problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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2. A ball of mass $\mathrm{m}_{1}$, sliding on a smooth horizontal surface, collides with a ball of mass $\mathrm{m}_{2}$, which was at rest on the same surface. After the central elastic collision, the balls slide in the same direction, with the speed of the ball of mass $\mathrm{m}_{2}$ being 4 times the speed of the... | 2. CSE: $\frac{m_{1} V_{0}^{2}}{2}=\frac{m_{1} V^{2}}{2}+\frac{m_{2}(4 V)^{2}}{2} ; \quad\left(\frac{V_{0}}{V}\right)^{2}=1+16 \frac{m_{2}}{m_{1}}$
$$
\text { CSI: } m_{1} V_{0}=m_{1} V+m_{2} \cdot 4 V ; \quad \frac{V_{0}}{V}=1+4 \frac{m_{2}}{m_{1}}
$$
$(2) \rightarrow(1):$
$1+8 \frac{m_{2}}{m_{1}}+16\left(\frac{m_{... | \frac{m_{2}}{m_{1}}=\frac{1}{2},\quad\frac{V_{0}}{V}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,999 |
3. A ball sliding on a smooth horizontal surface catches up with a block moving along the same surface. The ball's velocity is perpendicular to the edge of the block, against which it collides. The mass of the ball is much less than the mass of the block. After an elastic collision, the ball slides along the surface in... | 3. Law of velocity addition:
$$
\begin{array}{cl}
(V-U)-U=\frac{1}{2} V ; & V\left(1-\frac{1}{2}\right)=2 U \\
\frac{V}{2}=2 U ; & \frac{V}{U}=4
\end{array}
$$
 | \frac{V}{U}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,000 |
4. A vessel is divided by a partition into 2 unequal parts. In one part of the vessel, there is $1 / 3$ mole of a monatomic ideal gas with a temperature of $T_{1}=300 \mathrm{~K}$. In the other part, there is $1 / 5$ mole of another monatomic ideal gas with a temperature of $T_{2}=500$ K. The walls of the vessel and th... | 4. Since p-const, then $\frac{v_{1} T_{1}}{V_{1}}=\frac{v_{2} T_{2}}{V_{2}} ; \quad \frac{V_{2}}{V_{1}}=\frac{v_{2} T_{2}}{v_{1} T_{1}}$
$3 C E: v_{1} C_{\mathrm{V}} T_{1}+v_{2} C_{V} T_{2}=\left(v_{1}+v_{2}\right) C_{v} T \Rightarrow$ $\Rightarrow T=\frac{v_{1} T_{1}+v_{2} T_{2}}{v_{1}+v_{2}}=\frac{\frac{1}{3} \cdot ... | T=375K,\frac{P'}{P}=1 | Other | math-word-problem | Yes | Yes | olympiads | false | 23,001 |
1. V a r i a n t A. According to the condition, the function $y=\sin x+a-b x$ turns into zero at exactly two points $x_{1}$ and $x_{2}, x_{1}<x_{2}$.
These points divide the number line into 3 intervals $\left(-\infty, x_{1}\right]$, $\left(x_{1}, x_{2}\right],\left(x_{2},+\infty\right)$. Since $b \neq 0$, and $|\sin ... | The solution to problem B is analogous.
Note. Prove independently that the system has exactly one solution.
Comment. This problem is based on the following physical problem proposed by T. Golenishcheva-Kutuzova and I. Yashchenko.
Vasya is descending with a parachute at a constant speed of $3 \mathrm{~m} / \mathrm{c}... | 3\, | Calculus | math-word-problem | Yes | Yes | olympiads | false | 23,005 |
1. Given a strictly increasing function $f: \mathbb{N}_{0} \rightarrow \mathbb{N}_{0}$ (where $\mathbb{N}_{0}$ is the set of non-negative integers), which satisfies the relation $f(n+f(m))=f(n)+m+1$ for any $m, n \in \mathbb{N}_{0}$. Find all possible values that $f(2023)$ can take.
(T.A. Garmanova) | Solution. 1) Substituting $m=0, n=0$, we get $f(f(0))=f(0)+1$.
If $f(0)=0$, then we get $f(0)=f(0)+1$, which is impossible.
2) Let $f(0)=a$, then $a \in \mathbb{N}$. From the first point, we get that $f(a)=a+1$. If we substitute $m=0, n=a$, then we get that $f(2a)=f(a)+1=a+2$. Therefore, the values of the function at... | 2024 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,006 |
2. What is the smallest number of different integers that need to be taken so that among them one can choose both a geometric and an arithmetic progression of length $5$?
(M.A. Evdokimov) | Solution. Let's provide an example of six integers that satisfy the condition: $-8, -2, 1, 4, 10, 16$. The numbers $1, -2, 4, -8, 16$ form a geometric progression, while the numbers $-8, -2, 4, 10, 16$ form an arithmetic progression.
We will show that no five distinct integers satisfy the condition of the problem. Sup... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,007 |
3. In triangle $ABC$, the altitudes $BE$ and $CF$ intersect at point $H$, point $M$ is the midpoint of side $BC$, and $X$ is the intersection point of the internal tangents to the incircles of triangles $BMF$ and $CME$. Prove that points $X$, $M$, and $H$ lie on the same line.
(I. N. Mikhailov) | Solution.
Method 1. Let $S, T$ be the midpoints of the altitudes $B E$ and $C F$, and $L, N$ be the midpoints of the segments $B F$ and $C E$. Denote the incircles of triangles $B M F$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,008 |
4. We have perfectly accurate two-pan scales and a set of 50 weights, the weights of which are equal to $\operatorname{arctg} 1$, $\operatorname{arctg} \frac{1}{2}$, $\operatorname{arctg} \frac{1}{3}$, ..., $\operatorname{arctg} \frac{1}{50}$. Prove that it is possible to choose 10 of them and distribute 5 weights on e... | Solution. First, we will show that in this set, there are triples of weights, one of which balances the other two. All weights do not exceed $\pi / 4$, so the equality $\operatorname{tg}\left(\operatorname{arctg} \frac{1}{n}+\operatorname{arctg} \frac{1}{m}\right)=\operatorname{tg}\left(\operatorname{arctg} \frac{1}{k}... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 23,009 |
5. In a convex polyhedron, let B, P, and T denote the number of vertices, edges, and the maximum number of triangular faces that share a common vertex, respectively. Prove that $B \sqrt{P+T} \geqslant 2P$.
For example, for a tetrahedron ( $B=4, P=6, T=3$ ), the equality holds, while for a triangular prism ( $B=6, P=9,... | Solution. The degree of a vertex of a polyhedron is the number of edges emanating from it. Vertices are called adjacent if they are connected by an edge.
Let $A$ be an arbitrary vertex of the polyhedron, $k$ its degree, and $m_{j}$ the degrees of all vertices adjacent to it $(j=1,2, \ldots, k)$, numbered in any order.... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,010 |
3. Find the smallest natural number $N>9$, which is not divisible by 7, but if any of its digits is replaced by 7, the resulting number is divisible by 7. | (M. A. Evdokimov) Solution. Let the smallest such number be of the form $\overline{a_{1} a_{2} \ldots a_{n}}$. From the condition, it follows that among its digits there are no 0 and 7. If the number contains digits 8 or 9, then they can be replaced by 1 or 2, respectively, to obtain a smaller number with the same prop... | 13264513 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,012 |
5. In the laboratory, 120 visually indistinguishable test tubes are standing on a shelf, 118 of which contain a neutral substance, one contains poison, and one contains an antidote. The test tubes have been randomly mixed, and it is necessary to find the test tube with the poison and the test tube with the antidote. Fo... | (S. D. Bragin, D. V. Galatenko) Solution.
To describe the mixtures sent to the laboratory, we will create a table consisting of 120 rows and 19 columns. Each column of the table is a description of the composition of the mixture sent to the laboratory. At the intersection of the $i$-th row and the $j$-th column, there... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 23,013 |
1. For four integers $a, b, c, d$ it is known that
$$
a+b+c+d=ab+bc+cd+da+1
$$
Prove that the absolute values of some two of these numbers differ by one.
$$
\text { (2023-64, A. Dolodenok) }
$$ | Solution. Transform the equation to the form
$$
a+b+c+d=(a+c)(b+d)+1
$$
After substituting $x=a+c$ and $y=b+d$, we get
$$
x+y=x y+1
$$
Move all terms to one side and factorize
$$
(x-1)(y-1)=0
$$
Thus, $a+c=1$ or $b+d=1$. Suppose $a+c=1$. The sum of two positive integers is no less than two, so $a \leqslant 0$ or ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 23,014 |
2. In the relay race Moscow - Petushki, two teams of 20 people each participated. Each team divided the distance into 20 not necessarily equal segments and distributed them among the participants so that each ran exactly one segment (the speed of each participant is constant, but the speeds of different participants ma... | Answer: 38 overtakes.
Solution. First, let's prove that no more than 38 overtakes occurred. Note that between the start and the first overtake, and between two consecutive overtakes, at least one of the teams must have changed the runner. There were 19 changes of runners in each team, meaning a total of 38 changes, so... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,015 |
4. On the screen of a supercomputer, the number $11 \ldots 1$ (900 ones) is printed. Every second, the supercomputer replaces it according to the following rule. The number is written in the form $\overline{A B}$, where $B$ consists of its last two digits, and it is replaced by $2 \cdot A + 8 \cdot B$ (if $B$ starts wi... | Answer. No, it will not stop.
Solution. The condition can be rephrased as follows: if the number on the screen is of the form $100 A + B$, where $0 \leqslant B < 100$, then it is replaced by $2 A + 8 B$. Since $B < 100$, it follows that $8 B < 800$, so all numbers starting from 800 will necessarily decrease. A natural... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,017 |
5. On the plane, two circles $\omega_{1}$ and $\omega_{2}$ are given, touching each other externally. On circle $\omega_{1}$, a diameter $A B$ is chosen, and on circle $\omega_{2}$, a diameter $C D$ is chosen. Consider all possible positions of points $A, B, C$, and $D$ such that $A B C D$ is a convex circumscribed qua... | Answer. Point $I$ is the point of tangency of the circles.
Solution. So, suppose the diameters are chosen such that the quadrilateral $A B C D$ is circumscribed. Then $A B+C D=B C+A D$. Let $M$ and $N$ be the centers of circles $\omega_{1}$ and $\omega_{2}$ respectively, and $P$ be the point of tangency. $\angle A P B... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,018 |
6. On an island, there live chameleons of five colors. When one chameleon bites another, the color of the bitten chameleon changes according to some rule, and the new color depends only on the color of the biter and the color of the bitten. It is known that 2023 red chameleons can agree on a sequence of bites, after wh... | # Answer. For $k=5$. Solution.
First, let's provide an example of rules under which the described recoloring would require at least 5 red chameleons. Let's number the colors so that red is the first color and blue is the last. Then, let the rules be as follows: if a chameleon of color $k1$ can reduce the number of its... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,019 |
1. Answer: $2011^{2011}+2009^{2009}>2011^{2009}+2009^{2011}$. | Solution. Transform the difference between the left and right parts:
$$
\begin{aligned}
& 2011^{2011}+2009^{2009}-\left(2011^{2009}+2009^{2011}\right)= \\
& =2011^{2011}-2011^{2009}-\left(2009^{2011}-2009^{2009}\right)= \\
& =2011^{2009}\left(2011^{2}-1\right)-2009^{2009}\left(2009^{2}-1\right)
\end{aligned}
$$
Notic... | 2011^{2011}+2009^{2009}>2011^{2009}+2009^{2011} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 23,037 |
5. First solution. We will prove that this is true for the partition of any rectangular parallelepiped. For brevity, we will call the encompassing parallelepiped a box, its three non-parallel faces - the left, the bottom, and the front, the parallelepipeds of the partition - bricks, and the partitions that satisfy the ... | The second solution (proposed by Alexander Skutin, an Olympiad participant). Let's introduce a coordinate system with the origin at the vertex of the cube and axes parallel to its edges. Instead of projections onto the faces, we will consider projections onto the coordinate planes (let's call them the bottom, front, an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,041 |
1. Answer: Yes, it can. For example, a new coordinate system would work where the center has coordinates $\left(-\frac{\pi}{2},-1\right)$ in the old system, and the unit of measurement on each axis of the new system is twice as large compared to the old one. | Solution. Note that to solve the problem, it is sufficient to present at least one desired coordinate system. Let $y=\sin x$. Since
$$
\sin ^{2} t=\frac{1-\cos 2 t}{2}=\frac{1+\sin \left(2 t-\frac{\pi}{2}\right)}{2}
$$
then for $x=2 t-\frac{\pi}{2}$ and $z=\frac{y+1}{2}$ (or, in other words, $y=2 z-1$), we get $z=\si... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,044 |
6. Answer: a) yes; b) no.
a) First solution. Divide the chests into three pairs. The total number of coins in each pair of chests is even, so the number of coins in all six chests is even as well. Now divide the chests into two triplets. The number of coins in each triplet is divisible by three, so the total number of... | Second solution. First, note that the number of coins in all chests has the same parity. After all, it is impossible to evenly divide the contents of two chests with different parities of coins.
Then, pay attention to the fact that the total number of coins in the first three chests is divisible by three. If we replac... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,046 |
3. A n s w e r: $p+1$.
If $\left(p^{2}\right)!$ is divisible by $(p!)^{n}, n \in \mathbb{N}$, then $n \leqslant p+1$, because $p$ enters the factorization of the number $p!$ into prime factors with an exponent of 1 (and thus in the factorization of the number $(p!)^{n}$ - with an exponent of $n$), while in the factori... | First solution. Write down $p^{2}$ different elements in the form of a $p \times p$ table. Two such tables are called equivalent if one can be obtained from the other by some permutations of elements within rows, as well as some permutation of the rows themselves (a total of $p+1$ permutations of $p$ objects). Since $m... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,050 |
1. Answer. $\log _{2} 3, \log _{3} 5, \log _{5} 2$. | Solution. In the notation $a=\log _{2} 3, b=\log _{3} 5, c=\log _{5} 2$, the original equation takes the form
$$
x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c=0
$$
which is equivalent to the equation $(x-a)(x-b)(x-c)=0$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,051 | |
5. First solution. Take two nodes $X$ and $Y$ inside the triangle and draw a line $s$ through them. If this line is parallel to one of the sides of the triangle, or passes through one of the vertices of the triangle, then the proof is complete. Otherwise, draw a line parallel to $s$ through each vertex of the triangle.... | The second solution. Consider the triangle $ABC$ and a node $X$ inside it. Draw three lines through $X$ parallel to the sides of the triangle. Thus, the triangle $ABC$ is divided into three triangles and three parallelograms (Fig. 22). Each of the remaining nodes inside the triangle can either lie on one of the three d... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,060 |
5. First solution. We will prove by induction on $k$ a more general fact: if in the conditions of the problem for each $p$ the polynomial $Q_{p}(x)$ has a degree not greater than $k$, then $f(x)$ at integer points coincides with some polynomial, the degree of which is also not greater than $k$.
Base: $k=0$. In this ca... | The second solution. Let's try to approach the solution of the problem: we will find a polynomial $f_{0}(x)$ of degree not higher than 2013, which coincides with $f(x)$ at the points $1,2, \ldots, 2014$. Such a polynomial is known and is called the Lagrange interpolation polynomial. In our case, it is given by the form... | proof | Algebra | proof | Yes | Yes | olympiads | false | 23,064 |
6. First solution. Let the intersection point of the line $A_{1} A^{\prime}$ with the line $A_{2} I$ be denoted as $X_{A}$, and the circumcircle of $\triangle A B C$ as $\omega$ (Fig. 25). By the problem's condition, $\angle A_{2} X_{A} A_{1}=90^{\circ}$. Since $A_{2} A_{1}$ is the diameter of $\omega$, the point $X_{A... | The second solution (only for part a) ${ }^{2}$. Let $A_{0}, B_{0}$, and $C_{0}$ be the feet of the angle bisectors of angles $A, B$, and $C$ of the triangle, respectively. Let $X_{A}$ be the intersection of $A_{1} A^{\prime}$ with $A_{2} I$, and let $\omega$ be the circumcircle of triangle $ABC$. Since $\angle A_{1} X... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,065 |
1. First solution. Let $a$ be the common root of the described quadratic trinomials. Then they can be represented as $(x-a)(x-b)$ and $(x-a)(x-c)$, and their sum is
$$
\begin{aligned}
&(x-a)(x-b)+(x-a)(x-c)=(x-a)(2 x-b-c)= \\
&=2(x-a)\left(x-\frac{b+c}{2}\right)
\end{aligned}
$$
Note that the discriminant of the quad... | Second solution. Let $x^{2}+p_{1} x+q_{1}, x^{2}+p_{2} x+q_{2}$ be the quadratic trinomials given in the problem, $D_{1}=p_{1}^{2}-4 q_{1}$ and $D_{2}=p_{2}^{2}-4 q_{2}$ be their discriminants. Then the discriminant of their sum
$$
2 x^{2}+\left(p_{1}+p_{2}\right) x+\left(q_{1}+q_{2}\right)
$$
is
$$
D=\left(p_{1}+p_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 23,066 |
5. First solution. Let's estimate the number of 2016-digit squares. There are no fewer than
$$
10^{2016 / 2}-10^{2015 / 2}-1>10^{1000}
$$
The number of different sets of 2016 digits is no more than the number of sets where each digit appears no more than 2016 times, which is $2017^{10}$. Therefore, there exists a set... | The second solution. Let's present such a 2016-digit number explicitly. Consider all 1008-digit numbers of the form $x_{a, b}=4 \cdot 10^{1007}+10^{a}+10^{b}$, where $1007>a>b \geqslant 0, 2a \neq 1007+b$. Note that
\[
\begin{aligned}
x_{a, b}^{2}=16 \cdot 10^{2014}+8 \cdot 10^{1007+a}+ & 8 \cdot 10^{1007+b}+ \\
& +10... | 1007\cdot502>2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,079 |
1. Answer. $16^{5}=32^{4}$. | Solution. This equality is true because $\left(2^{4}\right)^{5}=2^{20}=\left(2^{5}\right)^{4}$.
Comment. It is not difficult to prove that there are no other solutions (not counting the swapping of the right and left sides of the equality). | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,085 |
2. First solution. The graphs of the functions $y=2 a+\frac{1}{x-b}$ and $y=2 c+\frac{1}{x-d}$ are centrally symmetric with respect to the point with coordinates $(\frac{b+d}{2}, a+c)$ and, therefore, have exactly one common point if and only if
$$
2 a+\frac{1}{x-b}=2 c+\frac{1}{x-d}=a+c
$$
when $x=\frac{b+d}{2}$. Th... | The second solution. Let's rephrase the problem statement as follows: the equation $2 a+\frac{1}{x-b}=2 c+\frac{1}{x-d}$ has exactly one solution. It is not hard to see that if $a=c$ or $b=d$, this equation cannot have exactly one solution. When $a \neq c$ and $b \neq d$, this equation is equivalent to the quadratic eq... | (-)(b-)=2 | Algebra | proof | Yes | Yes | olympiads | false | 23,109 |
5. First solution. Let $a_{1}, a_{2}, a_{3}, \ldots$ be a non-increasing sequence of natural numbers. For $k \geqslant 2$, denote by $S_{k}=a_{1}+\ldots+a_{k-1}$ the sum of the first $k-1$ terms of this sequence, and set $S_{1}=0$.
We will prove that any natural number can be represented as the sum of distinct terms o... | The solution to the original problem applies the proven statement to the sequence $\left\{a_{k}\right\}$, which consists of the terms of the geometric progressions $1,3,3^{2}, 3^{3}, \ldots, 1,4,4^{2}$, $4^{3}, \ldots$ and $1,5,5^{2}, 5^{3}, \ldots$, arranged in non-decreasing order: $1,1,1,3,4,5,9,16,25,27,64,81,125,2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,111 |
Problem 1. Positive numbers $a$ and $b$ are such that $a-b=a / b$. Which is greater, $a+b$ or $a b ?$
(V. Kleptsyn, A. Gusev) | Answer: $a b > a + b$.
Solution 1. Consider the desired inequality
$$
a b \vee a + b
$$
Multiply it by the equality $a / b = a - b$ (left side by left side, right side by right side). When multiplying by a positive number (and $a / b$ is positive), the inequality is preserved. We get
$$
a^{2} \vee a^{2} - b^{2}
$$
... | >b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,135 |
Problem 2. The cells of a paper square $8 \times 8$ are painted in two colors. Prove that Arseny can cut out from it two $2 \times 2$ squares along the grid lines, which do not share any cells, and whose colorings are identical. (Colorings that differ by a rotation are considered different.)
(L. Popov, A. Gushev) | Solution 1. Suppose that Arseny will not be able to cut out two such squares. This means that any two squares that match in coloring intersect in at least one cell.
Let's divide our square into 16 squares $2 \times 2$, as shown in Fig. 1a; all of them must be colored differently. Since there are exactly 16 different w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,136 |
Problem 3. In the nodes of a grid of a $4 \times 5$ rectangular grid, there are 30 light bulbs, all initially turned off. In one move, it is allowed to draw any straight line that does not touch the light bulbs (the size of the light bulbs should be neglected, considering them as points), such that on one side of it no... | Answer: Yes.
Solution. Draw an auxiliary line through two bulbs in opposite corners of the rectangle. Then no other bulb will fall on this line. Now draw the first line parallel to the auxiliary one, slightly below, so that these two bulbs are above it, while all other bulbs remain on the same side as before; turn on ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,137 |
Problem 4. Point $M$ is the midpoint of side $BC$ of triangle $ABC$. Circle $\omega$ passes through point $A$, is tangent to line $BC$ at point $M$, and intersects side $AB$ at point $D$ and side $AC$ at point $E$. Let $X$ and $Y$ be the midpoints of segments $BE$ and $CD$ respectively. Prove that the circle circumscri... | Solution. Note that $M X$ and $M Y$ are the midlines of triangles $B C E$ and $B C D$ (Fig. 3), so $\angle X M B = \angle C$ and $\angle C M Y = \angle B$. Then
$$
\angle Y M X = 180^{\circ} - \angle X M B - \angle C M Y = \angle A
$$
By the property of a tangent and a secant to a circle, we have $B M^{2} = B D \cdot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,138 |
Problem 5. In a row, there are $100 N$ sandwiches, each with sausage and cheese. Uncle Fyodor and the cat Matroskin are playing a game. Uncle Fyodor, in one move, eats one sandwich from one of the ends. Matroskin, in one move, can remove the sausage from one sandwich (or do nothing). Uncle Fyodor makes 100 moves in a r... | Answer: No.
Solution. We will prove that with $N=3^{100}$, the cat Matroskin will win. For this to happen, all 100 remaining sandwiches must be without sausage by the last move of Uncle Fyodor (otherwise, he can choose a sequence of actions to end on a sandwich with sausage).
Let's number the sandwiches in order. We ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,139 |
Problem 6. Let $p$ and $q$ be coprime natural numbers. A frog jumps along the number line, starting at point 0, each time either $p$ to the right or $q$ to the left. One day, the frog returned to 0. Prove that for any natural number $d < p+q$, there are two numbers visited by the frog that differ by $d$.
( | Solution 1. Suppose, for some $d1$.
Let the frog return to 0, having made $n$ steps to the right and $m$ steps to the left. Then $n p = m q$, which means $\frac{n}{m} = \frac{q}{p}$. We will assume that the frog repeats this sequence of $m+n$ steps infinitely many times in a cycle; it is clear that the points it visit... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,140 |
Problem 4. On the segment $AC$, point $B$ is marked. On the segments $AB$, $BC$, and $AC$, as diameters, semicircles are constructed in the same half-plane. A circle with center at point $O$ is tangent to all these semicircles (see figure). Find the radius of this circle if $AB=4$, $BC=2$. If necessary, round the answe... | Answer: $6 / 7 \approx 0.86$. | 0.86 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,142 |
Problem 6. The radius of the base of a right circular cylinder is 6, and its height is 8. On the circle bounding the upper base, points $X$ and $Y$ are marked such that one of the arcs with endpoints at points $X$ and $Y$ equals $120^{\circ}$. Find the area of the section of the cylinder by a plane passing through poin... | Answer: $20 \pi+30 \sqrt{3} \approx 114.79$. | 20\pi+30\sqrt{3}\approx114.79 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,143 |
5. Solution. First, let's prove that building towers taller than 2 is meaningless.
Indeed, suppose one of the towers has a height of \( h > 2 \). Let's remove the top floor from it, turning it into a new tower of height 1. How will this change the inspector's sum? The shortened tower no longer sees towers of height \(... | Answer: Either 16 single-story and 7 double-story, or 14 single-story and 8 double-story towers need to be built. | 112 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,149 |
5. Solution. First, let's prove that building towers taller than 2 is meaningless.
Indeed, suppose one of the towers has a height of \( h > 2 \). Let's remove the top floor from it, turning it into a new tower of height 1. How will this change the inspector's sum? The shortened tower no longer sees towers of height \(... | Answer: Either 16 single-story and 7 double-story, or 14 single-story and 8 double-story towers need to be built. | 112 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,163 |
1. (M. Evdokimov) Given three distinct non-zero numbers. Petya and Vasya form quadratic equations by substituting these numbers as coefficients, but each time in a new order. If the equation has at least one root, Petya gets a candy wrapper, and if it has none, the wrapper goes to Vasya. The first three wrappers went t... | Solution: A quadratic equation has at least one root if and only if its discriminant is non-negative. If the first and last coefficients are swapped, the discriminant does not change. Therefore, the six equations can be divided into pairs, in each of which either both equations have roots or both do not. Thus, the numb... | Pete | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,166 |
2. (M. Evdokimov) On the table, there are 23 boxes arranged in a row, one of which contains a prize. On each box, it is either written "There is no prize here" or "The prize is in the adjacent box." It is known that exactly one of these statements is true. What is written on the middle box? | Answer: "The prize is in the adjacent casket."
Solution: Note that on the casket adjacent to the one containing the prize, a true statement is always written. Since we have only one true statement, there is only one such "adjacent" casket. However, this means the prize must be in one of the outer caskets, and the true... | Theprizeisintheadjacentcasket | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 23,167 |
3. (E. Bakayev) Prove that in a right-angled triangle with a 30-degree angle, one bisector is twice as short as the other. | Solution 1: Let there be a right-angled triangle $ABC$, $\angle A=30^{\circ}$, $\angle C=90^{\circ}$. We will prove that the bisector $AK$ is twice as long as the bisector $CL$ of the given triangle. Extend $ABC$ to an equilateral triangle $ABB'$. In this case, $AK$ forms an angle of $15^{\circ}$ with the side of the e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,168 |
4. (V. Kleptsyn, K. Knop) We will call a natural number good if its decimal representation contains only zeros and ones. Suppose the product of two good numbers is also a good number. Is it true that then the sum of the digits of the product equals the product of the sums of the digits of the factors? | Answer: Not necessarily.
Solution: Consider the product of two good numbers
$$
\left(10^{2}+10^{4}+10^{8}+\cdots+10^{1024}\right)\left(10^{N-2}+10^{N-4}+10^{N-8}+\cdots+10^{N-1024}\right)
$$
where $N$ is a large even number (for example, a million). When we expand all the brackets, we get many terms, each of which i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,169 |
5. (D. Brodsky) On the sides of an equilateral triangle $ABC$, triangles $AB'C$, $CA'B$, $BC'A$ are constructed outward from the triangle such that a hexagon $AB'C A' B C'$ is formed, in which each of the angles $A'BC'$, $C'AB'$, $B'CA'$ is greater than 120 degrees, and for the sides, the equalities $AB' = AC'$, $BC' =... | Solution: To form a triangle from the given segments, it is sufficient to prove that the largest of the segments $A B^{\prime}, B C^{\prime}$, $C A^{\prime}$ is less than the sum of the other two. We can assume that the largest of these segments is $A B^{\prime}$. Then, rotate the triangle $A B^{\prime} C$ by 60 degree... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,170 |
6. (B. Novikov) On each cell of an $8 \times 8$ board, a guard is placed. Each guard can look in one of four directions (along the lines of the board) and watch all the guards on the line of his sight. For what largest $k$ can the guards' gazes be directed so that each guard is watched by at least $k$ other guards? | Solution: Answer: 5. We will prove that $k \leq 5$. For this, assume that $k \geqslant 6$. Consider the guards standing at the corners of the board. Each of them is watched by at least 6 guards, and these guards must stand at the edge of the board. Moreover, if a guard sees one of the corner guards, they do not see the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,171 |
1. Tangents were drawn to the graphs of the functions $y=\cos x$ and $y=a \operatorname{tg} x$ at some point of their intersection. Prove that these tangents are perpendicular to each other for any $a \neq 0$.
(V.A. Klypin, G.A. Merzon) | Solution. The abscissa $x_{0}$ of any intersection point of the given functions' graphs satisfies the equation $\cos x_{0}=a \operatorname{tg} x_{0}$. At this point, the tangent to the graph of the function $y=\cos x$ has a slope $k_{1}=-\sin x_{0}$, and the tangent to the graph of the function $y=a \operatorname{tg} x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 23,175 |
2. Let $A B C D$ be a parallelogram different from a rectangle, and let point $P$ be chosen inside it such that the circumcircles of triangles $P A B$ and $P C D$ have a common chord perpendicular to $A D$. Prove that the radii of these circles are equal.
(A. A. Zaslavsky) | Solution.
Method 1. Notice that the line of centers $O_{1} O_{2}$ is perpendicular to the common chord of the given circles, and therefore parallel to the lines $A D$ and $B C$. Let $M$ be th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,176 |
4. In a tennis tournament (where there are no draws), more than 4 athletes participated. Each game day, each tennis player participated in exactly one game. By the end of the tournament, everyone played against each other exactly once. We will call a player persistent if they won at least one match and never lost after... | (B.R. Frenkin) Solution. On the last day, all the persistent players won. Therefore, there are no more than half of them. If there are fewer than half, then every day there was a meeting between non-persistent players. It remains to consider the case where the number of persistent players $k$ is half of the total numbe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,178 |
5. The midpoints of all altitudes of a certain tetrahedron lie on its inscribed sphere. Is it true that the tetrahedron is regular?
(M.A. Evdokimov)
# | # Solution.
Consider the tetrahedron \(ABCD\) satisfying the condition of the problem. Note that by the condition, for any height \(h_i\) of the given tetrahedron, the inequality \(\frac{h_i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,179 |
6. Let's call a triplet of numbers a "triplet" if one of them is equal to the arithmetic mean of the other two. Given an infinite sequence $\left(a_{n}\right)$ consisting of natural numbers, it is known that $a_{1}=a_{2}=1$ and for $n>2$, the number $a_{n}$ is the smallest natural number such that among the numbers $a_... | Solution. It is obvious that the sequence is non-decreasing. Indeed, the inequality $a_{n}>a_{n+1}$ would contradict the choice of $a_{n}$. It is also clear that no number repeats more than twice, otherwise there would be three identical numbers in the sequence, forming a triplet. Now it is easy to see that if a number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,180 |
1. Find the largest natural number $n$ with the following property: for any odd prime $p$ less than $n$, the difference $n-p$ is also a prime number.
(I. Akulich) | Answer: 10
Solution: Suppose that $n>10$. Notice that the numbers $n-3, n-5, n-7$ are all greater than three, and one of them is divisible by three, and consequently composite. Contradiction. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,181 |
2. Points $M$ and $N$ are the midpoints of sides $AB$ and $AC$ of triangle $ABC$. The tangent $\ell$ to the circumcircle of triangle $ABC$ at point $A$ intersects line $BC$ at point $K$. Prove that the circumcircle of triangle $MKN$ is tangent to $\ell$.
(D. Brodsky) | Solution. By the property of the tangent to the circumscribed circle, we have $\angle B A K=$ $\angle A C K$. Therefore, triangles $B A K$ and $A C K$ are similar by two angles. Since $K M$ and $K N$ are corresponding medians in these similar triangles, we have $\angle A K M=\angle C K N$. Finally, since $M N \| K C$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,182 |
3. Among any five nodes of a regular grid paper, there will definitely be two nodes, the midpoint of the segment between which is also a node of the grid paper. What is the minimum number of nodes of a grid made of regular hexagons that need to be taken so that among them, there will definitely be two nodes, the midpoi... | # Answer. 9
Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof. Introduce the origin at one of the grid nodes and denote by $\vec{a}$ and $\vec{b}$ the radius vectors to the two nearest nodes (see picture). Th... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,183 |
4. Given a polynomial of degree 2022 with integer coefficients and leading coefficient 1. What is the maximum number of roots it can have on the interval $(0,1)$?
(A. Kanel-Belov) | Answer: 2021
Solution. If the interval $(0,1)$ contains all 2022 roots of the polynomial, then by Vieta's theorem, the constant term of the polynomial must be equal to their product, and thus will also lie in the interval $(0,1)$ and will not be an integer. We will prove that the polynomial $P(x)=x^{2022}+(1-$ $4042 x... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,184 |
5. Two triangles intersect along a hexagon, which cuts off 6 small triangles from them. The radii of the inscribed circles of these six triangles are equal. Prove that the radii of the inscribed circles of the two original triangles are also equal.
(A. Kushnir) | Solution. Let the radii of the six equal circles be denoted by $r$, and their centers by $A, B, C, D, E, F$ in counterclockwise order. Consider the triangle $A C E$. Note that if we consider the circle inscribed in the original triangle containing triangle $A C E$, and then reduce its radius by $r$ while keeping the ce... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,185 |
6. Andrei Mikhailovich wrote on the board all possible sequences of length 2022, consisting of 1011 zeros and 1011 ones. Let's call two sequences compatible if they coincide in exactly 4 positions. Prove that Andrei Mikhailovich can divide all the sequences into 20 groups so that no two compatible sequences end up in t... | Solution. It is clear that compatible sequences coincide in two positions by ones, and in two - by zeros. Consider the first five positions. There are $C_{5}^{3}=10$ ways to place three ones in these five positions. For each of these ten ways, Andrey Mikhailovich groups the sequences written on the board, the first thr... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,186 |
1. M. Evdokimov
}
A natural number is written on the board. If the last digit (in the units place) is erased, the remaining non-zero number will be divisible by 20, and if the first digit is erased, the remaining number will be divisible by 21. What is the smallest number that can be written on the board if its second... | Answer: 1609.
Solution.
The second to last digit of the number is 0, since the number without the last digit is divisible by 20. Therefore, the number is at least four digits. Note that the number remaining after erasing the last digit cannot be 100 according to the condition. Also, this number cannot be 120 or 140, ... | 1609 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,187 |
# 2. A. Blinkov
Given an isosceles trapezoid, the sum of the lateral sides of which is equal to the larger base. Prove that the acute angle between the diagonals is no more than 60 degrees. | Solution 1.
Consider trapezoid $ABCD$ with bases $BC$ and $AD$, where $AD = AB + CD = 2AB$. In triangle $ABD$, the angle $\sin \angle ADB = \frac{AH}{AD} \leqslant \frac{AB}{AD} = \frac{1}{2}$, where $AD$ is the height from point $A$. Therefore, $\angle ADB \leqslant 30^{\circ}$, similarly $\angle CAD \leqslant 30^{\c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,188 |
3. proposed by A. Shen
There is an infinite one-way strip of cells, numbered with natural numbers, and a bag with ten stones. Initially, there are no stones in the cells of the strip. The following actions are allowed:
- moving a stone from the bag to the first cell of the strip or back;
- if there is a stone in the ... | Answer: $\partial a$.
Solution. Note that for each action there is an inverse to it. Therefore, if we get from situation $A$ to situation $B$ by following the rules, we can also get from situation $B$ to situation $A$ by following the rules.
We will show by induction that if there is a reserve of $n$ stones, then, ac... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 23,189 |
4. N. Nilov
Inside the quadrilateral $ABCD$, a point $P$ is taken. The lines $BC$ and $AD$ intersect at point $X$. It turns out that the line $XP$ is the external bisector of angles $APD$ and $BPC$. Let $PY$ and $PZ$ be the bisectors of triangles $APB$ and $DPC$. Prove that the points $X, Y$, and $Z$ lie on the same l... | The first solution.
Consider $\triangle P B C$ and the external bisector $X P$ of angle $B P C$, $\triangle A P B$ and the bisector $P Y$ of angle $A P B$, $\triangle P C D$ and the bisector ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,190 |
# 5. N. Beluhov
Let $p$ and $q$ be coprime natural numbers. A frog jumps along the number line, starting at point 0. Each time, it jumps either $p$ to the right or $q$ to the left. One day, the frog returned to 0. Prove that for any natural number $d < p+q$, there exist two numbers visited by the frog that differ by $... | Solution. Let at time $k$ the frog is at point $a_{k}$, $a_{0}=a_{N}=0$. Extend the sequence $\left(a_{i}\right)$ periodically by the rule $a_{i+N}=a_{i}$. Denote $A=p+q$. Note that $-q \equiv p \bmod A$, so for any $n a_{n} \equiv n p \bmod A$.
Since $p$ and $q$ are coprime, then $p$ and $A$ are coprime. We will prov... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,191 |
6. D. Krekov
Prove that there exists a real number $A$, such that the distance from the upper integer part of $A^{n}$ to the nearest square is two for any $n$. | Solution.
Let $t$ be the larger root of the polynomial $x^{2}-10 x+1$, then $t+\frac{1}{t}=10$. We will prove by induction that the number $t^{n}+\frac{1}{t^{n}}$ is an integer for any non-negative integer $n$. Indeed, this is true for $n=0,1$. Moreover,
$$
t^{n+1}+\frac{1}{t^{n+1}}=\left(t^{n}+\frac{1}{t^{n}}\right)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,192 |
2. Does there exist a function $f$, defined on the interval $[-1 ; 1]$, which for all real $x$ satisfies the equation $2 f(\cos x)=f(\sin x)+\sin x$? | Solution. Suppose such a function exists. Then, substituting $\pi-x$ for $x$ in the given equation, we get
$$
2 f(-\cos x)=f(\sin x)+\sin x
$$
Therefore, $f(-\cos x)=f(\cos x)$ for all $x$, so $f(-t)=f(t)$ for all $t \in[-1 ; 1]$, i.e., the function $f$ is even.
On the other hand, substituting $-x$ for $x$ in the or... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,193 |
4. In a certain country, there are 100 cities connected by a road network such that from any city to any other, there is only one way to travel without turning back. The network scheme is known, intersections and crossroads of the network are not necessarily cities, and every dead-end branch of the network must end at ... | Solution. Let's represent the road network described in the problem as a graph, where the cities, forks, and intersections are the vertices, and the roads are the edges. We will show that this graph is a tree, i.e., a connected graph without cycles. The connectivity follows from the fact that from any city, you can dri... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 23,195 |
5. A polyhedron with vertices at the midpoints of the edges of a certain cube is called a cuboctahedron. When the cuboctahedron is intersected by a plane, a regular polygon is obtained. What is the maximum number of sides this polygon can have?
(M. A. Evdokimov) | Solution.
Let the edge of the original cube, from which the cuboctahedron is obtained, be 1. Consider the sections of the cuboctahedron by a plane parallel to the base of the cube at a distan... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,196 |
1. (N. Strelkova) Nезнayka does not know about the operations of multiplication and exponentiation. However, he has well mastered addition, subtraction, division, and square root extraction, and he also knows how to use parentheses. Practicing, Nезнayka chose three numbers 20, 2, and 2 and formed the expression:
$$
\s... | Solution: $\frac{20}{2-\sqrt{2}}=\frac{20(2+\sqrt{2})}{2}=20+10 \sqrt{2}>20+10$. There are other solutions. | 20+10\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,199 |
2. (I. Akulich) Find the largest natural $n$ with the following property: for any odd prime $p$ less than $n$, the difference $n-p$ is also a prime number. | Solution: Answer: 10. Indeed, $10=3+7=5+5$. We will prove that numbers greater than 10 do not work. Let $n$ be a number greater than 10. Note that the numbers $3, 5, 7$ give different remainders when divided by 3. Then the numbers $n-3, n-5, n-7$ give different remainders when divided by 3, meaning one of them is divis... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,200 |
3. (K. Knop) On the side of a regular octagon, a square is constructed outward. In the octagon, two diagonals intersecting at point $B$ are drawn (see figure). Find the measure of angle $A B C$.
Note. A polygon is called regular if all its sides are equal and all its angles are equal. | Solution: First, note that the angle of a regular octagon is 6 * $180^{\circ} / 8=135^{\circ}$. Let's denote the vertices of the octagon as shown in the figure. Note that $K L D E$ is an isosceles trapezoid, so the angle $B E D$ is $45^{\circ}$, and the angle $F E B$ is $135^{\circ}-45^{\circ}=90^{\circ}$. Further, $H ... | 22.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,201 |
6. (B. Frenkin) A rook has traversed an $n \times n$ board, visiting each cell exactly once, with each move being exactly one cell. The cells are numbered from 1 to $n^{2}$ in the order of the rook's traversal. Let $M$ be the maximum difference between the numbers of adjacent (by side) cells. What is the smallest possi... | Solution: Answer: $2 N-1$.
Example: traverse the board in a "snake" pattern, starting from the bottom left corner: to the right to the end, up by 1, to the left to the end, up by 1, ...
Estimate: Assume the opposite: $MB$ (by columns), on the other hand, $A<B$ (by rows). Contradiction | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,204 |
Task 1. Sasha writes down the numbers 1, 2, 3, 4, 5 in some order, places the arithmetic operation signs "++", "-", "×" and parentheses, and looks at the result of the obtained expression. For example, he can get the number 8 using the expression $(4-3) \times (2+5)+1$. Can he get the number 123?
Forming numbers from ... | Answer: 2 yes
Solution. For example, $3 \times(2 \times 4 \times 5+1)=123$. | 123 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 23,216 |
Problem 2. Given two sequences of letters A and B, each containing 100 letters. In one operation, it is allowed to insert one or more identical letters into some position in the sequence (possibly at the beginning or end) or to remove one or more consecutive identical letters from the sequence. Prove that the first seq... | Solution 1. First, let's solve a similar problem for sequences of two letters, namely, prove that one sequence can be transformed into another in no more than 2 operations.
If one of the sequences is AA and the other is BB, then we remove all the letters of the first sequence and then add the letters of the second seq... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,217 |
Problem 3. The perimeter of triangle $ABC$ is 1. Circle $\omega$ touches side $BC$, the extension of side $AB$ at point $P$, and the extension of side $AC$ at point $Q$. The line passing through the midpoints of $AB$ and $AC$ intersects the circumcircle of triangle $APQ$ at points $X$ and $Y$. Find the length of segmen... | Answer: $\frac{1}{2}$.
Solution 1. Let the point of tangency of $\omega$ with segment $B C$ be $R$. Consider segments $A P$ and $A Q$. On one hand, they are equal. On the other hand, $B R = B P$ and $C R = C Q$, so
$$
A P + A Q = A B + B P + C Q + A C = A B + B C + A C
$$
Therefore, the lengths of segments $A P$ and... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,219 |
Problem 4. Given a natural number $n>1$. We will call a positive common fraction (not necessarily irreducible) good if the sum of its numerator and denominator equals $n$. Prove that any positive common fraction, the denominator of which is less than $n$, can be expressed through good fractions (not necessarily distinc... | Solution. First, we prove that if $n$ is composite, then not every fraction with a denominator less than $n$ can be expressed. If among the good fractions there are reducible ones, then reduce them. Let the least common multiple (LCM) of the denominators of all the fractions obtained after reduction be denoted by $N$. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,220 |
Problem 5. A regular 100-gon was cut into several parallelograms and two triangles. Prove that these triangles are equal.
(A. Yuran) | Solution. For each figure - the parallelogram and the triangle - consider all the vertices of the figures on the boundary. For each figure, draw vectors between adjacent points on the boundary so that their direction corresponds to the traversal of the boundary of the figure counterclockwise.
 factor, without changing the number of badges, then his collection will have 100 items. And if, conversely, ... | Solution. Let Alik have $x$ badges and $y$ bracelets, and the increase occurs by a factor of $n$. Then we get the system
$$
x+n y=100, \quad n x+y=101
$$
By adding and subtracting the equations and eliminating $n$, we obtain
$$
\frac{201}{x+y}-\frac{1}{x-y}=2
$$
We transform this equation to the form
$$
(201-2 u)(... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,226 |
3. In an acute-angled triangle $ABC$, the bisector $AL$ is drawn. On the extension of the segment $LA$ beyond point $A$, a point $K$ is chosen such that $AK = AL$. The circumcircles of triangles $BLK$ and $CLK$ intersect segments $AC$ and $AB$ at points $P$ and $Q$ respectively. Prove that the lines $PQ$ and $BC$ are p... | Solution.
First method. Consider the segment $K L$, which is the common chord of the circumcircles of triangles $B L K$ and $C L K$. Point $A$ is the midpoint of $K L$, so it lies on the line... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,227 |
4. A spaceship is located in a half-space at a distance $a$ from its boundary. The crew knows this but has no idea in which direction to move to reach the boundary plane. The spaceship can fly along any trajectory in space, measuring the length of the path traveled, and has a sensor that signals when the boundary is re... | (M. A. Evdokimov) Solution.
Let the ship be at some point $O$. Consider a regular octahedron $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ circumscribed around a sphere of radius $a$ with center at p... | Yes,itcan | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 23,228 |
6. The Sultan gathered 300 court wise men and proposed a test. There are 25 different colors of hats, known in advance to the wise men. The Sultan informed them that one of these hats would be placed on each of the wise men, and if the number of hats of each color is written down, all the numbers will be different. Eac... | (A. V. Grialko) Solution. Since $0+1+2+\ldots+24=300$, the quantities of caps of different colors take all values from 0 to 24.
Next, each sage counts the number of caps of each color. For two colors, the quantities of caps coincide, and the sage understands that he is wearing a cap of one of these two colors. It only... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,229 |
4. First solution. Mark the pears with the smallest and the largest weights (we will call them pears $X$ and $Y$ respectively). Obviously, if the pears placed between them are sorted by weight (separately in a clockwise direction from $X$ to $Y$ and separately in a counterclockwise direction from $X$ to $Y$, thus divid... | Second solution. We will prove the following statement by mathematical induction.
$A_{n}:$ if the number of pears laid out is $2 n+2$, then they can be paired as required by the condition.
B a s e. The number of pears is 4. We combine the lightest and heaviest pears into one pair, and the remaining two into another. I... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 23,231 |
Problem 4. Given a natural number $n>1$. We will call a positive common fraction (not necessarily irreducible) good if the sum of its numerator and denominator equals $n$. Prove that any positive common fraction, the denominator of which is less than $n$, can be expressed through good fractions (not necessarily distinc... | Solution. First, we prove that if $n$ is composite, then not every fraction with a denominator less than $n$ can be expressed. If among the good fractions there are reducible ones, then reduce them. Let the least common multiple (LCM) of the denominators of all the fractions obtained after reduction be denoted by $N$. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,232 |
1. Answer. $\frac{x^{2}-10 x+15}{11}$. | Solution. Let the desired polynomial be $P(x)=a x^{2}+b x+c$. Then
$$
\begin{aligned}
P(x)+ & P(x+1)+\ldots+P(x+10)= \\
= & a\left(x^{2}+(x+1)^{2}+\ldots+(x+10)^{2}\right)+ \\
& \quad+b(x+(x+1)+\ldots+(x+10))+11 c= \\
= & a\left(11 x^{2}+(2+4+\ldots+20) x+\left(1+2^{2}+\ldots+10^{2}\right)\right)+ \\
& \quad+b(11 x+1+... | \frac{x^{2}-10x+15}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,234 |
Problem 1. For each of the nine natural numbers $n, 2 n, 3 n, \ldots, 9 n$, the leftmost digit was written down. Can it be that for some natural $n$, among the nine written digits, there are no more than four different ones?
$(2022-69$ / Tolpygo A.) | Answer: Yes, it can.
Solution. For example, $n=25$ works. For this value of $n$, we get the numbers
$$
25,50,75,100,125,150,175,200,225
$$
Each of these numbers starts with one of the four digits $1,2,5,7$.
Comment. It can be proven that it is impossible to have fewer than four different digits. Exactly four differ... | Yes,itcan | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,236 |
Problem 2. Rectangles $A B C D$ and $D E F G$ are positioned such that point $D$ lies on segment $B F$, and points $B, C, E, F$ lie on the same circle (see figure). Prove that $\angle A C E = \angle C E G$.
-\angle A C B=180^{\circ}-\angle D F E-\angle A C B
$$
Similarly,
$$
\angle C E G=\angle C E F-\angle F E G=\left(180^{\circ}-\angle C B F\right)-\angle F E G=180^{\circ}-\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,237 |
Problem 3. Sasha's collection consists of coins and stickers, with fewer coins than stickers, but at least one coin is present. Sasha chose a positive number $t>1$ (not necessarily an integer). If he increases the number of coins by a factor of $t$ and leaves the number of stickers the same, his collection will have 10... | Answer: 34 or 66.
Solution. Let $m$ be the number of coins, and $n$ be the number of stickers. Then the condition can be rewritten as the system of equations
$$
\left\{\begin{array}{l}
m t+n=100 \\
m+n t=101
\end{array}\right.
$$
Subtract the first equation from the second:
$$
1=101-100=(m+n t)-(m t+n)=(n-m)(t-1)
$... | 34or66 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,238 |
Problem 4. Some cells of a $100 \times 100$ board are painted black. In all rows and columns where there are black cells, their number is odd. In each row where there are black cells, we place a red chip in the middle black cell. In each column where there are black cells, we place a blue chip in the middle black cell.... | Solution. Remove all rows and columns from the board that do not contain black cells. Note that after this, the condition of the problem will still be satisfied. Now all black cells lie within some rectangle $n \times m$, in each row and each column of which there is at least one black cell.
By the condition, there is... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,239 |
Problem 5. Two triangles intersect along a hexagon, which cuts off 6 small triangles from them. The radii of the inscribed circles of these six triangles are equal. Prove that the radii of the inscribed circles of the two original triangles are also equal.
 On the sides of an equilateral triangle $ABC$, triangles $AB'C$, $CA'B$, $BC'A$ are constructed outward from the triangle such that a hexagon $AB'C A' B C'$ is formed, in which each of the angles $A'BC'$, $C'AB'$, $B'CA'$ is greater than 120 degrees, and for the sides, the equalities $AB' = AC'$, $BC' =... | Solution: To form a triangle from the given segments, it is sufficient to prove that the largest of the segments $A B^{\prime}, B C^{\prime}$, $C A^{\prime}$ is less than the sum of the other two. We can assume that the largest of these segments is $A B^{\prime}$. Then, rotate the triangle $A B^{\prime} C$ by 60 degree... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,249 |
4. First solution. If one of these numbers (for example, $n$) equals 1, then the inequality takes the form $|m-1|>\frac{1}{m}$ and is satisfied for any $m \geqslant 2$. Further, without loss of generality, we will assume that $m>n \geqslant 2$. Then $m \geqslant n+1$, so
$$
\sqrt[n]{m}-\sqrt[m]{n}>\sqrt[n]{n+1}-\sqrt[... | The second solution. First, let's prove an auxiliary statement: if the derivative \( f'(x) \) of the function \( f(x) \) is positive and increasing on the interval \([a; b]\), then \( f(b) - f(a) > f'(a)(b - a) \). Indeed, in this case, the tangent at the point \( a \) to the graph of the function \( y = f(x) \) lies b... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,255 |
Task 1. Provide an example of a nine-digit natural number that is divisible by 2 if the second (from the left) digit is erased, by 3 if the third digit is erased in the original number, ..., and by 9 if the ninth digit is erased in the original number. | 1. Answer. For example, 900900000.
Note. In fact, there are 28573 numbers that satisfy the conditions of the problem, the smallest of which is 100006020, and the largest is 999993240. | 900900000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,262 |
Problem 4. Each point on the plane is painted in one of three colors. Is it necessarily true that there exists a triangle of area 1, all vertices of which have the same color? | 4. Answer. Yes.
First solution. Suppose that such a triangle does not exist, and we will prove that there exists a line, all points of which have the same color.
Let there be two points $A, B$ of the same color (denote this color 1) on some line $l$, with the distance between them being $d$. Let $l_{1}, l_{2}$ be two... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,265 |
Problem 5. Petya and Vasya are playing a game. For each set of five different variables from the set $x_{1}, \ldots, x_{10}$, there is a unique card on which their product is written. Petya and Vasya take turns picking a card, with Petya starting first. When all the cards are taken, Vasya assigns values to the variable... | 5. Answer. Yes.
Solution. We will show how Vasya can act to make the sum of the products on his cards greater than Petya's.
Suppose Petya did not take the card with $x_{6} x_{7} x_{8} x_{9} x_{10}$. Then Vasya can take this card and then take any cards. If
$$
x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0, \quad x_{6}=x_{7}=x_{8}=... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,266 |
5. First solution. First, we will show that the distance to the nearest integer from a number of the form $c - m q$ (where $m \in \mathbb{N}$, $q$ is irrational, and $c$ is any fixed number) can be made arbitrarily small by choosing $m$. Consider $n+1$ numbers $q, 2q, 3q, \ldots, (n+1)q$. Their fractional parts fall in... | The second solution. Regardless of its initial position $x_{0}$, the grasshopper will eventually end up in the interval $\Delta=\left[-\frac{a+b}{2} ; \frac{a+b}{2}\right)$. Indeed, if $x_{0}<\frac{b-a}{2}$, it will jump to the right by $a$ until it jumps past the point $\frac{b-a}{2}$ and ends up in the interval $\Del... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,274 |
5. First solution. Let's divide all guests into ordered triples; we will put a hat with the letter A on the first person in the triple, a hat with the letter B on the second, and a hat with the letter C on the third. To do this, we will line up the guests (which can be done in $(3n)!$ ways), combine the first three int... | The second solution. Number the people from 1 to $3 n$. There are exactly (3n)! ways to arrange these people in a row, so it is sufficient to establish a one-to-one correspondence between such arrangements and partitions into circles.
Take any arrangement, put top hats on everyone in the order ABABAB...ABAB from left ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,277 |
1. Some non-negative numbers $a, b, c$ satisfy the equation $a+b+c=2 \sqrt{abc}$. Prove that $bc \geqslant b+c$.
(D. V. Goryashin) | Solution. Method 1. Move all terms to the left side of the equation and complete the square:
$$
a-2 \sqrt{a} \sqrt{b c}+b+c=(\sqrt{a}-\sqrt{b c})^{2}+b+c-b c
$$
Therefore, $b c=b+c+(\sqrt{a}-\sqrt{b c})^{2} \geqslant b+c$.
Method 2. The numbers $a, b$, and $c$ are non-negative, so the original equation can be consid... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,278 |
3. In a convex 12-sided polygon, all angles are equal. It is known that the lengths of some ten of its sides are 1, and the length of one more is 2. What can the area of this 12-sided polygon be? (M. A. Evdokimov) Solution. Consider the 12-sided polygon $A_{1} A_{2} \ldots A_{12}$, satisfying the condition of the probl... | Answer: $8+4 \sqrt{3}$.
 | 8+4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,279 |
4. In an isosceles trapezoid, a diagonal is drawn. A beetle crawls along the perimeter of each of the two resulting triangles. The speeds of the beetles are constant and equal. The beetles do not change the direction of their traversal of their respective perimeters, and they crawl along the diagonal of the trapezoid i... | Solution. Let in an isosceles trapezoid $ABCD$ with bases $AB > CD$, a diagonal $AC$ is drawn, so that the first beetle crawls along the cycle $A \rightarrow C \rightarrow D \rightarrow A$, and the second one - along the cycle $A \rightarrow B \rightarrow C \rightarrow A$. Consider the moments when the first beetle is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,280 |
5. Tanya was sequentially writing down numbers of the form $n^{7}-1$ for natural numbers $n=2,3, \ldots$ and noticed that for $n=8$ the resulting number is divisible by 337. For what smallest $n>1$ will she get a number divisible by $2022?$ | (T. A. Garmanova) Solution. Let the natural number $n$ be such that $n^{7}-1$ is divisible by $2022=2 \cdot 3 \cdot 337$. Then $n^{7}-1$ is divisible by 2 and 3, so $n$ is an odd number, having a remainder of 1 when divided by 3. In addition, $n^{7}-1$ is divisible by 337. Note that if two numbers are congruent modulo ... | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,281 |
Problem 1. (M. Evdokimov) Baron Munchausen claims that to any two-digit number, one can append two more digits on the right to form a perfect square (for example, if the number 10 is given, we append 24 and get $1024 = 32^2$). Is the baron correct? | Answer. The Baron is wrong.
First solution. Note that $99^{2}=(100-1)^{2}=100^{2}-2 \cdot 100+1=98019999$. Thus, there are no four-digit perfect squares starting with 99, so it is impossible to append two digits to the number 99 to form a perfect square.
Second solution. Suppose the Baron is right. There are 90 two-d... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,286 |
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