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1. Find all solutions of the equation
$$
\frac{x-2}{y}+\frac{5}{x y}=\frac{4-y}{x}-\frac{|y-2 x|}{x y}
$$ | Answer: $x=1, y=2$.
Solution. This equation can be transformed into
$$
\begin{gathered}
x^{2}-2 x+5=4 y-y^{2}-|y-2 x|, \quad x y \neq 0 \\
(x-1)^{2}+(y-2)^{2}+|y-2 x|=0
\end{gathered}
$$
Each term on the left is non-negative, so the sum will be zero only if each term is zero: $x-1=0, y-2=0, y-2 x=0, x y \neq 0$. Fro... | 1,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,475 |
2. Chord $CD$ of a circle with center at point $O$ is perpendicular to its diameter $AB$, chord $AE$ intersects radius $OC$ at point $M$, and chord $DE$ - chord $BC$ at point $N$. Prove that line $MN$ is parallel to line $AB$. | Solution. $\angle C A B=\angle C D B \Longrightarrow \triangle A O C \sim \triangle D B C$ (since they are isosceles). $\angle C A E=$ $\angle C D E$ and $\angle A C M=\angle D C N \Longrightarrow \triangle A C M \sim \triangle D C N$. From the similarities, it follows that $\frac{C M}{C N}=$ $\frac{C A}{C D}=\frac{C O... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,476 |
3. For natural numbers $a, b, c$ and $x, y, z$, the equalities $a^{2}+b^{2}=c^{2}$ and $x^{2}+y^{2}=z^{2}$ are satisfied. Prove the inequality
$$
(a+x)^{2}+(b+y)^{2} \leqslant(c+z)^{2}
$$
When is equality achieved | Solution. Let's define vectors $\vec{c}\{a ; b\}$ and $\vec{z}\{x ; y\}$. Then $|\vec{c}+\vec{z}| \leqslant|\vec{c}|+|\vec{z}|=c+z$ by the triangle inequality. In coordinates, this inequality takes the form:
$$
\sqrt{(a+x)^{2}+(b+y)^{2}} \leqslant c+z
$$
Equality is achieved when the vectors $\vec{c}\{a ; b\}$ and $\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,477 |
4. Cut a regular hexagon into 5 pieces and assemble them into a square. | Solution. Segment $A G$ is equal to the side of the square.
 | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,478 |
5. Find all values of the parameter $a$, for each of which the common solutions of the inequalities $x^{2}-4 x+2-a \leqslant 0$ and $x^{2}-5 x+2 a+8 \leqslant 0$ form a segment on the number line with a length of one. | Answer: $a=-1 ; a=-1.75$
Solution. In the plane $O x a$ we construct the sets of solutions of the inequalities (the part of the plane bounded by parabolas) $a \geqslant x^{2}-4 x+2$ and $a \leqslant-0.5 x^{2}+2.5 x-4$. In the common part of the solutions, we find the segments $A B$ and $C D$ of length one, correspondi... | =-1;=-1.75 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,479 |
9. (20 points) $M$ and $N$ are the midpoints of edges $AB$ and $CD$ of the tetrahedron $ABCD$ respectively. A section is drawn through points $M$ and $N$, intersecting edges $AC$ and $BD$ at points $P$ and $Q$ respectively. Prove that
$$
AP: AC = BQ: BD
$$ | Solution. Let the section from the problem condition intersect the line $B C$ at point $K$. From Menelaus' theorem for triangles $A B C$ and $D B C$, we obtain
$$
\begin{aligned}
& \frac{A M}{M B} \cdot \frac{B K}{K C} \cdot \frac{C P}{P A}=1 \\
& \frac{D Q}{Q B} \cdot \frac{B K}{K C} \cdot \frac{C N}{N D}=1
\end{alig... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,482 |
10. (20 points) Find the smallest value of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
8(x+7)^{4}+(y-4)^{4}=c \\
(x+4)^{4}+8(y-7)^{4}=c
\end{array}\right.
$$ | Answer: $c=24$.
Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have
(1) $\left(\frac{1}{2}+1\right)^{3}\left(8(x+\alpha)^{4}+(y-\beta)^{4}\right) \geqslant$
$$
\geqslant\left(\left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right)\right)^{2} \geqslant
$$
$$
\geqslant(|x+\alpha|+|y-\beta|)^{... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,483 |
8. (10 points) On a horizontal plane, 4 spheres of radius $20-10 \sqrt{2}$ are touching each other, and their centers form a square. Above the center of the square, on top of the four spheres, a fifth sphere of the same radius is placed. Find the distance from its highest point to the plane. | Answer: 20
## Solve the following problems with justification of the answer
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,484 | |
8. (10 points) On a horizontal plane, 4 spheres of radius $22-11 \sqrt{2}$ are touching each other, and their centers form a square. Above the center of the square, on top of the four spheres, a fifth sphere of the same radius is placed. Find the distance from its highest point to the plane. | Answer: 22
## Solve the following problems with justification of the answer
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,485 | |
8. (10 points) On a horizontal plane, 4 spheres of radius $24-12 \sqrt{2}$ are touching each other, and their centers form a square. Above the center of the square, on top of the four spheres, a fifth sphere of the same radius is placed. Find the distance from its highest point to the plane. | Answer: 24
## Following tasks should be solved with justification of the answer
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,486 | |
2. Given two polynomials $G(x)=x^{4}+A x^{3}+B x^{2}+C x+D$ and $H(x)=x^{2}+A x+B$. Find the value of $G\left(x_{1}\right)+G\left(x_{2}\right)$, where $x_{1}, x_{2}$ are the roots of the polynomial $H(x)$. | Answer: $2 D-A C$.
Solution: Notice that
$$
G(x)=x^{2} \cdot H(x)+C x+D
$$
therefore
$$
G\left(x_{1}\right)+G\left(x_{2}\right)=\left(x_{1}^{2} \cdot H\left(x_{1}\right)+C x_{1}+D\right)+\left(x_{2}^{2} \cdot H\left(x_{2}\right)+C x_{2}+D\right)=C\left(x_{1}+x_{2}\right)+2 D
$$
It remains to notice that by Vieta's... | 2D-AC | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,487 |
3. (7 points) Find all pairs of natural numbers $x$ and $y$ such that
$$
\log _{2} a x+\log _{2} b y=\log _{2}\left(b x+a y+p_{1} p_{2}-1\right), \text { where } p_{1}, p_{2} \in \mathbb{P} \quad a, b>2
$$
In your answer, write the smallest possible value of $x+y$. | Solution: Let's get rid of the logarithms in the equation
$$
a x b y=b x+a y+p_{1} p_{2}-1
$$
Transform and factorize
$$
(a y-1)(b x-1)=p_{1} p_{2}
$$
Since $p_{1}, p_{2} \in \mathbb{P}$, we get four possible solutions
$$
\left\{\begin{array} { l }
{ x = \frac { p _ { 1 } p _ { 2 } + 1 } { b } } \\
{ y = \frac { ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,488 |
4. (7 points) A fair coin is tossed $n$ times. Find the probability that no two consecutive tosses result in heads? | Solution: Let O denote heads and P denote tails. Let $S_{n}$ be the number of sequences of O and P of length $n$ in which there are no two consecutive O symbols (we will call such sequences correct). Obviously, $S_{1}=2$ and $S_{2}=3$. We will find a recurrence formula for calculating $S_{n}$ for $n>2$.
Consider all p... | \frac{f_{n+2}}{2^n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,489 |
5. (8 points) Find the smallest natural value of $n$ such that for all $x$ the inequality $|x-1|+|x-2|+\ldots+|x-n| \geqslant m$ holds. | Solution: $f(x)=|x-1|+|x-2|+\ldots+|x-n|-$ is a continuous piecewise linear function. We need to find the minimum value of $f(x)$.
If $k+1>x \geqslant k$, then the first $k$ absolute values are opened with a plus sign, and the remaining $n-k$ with a minus sign. Therefore, we get
$$
\begin{gathered}
f(x)=(x-1)+(x-2)+\... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,490 |
6. (8 points) Quadrilateral $ABCD$ is inscribed in a circle of radius $r$ such that diagonal $AC$ is the diameter of the circle. The diagonals of the quadrilateral intersect at point $P$. It is known that $BD = AB$ and $PC = \alpha < r$. Find the length of side $CD$.
# | # Solution:
Let $O$ be the center of the circumscribed circle. Clearly, $A O = D O$ and $A B = D B$, so points $O$ and $B$ lie on the perpendicular bisector of segment $A D$. Since $A C$ is ... | \frac{r\alpha}{r-\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,491 |
7. (10 points) A group of $m$ students is called unfriendly if any two of its members do not friend each other. A student is called super-social if there exist at least $n$ other students with whom they are friends. What is the maximum number of students that can study at Innopolis University, given that there is no un... | Solution: We will construct a graph where the vertices are students, and edges connect only those students who are friends. We will prove by induction (on the size of the unfriendly group) that in our graph, there are no more than \( n(m-1) \) students.
Base case \( m=2 \): In our graph, there should be an edge betwee... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,492 |
8. (10 points) Find the sum of the real roots of the equation
$$
2 \cdot 3^{3 x}-a \cdot 3^{2 x}-3(a+4) \cdot 3^{x}+18=0
$$ | Solution: Let's make the substitution $t=3^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation
$$
2 t^{3}-a t^{2}-3(a+4) t+18=0
$$
It is obvious that the number $t_{1}=-3$ is a root of the equation and the corresponding $x_{1}$ is not real. We get
$$
(t+3)\left(2 t^{2}-(a+6) t+6\right)=0... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,493 |
9. (20 points) In triangle $A B C$, an inscribed circle with center at point $I$ touches sides $A C$ and $B C$ at points $B_{1}$ and $A_{1}$, respectively. Let $O$ be the center of the circumcircle of triangle $A I B$. Prove that $\angle O B_{1} A=\angle O A_{1} B$.
# | # Solution:
Consider the point $O^{\prime}$ - the midpoint of the arc $A B$ (the arc does not contain point $C$) of the circumcircle of triangle $A B C$. By the trident lemma, we get $O^{\pr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,494 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2537^{\circ}+\sin 2538^{\circ}+\cdots+\sin 6137^{\circ}\right)^{\cos } 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6120^{\circ}\right)
$$ | Answer: $73^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6119^{\circ}=0$ and in the exponent only $\cos ... | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,496 |
2. (5 points) Two different natural numbers end with 5 zeros and have exactly 42 divisors. Find their sum. | Answer: 700000
Solution: Since the number ends with 5 zeros, it has the form $N=10^{5} k$. The smallest number of this form $10^{5}$ has 36 divisors: all divisors have the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 5. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has othe... | 700000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,497 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $15$, and $\sin \alpha = \frac{\sqrt{21}}{5}$? | Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,498 |
4. (7 points) On the board, 40 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 40 minutes? | Answer: 780.
Solution: Let's represent 40 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... | 780 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,499 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=53$.
 | Answer: 2862
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=53(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{53 S}{k}+\frac{53 S}{l}$... | 2862 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,500 |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=91
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 40
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=16, A C^{2}=91$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,501 |
8. (10 points) In the expression $(x+y+z)^{2018}+(x-y-z)^{2018}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1020100
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2018}+(x-t)^{2018}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2018}=x^{2018}+a_{1} x^{2017} t+\ldots+a_{2017} x t^{2017}+t^{2018} \\
& (x-t)^{2018}=x^{2018}-a_{1} x^{2017} t+\ldots-a_{2... | 1020100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,503 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-5)^{2}+(y-6)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=5 \frac{1}{3}, y=5 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-5)^{2}+(6-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-5|+|6-y|+|x-y|}{3} \geqslant \frac{x-5+6-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-5)^{2}+(6-y)^{2}+(x-y)^{2} \geqslant \frac... | 5\frac{1}{3},5\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,504 |
10. (20 points) Let $x, y, z$ be natural numbers satisfying the condition $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Prove that $\text{GCD}(x, y, z) \cdot x y z$ 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 $\gcd(x_{1}, y_{1}, z_{1}) = 1$. Substitute and we get $\frac{1}{x_{1}} = \frac{1}{y_{1}} + \frac{1}{z_{1}}$ and we need to prove that $d^{4} x_{1} y_{1} z_{1}$ is a square of a natural number, hence it is sufficient to prove that $x_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,505 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2539^{\circ}+\sin 2540^{\circ}+\cdots+\sin 6139^{\circ}\right)^{\cos } 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6120^{\circ}\right)
$$ | Answer: $71^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6119^{\circ}=0$ and in the exponent only $\cos ... | 71 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,506 |
2. (5 points) Two different natural numbers end with 7 zeros and have exactly 72 divisors. Find their sum. | Answer: 70000000
Solution: Since the number ends with 7 zeros, it has the form $N=10^{7} k$. The smallest number of this form $10^{7}$ has 64 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 7. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has ... | 70000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,507 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\sin \alpha = \frac{\sqrt{24}}{5}$? | Answer: 4.
## Solution:
Consider the point $B_{1}$ symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,508 |
4. (7 points) On the board, 38 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 38 minutes? | Answer: 703.
Solution: Let's represent 38 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... | 703 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,509 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=47$.
 | Answer: 2256
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=47(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{47 S}{k}+\frac{47 S}{l}$... | 2256 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,510 |
6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=124
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 48
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, $A B^{2}=108$, $B C^{2}=16, A C^{2}=124$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,511 |
8. (10 points) In the expression $(x+y+z)^{2020}+(x-y-z)^{2020}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1022121
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2020}+(x-t)^{2020}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2020}=x^{2020}+a_{1} x^{2019} t+\ldots+a_{2019} x t^{2019}+t^{2020} \\
& (x-t)^{2020}=x^{2020}-a_{1} x^{2019} t+\ldots-a_{2... | 1022121 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,513 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-6)^{2}+(y-7)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=6 \frac{1}{3}, y=6 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-6)^{2}+(7-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-6|+|7-y|+|x-y|}{3} \geqslant \frac{x-6+7-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-6)^{2}+(7-y)^{2}+(x-y)^{2} \geqslant \frac... | 6\frac{1}{3},6\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,514 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2541^{\circ}+\sin 2542^{\circ}+\cdots+\sin 6141^{\circ}\right)^{\cos } 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6120^{\circ}\right)
$$ | Answer: $69^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6119^{\circ}=0$ and in the exponent only $\cos ... | 69 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,515 |
2. (5 points) Two different natural numbers end with 9 zeros and have exactly 110 divisors. Find their sum. | Answer: 7000000000
Solution: Since the number ends with 9 zeros, it has the form $N=10^{9} k$. The smallest number of this form $10^{9}$ has 100 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 9. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ h... | 7000000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,516 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\sin \alpha = \frac{\sqrt{21}}{5}$? | Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,517 |
4. (7 points) On the board, 37 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 37 minutes? | Answer: 666.
Solution: Let's represent 37 units as points on a plane. Each time we combine 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 connected ... | 666 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,518 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=43$.
 | Answer: 1892
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=43(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{43 S}{k}+\frac{43 S}{l}$... | 1892 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,519 |
6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=147 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=163
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 56
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=147$, $B C^{2}=16, A C^{2}=163$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,520 |
8. (10 points) In the expression $(x+y+z)^{2022}+(x-y-z)^{2022}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1024144
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2022}+(x-t)^{2022}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2022}=x^{2022}+a_{1} x^{2021} t+\ldots+a_{2021} x t^{2021}+t^{2022} \\
& (x-t)^{2022}=x^{2022}-a_{1} x^{2021} t+\ldots-a_{2... | 1024144 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,522 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-7)^{2}+(y-8)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=7 \frac{1}{3}, y=7 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-7)^{2}+(8-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-7|+|8-y|+|x-y|}{3} \geqslant \frac{x-7+8-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-7)^{2}+(8-y)^{2}+(x-y)^{2} \geqslant \frac... | 7\frac{1}{3},7\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,523 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2903^{\circ}+\sin 2904^{\circ}+\cdots+\sin 6503^{\circ}\right)^{\cos } 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}\right)
$$ | # Answer: $67^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$, it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6479^{\circ}=0$ and in the exponent only $\c... | 67 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,524 |
2. (5 points) Two different natural numbers end with 8 zeros and have exactly 90 divisors. Find their sum. | Answer: 700000000
Solution: Since the number ends with 8 zeros, it has the form $N=10^{8} k$. The smallest number of this form $10^{8}$ has 81 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 8. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has... | 700000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,525 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\sin \alpha = \frac{\sqrt{35}}{6}$? | Answer: 4.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,526 |
4. (7 points) On the board, 39 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 39 minutes? | Answer: 741.
Solution: Let's represent 39 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... | 741 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,527 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=37$.
 | Answer: 1406
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=37(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{37 S}{k}+\frac{37 S}{l}$... | 1406 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,528 |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=36 \\
z^{2}+x z+x^{2}=111
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 60
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, $A B^{2}=75$, $B C^{2}=36, A C^{2}=111$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,529 |
8. (10 points) In the expression $(x+y+z)^{2024}+(x-y-z)^{2024}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1026169
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2024}+(x-t)^{2024}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2024}=x^{2024}+a_{1} x^{2023} t+\ldots+a_{2023} x t^{2023}+t^{2024} \\
& (x-t)^{2024}=x^{2024}-a_{1} x^{2023} t+\ldots-a_{2... | 1026169 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,531 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-8)^{2}+(y-9)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=8 \frac{1}{3}, y=8 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-8)^{2}+(9-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-8|+|9-y|+|x-y|}{3} \geqslant \frac{x-8+9-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-8)^{2}+(9-y)^{2}+(x-y)^{2} \geqslant \frac... | 8\frac{1}{3},8\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,532 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2905^{\circ}+\sin 2906^{\circ}+\cdots+\sin 6505^{\circ}\right)^{\cos } 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}\right)
$$ | Answer: $65^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6479^{\circ}=0$ and in the exponent only $\cos ... | 65 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,533 |
2. (5 points) Two different natural numbers end with 6 zeros and have exactly 56 divisors. Find their sum. | Answer: 7000000
Solution: Since the number ends with 6 zeros, it has the form $N=10^{6} k$. The smallest number of this form $10^{6}$ has 49 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 6. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has o... | 7000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,534 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\sin \alpha = \frac{\sqrt{11}}{6}$? | Answer: 20.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,535 |
4. (7 points) On the board, 45 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 45 minutes? | Answer: 990.
Solution: Let's represent 45 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... | 990 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,536 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=31$.
 | Answer: 992
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=31(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{31 S}{k}+\frac{31 S}{l}$.... | 992 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,537 |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=49 \\
z^{2}+x z+x^{2}=124
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 70
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=49, A C^{2}=124$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,538 |
8. (10 points) In the expression $(x+y+z)^{2026}+(x-y-z)^{2026}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1028196
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2026}+(x-t)^{2026}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2026}=x^{2026}+a_{1} x^{2025} t+\ldots+a_{2025} x t^{2025}+t^{2026} \\
& (x-t)^{2026}=x^{2026}-a_{1} x^{2025} t+\ldots-a_{2... | 1028196 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,540 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-9)^{2}+(y-10)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=9 \frac{1}{3}, y=9 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-9)^{2}+(10-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-9|+|10-y|+|x-y|}{3} \geqslant \frac{x-9+10-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-9)^{2}+(10-y)^{2}+(x-y)^{2} \geqslant \... | 9\frac{1}{3},9\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,541 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2907^{\circ}+\sin 2908^{\circ}+\cdots+\sin 6507^{\circ}\right)^{\cos } 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}\right)
$$ | Answer: $63^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6479^{\circ}=0$ and in the exponent only $\cos ... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,542 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $28$, and $\sin \alpha = \frac{\sqrt{45}}{7}$? | Answer: 16.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,543 |
4. (7 points) On the board, 46 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 46 minutes? | Answer: 1035.
Solution: Let's represent 46 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... | 1035 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,544 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=29$.
 | Answer: 870
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=29(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{29 S}{k}+\frac{29 S}{l}$.... | 870 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,545 |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=64 \\
z^{2}+x z+x^{2}=139
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 80
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, $A B^{2}=75$, $B C^{2}=64, A C^{2}=139$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,546 |
8. (10 points) In the expression $(x+y+z)^{2028}+(x-y-z)^{2028}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1030225
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2028}+(x-t)^{2028}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2028}=x^{2028}+a_{1} x^{2027} t+\ldots+a_{2027} x t^{2027}+t^{2028} \\
& (x-t)^{2028}=x^{2028}-a_{1} x^{2027} t+\ldots-a_{2... | 1030225 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,548 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-10)^{2}+(y-11)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=10 \frac{1}{3}, y=10 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-10)^{2}+(11-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-10|+|11-y|+|x-y|}{3} \geqslant \frac{x-10+11-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-10)^{2}+(11-y)^{2}+(x-y)^{2} \geqs... | 10\frac{1}{3},10\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,549 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 3269^{\circ}+\sin 3270^{\circ}+\cdots+\sin 6869^{\circ}\right)^{\cos } 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6840^{\circ}\right)
$$ | Answer: $61^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6839^{\circ}=0$ and in the exponent only $\cos ... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,550 |
2. (5 points) Two different natural numbers end with 7 zeros and have exactly 72 divisors. Find their sum.
# | # Answer: 70000000
Solution: Since the number ends with 7 zeros, it has the form $N=10^{7} k$. The smallest number of this form $10^{7}$ has 64 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 7. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ ha... | 70000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,551 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $21$, and $\sin \alpha = \frac{\sqrt{40}}{7}$? | Answer: 18.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,552 |
4. (7 points) On the board, 47 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 47 minutes? | Answer: 1081.
Solution: Let's represent 47 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... | 1081 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,553 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=23$.
 | Answer: 552
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=23(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{23 S}{k}+\frac{23 S}{l}$.... | 552 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,554 |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=64 \\
z^{2}+x z+x^{2}=172
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 96
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=64, A C^{2}=172$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,555 |
8. (10 points) In the expression $(x+y+z)^{2030}+(x-y-z)^{2030}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1032256
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2030}+(x-t)^{2030}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2030}=x^{2030}+a_{1} x^{2029} t+\ldots+a_{2029} x t^{2029}+t^{2030} \\
& (x-t)^{2030}=x^{2030}-a_{1} x^{2029} t+\ldots-a_{2... | 1032256 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,557 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-11)^{2}+(y-12)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=11 \frac{1}{3}, y=11 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-11)^{2}+(12-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-11|+|12-y|+|x-y|}{3} \geqslant \frac{x-11+12-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-11)^{2}+(12-y)^{2}+(x-y)^{2} \geqs... | 11\frac{1}{3},11\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,558 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 3271^{\circ}+\sin 3272^{\circ}+\cdots+\sin 6871^{\circ}\right)^{\cos } 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6840^{\circ}\right)
$$ | Answer: $59^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6839^{\circ}=0$ and in the exponent only $\cos ... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,559 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $14$, and $\sin \alpha = \frac{\sqrt{33}}{7}$? | Answer: 16
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,560 |
4. (7 points) On the board, 48 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 48 minutes? | Answer: 1128.
Solution: Let's represent 48 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 s... | 1128 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,561 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=19$.
 | Answer: 380
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=19(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{19 S}{k}+\frac{19 S}{l}$.... | 380 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,562 |
6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=49 \\
z^{2}+x z+x^{2}=157
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 84
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, $A B^{2}=108$, $B C^{2}=49, A C^{2}=157$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,563 |
8. (10 points) In the expression $(x+y+z)^{2032}+(x-y-z)^{2032}$, the parentheses were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1034289
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2032}+(x-t)^{2032}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2032}=x^{2032}+a_{1} x^{2031} t+\ldots+a_{2031} x t^{2031}+t^{2032} \\
& (x-t)^{2032}=x^{2032}-a_{1} x^{2031} t+\ldots-a_{2... | 1034289 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,565 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-12)^{2}+(y-13)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=12 \frac{1}{3}, y=12 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-12)^{2}+(13-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-12|+|13-y|+|x-y|}{3} \geqslant \frac{x-12+13-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-12)^{2}+(13-y)^{2}+(x-y)^{2} \geqs... | 12\frac{1}{3},12\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,566 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2193^{\circ}+\sin 2194^{\circ}+\cdots+\sin 5793^{\circ}\right)^{\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5760^{\circ}}\right)
$$ | # Answer: $57^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$, it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5759^{\circ}=0$ and in the exponent only $\c... | 57 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,567 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $16$, and $\sin \alpha = \frac{\sqrt{55}}{8}$? | Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,568 |
4. (7 points) On the board, 49 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 49 minutes? | Answer: 1176.
Solution: Let's represent 49 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... | 1176 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,569 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=17$.
 | Answer: 306
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=17(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{17 S}{k}+\frac{17 S}{l}$.... | 306 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,570 |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=117
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 36
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=9, A C^{2}=117$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,571 |
8. (10 points) In the expression $(x+y+z)^{2034}+(x-y-z)^{2034}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1036324
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2034}+(x-t)^{2034}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2034}=x^{2034}+a_{1} x^{2033} t+\ldots+a_{2033} x t^{2033}+t^{2034} \\
& (x-t)^{2034}=x^{2034}-a_{1} x^{2033} t+\ldots-a_{2... | 1036324 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,573 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-13)^{2}+(y-14)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=13 \frac{1}{3}, y=13 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-13)^{2}+(14-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-13|+|14-y|+|x-y|}{3} \geqslant \frac{x-13+14-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-13)^{2}+(14-y)^{2}+(x-y)^{2} \geqs... | 13\frac{1}{3},13\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,574 |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2195^{\circ}+\sin 2196^{\circ}+\cdots+\sin 5795^{\circ}\right)^{\cos } 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5760^{\circ}\right)
$$ | Answer: $55^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5759^{\circ}=0$ and in the exponent only $\cos ... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,575 |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $16$, and $\sin \alpha = \frac{\sqrt{39}}{8}$? | Answer: 20.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,576 |
4. (7 points) On the board, 50 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 50 minutes? | Answer: 1225.
Solution: Let's represent 50 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 conne... | 1225 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,577 |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=13$.
 | Answer: 182
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=13(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{13 S}{k}+\frac{13 S}{l}$.... | 182 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,578 |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=147 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=156
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 42
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, $A B^{2}=147$, $B C^{2}=9, A C^{2}=156$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,579 |
8. (10 points) In the expression $(x+y+z)^{2036}+(x-y-z)^{2036}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1038361
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2036}+(x-t)^{2036}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2036}=x^{2036}+a_{1} x^{2035} t+\ldots+a_{2035} x t^{2035}+t^{2036} \\
& (x-t)^{2036}=x^{2036}-a_{1} x^{2035} t+\ldots-a_{2... | 1038361 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,581 |
9. (20 points) Find all values of $x$ and $y$ for which the following equality holds:
$$
(x-14)^{2}+(y-15)^{2}+(x-y)^{2}=\frac{1}{3}
$$ | Answer: $x=14 \frac{1}{3}, y=14 \frac{2}{3}$
Solution: We use the inequality between the quadratic mean and the arithmetic mean:
$$
\sqrt{\frac{(x-14)^{2}+(15-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x-14|+|15-y|+|x-y|}{3} \geqslant \frac{x-14+15-y+y-x}{3}=\frac{1}{3}
$$
That is, $(x-14)^{2}+(15-y)^{2}+(x-y)^{2} \geqs... | 14\frac{1}{3},14\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,582 |
1. In a $4 \times 4$ table, 16 different natural numbers are arranged. For each row and each column of the table, the greatest common divisor (GCD) of the numbers located in it was found. It turned out that all eight found numbers are different. For what largest $n$ can we assert that there is a number in such a table ... | # Answer: 32.
Solution. If the greatest common divisor (GCD) in some row is $n$, then there are four numbers in that row that are divisible by $n$, meaning there is a number no less than $4n$. Since the greatest common divisors in all rows are distinct, one of them is certainly no less than 8. Then, in the correspondi... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,583 |
2. For positive numbers $a, b, c$ and $d$, prove the inequality
$$
a+b+c+d+\frac{8}{a b+b c+c d+d a} \geqslant 6
$$
(A. Khryabrov) | First solution. Let $x=a+b+c+d$ and $y^{2}=a b+b c+$ $c d+d a$. Since
$$
x^{2}=((a+c)+(b+d))^{2} \geqslant 4(a+c)(b+d)=4 y^{2}
$$
we have the inequality $x \geqslant 2 y$. Write the inequality to be proved in the new notation: $x+\frac{8}{y^{2}} \geqslant 6$. It is sufficient to check that
$$
2 y+\frac{8}{y^{2}} \ge... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 19,584 |
3. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{2}-6 \sqrt{3-2 x}-y+11=0 \\
y^{2}-4 \sqrt{3 y-2}+4 x+16=0
\end{array}\right.
$$
(P. Alishev) | Answer: $x=-3, y=2$.
Solution. Add these equations and gather complete squares: $x^{2}+4 x-6 \sqrt{3-2 x}+y^{2}-y-4 \sqrt{3 y-2}+27=x^{2}+$ $6 x+9+3-2 x-6 \sqrt{3-2 x}+9+y^{2}-4 y+4+3 y-2-4 \sqrt{3 y-2}+4=$ $(x+3)^{2}+(\sqrt{3-2 x}-3)^{2}+(y-2)^{2}+(\sqrt{3 y-2}-2)^{2}=0$. It is clear that only when $x=-3$ and $y=2$ d... | -3,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,585 |
5. Two bandits are dividing a loot consisting of coins with odd values from 1 to 2017 tugriks. It is known that the total amount of tugriks they have is even, and for any odd $k$ not exceeding 2017, there is a coin worth $k$ tugriks in their loot. Prove that the bandits can divide the loot equally.
(A. Khryabrov based... | Solution. Let the bandits divide a sum of $2 n$ coins. Suppose they cannot divide them equally. Give the first bandit the largest possible sum $s$ that is less than $n$. Then he will receive coins with a total value of $s \leqslant n-1$ tugriks. Give the remaining coins to the second bandit. If the second bandit had a ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 19,587 |
2. (5 points) Solve the equation in integers
$$
x^{2}-y^{2}+2(a+b) x+2(a-b) y=p-4 a b \quad p \in \mathbb{P}
$$
In your answer, write the smallest value of the variable $x$. | Solution: We transform our equation to the form
$$
\left(x^{2}+2(a+b) x+(a+b)^{2}\right)-\left(y^{2}-2(a-b) y+(a-b)^{2}\right)=p
$$
We get
$$
(x+a+b)^{2}-(y-a+b)^{2}=p
$$
or
$$
(x-y+2 a) \cdot(x+y+2 b)=p
$$
We obtain four possible cases
$$
\left\{\begin{array} { l }
{ x - y + 2 a = 1 } \\
{ x + y + 2 b = p }
\e... | -\frac{p+1}{2}-b | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,588 |
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