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 |
|---|---|---|---|---|---|---|---|---|---|
6. Find all pairs of prime numbers $p$ and $q$ such that $p^{2}+5 p q+4 q^{2}$ is a square of a natural number. | Answer: $(13,3),(7,5),(5,11)$
Solution. Let $p^{2}+5 p q+4 q^{2}=n^{2}$ for some natural number $n$. Then
$$
p q=n^{2}-(p+2 q)^{2}=(n-p-2 q)(n+p+2 q)
$$
The left side can be factored into the product of two numbers, one of which is an integer and the other a natural number, in four ways: $p q \cdot 1, p \cdot q, q \... | (13,3),(7,5),(5,11) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,090 |
1. Prove that for any real numbers $a$ and $b$, the equation
$$
\left(a^{6}-b^{6}\right) x^{2}+2\left(a^{5}-b^{5}\right) x+\left(a^{4}-b^{4}\right)=0
$$
has a solution. | Solution. If $a=b$, then any $x$ is a solution to the equation. If $a \neq b$, we need to prove that the quadratic trinomial has a root. For this, it is sufficient to check that its discriminant is non-negative. The discriminant, divided by 4, is
$$
\left(a^{5}-b^{5}\right)^{2}-\left(a^{4}-b^{4}\right)\left(a^{6}-b^{6... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,091 |
2. On an island, there live liars and knights. Knights always tell the truth, and liars always lie. Each islander knows whether every other islander is a knight or a liar. One day, 28 islanders met. Two of them said: "Exactly two of us are liars," then four of the others said: "Exactly four of us are liars," then eight... | Answer: 14 or 28.
Solution. Let the last statement be true. Then all who said it are knights, and there must be exactly 14 liars. Therefore, all who said the other statements are liars, and such a situation is possible. Let the last statement be false. Then those who said it are liars and there are no fewer than 14 li... | 14or28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,092 |
3. The sum of non-negative numbers $a, b$, and $c$ is 3. Find the maximum value of the expression $a b + b c + 2 c a$. | Answer: $\frac{9}{2}$
First solution. By the condition
$9=(a+b+c)^{2}=2(a b+b c+c a)+a^{2}+b^{2}+c^{2} \geqslant 2(a b+b c+c a)+b^{2}+2 c a \geqslant 2(a b+b c+2 c a)$.
Equality is achieved when $b=0$ and $a=c=\frac{3}{2}$. Therefore, the maximum value will be $\frac{9}{2}$.
Second solution. Notice that
$a b+b c+2... | \frac{9}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,093 |
4. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $K$ and $L$. Line $\ell$ intersects circle $\omega_{1}$ at points $A$ and $C$, and circle $\omega_{2}$ at points $B$ and $D$, with the points appearing on line $\ell$ in alphabetical order. Let $P$ and $Q$ be the projections of points $B$ and $C$ onto line $K... | Solution. Let the intersection point of lines $K L$ and $A D$ be denoted as $M$. From the cyclic quadrilateral $B K D L$, we conclude that $\angle B K L = \angle B D L$. Therefore, triangles $B K M$ and $L D M$ are similar, and thus $\frac{B M}{K M} = \frac{L M}{D M}$. From the cyclic quadrilateral $A K C L$, we conclu... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,094 |
5. Given a $2021 \times 2021$ grid. Petya and Vasya are playing the following game. They take turns placing chips in free cells of the board. The player who, after their move, ensures that every $3 \times 5$ and $5 \times 3$ rectangle contains a chip, wins. Petya starts. Which player can guarantee a win regardless of t... | Answer: Petya
Solution. Petya should move in such a way that Vasya cannot win on the next move. We will show that he can always achieve this. If there is a pair of non-intersecting empty rectangles $3 \times 5$ on the board, Petya should place a chip so that they remain even after his move. If such a move is impossibl... | Petya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,095 |
6. Find all natural numbers $n$ such that the number $2^{n}+n^{2}+25$ is a cube of a prime number. | Answer: $n=6$
Solution. Let $2^{n}+n^{2}+25=p^{3}$ for some prime number $p$. Since $p>3$, $p$ is an odd prime number. Then $n$ is an even number, and $2^{n}$ gives a remainder of 1 when divided by three. If $n$ is not divisible by three, then $n^{2}$ gives a remainder of 1 when divided by three, and then $2^{n}+n^{2}... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,096 |
1. Find all values of $a$ for which the quadratic trinomials $x^{2}+2 x+a$ and $x^{2}+a x+2=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial. | Answer: $a=-4$
Solution. If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $2^{2}-2 a=a^{2}-2 \cdot 2$. Thus, we need to solve the... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,097 |
2. Each of the islanders is either a knight, who always tells the truth, or a liar, who always lies (both types are present on the island). Every resident knows whether each other is a knight or a liar. Some of the islanders stated that there is an even number of knights on the island, while all the others claimed that... | Answer: No
Solution. Since 2021 is an odd number, on the island there is either an odd number of knights and an even number of liars, or an odd number of liars and an even number of knights. The first case is impossible, since both statements are false and, therefore, no knight could have made either the first or the ... | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,098 |
3. For arbitrary real numbers $a$ and $b(b \neq 0)$, find the minimum value of the expression $a^{2}+b^{2}+\frac{a}{b}+\frac{1}{b^{2}}$. | Answer: $\sqrt{3}$
Solution. Since
$$
a^{2}+b^{2}+\frac{a}{b}+\frac{1}{b^{2}}=\left(a+\frac{1}{2 b}\right)^{2}+b^{2}+\frac{3}{4 b^{2}} \geqslant b^{2}+\frac{3}{4 b^{2}} \geqslant 2 \sqrt{b^{2} \cdot \frac{3}{4 b^{2}}}=\sqrt{3}
$$
the expression $a^{2}+b^{2}+\frac{a}{b}+\frac{1}{b^{2}}$ is always no less than $\sqrt{... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,099 |
4. Points $B_{1}$ and $C_{1}$ are the midpoints of sides $A C$ and $A B$ of triangle $A B C$. Circles $\omega_{1}$ and $\omega_{2}$ are constructed on sides $A B$ and $A C$ as diameters. Let $D$ be the point of intersection of line $B_{1} C_{1}$ with circle $\omega_{1}$, lying on the other side of $C$ relative to line ... | Solution. Let $M$ be the midpoint of segment $DE$, and lines $AM$ and $BC$ intersect at point $L$. Since $B_1C_1$ is the midline of triangle $ABC$, lines $B_1C_1$ and $BC$ are parallel. Therefore, $B_1M$ is the midline of triangle $ACL$, and in particular, $M$ is the midpoint of segment $AL$. Thus, the diagonals of qua... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,100 |
5. A chip is placed on the central cell of an $11 \times 11$ board. Petya and Vasya play the following game. On each of his turns, Petya moves the chip one cell vertically or horizontally. On each of his turns, Vasya places a wall on one of the sides of any cell. Petya cannot move the chip through a wall. The players t... | Answer: No
Solution. With his first move, Vasya chooses one of the corner cells and builds a wall along one of its two outer sides. With his next three moves, he builds similar walls for the three remaining corner cells. In these four moves, Petya will not have time to reach the edge of the board. If, with some move, ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,101 |
6. Prove that there are infinitely many natural numbers $n$ such that the number of distinct odd prime divisors of the number $n(n+3)$ is divisible by three. | Solution. Let $a_{n}$ denote the number of distinct odd prime divisors of the number $n(n+3)$. Suppose that there are only finitely many numbers for which $a_{n}$ is divisible by three. Then for some $m$, when $n \geqslant m$, the number $a_{n}$ will not be divisible by three.
Consider the product
$$
n(n+1)(n+3)(n+4)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,102 |
1. Find all values of $a$ for which the quadratic trinomials $x^{2}-6 x+4 a$ and $x^{2}+a x+6=0$ have two roots each, and the sum of the squares of the roots of the first trinomial is equal to the sum of the squares of the roots of the second trinomial. | Answer: $a=-12$
If $x_{1}$ and $x_{2}$ are the roots of the quadratic polynomial $x^{2}+p x+q$, then by Vieta's theorem
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=p^{2}-2 q
$$
Therefore, we need to find such numbers $a$ for which $6^{2}-8 a=a^{2}-2 \cdot 6$. Thus, we need to solve the equation... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,103 |
2. On an island, there live liars and knights, a total of 2021 people. Knights always tell the truth, and liars always lie. Every resident of the island knows whether each person is a knight or a liar. One fine day, all the residents of the island lined up. After that, each resident of the island stated: "The number of... | Answer: 1010
Solution. The rightmost islander cannot be telling the truth, since there is no one to the left of him, and in particular, there are no liars. Therefore, he is a liar. The leftmost islander cannot be lying, since there is at least one liar to the left of him, and there is no one to the right of him, and i... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,104 |
3. For arbitrary real numbers $a$ and $b (a \neq 0)$, find the minimum value of the expression $\frac{1}{a^{2}}+2 a^{2}+3 b^{2}+4 a b$. | Answer: $\sqrt{\frac{8}{3}}$
Since
$$
\frac{1}{a^{2}}+2 a^{2}+3 b^{2}+4 a b=\left(\sqrt{3} b+\frac{2 a}{\sqrt{3}}\right)^{2}+\frac{2 a^{2}}{3}+\frac{1}{a^{2}} \geqslant \frac{2 a^{2}}{3}+\frac{1}{a^{2}} \geqslant 2 \sqrt{\frac{2 a^{2}}{3} \cdot \frac{1}{a^{2}}}=\sqrt{\frac{8}{3}}
$$
the expression $\frac{1}{a^{2}}+2... | \sqrt{\frac{8}{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,105 |
5. Given a grid board $3 \times 2021$ (3 cells vertically and 2021 cells horizontally) and an unlimited supply of cardboard strips $1 \times 3$. Petya and Vasya play the following game. They take turns placing the strips on the board (cell by cell) without overlapping, one strip per turn. Petya places his strips horizo... | # Answer: Petya
Solution. Since there are a total of $3 \cdot 2021$ cells on the board, both players together will make no more than 2021 moves. We will show how Petya should act to win. He will mentally divide the board into 673 squares of $3 \times 3$ and one rectangle of $2 \times 3$. With his first moves, Petya wi... | Petya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,107 |
6. Find all such prime numbers $p$ and $q$ that $p^{q+1}+q^{p+1}$ is a square of a natural number. | Answer: $p=q=2$
Solution. It is clear that $p=q=2$ works. Let $p$ be odd and $p^{q+1}+q^{p+1}=n^{2}$. Then
$$
p=2 k-1 \quad \text { and } \quad p^{q+1}=n^{2}-q^{2 k}=\left(n-q^{k}\right)\left(n+q^{k}\right) .
$$
Let the greatest common divisor of the numbers $n-q^{k}$ and $n+q^{k}$ be denoted by $d$. Then $d$ is a p... | p=q=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,108 |
1. (15 points) If all the trees on one hectare of forest are cut down, then 100 cubic meters of boards can be produced. Assuming that all the trees in the forest are the same, are evenly distributed, and that 0.4 m$^{3}$ of boards can be obtained from each tree, determine the area in square meters on which one tree gro... | Answer: 40.
Solution. Let's find out how many trees grow on one hectare of forest: $\frac{100 \mathrm{m}^{3}}{0.4 \mathrm{M}^{3}}=250$. Let's recall that 1 ha is $100 \mathrm{~m} \times 100 \mathrm{~m}=10000 \mathrm{~m}^{2}$. Thus, one tree grows on $\frac{10000 \mathrm{M}^{2}}{250}=40 \mathrm{M}^{2}$. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,109 |
2. (15 points) Witch Gingema enchanted the wall clock so that the minute hand moves in the correct direction for five minutes, then three minutes in the opposite direction, then again five minutes in the correct direction, and so on. How many minutes will the hand show after 2022 minutes from the moment it pointed exac... | Answer: 28.
Solution. In 8 minutes of magical time, the hand will move 2 minutes in the clockwise direction. Therefore, in 2022 minutes, it will complete 252 full eight-minute cycles and have 6 minutes left. Since $252 \cdot 2=60 \cdot 8+24$, the hand will travel 8 full circles, plus 24 minutes, and then 5 minutes in ... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,110 |
3. (35 points) In country $B$, there are 4 southern cities and 5 northern cities. Transportation links connect each southern city with each northern city in both directions; there is no transportation between other pairs of cities. A one-way ticket from any city $A$ to any city $B$ costs $N$ rubles. A round-trip ticket... | Answer: $4 \times 1.6 N$ rubles.
Solution. Since the southern cities are not connected by transportation, to visit all of them and return, it will require $2 \times 4$ trips. Any two trips cost at least $1.6 N$ rubles in total, so the total expenses will be at least $4 \times 1.6 N$ rubles.
Let's provide an example o... | 4\times1.6N | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,111 |
4. (35 points) Magician David Copperfield secretly told the same natural number to schoolchildren Misha and Masha. Masha multiplied this number by 3 several times, while Misha added 2500 to this number the same number of times. After the children compared the results of their calculations, it turned out that they had t... | Answer: $1250, 625, 125$.
Solution. Let $x$ be Copperfield's number, and $k$ be the number of times Masha multiplied the number by 3. Then the equation $x + 2500k = x \cdot 3^{k}$ holds. From this, we get $x = \frac{2500k}{3^{k} - 1}$. By substitution, we find that $k=1, 2, 4$ work, corresponding to the values $x_{1}=... | 1250,625,125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,112 |
5. (50 points) Vasya thought of a natural number not greater than 100. Fedya asks Vasya questions of the form “What is the quotient of the division of the thought number by $m$?”, where $m$ is a natural number not less than 2, and Fedya chooses it himself. What is the minimum number of such questions Fedya can guarante... | Answer: 2.
Solution. Let Vasya think of a number $V$. Then $V=k \cdot m+d$, where $d$ can take values from 0 to $m-1$ inclusive, and Vasya's answer to Fyodor's question will be the number $k$. Since by the condition $m>1$, there are at least two variants for the remainder $d$ (the exception is the case when Vasya thou... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,113 |
6. (50 points) In the $B$ football championship, 16 teams participate. According to the rules of the championship, in each round, all participants are divided into pairs, and in each pair, a match is played until one of the teams wins. The winning team advances to the next round, while the losing team is eliminated fro... | Answer: Not always. It is impossible if the teams of equal strength are the strongest among the participants.
Solution. Note that if in some round only teams of different strength participate, then the winner of each match in this and all subsequent rounds can be predicted. This happens because, with any pairing, the ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,114 |
7. (50 points) Two multiplication examples of two natural numbers were written on the board, and in one of the examples, the numbers were the same. The troublemaker Petya erased both the original numbers and almost all the digits of the multiplication results, leaving only the last three digits: ... 689 and ... 759. Ca... | Answer: It could be 689.
First solution. Statement: at least one of the two last digits of the square of a natural number is necessarily even. Then the second number is eliminated. Example for the first number: $133 \cdot 133=17689$.
Proof of the statement: the statement is true for even numbers, let's consider the s... | 689 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,115 |
5. How many ways are there to color the cells of a $10 \times 10$ board in blue and green such that in each $2 \times 2$ square there are two blue and two green cells? | Answer: $2^{11}-2=2046$.
Solution. Note that if the colors of three cells in a $2 \times 2$ square are known, the color of the remaining cell is uniquely determined. Also, the colors of the two remaining cells are uniquely determined if there are two adjacent cells of the same color in a $2 \times 2$ square.
Consider... | 2046 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,116 |
1. (10 points) Evaluate the value of the ratio of numbers $A$ and $B$, if
$$
\begin{aligned}
& A=1 \cdot 2 \cdot 7+2 \cdot 4 \cdot 14+3 \cdot 6 \cdot 21+4 \cdot 8 \cdot 28 \\
& B=1 \cdot 3 \cdot 5+2 \cdot 6 \cdot 10+3 \cdot 9 \cdot 15+4 \cdot 12 \cdot 20
\end{aligned}
$$
a) $0<A / B<1$;
b) $1<A / B<10$;
c) $10<A / ... | Answer: a).
Solution. It is not hard to see that in both number $A$ and number $B$, the second term is twice the first, the third is three times, and the fourth is four times. Thus, the ratio of $A$ to $B$ is equal to the ratio of their first terms. This ratio is $\frac{14}{15}$, which is less than 1. | \frac{14}{15} | Algebra | MCQ | Yes | Yes | olympiads | false | 8,117 |
2. (10 points) Determine how many prime divisors the number 17! - 15! has (here $n!=1 \cdot 2 \cdots n-$ is the factorial of the number $n$, i.e., the product of all natural numbers from 1 to n inclusive).
a) 6 ;
b) 7 ;
c) 8 ;
d) none of the listed answers is correct.
Mark the correct answer, no need to provide the... | Answer: b).
Solution. Transform: $17!-15!=15!(16 \cdot 17-1)=15! \cdot 271$. The prime divisors of this number are $2,3,5,7,11,13,271$. | b) | Number Theory | MCQ | Yes | Yes | olympiads | false | 8,118 |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second digit does not have to be different. For example, 101 and 222 have this form, while 220 and 123 do not. Similarly, we define the form of the number $\overline{\overline{b a b c}}$. How many nu... | Answer: 180.
Solution. Numbers divisible by $5$ are those ending in 0 or 5, so we have two options for $c$. For $a$, we have 9 options, as the number cannot start with zero, and the value of $b$ can be anything.
Thus, we get that the total number of such numbers is $2 \cdot 9 \cdot 10=180$. | 180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,119 |
4. (20 points) Nезнayka came up with a password for his email consisting of five characters. Deciding to check the reliability of this password, he calculated all possible combinations that can be formed from these five characters. In the end, he got 20 different combinations. Is a password with this number of combinat... | For example, error.
Solution. The maximum number of different combinations that can be formed from 5 symbols equals $5!(n!=1 \cdot 2 \cdots n$ - the factorial of number $n$, i.e., the product of all natural numbers from 1 to $n$ inclusive). This number is obtained when all 5 symbols are distinct.
Suppose among the gi... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,120 |
6. (30 points) $S M$ is the angle bisector in triangle $S Q T$. Point $O$ on side $S T$ is such that angle $O Q T$ is equal to the sum of angles $Q T S$ and $Q S T$. Prove that $O M$ is the angle bisector of angle $Q O T$.
 Let $R E$ be the bisector of triangle $R S T$. Point $D$ on side $R S$ is such that $E D \| R T, F$ is the intersection point of $T D$ and $R E$. Prove that if $S D = R T$, then $T E = T F$.
 Given a quadrilateral ELMI. It is known that the sum of angles LME and $MEI$ is 180 degrees and $EL = EI + LM$. Prove that the sum of angles LEM and EMI is equal to angle MIE. | Solution. Mark a point $K$ on the ray $L M$ beyond point $M$ such that $M K=I E$, and consider triangles $IEM$ and $K E M$. By construction, $\angle L M E + \angle E M K = 180^{\circ}$, and by the problem's condition, $\angle L M E + \angle M E I = 180^{\circ}$, hence $\angle K M E = \angle M E I$; sides $M K$ and $E I... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,124 |
9. (40 points) Vasya and Misha write down natural numbers on the board and compute their squares. At some point, it turns out that for three numbers $n, k, l$ the equality $n^{2}+k^{2}=2 l^{2}$ holds. Prove that the number
$$
\frac{(2 l-n-k)(2 l-n+k)}{2}
$$
is a perfect square. | Solution. Transform the numerator:
$$
(2 l-n-k)(2 l-n+k)=(2 l-n)^{2}-k^{2}=4 l^{2}-4 l n+n^{2}-k^{2}
$$
By the condition $k^{2}=2 l^{2}-n^{2}$, we obtain
$$
4 l^{2}-4 l n+n^{2}-\left(2 l^{2}-n^{2}\right)=2 l^{2}-4 l n+2 n^{2}=2(l-n)^{2}
$$
Thus, the original expression is the square of the number $l-n$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,125 |
11. (40 points) Five different positive numbers are written on the board. It turns out that for any three different numbers $a, b$, and $c$ on the board, the number $ab + bc + ca$ is rational. Prove that the ratio of any two numbers on the board is rational. | Solution. Let $a, b, c$ and $d$ be some four numbers written on the board. Then $a b + b c + c a$ and $a b + b d + d a$ are rational, hence their difference $b c + c a - b d - d a = (a + b)(c - d)$ is also rational.
Similarly, the difference of the rational numbers $a c + c d + d a$ and $b c + c d + d b$, which is $a ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,127 |
3. A circle $\omega_{1}$ is circumscribed around quadrilateral $A B C D$. A circle $\omega_{2}$ is drawn through points $A$ and $B$, intersecting ray $D B$ at point $E \neq B$. Ray $C A$ intersects circle $\omega_{2}$ at point $F \neq A$. Prove that if the tangent to circle $\omega_{1}$ at point $C$ is parallel to line... | Solution. Let $G \neq A$ be the intersection point of line $D A$ with circle $\omega_{2}$. Mark a point $X$ on the tangent to $\omega_{1}$ at point $C$ such that $\angle D C X \leqslant 90^{\circ}$, and a point $Y$ on the tangent to $\omega_{2}$ at point $F$ such that $\angle G F Y \leqslant 90^{\circ}$.
By the proper... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,128 |
3. The circle $\omega$ is circumscribed around an acute-angled triangle $ABC$. The tangent to the circle $\omega$ at point $C$ intersects the line $AB$ at point $K$. The point $M$ is the midpoint of the segment $CK$. The line $BM$ intersects the circle $\omega$ again at point $L$, and the line $KL$ intersects the circl... | Solution. Since $K C$ is tangent to the circle $\omega, \angle M C B=$ $\angle M L C$ and, therefore, triangles $M C B$ and $M L C$ are similar. Hence, $\frac{M B}{M C}=$ $\frac{M C}{M L}$. Consequently, $\frac{M B}{M K}=\frac{M K}{M L}$. This means that triangles $M K B$ and $M L K$ are similar. Thus,
$$
\angle C K N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,129 |
1. Petya and Vasya simultaneously set off on scooters towards each other. Exactly in the middle between them is a bridge. The road from Petya to the bridge is paved, while the road from Vasya to the bridge is dirt. It is known that they travel at the same speed on the dirt road, and on the paved road, Petya moves 3 tim... | Answer: in 2 hours.
Solution. Let the distance to the bridge from the starting positions be $x$. Petya reached the bridge in one hour. Vasya was traveling three times slower, so in one hour he covered the distance $x / 3$. At this moment, the distance between Petya and Vasya became $2 x / 3$, and then they moved at th... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,130 |
2. Given the quadratic trinomial $2 x^{2}-x-36$. Find all integer $x$ for which the values of this trinomial are equal to the square of a prime number. | Answer: $x=5$ or $x=13$.
Solution. We will find all pairs $(x, p)$ satisfying the equation $2 x^{2}-x-36=p^{2}$, where $x$ is an integer and $p$ is a prime. Factorize the quadratic polynomial:
$$
p^{2}=2 x^{2}-x-36=(2 x-9)(x+4) \text {. }
$$
If both brackets on the right-hand side are divisible by $p$, then $p$ will... | 5or13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,131 |
3. Positive numbers $a, b$ and $c$ satisfy the condition $a b c(a+b+c)=a b+b c+c a$. Prove the inequality $5(a+b+c) \geqslant 7+8 a b c$.
---
The translation is provided as requested, maintaining the original formatting and structure. | Solution. Since
$$
a b c=\frac{a b+b c+c a}{a+b+c} \leqslant \frac{(a+b+c)^{2}}{3(a+b+c)}=\frac{a+b+c}{3}
$$
it suffices to prove the inequality $5(a+b+c) \geqslant 7+\frac{8}{3}(a+b+c)$, which is equivalent to $a+b+c \geqslant 3$. But this is true, since $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{9}{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,132 |
4. Masha has 1000 beads in 50 different colors, 20 beads of each color. What is the smallest $n$ such that for any way of making a necklace from all the beads, one can choose $n$ consecutive beads, among which there are beads of 25 different colors? | Answer: $n=462$.
First solution. Let's call a segment of the necklace of length $m$ a set of $m$ consecutive beads. If the beads in the necklace are arranged in groups of 20 of the same color, then a segment of length 461 cannot contain more than 24 different colors. Therefore, $n \geqslant 462$. Consider a segment of... | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,133 |
5. Points $A_{1}$ and $B_{1}$ are the midpoints of sides $B C$ and $A C$ of an acute triangle $A B C$, and point $M$ is the midpoint of segment $A_{1} B_{1}$. Point $H$ is the foot of the altitude dropped from vertex $C$ to side $A B$. Circles are drawn through point $M$, touching sides $B C$ and $A C$ at points $A_{1}... | Solution. Let $N^{\prime}$ be the point symmetric to point $N$ with respect to the line $A_{1} B_{1}$, and the line $M N^{\prime}$ intersects the lines $A C$ and $B C$ at points $C_{1}$ and $C_{2}$, respectively. Since the angle between the tangent and the secant is equal to the inscribed angle subtended by the secant,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,134 |
1. Kolya went to the store in the neighboring village on an electric scooter at a speed of 10 km/h. Having traveled exactly one third of the entire distance, he realized that with the previous speed, he would arrive exactly when the store closes, and he doubled his speed. But when he had traveled exactly $2 / 3$ of the... | Answer: $6 \frac{2}{3}$ km/h.
Solution. Let the distance to the village be $3 x$ km. If Kolya had traveled the entire time at a speed of $10 \mathrm{km} /$ h, he would have spent $3 x / 10$ hours on the road. This is the time from his departure to the store's closing. The first third of the distance Kolya traveled in ... | 6\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,136 |
2. Find all integers $a$ for which the quadratic trinomial $x^{2}+a x+2 a$ has two distinct integer roots. | Answer: $a=-1$ and $a=9$.
First solution. Let $u$ and $v$ be the roots of the quadratic polynomial, with $u|v|$. Moreover, $a>0$ and
$$
|u| \cdot|v|=2 a, \quad|u|+|v|=a \Rightarrow \frac{1}{|u|}+\frac{1}{|v|}=\frac{1}{2}
$$
The last equality is possible only if $|v| \geqslant 3$. However, if $|v| \geqslant 4$, then ... | =-1=9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,137 |
3. Positive numbers $a, b$ and $c$ satisfy the condition $a b c(a+b+c)=3$. Prove the inequality $(a+b)(b+c)(c+a) \geqslant 8$. | First solution. Note that
$$
(a+b)(b+c)=b(a+b+c)+a c=\frac{3}{a c}+a c \geqslant \frac{2}{a c}+2
$$
since $x+\frac{1}{x} \geqslant 2$ for $x>0$. Therefore,
$$
(a+b)(b+c)(c+a) \geqslant\left(\frac{2}{a c}+2\right)(c+a) \geqslant 2 \sqrt{2 \cdot \frac{2}{a c}} \cdot 2 \sqrt{c a}=8
$$
Second solution. Note that
$$
(a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,138 |
4. What is the minimum number of chips that can be placed in the cells of a $99 \times 99$ table so that in each $4 \times 4$ square there are at least eight chips? | Answer: 4801.
Solution. Add a row and a column with number 100 to the table. Place a chip in each of their cells. Divide the expanded table into 625 squares $4 \times 4$. In each square, there can be no more than eight empty cells, so there are no more than 5000 empty cells in the entire table. Therefore, the total nu... | 4801 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,139 |
5. The diagonals of quadrilateral $A B C D$ intersect at point $O$. Diagonal $A C$ is the bisector of angle $\angle B A D$, point $M$ is the midpoint of side $B C$, and point $N$ is the midpoint of segment $D O$. Prove that quadrilateral $A B C D$ is cyclic if and only if quadrilateral $A B M N$ is cyclic.
 / k=n!/ k+1$. If the numbers $m_{k}$ and $n!$ are coprime, then take any prime divisor $p_{k}$ of the number $m_{k}$. Then $p_{k}>n$ and $p_{k}$ does not divide any of the numb... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,141 |
1. Boy Tolya loves to swim. When he comes to one grandmother's cottage, he swims in the Volkhov and floats downstream from one beach to another in 18 minutes. On the way back, it takes him exactly 1 hour. When he came to another grandmother's cottage, he swam the same distance downstream in the Luga River in 20 minutes... | Answer: 45 minutes.
Solution. Let the distance between the beaches be $x$ km. Then in Volkhov, Tolya swims downstream at a speed of $x / 18$ km/min, and upstream at a speed of $x / 60$ km/min. Therefore, Tolya's own speed is $\frac{1}{2}(x / 60 + x / 18) = 13 x / 360$ km/min. In Luga, Tolya swims downstream at a speed... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,142 |
3. The sum of the squares of positive numbers $a, b$, and $c$ is three. Prove the inequality $a+b+c \geqslant a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}$. | Solution. Note that
$$
2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)=\left(a^{2}+b^{2}+c^{2}\right)^{2}-\left(a^{4}+b^{4}+c^{4}\right)=3\left(a^{2}+b^{2}+c^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)
$$
Therefore, we need to prove the inequality
$$
a^{4}-3 a^{2}+2 a+b^{4}-3 b^{2}+2 b+c^{4}-3 c^{2}+2 c \geqslant 0
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,144 |
4. The cells of a $20 \times 20$ table are painted in $n$ colors, and there are cells of each color. In each row and each column of the table, no more than six different colors are used. What is the largest $n$ for which this is possible? | Answer: 101.
Solution. Suppose there is a coloring with 102 colors. Then there are two rows that together use at least 12 colors (otherwise, the total number of colors does not exceed $11+5 \cdot 18=101$). Let's assume for definiteness that these are the first and second rows. By the condition, no more than six differ... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,145 |
5. The incircle of triangle $ABC$ has center $I$ and touches sides $BC$ and $AC$ at points $A_1$ and $B_1$ respectively. The perpendicular bisector of segment $CI$ intersects side $BC$ at point $K$. A line through point $I$, perpendicular to $KB_1$, intersects side $AC$ at point $L$. Prove that lines $AC$ and $A_1L$ ar... | First solution. Let $N$ be the point of intersection of the perpendicular bisector of segment $C I$ with side $A C$, and $P$ be the point of intersection of lines $I L$ and $K B_{1}$. Since $C I$ is the angle bisector of $\angle A C B$, by the construction of points $K$ and $N$, as well as points $A_{1}$ and $B_{1}$, t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,146 |
6. Given two such prime numbers $p$ and $q$, that $p<q<2p$. Prove that there exist two consecutive natural numbers such that the greatest prime divisor of one of them is $p$, and the greatest prime divisor of the other is $q$. | Solution. Consider the numbers $ap$ for all integers $a$ satisfying the inequality $-\frac{1}{2}(q-1) \leqslant a \leqslant \frac{1}{2}(q-1)$. All of them give different remainders when divided by $q$. Since there are exactly $q$ such remainders, one of them is equal to 1. Therefore, there exists an $a$ such that $ap =... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,147 |
6. (30 points) $S M$ is the angle bisector in triangle $S Q T$. Point $O$ on side $S T$ is such that angle $O Q T$ is equal to the sum of angles $Q T S$ and $Q S T$. Prove that $O M$ is the angle bisector of angle $Q O T$.
 Let $R E$ be the bisector of triangle $R S T$. Point $D$ on side $R S$ is such that $E D \| R T, F$ is the intersection point of $T D$ and $R E$. Prove that if $S D = R T$, then $T E = T F$.
 Given a quadrilateral ELMI. It is known that the sum of angles LME and $M E I$ is 180 degrees and $E L=E I+L M$. Prove that the sum of angles LEM and EMI is equal to angle MIE. | Solution. Mark a point $K$ on the ray $L M$ beyond point $M$ such that $M K=I E$, and consider triangles $IEM$ and $K E M$. By construction, $\angle L M E + \angle E M K = 180^{\circ}$, and by the problem's condition, $\angle L M E + \angle M E I = 180^{\circ}$, hence $\angle K M E = \angle M E I$; sides $M K$ and $E I... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,150 |
5. Points $A_{1}$ and $B_{1}$ are the midpoints of sides $B C$ and $A C$ of an acute triangle $A B C$, and point $M$ is the midpoint of segment $A_{1} B_{1}$. Point $H$ is the foot of the altitude dropped from vertex $C$ to side $A B$. Circles are drawn through point $M$, touching sides $B C$ and $A C$ at points $A_{1}... | Solution. Let $N^{\prime}$ be the point symmetric to point $N$ with respect to the line $A_{1} B_{1}$, and the line $M N^{\prime}$ intersects the lines $A C$ and $B C$ at points $C_{1}$ and $C_{2}$, respectively. Since the angle between the tangent and the secant is equal to the inscribed angle subtended by the secant,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,151 |
5. The diagonals of quadrilateral $A B C D$ intersect at point $O$. Diagonal $A C$ is the bisector of angle $\angle B A D$, point $M$ is the midpoint of side $B C$, and point $N$ is the midpoint of segment $D O$. Prove that quadrilateral $A B C D$ is cyclic if and only if quadrilateral $A B M N$ is cyclic.
 We will call a date diverse if in its representation in the form DD/MM/YY (day-month-year) all digits from 0 to 5 are present. How many diverse dates are there in the year 2013? | Answer: 2.
Solution. Note that in any date of 2013 in the specified format, the digits 1 and 3 are used, so for the day and month of a diverse date, the digits left are 0, 2, 4, and 5.
Let $Д_{1}$ and $Д_{2}$ be the first and second digits in the day's notation, and $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ be the first ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,154 |
3. (10 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different six-letter words can be formed from the letters of the word СКАЛКА? And how many seven-letter words from the letters of the word ТЕФТЕЛЬ? In your answer, indicate the quotient of dividing the larger of the foun... | # Answer: 7.
Solution. The word СКАЛКА has six letters, but the letter А and the letter K each appear twice. Therefore, the number of different words will be $\frac{6!}{2!\cdot 2!}$. In
the word ТЕФТЕЛЬ, there are seven letters, and the letters $\mathrm{T}$ and $\mathrm{E}$ each appear twice, so the number of differen... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,156 |
4. (20 points) Several positive numbers are written on the board. The sum of the five largest of them is 0.29 of the sum of all the numbers, and the sum of the five smallest is 0.26 of the sum of all the numbers. How many numbers are written on the board? | Answer: 18.
Solution. Let the total number of numbers be $k+10$. From the condition, it is clear that $k>0$. We can assume that the sum of all numbers is 1 (otherwise, we divide each number by this sum, and the condition of the problem remains). We order the numbers in ascending order:
$$
a_{1} \leqslant a_{2} \leqsl... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,157 |
5. (20 points) Students' written work is evaluated on a two-point grading system, i.e., a work will either be credited if it is done well, or not credited if it is done poorly. The works are first checked by a neural network, which gives an incorrect answer in $10 \%$ of cases, and then all works deemed uncredited are ... | Answer: 66.
Solution. We will represent the students' works in the diagram below.
Neural network errors can:
a) classify all 10% of good works as bad;
b) classify part of the good works as bad and part of the bad works as good;
c) classify all 10% of bad works as good.
 Let a sequence of non-negative integers be given
$$
k, k+1, k+2, \ldots, k+n
$$
Find the smallest $k$, for which the sum of all numbers in the sequence is equal to 100.
# | # Answer: 9.
Solution. The given sequence of numbers is an arithmetic progression with $n+1$ terms. Its sum is $\frac{(n+1)(2 k+n)}{2}$. Therefore, the condition can be rewritten as $(n+1)(2 k+n)=200$, where $n$ is a non-negative integer. It is clear that $k$ decreases as $n$ increases. Thus, we need to find the large... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,159 |
7. (30 points) Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.
 On the extension of side $A B$ of parallelogram $A B C D$ beyond point $B$, point $K$ is marked, and on the extension of side $A D$ beyond point $D$, point $L$ is marked. It turns out that $B K = D L$. Segments $B L$ and $D K$ intersect at point $M$. Prove that $C M$ is the bisector of angle $B C D$.
Solution. Point $O$ is the midpoint of $A C$. Since the center of the circumscribed circle of triangle $A B C$ lies on its median $B O$, this median coincid... | \frac{4\sqrt{55}}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,162 |
10. (40 points) At a ball, 100 boys and 100 girls, numbered from 1 to 100, arrived. Sometimes boys and girls danced (in a dance, pairs of a boy and a girl participated). After all the dances, it turned out that for each girl, all the boys could be divided into two groups such that the girl danced with the boys from the... | Solution. Suppose that for any $k$, boy $k$ danced with girl $k$ more than half of his dances. Then, summing over all boys, we get that the number of dances between boys and girls with the same number was greater than the number of dances between boys and girls with different numbers. Now let's look at the situation fr... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,163 |
11. (40 points) Let's call the tail of a natural number any number that is obtained from it by discarding one or several of its first digits. For example, 234, 34, and 4 are tails of the number 1234. Does there exist a six-digit number without zeros in its decimal representation that is divisible by each of its tails? | Answer: Yes (721875 fits).
Solution. Suppose the required number exists. Let's write it as $A=$ $=\overline{a_{5} a_{4} \ldots a_{0}}$. Then $A$ is divisible by its five-digit tail, that is, by $\overline{a_{4} \ldots a_{0}}$. Therefore, $\overline{a_{4} \ldots a_{0}}$ divides the difference between $A$ and its tail, ... | 721875 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,164 |
12. (40 points) Petya thought of some natural number. In one move, he can multiply it by any rational number from $1 / 3$ to 2. Prove that in several such operations, Petya can obtain the number 1001. | The first solution. Note that any rational number from the interval $[1001 / 2 ; 1001 \cdot 3]$ can be multiplied by a rational number from the interval $[1 / 3 ; 2]$, to ultimately get the number 1001. We will prove that any positive number can be placed in this interval by multiplying it by allowed numbers. If we hav... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,165 |
14. (40 points) Determine the sign of the sum
$$
(\sqrt{2021}-\sqrt{2020})^{3}+(\sqrt{2020}-\sqrt{2019})^{3}+(\sqrt{2019}-\sqrt{2021})^{3}
$$
without raising the addends to the third power. | Answer: The expression is negative.
Solution. If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$, since
$$
(a+b+c)^{3}=a^{3}+b^{3}+c^{3}-3 a b c+(a b+b c+a c)(a+b+c)
$$
We have $a=\sqrt{2021}-\sqrt{2020}>0, b=\sqrt{2020}-\sqrt{2019}>0, c=\sqrt{2019}-\sqrt{2021}<0$. | The\expression\is\negative | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,167 |
1. (15 points) In a bookstore, there is a rule for "summing" discounts: "if different types of discounts apply to an item, they are applied sequentially one after another." For example, if two discounts A% and B% apply to an item, the first discount is applied to the original price, and the second discount is applied t... | Answer: $50 \%$.
Solution. Note that the sequence of applying discounts is irrelevant, since applying a discount of A\% is equivalent to multiplying the cost of the item by ( $1-\frac{A}{100}$ ). Let $R$ be the magnitude of the "random" discount. Then, as a result of "summing" the three discounts, we get
$$
230 \cdot... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,168 |
2. (15 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different four-letter words can be formed from the letters of the word КАША? And from the letters of the word ХЛЕБ? In your answer, indicate the sum of the found numbers. | Answer: 36.
Solution. In the word ХЛЕБ, all letters are different. Therefore, by rearranging the letters, we get $4 \cdot 3 \cdot 2 \cdot 1=24$ words. From the word КАША, we can form 12 words. Indeed, for the letters K and Ш, there are $4 \cdot 3=12$ positions, and we write the letters А in the remaining places. Thus,... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,169 |
3. (35 points) Journalists have found out that
a) in the lineage of Tsar Pafnuty, all descendants are male: the tsar himself had 2 sons, 60 of his descendants also had 2 sons each, and 20 had 1 son each;
b) in the lineage of Tsar Zinovy, all descendants are female: the tsar himself had 4 daughters, 35 of his descenda... | Answer: At Zinovy.
Solution. We need to find the total number of children in the lineage of Pafnuty, including the children of the king himself. It is equal to $60 \cdot 2+20 \cdot 1+2=142$. The total number of children in the lineage of King Zinovy is $4+35 \cdot 3+35 \cdot 1=144$. | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,170 |
4. (35 points) In a state, there are 11 cities and a capital. There are two-way bus routes between the capital and the other cities, each of which (in both directions) takes 7 hours. Non-capital cities are connected by two-way roads in a cyclic manner: the 1st city with the 2nd, the 2nd with the 3rd, ..., the 10th with... | # Answer: No.
Solution. Let $x$ be the time required for a transfer, $t$ be the minimum travel time from A to B. To travel optimally from the 1st non-capital city to the 6th, one can either transit through the capital or travel "around the ring." In the first case, the travel time will be $2 \cdot 7 + x = 14 + x$ hour... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,171 |
5. (35 points) $B$ In the class, an example of adding two integers was written on the board. Top student Petya mentally added these numbers during the break and got 2021. Troublemaker Vasya added a 0 at the end of one of the numbers. When the lesson started, Vasya was called to solve the problem on the board. As a resu... | Answer: Vasya calculated incorrectly
Solution. Let the integers written on the board be denoted by $x$ and $y$. Then $x+y=2021$. Suppose Vasya added a 0 to the number $x$. If Vasya's subsequent calculations are correct, then $10x + y = 2221$. Subtracting the recorded equations, we get $9x = 200$, which is impossible f... | Vasyacalculatedincorrectly | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,172 |
6. (50 points) A $5 \times 5$ square of cells was cut into several pieces of different areas, each consisting of an integer number of cells. What is the maximum number of pieces that could result from such a cutting? | Answer: 6.
Solution. We will show that there cannot be more than 6 parts. Indeed, the total area of 7 parts cannot be less than $1+2+3+4+5+6+7=28$, which exceeds the area of the square.
Now we will show how a $5 \times 5$ square can be cut into 6 parts of different sizes:
 From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property? | Answer: 40.
Solution. Suppose that the group selected a number $A$, which gives a remainder $i \neq 0$ when divided by 5. If there is another number $B$ in the group, then it must give a remainder $5-i$ when divided by 5, so that $A+B$ is divisible by 5. We will show that there cannot be any other numbers in this grou... | 40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,174 |
8. (50 points) Each of the 12 knights sitting around a round table has thought of a number, and all the numbers are different. Each knight claims that the number they thought of is greater than the numbers thought of by their neighbors to the right and left. What is the maximum number of these claims that can be true? | Answer: 6.
Solution. Let's renumber the knights in order with numbers from 1 to 12. In the pairs $(1,2),(3,4)$, $\ldots,(11,12)$, at least one of the knights is lying (specifically, the one who guessed the smaller number). Therefore, there can be no more than 6 true statements. Now let's provide an example where exact... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,175 |
9. (50 points) In the kindergarten's dressing room, in the lost items basket, there are 30 mittens, of which 10 are blue, 10 are green, 10 are red, 15 are right-handed, and 15 are left-handed. Is it always possible to form sets of right and left mittens of the same color for 5 children? | Answer: It is always possible
Solution. Suppose that for all colors, the number of left and right mittens is different (otherwise, we immediately have 5 required pairs). Then, for each color, we determine which type of mittens is less - left or right. In at least two out of three colors, the lesser number will be of o... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,176 |
10. (50 points) In the kindergarten's changing room, in the lost and found basket, there are 30 mittens, of which 10 are blue, 10 are green, 10 are red, 15 are right-handed, and 15 are left-handed. Is it always possible to form sets of right and left mittens of the same color for 6 children? | Answer: No, not always.
Solution. There could have been 10 right red mittens, 10 left green mittens, and 5 right and 5 left blue mittens. From these, only 5 pairs of the same color (specifically, blue) can be formed. | No,notalways | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,177 |
4. On the side $B C$ of triangle $A B C$, a point $K$ is marked. A tangent $\ell_{1}$ to the circumcircle of triangle $A K C$ is drawn, parallel to the line $A B$ and closest to it. It touches the circle at point $L$. The line $A L$ intersects the circumcircle of triangle $A B K$ at point $M (M \neq A)$. A tangent $\el... | Solution. Let $N$ be the intersection point of lines $B C$ and $\ell_{1}$. If we prove that $N M$ is a tangent to the circumcircle of triangle $A B K$, then point $N$ will lie on line $\ell_{2}$, which means that lines $B K, \ell_{1}$, and $\ell_{2}$ intersect at one point.
Since lines $A B$ and $L N$ are parallel, an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,178 |
4. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $K$ and $L$. Line $\ell$ intersects circle $\omega_{1}$ at points $A$ and $C$, and circle $\omega_{2}$ at points $B$ and $D$, with the points appearing on line $\ell$ in alphabetical order. Let $P$ and $Q$ be the projections of points $B$ and $C$ onto line $K... | Solution. Let the intersection point of lines $K L$ and $A D$ be denoted as $M$. From the cyclic quadrilateral $B K D L$, we conclude that $\angle B K L = \angle B D L$. Therefore, triangles $B K M$ and $L D M$ are similar, and thus $\frac{B M}{K M} = \frac{L M}{D M}$. From the cyclic quadrilateral $A K C L$, we conclu... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,179 |
4. Points $B_{1}$ and $C_{1}$ are the midpoints of sides $A C$ and $A B$ of triangle $A B C$. Circles $\omega_{1}$ and $\omega_{2}$ are constructed on sides $A B$ and $A C$ as diameters. Let $D$ be the point of intersection of line $B_{1} C_{1}$ with circle $\omega_{1}$, lying on the other side of $C$ relative to line ... | Solution. Let $M$ be the midpoint of segment $DE$, and lines $AM$ and $BC$ intersect at point $L$. Since $B_1C_1$ is the midline of triangle $ABC$, lines $B_1C_1$ and $BC$ are parallel. Therefore, $B_1M$ is the midline of triangle $ACL$, and in particular, $M$ is the midpoint of segment $AL$. Thus, the diagonals of qua... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,180 |
3. (35 points) In country $B$, there are 4 southern cities and 5 northern cities. Transportation links connect each southern city with each northern city in both directions; there is no transportation link between other pairs of cities. A one-way ticket from any city $A$ to any city $B$ costs $N$ rubles. A round-trip t... | Answer: $4 \times 1.6 N$ rubles.
Solution. Since the southern cities are not connected by transportation, to visit all of them and return, it will require $2 \times 4$ trips. Any two trips cost at least $1.6 N$ rubles in total, so the total expenses will be at least $4 \times 1.6 N$ rubles.
Let's provide an example o... | 4\times1.6N | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,182 |
1. On a $5 \times 7$ grid, 9 cells are marked. We will call a pair of cells with a common side interesting if exactly one cell in the pair is marked. What is the maximum number of interesting pairs that can be? | Answer: 35.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, it is no more than 3. Then the total number of interesting pairs does not exceed $9 \cdot 4 = 36$. At the same time, if there are two adja... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,183 |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a+b}{c}\right)^{4}+\left(\frac{b+c}{d}\right)^{4}+\left(\frac{c+d}{a}\right)^{4}+\left(\frac{d+a}{b}\right)^{4}
$$ | Answer: 64.
Solution. By Cauchy's inequality for means,
$$
A \geqslant 4 \cdot \frac{(a+b)(b+c)(c+d)(d+a)}{a b c d}=64 \cdot \frac{a+b}{2 \sqrt{a b}} \cdot \frac{b+c}{2 \sqrt{b c}} \cdot \frac{c+d}{2 \sqrt{c d}} \cdot \frac{d+a}{2 \sqrt{d a}} \geqslant 64
$$
Equality is achieved when $a=b=c=d=1$. | 64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,184 |
3. Quadrilateral $A B C D$ is inscribed in a circle. A tangent line $\ell$ is drawn to this circle at point $C$. Circle $\omega$ passes through points $A$ and $B$ and is tangent to line $\ell$ at point $P$. Line $P B$ intersects segment $C D$ at point $Q$. Find the ratio $\frac{B C}{C Q}$, given that $B$ is the point o... | Answer: 1.
Solution. The angle between the tangent $BD$ and the chord $AB$ of the circle $\omega$ is equal to the inscribed angle that subtends $AB$, so $\angle APB = \angle ABD = \angle AC... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,185 |
4. On the board, the product of the numbers $\overline{\text{IKS}}$ and $\overline{\text{KSI}}$ is written, where the letters correspond to different non-zero decimal digits. This product is a six-digit number and ends with S. Vasya erased all zeros from the board, after which IKS remained. What was written on the boar... | Answer: $100602=162 \cdot 621$.
Solution. Let $p=\overline{\text { IKS }} \cdot \overline{\text { KSI }}$. In the decimal representation of $p$, there are three zeros, I, K, and S. Then the numbers $(I+K+S)^{2}$ and $(I+K+S)$ give the same remainders when divided by 9, from which $(I+K \text{ or } (I+K+S-1)) \vdots$. ... | 100602=162\cdot621 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,186 |
5. On a circle, $n$ points are marked ( $n \geqslant 5$ ). Petya and Vasya take turns (starting with Petya) to draw one chord connecting pairs of these points. Any two drawn chords must intersect (possibly at the endpoints). The player who cannot make a move loses. For which $n$ will Petya win regardless of Vasya's act... | Answer: for odd $n$.
Solution. First, let's prove the following statement: if in the final set of chords a triangle can be identified, two sides of which are adjacent points, then for odd $n$ Petya will win, and for even $n$ Vasya will win. Indeed, let the triangle be denoted as $A B C$, where $A$ and $B$ are adjacent... | for\odd\n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,187 |
6. On the table, there are three spheres and a cone (with its base $k$ on the table), touching each other externally. The radii of the spheres are 5, 4, and 4, and the height of the cone is to the radius of its base as 4:3. Find the radius of the base of the cone. | # Answer: $\frac{169}{60}$.
Solution. Let $O$ be the cente... | \frac{169}{60} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,188 |
1. On a $7 \times 7$ checkerboard, 14 cells are marked. We will call a pair of cells with a common side interesting if at least one cell in the pair is marked. What is the maximum number of interesting pairs that can be? | # Answer: 55.
Solution. Let's call two cells adjacent if they share a side. The number of interesting pairs containing a given marked cell is no more than 4, and for a boundary cell, no more than 3. Then the total number of interesting pairs does not exceed $14 \cdot 4 = 56$. At the same time, if there are two adjacen... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,189 |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a^{2}+b^{2}}{c d}\right)^{4}+\left(\frac{b^{2}+c^{2}}{a d}\right)^{4}+\left(\frac{c^{2}+d^{2}}{a b}\right)^{4}+\left(\frac{d^{2}+a^{2}}{b c}\right)^{4}
$$ | Answer: 64.
Solution. By the Cauchy inequalities for means,
$$
A \geqslant 4 \cdot \frac{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+d^{2}\right)\left(d^{2}+a^{2}\right)}{c d \cdot a d \cdot a b \cdot b c}=64 \cdot \frac{a^{2}+b^{2}}{2 a b} \cdot \frac{b^{2}+c^{2}}{2 b c} \cdot \frac{c^{2}+d^{2}}{2 c ... | 64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,190 |
3. A circle $\omega$ is circumscribed around triangle $A B C$. Tangents to the circle, drawn at points $A$ and $B$, intersect at point $K$. Point $M$ is the midpoint of side $A C$. A line passing through point $K$ parallel to $A C$ intersects side $B C$ at point $L$. Find the angle $A M L$. | Answer: $90^{\circ}$.
Solution. Let $\alpha=\angle A C B$. The angle between the tangent $A K$ and the chord $A B$ of the circle $\omega$ is equal to the inscribed angle that subtends $A B$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,191 |
4. On the board, the product of the numbers $\overline{\text{IKS}}$ and $\overline{\text{KSI}}$ is written, where the letters correspond to different non-zero decimal digits. This product is a six-digit number and does not end in zero. Petya erased all zeros and one digit $C$ from the board, after which $\text{IKS}$ re... | Answer: $206032=632 \cdot 326$
Solution. Let $p=\overline{\text { IKS }} \cdot \overline{\mathrm{KSI}}$. In the decimal representation of $p$, the digits I and K appear once, while the digits S and 0 appear twice. Then the numbers $(I+K+S)^{2}$ and ( $I+K+2S$ ) give the same remainders when divided by 9, from which $S... | 206032=632\cdot326 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,192 |
5. On a circle, $n$ points are marked ($n \geqslant 5$). Petya and Vasya take turns (starting with Petya) to draw one chord connecting pairs of these points that are not adjacent. Any two drawn chords can only intersect at their endpoints. The player who cannot make a move loses. For which $n$ will Petya win regardless... | Answer: when $n$ is even.
Solution 1. The actions of the players can be described as follows: they draw diagonals in the $n$-gon that intersect only at the ends. For convenience, we can assume this $n$-gon is regular. Let $n$ be even, that is, $n=2 m$. A winning strategy for Pete is as follows. His first move should b... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,193 |
6. On the table, there are three spheres and a cone (with its base $k$ on the table), touching each other externally. The radii of the spheres are 20, 40, and 40, and the radius of the base of the cone is 21. Find the height of the cone. | Answer: 28.
Solution. Let $O$ be the center of the base of the cone, $h$ be its height, and $2 \alpha$ be the angle of inclination of the cone's generators to the table. Consider the sectio... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,194 |
1. Each cell of a $5 \times 6$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of red cells is not less than the number of blue cells and not less than the number of yellow cells, and in each column of the table, the number of blue cells is not less than the number of... | Answer: 6.
| $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{C}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,195 |
2. Given numbers $x, y \in\left(0, \frac{\pi}{2}\right)$. Find the maximum value of the expression
$$
A=\frac{\sqrt{\cos x \cos y}}{\sqrt{\operatorname{ctg} x}+\sqrt{\operatorname{ctg} y}}
$$ | Answer: $\frac{\sqrt{2}}{4}$.
Solution. Note that $\sqrt{a}+\sqrt{b} \leqslant \sqrt{2(a+b)}$ for any $a, b \geqslant 0$. Applying the inequality for the harmonic mean and the arithmetic mean, we get
$$
\begin{aligned}
A \leqslant \frac{1}{4} \sqrt{\cos x \cos y}(\sqrt{\operatorname{tg} x}+\sqrt{\operatorname{tg} y})... | \frac{\sqrt{2}}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,196 |
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BC$ intersects circle $\omega_{1}$ agai... | Answer: $180^{\circ}$.
Solution 1. Since $\angle B C O=\angle P C O_{1}$, the isosceles triangles $B O C$ and $P O_{1} C$ are similar, from which $\angle B O C=\angle P O_{1} C$. Similarly, it can be verified that $\angle A O D=\angle Q O_{2} D$. Then the segments $O_{1} P$ and $O_{2} Q$ are parallel to the line $A B$... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,197 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.