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3. (7 points) A bus ticket is numbered with six digits: from 000000 to 999999. You buy one ticket. What is the probability that you will get a ticket where the digits are in ascending (descending) order? | Solution: There are a total of $10^{6}$ different tickets, of which $C_{10}^{6}$ have digits in ascending (or descending) order (any 6 digits out of 10 form exactly one number we need). Then the probability is $\frac{C_{10}^{6}}{10^{6}}=0.00021$.
(7 points) A bus ticket is numbered with six digits: from 000000 to 9999... | 0.151 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,589 |
4. (7 points) Solve the equation
$$
x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=n^{2} \quad n>1
$$ | Solution: Let's complete the square in the expression under the root
$$
x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}}=n^{2}
$$
The root can be removed without the absolute value due to the fact that in the domain of definition (ODZ) $\sqrt{x+\frac{1}{4}}+\frac{1}{2}>0$. Therefore,
$$
x+\sqrt{x+\frac{1}{... | n^2-n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,590 |
5. (8 points) Through the sides of a regular $2 n$-gon, lines are drawn. Into how many parts do these lines divide the plane | Solution: We will make two remarks: 1) no three lines pass through one point; 2) lines will be parallel if and only if they contain opposite sides of a regular $2 n$-gon.
Lemma: Suppose several lines are drawn on a plane, and no three of them intersect at one point. Draw another line that intersects $m$ already drawn ... | 2n^ | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,591 |
6. (8 points) At Innopolis University, $2 n(n>1)$ students are enrolled. It turned out that any $2 n-2$ of them can be divided into $n-1$ pairs of acquaintances. What is the smallest number of pairs of acquaintances that can exist among all students. | Solution: Let there be a student (let's call him Bob) who has no more than two acquaintances. Then consider a group consisting of Bob and another $2 n-3$ students who are not acquainted with Bob. We have found a group that contradicts the problem's condition (no one can form a pair with Bob). Therefore, we have proven ... | 3n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,592 |
7. (10 points) On the hypotenuse $A B$ of a right triangle $A B C$, points $X$ and $Y$ are taken such that $A X=A C$ and $B Y=B C$. It turns out that $X Y=p$. Find the product $A Y \cdot B X$.
# | # Solution:
By the triangle inequality $A C+C B>A B$, so the placement of points $X$ and $Y$ on the hypotenuse will be exactly as shown (see figure).
Let $A Y=a$ and $B X=b$, then $A C=a+p$... | \frac{p^2}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,593 |
8. (10 points) Find the coefficient of $x^{m}$ in the polynomial
$$
P(x)=(1+x) \cdot\left(2+x^{2}\right) \cdot\left(4+x^{4}\right) \cdot\left(8+x^{8}\right) \cdot \ldots \cdot\left(1024+x^{1024}\right)
$$ | Solution: Notice that in all parentheses, the variables $x$ enter in powers of $2^{k}$, where $k \in\{0,1, \ldots, 10\}$. Therefore, to form a term $x^{m}$, we need to take $x^{2^{k}}$ in all parentheses, where $k$ is a non-zero bit in the binary representation of $m$, and the number $2^{k}$ if $k$ is a zero bit.
In t... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,594 |
9. (20 points) On the sides of a triangle, isosceles triangles are constructed outward with angles at the vertices $\alpha, \beta$ and $\gamma$, where $\alpha+\beta+\gamma=360^{\circ}$. Find the angles of the triangle formed by the vertices of these isosceles triangles.
# | # Solution:
Since the sum of the interior angles of a hexagon is $720^{\circ}$, then
$$
\begin{gathered}
\angle Z A Y + \angle Y C X + \angle X B Z = 720^{\circ} - (\angle A Y C + \angle C... | \frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,595 |
10. (20 points) Find the minimum value of the function
$$
f(x)=\sqrt{(x-a)^{2}+b^{2}}+\sqrt{(x-b)^{2}+a^{2}} \quad a, b>0
$$ | Solution:
We will solve this using a geometric method. Consider three points \( A(a, b), B(b, -a) \), and \( C(x, 0) \). The distance from point \( C \) to point \( A \) is \( \sqrt{(x-a)^{2... | \sqrt{2^{2}+2b^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,596 |
5. (8 points) Given an isosceles right triangle with a leg of 10. An infinite number of equilateral triangles are inscribed in it as shown in the figure: the vertices lie on the hypotenuse, and the bases are sequentially laid out on one of the legs starting from the right angle vertex. Find the sum of the areas of the ... | Answer: 25 .
 | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,597 |
9. (20 points) Inside an acute triangle $A B C$, a point $M$ is marked. The lines $A M, B M$, $C M$ intersect the sides of the triangle at points $A_{1}, B_{1}$ and $C_{1}$ respectively. It is known that $M A_{1}=M B_{1}=M C_{1}=3$ and $A M+B M+C M=43$. Find $A M \cdot B M \cdot C M$. | Answer: 441.
Solution. Let $A M=x, B M=y, C M=z$. Note that $\frac{M C_{1}}{C C_{1}}=\frac{S_{A M B}}{S_{A B C}}$ and similarly for the other two segments. From the equality $\frac{S_{A M B}}{S_{A B C}}+\frac{S_{B M C}}{S_{A B C}}+\frac{S_{A M C}}{S_{A B C}}=1$ it follows that $\frac{3}{x+3}+\frac{3}{y+3}+\frac{3}{z+3... | 441 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,599 |
10. (20 points) Find all values of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2|x+7|+|y-4|=c \\
|x+4|+2|y-7|=c
\end{array}\right.
$$ | Answer: $c=3$.
Solution. Let $\left(x_{0} ; y_{0}\right)$ be the unique solution of the system. Then $\left(-y_{0} ;-x_{0}\right)$ also satisfies the conditions of the system. This solution coincides with the first, so $y_{0}=-x_{0}$. The equation $2|x+7|+|x+4|=$ $c$ has a unique solution $x_{0}=-7$ when $c=|7-4|$, si... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,600 |
1. Prove that from any infinite arithmetic progression of integers in both directions, an infinite geometric progression of integers can be extracted.
$\left(\right.$ Folklore $^{(2)}$ | Solution: Let the arithmetic progression have the form $c_{k}=a+b k$, where $a, b \in \mathbb{Z}$. Then consider the geometric progression $d_{k}=a(b+1)^{k}$ and prove that all its terms belong to our arithmetic progression
$$
\begin{aligned}
& d_{k}=a \cdot\left(1+C_{k}^{1} \cdot b+C_{k}^{2} \cdot b^{2}+\ldots+C_{k}^... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,601 |
2. Let $f(x)=x^{2}+6 x+6$. Solve the equation in real numbers
$$
\underbrace{f(f(\ldots f}_{2017}(x) \ldots))=2017
$$
(P. Alishev) | Answer: $x=-3 \pm \sqrt[2^{2017}]{2020}$.
Solution: Note that $f(x)=(x+3)^{2}-3$ or $f(x)+3=$ $=(x+3)^{2}$. Then
$$
\begin{gathered}
2020=\underbrace{f(f(\ldots f}_{2017}(x) \ldots))+3=(\underbrace{f(f(\ldots f}_{2016}(x) \ldots))+3)^{2}= \\
=(\underbrace{f(f(\ldots f}_{2015}(x) \ldots))+3)^{2^{2}}=\ldots=(f(x)+3)^{2... | x_{1,2}=-3\\sqrt[2^{2017}]{2020} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,602 |
3. What is the minimum number of cells on a $3 \times 2016$ board that can be painted so that each cell has a side-adjacent painted cell?
(A. Khryabrov) | Answer: 2016.
Solution: Let's divide our board as follows:
We have obtained two three-cell corners and 1007 D-hexomino figures. In our three-cell corners, at least one cell must be shaded ... | 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,603 |
4. Find all values of the parameter $a$ for which the equation
$$
\left(\sqrt{6 x-x^{2}-4}+a-2\right)((a-2) x-3 a+4)=0
$$
has exactly two distinct real roots.
(P. Alishev) | Answer: $a \in\{2-\sqrt{5}\} \cup\{0\} \cup\{1\} \cup\left(2-\frac{2}{\sqrt{5}} ; 2\right]$
Solution: We will solve the problem graphically. Replace $a$ with $y$ and draw the set of solutions of the equation in the $Oxy$ plane.
$ and $P_{\omega}(B)$ respectively, then
$$
P_{\omega}(M)=\frac{2 P_{\omega}(A)+2 P_{\omega}(B)-A B^{2}}{4}
$$
where $M$ is the midpoint of segment $A B$.
Proof: Let $\omega$ be a circle with center at point $O$ and radiu... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,605 |
2. (5 points) Find the remainder when $20^{16}+201^{6}$ is divided by 9. | Answer: 7
Solution. $201^{16}$ is divisible by 9. 20 gives a remainder of $2.2^{6}$ gives a remainder of $1,2^{16}=2^{6} \cdot 2^{6} \cdot 2^{4}$ gives the same remainder as 16. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,606 |
3. (7 points) Calculate the sum
$$
1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+2015 \cdot 2016
$$
Answer: 2731179360 | Solution. Notice that $1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n \cdot(n+1)=\frac{n \cdot(n+1) \cdot(n+2)}{3}$. | 2731179360 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,607 |
4. (10 points) If the sum of the digits of a natural number $n$ is subtracted from $n$, the result is 2016. Find the sum of all such natural numbers $n$.
Answer: 20245 | Solution. The sum of the digits decreases quickly, so the number $n$ can only be a four-digit number. The sum of the digits of a four-digit number does not exceed 36, therefore, $n=S(n)+2016 \leq 2052$, i.e., the first two digits are 2 and 0. Then $\overline{20 x y}=$ $2+x+y+2016 ; 9 x=18$, and $y-$ is any. Adding all ... | 20245 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,608 |
5. (10 points) A hare jumps in one direction along a strip divided into cells. In one jump, it can move either one or two cells. In how many ways can the hare reach the 12th cell from the 1st cell?
 On the coordinate plane, all points whose coordinates satisfy the condition
$$
|2 x-2|+|3 y-3| \leq 30
$$
are shaded. Find the area of the resulting figure. | Answer: 300
 | 300 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,610 |
9. (15 points) Find all values of the parameter $a$ for which the equation
$$
3 x^{2}-4(3 a-2) x+a^{2}+2 a=0
$$
has roots $x_{1}$ and $x_{2}$, satisfying the condition $x_{1}<a<x_{2}$. | Answer: $(-\infty ; 0) \cup(1,25 ;+\infty)$
Solution. $y(x)=3 x^{2}-4(3 a-2) x+a^{2}+2 a-$ is a parabola with branches upwards, so the solution to the inequality $y<0$ is the interval $\left(x_{1} ; x_{2}\right)$. Then
$$
a \in\left(x_{1} ; x_{2}\right) \Longleftrightarrow y(a)<0
$$
Solving the inequality, we get th... | (-\infty;0)\cup(1,25;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,611 |
10. (15 points) Provide an example of a non-zero polynomial with integer coefficients, one of whose roots is the number $\cos 18^{\circ}$. | Answer: $16 x^{4}-20 x^{2}+5$
Solution I. In an isosceles triangle with the angle at the vertex $36^{\circ}$ and the lateral side 1, the height $h$ is equal to $\cos 18^{\circ}$. Let the base be $x$, and the bisector of the angle at the base cuts off a similar triangle from this triangle with the ratio of correspondin... | 16x^{4}-20x^{2}+5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,612 |
4. A function $f$ is called periodic if it takes at least two different values, and there exists $p>0$ such that $f(x+p)=f(x)$ for any $x$. Each such number $p$ is called a period of the function $f$.
Do there exist periodic functions $g$ and $h$ with periods 6 and $2 \pi$ respectively, such that $g-h$ is also a perio... | 4. A function $f$ is called periodic if it takes at least two different values and there exists $p>0$ such that $f(x+p)=f(x)$ for any $x$. Each such number $p$ is called a period of the function $f$.
Do there exist periodic functions $g$ and $h$ with periods 6 and $2 \pi$ respectively, such that $g-h$ is also a period... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,616 |
1. For a natural number $n$, the smallest divisor $a$, different from 1, and the next largest divisor $b$ were taken. It turned out that $n=a^{a}+b^{b}$. Find $n$. | Answer: $n=2^{2}+4^{4}=260$.
Solution. If $n$ is odd, then all its divisors are also odd. Then $a$ and $b$ are odd and $a^{a}+b^{b}$ is even. Therefore, $n$ is even. Then its smallest divisor, different from 1, is 2, and thus $a=2$. Therefore, $n=2^{2}+b^{b}$. Consequently, $4=n-b^{b}$ is divisible by $b$. Then $b=4$,... | 260 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,617 |
2. Given natural numbers $m$ and $n$. On the table, there are two piles: the first pile has $n$ stones, and the second pile has $m$ stones. Petya and Vasya play the following game. Petya starts. In one move, one can split one of the existing piles on the table into several smaller piles. The player who cannot make a mo... | Answer: $m \neq n$.
Solution. If $m=n$, Vasya can repeat Petya's moves and, therefore, will always be able to make a move. Let $m>n$. Then Petya will break the pile with $m$ stones into a pile with $n$ stones and $m-n$ piles of one stone each. Nothing more can be done with the piles of one stone, so the game will cont... | \neqn | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,618 |
3. On the line $\ell$, points $A, B$, and $C$ are marked. Point $B$ lies between points $A$ and $C$ and $A B < B C$. Two circles $\omega_{1}$ and $\omega_{2}$, whose radii are greater than $A B$ but less than $B C$, lie on opposite sides of the line $\ell$ and touch it at point $B$. Let $K$ be the point of intersection... | # Solution.
Let the tangents from point $A$ to circles $\omega_{1}$ and $\omega_{2}$ touch them at points $A_{1}$ and $A_{2}$, respectively, and the tangents from point $C$ to circles $\omega... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,619 |
4. Find all values of parameters $a$ and $b$, for which the inequality
$$
\left|x^{2}+a x+b\right| \leqslant \frac{1}{8}
$$
holds for all values $x \in [0 ; 1]$.
(P. Alishev) | Answer: $a=-1, b=\frac{1}{8}$.
Solution. The maximum value of the function $f(x)=\left|x^{2}+a x+b\right|$ on the interval $[0 ; 1]$ is achieved at critical points within the interval or at the endpoints. $f^{\prime}(x)=\frac{x^{2}+a x+b}{\left|x^{2}+a x+b\right|} \cdot(2 x+a)$, the critical points are the roots of th... | =-1,b=\frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,620 |
5. On the plane, $4 n+2$ points are marked, no three of which lie on the same line. Half of the points are painted red, and half are painted green. Prove that there exists a line passing through one red and one green point, on one side of which there are exactly $n$ red and exactly $n$ green points.
(a discrete versio... | Solution. Consider an arbitrary line $\ell$ that is not parallel to any of the lines connecting the red points. Draw a line parallel to it such that all points lie on one side of this line. We will move this line parallelly towards the points.
Each time it passes through one red point, the number of points on one side... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,621 |
1. A non-constant function $f(x)$ for all real values of $x$ satisfies the equation
$$
f(x+1) + f(x-1) = \sqrt{3} f(x)
$$
Prove that $f(x)$ is periodic and provide an example of such a function. | Solution. $3 f(x)=\sqrt{3} f(x+1)+\sqrt{3} f(x-1)=f(x+2)+f(x)+f(x)+f(x-2)$. From this, $f(x)=f(x+2)+f(x-2)=f(x+4)+f(x)+f(x-2)$, i.e., $f(x+4)=-f(x-2)$ for all $x \in \mathbb{R}$. Then $f(x)=-f(x+6)=f(x+12)$. The function $f(x)=\cos \frac{\pi x}{6}+\sin \frac{\pi x}{6}$ satisfies the condition of the problem. | proof | Algebra | proof | Yes | Yes | olympiads | false | 19,622 |
2. Two sides of the quadrilateral $A B C D$ are parallel. Let $M$ and $N$ be the midpoints of sides $B C$ and $C D$ respectively, and $P$ be the intersection point of $A N$ and $D M$. Prove that if $A P=4 P N$, then $A B C D$ is a parallelogram. | Solution. 1) Let $A B \| C D$. Draw the midline $M K$, let $L-$ be the intersection point of $M K$ and $A N$. Let $P N=x$, then by the condition $A N=5 \cdot P N=5 x$. Since $K L-$ is the midline in $\triangle A D N$, then $L N=2.5 x$. Therefore, $L P=1.5 x$. Triangles $D P N$ and $L P M$ are obviously similar with a s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,623 |
3. It is known that the polynomial $P(x)=x^{3}+a x^{2}+b x-1$ has three distinct positive roots. Prove that $P(-1)<-8$.
---
The provided text has been translated into English while preserving the original formatting and line breaks. | Solution. By Vieta's theorem $x_{1}+x_{2}+x_{3}=-a, x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}=b$ and $x_{1} x_{2} x_{3}=1$. $P(-1)=-1+a-b-1=-2-\left(x_{1}+x_{2}+x_{3}\right)-\left(x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}\right)=-2-\left(x_{1}+x_{2}+\right.$ $\left.x_{3}+\frac{1}{x_{3}}+\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)=-2-\l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 19,624 |
4. On a sphere of radius 1, there are $n$ points. Prove that the sum of the squares of the pairwise distances between them is not greater than $n^{2}$.
| Solution. Let $A_{1}, \ldots, A_{n}$ be the given points, $O-$ the center of the sphere. Denote $\overrightarrow{x_{i}}=\overrightarrow{O A_{i}}$. Then the sum of the squares of the distances between the points $S=\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n}\left(\overrightarrow{x_{i}}-\overrightarrow{x_{j}}\right)^{2}=\f... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,625 |
5. Find all values of the parameter $a$ for which the equation
$$
3|x+3 a|+\left|x+a^{2}\right|+2 x=a
$$
has no solution. | Answer: $(-\infty ; 0) \cup(10 ;+\infty)$
Solution. Method I. Consider the function $f(x)=3|x+3 a|+\left|x+a^{2}\right|+2 x-a$. $f^{\prime}(x)=3 \cdot \frac{x+3 a}{|x+3 a|}+\frac{x+a^{2}}{\left|x+a^{2}\right|}+2, x_{1}=-3 a$ and $x_{2}=-a^{2}$ are critical points. When $x<x_{1} x<x_{2} f^{\prime}(x)=-4 \Longrightarrow... | (-\infty;0)\cup(10;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,626 |
3. (7 points) In the first stage of the mathematics Olympiad, $a \%$ of the participants were girls, and among the participants of the informatics Olympiad, $b \%$ were girls. It is also known that no one participated in both subjects, and among all the participants of the first stage, $c \%$ were girls. Find the ratio... | Solution: Let $x$ be the total number of participants in the mathematics Olympiad, and $y$ be the total number of participants in the informatics Olympiad. Then
$$
c=\frac{a x+b y}{x+y} \Longrightarrow \frac{y}{x}=\frac{a-c}{c-b}
$$
The desired ratio is
$$
\frac{b y}{a x}=\frac{b(a-c)}{a(c-b)}
$$
Answers:
| Option... | \frac{b(-)}{(-b)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,629 |
4. (7 points) Given a string of four numbers $a, b, c, d$, we will perform the following operation: between every two adjacent numbers, we will insert a number that is the result of subtracting the left number from the right number. We will perform the same operation on the new string and so on. Find the sum of the num... | Solution: Let's see how the sum of a row changes after the next operation. Let $a_{1}, a_{2}, \ldots, a_{n}$ be the row to which the operation is applied. Then the new row has the form $a_{1}, a_{2}-a_{1}, a_{2}, a_{3}-a_{2}, \ldots, a_{n-1}, a_{n}-a_{n-1}, a_{n}$. The sum of the numbers in the new row is
$$
\begin{ga... | +b+++\cdot(-) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,630 |
5. (8 points) Solve the equation $x^{4}-(a+b) \cdot x^{3}+(a b-2 c) \cdot x^{2}+c(a+b) \cdot x+c^{2}=0$. In your answer, write the sum of the squares of the real roots of the equation. | Solution: $x=0$ is not a solution to the equation, so we can divide by $x^{2}$. Grouping terms, we get
$$
\begin{gathered}
\left(x^{2}-2 c+\frac{c^{2}}{x^{2}}\right)-(a+b) \cdot\left(x-\frac{c}{x}\right)+a b=0 \\
\left(x-\frac{c}{x}\right)^{2}-(a+b) \cdot\left(x-\frac{c}{x}\right)+a b=0 \\
\left(x-\frac{c}{x}-a\right)... | ^{2}+b^{2}+4c | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,631 |
6. (8 points) On the circumscribed circle of isosceles triangle $A B C(A C=$ $B C$ ), a point $P$ is taken (point $P$ lies on the arc $A B$ not containing point $C$). Point $D$ is the foot of the perpendicular dropped from point $C$ to the line $B P$. Find the length of the segment $P D$, if $A P=a$ and $B P=b$.
# | # Solution:
In all cases, the parameters $a$ and $b$ were given such that $\angle C A C^{\prime}=90^{\circ}$, and therefore the height $C D^{\prime}$ falls on the extension of $A P$ beyond p... | \frac{+b}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,632 |
7. (10 points) A cubic polynomial $P(x)$ with real coefficients is divisible by two quadratic trinomials
$$
Q(x)=x^{2}+(k-a) x-k \quad \text { and } \quad R(x)=2 x^{2}+(2 k-(2 a-15)) x+k
$$
without a remainder. Find the largest value of $k$ for which this is possible. | Solution: Note that $R(x)=2 Q(x)+15 x+3 k$. Therefore, the polynomials cannot differ from each other by some real multiplier $c$. Thus, we obtain that our cubic polynomial $P(x)$ has three real roots $n, m$, and $l$. Without loss of generality, we can assume that $Q(x)=(x-n) \cdot(x-m)$ and $R(x)=2 \cdot(x-m) \cdot(x-l... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,633 | |
8. (10 points) Real numbers $a, b, x$ and $y$ are such that
$$
\left\{\begin{array}{l}
a x+b y=r-t \\
a x^{2}+b y^{2}=r \\
a x^{3}+b y^{3}=r+t \\
a x^{4}+b y^{4}=r+t^{2}
\end{array}\right.
$$
Find $a x^{2}+b y^{2}$. | Solution: Notice that
$$
\begin{gathered}
a x^{3}+b y^{3}=\left(a x^{2}+b y^{2}\right) \cdot(x+y)-x y \cdot(a x+b y) \\
a x^{4}+b y^{4}=\left(a x^{3}+b y^{3}\right) \cdot(x+y)-x y \cdot\left(a x^{2}+b y^{2}\right)
\end{gathered}
$$
We obtain a system of two equations:
$$
\left\{\begin{array}{l}
r \cdot(x+y)-(r-t) \c... | r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,634 |
9. (20 points) A four-digit number $\overline{a b c d}$ is called perfect if $a+b=c+d$. How many perfect numbers can be represented as the sum of two four-digit palindromes? | Solution: Let the number $\overline{a b c d}=\overline{n m m n}+\overline{x y y x}$, then
$$
\overline{a b c d}=1001(n+x)+110(m+y) \vdots 11
$$
From the divisibility rule by 11, it follows that $b+d=a+c$. Since the number $\overline{a b c d}$ is perfect, we get $a=d$ and $b=c$, hence the original number is a palindro... | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,635 |
10. (20 points) The diagonals $A C$ and $B D$ of a convex quadrilateral $A B C D$ intersect at point $P$. Find $\angle A P B$, if $\angle B A C=2 \alpha, \angle B C A=2 \beta, \angle C A D=90^{\circ}-\alpha$ and $\angle A C D=90^{\circ}-\beta$. | # Solution:
It is clear that
$$
\angle L C D=180^{\circ}-\left(90^{\circ}-\beta\right)-2 \beta=90^{\circ}-\beta=\angle A C D
$$
and
$$
\angle K A D=180^{\circ}-\left(90^{\circ}-\alpha\ri... | 90-\alpha+\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,636 |
9. (20 points) Inside triangle $A B C$, a point $P$ is marked such that $\angle P A C = \angle P B C$. Points $M$ and $N$ are the projections of point $P$ onto sides $A C$ and $B C$ respectively, and $D$ is the midpoint of $A B$. Prove that $D M = D N$. | Solution. It is clear that points $C, M, P, N$ lie on the circle $\omega$ with diameter $CP$. Let $B_{1}$ be the second intersection point of line $BP$ with $\omega$. We will prove that $B_{1}, M$, and $D$ lie on the same line. Let $K$ be the intersection point of lines $BP$ and $AC$. By Menelaus' theorem for triangle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,637 |
10. (20 points) The plane is divided into regular triangles with side length 1. Consider the set $M$ of all possible distances between the vertices of these triangles, i.e., $x \in M$ if and only if there exist two vertices at a distance $x$ from each other. Prove that if $x \in M$ and $y \in M$, then $x \cdot y \in M$... | Solution. Note that $x \in M$ if and only if there exist integers $a, b$ such that $x^{2}=a^{2}+a b+b^{2}$. Indeed, let $x \in M$. Then there exist nodes $A, B$ of the triangular grid such that $A B=x$. Consider the intersection $O$ of the lines forming the grid, containing points $A, B$, respectively. Let $a=O A$, $b=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,638 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=1000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+4 x+2$ holds. | Answer: 999001
Solution: In the equation $f(x+2)-f(x)=4 x+2$, we will substitute for $x$ the numbers $0,2,4, \ldots, 998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=4 \cdot 0+2 \\
& f(4)-f(2)=4 \cdot 2+2
\end{aligned}
$$
$$
f(1000)-f(998)=4 \cdot 998+2
$$
Adding the equations, we get: $f(1000)-f(0)=4 \cdot(0+2+4+\cdot... | 999001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,639 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 70 negative numbers were recorded. What is the ... | Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=70+62$, which means $x=12$.
2) Let there be $y$ people with "positive tempe... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,640 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 30 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-4.5,4.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 90
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-10 \leqslant y-1 \leqslant 8$ and $-7 \leqslant 2-x \leqslant 11$. Therefore, $(y-1)(2-x)+2 \leqslant 8 \cdot 11+2=90$. The m... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,642 |
5. (8 points) In $\triangle A B C, A B=86$, and $A C=97$. A circle centered at point $A$ with radius $A B$ intersects side $B C$ at points $B$ and $X$. Moreover, $B X$ and $C X$ have integer lengths. What is the length of $B C ?$ | Answer: 61
Solution: Let $x=B X$ and $y=C X$. We will calculate the power of point $C$ in two ways
$$
y(y+x)=97^{2}-86^{2}=2013
$$
Considering all divisors of the number 2013 and taking into account the triangle inequality $\triangle A C X$, we obtain the unique solution 61. | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,643 |
6. (8 points) On the board, 34 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 34 minutes? | Answer: 561.
Solution: Let's represent 34 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line ... | 561 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,644 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=25 \\
z^{2}+x z+x^{2}=37
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 20
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=12$, $B C^{2}=25$, and $A C^{2}=37$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The s... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,646 |
10. (20 points) Let $x, y, z$ be natural numbers satisfying the condition $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Prove that $\text{GCD}(x ; y ; z) \cdot (y-x)$ is a square of a natural number. | Solution: Let $\gcd(x, y, z) = d$ and $x = d x_{1}, y = d y_{1}, z = d z_{1}$, where $\gcd(x_{1}, y_{1}, z_{1}) = 1$. Substituting, we get $\frac{1}{x_{1}} = \frac{1}{y_{1}} + \frac{1}{z_{1}}$ and we need to prove that $d^{4}(y_{1} - x_{1})$ is a square of a natural number, so it is sufficient to prove that $y_{1} - x_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,648 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=2000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+4 x+2$ holds. | Answer: 3998001
Solution: In the equation $f(x+2)-f(x)=4 x+2$, we will substitute for $x$ the numbers $0, 2, 4, \ldots, 1998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=4 \cdot 0+2 \\
& f(4)-f(2)=4 \cdot 2+2
\end{aligned}
$$
$$
f(2000)-f(1998)=4 \cdot 1998+2
$$
Adding the equations, we get: $f(2000)-f(0)=4 \cdot(0+2+4... | 3998001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,649 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 68 positive and 64 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=64+68$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,650 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 28 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-5.5,5.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 132
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-12 \leqslant y-1 \leqslant 10$ and $-9 \leqslant 2-x \leqslant 13$. Therefore, $(y-1)(2-x)+2 \leqslant 10 \cdot 13+2=132$... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,652 |
5. (8 points) In $\triangle A B C, A B=86$, and $A C=97$. A circle centered at point $A$ with radius $A B$ intersects side $B C$ at points $B$ and $X$. Additionally, $B X$ and $C X$ have integer lengths. What is the length of $B C ?$ | Answer: 61
Solution: Let $x=B X$ and $y=C X$. We will calculate the power of point $C$ in two ways
$$
y(y+x)=97^{2}-86^{2}=2013
$$
Considering all divisors of the number 2013 and taking into account the triangle inequality $\triangle A C X$, we obtain the unique solution 61. | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,653 |
6. (8 points) On the board, 33 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 33 minutes? | Answer: 528.
Solution: Let's represent 33 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $xy$ line seg... | 528 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,654 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=28
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 16
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=12$, $B C^{2}=16, A C^{2}=28$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,656 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=3000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+3 x+2$ holds. | Answer: 6748501
Solution: In the equation $f(x+2)-f(x)=3 x+2$, we will substitute for $x$ the numbers $0, 2, 4, \ldots, 2998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=3 \cdot 0+2 \\
& f(4)-f(2)=3 \cdot 2+2
\end{aligned}
$$
$$
f(3000)-f(2998)=3 \cdot 2998+2
$$
Adding the equations, we get: $f(3000)-f(0)=3 \cdot(0+2+4... | 6748501 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,658 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 78 positive and 54 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=54+78$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,659 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 26 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-6.5,6.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 182
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-14 \leqslant y-1 \leqslant 12$ and $-11 \leqslant 2-x \leqslant 15$. Therefore, $(y-1)(2-x)+2 \leqslant 12 \cdot 15+2=182... | 182 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,661 |
6. (8 points) On the board, 32 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 32 minutes? | Answer: 496.
Solution: Let's represent 32 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line ... | 496 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,662 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=21
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 12
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=12$, $B C^{2}=9, A C^{2}=21$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,664 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=4000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+3 x+2$ holds. | Answer: 11998001
Solution: In the equation $f(x+2)-f(x)=3 x+2$, we will substitute for $x$ the numbers $0,2,4, \ldots, 3998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=3 \cdot 0+2 \\
& f(4)-f(2)=3 \cdot 2+2
\end{aligned}
$$
$$
f(4000)-f(3998)=3 \cdot 3998+2
$$
Adding the equations, we get: $f(4000)-f(0)=3 \cdot(0+2+4+... | 11998001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,666 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 92 positive and 40 negative numbers were recorded. What is the ... | Answer: 2
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=40+92$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,667 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 24 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-7.5,7.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 240
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-16 \leqslant y-1 \leqslant 14$ and $-13 \leqslant 2-x \leqslant 17$. Therefore, $(y-1)(2-x)+2 \leqslant 14 \cdot 17+2=240... | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,669 |
6. (8 points) On the board, 31 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could eat in 31 minutes? | Answer: 465.
Solution: Let's represent 31 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $xy$ line s... | 465 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,670 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=27 \\
y^{2}+y z+z^{2}=25 \\
z^{2}+x z+x^{2}=52
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 30
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=27$, $B C^{2}=25, A C^{2}=52$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,672 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=1500$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 750001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute the numbers $0,3,6, \ldots, 1497$ for $x$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(1500)-f(1497)=2 \cdot 1497+3
$$
Adding the equations, we get: $f(1500)-f(0)=2 \cdot(0+3+6+\c... | 750001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,674 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 50 positive and 60 negative numbers were recorded. What is the ... | Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=60+50$, which means $x=11$.
2) Let there be $y$ people with "positive tempe... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,675 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 22 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-8.5,8.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 306
Solution: Notice that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-18 \leqslant y-1 \leqslant 16$ and $-15 \leqslant 2-x \leqslant 19$. Therefore, $(y-1)(2-x)+2 \leqslant 16 \cdot 19+2=3... | 306 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,677 |
6. (8 points) On the board, 30 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 30 minutes? | Answer: 435.
Solution: Let's represent 30 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line se... | 435 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,678 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=27 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=43
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 24
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=27$, $B C^{2}=16, A C^{2}=43$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,680 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=3000$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 3000001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute for $x$ the numbers $0,3,6, \ldots, 2997$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(3000)-f(2997)=2 \cdot 2997+3
$$
Adding the equations, we get: $f(3000)-f(0)=2 \cdot(0+3+6+\... | 3000001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,682 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 54 positive and 56 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=56+54$, which means $x=11$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,683 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 20 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-9.5,9.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 380
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-20 \leqslant y-1 \leqslant 18$ and $-17 \leqslant 2-x \leqslant 21$. Therefore, $(y-1)(2-x)+2 \leqslant 18 \cdot 21+2=380... | 380 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,685 |
6. (8 points) On the board, 29 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 29 minutes? | Answer: 406.
Solution: Let's represent 29 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 406 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,686 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=27 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=36
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 18
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=27$, $B C^{2}=9, A C^{2}=36$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,688 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=4500$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 6750001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute for $x$ the numbers $0,3,6, \ldots, 4497$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(4500)-f(4497)=2 \cdot 4497+3
$$
Adding the equations, we get: $f(4500)-f(0)=2 \cdot(0+3+6+\... | 6750001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,690 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 48 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+62$, which means $x=11$.
2) Let there be $y$ people with "positive t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,691 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 18 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-10.5,10.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 462
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-22 \leqslant y-1 \leqslant 20$ and $-19 \leqslant 2-x \leqslant 23$. Therefore, $(y-1)(2-x)+2 \leqslant 20 \cdot 23+2=462... | 462 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,693 |
6. (8 points) On the board, 28 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 28 minutes? | Answer: 378.
Solution: Let's represent 28 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line se... | 378 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,694 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=48 \\
y^{2}+y z+z^{2}=25 \\
z^{2}+x z+x^{2}=73
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 40
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=48$, $B C^{2}=25, A C^{2}=73$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,696 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=6000$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 12000001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute for $x$ the numbers $0,3,6, \ldots, 5997$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(6000)-f(5997)=2 \cdot 5997+3
$$
Adding the equations, we get: $f(6000)-f(0)=2 \cdot(0+3+6+... | 12000001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,698 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 42 positive and 48 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+42$, which means $x=10$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,699 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 16 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-11.5,11.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 552
Solution: Notice that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-24 \leqslant y-1 \leqslant 22$ and $-21 \leqslant 2-x \leqslant 25$. Therefore, $(y-1)(2-x)+2 \leqslant 22 \cdot 25+2=552$... | 552 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,701 |
6. (8 points) On the board, 27 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 27 minutes? | Answer: 351.
Solution: Let's represent 27 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line se... | 351 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,702 |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=48 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=64
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 32
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=48$, $B C^{2}=16, A C^{2}=64$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,704 |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=2000$, if $f(0)=1$ and for any $x$ the equality $f(x+4)=f(x)+3 x+4$ holds. | Answer: 1499001
Solution: In the equation $f(x+4)-f(x)=3 x+4$, we will substitute for $x$ the numbers $0, 4, 8, \ldots, 1996$. We get:
$$
\begin{aligned}
& f(4)-f(0)=3 \cdot 0+4 \\
& f(8)-f(4)=3 \cdot 4+4
\end{aligned}
$$
$$
f(2000)-f(1996)=3 \cdot 1996+4
$$
Adding the equations, we get: $f(2000)-f(0)=3 \cdot(0+4+8... | 1499001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,706 |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 48 positive and 42 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each of them gave $x-1$ answers, so $x(x-1)=42+48$, which means $x=10$.
2) Let there be $y$ people with "posi... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,707 |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 14 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-12.5,12.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 650
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-26 \leqslant y-1 \leqslant 24$ and $-23 \leqslant 2-x \leqslant 27$. Therefore, $(y-1)(2-x)+2 \leqslant 24 \cdot 27+2=650$. ... | 650 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,709 |
6. (8 points) On the board, 26 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 26 minutes? | Answer: 325.
Solution: Let's represent 26 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 325 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,710 |
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