problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. A rectangle is drawn on the board, and it is cut into 4 smaller rectangles using two straight cuts. If Andrey approaches the board, he will name the area of each of the smaller rectangles (in some random order), and if his twin brother Kirill approaches, he will name the perimeters. Someone of the brothers approache... | Solution. Note that the product of the areas of the lower left and upper right rectangles in the partition is equal to the product of the areas of the upper left and lower right rectangles. Also note that the sum of the perimeters of the lower left and upper right rectangles in the partition is equal to the sum of the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,703 |
6. There is a strip of 101 cells, on which a chip can move: forward by any even number of cells, and backward by any odd number of cells. Vasya and Petya want to visit all the cells of the board once with their chips: Vasya starting from the first cell, and Petya starting from the fiftieth. Who has more ways to do this... | Solution. The number of ways is the same. Imagine that this is not a strip but a ring (connect the beginning with the end), and you can only move forward and only by an even number of cells. Making a move backward by an odd number of cells is equivalent to making a move forward by an even number of cells on the ring. T... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,704 |
7. Each cell of a $10 \times 10$ board is painted black or white. A cell is said to be out of place if it has at least seven neighbors of a different color than itself. (Neighbors are cells that share a common side or corner.) What is the maximum number of white cells on the board that can be out of place at the same t... | Solution. Answer: 26
Example: 13 dominoes, not touching each other or the edge of the board.
Estimate. Let's call a cell that is not in its own place an OOP cell. For each OOP cell, we draw ... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,705 |
3. Masha bought 2021 weights of pairwise distinct masses. Now Masha places one weight on each pan of a two-pan balance (weights placed on the balance previously are not removed). Each time the balance is in equilibrium, Masha rejoices. What is the maximum number of times she can find a reason to be happy | Solution. Answer: 673 times.
Example. Let Masha have bought 673 triples of the form $x, y, x+y$ (we will choose the weights of the new triples so that they do not duplicate the old ones). The last two weights are any. Masha puts $x$ on the left pan, then $-y$ on the same left pan, and finally $x+y$ on the right pan an... | 673 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,708 |
4. Baron Munchausen claims that in childhood, he was able to draw a non-rectangular cellular figure that could be cut into 2, 3, ..., and 7 equal parts along the cell lines. Could his words be true? | Solution. Method 1: Let's call a figure made up of $2 n$ cells, which is glued from two $1 \times n$ rectangles along a boundary piece of length $n-1$, a zigzag of length $n$. For example, a zigzag of length $5$ is shown in the figure:
 - how many times the number mentioned by the fisherman is greater than the true value. For example, if a fisherman with a lying measure of 3 catch... | Solution. Note that the product of all the numbers named by the fishermen in each of the surveys is the product of the number of fish caught, multiplied by the product of all the measures of lying of these fishermen. Therefore, the seventh answered $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cd... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,710 |
6. At the tournament, there were 1000 volleyball teams, some of which played one match against each other. Each team had an equal number of wins and losses (there are no draws in volleyball). It is known that if team A won against team B, then A also won against all the teams that B did not play against. Prove that som... | Solution. First, note that from the condition, the following easily follows: if A did not play with B, and B did not play with C, then A did not play with C, otherwise the winner would have to play with B.
Now, take an arbitrary team and color all those it did not play with in one color. Take one of the unpainted and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,711 |
7. The lye trader Matti wrote several three-digit palindromic numbers on the board. It turned out that no two of them add up to a palindrome. Could he have written more than half of all three-digit palindromes? Recall that a palindrome is a number that remains the same when its digits are reversed. | Solution. No. There are $9 \cdot 10$ three-digit palindromes in total (9 options for the first digit, 10 for the second, and the third is uniquely determined by the first). We will prove that at least 45 palindromes are missing from the board.
If $a, b, c, d$ are such digits that $a+c=b+d=11$, then
$$
\overline{a b a... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,712 |
7.1. Let the table have size $2 \times 10$. Is it true that sooner or later the numbers will stop changing? | Solution. After the first evening, in each column, the numbers $a_{i}$ and $-a_{i}$ will be written, so the arithmetic means in the rows will be equal to $a$ and $-a$, which means the numbers in the column will turn into $a_{i}+s$ and $-a_{i}-s$ by morning. Now the sums in both rows and columns are all equal to zero, s... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,713 |
7.3. Prove that there will be a morning when all the numbers in the table will change by no more than 0.01. | Solution. Let's look at the sum of the squares of the numbers in the table. When subtracting the average $a=\frac{1}{k} \sum a_{i}$ from the set $a_{1}, \ldots, a_{k}$, for example, the sum of the squares becomes
$\sum\left(a_{i}-a\right)^{2}=\sum a_{i}^{2}+k a^{2}-2 a \sum a_{i}=\sum a_{i}^{2}+k a^{2}-2 k a^{2}=\sum ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 21,715 |
1. Andrey named all natural numbers from 180 to 220 inclusive ("one hundred eighty", "one hundred eighty one", etc.). How many words did he say? | Solution. A total of 41 numbers are named. In each, the hundreds place is named (a total of 41 words). In addition to the hundreds place, the numbers 180, 190, 201-220 each contain one more word (this is 22 words). The number 200 does not contribute any more words, while the remaining 18 numbers each contain two more w... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,717 |
3. Around a campfire, eight natives from four tribes are sitting in a circle. Each one says to the person on their left: “if we don’t count us, there is no one else from my tribe here.” It is known that the natives lie to strangers and tell the truth to their own. How many natives can there be from each tribe? | Solution. There must be at least two natives from one tribe; a unique native could not lie to a neighbor that there is no one else from their tribe. Since there are only four tribes, there are two from each tribe. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,719 |
4. A girl is standing on the 1st floor of a 14-story building, where her friends live on the 12th and 14th floors. She can only move around the building using an elevator that can only move 3 or 7 floors up or down. Can the girl visit all her friends by making no more than 6 elevator trips? | Solution (1).
In the diagram, the possible moves are shown - the floors are labeled in circles, and lines connect the floors that can be reached by pressing one button in the elevator.
From ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,720 |
5. Yesterday at the market, with one hundred tugriks you could buy 9 gingerbreads and 7 pastries (and even get some change), but today this amount is no longer enough. However, with the same one hundred tugriks today, you can buy two gingerbreads and 11 pastries (also with some change), but yesterday this amount would ... | Solution. Note that something has become more expensive (otherwise, the situation where we could buy a set yesterday but not today could not have arisen), and another sweet has become cheaper for similar reasons. Nine cookies and seven cakes together have become more expensive, so the cookie has become more expensive, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,721 |
6. The set contained weights of 43, 70, and 57 grams, with an equal number of each type. Maly lost several weights (less than half), weighed the remaining on the scales, and got 20172 grams. How many and which weights were lost? | Solution (1). If the weights had not been lost, the total weight would end in 0. Therefore, the weight of the lost weights ends in 8. This can only be the case if 4 weights of 57 g are lost (losing 70 g or the pair $43+57$ does not affect the last digit of the total weight).
Solution (2). Let's divide the initial set ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,722 |
7. 22 football players played three training matches (each time they were divided into two teams of 11 players). Prove that some two players played on different teams all three times. | Solution (1). After the first match, we will paint the players who played on the same team red, and those who played on the other team blue. Then, in the second match, among each of the teams, there will be at least 6 players of the same color, and these six will be of different colors. In these two groups of six, ther... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,723 |
8. In the foggy city of London, there are exactly $10^{10}$ clubs, and in each club, there are exactly 10 gentlemen. An enemy spy wants to kidnap several gentlemen so that among the kidnapped, there is at least one member of each club. It turns out that for any two clubs, there is a gentleman who is a member of both. P... | # Solution:
Consider an arbitrary club, for example, "Diogenes". Since for any club, there is a gentleman from "Diogenes" who is a member, one of the gentlemen from "Diogenes" (let's say Vasya) is a member of more than $10^{9}$ clubs.
Suppose it is not enough for the spy to kidnap only Vasya, then there exists a club... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,725 |
2. One night, a clock (with hands) broke. The hour hand started moving twice as fast, while the minute hand moved twice as slow as they did initially. Andrey woke up when the hour hand pointed at 6, and the minute hand pointed at 12. Could the clock have broken at midnight? | Solution. Suppose the clock broke at midnight. By the time Andrey woke up, an even number of hours had passed - since the minute hand had made a whole number of revolutions, and now it makes each revolution in two hours. But then the hour hand should point to a number divisible by 4, as it moves twice as fast as usual.... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,726 |
3. Is it possible to place a rook, knight, or bishop in each square of a chessboard so that the rooks attack only knights, knights attack only bishops, and bishops attack only rooks? | Solution. It is impossible. Let's consider three cases.
- A rook stands on square A1. Then a knight must be on A2, a bishop on C1, a rook on B2, a knight on C2, and this knight attacks the rook on A1, which should not happen.
- A knight stands on A1. Then a bishop stands on C2, a rook on B3, and this rook is attacked ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,727 |
4. On Valentine's Day at school, boys gave valentines to girls, and vice versa. Each boy gave valentines expressing love to five girls. The girls, however, were much more reserved: each gave valentines expressing love to only four boys. Five students (three Valentinias and two Valentins) received an equal number of val... | Solution. Let there be $m$ boys and $d$ girls in the class, and let the Valentines received by each girl be $k$. Then, since the number of sent valentines equals the number of received valentines,
$$
\left\{\begin{array}{l}
5 m=3 k+2(d-3) \\
4 d=2 k+2(m-2)
\end{array}\right.
$$
Multiply the first equation by 3, the s... | Combinatorics | proof | Yes | Yes | olympiads | false | 21,728 | |
5. A natural number $n$ is such that the sum of four of its distinct natural divisors (possibly including the number itself) is equal to $2 n$. What can the sum of the four smallest natural divisors of this number be? List all possible answers and prove that there are no others. | Solution. The sums can be 10, 11, and 12.
We will prove that the divisors that sum up to $2n$ are $n, n/2, n/3$, and $n/6$.
Indeed, if there is no divisor $n$, then the largest possible sum is
$$
\frac{n}{2}+\frac{n}{3}+\frac{n}{4}+\frac{n}{5}<2n
$$
If there is no divisor $n/2$, then the maximum sum is
$$
n+\frac{... | 10,11,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,729 |
6. Each boy is friends with 5 girls, while all girls are friends with a different number of boys. What is the minimum number of children that can be in this group? | Solution. If there is a girl who does not have any friends, we can exclude her (the number of children will decrease, but the condition of the problem will remain).
Let us have $m$ boys and $d$ girls. Draw a segment between a boy and a girl if they are friends. Then $5m$ segments are drawn from the boys, and at least ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,730 |
7. On an eleven-kilometer circular track, many cars are driving at a constant speed of 120 km/h. At one unfortunate spot on the road, there is an electronic police officer. At any moment, his baton can be in one of two positions: raised or lowered. If a car passes the electronic police officer with the baton lowered, i... | Solution. If the police officer stands with the baton lowered, any car completes a full circle in 6 minutes: one minute it drives at a speed of 60 km/h (i.e., it travels 1 km), and the rest of the time at a speed of 120 km/h (i.e., it travels 10 km in 5 minutes).
Consider the procedure: the electronic police officer r... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 21,731 |
1. In a volleyball tournament, $n$ teams from city $A$ and $2 n$ teams from city $B$ participated. Each team played exactly one game against each other team. The ratio of the number of wins by teams from city $B$ to the number of wins by teams from city $A$ is $3: 4$. Find $n$, given that there were no draws in the tou... | (12 points) Solution. The number of games in which only teams from city $A$ participated is $\frac{(n-1) n}{2}$. In these games, teams from city $A$ won $\frac{(n-1) n}{2}$ games. The number of games in which only teams from city $B$ participated is $\frac{(2 n-1) 2 n}{2}$. In these games, teams from city $B$ won $(2 n... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,732 |
2. What is the smallest area that a figure on the plane $Oxy$ can have, located between the lines $x=-11$ and $x=1$, bounded below by the $x$-axis, and above by the tangent to the graph of the function $y=3+\sqrt{4-x}$ at the point of tangency with abscissa $x_{0}$, where $-11 \leq x_{0} \leq 1$? | Solution. Let's write the equation of the tangent line to the graph of the function $y=3+\sqrt{4-x}$ at the point with abscissa $x_{0}$:
$y_{\text {tan }}=-\frac{1}{2 \sqrt{4-x_{0}}}\left(x-x_{0}\right)+3+\sqrt{4-x_{0}}$.
The figure is a trapezoid with bases equal to $y_{\text {tan }}(-11)$ and $y_{\text {tan }}(1)$.... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 21,733 | |
5. Find the set of values of the expression $\frac{a \cos x + b \sin x + c}{\sqrt{a^{2} + b^{2} + c^{2}}}$, where $x, a, b, c$ are arbitrary numbers such that $a^{2} + b^{2} + c^{2} \neq 0$. (20 points) | Solution: Let $\vec{n}=(a ; b ; c), \vec{m}=(\cos x ; \sin x ; 1)$. Then $|\vec{n}|=\sqrt{a^{2}+b^{2}+c^{2}}, \quad|\vec{m}|=\sqrt{2}$. Consider the function $f(x, a, b, c)=\frac{a \cos x+b \sin x+c}{\sqrt{a^{2}+b^{2}+c^{2}}}=\frac{\vec{n} \cdot \vec{m}}{|\vec{n}|}=|\vec{m}| \cos \alpha=\sqrt{2} \cos \alpha$, where $\a... | [-\sqrt{2};\sqrt{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,734 |
6. In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side $a$, point $K$ is the midpoint of edge $B_{1} C_{1}$, point $L$ lies on edge $C_{1} D_{1}$, such that $D_{1} L=2 C_{1} L$, and point $N$ is the midpoint of edge $A A_{1}$. Construct the section of the cube by the plane passing through points $K, L, N$. Describe... | # Solution:
Algorithm:
1) Connect $K$ and $L$. $E$ - the intersection point of lines $K L$ and $A_{1} B_{1}$.
2) Connect $E$ and $N$. $F$ - the intersection point of lines $E N$ and $B B_{1}$.
3) $G$ - the intersection point of lines $K L$ and $A_{1} D_{1}$.
4) $M$ - the intersection point of lines $G N$ and $D D_{1}... | \frac{11^{2}\sqrt{77}}{96} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,735 |
2. What is the smallest area that a figure on the plane $Oxy$ can have, located between the lines $x=-26$ and $x=2$, bounded below by the $x$-axis, and above by the tangent to the graph of the function $y=5+\sqrt{4-x}$ at the point of tangency with abscissa $x_{0}$, where $x_{0}$ lies in the interval $-26 \leq x_{0} \l... | Solution. Solution. We will write the equation of the tangent line to the graph of the function $y=5+\sqrt{4-x}$ at the point with abscissa $x_{0}$
$$
y_{\text {tan }}=-\frac{1}{2 \sqrt{4-x_{0}}}\left(x-x_{0}\right)+5+\sqrt{4-x_{0}}
$$
 \cdot n, y_{n}=1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2}$.
(20 points) | Solution: Let $z_{n}=1+2+\cdots+n=\frac{(n+1) n}{2}=\frac{n^{2}+n}{2}$. We have
$$
\begin{aligned}
& x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n=(2+3+\cdots n)+(3+4+\cdots n)+\cdots((n-1)+n)+n= \\
& =\left(z_{n}-z_{1}\right)+\left(z_{n}-z_{2}\right)+\cdots+\left(z_{n}-z_{n-1}\right)=(n-1) z_{n}-\left(z_{1}+z_{2}+\c... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,737 |
4. The area of a convex quadrilateral $A B C D$ is 40. Points $K, L, M, N$ are chosen on sides $A B, B C, C D, A D$ respectively, such that $A K: A B=B L: B C=C M: C D=D N: A D$. Find the ratio $A K: K B$ if the area of quadrilateral $K L M N$ is 25. | Solution: $A K: A B=x, D N: A D=x$, $A N: A D=(A D-N D): A D=1-x$, $S_{A K N}=x(1-x) S_{A B D}$. Similarly, we get $S_{C L M}=x(1-x) S_{\text {BCD }}$. Then $S_{A K N}+S_{C L M}=x(1-x) S_{A B C D}=100 x(1-x)$.
Similarly $\quad S_{B L K}+S_{D N M}=x(1-x) S_{A B C D}=40 x(1-x) . \quad$ From this, $S_{\text {KLMN }}=40-8... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,738 |
5. Find the set of values of the expression $\frac{a \cos x - b \sin x + 2c}{\sqrt{a^{2} + b^{2} + c^{2}}}$ where $x, a, b, c$ are arbitrary numbers such that $a^{2} + b^{2} + c^{2} \neq 0$. | Solution: Let $\vec{n}=(a ; b ; c), \vec{m}=(\cos x ;-\sin x ; 2)$. Then $|\vec{n}|=\sqrt{a^{2}+b^{2}+c^{2}},|\vec{m}|=\sqrt{5}$. Consider the function $\quad f(x, a, b, c)=\frac{a \cos x-b \sin x+2 c}{\sqrt{a^{2}+b^{2}+c^{2}}}=\frac{\vec{n} \cdot \vec{m}}{|\vec{n}|}=|\vec{m}| \cos \alpha=\sqrt{5} \cos \alpha, \quad$ w... | [-\sqrt{5};\sqrt{5}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,739 |
6. In the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side $a$, point $K$ is the midpoint of edge $B_{1} C_{1}$, point $L$ lies on edge $C_{1} D_{1}$, such that $D_{1} L=3 C_{1} L$, and point $N$ is the midpoint of edge $A A_{1}$. Construct the section of the cube by the plane passing through points $K, L, N$. Describe... | # Solution:
Algorithm:
1) Connect $K$ and $L$. $E$ is the intersection point of lines $K L$ and $A_{1} B_{1}$.
2) Connect $E$ and N. F is the intersection point of lines $E N$ and $B B_{1}$.
3) $G$ is the intersection point of lines $K L$ and $A_{1} D_{1}$.
4) $M$ is the intersection point of lines $G N$ and $D D_{1}... | \frac{3^{2}\sqrt{30}}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,740 |
1. Compute the coefficient of $x^{80}$ in the polynomial $\left(1+x+x^{2}+\cdots+x^{80}\right)^{3}$ after combining like terms. | Solution: $\left(1+x+x^{2}+\cdots+x^{80}\right)^{3}=\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)$
Let after expanding the brackets, the term $x^{p} \cdot x^{q} \cdot x^{r}$ is obtained if the factor $x^{p}$ is chosen from the first bracket, $x^{q}$ - from... | 3321 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,741 |
2. Find the area of the figure, the coordinates $(x ; y)$ of the points of which satisfy the system of inequalities
$$
\left\{\begin{array}{c}
|x+1|-|y+3| \geq 0 \\
\log _{4}\left(x^{2}+y^{2}+6 y+2 x-26\right) \leq 3
\end{array}\right.
$$ | Solution. $\left\{\begin{array}{c}|x+1|-|y+3| \geq 0 \\ \log _{4}\left((x+1)^{2}+(y+3)^{2}-36\right) \leq 3\end{array}\right.$
Substitution: $\mathrm{u}=\mathrm{x}+1, \mathrm{v}=\mathrm{y}+3 /$
We have: $\left\{\begin{array}{c}|u|-|v| \geq 0 \\ 36 \geq v^{2}+u^{2} \leq 100 .\end{array}\right.$
The area of the figure... | 32\pi | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,742 |
3. What is the smallest area that a circle centered at the origin can have, which has common points with the graph of the function $y=\frac{6}{x}-\frac{4 x}{3}$?
(16 points) | Solution. $(x ; f(x))$ belongs to a circle with radius $r$
$r^{2}=f^{2}(x)+x^{2}=\left(\frac{6}{x}-\frac{4 x}{3}\right)^{2}+x^{2}=\frac{36}{x^{2}}+\frac{25 x^{2}}{9}-16\left(r^{2}\right)^{\prime}=-\frac{72}{x^{3}}+\frac{50 x}{9}=\frac{2\left(25 x^{4}-36 \cdot 9\right)}{16 x^{3}}$,
$x_{\min }=\sqrt{18 / 5}, \quad S_{\... | 4\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,743 |
4. A tangent is drawn from point $C$ to a circle with radius $2 \sqrt{5}$ and center at point $O$. Point $A$ is the point of tangency. Segment $C O$ intersects the circle at point $B$. A perpendicular is dropped from point $B$ to line $B C$ until it intersects line $A C$ at point $F$. Find the radius of the circumscrib... | # Solution:
Since $B F$ and $A F$ are segments of tangents, $B F=A F=2$.
Right triangles $A O C$ and $B F C$ are similar, $\frac{O C}{F C}=\frac{A C}{B C}=\frac{O A}{B F}$.
Let $x=C F, \qua... | \frac{\sqrt{30}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,744 |
5. Find all values of $x$ for which the inequality $(a+2) x-(1+2 a) \sqrt[3]{x^{2}}-6 \sqrt[3]{x}+a^{2}+4 a-5>0$ holds for at least one $a \in[-2 ; 1]$.
# | # Solution:
$$
a^{2}+\left(x-2 \sqrt[3]{x^{2}}+4\right) a+2 x-\sqrt[3]{x^{2}}-6 \sqrt[3]{x}-5>0
$$
Find $x$, for which the inequality
$a^{2}+\left(x-2 \sqrt[3]{x^{2}}+4\right) a+2 x-\sqrt[3]{x^{2}}-6 \sqrt[3]{x}-5 \leq 0$ holds for all $a \in[-2 ; 1]$.
Let $f(a)=a^{2}+\left(x-2 \sqrt[3]{x^{2}}+4\right) a+2 x-\sqrt[... | (-\infty;-1)\cup(-1;0)\cup(8;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,745 |
6. The base of the pyramid $S A B C$ is an isosceles triangle $A B C$, with $A B = B C = 3$, and $A C = 5$. The height of the pyramid $S A B C$ is the segment $S O$, where $O$ is the intersection point of the line passing through vertex $B$ parallel to side $A C$ and the line passing through $C$ perpendicular to side $... | # Solution:
$$
A B=B C=b=3, \quad A C=a=5, d=\rho\left(O_{1}, B S C\right)
$$
$$
O B=a / 2, \quad O C=\sqrt{b^{2}-a^{2} / 4}, O K=\frac{O B \cdot O C}{B C}=\frac{a \sqrt{4 b^{2}-a^{2}}}{4 b}
$$
$r=\frac{2 S_{A B C}}{P_{A B C}}=\frac{a \sqrt{4 b^{2}-a^{2}}}{2(a+2 b)}, L K=r$, in triangle OSK we have $\frac{O T}{d}=\f... | \frac{35\sqrt{11}}{396} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,746 |
1. Shooting at a target, with each shot the athlete scored only eight, nine, or ten points (all these points were scored at least once). Making more than 11 shots, in total he scored 100 points. How many 8-point shots did the athlete make? | Solution. $8 x+9 y+10 z=100, x, y, z \in \mathbb{N} \Rightarrow 8(x+y+z)<100$
$$
\begin{gathered}
\Rightarrow 11<x+y+z<\frac{100}{8}=12.5 \Rightarrow x+y+z=12 \\
\Rightarrow\left\{\begin{array} { c }
{ x + y + z = 1 2 } \\
{ 8 x + 9 y + 1 0 z = 1 0 0 }
\end{array} \Rightarrow \left\{\begin{array} { c }
{ x + y + z =... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,747 |
2. For any natural values $\quad m$ and $n(n>1)$, the function $f\left(\frac{m}{n}\right)=\frac{\sqrt[n]{3^{m}}}{\sqrt[n]{3^{m}}+3}$ is defined. Calculate the sum
$f\left(\frac{1}{2020}\right)+f\left(\frac{2}{2020}\right)+f\left(\frac{3}{2020}\right)+\cdots+f\left(\frac{4039}{2020}\right)+f\left(\frac{4040}{2020}\righ... | Solution. Consider the function $f(x)=\frac{3^{x}}{3^{x}+3}$. We have
$$
f(x)+f(2-x)=\frac{3^{x}}{3^{x}+3}+\frac{3^{2-x}}{3^{2-x}+3}=\frac{3^{x}}{3^{x}+3}+\frac{9}{9+3 \cdot 3^{x}}=\frac{3^{x}}{3^{x}+3}+\frac{3}{3+3^{x}}=1
$$
Then $\quad f\left(\frac{1}{2020}\right)+f\left(\frac{4039}{2020}\right)=f\left(\frac{2}{202... | 2020.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,748 |
3. Find all pairs of real numbers $(x ; y)$ that satisfy the inequality $\sqrt{x+y-1}+x^{4}+y^{4}-\frac{1}{8} \leq 0$. In your answer, write the greatest value of the product $xy$ for all found pairs $(x ; y)$.
(6 points) | Solution. Domain of definition: $x+y \geq 1$. We will prove the validity of the inequality $x^{4}+y^{4} \geq 1 / 8$ under this condition. For any $x$ and $y$, the inequality
$$
(x+y)^{2} \geq 4 x y \quad \Rightarrow \quad \frac{(x+y)^{2}}{2} \geq 2 x y
$$
holds. Then $x^{2}+y^{2}=(x+y)^{2}-2 x y \geq \frac{(x+y)^{2}}... | 0.25 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,749 |
5. The Fibonacci sequence is defined recursively: $a_{1}=a_{2}=1$, $a_{n+2}=a_{n+1}+a_{n}$ for all natural $n$. What is the last digit of $a_{2020}$? | Solution. By induction, it is proved that
1) the parity of the numbers alternates as: OOEEOOEEO...
2) the number with an index divisible by 5 is divisible by 5.
Since 2020 is divisible by 5 and leaves a remainder of 1 when divided by 3, the number $a_{2020}$ is odd and divisible by 5.
Answer: 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,750 |
6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1 ; 0) \), and its legs lying on the lines \( y = -2 \) and \( x = 0 \)?
(12 points) | Solution. $A B: \quad y=k x+b, \quad M \in A B \Rightarrow b=-k$
$$
\begin{aligned}
& A(a ;-2) \in A B \Rightarrow-2=k a+b \Rightarrow a=1-\frac{2}{k} \\
& S_{A B C}=\frac{1}{2} B C \cdot C A=\frac{1}{2}(b+2) a=\frac{1}{2}(2-k)(1-2 / k) \\
& S^{\prime}=\frac{(2-k)(2+k)}{2 k^{2}}=0, k_{\min }=-2, \quad S_{\min }=4
\end... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,751 |
8. Specify the greatest value of the parameter $a$ for which there is a unique solution to the system
$$
\left\{\begin{array}{l}
y=1-\sqrt{x} \\
a-2(a-y)^{2}=\sqrt{x}
\end{array}\right.
$$ | # Solution:
Equation: $\left\{\begin{array}{l}\sqrt{x}=1-y \\ a-2(a-y)^{2}=1-y\end{array}\right.$, considering that $1-y \geq 0$, we solve the resulting quadratic equation:
$$
\begin{align... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,752 |
9. The base of the pyramid $T A B C D$ is a trapezoid $A B C D$, the length of the shorter base $B C$ of which is $4 \sqrt{11}$. The diagonals of the trapezoid $A C$ and $B D$ intersect at point $O$, the ratio of the areas of triangles $A O B$ and $B O C$ is $3: 2$, and the area of triangle $D O C$ is $132 \sqrt{2 / 25... | # Solution.
Let $T H$ be the height of the pyramid. Since all lateral edges are inclined to the base at the same angle, $H$ is the center of the circle circumscribed around the base. It follows that the trapezoid is isosceles. From the condition, $\frac{S_{A O B}}{S_{\text {ВОС }}}=\frac{3}{2}=\frac{A O}{O C}$. Triang... | 105.875 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,753 |
1. The resistance to bending of a beam with a rectangular cross-section is proportional to the product of the width of this section and the square of its height. What should be the width of the cross-section of a beam cut from a round log with a diameter of $15 \sqrt{3}$ so that its resistance to bending is maximized?
... | # Solution.
$F(a)=k a b^{2}, a-$ width of the beam, $b$ - height of the beam. Since $b^{2}=225 \cdot 3-a^{2}$, then
$F(a)=k a\left(225 \cdot 3-a^{2}\right)=k\left(225 \cdot 3 a-a^{3}\right), F^{\prime}(a)=k\left(225 \cdot 3-3 a^{2}\right)=3 k\left(225-a^{2}\right)$.
The derivative equals zero and changes sign from p... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,754 |
2. Find all pairs of integers $(x, y)$ that satisfy the equation $x y=20-3 x+y$. For each found pair $(x, y)$, calculate the product $x y$. In the answer, write the sum of these products. | Solution. $\quad x y=20-3 x+y, \quad x y=17+3(1-x)+y, \quad(x-1) y+3(x-1)=17$, $(x-1)(y+3)=17$. Since x and y are integers, we have four cases:
1) $\left\{\begin{array}{c}y+3=17, \\ x-1=1,\end{array} \Leftrightarrow\left\{\begin{array}{c}y=14 \\ x=2\end{array}\right.\right.$
2) $\left\{\begin{array}{c}y+3=-17, \\ x-1=... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,755 |
3. Find all natural solutions to the inequality
$$
\frac{4}{5}+\frac{4}{45}+\frac{4}{117}+\cdots+\frac{4}{16 n^{2}-8 n-3}>n-5
$$
In your answer, write the sum of all found solutions. | Solution. We have $16 n^{2}-8 n-3=(4 n-3)(4 n+1), \quad \frac{1}{4 n-3}-\frac{1}{4 n+1}=\frac{4}{16 n^{2}-8 n-3}$,
$\frac{4}{1 \cdot 5}+\frac{4}{5 \cdot 9}+\frac{4}{9 \cdot 13}+\cdots+\frac{4}{(4 n-3)(4 n+1)}>n-5$,
$1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\cdots+\frac{1}{4 n-3}-\frac{1}{4 n+1}>n-5,1-\frac{1}{4 n+1}>n-5... | 15 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,756 |
4. There is an unlimited supply of square glasses in 10 colors. In how many ways can 4 glasses be inserted into a $2 \times 2$ window frame so that some color appears in both the upper and lower halves of the window. | Solution. If the upper half of the glass is of one color (one of 10), then the colors of the lower glasses can be chosen in $10^{2}-9^{2}=19$ ways. If the upper glasses are different $\left(10 * 9=90\right.$ ways to choose), then the colors of the lower ones can be chosen in $10^{2}-8^{2}=36$ ways.
In total $10 * 19+9... | 3430 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,757 |
5. Let $x, y, z$ be the roots of the equation $t^{3}-5 t-3=0$. Find $x^{3} y^{3}+x^{3} z^{3}+y^{3} z^{3}$. | Solution. The polynomial has 3 different real roots, since $\mathrm{P}(-100)0, \mathrm{P}(0)0$. By Vieta's theorem, $\mathrm{x}+\mathrm{y}+\mathrm{z}=0, \mathrm{xy}+\mathrm{xz}+\mathrm{yz}=-5, \mathrm{xyz}=3$.
$$
\begin{gathered}
x^{3} y^{3}+x^{3} z^{3}+y^{3} z^{3}=(5 x+3)(5 y+3)+(5 x+3)(5 z+3)+(5 y+3)(5 z+3) \\
=25(x... | -98 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,758 |
6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1; 3) \), and its legs lying on the lines \( y = x \) and \( y = -x \)?
(12 points) | Solution.
$A B: \quad y=k x+d, \quad M \in A B \Rightarrow d=3-k$
$A(a ;-a) \in A B \Rightarrow-a=k a+3-k \Rightarrow a=\frac{3-k}{k+1}$,
$B(b ; b) \in A B \Rightarrow b=k b+3-k \Rightarrow b=\frac{k-3}{k-1}$,
 ^ { 2 } } \\
{ ( y + 1 ) ^ { 2 } + ( x - 0.5 a ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x-1=(y+1)^{2} \\
x-2+(x-0.5 a)^{2}=0
\end{array}\right.\right.
$$
Consider the second equation of the system
$x^{2}+x(1-a)+\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,760 |
9. The base of the pyramid $T A B C D$ is an isosceles trapezoid $A B C D$, the length of the smaller base $B C$ of which is $\sqrt{3}$. The ratio of the areas of the parts of the trapezoid $A B C D$, into which it is divided by the midline, is $5: 7$. All lateral faces of the pyramid $T A B C D$ are inclined to the ba... | # Solution.
Let $TO$ be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, $O$ is the center of the circle inscribed in the base. Let $MP$ be the midline of the trapezoid, $AD=a, BC=b=\sqrt{3}$. According to the problem, we have $S_{MB CP}=5x, S_{\text{AMPD}}=7x, \quad \frac... | 0.875 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,761 |
1. There are 5 pieces of transparent glass of the same square shape and size. Each piece of glass is conditionally divided into 4 equal parts (right triangles) by its diagonals, and one of these triangles is painted with an opaque paint of its individual color, different from the colors of the painted parts of the othe... | Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then... | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,762 |
2. Solve the equation $\sin ^{4}(2025 x)+\cos ^{2019}(2016 x) \cdot \cos ^{2018}(2025 x)=1$.
(12 points) | # Solution:
$\sin ^{4} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1 \Leftrightarrow\left(1-\cos ^{2} 2025 x\right)^{2}+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1$
$\Leftrightarrow \cos ^{4} 2025 x-2 \cos ^{2} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=0 \Leftrightarrow$
$\cos ^{2} 2025 x\lef... | \frac{\pi}{4050}+\frac{\pin}{2025},n\inZ,\frac{\pik}{9},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,763 |
3. All members of an infinite geometric progression are natural numbers. The sum of the third, fifth, and seventh terms of this progression is equal to $819 \cdot 6^{2016}$. Find the common ratio of the progression. (16 points) | Solution: We have a geometric progression $b_{1}, b_{1} q, b_{1} q^{2}, \cdots, b_{1} q^{n-1}, \cdots$, and $b_{1} q^{n-1} \in N$ for any number $n \in N$. Thus, $b_{1}$ and $q$ are natural numbers. According to the condition $b_{3}+b_{5}+b_{7}=819 \cdot 6^{2016}$, or $b_{1} q^{2}+b_{1} q^{4}+b_{1} q^{6}=2^{2016} \cdot... | q=1,q=2,q=3,q=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,764 |
4. On the sides $AB$ and $AC$ of triangle $ABC$, points $D$ and $E$ are chosen such that the area of triangle $ADE$ is 0.5. The incircle of quadrilateral $BDEC$ touches side $AB$ at point $K$, and $AK=3$. Find the tangent of angle $BAC$, if a circle can be circumscribed around quadrilateral $BDEC$, and $BC=15$. (20 poi... | Solution: Let $O_{1}$ be the center of the incircle of triangle $ABC$, $R$ its radius, $O_{2}$ the center of the incircle of triangle $ADE$, $r$ its radius. Denote $p_{1}$ as the semiperimeter of triangle $ABC$, and $p_{2}$ as the semiperimeter of triangle $ADE$. Then $A K=p_{2}, A K=p_{1}-BC$. Indeed, by the propertie... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,765 |
6. Find the volumes of the parts into which the plane, parallel to the diagonal $A C_{1}$ of the lateral face $A A_{1} C_{1} C$, passing through vertex $C$ and the center of symmetry of the lateral face $A A_{1} B_{1} B$, divides the regular triangular prism $A B C A_{1} B_{1} C_{1}$, if the area of the cross-section o... | # Solution:
1) Construction of the section.
In the lateral face $A A_{1} C_{1} C$, draw a line $A_{2} C$ through point $C$ parallel to $A C_{1}$, where point $A_{2}$ is the intersection of $A A_{1}$ and $A_{2} C$, and $A A_{1}=A A_{2}$.
Let $D$ be the center of symmetry of the lateral face $A A_{1} B_{1} B$. Draw a ... | \frac{112}{3},\frac{154}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,767 |
1. A student wrote a program to recolor a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 19, green - 23, blue - 53, blue - 55, purple - 83, black - 128. If ... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$, where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text { times }}(n) \ldots)-k$ is the $k$-fold composition of the function $f(n)$, which is equal to $3 n-2$ for $n \leq 17$, and equal to $|129-2 n|$ for $n \geq 18$. Let's compute and write down the fi... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,768 |
6. The base of the pyramid $TABC$ is the triangle $ABC$, all sides of which are equal to 3, and the height of the pyramid, equal to $\sqrt{3}$, coincides with the lateral edge $TA$. Find the area of the section of the pyramid by a plane that passes through the center of the sphere circumscribed around the pyramid, is p... | # Solution for Variant №6 (11th grade) | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,769 |
3. Find the set of values of the function $y=f^{[2019]}(x)$, where $f(x)=\log _{0.5}\left(\frac{\sin x}{\sin x+15}\right)$,
 | Solution: First, let's find the range of the function \( y_{1}=f(x) \). Let \( t=\sin x, t \in[-1 ; 1] \Rightarrow y_{1}=\log _{0.5}(z), \quad z=\frac{t}{t+15}=1-\frac{15}{t+15} \quad \) (the function \( z(t) \) is increasing). For \( t \in[-1 ; 1] \) we have \( z \in\left[-\frac{1}{14} ; \frac{1}{16}\right] \), but by... | [4;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,770 |
4. The lateral sides $AB$ and $CD$ of trapezoid $ABCD$ are equal to 2 and 3, respectively, and angles $A$ and $D$ are acute. The bisectors of angles $A$ and $B$ of the trapezoid intersect at point $M$, and the bisectors of angles $C$ and $D$ intersect at point $N$.
The length of segment $MN$ is 4. Find the radius of t... | Solution: Draw the angle bisectors of angles $A M$ and $B M$ of angles $A$ and $B$ respectively. Since $\angle A + \angle B = 180^{\circ}$, triangle $A B M$ is a right triangle, $\angle A M B = 90^{\circ}$. Let $L$ be the intersection point of $\mathrm{BM}$ with the base $A D$. Then triangle $A B L$ is isosceles,
^{3}-(\cos 2 x+3 a \sin x+11)^{3}=12-2 \sin ^{2} x+(3 a-2) \sin x-a^{2}-a$ has two distinct solutions on the interval $\left[-\frac{\pi}{6} ; \frac{3 \pi}{2}\right]$. Specify these solutions for each found $a$.
Solution: $\lef... | Answer: $a \in[2.5, 4), x_{1}=\arcsin (a-3), x_{2}=\pi-\arcsin (a-3)$;
$$
a \in[-5, -2), x_{1}=\arcsin (0.5 a+2), x_{2}=\pi-\arcsin (0.5 a+2)
$$ | \in[2.5,4),x_{1}=\arcsin(-3),x_{2}=\pi-\arcsin(-3);\in[-5,-2),x_{1}=\arcsin(0.5+2),x_{2}=\pi-\arcsin(0.5+2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,772 |
1. The wolf saw a roe deer several meters away from him and chased after her along a straight forest path. The wolf's jump is $22\%$ shorter than the roe deer's jump. Both animals jump at a constant speed. All the roe deer's jumps are of the same length, and the wolf's jumps are also equal to each other. There is a per... | Solution: Let x be the length of the roe deer's jump, then $0.78x$ is the length of the wolf's jump; y - the number of jumps the roe deer makes in a unit of time specified in the condition, $y\left(1+\frac{t}{100}\right)$ - the number of jumps the wolf makes in the same unit of time. The wolf will not be able to catch ... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,776 |
3. In the arithmetic progression $\left(a_{n}\right) a_{1000}=75, d=0.25$.
Calculate: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)$. | Solution: The expression in parentheses consists of several terms of the form $\frac{1}{x \cdot(x+d)}$, which can be decomposed into the sum of simpler fractions: $\frac{1}{x \cdot(x+d)}=\frac{1}{d}\left(\frac{1}{x}-\frac{1}{x+d}\right)$. Transform the original expression: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cd... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,778 |
4. In $\triangle A B C$ with $\angle B=120^{0}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn.
Segment $A_{1} B_{1}$ intersects the bisector $C C_{1}$ at point M. Find the angle $\Delta B_{1} C_{1} M$.
# | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and thus is equidistant from its sides, we get that $A_{1}$ is equid... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,779 |
5. Two tribes share one shaman. One tribe lives in the river valley, the other - on a hill, on the slope of a mountain. Due to the spread of the internet and rumors, the shaman had to lower the prices for his services. For the tribe living in the valley, the price "in furs" was reduced by $10 \%$ and "in fish" by $15 \... | Solution. For the tribe living in the valley, the price reduction in "furs" is $0.9 x$; in "fish" it is $0.85 x$. The price ratio in the local market is constant $\frac{0.9}{0.85}=\frac{18}{17}$. For the tribe on the hill, the reduction in "furs" is $0.8 x$, and let the price in "fish" be $y$. $\frac{18}{17}=\frac{0.8 ... | 24.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,780 |
6. What is the minimum value that the function $F(x ; y)=6 y+8 x-9$ can take, given that $x^{2}+y^{2}+25=10(x+y)$.
# | # Solution:
$x^{2}+y^{2}+25=10(x+y) \Leftrightarrow (x-5)^{2}+(y-5)^{2}=5^{2}$ - this is a circle with center $(5 ; 5)$ and radius 5. Let $F(x ; y)=\mathrm{M}$, then $\mathrm{M}=6 y+8 x-9$ - this is a straight line.
The condition for the minimum of the function is equivalent to the condition for the minimum of $\math... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,781 |
7. What is the minimum number of cells that need to be painted in a square with a side of 65 cells (65x65 - there are a total of 4225 cells in the square), so that among any four of its cells forming a corner (an "L" shape), there is at least one painted cell.
# | # Solution.
Shading should be done diagonally every 3rd (see fig.). Thus, $\left[\frac{N^{2}}{3}\right]$ cells will be shaded. This is the minimum possible number, as within any $3 \times 3$ square, at least three cells need to be shaded. $\left[\frac{65^{2}}{3}\right]=1408$.
=10|\mathrm{p}-3|+|2 \mathrm{p}-| \mathrm{p}+\mathrm{c}||-6 \mathrm{p}$. The coefficient of the first modulus in absolute value is greater than the sum of the other coefficients of $\mathrm{p}$. $10>2+1+6$. Therefore, on all intervals up to $\mathrm{p}=3$, the c... | -147 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,783 |
1. Ivan Ivanovich approached a source with two empty cans; one held 10 liters, and the other held 8 liters. Water from the source flowed in two streams - one stronger, the other weaker. Ivan Ivanovich simultaneously placed the cans under the streams and, when half of the smaller can was filled, he switched the cans. To... | Solution: Let x liters of water fill the larger can while 4 liters fill the smaller can. After the switch, (10-x) liters fill the larger can while 4 liters again fill the smaller can. Since the flow rates are constant, the ratio of the volumes of water filled in the same time is also constant. We can set up the equatio... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,785 |
3. In the arithmetic progression $\left(a_{n}\right) a_{1}=1, d=4$.
$$
\text { Calculate } A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}
$$
In the answer, write the smallest integer greater than $A$. | Solution: We transform the expression by multiplying the numerator and denominator of each fraction by the expression conjugate to the denominator:
$$
\begin{aligned}
& A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}= \\
& =\frac{\sqrt{a_{2}}-\... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,787 |
4. Right triangles $\triangle M D C$ and $\triangle A D K$ have a common right angle $\angle D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$ from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of $\angle A K D$ and $\angle M C D$, if $A D: C D=2: 5$.
$ satisfies the condition $f(x+1)+1=f(x)+\frac{20}{(x+1)(x+2)}$. Calculate $\frac{2019}{f(2019)}$, if $\quad f(0)=2019$. | # Solution.
Notice that
$f(x+2019)-f(x)=(f(x+2019)-f(x+2018))+(f(x+2018)-f(x+2017))+\cdots+(f(x+$ 1) $-f(x))=\frac{20}{(x+2019)(x+2020)}-1+\frac{20}{(x+2018)(x+2019)}-1+\cdots+\frac{20}{(x+1)(x+2)}-1$. Therefore, $f(2019)-f(0)=20\left(\frac{1}{2019}-\frac{1}{2020}+\ldots+1-\frac{1}{2}\right)-2019=20\left(1-\frac{1}{2... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,790 |
7. What is the minimum number of cells that need to be painted in a square with a side of 65 cells (a total of $65 \times 65$ cells, which is 4225 cells in the square), so that from any unpainted cell it is impossible to move to another unpainted cell with a knight's move in chess? | # Solution
Cells should be colored in a checkerboard pattern. Thus, $\left[\frac{N^{2}}{2}\right]$ cells will be colored. Since any "knight's move" lands on a cell of a different color, there is no move to a cell of the same color. A "knight's move" can traverse any square table larger than 4x4 such that the knight vi... | 2112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,791 |
8. Find the sum of all integer values of $\mathrm{h}$ for which the equation ||$r+h|-r|-4 r=9|r-3|$ in terms of $r$ has no more than one solution.
# | # Solution
Consider the function $\mathrm{f}(\mathrm{r})=9|\mathrm{r}-3|-|| \mathrm{r}+\mathrm{h}|-\mathrm{r}|+4 \mathrm{r}$. The coefficient of the first modulus in absolute value is greater than the sum of the other coefficients of $\mathrm{r}$. $9>1+1+4$. Therefore, on all intervals up to $r=3$, the coefficient of ... | -93 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,792 |
9. A table consisting of 1861 rows and 1861 columns is filled with natural numbers from 1 to 1861 such that each row contains all numbers from 1 to 1861. Find the sum of the numbers on the diagonal that connects the top left and bottom right corners of the table, if the table is filled symmetrically with respect to thi... | # Solution:
We will show that all numbers from 1 to 1861 are present on the diagonal. Suppose the number $a \in\{1,2,3 \ldots, 1861\}$ is not on the diagonal. Then, due to the symmetry of the table, the number $a$ appears an even number of times. On the other hand, since the number $a$ appears once in each row, the to... | 1732591 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,793 |
1. Ilya takes a triplet of numbers and transforms it according to the rule: at each step, each number changes to the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 1989 applications of this rule, if the initial triplet of numbers was $\{70 ; 61 ; 20\}$? If... | # Solution:
Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b . \quad b=70-20=50$.
Answer: 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,794 |
2. Two points are moving along a circle. When moving in the same direction, the faster point catches up with the slower point every 16 seconds. If they move in opposite directions at the same speeds, they meet every 4 seconds. It is known that when moving towards each other along the circle, the distance between the ap... | # Solution:
Let $v-$ be the speed of the faster point, and $u-$ be the speed of the slower point. According to the problem, we set up the system of equations: $\left\{\begin{array}{l}(u+v) \cdot 4=(v-u) \cdot 16 \\ (u+v) \cdot 2=64\end{array} ;\left\{\begin{array}{l}3 v=5 u \\ u+v=32\end{array} ;\left\{\begin{array}{l... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,795 |
3. In triangle $A B C$, side $B C$ is 18 cm. The perpendicular $D F$, drawn from the midpoint of side $A B$ - point $D$, intersects side $B C$ at point $F$. Find the perimeter of triangle $A F C$, if side $A C$ is $9 \, \text{cm}$. | # Solution:
Triangle $ABF (BF = AF)$ is isosceles, since $DF \perp AB$, and $D$ is the midpoint of $AB$. $P_{AFC} = AF + FC + AC = BF + FC + AC = BC + AC = 27 \text{ cm}$.
Answer: 27 cm. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,796 |
4. There are two types of containers: 27 kg and 65 kg. How many containers of the first and second type were there in total, if the cargo in the first type of containers exceeds the cargo in the second type of containers by 34 kg, and the number of 65 kg containers does not exceed 44 units?
# | # Solution:
Let $x$ be the number of containers weighing 27 kg, and $y$ be the number of containers weighing 65 kg. We get the equation $27 x - 65 y = 34$.
$27(x - 2y) - 11y = 34$, let $x - 2y = k$
$27k - 11y = 34$
$11(2k - y) + 5k = 34$, let $2k - y = t(2)$
$11t + 5k = 34$
$5(2t + k) + t = 34$, let $2t + k = n(3... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,797 |
6. A farmer presented 6 types of sour cream in barrels of $9,13,17,19,20,38$ liters at the market. On the first day, he sold sour cream from three barrels completely, and on the second day, from two more barrels completely. The volume of sour cream sold on the first day was twice the volume of sour cream sold on the se... | # Solution:
A total of 116 liters were delivered; $116=3 \cdot 38+2$; therefore, the unsold barrel, when divided by 3, gives a remainder of 2.
1) If it is 17 liters, then $116-17=99$, then on the second day, one third of 99 liters were sold, which is $33=13+20$, then on the first day, barrels with capacities of $9, 1... | 66 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,799 |
7. In rectangle $A B C D$, point $E$ is the midpoint of side $C D$. On side $B C$, point $F$ is taken such that angle $A E F$ is a right angle. Find the length of segment $F C$, if $A F=9, B F=6$. | # Solution:
Let the lines $A E$ and $B C$ intersect at point $K$, then triangles $A E D$ and $K C E$ are congruent ( $\angle A E D = \angle C E D$ as vertical angles, $C E = E D, \angle A D ... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,800 |
8. In triangle $A B C$ with $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find $\angle C B M$.
# | # Solution:
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and therefore is equidistant from its sides.
We obtain that $A_{1}$ i... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,801 |
9. For what values of the parameter $\boldsymbol{a}$ does the equation $|f(x)-5|=p(x)$, where $f(x)=\left|\frac{x^{2}-10 x+25}{x-5}-\frac{x^{2}-3 x}{3-x}\right|$, $p(x)=a \quad$ have three solutions? If there is more than one value of the parameter, indicate their product in the answer. | # Solution:
Simplify $f(x)=\left|\frac{x^{2}-10 x+25}{x-5}-\frac{x^{2}-3 x}{3-x}\right|$, we get $f(x)=$ $|2 x-5|$, where $x \neq 5, x \neq 3$.
Solve the equation || $2 x-5|-5|=a$, where $x \neq 5, x \neq 3$ graphically in the system $x O a$.
The equation has three solutions when $a=4, a=5$.
The product is 20.
Ans... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,802 |
1. The clock shows $00:00$, at which the hour and minute hands of the clock coincide. Considering this coincidence as number 0, determine after what interval of time (in minutes) they will coincide for the 21st time. If the answer is not an integer, round the result to the hundredths according to rounding rules. | # Solution:
The minute hand passes 1 circle in an hour, while the hour hand passes $1 / 12$ of a circle, so their closing speed is $11 / 12$ of a circle per hour, and one closing takes $1 /(11 / 12)=12 / 11$ hours or 720/11 minutes. 19 closings is $21 * 720 / 11=15120 / 11=1374.55$ minutes.
Answer: 1374.55. | 1374.55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,803 |
2. To get to the Solovyov family's cottage, you first need to walk 3 km along the highway and then 2 km along a path. Upon arriving at the station, Mom called her son Vasya at the cottage and asked him to meet her on a bicycle. They started moving towards each other at the same time. Mom walks at a constant speed of 4 ... | # Solution:
Vasya rode 2 kilometers along the path in $2: 20=0.1$ hours. During this time, his mother walked $4 \cdot 0.1=0.4$ km. After this, Vasya and his mother started moving towards each other on the road at constant speeds. $3-0.4=2.6$ km. We can set up the equation $(22+4) \cdot t=2.6$; from which $t=0.1$ hours... | 800 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,804 |
3. On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $N$ are marked respectively such that $\angle CMA = \angle ANC$. Segments $MC$ and $AN$ intersect at point $O$, and $ON = OM$. Find $BC$, if $AM = 3 \, \text{cm}, BM = 4 \, \text{cm}$. | # Solution:
Triangles $A M O$ and $C N O$ are congruent ($\angle A M O=\angle C N O, O N=O M, \angle M O A=\angle N O C$, as vertical angles). Therefore, $A M=N C, \angle M A O=\angle N C O$... | 8\, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,805 |
4. If a two-digit number is decreased by 54, the result is a number written with the same digits but in reverse order. In the answer, specify the arithmetic mean of the resulting numbers.
# | # Solution:
$\overline{x y}=10 x+y-$ the original two-digit number, then $\overline{y x}=10 y+x$ - the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+54$.
From the equation, it is clear that the two-digit number is greater than 54. Let's start the investigation with the t... | 82 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,806 |
5. During the shooting practice, each soldier fired 10 times. One of them successfully completed the task and scored 90 points. How many times did he score 9 points, if there were 4 tens, and the results of the other hits were sevens, eights, and nines. There were no misses at all. | # Solution:
Since the soldier scored 90 points and 40 of them were scored in 4 shots, he scored 50 points with the remaining 6 shots. As the soldier only hit the 7, 8, and 9, let's assume he scored 24 points with three shots (one each in 7, 8, and 9). Then, for the remaining three shots, he scored 26 points, which is ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,807 |
6. Players divide the chips. The first player takes $m$ chips and a sixth of the remainder; the second $-2 m$ chips and a sixth of the new remainder; the third $-3 m$ chips and a sixth of the new remainder, and so on. It turned out that the chips were divided equally in this way. How many players were there? | # Solution:
Let $x$ be the number of players, and $y$ be the number of chips each has. The last player took $y=m x$ chips, with no remainder, otherwise the condition of equal distribution would not be met. The second-to-last player took $y=(x-$ 1)m plus one-sixth of the remainder, and $\frac{5}{6}$ of the remainder eq... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,808 |
7. In rectangle $A B C D$, point $E$ is located on diagonal $A C$ such that $B C=E C$, point $M$ is on side $B C$ such that $E M=M C$. Find the length of segment $M C$, if $B M=6, A E=3$. If the answer is not an integer, round the result to tenths according to rounding rules. | # Solution:
Draw $A F$ parallel to $B E$ (point $F$ lies on line $B C$), then $\angle C B E=\angle C F A, \angle C E B=$ $\angle C A F$. Considering that $B C=C E$, we get that triangle $F C ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,809 |
8. Given triangle $A B C . \angle A=\alpha, \angle B=\beta$. Lines $O_{1} O_{2}, O_{2} O_{3}, O_{1} O_{3}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point O is the center of the inscribed circle of triangle $A B C$. Find the angle between the lines $\mathrm{CO}_{1}$ and $\mat... | # Solution:
We will prove that the bisectors of two external angles and one internal angle intersect at one point. Let \( O_{1} G, O_{1} H, O_{1} F \) be the perpendiculars to \( B C, A C \)... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,810 |
9. For what values of the parameter a does the equation $f(x)=p(x)$ have one solution, if $f(x)=$ $\left|\frac{2 x^{3}-5 x^{2}-2 x+5}{(1.5 x-3)^{2}-(0.5 x-2)^{2}}\right|, \quad p(x)=|2 x+5|+a$. If there are more than one value of the parameter, then in the answer, specify their sum.
# | # Solution:
Simplify $f(x)=\left|\frac{2 x^{3}-5 x^{2}-2 x+5}{(1.5 x-3)^{2}-(0.5 x-2)^{2}}\right|$, we get
$$
f(x)=|x+1|, \text { where } x \neq 1, x \neq 2.5 \text {. }
$$
Solve the equation $|x+1|=|5+2 x|+a$, where $x \neq 1, x \neq 2.5$ graphically in the system $x O a$.
1) $\left\{\begin{array}{c}x<-2.5 \\ -x-1... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,811 |
1. In how many ways can two knights - a white one and a black one - be placed on a chessboard consisting of $16 \times 16$ cells so that they threaten each other? (A knight moves in an "L" shape, i.e., it can move to one of
$ rectangular coordinate system to the board. Denote the coordinates of the two knights as $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$, where $x_{k}, y_{k} \in\{1,2, \ldots, n\}, k=1,2$. The knights threaten each other if 1) $\left|x_{1}-x_{2}\right|=1, \qua... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,812 |
2. Vasya and Petya, participating in a school sports and entertainment game, need to cover a distance of 3 km as quickly as possible with only one pair of roller skates between them. They start simultaneously, one just running, the other running on roller skates. At any time, the one running on roller skates can leave ... | Solution: The entire distance is divided into several stages, where one of the schoolchildren runs without roller skates, and the other on roller skates. Let $x$ be the sum of the lengths of the stages that Vasya runs on roller skates, $t_{1}$ be the time he spends running the entire distance, and $t_{2}$ be the time t... | 0.5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,813 |
3. Find the smallest natural number that has exactly 70 natural divisors (including 1 and the number itself).
$(16$ points $)$ | Solution: Let $n$ be the required natural number, $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$ - the prime factorization of the number $n$. Any natural divisor of this number has the form $d=p_{1}^{h_{1}} \cdot p_{2}^{l_{2}} \cdot \ldots \cdot p_{m}^{l_{m}}$, where $l_{i} \in\left\{0,1, \ldots... | 25920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,814 |
5. Specify all values of $a$ for which the system of equations $3(x-a)^{2}+y=2-a$, $y^{2}+\left(\frac{x-2}{|x|-2}\right)^{2}=1$ has at least one solution, and solve it for each $a$. | Solution. Transform the second equation: $y^{2}=\frac{x^{2}-4|x|+4-x^{2}+4 x-4}{(|x|-2)^{2}} ; y^{2}=-\frac{4(|x|-x)}{(|x|-2)^{2}}$.
For $x<0$ the equation has no solutions, as the right side is less than zero; for $x \geq 0$ it is equivalent to the system: $x \geq 0, x \neq 2, y=0$. Substituting $y=0$ into the first ... | \frac{5}{3},solution:(x,y)=(\frac{4}{3},0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,815 |
6. Two circles touch each other and the sides of triangle $A B C$. The first circle with radius $\frac{1}{18}$ touches sides $A B$ and $A C$ at points $L$ and $K$, the second circle with radius $\frac{2}{9}$ touches sides $A C$ and $B C$ at points $N$ and $M$. Find the area of triangle $A B C$, if $A L=\frac{1}{9}, C M... | Solution: $a=\frac{1}{9}, \quad b=\frac{1}{6}, \quad r=\frac{1}{18}, \quad R=\frac{2}{9}$ $K N=2 \sqrt{r R}=\frac{2}{9}, \quad A C=a+K N+b=\frac{1}{2}$
$\operatorname{tg} \frac{A}{2}=\frac{r}{a}, \quad \operatorname{tg} \frac{C}{2}=\frac{R}{b}$
$\cos ^{2} \frac{A}{2}=\frac{a^{2}}{a^{2}+r^{2}}, \quad \cos ^{2} \frac{C... | \frac{3}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,817 |
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