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5. In a class, there are $n$ boys and $n$ girls $(n \geqslant 3)$. They sat around a round table such that no two boys and no two girls sit next to each other. The teacher has $2 n$ cards, on which the numbers $1,2,3, \ldots, 2 n$ are written, each one appearing once. He distributed one card to each student such that t... | Answer: for odd $n$
Solution. According to the problem, the boys received cards with numbers from 1 to $n$, and the girls received cards with numbers from $n+1$ to $2 n$. Suppose that all the boys ended up with the number $m$ on their slips of paper. Then the sum of all the numbers on the slips is $m n$, which can als... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,304 |
1. What is the minimum number of cells that need to be marked in a $15 \times 15$ table so that each vertical or horizontal strip $1 \times 10$ contains at least one marked cell. | Answer: 20
Solution. Let's cut the $15 \times 15$ table without the central $5 \times 5$ square into 20 rectangles of $1 \times 10$ (see the left figure). Therefore, we will need to mark at least 20 cells. An example with 20 cells: all cells of two parallel diagonals of length 10 are marked (see the right figure).
![]... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,305 |
2. Find all such triples of prime numbers $p, q$ and $r$, that $\frac{p}{q}=\frac{4}{r+1}+1$. | Answer: $(7,3,2) ;(5,3,5) ;(3,2,7)$.
First solution. Rewrite the relation as
$$
4 q=(p-q)(r+1)
$$
If $q=2$, it becomes $8=(p-2)(r+1)$. Then $p-2$ is a power of two, so $p=3$ and $r=7$. If $r=2$, the relation $(*)$ becomes $4 q=3(p-q)$, from which $q=3$ and $p=7$. We will further assume that $q$ and $r$ are odd prime... | (7,3,2);(5,3,5);(3,2,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,306 |
3. The sum of positive numbers $a, b, c$ and $d$ does not exceed 4. Find the maximum value of the expression
$$
\sqrt[4]{a^{2}(a+b)}+\sqrt[4]{b^{2}(b+c)}+\sqrt[4]{c^{2}(c+d)}+\sqrt[4]{d^{2}(d+a)}
$$ | Answer: $4 \sqrt[4]{2}$
First solution. By the inequality of means for four numbers, we have
$$
\sqrt[4]{a^{2}(a+b)}=\frac{\sqrt[4]{2 a \cdot 2 a \cdot(a+b) \cdot 2}}{\sqrt[4]{8}} \leqslant \frac{1}{\sqrt[4]{8}} \cdot \frac{2 a+2 a+(a+b)+2}{4}=\frac{5 a+b+2}{4 \sqrt[4]{8}}
$$
Summing this inequality with three simil... | 4\sqrt[4]{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,307 |
4. On the side $AB$ of triangle $ABC$, a point $K$ is marked, and on the sides $AC$ and $BC$, points $L$ and $M$ are chosen respectively such that $AK = AL$ and $BK = BM$. It turns out that the lines $LM$ and $AB$ are parallel. The tangent at point $L$ to the circumcircle of triangle $KLM$ intersects the segment $CK$ a... | Answer: $\angle D E O=90^{\circ}$
First solution. By the problem's condition, $O K=O L=O M$ and triangles $A K O$ and $A L O$ are equal by three sides, so $\angle A O K=\angle A O L=\angle K... | \angleDEO=90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,308 |
5. Around a round table, $n-1$ schoolchildren are sitting ($n \geqslant 4$). The teacher has $n$ cards, on which the numbers $1, 2, 3, \ldots, n$ are written, each one time. He gave each schoolchild one card and kept one card for himself. For each pair of adjacent schoolchildren, the sum of the numbers on their cards w... | Answer: for even $n$
Solution. Since we are only interested in the remainders when divided by $n$, for convenience, we will replace the card with the number $n$ with a card with the number 0. Suppose there exists a suitable way to distribute the cards. We will number the students in a circle from 0 to $n-2$ (starting ... | forevenn | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,309 |
1. What is the minimum number of cells that need to be marked in a $20 \times 20$ table so that each vertical or horizontal strip of $1 \times 12$ contains at least one marked cell. | Answer: 32
Solution. Let's cut the $20 \times 20$ table without the central $4 \times 4$ square into 32 rectangles of $1 \times 12$ (see the left figure). Therefore, we will need to mark at least 32 cells. An example with 32 cells: all cells of three parallel diagonals of lengths 4, 16, and 12 are marked (see the righ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,310 |
2. Find all such triples of prime numbers $p, q$ and $r$, that $\frac{p}{q}=\frac{8}{r-1}+1$. | Answer: $(7,3,7); (5,3,13); (3,2,17)$.
First solution. Rewrite the relation as
$$
8 q=(p-q)(r-1)
$$
If $q=2$, it becomes $16=(p-2)(r-1)$. Then $p-2$ is a power of two, so $p=3$ and $r=17$. If $r=2$, the relation $(*)$ becomes $8 q=p-q$, from which $p=9 q$, which is impossible. We will further assume that $q$ and $r$... | (7,3,7);(5,3,13);(3,2,17) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,311 |
3. The sum of positive numbers $a, b, c$ and $d$ does not exceed 4. Find the maximum value of the expression
$$
\sqrt[4]{2 a^{2}+a^{2} b}+\sqrt[4]{2 b^{2}+b^{2} c}+\sqrt[4]{2 c^{2}+c^{2} d}+\sqrt[4]{2 d^{2}+d^{2} a}
$$ | Answer: $4 \sqrt[4]{3}$
First solution. By the inequality of means for four numbers, we have
$$
\sqrt[4]{2 a^{2}+a^{2} b}=\frac{\sqrt[4]{3 a \cdot 3 a \cdot(2+b) \cdot 3}}{\sqrt[4]{27}} \leqslant \frac{1}{\sqrt[4]{27}} \cdot \frac{3 a+3 a+(2+b)+3}{4}=\frac{6 a+b+5}{4 \sqrt[4]{27}}
$$
Summing this inequality with thr... | 4\sqrt[4]{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,312 |
4. On the leg $AC$ of the right triangle $ABC$ with hypotenuse $AB$, a point $P$ is marked. Point $D$ is the foot of the perpendicular dropped from vertex $A$ to the line $BP$, and point $E$ is the foot of the perpendicular dropped from point $P$ to the side $AB$. On the plane, a point $T$ is chosen such that the lines... | Answer: $90^{\circ}$
Solution. Let $\angle A B P=\varphi$ and $\angle C B P=\psi$. Triangles $A O T$ and $P O T$ are equal by three sides, and the inscribed angle $\angle A B P$ subtends ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,313 |
5. In a math club, there are $m$ boys and $n$ girls. The teacher has $m n$ cards, on which the numbers $1, 2, 3, \ldots, m n$ are written, each appearing once. He distributes the cards to the club members (until the cards or the members run out). In the end, each club member will have one card or none, if someone did n... | Answer: when $n=1$ or when $m=1$
Solution. We can assume that $m \leqslant n$. If $m=1$, then each girl must receive a card, and the boy will not get anything, which satisfies the condition.
Let $m \geqslant 2$. In this case, $m n \geqslant m+n$, so each club member will receive exactly one card. Suppose there exists... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,314 |
1. In each cell of an $n \times n$ table, an integer is written. It turns out that for all $k$ from 1 to $n$, the sum of the numbers in the $k$-th column is either one less or two more than the sum of the numbers in the $k$-th row. For which $n$ is this possible? | Answer: for $n$ divisible by three
Solution. Let the number of rows, the sum of the numbers in which is one less than in the columns with the same number, be $m$. Then the number of rows, the sum of the numbers in which is two more than in the columns with the same number, is $n-m$. Therefore, the sum of all numbers, ... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,315 |
2. Find all triples of natural numbers $a, b$, and $c$, for which the numbers $a^{2}+1$ and $b^{2}+1$ are prime and $\left(a^{2}+1\right)\left(b^{2}+1\right)=c^{2}+1$. | Answer: $(1,2,3)$ and $(2,1,3)$.
Solution. We can assume that $a \geqslant b$. Then $c^{2}+1=\left(a^{2}+1\right)\left(b^{2}+1\right) \leqslant\left(a^{2}+1\right)^{2}$, so $c^{2}<\left(a^{2}+1\right)^{2}$ and, therefore, $c<a^{2}+1$. On the other hand,
$$
(c+a)(c-a)=\left(c^{2}+1\right)-\left(a^{2}+1\right)=b^{2}\le... | (1,2,3)(2,1,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,316 |
3. The sum of positive numbers $a, b, c$ and $d$ is not less than 8. Find the minimum value of the expression
$\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{b^{4}}{(b+c)(b+d)(b+a)}+\frac{c^{4}}{(c+d)(c+a)(c+b)}+\frac{d^{4}}{(d+a)(d+b)(d+c)}$. | Answer: 1
First solution. By the inequality of means for four numbers, we have
$$
\begin{aligned}
\frac{a^{4}}{(a+b)(a+c)(a+d)}+\frac{a+b}{16}+\frac{a+c}{16} & +\frac{a+d}{16} \geqslant \\
& \geqslant 4 \sqrt[4]{\frac{a^{4}}{(a+b)(a+c)(a+d)} \cdot \frac{a+b}{16} \cdot \frac{a+c}{16} \cdot \frac{a+d}{16}}=\frac{a}{2}
... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,317 |
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and triangle $A P D$ is acute-angled. Points $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. A perpendicular is drawn from point $E$ to line $A C$, and a perpendicular is drawn from point $F$ to line $B D$, these perp... | Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $T$ be the point of inte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,318 |
5. In a class of $n \geqslant 3$ students, the teacher has $m=\frac{1}{2} n(n-1)$ cards, on which the numbers $1,2,3, \ldots, m$ are written, each appearing once. He gave each student one card. For each pair of students, the sum of the numbers on their cards was calculated, and all the resulting numbers were written on... | Solution. Let's number the schoolchildren from 1 to $n$. Let $a_{j}$ be the number written on the card of the $j$-th schoolchild. Then all pairwise sums $a_{i}+a_{j}$ give different remainders when divided by $m=\frac{1}{2} n(n-1)$. Let $k$ of the numbers $a_{1}, a_{2}, \ldots, a_{n}$ be even, and $n-k$ numbers be odd.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,319 |
1. In each cell of an $n \times n$ table, a natural number is written. It turns out that for all $k$ from 1 to $n$, the sum of the numbers in the $k$-th column differs by one from the sum of the numbers in the $k$-th row. For which $n$ is this possible? | Answer: for even $n$
Solution. If $n$ is odd, then in the table, the sum of all numbers, calculated by rows, differs from the sum of numbers, calculated by columns, by an odd number. But this is impossible. In a table with even $n$, the upper half of the diagonal running from the bottom left corner to the top right co... | forevenn | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,320 |
2. Find all triples of natural numbers $a, b$, and $c$, for which the numbers $a^{2}-23$ and $b^{2}-23$ are prime and $\left(a^{2}-23\right)\left(b^{2}-23\right)=c^{2}-23$. | Answer: $(5,6,7)$ and $(6,5,7)$.
Solution. Note that $a, b \geqslant 5$. If $a=b=5$, then $c^{2}-23=\left(a^{2}-23\right)\left(b^{2}-23\right)=2 \cdot 2=4$, which is impossible. Therefore, we can assume that $a \geqslant b$ and $a \geqslant 6$. Then
$$
c^{2}-23=\left(a^{2}-23\right)\left(b^{2}-23\right) \leqslant\lef... | (5,6,7)(6,5,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,321 |
3. The sum of positive numbers $a, b, c$ and $d$ is 4. Find the minimum value of the expression
$$
\frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)}+\frac{b^{8}}{\left(b^{2}+c\right)\left(b^{2}+d\right)\left(b^{2}+a\right)}+\frac{c^{8}}{\left(c^{2}+d\right)\left(c^{2}+a\right)\left(c^{2}+b\rig... | Answer: $\frac{1}{2}$
First solution. By the inequality of means for four numbers, we have
$$
\begin{aligned}
& \frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)}+\frac{a^{2}+b}{16}+\frac{a^{2}+c}{16}+\frac{a^{2}+d}{16} \geqslant \\
& \quad \geqslant 4 \sqrt[4]{\frac{a^{8}}{\left(a^{2}+b\right... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,322 |
4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and the angle $A P B$ is obtuse. Points $E$ and $F$ are the midpoints of sides $A D$ and $B C$ respectively. A perpendicular is drawn from point $E$ to the line $A C$, and a perpendicular is drawn from point $F$ to the line $B D$, these p... | Answer: $90^{\circ}$
First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $B P$, respectively, and let $T$ be the point of inte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,323 |
5. In a class of $n \geqslant 3$ students, the teacher has $m=\frac{1}{2} n(n-1)$ cards, on which the numbers $1,2,3, \ldots, m$ are written, each appearing once. He gave each student one card. For each pair of students, the sum of the numbers on their cards was calculated and all the resulting numbers were written on ... | Solution. Let's number the schoolchildren from 1 to $n$. Let $a_{j}$ be the number written on the card of the $j$-th schoolchild. Then all pairwise sums $a_{i}+a_{j}$ give different remainders when divided by $m=\frac{1}{2} n(n-1)$.
Consider the remainders of the pairwise differences $a_{i}-a_{j}$ modulo $m$ for diffe... | n-2 | Number Theory | proof | Yes | Yes | olympiads | false | 8,324 |
7. (30 points) The bisector $Q K$ of triangle $P Q R$ intersects its circumcircle at point $M$ (different from $Q$). The circumcircle of triangle $P K M$ intersects the extension of side $P Q$ beyond point $P$ at point $N$. Prove that $N R$ and $Q M$ are perpendicular.
 A circle with center $O$, inscribed in angle $Q P R$, touches side $P R$ at point $L$. A tangent to the circle, parallel to $P O$, intersects ray $P Q$ at point $S$, and ray $L P$ at point $D$. Prove that $D S=2 P L$.
 Circles $K_{1}$ and $K_{2}$ touch a line at points $A$ and $B$ respectively and, in addition, intersect at points $X$ and $Y$, with point $X$ being closer to the line $AB$. Line $AX$ intersects $K_{2}$ again at point $P$. The tangent to $K_{2}$ at point $P$ intersects line $AB$ at point $Q$. Prove that a... | Solution: Let $Q Y$ intersect the second circle at point $S$, and the line $Y X$ intersect $A B$ at point $T$. From the similarity of triangles $Q S B$ and $Q B Y$, we have $B S: B Y=Q B: Q Y$. Similarly, $P S: P Y=Q P: Q Y$, hence $B S: S P=B Y: Y P$.
By the Law of Sines in triangle $X Y B$, we have: $B Y: X Y = \sin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,328 |
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. Denote the angles as indicated in the figure. The angle $A D C$ is an exterior angle for triangle $A D B$; hence, $\beta+18^{\circ}=\alpha+x$. Angles $A B C$ ... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,336 |
5. (30 points) Point $M$ is the midpoint of side $BC$ of triangle $ABC$. On segment $AC$, there is a point $D$ such that $DM$ and $BC$ are perpendicular. Segments $AM$ and $BD$ intersect at point $X$. It turns out that $AC = 2BX$. Prove that $X$ is the midpoint of segment $AM$.
 Masha marks a point $E$ on the side $CD$ of the square $ABCD$ and finds the length of the segment $AE$. Misha draws the angle bisector of $\angle BAE$, marks the point $F$ where this bisector intersects the side $BC$ of the square, and finds the sum of the lengths of segments $BF$ and $ED$. Can Masha cho... | Answer: It cannot, their results are always equal.
Solution. On the extension of side $CD$ of the square beyond point $D$, we lay off segment $DG$, equal to segment $BF$. We compare the leng... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,338 |
7. (40 points) On the board, the numbers $1^{2}, 2^{2}, 3^{2}, \ldots, 1000^{2}$ are written. Two players take turns erasing one of the numbers on the board until only two numbers remain. If the difference between these two numbers is divisible by 13, the second player wins; otherwise, the first player wins. How should... | Answer: The second player should erase the number whose base, when added to the base of the number erased by the first player, equals 1001.
Solution. Let the numbers left on the board be $x^{2}$ and $y^{2}$. Then $x^{2}-y^{2}=(x+y)(x-y)$. For this value to be divisible by 13, either $x+y$ or $x-y$ must be divisible by... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,339 |
8. (40 points) At the beginning of each physical education lesson, 30 students are divided into 3 teams of 10 each. Prove that there will be two students who were on the same team for three consecutive lessons. | The first solution. Take all the students from any team from the first lesson and see how they will be distributed among the three teams in the second lesson. Since 10 people are distributed among three teams, by the Pigeonhole Principle, there will be a team that will have at least 4 people.
Now, take these four and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,340 |
9. (40 points) For real numbers $a, b$ and $c$ it is known that $a b + b c + c a = 3$. What values can the expression $\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}$ take? | Answer: 6.
Solution. Consider the first term of the desired expression. Using the condition that $a b+b c+c a=3$. Then
$$
b^{2}+3=b^{2}+a b+b c+c a=(b+a)(b+c)
$$
Therefore,
$$
\frac{a\left(b^{2}+3\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c) \text {. }
$$
Similarly for the second and third terms, we get
$$
\frac{b... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,341 |
2. We will call a cascade generated by a number $r$, a set of 12 natural numbers: $r, 2r, \ldots, 12r$.
a) Can some pair of numbers $(a, b)$ be contained in six different cascades? If yes, provide an example of such numbers; if no, explain why it cannot be.
b) Is it true that the set of natural numbers can be colored... | 2. a) It can. Let $a < b$, then $a / b$ is a proper fraction with a denominator no greater than 12. It is not difficult to find 6 equal fractions with denominators no greater than 12: $1 / 2 = 2 / 4 = 3 / 6 = \cdots = 6 / 12$. Each such fraction determines the positions of the pair of numbers $a, b$ in the cascade: for... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,342 |
3. In triangle $ABC$, point $D$ is the midpoint of side $AB$, $E$ is the midpoint of side $AC$, and $F$ is the midpoint of the angle bisector $AL$, with $DF=1$ and $EF=2$. On the plane, points $D$, $E$, and $F$ are depicted such that line $DF$ is horizontal, and the other elements of the drawing have been erased. Can a... | 3. No. Note that if we specify a point $A$ for which $\angle D A F=\angle F A E$, then by laying off the segment $A B=2 A F$ on the ray $A D$, and the segment $A C=2 A E$ on the ray $A E$, we will obtain a triangle $A B C$ that satisfies the problem's condition. In this case, the positions of vertices $B$ and $C$ are u... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,343 |
4. The city has the shape of a grid rectangle $5 \times 10$ cells: lines are streets, cells are residential blocks. The distance between intersections is measured as the length of the shortest path along the city streets, passing from one intersection to another. For example, for intersections $A, B$, and $C$ in the pi... | 4. а) It is possible. We can construct the required configuration. It is convenient to choose intersections either only on the very bottom or only on the very right street - then it is easy to calculate distances between them by moving along these streets. In the image below, $1=A B, 2=D E, 3=B C, 4=A C, 9=C D$, $11=C ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,344 |
1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $2 f(2 x-3)-f(3 x+1)$ also has exactly one root. Find the root of the trinomial $f(x)$. | Answer: -11.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=2 f(2 x-3)-f(3 x+1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$... | -11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,345 |
2. In the cells of an $11 \times 11$ square, zeros and ones are arranged in such a way that in any figure of four cells in the shape of $\square$, the sum of the numbers is odd. (The figure can be rotated and flipped). What is the minimum number of ones that can be in such an arrangement? | Answer: 25
Solution. Place 25 figures in the square without any common cells (see the left figure). Each of them contains at least one unit, so the total number of units is no less than 25. A suitable arrangement of 25 units: in all cells with even coordinates (see the right figure).
(b+1)$ are squares of some natural numbers. Prove that for some natural number $n>1$ the number $(a+n)(b+n)$ is also a square of some natural number. | Solution. Let $n=a b$, then
$$
(a+n)(b+n)=(a+a b)(b+a b)=a b \cdot(a+1)(b+1)
$$
which is the product of two perfect squares. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,347 |
4. Point $M$ is the midpoint of the leg $A B$ of an isosceles right triangle $A B C$ with a right angle $\angle B$. The bisector of angle $\angle A$ intersects the circumcircle of the triangle at point $L$. Point $H$ is the foot of the perpendicular dropped from $L$ to the line $A C$. Find the angle $\angle A M H$. | Answer: $112.5^{\circ}$
First solution. Let $D$ be the intersection point of lines $A C$ and $B L$. Since quadrilateral $A B L C$ is cyclic, $\angle A L B = \angle A C B = 45^{\circ}$ and $\... | 112.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,348 |
5. The sum of positive numbers $a, b$, and $c$ is three. Prove the inequality
$$
\frac{a b}{\sqrt{c^{2}+3}}+\frac{b c}{\sqrt{a^{2}+3}}+\frac{c a}{\sqrt{b^{2}+3}} \leqslant \frac{3}{2}
$$ | Solution. It is not difficult to verify by expanding the brackets that $3(a b+b c+c a) \leqslant (a+b+c)^{2}=9$. Therefore,
$$
c^{2}+3 \geqslant c^{2}+a b+b c+c a=(c+a)(c+b)
$$
Therefore, by the inequality of means for two numbers,
$$
\frac{a b}{\sqrt{c^{2}+3}} \leqslant \frac{a b}{\sqrt{(c+a)(c+b)}} \leqslant \frac... | \frac{3}{2} | Inequalities | proof | Yes | Yes | olympiads | false | 8,349 |
6. On the plane, $2 n+1$ blue and $n-1$ red lines are drawn, no two of which are parallel and no three intersect at the same point. Prove that among the parts into which they divide the plane, there are at least $4 n+2$ parts bounded only by blue lines. | Solution. By induction, it is not difficult to verify that a $2n+1$ blue line divides the plane into $(2n+1)(n+1)+1=2n^2+3n+2$ parts. Each red line intersects no more than $2n+2$ areas with blue boundaries. Therefore, after drawing the first red line, there will be no fewer than $\left(2n^2+3n+2\right)-(2n+2)=2n^2+n$ a... | 4n+2 | Combinatorics | proof | Yes | Yes | olympiads | false | 8,350 |
1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $f(3 x+2)-2 f(2 x-1)$ also has exactly one root. Find the root of the trinomial $f(x)$. | Answer: -7.
Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=f(3 x+2)-2 f(2 x-1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$ ... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,351 |
3. Natural numbers $a$ and $b$ are such that the numbers $ab$ and $(2a+1)(2b+1)$ are squares of some natural numbers. Prove that for some even number $n>2$, the number $(a+n)(b+n)$ is also a square of some natural number. | Solution. Let $n=2 a b$, then
$$
(a+n)(b+n)=(a+2 a b)(b+2 a b)=a b \cdot(2 a+1)(2 b+1)
$$
which is the product of two perfect squares. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,353 |
4. Given an isosceles right triangle \(ABC\) with hypotenuse \(AB\). Point \(M\) is the midpoint of side \(BC\). On the smaller arc \(AC\) of the circumcircle of triangle \(ABC\), point \(K\) is chosen. Point \(H\) is the foot of the perpendicular dropped from \(K\) to line \(AB\). Find the angle \(\angle CAK\), given ... | Answer: $22.5^{\circ}$
First solution. Let $D$ be the intersection point of lines $A B$ and $C K$. Since quadrilateral $A B C K$ is cyclic, $\angle A K D = \angle A B C = 45^{\circ}$ and $\a... | 22.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,354 |
5. Positive numbers $a, b$, and $c$ satisfy the condition $a b + b c + c a = 1$. Prove the inequality
$$
\frac{a}{\sqrt{a^{2}+1}}+\frac{b}{\sqrt{b^{2}+1}}+\frac{c}{\sqrt{c^{2}+1}} \leqslant \frac{3}{2}
$$ | The first solution. By the inequality of means for two numbers
$$
\frac{a}{\sqrt{a^{2}+1}}=\frac{a}{\sqrt{a^{2}+a b+b c+c a}}=\sqrt{\frac{a}{a+b} \cdot \frac{a}{a+c}} \leqslant \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)
$$
Therefore,
$$
\frac{a}{\sqrt{a^{2}+1}}+\frac{b}{\sqrt{b^{2}+1}}+\frac{c}{\sqrt{c^{2}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,355 |
6. On a plane, $2 n$ red and $n$ blue lines are drawn, no two of which are parallel and no three intersect at the same point. Prove that among the parts into which they divide the plane, there are at least $n$ parts bounded only by red lines. | Solution. It is not difficult to verify by induction that $2 n$ red lines divide the plane into $2 n^{2}+n+1$ parts. Each blue line intersects no more than $2 n+1$ regions with red boundaries. Therefore, after drawing the first blue line, there will be no less than $\left(2 n^{2}+n+\right.$ 1) $-(2 n+1)=2 n^{2}-n$ regi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,356 |
1. Given the quadratic trinomial $f(x)=a x^{2}-a x+1$. It is known that $|f(x)| \leqslant 1$ for all $x \in[0,1]$. What is the greatest value that $a$ can take? | Answer: 8
Solution. It is not difficult to check that $a=8$ works. Indeed, $|2 x-1| \leqslant 1$ for $x \in[0,1]$, so $f(x)=8 x^{2}-8 x+1=2(2 x-1)^{2}-1 \leqslant 1$, and the inequality $f(x) \geqslant-1$ holds for all $x$.
Suppose that $a>8$. Then
$$
f\left(\frac{1}{2}\right)=\frac{a}{4}-\frac{a}{2}+1=1-\frac{a}{4}... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,357 |
2. In each cell of a $15 \times 15$ square, there is a natural number not exceeding 4, and the sum of the numbers in each $2 \times 2$ square is 7. What is the maximum value that the sum of the numbers in the entire table can take? | Answer: 417
Solution. Note that the sum of the numbers in two adjacent cells does not exceed five, since otherwise the sum of the numbers in the $2 \times 2$ square containing these two cells would be at least eight, which is impossible according to the condition.
Divide the table into 49 squares of $2 \times 2$ and ... | 417 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,358 |
3. Given natural numbers $a$ and $b$. It turns out that for any natural number $n$, the numbers $a+n$ and $b+n$ are not coprime. Prove that $a=b$. | First solution. Take some prime number $p>a$ and set $n=p-a$. Then $a+n=p$ is a prime number and by the condition $b+n$ must be divisible by $p$, in particular, $b+n \geqslant p=a+n$. Thus, $b \geqslant a$. Similarly, we get that $a \geqslant b$ and, therefore, $a=b$.
Second solution. Suppose that $a1$. Consequently, ... | b | Number Theory | proof | Yes | Yes | olympiads | false | 8,359 |
4. For any positive numbers $a, b$ and $c$, prove the inequality
$$
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2} \geqslant \frac{3}{2}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right) .
$$ | Solution. Let $x=\frac{a}{b}, y=\frac{b}{c}$ and $z=\frac{c}{a}$, then we need to prove that
$$
(x+y+z)^{2} \geqslant \frac{3}{2}\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)
$$
Notice that
$$
x^{2}+y^{2}+z^{2}+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=(x+y+z)^{2}
$$
so the inequality to be proved ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,360 |
5. Given trapezoid $A B C D$ with bases $A B$ and $C D$, angles $\angle C=30^{\circ}$ and $\angle D=80^{\circ}$. Find $\angle A C B$, if it is known that $D B$ is the bisector of angle $\angle D$. | Answer: $10^{\circ}$
Let $E$ be the intersection point of lines $A D$ and $B C$, and $D^{\prime}$ be the point symmetric to point $D$ with respect to line $B C$. Then $C D = C D^{\prime}$ a... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,361 |
6. On a plane, $n>2$ lines are drawn, no two of which are parallel and no three intersect at the same point. They divide the plane into parts. Prove that the number of parts bounded by exactly three lines is at least 4 more than the number of parts bounded by more than four lines. | Solution. Let $a_{k}$ denote the number of regions bounded by $k$ lines. Note that $a_{2} \leqslant n$. Indeed, being bounded by exactly two lines means that the boundary of the region consists of two rays, each ray can be the boundary of no more than two such regions, so the total number of such regions is no more tha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,362 |
3. There are 2021 minuses written on the board. Petya and Vasya are playing a game. A move consists of either replacing one minus with a plus, or erasing one plus and one minus, or replacing two minuses with three pluses. They take turns, with Petya going first, and the one who cannot make a move loses. Who will win wi... | Answer: Petya will win.
Solution. Petya's first move should be to replace two minuses with three pluses. As a result of this move, the number of minuses on the board will become divisible by 3. After any of Vasya's moves, the number of minuses will change by 1 or 2, and Petya will be able to move in such a way that th... | Petyawillwin | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,365 |
4. a) There is a large group of people - more than 100 people, in which some people are friends. Is it true that any such group can be divided into two groups in a "friendly manner," i.e., such that each person has more or an equal number of friends in their own group compared to the opposite group?
b) Prove that any ... | Answer: No.
Solution. For example, if the company has an odd number of people and everyone is friends with everyone else, then in the group that contains fewer people, this condition will not be met.
b) Let's draw a friendship graph: people are vertices, and friendship relations are edges. We will color vertices of t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,366 |
3. There are 865 plus signs written on the board. Petya and Vasya are playing such a game. A move consists of either replacing one plus with a minus, or erasing one plus and two minuses, or replacing two pluses with two minuses. They take turns, with Petya going first, and the one who cannot make a move loses. Who will... | Answer: Petya will win.
Solution. Petya's first move should be to replace one plus with a minus. As a result, the number of pluses on the board will become divisible by 3. After any of Vasya's moves, the number of pluses will change by 1 or 2, and Petya will be able to move in such a way that the number of pluses is a... | Petyawillwin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,369 |
4. a) There is a large company of people - more than 100 people, some of whom are friends. Is it true that any such company can be divided into three groups in a "friendly way," i.e., such that each person has at least $1 / 3$ of their friends in their own group?
b) Prove that any company consisting of 2022 people can... | Answer: No.
Solution. For example, if the number of people in the company is not divisible by 3 and everyone is friends with everyone else, then one of the groups will end up with fewer than $1 / 3$ of the total number of people, and the condition will not be met in this group.
b) Let's draw a friendship graph: peopl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,370 |
5. Given a triangle $A B C$. On its sides $B C, C A$ and $A B$, points $A_{1}, B_{1}$ and $C_{1}$ are chosen respectively such that the quadrilateral $A B_{1} A_{1} C_{1}$ is cyclic. Prove that
$$
\frac{S_{A_{1} B_{1} C_{1}}}{S_{A B C}} \leqslant\left(\frac{B_{1} C_{1}}{A A_{1}}\right)^{2}
$$
 and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime with... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among seven numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,375 |
1. The picture shows several circles connected by segments. Sasha chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a+b$ must be coprime with $n$; if connected... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among eight numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,376 |
1. The picture shows several circles connected by segments. Tanya chooses a natural number n and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers a and b are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with n; if connected... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, t... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,377 |
1. The picture shows several circles connected by segments. Tanya chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with $n$; if $... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, th... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,378 |
4. The board has the number 5555 written in an even base $r$ ($r \geqslant 18$). Petya found out that the $r$-ary representation of $x^{2}$ is an eight-digit palindrome, where the difference between the fourth and third digits is 2. (A palindrome is a number that reads the same from left to right and from right to left... | Answer: $r=24$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. According to the condition, there exist such $r$-ary digits $a, b, c, d$ that $d-c=2$ and
$$
25(r+1)^{2}\left(r^{2}+1\right)^{2}=a\left(r^{7}+1\right)+b\left(r^{6}+r\right)+c\left(r^{5}+r^{2}\right)+d\left(r^{4}+r^{3}\right)
$$
... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,379 |
1. The picture shows several circles connected by segments. Nastya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must be copr... | Answer: $n=5 \cdot 7 \cdot 11=385$.
Solution. We will make two observations.
1) $n$ is not divisible by 2 and 3. Among seven numbers, there are always three numbers of the same parity. If $n$ is even, then they must be pairwise connected. Moreover, among seven numbers, there will be three numbers that give the same r... | 385 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,380 |
1. In the picture, several circles are drawn, connected by segments. Nastl chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the difference \( a - b \) must... | Answer: $n=3 \cdot 5 \cdot 7=105$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among the six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the diagram.
2) If $p$ is a pr... | 105 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,381 |
1. The picture shows several circles connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime wit... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among any five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) $n$ has at least two distinct pr... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,382 |
1. In the picture, several circles are drawn, connected by segments. Kostya chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be co... | Answer: $n=15$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, suppose $n$ is even. If there are three even and three odd numbers, then each of these triples forms a cycle. Suppose there are four numbers $a, b, c, d$ of the same parity. Then all of them are pairwise connected, and we again get two th... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,383 |
4. a) There is a large company of people - more than 100 people, some of whom are friends. Is it true that any such company can be divided into three groups in a "friendly manner," i.e., such that each person has at least $1 / 3$ of their friends in their own group?
b) Prove that any company consisting of 2022 people ... | Answer: No.
Solution. For example, if the number of people in the company is not divisible by 3 and everyone is friends with everyone else, then one of the groups will end up with fewer than $1 / 3$ of the total number of people, and the condition will not be met in this group.
b) Let's draw a friendship graph: peopl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,387 |
5. (30 points) Point $M$ is the midpoint of side $BC$ of triangle $ABC$. On segment $AC$, there is a point $D$ such that $DM$ and $BC$ are perpendicular. Segments $AM$ and $BD$ intersect at point $X$. It turns out that $AC = 2BX$. Prove that $X$ is the midpoint of segment $AM$.
 At the beginning of each physical education class, 30 students are divided into 3 teams of 10 each. Prove that there will be two students who were on the same team for three consecutive classes. | The first solution. Take all the students from any team from the first lesson and see how they will be distributed among the three teams in the second lesson. Since 10 people are distributed among three teams, by the Pigeonhole Principle, there will be a team that will have at least 4 people.
Now, take these four and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,389 |
4. The city has the shape of a grid rectangle $5 \times 10$ cells: lines are streets, cells are residential blocks. The distance between intersections is measured as the length of the shortest path along the city streets, passing from one intersection to another. For example, for intersections $A, B$, and $C$ in the pi... | 4. а) It is possible. We can construct the required configuration. It is convenient to choose intersections either only on the very bottom or only on the very right street - then it is easy to calculate distances between them by moving along these streets. In the image below, $1=A B, 2=D E, 3=B C, 4=A C, 9=C D$, $11=C ... | possible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,390 |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,394 |
1. Pete and Vasya are playing the following game. Pete marks $k$ cells on a $13 \times 13$ board, after which Vasya places a $1 \times 6$ rectangle on the board and tells Pete which of the marked cells are covered by the rectangle (the rectangle can be rotated). Vasya wins if Pete cannot uniquely determine the position... | Answer: 84.
Solution. We will show that it is necessary to mark at least 84 cells. Consider 7 consecutive cells: $A|B| C|D| E|F| G$. Petya must mark at least one of the cells $A$ and $G$. Indeed, if Petya does not mark cells $A$ and $G$, then he will not be able to distinguish the placement of the rectangle on cells $... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,395 |
1. (10 points) Chess clubs from Moscow, Saint Petersburg, and Kazan agreed to hold a tournament. Each Muscovite played exactly 9 Saint Petersburg residents and $n$ Kazan residents. Each Saint Petersburg resident played exactly 6 Muscovites and 2 Kazan residents. Each Kazan resident played exactly 8 Muscovites and 6 Sai... | # Answer: 4.
Solution. Let the team from Moscow consist of $m$ participants, the team from Saint Petersburg - of $p$ participants, and the team from Kazan - of $k$ participants.
According to the problem, each Muscovite, i.e., each of the $m$ people, played exactly 9 games with the Saint Petersburg residents; and each... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,396 |
2. (10 points) In each cell of a $50 \times 50$ square, a number is written that is equal to the number of $1 \times 16$ rectangles (both vertical and horizontal) in which this cell is an end cell. In how many cells are numbers greater than or equal to 3 written? | Answer: 1600.
Solution. We will denote the cells of the square by pairs $(i, j)$, where $i=1, \ldots, 50, j=$ $=1, \ldots, 50$. We will start the numbering from the bottom left corner of the square.
The cell $(i, j)$ is the rightmost for a horizontal rectangle if $16 \leqslant$ $\leqslant i-$ inequality (1), and the ... | 1600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,397 |
3. (20 points) At the Faculty of Journalism of the University of Enchanted Commonwealth, there are 5 targeted places: 2 for the daytime department and 3 for the evening department. Four chickens have applied for these places: three black and one white. A place is reserved for each chicken immediately after they submit ... | Answer: 0.922.
Solution. The faculty has a total of 5 places, to which 4 applicants are applying. Since all applicants will be admitted to the faculty, the outcome is determined by the selection of those who will be admitted to the daytime department. This can be either two applicants (who can be chosen in six ways) o... | 0.922 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,398 |
4. (20 points) For how many natural numbers $n$, not exceeding 600, are the triples of numbers
$$
\left[\frac{n}{2}\right],\left[\frac{n}{3}\right],\left[\frac{n}{5}\right] \quad \text{and} \quad \left[\frac{n+1}{2}\right],\left[\frac{n+1}{3}\right],\left[\frac{n+1}{5}\right]
$$
different? As always, $[x]$ denotes th... | Answer: 440.
## Solution.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,399 | |
5. (20 points) At a market in Egypt, a tourist is bargaining with a seller over a souvenir worth 10000 Egyptian pounds. The tourist first reduces the price by x percent $(0<x<100)$, then the seller increases the price by $x$ percent, and so on. The number $x$ remains constant throughout the bargaining, and the seller i... | Answer: 5.
Solution. The final cost of the souvenir can be found using one of two formulas (depending on who had the last word): $10000 \cdot\left(1-\frac{x}{100}\right)^{n} \cdot\left(1+\frac{x}{100}\right)^{n}$ or $10000 \cdot\left(1-\frac{x}{100}\right)^{n+1} \cdot\left(1+\frac{x}{100}\right)^{n}$.
After some tran... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,400 |
6. (30 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are 15 and 19, respectively. $AH$ and $BG$ are heights to the line $DC$, and $CF$ is a height to the line $AB$. Points $K, L, M$, and $N$ are the midpoints of segments $AB, CF, CD$, and $AH$ respectively. Find the ratio of the area of trapezoid $ABCD$ to the ar... | Answer: 2 or $\frac{2}{3}$.
Solution. Let the trapezoid $ABCD$ be labeled in a clockwise direction, with the bases assumed to be horizontal. Since the problem does not specify where point $G... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,401 |
7. (30 points) In an acute isosceles triangle \(ABC (AB = BC)\), altitudes \(AD\) and \(CE\) are drawn, intersecting at point \(H\). The circumcircle of triangle \(ACH\) intersects segment \(BC\) at point \(F\). Prove that \(\angle ABH = \angle BFH\).
 On the sides $A B$ and $A D$ of the square $A B C D$, points $E$ and $F$ are marked such that $BE : EA = AF : FD = 2022 : 2023$. Segments $E C$ and $F C$ intersect the diagonal of the square $B D$ at points $G$ and $H$ respectively. Find $GH : BD$. | Answer: $G H: B D=12271519: 36814556$.
First solution. We will solve the problem in a general form: $B E: E A=A F: F D=2022: 2023=$ $=m: n$. Let $B E=A F=m x, E A=F D=n x$, then $B C=C D=(m+n) x$, $B D=(m+n) x \sqrt{2}, E C^{2}=\left(m^{2}+(m+n)^{2}\right) x^{2}, F C^{2}=\left(n^{2}+(m+n)^{2}\right) x^{2}$. Note that ... | 12271519:36814556 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,403 |
9. (40 points) What is the maximum number of numbers that can be chosen among the natural numbers from 1 to 3000 such that the difference between any two of them is different from 1, 4, and 5? | Answer: 1000.
Solution. Let's provide an example. We can choose all numbers divisible by 3. Then the difference between any two numbers will also be divisible by 3, while the numbers 1, 4, and 5 are not divisible by 3.
The estimate is based on the consideration that among 6 consecutive numbers, 3 numbers cannot be ch... | 1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,404 |
10. (40 points) On a board, 1000 different natural numbers are written. It turns out that for each written number \( a \) on the board, there is at least one other number \( b \) such that \( |a-b| \) is a prime number. Prove that it is possible to underline no more than 500 numbers so that for each non-underlined numb... | First solution. Consider a graph where vertices are numbers, and an edge is drawn if the absolute difference between these numbers is a prime number. We will color each connected component of this graph in two colors: first, we will color any vertex red. Then we will color all its neighbors blue. Then we will color all... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,405 |
11. (40 points) Find all pairs of rational numbers $(a, b)$, for which $\sqrt{a}+\sqrt{b}=$ $=\sqrt{2+\sqrt{3}}$. | Answer: $(0.5 ; 1.5)$ and $(1.5 ; 0.5)$.
Solution. Squaring the equality, we get $a+b+2 \sqrt{a b}=2+\sqrt{3}$. Transferring $a+b$ to the right side and squaring again, we get $4 a b=(2-a-b)^{2}+2(2-a-b) \sqrt{3}+$ +3. In this expression, all terms except $2(2-a-b) \sqrt{3}$ are rational, and therefore this term must ... | (0.5;1.5)(1.5;0.5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,406 |
12. (40 points) Solve the equation
$$
2024\left(\frac{1}{n!}-\frac{1}{m!}\right)=\frac{m}{n!}-\frac{1}{m!}
$$
in natural numbers. | Answer: $m=n=1$ or $n=2022, m=2023$.
Solution. Rewrite the equation as $2024(m!-n!)=m \cdot m!-n!,(2024-m) m!=$ $=2023 n!$. Next, consider several scenarios.
If $m=n$, then $2024-m=2023$, that is, $m=n=1$.
If $m>n$, then $m=n+k$, where $k$ is a natural number. In this case, the equation takes the form $(2024-(n+k))(... | =n=1orn=2022,=2023 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,407 |
3. Points $A, B, C, D$ are located on a line (in that order). On the plane containing this line, a point $E$ is marked. It turns out that $A B=B E$ and $E C=C D$. Prove that angle $A E D$ is right.
^{2}=10^{18}+16 \cdot 10^{9}+64=1 \cdot 10^{18}+1 \cdot 10^{10}+6 \cdot 10^{9}+6 \cdot 10+4
$$
and the sum of its digits is 18. The square of the second number ... | doesnotlie | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,412 |
6. (40 points) On a circle with center $O$, points $A$ and $B$ are taken such that the angle $A O B$ is $60^{\circ}$. From an arbitrary point $R$ on the smaller arc $A B$, segments $R X$ and $R Y$ are drawn such that point $X$ lies on segment $O A$ and point $Y$ lies on segment $O B$. It turns out that the angle $R X O... | First solution: Let $r_{1}$ be the radius of the circle centered at point $O$. Since $\angle R X O + \angle R Y O = 180^{\circ}$, the quadrilateral $R X O Y$ is cyclic; denote the radius of its circumscribed circle by $r_{2}$. The diagonal $O R$ equals $r_{1}$ and subtends an arc on which the inscribed angle $R X O$ is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,413 |
6. (40 points) On a circle with center $O$, points $A$ and $B$ are taken such that the angle $A O B$ is $60^{\circ}$. From an arbitrary point $R$ on the smaller arc $A B$, segments $R X$ and $R Y$ are drawn such that point $X$ lies on segment $O A$ and point $Y$ lies on segment $O B$. It turns out that the angle $R X O... | First solution: Let $r_{1}$ be the radius of the circle centered at point $O$. Since $\angle R X O + \angle R Y O = 180^{\circ}$, the quadrilateral $R X O Y$ is cyclic; denote the radius of its circumscribed circle by $r_{2}$. The diagonal $O R$ equals $r_{1}$ and subtends an arc on which the inscribed angle $R X O$ is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,414 |
1. (10 points) 16 children of different heights stood in a circle, all facing the center. Each of them said: “My right neighbor is taller than my left neighbor.” What is the minimum number of children who could be lying?
.
Solution: Let's number the positions of the children in the circle from 1 to 16 counterclockwise. We will divide the children into 2 groups: those standing in positions with odd numbers - the first group, and those standing in positions with even numbers - the second group. Note that the statements of chil... | b)1 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 8,415 |
7. (30 points) On the side $AC$ of triangle $ABC$, where $\angle ACB=45^{\circ}$, a point $K$ is marked such that $AK=2KC$. On the segment $BK$, there is a point $S$ such that $AS \perp BK$ and $\angle AKS=60^{\circ}$. Prove that $AS=BS$.
 The altitudes of an acute, non-isosceles triangle \(ABC\), dropped from vertices \(A\) and \(C\), intersect at point \(H\), and also intersect the angle bisector of \(\angle ABC\) at points \(F\) and \(G\) respectively. Prove that triangle \(FGH\) is isosceles.
 or point $F$ lies between points $B$ and $G$ (right figure).
Let $Q$ and $S$ be the feet of the altitudes... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,417 |
2. In a bag, there are 21 regular dice, each with numbers from 1 to 6 on their faces. All dice are numbered. The audience member takes 3 dice from the bag, shows them to the magician, and puts one die in their pocket. The magician then places the two remaining dice on the table. The audience member writes down the numb... | Solution. The spectator tells the second magician two numbers, each from 1 to 6, and the order in which the numbers are given carries no information. There are 21 such pairs of numbers:
| 11 | 12 | 13 | 14 | 15 | 16 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| | 22 | 23 | 24 | 25 | 26 |
| | | 33 | 34 | 35 | 36 |... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,418 |
4. In the maze diagram in Fig. 1, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, d3, a5.
## Solution.
Fig. 2: Parts of the maze
![](https://cdn.mathpix.com/cropped/2024_05_06_7a66... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,419 |
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