 \c... | 93775=31\cdot3025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,758 |
6. (3 points) Given the cryptarithm: RIVER + SQUARE = ABVAD. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter B. | Answer: 2
Solution: The rebus can be rewritten as ABVAD $=(\mathrm{KA}+\mathrm{PE}) \cdot 101$. First, this means that the last digit of $\mathrm{KA}+\mathrm{PE}$ is D. Second, if $\mathrm{KA}+\mathrm{PE}<100$, the result will be a four-digit number. Let $\mathrm{KA}+\mathrm{PE}=1X$D, where $X$ is some digit.
Then AB... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,759 |
8. (4 points) A triangle is divided into 1000 triangles. What is the minimum number of distinct points at which the vertices of these triangles can be located? | Answer: 503
Solution: The sum of the angles of a triangle is $180^{\circ}$, so for a thousand triangles, it is $180000^{\circ}$. Where did the extra $179820^{\circ}$ come from? Each internal vertex where only the angles of triangles meet adds $360^{\circ}$. Each vertex on the side of the original triangle or on the si... | 503 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,760 |
# Problem 1. (2 points)
Petya came up with a quadratic equation $x^{2}+p x+q$, the roots of which are numbers $x_{1}$ and $x_{2}$. He told Vasya three out of the four numbers $p, q, x_{1}, x_{2}$, without specifying which was which. These turned out to be the numbers $1, 2, -6$. What was the fourth number? | # Answer: -3
Solution:
Notice that by Vieta's theorem $x_{1}+x_{2}+p=0$. However, among the three reported numbers, there are no three that sum up to 0. Therefore, the fourth number must sum to 0 with any two of the existing ones, which means it is either -3, 5, or 4.
Moreover, the condition $q=x_{1} x_{2}$ must be ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,761 |
Problem 2. (2 points)
Prove that the equation $16^{x}+21^{y}+26^{z}=t^{2}$ has no solutions in natural numbers. | # Solution:
Let's note that all numbers on the left side of the condition give a remainder of 1 when divided by 5. This means that \( t^2 \) gives a remainder of 3 when divided by 5. By checking all possible remainders, it is easy to see that this is impossible.
Instead of remainders when divided by 5, we can conside... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,762 |
Problem 3. (3 points)
Can a square be divided into 14 equal-area triangles, with a common vertex $O$ and the other vertices on the boundary of the square? | Answer: Yes
Solution:
Place point $O$ inside the square such that the distances from it to the left and right sides are in the ratio $3:2$, and the distances to the bottom and top sides are ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,763 |
# Problem 4. (3 points)
Three runners are moving along a circular track at constant equal speeds. When two runners meet, they instantly turn around and start running in opposite directions.
At some point, the first runner meets the second. Twenty minutes later, the second runner meets the third for the first time. An... | # Answer: 100
Solution: (in general form)
Let the first runner meet the second, then after $a$ minutes the second runner meets the third for the first time, and after another $b$ minutes the third runner meets the first for the first time.
Let the first and second runners meet at point $A$, the second and third at p... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,764 |
# Problem 5. (3 points)
In an isosceles trapezoid $A B C D$, the bisectors of angles $B$ and $C$ intersect on the base $A D$. $A B=50, B C=128$. Find the area of the trapezoid. | Answer: 5472
## Solution:
Let $K$ be the point of intersection of the angle bisectors. Angles $\angle B=\angle C$ are equal as the base angles of an isosceles trapezoid, so their halves are... | 5472 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,765 |
# Problem 6. (3 points)
Natural numbers $x, y, z$ are such that $\operatorname{GCD}(\operatorname{LCM}(x, y), z) \cdot \operatorname{LCM}(\operatorname{GCD}(x, y), z)=1400$.
What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$ can take?
# | # Answer: 10
## Solution:
Notice that $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $z$, and $z$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$, so $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.
$1400=2^{3} \... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,766 |
# Problem 7. (3 points)
On an island, there live vegetarians who always tell the truth, and cannibals who always lie. 50 residents of the island, including both women and men, gathered around a campfire. Each of them said either "All men at this campfire are cannibals" or "All women at this campfire are vegetarians," ... | Answer: 48
## Solution:
We will prove that if we have $n$ people, the maximum number of vegetarian women is $n-2$.
First, note that if someone has called someone a cannibal, then we definitely have at least one cannibal.
Second, according to the condition, there is at least one man. Therefore, the only case where w... | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,767 |
# Problem 8. (5 points)
Can the natural numbers from 1 to 64 (each used exactly once) be arranged in an $8 \times 8$ rectangular table so that in every $1 \times 3$ rectangle (vertical or horizontal) the sum of the numbers is even? | Answer: No
## Solution:
First, note the following: numbers in cells that are separated horizontally or vertically by exactly two cells must have the same parity, because together with the numbers written in the two intermediate cells, they must sum to an even number. Thus, in each row and each column, the remainders ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,768 |
3. A spider has woven a web that consists of the x-axis, the y-axis, as well as the following curves: $y=x$, $y=-x$, $x^{2}+y^{2}=9$, $x^{2}+y^{2}=49$, $x^{2}+y^{2}=81$. In one day, flies got caught in all the nodes. The spider is sitting at the point $(0,0)$ and plans to eat all the flies. What is the minimum distance... | Do not round the answer. To write the number $\pi$, use the Russian letter п or the English $\mathrm{p}$. Examples of writing the answer:
$3 \mathrm{p}+10$
$1.5 \pi+7.5$ | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,769 |
3. On a piece of paper, a square with a side length of 1 was drawn, next to it a square with a side length of 2, then a square with a side length of 3, and so on. It turned out that the area of the entire resulting figure is 42925. How many squares were drawn? | Answer: 50
## Examples of answer notation: 45
# | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,771 |
3. In the country of Aviania, there are 40 cities, some of which are connected by two-way flights. Moreover, between any two cities, there is only one reasonable air route (i.e., a route where the same flight is not used in different directions).
For each city, the air distance to the capital was calculated. It is cal... | Answer: 780.
## Examples of answer recording: 45
## Problem 3 (3 points).
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,772 | |
1. It is known that the natural number $n$ is divisible by 3 and 4. Find all such possible $n$, if it is known that the number of all its divisors (including 1 and $n$) is 15? If there are several possible values, list them in any order separated by a semicolon. | Answer: 144,324 || 144; 324 || 324; 144 || 324, 144 | 144;324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,777 |
3. $\mathrm{ABCD}$ is an isosceles trapezoid, $\mathrm{AB}=\mathrm{CD}=25, \mathrm{BC}=40, \mathrm{AD}=60$. $\mathrm{BCDE}$ is also an isosceles trapezoid. Find AE. (Points A and E do not coincide)
If there are multiple possible values, list them in any order separated by a semicolon. | Answer: 44.
## Examples of answer recording:
45
$45 ; 56$
## Problem 7 (3 points). | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,779 |
3. Let $f(x)=\frac{34^{x}}{4^{x}+2}$. Find the sum $f(0)+f\left(\frac{1}{2017}\right) \ldots f\left(\frac{2}{2017}\right) f(1)$. | Answer: 3027
## Examples of answer recording:
45
## Problem 9 (4 points). | 3027 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,780 |
3. Two tangents are drawn from point A to a circle. The distance from point A to the point of tangency is 10, and the distance between the points of tangency is 16. Find the greatest possible distance from point A to a point on the circle. | Answer: 30
## Examples of answer recording:
## Problem 10 (2 points). | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,781 |
1. The graph of the fractional-linear function $\frac{3 x-3}{x-2}$ was rotated around some point by some angle, resulting in the graph of the fractional-linear function $\frac{-x+a}{2 x-4}$. What can $a$ be? If there are multiple possible values, list them in any order separated by a semicolon. | Answer: $-4,8\|-4 ; 8\| 8 ;-4 \| 8,-4$ | -4;8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,782 |
3. The graph of the fractional-linear function $\frac{3 x-11}{x-3}$ was rotated around some point by some angle, resulting in the graph of the fractional-linear function $\frac{12 x-a}{4 x-1}$. What can $a$ be? If there are multiple possible values, list them in any order separated by a semicolon. | Answer: $-5 ; 11\|-5,11\| 11 ;-5|| 11,-5$
## Examples of answer notation:
$-1$
$-1 ; 2$ | -5;11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,784 |
# Problem 10. (4 points)
Given 11 natural numbers with a sum of 40. What is the smallest value that the sum of all 55 numbers, which are the reciprocals of their pairwise products, can take? | Write the answer in the form of a proper or improper fraction.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,785 |
3. (3 points) From cards with letters, you can form the word KARAKATITSA. How many words (not necessarily meaningful) can be formed from these cards, in which the letters R and T are adjacent? | Answer: $\frac{9!}{4!}=9 \cdot 8 \cdot 7 \cdot 6 \cdot 5=15120$
## Solution:
Let's replace the adjacent letters R and T with one letter, for example, the letter Sh, since they stand together. Then we are left with 9 letters, and among them, 4 letters A and 2 letters K.
Thus, the number of ways to rearrange the lette... | 15120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,788 |
4. (3 points) Given a triangle $A B C$ with angle $C=120^{\circ}$. Point $D$ is the foot of the perpendicular dropped from point $C$ to side $A B$; points $E$ and $F$ are the feet of the perpendiculars dropped from point $D$ to sides $A C$ and $B C$ respectively. Find the perimeter of triangle $A B C$, if it is known t... | Answer: $16+8 \sqrt{3}$
## Solution:
Triangle $E F C$ is isosceles. Since $\angle C$ in it is obtuse, point $C$ is its vertex, which means $E C=$ $F C$. But then the right triangles $E C D$ and $F C D$ are equal by the leg and hypotenuse, so $E D=$ $F D$, that is, point $D$ lies on the bisector of angle $C$, from whi... | 16+8\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,789 |
5. (3 points) Given one hundred quadratic trinomials, all of which have distinct leading coefficients. It turns out that the graphs of any two of them have exactly one common point. Prove that the graphs of all the trinomials have a common point
# | # Solution:
If the graphs of two quadratic trinomials have exactly one common point, it means that the difference of these trinomials has exactly one root, which means it is either a linear function, which cannot be the case since the leading coefficients of the trinomials are different, or a perfect square.
Consider... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,790 |
6. (4 points) With the number written on the board, one of the following operations is allowed:
1) Replace the original number with the difference between the number obtained by removing the last three digits and the number formed by the last three digits (which may be written in an improper form - with leading zeros; ... | # Answer: 88 or 94
## Solution:
Let the number be $1000a + b$, where $b < 1000$, and the first operation is performed on it. Then it transforms into $a - b$ or $b - a$. In the first case, we add these numbers, and in the second case, we subtract them - the result is $1001a$, which is divisible by 91. This means that ... | 88or94 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,791 |
7. (4 points) It is known that $x, y, z, t$ are non-negative numbers such that $x y z=1, y+z+t=2$. Prove that $x^{2}+y^{2}+z^{2}+t^{2} \geqslant 3$. | Solution:
$x^{2}+y^{2}+z^{2}+t^{2}=x^{2}+2 y z+y^{2}-2 y z+z^{2}+t^{2}=x^{2}+\frac{2}{x}+(y-z)^{2}+t^{2} \geqslant x^{2}+\frac{2}{x}$, which is at least 3 by the inequality of the arithmetic mean and geometric mean for three numbers. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 20,792 |
8. (5 points) Given a triangle $A B C$, point $I$ is the center of the inscribed circle, and point $A_{1}$ is taken such that point $A$ is the midpoint of segment $A I$. Prove that point $A_{1}$ and the centers of the excircles of triangle $A B C$ lie on the same circle.
# | # Solution:
Let point $O$ be the center of the circumcircle of triangle $ABC$; $O_{1}$ be such a point that $O$ is the midpoint of $O_{1}I$; points $D, E$, and $F$ be the midpoints of the arcs $AB, BC$, and $AC$ of the circumcircle of triangle $ABC$, and points $D_{1}, E_{1}, F_{1}$ be the centers of the excircles of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,793 |
3. (3 points) From cards with letters, you can form the word WATERPIPE. How many words (not necessarily meaningful) can be formed from these cards, in which the letters R and P are adjacent? | Answer: $\frac{9!}{4!\cdot 2}=\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{2}=7560$
## Solution:
Let's replace the adjacent letters P and R with one letter, for example, the letter Sh, which is possible since they stand next to each other. Then we have 9 letters, and among them, 4 letters O, 2 letters B, and 2 letters D.... | 2\sqrt{3}+3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,796 |
5. (3 points) Given one hundred quadratic trinomials. It turns out that the graphs of any two of them have exactly one common point, but the graphs of no three of them have a common point. Prove that at least fifty of these trinomials have the same leading coefficient.
# | # Solution:
If the graphs of two quadratic trinomials have exactly one common point, it means that the difference between these trinomials has exactly one root, that is, it is either a linear function or a perfect square.
Consider three quadratic trinomials $f, g$, and $h$ with different leading coefficients. Their d... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,797 |
6. (4 points) With the number written on the board, one of the following operations is allowed:
1) Replace the original number with the difference between the number obtained by removing the last four digits and the number formed by the last four digits (possibly written in an improper form - with leading zeros; the di... | Answer: 80 or 66.
## Solution:
Let the number $10000 a+b$, where $b<10000$, undergo the first operation. Then it turns into $a-b$ or $b-a$. In the first case, we add these numbers, and in the second, we subtract - the result is $10001 a$, which is divisible by 73. This means that if the remainder of the division of o... | 80or66 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,798 |
# 7. (4 points)
It is known that $x, y, z, t$ are non-negative numbers such that $x y z=2, y+z+t=2 \sqrt{2}$. Prove that $2 x^{2}+y^{2}+$ $z^{2}+t^{2} \geq 6$. | Solution:
$2 x^{2}+y^{2}+z^{2}+t^{2}=2 x^{2}+2 y z+y^{2}-2 y z+z^{2}+t^{2}=2 x^{2}+\frac{4}{x}+(y-z)^{2}+t^{2} \geqslant 2\left(x^{2}+\frac{2}{x}\right)=2\left(x^{2}+\frac{1}{x}+\frac{1}{x}\right)$, and $\left(x^{2}+\frac{1}{x}+\frac{1}{x}\right) \geqslant 3$ by the inequality of arithmetic mean and geometric mean for... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 20,799 |
8. (5 points) Given a triangle $A B C$, point $I$ is the incenter, and points $A_{1}, B_{1}, C_{1}$ are taken such that points $A, B, C$ are the midpoints of segments $A I, B I$, and $C I$ respectively. Prove that points $A_{1}, B_{1}, C_{1}$ and the excenter of triangle $A B C$, touching side $A B$, lie on the same ci... | # Solution:
Let point $O$ be the center of the circumcircle of triangle $ABC$; $O_{1}$ be such a point that $O$ is the midpoint of $O_{1}I$; point $D$ be the midpoint of the arc $AB$ of the circumcircle of triangle $ABC$, and point $D_{1}$ be the center of the excircle of triangle $ABC$, touching sides $AB$, $BC$, and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,800 |
3. Given a convex hexagon ABCDEF such that AB $\|\mathrm{CF}\| \mathrm{DE}, \mathrm{BC}\|\mathrm{AD}\| \mathrm{EF}$ and $\mathrm{CD} \| \mathrm{BE}$ $\| \mathrm{FA}$ and $\mathrm{AB}=\mathrm{DE}=\mathrm{CD}=\mathrm{AF}=13, \mathrm{BD}=24$. Find the area of the union of triangles $\mathrm{ACE}$ and BDF.
. What is the maximum number of chips that can be on the board? | Answer: 21
## Examples of answer recording:
14
## 9th grade.
# | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,810 |
# Problem 1. (2 points)
Let $x, y, z$ be pairwise coprime three-digit natural numbers. What is the greatest value that the GCD $(x+y+z, x y z)$ can take? | Answer: 2994
Solution:
The GCD of two numbers cannot be greater than either of them. The maximum possible value of $x+$ $y+z=997+998+999=2994$ and for these numbers, $x y z$ is indeed divisible by $x+y+z=2994=3 \cdot 998$. | 2994 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,811 |
Problem 2. (2 points)
On a circle, 10 points are marked. Any three of them form three inscribed angles. Petya calculated the number of different values that these angles can take. What is the maximum number he could have obtained? | Answer: 80
Solution:
Any two points form two arcs. All inscribed angles subtending the same arc are equal. For two adjacent points, on one of the two arcs between them, no other point lies, meaning a pair of adjacent points gives us one possible angle value, while a pair of non-adjacent points gives two values.
In t... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,812 |
Problem 3. (3 points)
Let $f(x)$ be a quadratic trinomial with integer coefficients. Given that $f(\sqrt{3}) - f(\sqrt{2}) = 4$. Find $f(\sqrt{10}) - f(\sqrt{7})$. | # Answer: 12
## Solution:
Let $f(x)=c x^{2}+d x+e$. Then $f(\sqrt{3})-f(\sqrt{2})=3 c+\sqrt{3} d+e-(2 c+\sqrt{2} d+e)=c+d(\sqrt{3}-\sqrt{2})$. This number can only be an integer if $d=0$. Therefore, $f(x)=c x^{2}+e$ and $f(\sqrt{3})-f(\sqrt{2})=c$.
Then $f(\sqrt{10})-f(\sqrt{7})=10 c+e-(7 c+e)=3 c=12$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,813 |
Problem 4. (3 points)
Prove that the equation $15^{x}+29^{y}+43^{z}=t^{2}$ has no solutions in natural numbers. | # Solution:
Let's note that all numbers on the left side of the condition give a remainder of 1 when divided by 7. Therefore, $t^{2}$ gives a remainder of 3 when divided by 7. By checking all possible remainders, it is easy to see that this never happens.
# | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,814 |
# Problem 5. (3 points)
In triangle $A B C$, the midpoints of sides $A B=40$ and $B C=26$ are marked as points $K$ and $L$ respectively. It turns out that the quadrilateral $A K L C$ is a tangential quadrilateral. Find the area of triangle $A B C$. | # Answer: 264
## Solution:
According to the Midline Theorem, $K L=\frac{1}{2} A C$. In a cyclic quadrilateral, the sums of the opposite sides are equal, that is, $K L+A C=A K+C L=\frac{A B+B C}{2}$, so $\frac{3 A C}{2}=\frac{66}{2}$ and $A C=22$. Knowing the sides of the triangle, the area can be calculated using Her... | 264 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,815 |
# Problem 6. (3 points)
Positive numbers $x, y, z$ are such that $x y + y z + x z = 12$.
Find the smallest possible value of $x + y + z$.
# | # Answer: 6
## Solution:
By adding the inequalities $x^{2}+y^{2} \geqslant 2 x y, x^{2}+z^{2} \geqslant 2 x z$ and $y^{2}+z^{2} \geqslant 2 y z$ and dividing by 2, we get $x^{2}+y^{2}+z^{2} \geqslant x y+y z+x z$.
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+x z) \geqslant 3(x y+y z+x z)=36$, from which $x+y+z \geqslant... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,816 |
# Problem 7. (3 points)
From point $K$ on side $A C$ of triangle $A B C$, perpendiculars $K L_{1}$ and $K M_{1}$ were dropped to sides $A B$ and $B C$ respectively. From point $L_{1}$, a perpendicular $L_{1} L_{2}$ was dropped to $B C$, and from point $M_{1}$, a perpendicular $M_{1} M_{2}$ was dropped to $A B$.
It tu... | # Answer: 8
## Solution:
Notice that the quadrilateral $L_{1} M_{2} L_{2} M_{1}$ is cyclic, since $\angle L_{1} M_{2} M_{1}=\angle L_{1} L_{2} M_{1}=$ $90^{\circ}$. Therefore, $\angle B M_{2} L_{2}=180^{\circ}-\angle L_{1} M_{2} L_{2}=\angle L_{2} M_{1} L_{1}=\angle B M_{1} L_{1}$. Similarly, $\angle B L_{2} M_{2}=\a... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,817 |
# Problem 8. (2 points)
Can the natural numbers from 1 to 48 (each used exactly once) be arranged in a $6 \times 8$ rectangular table so that in every $1 \times 3$ rectangle (vertical or horizontal) the sum of the numbers is even? | Answer: Yes
## Solution:
To come up with an example, first note the following: numbers in cells that are two cells apart horizontally or vertically must have the same parity, because together with the numbers written in the two intermediate cells, they must sum to an even number. Thus, in each row and each column, th... | Yes | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,818 |
# Problem 1. (2 points)
A cubic polynomial has three roots. The greatest value of the polynomial on the interval $[4 ; 9]$ is achieved at $x=5$, and the smallest value is achieved at $x=7$. Find the sum of the roots of the polynomial.
# | # Answer: 18
## Solution:
Since in all options the minimum and maximum on the interval are not reached at its ends, they are reached at the roots of the polynomial's derivative. Let the polynomial be of the form $a x^{3}+b x^{2}+c x+d$, and its derivative, respectively, $3 a x^{2}+2 b x+c$. In this case, the sum of t... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,821 |
# Problem 2. (3 points)
Find the sum of natural numbers from 1 to 3000 inclusive that have common divisors with the number 3000, greater than 1. | # Answer: 3301500
## Solution:
$3000=2^{3} \cdot 3 \cdot 5^{3}$, so we are interested in numbers divisible by 2, 3, or 5. First, let's find the number of such numbers. For this, we will use the principle of inclusion and exclusion. There are exactly $\frac{3000}{2}=1500$ even numbers from 1 to 3000, $\frac{3000}{3}=1... | 3301500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,822 |
# Problem 5. (3 points)
Vasya chose four numbers and for each pair, he calculated the logarithm of the larger number to the base of the smaller one. This resulted in six logarithms. Four of them are 15, 20, 21, and 28. What values can the largest of all six logarithms take? | # Answer: $28 ; 420$
## Solution:
Let the four original numbers be $x \leqslant y \leqslant z \leqslant t$. Denote $a=\log _{x} y, b=\log _{y} z, c=\log _{z} t$. Then $\log _{x} z=a b, \log _{y} t=b c, \log _{x} t=a b c$, which means our six logarithms are $a, b, c, a b, b c$ and $a b c$. The largest of these is $a b... | 28;420 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,824 |
# Problem 6. (4 points)
Quadrilateral $A B C D$ is circumscribed around a circle with center at point $O, K, L, M, N$ - the points of tangency of sides $A B, B C, C D$ and $A D$ respectively, $K P, L Q, M R$ and $N S$ - the altitudes in triangles $O K B, O L C, O M D, O N A . O P=15, O A=32, O B=64$.
Find the length ... | # Answer: 30
## Solution:
Triangles $O K A$ and $O N A$ are right triangles with a common hypotenuse and a leg equal to the radius of the circle, so they are congruent. Therefore, their altitudes fall on the same point of the common hypotenuse, meaning $K S$ is the altitude in triangle $O K A$. Thus, points $S$ and $... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,825 |
# Problem 7. (4 points)
Two cubes with an edge of $12 \sqrt[4]{\frac{8}{11}}$ share a common face. A section of one of these cubes by a certain plane is a triangle with an area of 16. The section of the other by the same plane is a quadrilateral. What is the maximum value that its area can take? | # Answer: 128
## Solution:
Let our cubes be $A B C D A_{1} B_{1} C_{1} D_{1}$ and $A B C D A_{2} B_{2} C_{2} D_{2}$ with a common face $A B C D$. Let the triangular section of the first cube be $K L M$, where point $K$ lies on $A A_{1}$, point $L$ on $A B$, and point $M$ on $A D$. One side of the quadrilateral sectio... | 128 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,826 |
# Problem 8. (4 points)
Hansel and Gretel are playing a game, with Hansel going first. They take turns placing tokens on a $7 \times 8$ grid (7 rows and 8 columns). Each time Gretel places a token, she earns 4 points for each token already in the same row and 3 points for each token already in the same column.
Only o... | # Answer: 700
## Solution:
Let's say that Hansel also earns points according to the same principle as Gretel. In this case, each pair of cells in the same row will ultimately give one of the players 4 points, and each pair of cells in the same column will give 3 points. In one row, there are $\frac{8 \cdot 7}{2}=28$ ... | 700 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,827 |
# Problem 2. (2 points)
The sum of the sines of five angles from the interval $\left[0 ; \frac{\pi}{2}\right]$ is 3. What are the greatest and least integer values that the sum of their cosines can take?
# | # Answer: $2 ; 4$
## Solution:
Notice that $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$. The argument of the sine function ranges from $\frac{\pi}{4}$ to $\frac{3 \pi}{4}$, so $\frac{\sqrt{2}}{2} \leqslant \sin \left(x+\frac{\pi}{4}\right) \leqslant 1$. Therefore, $1 \leqslant \sin x+\cos x \leqslant \s... | 2;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,828 |
# Problem 1. (2 points)
The graphs of the quadratic trinomials $f(x)$ and $g(x)$ intersect at the point $(3 ; 8)$. The trinomial $f(x)+g(x)$ has a single root at 5. Find the leading coefficient of the trinomial $f(x)+g(x)$.
# | # Answer: 4
## Solution:
Since the quadratic polynomial $f(x)+g(x)$ has a unique root 5, it can be represented as $a(x-5)^{2}$, where $a$ is precisely the leading coefficient we are looking for. Additionally, the value of this quadratic polynomial at the point 3 is equal to the sum of the values of the polynomials $f... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,829 |
# Problem 2. (2 points)
A natural number $n$ when divided by 12 gives a remainder $a$ and an incomplete quotient $b$, and when divided by 10, it gives a remainder $b$ and an incomplete quotient $a$. Find $n$. | Answer: 119
## Solution:
From the condition, it follows that $n=12 b+a=10 a+b$, from which $11 b=9 a$. Therefore, $b$ is divisible by 9, and $a$ is divisible by 11. On the other hand, $a$ and $b$ are remainders from division by 12 and 10, respectively, so $0 \leqslant a \leqslant 11$ and $0 \leqslant b \leqslant 9$. ... | 119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,830 |
# Problem 3. (3 points)
It is known that $a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c=30$ and $a^{2}+b^{2}+c^{2}=13$.
Find $a+b+c$. | Answer: 5
## Solution:
$(a+b+c)^{3}-(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)=2\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c\right)$. Let $a+b+c$ be $x$, substitute the known values of the expressions from the condition, and we get the equation $x^{3}-13 x-60=0$.
It is not hard to notice that the number ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,831 |
# Problem 4. (3 points)
Circles $O_{1}$ and $O_{2}$ touch circle $O_{3}$ with radius 13 at points $A$ and $B$ respectively and pass through its center $O$. These circles intersect again at point $C$. It is known that $O C=12$. Find $A B$. | # Answer: 10
## Solution:
Since circles $O_{1}$ and $O_{2}$ touch circle $O_{3}$ at points $A$ and $B$ respectively and pass through its center $O$, $AO$ and $BO$ are their diameters. Therefore, angles $\angle OCA$ and $\angle OCB$ are right angles. These cannot be the same angle, as this would mean that circles $O_{... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,832 |
# Problem 6. (4 points)
Six positive numbers, not exceeding 3, satisfy the equations $a+b+c+d=6$ and $e+f=2$. What is the smallest value that the expression
$$
\left(\sqrt{a^{2}+4}+\sqrt{b^{2}+e^{2}}+\sqrt{c^{2}+f^{2}}+\sqrt{d^{2}+4}\right)^{2}
$$
can take? | Answer: 72
In the image, there are three rectangles $2 \times (a+b)$ and three rectangles $2 \times (c+d)$, forming a $6 \times 6$ square, since $a+b+c+d=6$. The segment of length 2 in the c... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,833 |
# Problem 7. (4 points)
In an $8 \times 8$ table, some cells are black, and the rest are white. In each white cell, the total number of black cells on the same row or column is written; nothing is written in the black cells. What is the maximum value that the sum of the numbers in the entire table can take? | Answer: 256
## Solution:
The number in the white cell consists of two addends: a "horizontal" and a "vertical" one. Consider the sum of all "horizontal" addends and the sum of all "vertical" addends separately across the entire table. If we maximize each of these two sums separately, the total sum will also be the gr... | 256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,834 |
# Problem 8. (5 points)
32 volleyball teams participate in a tournament according to the following scheme. In each round, all remaining teams are randomly paired; if the number of teams is odd, one team skips this round. In each pair, one team wins and the other loses, as there are no draws in volleyball. After three ... | # Solution:
For all teams to be eliminated except one, they must suffer at least 93 losses, meaning at least 93 matches must be played.
We also note that after each round, the number of teams decreases by at most half, because no more than half of the teams lose. In particular, in the final round, only 2 teams could ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,835 |
# Task 1. (2 points)
In a class, each student has either 5 or 6 friends (friendship is mutual), and any two friends have a different number of friends in the class. What is the smallest number of students, greater than 0, that can be in the class? | Answer: 11
## Solution:
Let's look at some person. Suppose he has five friends. Then each of these five people has six friends. Similarly, there are at least another 6 people with five friends each. In total, there are 11 people.
It is quite easy to construct an example: two groups of 5 and 6 people, people from dif... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,836 |
# Problem 2. (2 points)
In a positive non-constant geometric progression, the arithmetic mean of the second, seventh, and ninth terms is equal to some term of this progression. What is the minimum possible number of this term? | Answer: 3
Solution:
The second element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first element is even less suitable.
To prove that the answer "3" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we need to solv... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,837 |
# Problem 3. (2 points)
Can the number $n^{n^{n}}-4 n^{n}+3$ be prime for a natural number $n>2$? | Answer: No.
Solution:
It can be noticed that the given number is always divisible by $n-1$. This is easily proven using the formula for the difference of powers or by taking advantage of the fact that $n \equiv 1(\bmod n-1)$.
Moreover, the quotient is also greater than one for $n>2$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,838 |
# Problem 4. (2 points)
How many negative numbers are there among the numbers of the form $\operatorname{tg}\left(\left(15^{n}\right)^{\circ}\right)$, where $\mathrm{n}$ is a natural number from 1 to 2019? | Answer: 1009
Solution:
$\operatorname{tg} 15^{\circ}>0$.
$15^{2}=225 ; \operatorname{tg} 225^{\circ}>0$.
Further, $225 \cdot 15=3375$, this number gives a remainder of 135 when divided by $360 . \operatorname{tg} 135^{\circ}<0$.
$135 \cdot 15=2025$, this number gives a remainder of 225 when divided by 360. The sequ... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,839 |
# Problem 5. (3 points)
The height of the rhombus, drawn from the vertex of its obtuse angle, divides the side of the rhombus in the ratio $1: 3$, counting from the vertex of its acute angle. What fraction of the area of the rhombus is the area of the circle inscribed in it? | Answer: $\frac{\pi \sqrt{15}}{16}$
Solution:
The problem is solved up to similarity, so we can assume that the side of the rhombus is 4. Then the height divides it into segments of 1 and 3.
The height of the rhombus can be found using the Pythagorean theorem: $\sqrt{4^{2}-1^{2}}=\sqrt{15}$. This height is also the d... | \frac{\pi\sqrt{15}}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,840 |
# Problem 7. (4 points)
Can the numbers from 0 to 9999 (each used exactly once) be arranged in a $100 \times 100$ square table so that the sum of the numbers in each $2 \times 2$ square is the same? | Answer: Yes, it can.
Solution:
The desired table is obtained as the sum of two tables.
The first table looks like this:
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 98 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 9... | 19998 | Combinatorics | proof | Yes | Yes | olympiads | false | 20,841 |
# Problem 1. (2 points)
In a class, each student has either 5 or 7 friends (friendship is mutual), and any two friends have a different number of friends in the class. What is the smallest number of students, greater than 0, that can be in the class?
Answer: 12 | Solution:
Let's look at some person. Suppose he has five friends. Then each of these five people has seven friends. Similarly, there are at least seven more people with five friends each. In total, there are 12 people.
It is quite easy to construct an example: two groups of 5 and 7 people, people from different group... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,842 |
# Problem 2. (2 points)
In a positive non-constant geometric progression, the arithmetic mean of the third, fourth, and eighth terms is equal to some term of this progression. What is the minimum possible number of this term? | Answer: 4
Solution:
The third element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first and second elements are even less suitable.
To prove that the answer "4" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,843 |
# Problem 3. (2 points)
Can the number $n^{n^{n}}-6 n^{n}+5$ be prime for a natural number $n>2$? | Answer: No.
## Solution:
We can observe that the given number is always divisible by $n-1$. This can be easily proven using the formula for the difference of powers or by noting that $n \equiv 1(\bmod n-1)$.
Moreover, the quotient is also greater than one for $n>2$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,844 |
# Problem 4. (2 points)
How many positive numbers are there among the numbers of the form $\operatorname{ctg}\left(\left(15^{n}\right)^{\circ}\right)$, where $\mathrm{n}$ is a natural number from 1 to 2019? | Answer: 1010
Solution:
$\operatorname{ctg} 15^{\circ}>0$.
$15^{2}=225 ; \operatorname{ctg} 225^{\circ}>0$.
Next, $225 \cdot 15=3375$, this number gives a remainder of 135 when divided by $360 . \operatorname{ctg} 135^{\circ}<0$.
$135 \cdot 15=2025$, this number gives a remainder of 225 when divided by 360. The seq... | 1010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,845 |
# Problem 5. (3 points)
The height of a rhombus, drawn from the vertex of its obtuse angle, divides the side of the rhombus in the ratio $1: 3$, counting from the vertex of its acute angle. What part of the area of the rhombus is the area of the circle inscribed in it?
Answer: $\frac{\pi \sqrt{15}}{16}$ | Solution:
The problem is solved up to similarity, so we can assume that the side of the rhombus is 4. Then the height divides it into segments of 1 and 3.
The height of the rhombus can be found using the Pythagorean theorem: $\sqrt{4^{2}-1^{2}}=\sqrt{15}$. This height is also the diameter of the circle.
From this, w... | \frac{\pi\sqrt{15}}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,846 |
# Problem 7. (4 points)
Can the numbers from 0 to 999 (each used exactly once) be arranged in a $100 \times 10$ rectangular table so that the sum of the numbers in each $2 \times 2$ square is the same? | Answer: Yes, it can.
Solution:
The desired table is obtained as the sum of two tables.
The first table looks like this:
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 98 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 9... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,847 |
3. $\mathrm{ABCD}$ is a trapezoid with bases $\mathrm{AD}=6$ and $\mathrm{BC}=10$. It turns out that the midpoints of all four sides of the trapezoid lie on the same circle. Find its radius.
If there are multiple correct answers, list them in any order separated by a semicolon. | Answer: 4.
## Examples of answer notation:
45
$4 ; 5$
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,850 |
3. Given a regular tetrahedron with edge lengths that are integers. Two of these edges have lengths 9 and 11. What is the greatest possible value of the perimeter of the tetrahedron? | Answer: 68.
## Examples of answer recording: 45
# | 68 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,851 |
3. Find all positive solutions of the system of equations.
$$
\left\{\begin{array}{c}
x_{1}+x_{2}=4 x_{3}^{2} \\
x_{2}+x_{3}=4 x_{4}^{2} \\
\cdots \\
x_{2015}+x_{2016}=4 x_{2017}^{2} \\
x_{2016}+x_{2017}=4 x_{1}^{2} \\
x_{2017}+x_{1}=4 x_{2}^{2}
\end{array}\right.
$$
In your answer, specify the value of \( x_{1} \). ... | Answer: $0.5\|0.5\| 1 / 2$
## Examples of answer notation: 45 4.5 $4 / 5$ | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,854 |
3. Given a cubic polynomial $p(x)$. It is known that $p(-6)=30, p(-3)=45, p(-1)=15, p(2)=30$.
Find the area of the figure bounded by the lines $x=-6, y=0, x=2$ and the graph of the given polynomial, if it is also known that on the interval from -6 to 2 the given polynomial takes only positive values. | Answer: 240.
## Examples of answer recording:
45 | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,855 |
2. Solve the inequality:
$$
\frac{1}{\sqrt{n}+\sqrt{n+1}}+\frac{1}{\sqrt{n+1}+\sqrt{n+2}}+\ldots+\frac{1}{\sqrt{n+119}+\sqrt{n+120}}>4
$$ | Write the answer in the form of an interval. For example, the interval (-1;2] means that $1<x \leq 2$. If the boundary of the interval is "infinity," use the letter B.
Answer: $[0 ; 169)$ | [0;169) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 20,856 |
3. In triangle $\mathrm{ABC}$, angle $\mathrm{C}$ is twice as small as angle $\mathrm{B}$. $\mathrm{BB}_{1}$ is the bisector of angle $\mathrm{B}, \mathrm{D}$ is the point of intersection of the circumcircle of triangle $\mathrm{ABB}_{1}$ and side $\mathrm{AB} . \mathrm{BD}=14$, $\mathrm{CD}=18$. Find the length of $\m... | Answer: 30.
## Examples of answer notation:
45
# | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,858 |
3. On the planet Sharp Teeth, animals reproduce in a special way, specifically: every hamster gives birth to four hamsters every 4 months; every groundhog gives birth to one small groundhog every 4 months; and rabbits, mysterious creatures, reproduce faster the more time passes, specifically, if a person does not give ... | Answer: $13,41,130 \mid 13 ; 41 ; 130$
## Examples of answer notation:
## 9
$9 ; 23$
# | 13;41;130 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,864 |
3. Three consecutive terms of a geometric progression with a common ratio $q$ were used as coefficients of a quadratic trinomial, with the middle term being the leading coefficient. For what largest integer $q$ will the resulting trinomial have two distinct roots regardless of how the other two coefficients are arrange... | Answer: -1
## Examples of answer notation:
2
5;9
# | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,866 |
3. The sequence $c_{n}$ is the sum of a geometric progression $b_{n}$ with the first term 3 and some arithmetic progression. It is known that $c_{1}+c_{3}=141, c_{2}=33$. Find the common ratio of the progression $b_{n}$. If there are multiple possible answers, write them in any order separated by a semicolon. | Answer: $-4 ; 6|6 ;-4| 6,-4$
## Examples of answer notation: $1 / 4$ 0,$25 ;-10$ | 6;-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,870 |
3. Given a point on the plane that does not coincide with the origin. How many different points can be obtained from it by sequentially applying symmetries relative to the $O y$ axis and the line $y=-x$ (in any order and any number of times)? If the point itself can also be obtained, it should be counted. If different ... | Answer: $4 ; 8|8,4| 8 ; 4 \mid 4,8$.
## Examples of answer notation:
9
$9 ; 23$ | 4;8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,873 |
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