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Problem 7.4. The figure shows 5 lines intersecting at one point. One of the resulting angles is $34^{\circ}$. How many degrees is the sum of the four angles shaded in gray?
 | Answer: $146^{\circ}$.
Solution. Replace the two "upper" gray angles with their equal vertical angles, as shown in the picture:
Now it is clear that the gray angles, together with the $34^{\... | 146 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,944 |
Problem 7.5. On an island, there live knights who always tell the truth, and liars who always lie. One day, 35 inhabitants of the island sat down at 7 tables, with 5 people at each table. Each of these 35 inhabitants was asked: "Are there more than three tables where at least 3 knights are sitting?"
(a) (1 point) What... | Answer:
(a) (1 point) 35.
(b) (3 points) 23.
Solution. (a) It is clear that all 35 people could have answered "Yes" if they were knights. More than 35 could not be.
(b) Consider two cases.
First case. The statement "There are more than three tables where at least 3 knights are sitting" is true.
Then there are at l... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,945 |
Problem 7.6. Natural numbers 13, 14, 15, 16, 17, 18, 19, 20 are placed in the circles on the diagram (each number in one circle) such that the sums of the three numbers along each line are equal. What number can be in the circle $X$? List all possible options.
+a=132+a$ is divisible by 3. Then $a$ is divisible by 3, so $a=15$ or $a=18$.
The value $a=15$ is possible, for example... | 15,18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,946 |
Problem 7.7. On the side $AC$ of an equilateral triangle $ABC$, a point $D$ is marked. On the segments $AD$ and $DC$, equilateral triangles $ADE$ and $DCF$ are constructed outward from the original triangle. It is known that the perimeter of triangle $DEF$ is 19, and the perimeter of pentagon $ABCFE$ is 43.
 $(2$ points) $A B=12$.
(b) $(2$ points) $E F=7$.
Solution. Let $A E=E D=D A=a, D F=F C=C A=b$, then $A B=B C=C A=a+b$. According to the condition, $19=a+b+E F$ and $43=(a+b)+(a+b)+b+E F+a$. Subtract the first equation from the second and divide by two, we get $12=\frac{43-19}{2}=a+b=A B$. Then from the... | AB=12,EF=7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,947 |
Problem 7.8. In a chess tournament, 30 chess players participated, each playing against each other exactly once. A win was awarded 1 point, a draw - $1 / 2$, and a loss - 0. What is the maximum number of chess players who could have ended up with exactly 5 points by the end of the tournament? | Answer: 11.
Solution. Let $N-$ be the number of chess players with 5 points.
Assume $N \geqslant 12$. In each game between two playing chess players, 1 point is played, so the sum of points of all chess players who scored 5 points at the end of the tournament is no less than $\frac{N \cdot(N-1)}{2}$. Then, by the pig... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,948 |
11.5. Find all numbers $a$ such that for any natural $n$ the number $a n(n+2)(n+3)(n+4)$ is an integer.
(O. Podlipsky) | Answer. $a=\frac{k}{6}$, where $k$ is any integer.
Solution. Substituting $n=1, n=3$ and $n=4$, we get that the numbers $60a, 630a$, and $24 \cdot 56a$ are integers. Therefore, $a$ is a rational number, and the denominator $q$ in its irreducible form is a divisor of the numbers 60, 630, and $24 \cdot 56$. Consequently... | \frac{k}{6} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,949 |
11.6. An acute triangle $ABC$ is inscribed in a circle $\omega$. The tangents to $\omega$ drawn through points $B$ and $C$ intersect the tangent to $\omega$ drawn through point $A$ at points $K$ and $L$ respectively. The line through $K$ parallel to $AB$ intersects the line through $L$ parallel to $AC$ at point $P$. Pr... | Solution. First solution. We will prove that point $P$ lies on the perpendicular bisector of segment $B C$. Let $O$ be the center of circle $\omega$, and $X$ be the intersection point of lines $B C$ and $P L$ (see Fig. 3). Since $O$ lies on the perpendicular bisector of segment $B C$, it suffices to prove that $O P \pe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,950 |
11.7. Vasya drew several circles on the plane and drew all possible common tangents to each pair of these circles. It turned out that the drawn lines contain all sides of a certain regular 2011-gon. What is the smallest number of circles Vasya could have drawn?
(N. Agakhanov)
# | # Answer: 504.
Solution. Let the obtained regular 2011-gon be denoted by $M$, its vertices (in a clockwise direction) by $X_{1}, X_{2}, \ldots, X_{2011}$, its inscribed circle by $\omega$, and its center by $O$. We will call the lines containing the sides of the polygon the selected lines.
^{2011}(b+c)^{2011}(c-b)^{2011} \geqslant$
$$
\geqslant\left(b^{2011}-c^{2011}\right)\left(b^{2011}+c^{2011}\right)\left(c^{2011}-b^{2011}\right)
$$
(V. Senderov) | Lemma. For any real $x \geqslant y \geqslant 0$ and natural $n$, the inequality
$$
x^{n}-y^{n} \geqslant(x-y)^{n}
$$
holds.
Proof. Let $x=y+t, t \geqslant 0$. Expanding $x^{n}=(y+t)^{n}$ by the binomial theorem, we have $x^{n}=y^{n}+\ldots+t^{n} \geqslant y^{n}+t^{n}$, or $x^{n}-y^{n} \geqslant t^{n}$. The lemma is ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,952 |
6.2. Find all solutions to the puzzle: ARKA + RKA + KA + A = 2014. (Different letters correspond to different digits, and the same letters correspond to the same digits.) | Answer: $1471+471+71+1=2014$.
Solution. Since in the word ARKA, A is in the thousands place, then $\mathrm{A}=1$ or $\mathrm{A}=2$. Moreover, $\mathrm{A}+\mathrm{A}+\mathrm{A}+\mathrm{A}$ ends in 4, so $\mathbf{A}=\mathbf{1}$. Since $K+K+K$ ends in 1, then $K=7$. Since $\mathbf{P}+\mathbf{P}+2=10$, then $\mathbf{P}=\m... | 1471+471+71+1=2014 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,954 |
6.3. In each of the three chests, Ali-Baba found gold and silver coins; in total, there were 40 gold and 40 silver coins. In the first chest, there were 7 more gold coins than silver coins, in the second chest, there were 15 fewer silver coins than gold coins. Which type of coin is more in the third chest and by how ma... | Answer: There are 22 more silver coins.
Solution. In the first two chests, the total number of gold coins is $7+15=22$ more than the total number of silver coins. Since initially there were an equal number of gold and silver coins, there are 22 fewer gold coins than silver coins in the third chest.
## Evaluation crit... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,955 |
6.4. The frame for three square photographs has the same width everywhere (see figure). The perimeter of one opening is 60 cm, the perimeter of the entire frame is 180 cm. What is the width of the frame? | Answer: 5 cm.
Solution. From the condition of the problem, it follows that the length of one hole's side is 15 cm. Let $d$ cm be the width of the frame, then the perimeter of the rectangle is $8 \cdot 15 + 12 d = 120 + 12 d$ (cm). According to the condition, $120 + 12 d = 180$, which means $d = 5$.
Similar reasoning ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,956 |
6.5. The arithmetic mean of four numbers is 10. If one of these numbers is erased, the arithmetic mean of the remaining three increases by 1; if instead another number is erased, the arithmetic mean of the remaining numbers increases by 2; and if only the third number is erased, the arithmetic mean of the remaining inc... | Answer: will decrease by 6.
Solution. From the fact that the arithmetic mean of four numbers is 10, it follows that the sum of these numbers is 40. Similarly, the sum of three numbers (excluding the first) is 33, the sum of three numbers (excluding the second) is 36, and the sum of three numbers (excluding the third) ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,957 |
10.1. Two runners started simultaneously from the same point. First, they ran along the street to the stadium, and then to the finish - three laps around the stadium. Both ran the entire distance at constant speeds, and during the race, the first runner overtook the second twice. Prove that the first ran at least twice... | Solution. The first runner could only overtake the second on the circular track of the stadium. Since he entered the stadium first, he could not overtake the second runner on his first lap. Therefore, the overtakes must have occurred when the first runner was on his second and third laps. While the first runner was run... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,958 |
10.2. On the side $AC$ of an acute-angled triangle $ABC$, points $M$ and $K$ are chosen such that $\angle ABM = \angle CBK$. Prove that the centers of the circumcircles of triangles $ABM$, $ABK$, $CBM$, and $CBK$ lie on the same circle.
(T. Emelyanova) | Solution. Without loss of generality, we can assume that point $M$ lies between $A$ and $K$. Let $O_{1}, O_{2}, O_{3}$, and $O_{4}$ be the centers of the circumcircles of triangles $A B M, A B K, C B M$, and $C B K$, respectively. The lines $O_{1} O_{3}$ and $O_{1} O_{2}$ are the perpendicular bisectors of segments $B ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,959 |
10.3. Given various natural numbers $a_{1}, a_{2}, \ldots, a_{14}$. On the board, all 196 numbers of the form $a_{k}+a_{\ell}$ are written, where $1 \leqslant k, \ell \leqslant 14$. Can it be that for any combination of two digits among the numbers written on the board, there will be at least one number ending in this ... | Answer: It cannot.
Solution: Let among our 14 numbers there be $a$ even and $b=14-a$ odd numbers. An odd number on the board can only appear as the sum of an even and an odd number, i.e., there will be $a b$ such numbers (each will be written twice). But $a b=\frac{(a+b)^{2}-(a-b)^{2}}{4} \leqslant \frac{(a+b)^{2}}{4}... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,960 |
10.4. Non-zero numbers $a, b, c$ are such that any two of the three equations $a x^{11} + b x^{4} + c = 0, b x^{11} + c x^{4} + a = 0, c x^{11} + a x^{4} + b = 0$ have a common root. Prove that all three equations have a common root. | Solution. Note immediately that all roots of our equations are non-zero, since the free terms are not equal to zero.
Let $p$ be a common root of the first two equations. Then we have
$0=b\left(a p^{11}+b p^{4}+c\right)-a\left(b p^{11}+c p^{4}+a\right)=p^{4}\left(b^{2}-a c\right)-\left(a^{2}-b c\right)$, $0=b\left(b p... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,961 |
9.5. Given an infinite sequence of numbers $a_{1}, a_{2}, \ldots$, in which no two terms are equal. A segment $a_{i}, a_{i+1}, \ldots, a_{i+m-1}$ of this sequence is called a monotonic segment of length $m$ if the inequalities $a_{i}a_{i+1}>\ldots>a_{i+m-1}$ hold. It turns out that for each natural number $k$ the term ... | The first solution. We will call an index $k \geqslant 2$ bad if $a_{k-1}a_{k}k$ and consider a monotonic segment $I$ of length $n+1$ containing $a_{n}$. It cannot contain the terms $a_{k-1}$, $a_{k}$, and $a_{k+1}$ simultaneously; therefore, since $k+1 \leqslant n$, the segment $I$ definitely does not contain $a_{k-1}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,962 |
9.6. What is the smallest natural number $a$ for which there exist integers $b$ and $c$ such that the quadratic trinomial $a x^{2}+b x+c$ has two distinct positive roots, each not exceeding $\frac{1}{1000} ?$
(A. Khryabrov) | Answer. $a=1001000$.
First solution. We will prove that $a \geqslant 1001000$. Notice that if $y$ is a root of the quadratic polynomial $a x^{2}+b x+c$, then $1 / y$ is a root of the quadratic polynomial $c x^{2}+b x+a$. Therefore, in the problem, we need to find the smallest natural $a$ for which the roots $x_{1}$ an... | 1001000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,963 |
9.8. In triangle $ABC$, a circle $\omega$ is inscribed, touching side $BC$ at point $K$. Circle $\omega'$ is symmetric to circle $\omega$ with respect to point $A$. Point $A_{0}$ is chosen such that segments $BA_{0}$ and $CA_{0}$ are tangent to $\omega'$. Let $M$ be the midpoint of side $BC$. Prove that line $AM$ bisec... | Solution. Let points $B^{\prime}, C^{\prime}$, and $K^{\prime}$ be symmetric to points $B, C$, and $K$ with respect to $A$, respectively. Then the circle $\omega^{\prime}$ is inscribed in triangle $A B^{\prime} C^{\prime}$ and touches $B^{\prime} C^{\prime}$ at point $K^{\prime}$. The median $A M$ is the midline in tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,965 |
1. (7 points) 45 candies cost as many rubles as can be bought for 20 rubles. How many candies can be bought for 50 rubles?
Answer: 75 candies. | Solution. Let $x$ be the cost of one candy in rubles. Then $45 x=\frac{20}{x}$, from which $x=\frac{2}{3}$. Then with 50 rubles, one can buy $\frac{50}{x}=75$ candies.
Criteria. Any correct solution: 7 points.
The equation $45 x=\frac{20}{x}$ is correctly set up, but an arithmetic error is made in solving it or in su... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,966 |
2. (7 points) Zhenya arranged the numbers from 1 to 10 in a circle in some order, and Dima wrote the sum of each pair of adjacent numbers in the gaps between them. Could it happen that all the numbers written by Dima were distinct? | Answer: It could.
An example of arranging the numbers is shown below.
Criteria. Any correct solution: 7 points.
Only the correct answer or the correct answer and an incorrect example: 0 po... | Itcould | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,967 |
4. (7 points) Two squares share a common vertex. Find the ratio of segments \(AB\) and \(CD\), as shown in the figure.
 | Answer: $1: \sqrt{2}$.
Solution. Let point $O$ be the common vertex of two squares, and their sides are equal to $a$ and $b$. The diagonals of the squares are $\sqrt{2} a$ and $\sqrt{2} b$ respectively. Moreover, $\angle C O D=\angle C O B+\angle B O D=\angle C O B+45^{\circ}=\angle C O B+\angle A O C=$ $\angle A O B$... | 1:\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,969 |
5. (7 points) The numbers $a, b, c$ and $d$ are such that $a+b=c+d \neq 0, a c=b d$. Prove that $a+c=b+d$.
| Solution. If $a \neq 0$, then substitute $c=\frac{b d}{a}$, we get
$$
a+b=\frac{b d}{a}+d \Longleftrightarrow a+b=\frac{d}{a} \cdot(b+a) \Longleftrightarrow a=d
$$
From this, $b=c$ and $a+c=b+d$.
If $a=0$, then $b \neq 0$ (otherwise $a+b=0$), so $d=0$ (from $a c=b d$). But then the equality $a+b=c+d$ can be rewritte... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,970 |
6. (7 points) Along the road, there are 60 road signs. On each of them, the sum of the distances to the remaining 59 signs is written. Is it possible that 60 different natural numbers are written on the signs? (The distances between the signs are not necessarily integers.) | Answer: Impossible.
Solution. Let's number the signs sequentially from 1 to 60. We will prove that the numbers written on the signs numbered 30 and 31 are the same. Divide the remaining signs into pairs of the form \( k \) and \( k+31 \): 1 and 32, 2 and 33, ..., 29 and 60. Notice that the sum of the distances from si... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,971 |
11.1. When composing the options for the district mathematics olympiad for grades $7, 8, 9, 10, 11$, the jury aims to ensure that in the option for each grade, there are exactly 7 problems, of which exactly 4 do not appear in any other option. What is the maximum number of problems that can be included in the olympiad? | Answer: 27.
Solution. The number of non-repeating tasks will be 20, the number of repeating tasks does not exceed $3 \cdot 5 / 2<8$. We will provide an example of variants with exactly 7 repeating tasks: 7th grade: $1,2,3$; 8th grade: $1,2,3$; 9th grade: $4,5,6$; 10th grade: $4,5,7$; 11th grade: $4,6,7$. | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,972 |
11.2. Place six points on a plane so that any straight line drawn through any two of them divides the plane into two half-planes, each containing a different number of points. | Solution. Place three points at the vertices of an arbitrary triangle $A B C$. Place the remaining three points on the extensions of sides $A B, B C, C A$ beyond points $B, C$ and $A$ respectively. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,973 |
11.3. On the table, 10 cards are arranged in a circle, with one side white and the other black. Initially, all cards are lying with the white side up. In one move, it is allowed to either flip four consecutive cards or to select five consecutive cards and flip four of them - the first, second, fourth, and fifth. Is it ... | Answer: No.
Solution. Mark five cards every other one. Note that with each allowed operation, exactly two marked cards are flipped. Initially, among the marked chips, there were 5 white ones (the upper of the two colors). With each operation, the number of white cards either remains unchanged or changes by 2. Therefor... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,974 |
11.4. Solve the equation:
$$
\begin{gathered}
\frac{10}{x+10}+\frac{10 \cdot 9}{(x+10)(x+9)}+\frac{10 \cdot 9 \cdot 8}{(x+10)(x+9)(x+8)}+\cdots+ \\
+\frac{10 \cdot 9 \ldots 2 \cdot 1}{(x+10)(x+9) \ldots(x+1)}=11
\end{gathered}
$$ | Answer: $-\frac{1}{11}$.
Solution. Let's write the equation in the following form:
$$
\frac{10}{x+10}\left(1+\frac{9}{x+9}\left(1+\frac{8}{x+8}\left(\ldots\left(1+\frac{1}{x+1}\right) \ldots\right)\right)\right)=11
$$
By sequentially transforming the expressions inside the inner parentheses, we will have
$$
\begin{... | -\frac{1}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,975 |
11.5. Three edges of a tetrahedron, emanating from one vertex, are mutually perpendicular, and have lengths 3, 4, and 4. Find the radii of the inscribed and circumscribed spheres of the tetrahedron. | Answer: $R=\frac{\sqrt{41}}{2}, r=\frac{12(10-\sqrt{34})}{33}$.
Solution. a) The sphere circumscribed around the tetrahedron is determined by 4 points not lying in the same plane - the common point of three perpendicular edges and the ends of these edges. This sphere will also be the circumscribed sphere around the re... | R=\frac{\sqrt{41}}{2},r=\frac{12(10-\sqrt{34})}{33} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,976 |
9.5. One hundred integers are written in a circle. Each number is greater than the sum of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among the written ones
(S. Berlov) | Answer: 49.
Solution: Suppose that two non-negative numbers stand next to each other. Then the number preceding them is greater than their sum, meaning it is positive. Similarly, the number before it is also positive, and so on. In the end, we get that all numbers are non-negative; but then the smallest of them cannot... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,977 |
9.6. The field is a square grid of $41 \times 41$ cells, one of which conceals a tank. The fighter aircraft shoots at one cell per shot. If a hit occurs, the tank moves to an adjacent cell along a side; if not, it remains in place. After each shot, the pilot does not know whether a hit occurred. To destroy the tank, it... | Answer. $\frac{3 \cdot 41^{2}-1}{2}=2521$ shots.
Solution. Color the cells in a checkerboard pattern so that the corners of the field are black. Suppose the pilot first shoots at all the white cells, then at all the black cells, and then again at all the white cells. If the tank was on a white cell, the pilot will des... | 2521 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,978 |
9.7. An acute triangle $ABC (AB < AC)$ is inscribed in a circle $\Omega$. Let $M$ be the intersection point of its medians, and $AH$ be the altitude of this triangle. The ray $MH$ intersects $\Omega$ at point $A'$. Prove that the circle circumscribed around triangle $A'HB$ is tangent to $AB$.
(A. I. Golovanov, A. Yaku... | Solution. Choose a point $D$ on $\Omega$ such that $A D \| B C$ (see Fig. 1); then points $A$ and $D$ are symmetric with respect to the perpendicular bisector of $B C$. Let $H^{\prime}$ be the projection of point $D$ onto $B C$, and $K$ be the midpoint of $B C$. By symmetry, $K$ is also the midpoint of segment $H H^{\p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,979 |
9.8. On the board, there are $N \geqslant 9$ different non-negative numbers, each less than one. It turns out that for any eight different numbers on the board, there is a ninth, different from them, such that the sum of these nine numbers is an integer. For which $N$ is this possible?
(F. Nilov) | Answer. Only for $N=9$.
Solution. It is clear that for $N=9$ the required is possible - it is sufficient to write 9 different positive numbers on the board with a unit sum. We will show that for $N>9$ the required is impossible. Suppose the opposite; let $S$ be the sum of all numbers on the board.
Choose arbitrary nu... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,980 |
11.1. An infinite increasing arithmetic progression is such that the product of any two different terms is also a term of this progression. Prove that all its terms are integers.
(N. Agakhanov) | Solution. Let $a$ be one of the terms of the progression, and $d$ its difference. By the condition, the numbers $a(a+d)$ and $a(a+2d)$ are also terms of the progression; hence, their difference has the form $nd$ for some integer $n$, that is, $a(a+2d)-a(a+d)=nd$, or $ad=nd$. Since $d>0$, we get $a=n$, which means $a$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,981 |
11.2. Through the vertices of the base of a quadrilateral pyramid $S A B C D$, lines are drawn parallel to the opposite lateral edges (through vertex $A$ - parallel to $S C$, and so on). These four lines intersect at one point. Prove that the quadrilateral $A B C D$ is a parallelogram. | Solution. Let $P$ be the point of intersection of the given lines. Since $P A \parallel S C$ and $P C \parallel S A$, point $P$ lies in the plane $S A C$, and quadrilateral $A S C P$ is a parallelogram. Therefore, line $S P$ bisects segment $A C$. Similarly, line $S P$ bisects segment $B D$. Therefore, line $S P$ inter... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,982 |
11.3. On a plane, $n>2$ different vectors $\vec{a}_{1}, \vec{a}_{2}, \ldots, \vec{a}_{n}$ of equal lengths are drawn. It turns out that all vectors $-\vec{a}_{1}+\vec{a}_{2}+\ldots+\vec{a}_{n}, \quad \vec{a}_{1}-\vec{a}_{2}+\vec{a}_{3}+\ldots+\vec{a}_{n}$,
$$
\vec{a}_{1}+\vec{a}_{2}+\ldots+\vec{a}_{n-1}-\vec{a}_{n}
$$... | Solution. Let's lay out the vectors $2 \vec{a}_{1}, \ldots, 2 \vec{a}_{n}$ from one point: let $\overrightarrow{O A_{i}}=2 \vec{a}_{i}$. Then all points $A_{1}, \ldots, A_{n}$ are distinct and lie on the circle $\omega$ with center $O$.
Let $\vec{s}=\vec{a}_{1}+\vec{a}_{2}+\ldots+\vec{a}_{n}$. Consider the point $S$ s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,983 |
11.4. The main audience of the company "Horns and Hooves" is a square hall with eight rows of eight seats each. 64 employees of the company took a test in this hall, which had six questions with two answer options for each. Could it have been the case that among the answer sets of the employees, there were no identical... | # Answer. It could.
Solution. Let's denote the first answer to each question as zero, and the second as one. Then, each set of answers corresponds to a set of six zeros and ones. We will represent the audience as an $8 \times 8$ table, in which sets of six digits are placed in the cells. We need to fill the table so t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,984 |
11.5. Prove that for any natural $n$ the inequality $(n-1)^{n+1}(n+1)^{n-1}<n^{2n}$ holds.
(V. Senderov) | Solution. For $n=1$ the inequality holds, since $0<1$. Note that $0<n-1<n+1$, therefore $(n-1)^{n+1}(n+1)^{n-1}<(n-1)^{n}(n+1)^{n}=\left(n^{2}-1\right)^{n}<\left(n^{2}\right)^{n}=n^{2 n}$, which is what we needed to prove. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,985 |
11.6. In a volleyball tournament with 73 teams, each team played against each other once. At the end of the tournament, all teams were divided into two non-empty groups such that every team in the first group won exactly $n$ matches, and every team in the second group won exactly $m$ matches. Could it be that $m \neq n... | Answer: It could not.
Solution: Suppose $m \neq n$. In a tournament with 73 teams, a total of $\frac{73 \cdot 72}{2}=36 \cdot 73$ games are played. Let $x$ teams win $n$ games each, and the remaining $73-x$ teams win $m$ games each. Then we have the equation $x \cdot n+(73-x) \cdot m=36 \cdot 73$, from which $x \cdot(... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,986 |
11.7. Given different natural numbers $a, b$. On the coordinate plane, the graphs of the functions $y=\sin a x, y=\sin b x$ are drawn, and all their intersection points are marked. Prove that there exists a natural number $c$, different from $a, b$, such that the graph of the function $y=\sin c x$ passes through all th... | Solution. Let $a > b$ for definiteness.
If $\left(x_{0}, y_{0}\right)$ is one of the intersection points, then $\sin a x_{0} - \sin b x_{0} = 0$, or
$$
\sin \frac{a-b}{2} x_{0} \cdot \cos \frac{a+b}{2} x_{0} = 0
$$
Thus, $\frac{a-b}{2} x_{0} = k \pi$ or $(a+b) x_{0} = (2 k+1) \pi$ for some integer $k$, which implies... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,987 |
11.8. A convex quadrilateral $A B C D$ is such that $A B \cdot C D=$ $=A D \cdot B C$. Prove that $\angle B A C+\angle C B D+\angle D C A+\angle A D B=$ $=180^{\circ}$. | The first solution. Let $f(A B C D)$ denote the sum of the four angles in the condition. Note that if the quadrilateral $A B C D$ is cyclic, then the statement is true. Indeed, in this case $f(A B C D)=(\angle B A C+\angle C A D)+(\angle D C A+\angle A C B)=\angle B A D+$ $+\angle B C D=180^{\circ}$.
Now suppose the q... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,988 |
1. Answer: $2019^{2021} \cdot 2021^{2019}<2020^{2020} \cdot 2020^{2020}$. | Solution. Let's consider the quotient of these expressions:
$\frac{2019^{2021} \cdot 2021^{2019}}{2020^{2020} \cdot 2020^{2020}}=\frac{2019^{2020} \cdot 2021^{2020}}{2020^{2020} \cdot 2020^{2020}} \cdot \frac{2019}{2020}=\left(\frac{2019 \cdot 2021}{2020^{2}}\right)^{2020} \cdot \frac{2019}{2020}$.
Notice that the la... | 2019^{2021}\cdot2021^{2019}<2020^{2020}\cdot2020^{2020} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 17,989 |
# 1. CONDITION
How many integers between 100 and 10000 are there such that their representation contains exactly 3 identical digits? | Solution. If the digit that repeats three times is zero, then by adding one of the other 9 digits in front, we get 9 numbers. If the digit $a \neq 0$ repeats in the number, then the number of numbers with different $b, b \neq a$, will be $9 \cdot 4 \cdot 9$. Therefore, in total, there are 333 numbers.
Answer: 333. | 333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,990 |
# 4. CONDITION
Find the real solutions of the equation $(x+2)^{4}+x^{4}=82$.
Hint. Let $x+1=y$, then $(y+1)^{4}+(y-1)^{4}=82$, or
$$
y^{4}+6 y^{2}-40=0 \text {, hence } y^{2}=-10 \text { or } y^{2}=4 \text {. }
$$ | Answer. $x_{1}=-3, x_{2}=1$.
# | x_{1}=-3,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,991 |
7.1. The price for a ride on the "Voskhod" carousel in February 2020 was 300 rubles per person. In March, the price was reduced, and the number of visitors increased by $50 \%$, while the revenue increased by $25 \%$. By how many rubles was the price reduced? | Answer: 50.
Solution. The entry fee for two people in February 2020 was 600 rubles. In March, instead of every two people, the stadium is visited by three people, as the number of visitors increased by $50 \%$. Since the total collection increased by $25 \%$, they will pay $600+0.25 \cdot 600=750$ (rub.), and thus one... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,992 |
7.2. The external angles of a triangle are proportional to the numbers $5: 7: 8$. Find the angle between the altitudes of this triangle drawn from the vertices of its smaller angles. | Answer: $90^{0}$.
Solution. The sum of the exterior angles of a triangle is
$$
180^{\circ} \cdot 3-180^{\circ}=360^{\circ}
$$
From this, we get that one of the exterior angles of the triangle is $\frac{360^{\circ}}{5+7+8} \cdot 5=90^{0}$, and therefore the adjacent interior angle of the triangle is $90^{0}$. Consequ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,993 |
7.4. A rope was cut into 5 pieces, then some of them were cut into 5 parts each. Then, some of the resulting pieces were again cut into 5 pieces, and this was done several times. Could it result in 2019 pieces? | Answer: No.
Solution. Each cut increases the number of pieces by 4. Therefore, after $n$ cuts, the number of pieces will increase by $4 n$. Thus, after $n$ cuts, the number of pieces will be $1+4 n=2019$, from which $4 n=2018$. But 2018 is not divisible by 4.
Comment. An answer without justification - 0 points. | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,995 |
7.5. Miron and Varya are playing the following game. There is a pile of 32 stones on the table. The players take turns, with Miron starting first. On their turn, a player divides any pile that has more than one stone into several equal piles. The player who cannot make a move (when there is exactly one stone in each pi... | Answer: Varya.
Solution. Varya wins regardless of the moves the players make. Since the number 32 has no odd divisor, each player will divide one pile into an even number of piles on their turn, and the parity of the number of piles will change. Before Miron's move, there will always be an odd number of piles, and bef... | Varya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,996 |
1. Once in a physics and mathematics school, September was declared the month of information technology. Therefore, every Monday in September, there was 1 computer science lesson in each class, every Tuesday - 2 lessons, every Wednesday - 3 lessons, every Thursday - 4 lessons, and every Friday - 5 computer science less... | Answer: Wednesday.
Solution. In 1 full week, there are $1+2+3+4+5=15$ lessons. From September 2 to September 29 (28 days), 4 full weeks will pass, which is 60 lessons. Therefore, on September 30, there were $64-60=4$ lessons, meaning it was a Thursday. Thus, the Thursdays were $23, 16, 9$, and September 2, and Septemb... | Wednesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,997 |
2. On the table, there are 2 candles, each 20 cm long, but of different diameters. The candles burn uniformly, with the thin candle burning completely in 4 hours, and the thick one in 5 hours. After how much time will the thin candle be half the length of the thick one if they are lit simultaneously? | Answer: 3 hours 20 minutes ( $31 / 3$ hours).
Solution 1. In 1 hour, the thin candle burns down by $20 / 4=5$ cm, and the thick candle by $20 / 5=4$ cm. After t hours, the length of the thin candle will be $20-5 \mathrm{t}$, and the thick candle will be $20-4 \mathrm{t}$. From the condition, we get the equation $20-4 ... | \frac{31}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,998 |
4. On the extension of side $\mathrm{BC}$ of triangle $\mathrm{ABC}$ beyond point $\mathrm{C}$, mark point $\mathrm{D}$ such that $\mathrm{CD}=\mathrm{BC}$, and on the extension of side $\mathrm{CA}$ beyond point $\mathrm{A}$, mark point $\mathrm{E}$ such that $\mathrm{AE}=2 \mathrm{CA}$. Prove that if $\mathrm{AD}=\ma... | Solution. Extend side AC beyond point C and mark point F on the extension such that CF = AC. Then, $\triangle FBC = \triangle ADC$ by two sides and the included angle (FC = AC by construction, BC = DC by condition, $\angle FCB = \angle ACD$ as vertical angles). From the equality of the triangles, it follows that BF = A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,000 |
5. On the island, there are 217 residents, 17 of whom are knights, and the remaining 200 are tricksters. One day, a private detective arrived on the island and decided to find out who is who. For this, he asked each resident to write a list of 200 people they consider to be tricksters. The detective does not know who w... | Answer: Yes, it can always.
Solution. We will combine identical lists into groups. Let's call a group important if it contains at least 17 lists. Note that the number of important groups is no more than 12 (otherwise, the number of residents is at least $13 * 17 = 221$, which is more than 217).
Since all the knights ... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,001 |
8.1. Does there exist a ten-digit number, divisible by 11, in which all digits from 0 to 9 appear? | Answer. Yes, for example, 9576843210.
Solution. Let's consider a possible way to find the required number. Note that a number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 11. If we write all ten digits in descend... | 9576843210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,002 |
8.2. There is a balance scale without weights and 11 visually identical coins, among which one may be counterfeit, and it is unknown whether it is lighter or heavier than the genuine coins (genuine coins have the same weight). How can you find at least 8 genuine coins in two weighings? | Solution. Let's divide the coins into three piles of three coins each. Compare pile 1 and pile 2, and then compare pile 2 and pile 3. If all three piles weigh the same, then all the coins in them are genuine, and we have found 9 genuine coins. Otherwise, one of the piles differs in weight from the others, and the count... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,003 |
8.3. Given a convex quadrilateral $A B C D$. Prove that the distances from points $A$ and $B$ to the intersection point of the diagonals of the quadrilateral are equal, if it is known that the perimeters of triangles $A B C$ and $A B D$ are equal, as well as the perimeters of triangles $A C D$ and $B C D$ are equal. | Solution. Let $O$ be the point of intersection of the diagonals of the quadrilateral. By the condition, $A C+B C=A D+B D, A C+A D=B C+B D$. Writing the second equality as $A C-B C=B D-A D$ and adding it to the first, we get $2 A C=$ $2 B D$, that is, $A C=B D$. Therefore, $A D=B C$, hence triangles $A D B$ and $B C A$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,004 |
8.4. On each cell of a chessboard, there is a stack of several chips. In one move, one of the two operations is allowed:
- remove one chip from each stack in any column;
- double the number of chips in each stack in any row.
Is it possible to clear the board of chips in some number of such operations? | Answer: Yes.
Solution. Let's record the number of chips in each cell of the board in an $8 \times 8$ table. In this case, there are two types of operations possible: subtracting one from each cell in any column; doubling any row. Our goal is to obtain a table filled with zeros.
First, we will turn all the numbers in ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,005 |
8.5. Find all numbers of the form $22 \ldots 2$ that can be represented as the sum of two perfect squares. | Answer: 2.
Solution. Let $22 \ldots 2=a^{2}+b^{2}$ for some integers $a$ and $b$.
If the numbers $a$ and $b$ are even, then the sum of their squares is divisible by 4, but the number $22 \ldots 2$ is not.
Thus, the numbers $a$ and $b$ can only be odd:
$$
a=2 k+1, b=2 l+1(k, l \in Z)
$$
Therefore, the sum
$$
a^{2}... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,006 |
5. Let $x$ be the length of the smaller segment. On the upper side of the quadrilateral, which has a length of 1, there are 3 small segments and one large segment. Therefore, the length of the large segment is $(1-3x)$. On the lower side of the quadrilateral, which has a length of 2, there are 3 large segments and one ... | Answer: the ratio $5: 1$. | 5:1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,007 |
Task 10.1. Find any natural $x$ such that the value of the expression $2^{x}+2^{8}+2^{11}$ is a square of a natural number. | Answer: 12.
Solution. Note that $x=12$ works: $2^{12}+2^{8}+2^{11}=\left(2^{6}\right)^{2}+\left(2^{4}\right)^{2}+2 \cdot 2^{6} \cdot 2^{4}=\left(2^{6}+2^{4}\right)^{2}$. Remark. The number $x=12$ is the only one that fits the condition of the problem. | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,008 |
Problem 10.2. There are 36 balls lying in a circle, each of which is either red or blue (balls of each of these colors are present). It is known that:
- for any red ball, there is exactly one red ball such that there is exactly one ball between them;
- for any red ball, there is exactly one red ball such that there ar... | Answer:
(a) (2 points) 12.
(b) (2 points) 24.
Solution. By the condition, there will be two red balls, between which lies exactly one ball. Number the balls clockwise with numbers \(1, 2, \ldots, 36\) such that the 1st and 3rd balls are red. From the condition, it follows that the 35th ball is blue (otherwise, the f... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,009 |
Problem 10.3. Quadrilateral $ABCD$ is such that $\angle ABD = \angle ACD = 90^\circ$, $\angle CAD = 42^\circ$. Rays $CB$ and $DA$ intersect at point $K$. It is known that $BK = 3$, $AD = 6$.
... | # Answer:
(a) (2 points) $28^{\circ}$.
(b) (2 points) $34^{\circ}$.
Solution. From the condition, it follows that points $A, B, C, D$ lie on a circle with diameter $A D$. Connect point $B$ to the midpoint $O$ of segment $A D$, which is the center of this circle (Fig. 6). Clearly, $A O=O D=O B=B K=3$.
Triangles $K B... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,010 |
Problem 10.4. Given a rectangular grid of cells. It is known that:
- there are exactly 940 ways to cut out a $1 \times 2$ rectangle from it along the grid lines;
- there are exactly 894 ways to cut out a $1 \times 3$ rectangle from it along the grid lines.
How many ways are there to cut out a $1 \times 5$ rectangle f... | Answer: 802.
Solution. Let the table have $a$ rows and $b$ columns. Without loss of generality, $a \leqslant b$. It is clear that at least one of the numbers $a$ and $b$ is not less than 2.
- Suppose $a=1$. Then from the table $1 \times b$, a rectangle $1 \times 2$ can be cut in $b-1=940$ ways, and a rectangle $1 \ti... | 802 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,011 |
Problem 10.5. Several sweet-tooths participated in a candy eating contest. Each participant ate an integer number of candies, and any two participants ate a different number of candies. Summarizing the contest, the jury ranked all the people in descending order of the number of candies eaten (for example, the winner at... | Answer: 21.
Solution. Let there be $n$ sweet-tooth participants in total, and they ate $S$ candies in total. From the condition, it follows that the winner ate $\frac{S}{15}$ candies, the person in third place $-\frac{S}{21}$ candies, and the person in last place $-\frac{S}{22}$ candies.
All participants, except the ... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,012 |
Problem 10.6. Given trapezoid $ABCD$. Segments $AB$ and $CD$ - its lateral sides - are equal to 24 and 10, respectively. Points $X$ and $Y$ are marked on side $AB$ such that $AX=6, XY=8, YB=10$. It is known that the distances from points $X$ and $Y$ to line $CD$ are 23 and 27, respectively.
 (1 point) 100.
(b) (3 points) 260.
Fig. 7: to the solution of problem 10.6
Solution. We will prove that the distances from points $A$ and $B$ to the line $C D$ are 20 and 32,... | 260 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,013 |
Problem 10.7. On an island, there are 23 knights and 200 liars; all residents have different names. A tourist who knows this asked each of the 223 residents to write down 200 names of liars on a piece of paper. Each knight correctly wrote down 200 names of liars, while each liar wrote down an arbitrary list of 200 name... | Answer: 16.
Solution. First, let's outline a strategy for the liars that allows identifying no more than 16 of them. Let's call the knights $R_{1}, R_{2}, \ldots, R_{23}$, and the liars $-L_{1}, L_{2}, \ldots, L_{200}$. All knights will list $L_{1}, L_{2}, \ldots, L_{200}$ in their lists. For each integer $0 \leqslant... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,014 |
Problem 10.8. Non-negative numbers $a, b, c$ sum up to 1. Find the maximum possible value of the quantity
$$
(a+3 b+5 c) \cdot\left(a+\frac{b}{3}+\frac{c}{5}\right)
$$ | Answer: $\frac{9}{5}$.
Solution. It is not hard to verify that when $a=\frac{1}{2}, b=0, c=\frac{1}{2}$, the value $\frac{9}{5}$ is obtained. We will prove that it does not exceed $\frac{9}{5}$ for any non-negative $a, b, c$ with a sum of 1.
Notice that
$$
(a+3 b+5 c) \cdot\left(a+\frac{b}{3}+\frac{c}{5}\right)-\fra... | \frac{9}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,015 |
Problem 4. Prove that for any natural number, there exists a multiple of it that is written only with the digits 1 and 0. | Solution. Consider $n+1$ numbers $1,11,111, \ldots 11 \ldots 1$. Among these numbers, there exist at least two numbers that give the same remainder when divided by $n$. The difference between such numbers is written using digits 1 and 0.
Problem 5. (L. Emelyanov) The angle bisectors of angles $A$ and $C$ of triangle $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,016 |
1. It is known that $\sin (\alpha+\beta)=0.2$ and $\cos (\alpha-\beta)=0.3$. Calculate $\sin \left(\alpha+45^{\circ}\right) \cdot \sin (\beta+$ $\left.45^{\circ}\right)$. | Answer: 0.25.
Solution. Using the formula for the sine of the sum of angles, we have
$$
\sin \left(\varphi+45^{\circ}\right)=\frac{\sqrt{2}}{2}(\sin \varphi+\cos \varphi)
$$
and thus, $P=\sin \left(\alpha+45^{\circ}\right) \cdot \sin \left(\beta+45^{\circ}\right)=\frac{1}{2}(\sin \alpha+\cos \alpha)(\sin \beta+\cos ... | 0.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,017 |
2. For what values of $q$ is one of the roots of the equation $x^{2}-12 x+q=0$ the square of the other? | Answer: -64 or 27.
Solution. Let the roots of the equation be $a$ and $a^{2}$. By Vieta's theorem, $a + a^{2} = 12$, $a \cdot a^{2} = q$. Therefore, $a = -4$ or $a = 3, q = a^{3}$.
Criteria. Only the answer - 1 point. In the case of a complete justification, the score is 7 points. | -64or27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,018 |
3. Let $x_{1}, x_{2}, \ldots, x_{100}$ be some numbers belonging to the interval [0; 1]. Is it true that there exists a number $x$ on this interval such that $\left|x-x_{1}\right|+\left|x-x_{2}\right|+\ldots+\mid x-$ $\left|x_{100}\right|=50 ?$ | Answer. Correct.
Solution. Consider the function $f(x)=\left|x-x_{1}\right|+\left|x-x_{2}\right|+\ldots+\left|x-x_{100}\right|-$ 50, continuous on the interval [0;1]. We have $f(0)=x_{1}+x_{2}+\ldots+x_{100}-50, f(1)=$ $\left(1-x_{1}\right)+\left(1-x_{2}\right)+\ldots+\left(1-x_{100}\right)-50$. Therefore, $f(0)+f(1)=... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,019 |
4. Two circles, the radii of which are in the ratio 2:3, touch each other internally. A line is drawn through the center of the smaller circle, perpendicular to the line of centers, and tangents are drawn from the points of intersection of this line with the larger circle to the smaller circle. Find the angles between ... | Answer: All angles are right angles.
Solution. By the Pythagorean theorem, $O_{2} B=$ $\sqrt{9-1}=2 \sqrt{2}$. From the right triangle $O_{2} B M$, we find $B M=2$. Therefore, triangle $O_{2} B M$ is isosceles with an angle of $45^{\circ}$. Similarly, the angle $O_{2} B N$ is $45^{\circ}$.
Thus, the tangents drawn fr... | All\angles\\right\angles | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,020 |
Problem 4.1. On nine cards, the numbers from 1 to 9 are written (each one only once). These cards were laid out in a row such that there are no three consecutive cards with numbers in ascending order, and there are no three consecutive cards with numbers in descending order. Then, three cards were flipped over, as show... | Answer: The number 5 is written on card $A$, the number -2 on card $B$, and the number -9 on card $C$.
Solution. The missing numbers are $-2, 5$, and 9.
If the number 5 is on card $B$, then we get the consecutive numbers 3, 4, 5. If the number 5 is on card $C$, then we get the consecutive numbers 8, 7, 5. Therefore, ... | 5,-2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,021 |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,022 |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,023 |
Problem 4.8. The figure shows a road map between the houses of five children. The shortest distance by road from Asya to Galia is 12 km, from Galia to Borya - 10 km, from Asya to Borya 8 km, from Dasha to Galia - 15 km, from Vasya to Galia - 17 km. How many kilometers is the shortest distance by road from Dasha to Vasy... | Answer: 18.
Solution. Add the distances from Dasha to Gala and from Vasya to Gala: $15+17=32$. This will include the "main" road (from Dasha to Vasya) and twice the "branch" from it to Gala.
Add the distance from Asey to Bory to the obtained sum: $32+8=40$. Now all three branches (to Gala - twice) and the main road w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,024 |
Problem 5.3. The figure shows a plan of the road system of a certain city. In this city, there are 8 straight streets, and 11 intersections are named with Latin letters $A, B, C, \ldots, J, K$.
Three police officers need to be placed at some intersections so that at least one police officer is on each of the 8 streets... | Answer: $B, G, H$.
Solution. The option will work if police officers are placed at intersections $B, G, H$. It can be shown that this is the only possible option.
Since there are only three vertical streets, and each must have one police officer, there are definitely no police officers at intersections $C$ and $K$. T... | B,G,H | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,025 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,026 |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,027 |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,030 |
Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.
![](https://cdn.mathpix.com/c... | Answer: $52.5$.
Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.
Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.
![](https://cdn.mathpix.c... | 52.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,032 |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,033 |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,034 |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), the right triangles $X A B$ and $Y D A$ are similar by the acut... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,035 |
Problem 9.5. Leonid has a white checkered rectangle. First, he painted every other column gray, starting with the leftmost one, and then every other row, starting with the topmost one. All cells adjacent to the border of the rectangle ended up being painted.
How many painted cells could there be in the rectangle if 74... | Answer: 301 or 373.
Solution. From the condition, it follows that the rectangle has an odd number of both rows and columns. Let's number the rows from top to bottom with the numbers $1,2, \ldots, 2 k+1$, and the columns from left to right with the numbers $1,2, \ldots, 2 l+1$ (for non-negative integers $k$ and $l$). W... | 301or373 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,036 |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,037 |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,039 |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them as $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,040 |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,041 |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
 in a row so that in any pair of adjacent numbers, one divides the other. | Answer. For example, $9,3,6,2,4,8,1,5$.
Remark. There are other examples. However, in each correct example, the number 5 is at the edge, and next to it is the number 1.
Comment. Any correct example - 7 points. | 9,3,6,2,4,8,1,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,044 |
5.3. Can a rectangle be cut into 5 squares, at least four of which have different sizes? | Answer. Yes, it is possible.
Solution. An example is shown in Fig. 1.
Comment. Any correct drawing (even without explanations) - 7 points. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,045 |
5.4. At a round table, 10 people are sitting, some of them are knights, and the others are liars (knights always tell the truth, while liars always lie). Each of them said: "Both of my neighbors are liars." Then one person left the table. Could it have turned out that after this, each of the remaining people at the tab... | Answer: Could not.
First solution. Note that initially, there were at least 2 knights sitting at the table - otherwise, there would be 3 liars sitting next to each other, and the middle one would have told the truth, which is impossible. Since there is a knight sitting at the table, there are at least two liars sittin... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,046 |
1. Is it possible to draw four lines on a plane and mark seven points on them so that there are exactly three marked points on each line? | Answer: possible
Instructions. Example 1. Among the four lines, there are two parallel ones. The other two intersect them at different points and intersect each other. We mark five points of intersection of the lines and one more point on each of the parallel lines. Seven points are marked, and there are three marked ... | possible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,047 |
2. A palindrome is a natural number that reads the same from left to right as from right to left. Find all four-digit palindromes that are equal to the sum of two three-digit palindromes | Answer: 1111 and 1221.
Instructions. Since $1111=505+606$ and $1221=555+666$, the numbers specified in the answer are suitable. The sum of two three-digit numbers is a three-digit or four-digit number starting with 1. Since it is a palindrome, it ends with 1. The hundreds and tens digits must be the same in the four-d... | 11111221 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,048 |
3. On the line AB, point O is marked and from it rays OC, OD, OE, OF are drawn in the specified order into one half-plane of line AB (ray OC lies between rays OA and OD). Find the sum of all angles with vertex O, whose sides are rays OA, OC, OD, OE, OF, OB, if $\angle \mathrm{COF}=97^{\circ}, \angle \mathrm{DOE}=35^{\c... | Answer: $1226^{\circ}$.
Instructions. There are four pairs of adjacent angles and a straight angle, which sum up to $180^{\circ}$ or are equal to $180^{\circ}$. There are two pairs of angles whose sum equals $\angle \mathrm{COF}$, and the angle itself. And there is still $\angle \mathrm{DOE}$. Therefore, the sum of th... | 1226 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,049 |
4. There are twenty-seven weights weighing 1g, 2g, .., 27g. They were all placed on the scales so that equilibrium was achieved. Vasya claims that he can always remove three weights, one of which weighs 1g, such that the equilibrium is maintained. Is Vasya right? | Answer: incorrect.
Instruction. Equilibrium will be achieved when weights $1,20,21, \ldots, 27$ are placed on one pan, and the rest on the other. In this case, it is impossible to remove three weights to achieve equilibrium. If equilibrium is achieved after removing three weights, then it would also be achieved if the... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,050 |
5. In a room, 15 chairs are arranged in a circle. Three jewelers, when no one is watching, sit on three adjacent chairs, and the one sitting in the middle chair hides a diamond in the chair he is sitting on. The inspector has several detectors that show whether someone has sat on a chair or not. What is the minimum num... | Answer: 9 detectors.
Instructions. Evaluation. Consider five chairs such that between any two nearest ones there are two other chairs. If there is no detector in two nearest ones, then by adding to the two chairs between them any of them, we get two possible seating arrangements of jewelers, in which the detector read... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,051 |
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