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Blinkov A.d:
Each diagonal of a quadrilateral divides it into two isosceles triangles. Is it true that the quadrilateral is a rhombus? | Let $A B C$ be an isosceles triangle where angle $B$ is obtuse and not equal to $120^{\circ}, D$ - the center of the circumscribed circle of triangle $A B C$. Then the quadrilateral $A B C D$ satisfies the conditions of the problem and is not a rhombus.
## Answer
Incorrect. | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,477 |
Bakayev E.V.
Twenty children - ten boys and ten girls - stood in a row. Each boy said how many children were standing to the right of him, and each girl - how many children were standing to the left of her. Prove that the sum of the numbers named by the boys is equal to the sum of the numbers named by the girls. | The boy standing at the $k$-th place from the left will call out the number $20-k$, so the sum of the numbers called out by the boys is $200-S_{m}$, where $S_{m}$ is the sum of their positions. The girl standing at the $n$-th place from the left will call out the number $n-1$, so the sum of the numbers called out by th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,479 |
What is the maximum power of three that divides the number whose decimal representation consists of $3^{n}$ ones? | Use induction on $n$.
## Solution
Let $A_{n}$ denote the number consisting of $3^{n}$ ones. We will prove by induction that $A_{n}$ is divisible by $3^{n}$ and not divisible by $3^{n+1}$. The base case $(n=1)$ is obvious.
Inductive step. The number $A_{n+1}$ is obtained by multiplying the number $A_{n}$ by a number ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,482 |
8,9 |
All points on the circumference of a circle are arbitrarily colored in two colors.
Prove that there exists an isosceles triangle with vertices of the same color inscribed in this circle.
# | In a regular pentagon, any three vertices form an isosceles triangle.
## Solution
Consider a regular pentagon inscribed in a given circle. Among its five vertices, at least three are colored in the same color. These vertices form an isosceles triangle with vertices of the same color. | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,484 |
[ Central symmetry helps solve the task. ] [ Convex and non-convex figures (other). ]
A convex pancake with an area greater than $1 / 2$ was cooked on a round frying pan with an area of 1. Prove that the center of the pan is under the pancake. | Use the fact that through a point lying outside a convex figure, a line can be drawn that does not intersect this figure. Another approach is to show that under the pancake, there will be two points symmetric with respect to the center of the pan.
## Solution
First method. Let the center of the pan $O$ lie outside th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,485 |
Shapovalov A.V.
Does there exist in space a closed self-intersecting broken line that intersects each of its segments exactly once, and at its midpoint? | Consider the triangular prism $A B C A^{\prime} B^{\prime} C^{\prime}$ (see figure). The closed broken line $A B^{\prime} C A^{\prime} B C^{\prime} A$ satisfies the condition, since the lateral faces of the prism are parallelograms.
.]
On the extension of side $A D$ of rectangle $A B C D$ beyond point $D$, a point $E$ is taken such that $D E=0.5 A D, \angle B E C=$ $30^{\circ}$.
Find the ratio of the sides of rectangle $A B C D$. | Apply the formula for the tangent of the difference.
## Solution
Let $\angle B E A=\alpha, \angle C E A=\beta, AB / BC=t$. Then $\operatorname{tg} \alpha=2 t / 3, \operatorname{tg} \beta=2 t, \operatorname{tg} 30^{\circ}=\operatorname{tg}(\beta-\alpha)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operat... | \sqrt{3}:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,487 |
| |
a) The "Tower of Hanoi" puzzle consists of eight disks of decreasing size placed on one of three pegs. The goal is to move the entire tower to another peg, moving only one disk at a time and never placing a larger disk on a smaller one. Prove that the puzzle has a solution.
What method would be optimal (in terms... | a) Let the minimum number of steps to move a tower of $n$ disks be $K_{n}$. Express $K_{n+1}$ in terms of $K_{n}$. Consider a tower of $n+1$ disks. We can divide the entire procedure into three stages.
Stage 1 - from the beginning to the first move of the bottom disk. By the end of this stage, the tower of $n$ disks s... | 4373 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,490 |
[ Decimal number system ] $[$ Divisibility of numbers. General properties $]$
Let's call a six-digit number lucky if the sum of its first three digits equals the sum of its last three digits. Prove that the sum of all lucky numbers is divisible by 13. (Numbers that can be written with fewer than six digits are also c... | Let's divide the numbers of all lucky tickets into two groups. In the first group, we will include numbers that consist of two equal three-digit numbers (for example, 765765). All other numbers will be included in the second group. Since $\overline{a b c a b c}=\overline{a b c} \cdot 1001=\overline{a b c} \cdot 7 \cdot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,491 |
Find all values of $x$ that satisfy the inequality $(2-a) x^{3}+(1-2 a) x^{2}-6 x+5+4 a-a^{2}<0$ for at least one value of $a$ in the interval $[-1,2]$.
# | Let's consider $a$ as an unknown and $x$ as a parameter. Then the question of the problem becomes:
For which values of the parameter $x$ does the quadratic polynomial $f_{x}(a)=a^{2}+\left(x^{3}+2 x^{2}-4\right) a-2 x^{3}-x^{2}+6 x-5$ take at least one positive value on the interval
$[-1,2]?$
Obviously, this happens... | x<1orx>1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,492 |
[Methods for solving problems with parameters]
[Investigation of a quadratic trinomial]
For what positive value of $p$ do the equations $3 x^{2}-4 p x+9=0$ and $x^{2}-2 p x+5=0$ have a common root? | The common root of the given equations must also be a root of the equation $\left(3 x^{2}-4 p x+9\right)-3\left(x^{2}-2 p x+5\right)=0 \Leftrightarrow$ $2 p x-6=0$. Therefore, it equals $3 / p$. Substituting, for example, into the second equation, we get $9 / p^{2}=1$, from which $p=3$.
## Answer
For $p=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,493 |
[ [ordering by increasing (decreasing) ]
Artyom collects coins. In his collection, there are 27 coins, each with a different diameter, different mass, and issued in different years. Each coin is stored in a separate matchbox. Can Artyom stack these matchboxes into a $3 \times 3 \times 3$ parallelepiped so that any coi... | First, we will sort the coins by weight. The nine lightest coins will be arranged arbitrarily in the top layer, the nine medium-weight coins in the middle layer, and the nine heaviest coins in the bottom layer. Note that each coin is lighter than the coin below it. Thus, we are left to sort the coins by two parameters ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,498 |
[The Pigeonhole Principle (continued).]
Three people are playing table tennis in a knockout format, meaning the player who loses a match gives way to the player who did not participate in it. In the end, Nikanor played 10 matches, Filimon played 15, and Agafon played 17. Who among them lost the second match? | A total of $(10+15+17): 2=21$ games were played. Note that no one could skip two games in a row. This means that Nikanor played all the games with even numbers and only them. Since he did not play in the third game, he lost the second game.
## Answer
Nikanor. | Nikanor | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,499 |
[ Dirichlet's Principle (continued).]
Every day, from Monday to Friday, the old man went to the blue sea and cast his net into the sea. Each day, the net caught no more fish than the previous day. In total, over the five days, the old man caught exactly 100 fish. What is the smallest total number of fish he could have... | Evaluation. On Tuesday and Thursday, the old man caught no more fish than on Monday and Wednesday, which means that over the specified three days, he caught no less than half of 100, that is, no less than 50 fish.
Example. If the old man caught 25 fish each of the first four days and caught nothing on Friday, the cond... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,500 |
[ Pairing and Grouping; Bijections ]
From the natural numbers from 1 to 100, 50 different numbers are chosen. It turns out that the sum of no two of them is 100.
Is it true that among the chosen numbers, there will always be a square of some integer?
# | If among the selected numbers there is the number 100, then it will be the exact square. If 100 is not selected, then all numbers from 1 to 99, except for the number 50, can be divided into 49 pairs such that the sum of the numbers in each pair equals 100: $1+$ $99, 2+98, \ldots, 49+51$. According to the condition, exa... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,501 |
Akulyi I. F.
The carpet has the shape of a square with a side of 275 cm. A moth has eaten four holes in it. Can we guarantee that we can cut out a square piece with a side of 1 m from the carpet that does not contain any holes? Consider the holes as point-like. | A carpet can accommodate five non-overlapping squares with a side length of 1 m (see the figure; the side of the dashed square is $0.75 \sqrt{2} > 1$). At least one of them will remain without a hole.
 is obvious.
Induction step. Let $n=2m$. The number $m$ can be represented in the given form. By increasing all $v_{i}$ by 1, we obtain
a representation of the number $n$.
Let $n$ be odd. Take $v_{0}=0$. Consider the largest $u_{0}$ such that $3^{u_{0}} \leq n$. If the inequality is s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,503 |
A.K.
The sum of several positive numbers is 10, and the sum of the squares of these numbers is greater than 20. Prove that the sum of the cubes of these numbers is greater than 40.
# | This follows from the inequality $\left(a_{1}+\ldots+a_{n}\right)\left(a_{1}^{3}+\ldots+a_{n}^{3}\right) \geq\left(a_{1}^{2}+\ldots+a_{n}^{2}\right)^{2}$, which after expanding the brackets and mutual cancellation of terms of the form $a_{i}^{4}$ reduces to the sum of several inequalities of the form $b^{3} c+b c^{3} \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,505 |
Initially on the computer screen - some prime number. Every second, the number on the screen is replaced by the number obtained from the previous one by adding its last digit, increased by 1. What is the maximum time it will take for a composite number to appear on the screen?
# | Let the initial number on the screen be 2, then we get the following sequence: $2-5-11-13-17-25$. The sixth number is composite, so in this case, it will take 5 seconds. We will prove that in other cases, it will take no more than 5 seconds.
Indeed, if the number on the screen was an odd prime number not ending in 9, ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,506 |
In a certain class, in any distribution of 200 candies, there will be at least two students who receive the same number of candies (possibly none). What is the smallest number of students in such a class?
# | Let there be no more than 20 people in the class, then we will give nothing to the first student, one candy to the second student, two candies to the third student, and so on. In this case, no more than $0+1+2+\ldots+19=$ 190 candies will be distributed. The remaining candies will be given to the student with the most ... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,507 |
The Vellum Skin grants wishes, but after each wish, its area decreases: either by 1 dm² in the usual case, or by half - if the wish was a heartfelt one. Ten wishes reduced the area of the skin threefold, the next few - sevenfold, and after a few more wishes, the skin disappeared completely. What was the original area o... | Let the original area of the skin be $S$ dm² $(S>0)$. After the area was reduced threefold and then sevenfold, it became $S / 21$ dm². This number must be an integer; otherwise, it would be impossible to get 0 by dividing by 2 and subtracting 1.
Consider the first 10 wishes. Among them, there cannot be more than one u... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,508 |
Bogdanov I.I.
A cube was lying on a plane. It was rolled several times (over its edges) so that the cube returned to its original position with the same face up.
Could the top face have rotated by $90^{\circ}$ relative to its initial position? | Let's color the vertices of a cube so that each edge connects a blue vertex to a red one. Transform the plane into an infinite chessboard so that the bottom face of the cube exactly coincides with one of the black squares. With each roll, the two types of faces on top alternate: blue vertices are replaced by red ones a... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,509 |
[ Examples and counterexamples. Constructions ]
A palindrome is a natural number that reads the same from left to right and from right to left (for example, 1, 343, and 2002 are palindromes).
Can 2005 pairs of the form ( $n, n+110$ ) be found, where both numbers are palindromes? | For example, such are all numbers $n$ of the form 109... 901 with any number of nines.
## Answer
There are. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,510 |
The dentist forbade Sonya to eat more than ten caramels a day, and if on any day she eats more than seven caramels, then for the next two days she cannot eat more than five caramels per day. What is the maximum number of caramels Sonya will be able to eat in 25 days, following the dentist's instructions?
# | If Sonya eats more than 7 caramels on any day (except the last two days), then in the next two days she will eat no more than $5+5=10$, and over three days - no more than 20 candies. This means that it would be more beneficial for her to eat 7 candies per day over these three days.
Therefore, it is more beneficial for... | 178 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,514 |
$\underline{\text { Tanya's TV }}$
Tanya cut out a convex polygon from paper and folded it several times so that a two-layer quadrilateral was formed.
Could the cut-out polygon have been a heptagon? | Let's take a quadrilateral $A B C D$, where angle $B$ is obtuse, and the others are acute. Let $K$ be a point on side $C D$ such that $\angle C B K < 180^{\circ} - \angle B$. Points $B_{1}, K_{1}$ are symmetric to $B, K$ with respect to $A D$, and $K_{2}$ is symmetric to $K$ with respect to $B C$. Then the heptagon $A ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,515 |
Can three segments of even length be placed on a number line so that the common parts of each pair of them are segments of odd length? | Let the segments be arranged according to the condition. Place the origin of the axis so that the ends of the first segment fall on points with integer coordinates.
Notice that the second segment intersects the first, but does not lie within it and does not contain it (otherwise the intersection would have an even len... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,517 |
$\underline{\text { Novikov V. V. }}$.
Any number \$x\$, written on the board, can be replaced either by $3 \$ x \$+1$, or by $\left[{ }^{x} / 2\right]$.
Prove that if the number 1 is initially written, then by such operations any natural number can be obtained. | The number 1 is written. Let's show how to obtain a natural number $\$ n \$>1$, if we already know how to obtain all smaller numbers. The number $\$ \mathrm{n} \$$ can be represented in one of three forms: $3 \$ \mathrm{k} \$-1,3 \$ \mathrm{k} \$$ or $3 \$ \mathrm{k} \$+1$, where $\$ \mathrm{k} \$$ is a natural number.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,518 |
Zaslavsky A.A.
Three bogatyrs are fighting the Serpent Gorynych. Ilya Muromets with each blow cuts off half of the Serpent's heads and one more, Dobrynya Nikitich cuts off a third of the heads and two more, Alyosha Popovich cuts off a quarter of the heads and three more. The bogatyrs strike one at a time in any order ... | If the number of heads is even, the heroes can reduce it while preserving its parity. Indeed, if there are $\$ 4 n-$ $2 \$$ heads, then after Ilya's strike, there will be $\$ 2 n-2 \$$.
If there are $4 \$ n \$$ heads, then after Alyosha's strike, there will be \$3n - 3\$, and after the subsequent strike by Dobrynya, t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,519 |
[ Processes and operations ]
12 fields are arranged in a circle: on four adjacent fields, there are four differently colored chips: red, yellow, green, and blue. In one move, any chip can be moved from the field it occupies, over four fields to the fifth (if it is free) in either of the two possible directions. After ... | Answer: KZHS, SHKZ, ZKHS, HZSK. Let's make copies of our 12 fields and arrange them in a circle in the following order: $1,6,11,4,9,2,7,12,5,10,3,8$. On the new circle, the tokens move simply to the adjacent field (to the right or left). Therefore, any move in which the tokens swap places represents a movement around t... | SHKZ,HZSK,ZKHS,KZHS | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,522 |
Point $O$ lies inside a convex $n$-gon $A_{1} \ldots A_{n}$ and is connected to the vertices by segments. The sides of the $n$-gon are numbered from 1 to $n$, with different sides having different numbers. The same is done with the segments $O A_{1}, \ldots, O A_{n}$.
a) For $n=9$, find a numbering such that the sum o... | a) See fig.
b) The sum of all the side numbers of the triangles $A_{1} O A_{2}, \ldots, A_{n} O A_{1}$ is $3(1+2+\ldots+n)=3 / 2 n(n+1)$. If the sum of the side numbers for each of the $n$ t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,523 |
What is the maximum number of queens that can be placed on the black squares of an $8 \times 8$ chessboard so that each queen is attacked by at least one of the others?
# | Consider some arrangement of queens that satisfies the condition. A queen standing at the edge of the board cannot be attacked by any other queen, which means that all the queens are in the "inner" $6 \times 6$ square. Note that in none of the $3 \times 3$ squares can there be more than four queens. Indeed, if there ar... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,524 |
[ Examples and counterexamples. Constructions $\quad]$ [ Linear inequalities and systems of inequalities $]$
Can 20 numbers be written in a row so that the sum of any three consecutive numbers is positive, while the sum of all 20 numbers is negative? | Consider 20 numbers $a, b, b, a, b, b, \ldots, a, b, b, a, b$. The sum of three consecutive numbers is $a+2 b$. The sum of all numbers is $7 a+13 b$. Therefore, it is sufficient to choose $a$ and $b$ such that $a+2 b>0$ and $7 a+13 b<0$. For example, we can set $a=-25, b=13$.
## Answer
We can. | =-25,b=13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,525 |
In city $N$, from any metro station you can travel to any other station. Prove that one of the stations can be closed for repairs without the possibility of passing through it, so that from any of the remaining stations you can still travel to any other.
# | The first method. Let $S$ be some metro station, and $T$ be the farthest station from $S$, that is, such that the shortest path from $S$ to $T$ passes through a greater (or at least not fewer) number of stations than the shortest path from $S$ to any other station. Now, let's close station $T$. With this, we can still ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,526 |
On the board after the class, there was a note left:
"Calculate $t(0)-t(\pi / 5)+t\left((\pi / 5)-t(3 \pi / 5)+\ldots+t\left({ }^{8 \pi} / 5\right)-t(9 \pi / 5)\right.$, where $t(x)=\cos 5 x+* \cos 4 x+* \cos 3 x+* \cos 2 x+$ $*^{\cos x}+* "$.
Seeing it, a math student told a friend that he could calculate this sum e... | Let $t_{k}(x)=\cos k x$ for $k=0,1,2,3,4$. If we consider a similar sum for $t_{5}$ instead of $t$, we will get 10 as the result. We will check that for all other $t_{k}$ the sums are equal to 0. For $k=0$ this is obvious. For $k=1,2,3,4$ we arrive at the same equality
$\cos 0+\cos (2 \pi / 5)+\cos ((\pi / 5) \cos (6 ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,528 |
3+ [Geometry on grid paper]
Some cells of an infinite grid paper are painted red, the rest are white (not necessarily in a checkerboard pattern). A grasshopper jumps on the red cells, while a flea jumps on the white cells, and each jump can be made to any distance horizontally or vertically. Prove that the grasshopper... | Let's first prove that if the grasshopper and the flea are on the same line (vertical or horizontal), they can end up next to each other by making no more than two jumps in total. Indeed, on this line, there are cells of both colors, which means there are adjacent cells, one of which is painted red and the other white.... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,529 |
The numbers $1,2,3, \ldots, 99,100$ are written in a row. It is allowed to swap two numbers that have exactly one number between them.
Can the sequence $100,99,98, \ldots, 2,1$ be obtained? | Even numbers always remain in even positions, while odd numbers remain in odd positions. Therefore, the number 100 cannot move to the first position.
## Problem | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,532 |
Galinierin G.A.
Consider nine-digit numbers consisting of non-repeating digits from 1 to 9 in different orders. A pair of such numbers is called conditional if their sum equals 987654321.
a) Prove that there will be at least two conditional pairs $((a, b)$ and $(b, a)$ - the same pair).
b) Prove that the number of c... | a) $987654321-123456789=864197532$, therefore, ( 123456789,864197532 ) is a conditioning pair.
By swapping the last two digits in each of these numbers, we get another conditioning pair (123456798, $864197523)$.
b) The sum of the last digits of two numbers from a conditioning pair is 11, and the sum of the second-to-... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,533 |
Fomin C.B.
From a sheet of graph paper measuring $29 \times 29$ cells, 99 squares of size $2 \times 2$ have been cut out (cutting along the lines). Prove that at least one more such square can be cut from the remaining part of the sheet. | In the figure, 100 squares $2 \times 2$ are shaded. By cutting out any $2 \times 2$ square from the original $29 \times 29$ square, we "touch" exactly one shaded square. Therefore, after 99 such operations, one of them will remain untouched.
=p_{k} x+q_{k}(k=1,2, \ldots, 1000)$. We need to find the value of their composition $f(x)=$ $f_{1}\left(f_{2}\left(f_{3}\left(\ldots f_{1000}(x) \ldots\right)\right)\right)$ at the point $x_{0}$. Prove that this can be done in no more than 30 stages, if on each stage a... | $f(x)=p_{1} p_{2} \ldots p_{1000^{x}}^{0}+p_{1} p_{2} \ldots p_{999} q_{1000}+p_{1} p_{2} \ldots p_{998} q_{999}+\ldots+p_{1} q_{2}+q_{1}$.
The longest of the terms - the product of 1001 numbers $p_{1} p_{2} \ldots p_{1000} x_{0}$ - is computed in 10 stages,
since $1001<2^{10}:$ in the first stage, we compute 500 pro... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,535 |
[ $[$ Dirichlet's Principle (miscellaneous).]
Author: Andjensen $A$.
In a gang of 101 terrorists, none of them have participated in any raids together, but every pair of them has met in raids exactly once.
Prove that one of the terrorists participated in at least 11 different raids.
# | Let there be a terrorist $A$, who participated in no more than 10 raids. There are exactly 100 other terrorists, so in at least one of these raids (denote it as $Z$), no fewer than 10 terrorists (excluding $A$) participated.
Consider a terrorist $B$, who did not participate in $Z$. He participated in raids with all 11... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,536 |
Avor: Tokarev S.i.
Prove that there exists a set of 100 different natural numbers $c_{1}, c_{2}, \ldots, c_{100}$, such that for any two consecutive numbers $c_{i}$ and $c_{i+1}$ in this set, the sum $c_{i}^{2}+c_{i+1}^{2}$ is a square of an integer. | Consider, for example, the numbers $c_{i}=3^{100-i} \cdot 4^{i-1}$. Indeed,
$c_{i}^{2}+c_{i+1}^{2}=\left(3^{100-i} \cdot 4^{i-1}\right)^{2}+\left(3^{99-i} \cdot 4^{i}\right)^{2}=\left(3^{99-i} \cdot 4^{i-1}\right)^{2}\left(3^{2}+4^{2}\right)=\left(3^{99-i} \cdot 4^{i-1}\right)^{2} \cdot 5^{2}$.
Submit a comment | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,537 |
Tomong A.K.
10 chips are placed on the table in a circle. The top side of the chips is red, and the bottom side is blue. Two operations are allowed:
a) flip four chips in a row;
b) flip four chips arranged as follows: ××0×× (× - a chip included in the quartet, 0 - not included). Is it possible, using the allowed ope... | Mark five chips every other one. Note that with each allowed operation, exactly two marked chips are flipped. Initially, among the marked chips, there were 5 red ones (referring to the upper color). With each operation, the number of red chips either remains the same or changes by 2. Therefore, among the marked chips, ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,538 |
[ Quadratic Inequalities (Several Variables).]
Authors: Bugayenko V.O., Egorov A.A.
Positive numbers $a, b, c$ are such that $a^{2}+b^{2}-a b=c^{2}$. Prove that $(a-c)(b-c) \leq 0$.
# | Let's set segments $O A=a$ and $O B=b$ on the sides of an angle of $60^{\circ}$ with vertex $O$. By the cosine theorem and due to the equality $a^{2}+b^{2}-a b=c^{2}$, the length of segment $A B$ is $c$. Since in triangle $O A B$ the angle at vertex $O$ is the middle in size, the side $c$ is also the middle, that is, e... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,540 |
$[\quad$ The Pigeonhole Principle (angles and lengths). $\quad]$ [Integer and fractional parts. The Archimedean Principle ] A sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{1996}$ is written in a row. Prove that one can select one or several consecutive numbers such that their sum differs from an integer by... | We form the numbers $b_{1}=\left\{a_{1}\right\}, b_{2}=\left\{a_{1}+a_{2}\right\}, b_{3}=\left\{a_{1}+a_{2}+a_{3}\right\}, \ldots, b_{1996}=\left\{a_{1}+a_{2}+a_{3}+\ldots+a_{1996}\right\}$. We divide the segment $[0,1]$ into 1001 segments of length $\frac{1}{1001}$. Some two of the numbers $b_{1}, \ldots, b_{1996}$ wi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,541 |
Folklore.
Does there exist a set of 10 natural numbers such that none of them is divisible by any of the others, but the square of each is divisible by each of the others? | Example 1. $P p_{1}, P p_{2}, \ldots, P p_{10}$, where $p_{1}, p_{2}, \ldots, p_{10}-10$ are different prime numbers, and $P=p_{1} p_{2} \ldots p_{10}$.
Example 2. $2^{9} \cdot 3^{18}, 2^{10} \cdot 3^{17}, \ldots, 2^{18} \cdot 3^{9}$.
## Answer
There exists.
Send a comment | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,542 |
Zhendarov R.G.
On a chessboard of size $8 \times 8$, 17 cells are marked.
Prove that among them, two can be chosen such that a knight needs at least three moves to get from one to the other. | Let's divide the board into 16 identical four-cell figures: two of them are shaded in the diagram, two more can be obtained from them by shifting down by two cells, and the rest can be obtained from these four by shifting to the right.
 does not exceed 0.5. | Let's divide each face of the tetrahedron into four equilateral triangles using the midlines. We will call a "kulk" the union of three such triangles adjacent to one vertex of the tetrahedron. Now the surface of the tetrahedron is divided into 8 parts: four kulks and four remaining triangles. Since there are more than ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,544 |
Chebotarev, A.S.
2003 dollars were distributed into wallets, and the wallets were distributed into pockets. It is known that the total number of wallets is greater than the number of dollars in any pocket. Is it true that the number of pockets is greater than the number of dollars in any wallet? (Putting wallets insid... | Let there be $M$ wallets and $N$ pockets. In each pocket, there are fewer than $M$ dollars, so in total, there are fewer than $M N$ dollars. Therefore, in one of the wallets, there are fewer than $M N: M=N$ dollars.
## Answer
Correct. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,545 |
[ Coordinate method on the plane [ Ratio of areas of triangles with a common base or common height]
On the coordinate plane, points $A(1 ; 9)$, $C(5 ; 8)$, $D(8 ; 2)$, and $E(2 ; 2)$ are given. Find the area of the pentagon $A B C D E$, where $B$ is the intersection point of the lines $E C$ and $A D$. | If $y_{1} \neq y_{2}$ and $x_{1} \neq x_{2}$, then the equation of the line passing through the points $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ is
$$
\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}
$$
## Otвет
27.00 | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,548 |
[ Coordinate method on the plane [ The ratio of the areas of triangles with a common base or common height]
] On the coordinate plane, points $A(1 ; 3), B(1 ; 9), C(6 ; 8)$, and $E(5 ; 1)$ are given. Find the area of the pentagon $A B C D E$, where $D$ is the point of intersection of the lines $A C$ and $B E$. | If $y_{1} \neq y_{2}$ and $x_{1} \neq x_{2}$, then the equation of the line passing through the points ( $x_{1} ; y_{1}$ ) and $\left(x_{2} ; y_{2}\right)$ is
$$
\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}
$$
## Answer
21.00 | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,549 |
Authors: Yashchenko I.V., Botin D.A.
Gulliver found himself in the land of the Lilliputians with 7,000,000 rubles. He immediately spent all his money on bottles of kefir, which cost 7 ruble... | Conduct calculations in "hard" currency - empty bottles. Then there will be no inflation!
## Solution
Note that if all transactions are conducted in hard currency - empty bottles, - there will be no inflation. That is, a bottle of kefir will always cost 7 empty bottles, and the kefir in it -6 empty bottles. Gulliver,... | 1166666 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,550 |
Shapovalov A.V.
A triangle can be cut into three triangles similar to each other.
Prove that it can be cut into any number of triangles similar to each other.
# | Let triangle $ABC$ with the largest angle $C$ be cut into three similar triangles by segments $AX$, $BX$, and $CX$. Since $\angle AXB > \angle ACB$, the angle $\angle AXB$ can only equal the angles $\angle AXC$ and $\angle BXC$ in the other triangles. Therefore, $\angle AXB = \angle AXC = \angle BXC = 120^{\circ}$. But... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,551 |
Zaslavsky A.A.
Along the edges of a convex polyhedron with 2003 vertices, a closed broken line is drawn, passing through
each vertex exactly once. Prove that in each of the parts into which this broken line divides the surface of
the polyhedron, the number of faces with an odd number of sides is odd. | Let's choose any of the resulting parts. Consider the sum $a_{1}+a_{2}+\ldots+a_{n}$, where $a_{i}$ is the number of sides of the $i$-th face.
Each edge of the polyhedron that the broken line does not pass through is counted twice in this sum, and therefore the parity of the sum does not depend on the number of such e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,554 |
All natural numbers from 1 to 1000 inclusive are divided into two groups: even and odd.
In which of the groups is the sum of all digits used to write the numbers greater, and by how much?
# | The sum of the digits of the number 1 is equal to the sum of the digits of the number 1000; the remaining numbers will be divided into 499 pairs: $\{2,3\},\{4,5\}, \ldots$, $\{998,999\}$. In each pair, the sum of the digits of the odd number is 1 greater than the sum of the digits of the even number.
## Answer
The su... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,555 |
Yatsenko I.v.
From the volcano station to the summit of Stromboli, it takes 4 hours to walk along the road, and then another 4 hours along the path. At the summit, there are two craters. The first crater erupts for 1 hour, then remains silent for 17 hours, then erupts again for 1 hour, and so on. The second crater eru... | The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the eruption of the first crater, this crater will not be dangerous.
Movement along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the eruption of... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,556 |
[Coordinate method on the plane ] The midline of a trapezoid $\quad]$
Let $M\left(x_{0}, y_{0}\right)$ be the midpoint of the segment with endpoints at $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Prove that $x_{0}=1 / 2\left(x_{1}+x_{2}\right), y_{0}$ $=1 / 2\left(y_{1}+y_{2}\right)$. | Let different points $A_{x}\left(x_{1}, 0\right)$ and $B_{x}\left(x_{2}, 0\right)$ be located on the $O X$ axis, and $M_{x}\left(x_{0}, 0\right)$ be the midpoint of segment $A_{x} B_{x}$. Since $A_{x} M_{x}=B_{x} M_{x}$, then
$\left|x_{0}-x_{1}\right|=\left|x_{0}-x_{2}\right|$. Therefore, either $x_{0}-x_{1}=x_{0}-x_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,558 |
[Coordinate method on the plane]
Prove that the equation of a line with slope $k$, passing through the point $M_{0}\left(x_{0} ; y_{0}\right)$, has the form $y - y_{0}=k\left(x-x_{0}\right)$.
# | Let the line $y=k x+l$ pass through the point $M_{0}\left(x_{0} ; y_{0}\right)$. Then the coordinates of the point $M$ satisfy the equation $y=$ $k x+l$, i.e., $y_{0}=k x_{0}+l$ - a true numerical equality. The point $M(x ; y)$ belongs to the line if and only if $y=k x+l$ - also a true numerical equality. Subtracting t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,559 |
[ Thales' Theorem and the Theorem of Proportional Segments ] [ Orthogonal (Rectangular) Projection $\quad$ ]
Given points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ and a non-negative number $\lambda$. Find the coordinates of point $M$ on the ray $A B$, for which $A M: A B=\lambda$. | If the point $M\left(x_{0}, y_{0}\right)$ lies on the line $A B$ and $A M: A B=\lambda$, then by the theorem of proportional segments, the same ratio holds for the projection of point $M$ on the $O X$ axis, i.e., $\frac{x_{0}-x_{1}}{x_{2}-x_{1}}=\lambda$. From this, we find that $x_{0}=(1-\lambda) x_{1}+\lambda x_{2}$.... | ((1-\lambda)x_{1}+\lambdax_{2},(1-\lambda)y_{1}+\lambday_{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,560 |
8,9 |
| :---: | :---: | :---: |
| | [ Properties of medians. Centroid of a triangle | |
| | Vectors help solve the problem. | |
| | Average values | |
Prove that the coordinates of the intersection point of the medians of a triangle are the arithmetic means of the corresponding coordinates of the vertices of the... | Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ and $C\left(x_{3}, y_{3}\right)$ be the vertices of a triangle, and $M\left(x_{0}, y_{0}\right)$ be the point of intersection of its medians.
First method. It is known that for any point $O$ the equality $\overrightarrow{O M}=\frac{1}{3}(\overrightarrow{O A}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,561 |
$\left.\quad \begin{array}{ll}\text { Coordinate method on the plane }\end{array}\right]$
Given points $A(5 ; 5), B(8 ;-3)$ and $C(-4 ; 1)$. Find the coordinates of the intersection point of the medians of triangle $A B C$. | ## First method.
It is known that the medians of a triangle are divided by the point of intersection in the ratio 2:1, counting from the vertex. Therefore, if $D\left(x_{1} ; y_{1}\right)$ is the midpoint of segment $B C$, then $A M: M D=2: 1$. It is also known that the coordinates of the midpoint of a segment are the... | (3;1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,562 |
[ Scalar product. Relations ]
Compose the equation of the circle passing through the points $A(-2 ; 1), B(9 ; 3)$ and $C(1 ; 7)$. | Since
$$
\overrightarrow{A C}=\overrightarrow{(1-(-2) ; 7-1)}=\overrightarrow{(3 ; 6)}, \overrightarrow{B C}=\overrightarrow{(1-9) ; 7-3)}=\overrightarrow{(-8 ; 4)},
$$
then
$$
\overrightarrow{A C} \cdot \overrightarrow{B C}=3 \cdot(-8)+6 \cdot 4=0
$$
Therefore, $A C \perp B C$, i.e., triangle $A B C$ is a right tr... | (x-\frac{7}{2})^{2}+(y-2)^{2}=\frac{125}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,563 |
[Coordinate method on the plane $]$ $[\underline{\text { Orthocenter and orthotriangle }]}$
Given points $A(5, -1), B(4, -8), C(-4, -4)$. Find the coordinates of the intersection point of the altitudes of triangle $ABC$.
# | Applying the condition of perpendicularity of two lines ($k_{1} \cdot k_{2}=-1$), find the equations of the lines on which two altitudes of the triangle lie. Then find the coordinates of the intersection point of these lines by solving the corresponding system of equations.
## Solution
Let's find the equation of the ... | (3,-5) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,564 |
$3+$ [Arithmetic operations. Numerical identities]
Authors: Sadykov R., Cherapanov E.
Do there exist 10 such distinct integers that all sums, composed of nine of them, are perfect squares
# | Let the desired numbers and their sum be denoted by $x_{1}, \ldots, x_{10}$ and $S$, respectively. Then
$S-x_{1}=n_{1}^{2}, \ldots, S-x_{10}=n_{10}^{2}$. Therefore, $S=\frac{n_{1}^{2}+\ldots+n_{10}^{2}}{9}$. Let $n_{k}=3 k(k=1, \ldots, 10)$.
Then the sum of the squares is divisible by 9. It is clear that the numbers ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,565 |
X. Khanov
Prove that the numbers from 1 to 15 cannot be divided into two groups: $A$ consisting of two numbers and $B$ consisting of 13 numbers, such that the sum of the numbers in group $B$ is equal to the product of the numbers in group $A$. | Let such a partition be possible and the numbers $x$ and $y, x<y$, entered into $A$. Then the sum of the numbers in $B$ is $1+2+\ldots+15-x$ $-y=120-x-y$, that is, $xy=120-x-y$. Rewriting this equality as $(x+1)(y+1)=121$, we get that $x$ $+1=1, y+1=121$, which is impossible. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,566 |
Bogosnov I.I.
Numbers, either red or blue, are arranged in a circle. Each red number is equal to the sum of its neighboring numbers, and each blue number is equal to the half-sum of its neighboring numbers. Prove that the sum of the red numbers is zero.
# | Let $a, b, c$ - be three consecutive numbers. If $b$ is red, then $b=a+c$, and if $b$ is blue, then $2 b=a+c$. We will write such equations for all triples of consecutive numbers and add them. On the right side, we will get twice the sum of all numbers, and on the left side - the sum of red numbers plus twice the sum o... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 52,568 |
Baranov D.V.
A seven-digit code consisting of seven different digits is called good. The password of the safe is a good code. It is known that the safe will open if a good code is entered and one of the digits in the code matches the corresponding digit of the password. Can the safe be guaranteed to be opened in fewer... | Let's enter the codes 1234560, 2345610, 3456120, 4561230, 5612340, 6123450. Among the first six digits of the password, there is a digit from 1 to 6. Since we have entered each digit from 1 to 6 once in each of the first six positions, it will match at least once.
## Answer
It can be done. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,571 |
Avor: Tomomg A.K.
Baron Munchausen has 50 weights. The weights of these weights are different natural numbers not exceeding 100, and the total weight of the weights is an even number. The baron claims that it is impossible to place some of these weights on one pan of the scales and the rest on the other pan so that th... | Let the baron have weights of all even values from 2 to 100. Suppose we manage to balance all of them. Then the equilibrium will be maintained if all the weights are divided by 2. However, weights of 1, 2, ..., 50 cannot be balanced in this way, since the sum of their weights is odd.
## Answer
They can. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,572 |
Bakayev E.V.
On an $8 \times 8$ board, there are 8 non-attacking rooks. All cells of the board are distributed among the rooks according to the following rule. The cell on which a rook stands is assigned to that rook. A cell that is attacked by two rooks is obtained by the rook that is closer to this cell; if these tw... | Rook $A$ attacks a total of 15 cells - in its vertical and its horizontal. Consider another cell $C$ in this horizontal. It is attacked by exactly one more rook $B$, which is in the same vertical as $C$. This rook also attacks one cell $D$, which is in the same vertical as $A$. $A, B, C, D$ are the corner cells of a re... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,574 |
[Divisibility of numbers. General properties] $[$ The extremal principle (stronger). $]$
In the decimal representation of a certain number, the digits are arranged from left to right in descending order. Can this number be divisible by 111? | Suppose that the numbers indicated in the condition and divisible by 111 exist. Let $A$ be the smallest among them. There are two cases.
1) The number $A$ ends in zero. By erasing it, we obtain a smaller number with digits in descending order, also divisible by 111. Contradiction.
2) $A$ ends in a non-zero digit. Then... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,575 |
Rushkin C.
On the surface of a cube, a closed eight-segment broken line is drawn, the vertices of which coincide with the vertices of the cube.
What is the minimum number of segments of this broken line that can coincide with the edges of the cube?
# | The segments of the broken line are either edges of the cube or diagonals of its faces. To do this, let's color the vertices of the cube in two colors in a checkerboard pattern (see fig.). Note that an edge of the cube connects vertices of different colors, while a diagonal connects vertices of the same color. For the ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,577 |
Kohas K.P.
Two magicians show the audience such a trick. The audience has 24 cards numbered from 1 to 24. They choose 13 cards and give them to the first magician. The first magician returns two of them to the audience. The audience
adds one of the remaining 11 cards to these two and, after shuffling, gives these thre... | Let's divide the natural numbers from 1 to 24 into 12 pairs. The magicians can agree in advance on how to do this. For example, \{1,24\}, \{2,23\}, \{3,22\}, and so on. Among the thirteen cards chosen by the audience, there will be two cards with numbers from the same pair. These are the ones the first magician should ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,578 |
Rubanov I.S.
In five pots standing in a row, Rabbit poured three kilograms of honey (not necessarily into each and not necessarily equally). Winnie-the-Pooh can take any two adjacent pots. What is the maximum amount of honey that Winnie-the-Pooh can guarantee to eat?
# | Evaluation. Let's assume Winnie-the-Pooh cannot take at least a kilogram of honey. This means that in any pair of adjacent pots, there is less than a kilogram of honey. This is true for both the two rightmost pots and the two leftmost pots. However, then in the middle pot, there must be more than a kilogram of honey (o... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,580 |
[ Systems of points and segments. Examples and counterexamples ] Proof by contradiction
Each point in space is colored in one of five colors, and each of these five colors is used to color at least one point. Prove that there exists a plane, all points of which are colored in at least 4 colors. | If some line is colored in 3 colors, the problem is solved easily.
## Solution
If some line is colored in 3 or more colors, we draw a plane through this line and a point of one of the missing colors. This plane will be colored in at least 4 colors. Next, assume that each line is colored in no more than 2 colors. We f... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,581 |
The sequence $\mathrm{f}(\mathrm{n})(\mathrm{n}=1,2, \ldots)$, consisting of natural numbers, is such that $\mathrm{f}(\mathrm{f}(\mathrm{n}))=\mathrm{f}(\mathrm{n}+1)+\mathrm{f}(\mathrm{n})$ for all natural n. Prove that all terms of this sequence are distinct. | Assume the opposite: for some distinct natural numbers $\mathrm{k}$ and $\mathrm{m}$, $\mathrm{f}(\mathrm{k})=\mathrm{f}(\mathrm{m})$. Then $f(f(k))=f(k+1)+f(k)$ and $f(f(m))=f(m+1)+f(m)$, from which $f(k+1)=f(f(k))-f(k)$ and $f(m+1)=f(f(m))-f(m)$. The right-hand sides of these equations are equal by our assumption, he... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,582 |
Auto: : Bogosnov I.I.
All integers from -33 to 100 inclusive were arranged in some order, and the sums of each pair of adjacent numbers were considered. It turned out that there were no zeros among them. Then, for each such sum, the number reciprocal to it was found. The obtained numbers were added. Could the result b... | Let's consider an example of such an arrangement. Consider the sequence $100, -33, 99, -32, \ldots, 34, 33$. Then, if the first number of the pair is in an odd position, the sum is 67. If it is in an even position, the sum is 66. The reciprocals will be $\frac{1}{67}$ and $\frac{1}{66}$, respectively, with the first oc... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,583 |
[ Pairing and grouping; bijections ] [ Counting in two ways ]
In a brigade of 7 people, their total age is 332 years. Prove that it is possible to select three people whose combined age is not less than 142 years.
# | Let's consider all possible triples of the work team. The sum of their ages, as is easy to calculate, is equal to 15 $\cdot$ 332, and there are 35 such triples in total. Therefore, there is a triple whose total age is not less than $15 \cdot 332 / 35$, which is more than 142. | 142 | Combinatorics | proof | Yes | Yes | olympiads | false | 52,584 |
[ Extremal Properties (other) ] [ Examples and Counterexamples. Constructions ]
The number of edges of a convex polyhedron is 99. What is the maximum number of edges that a plane, not passing through its vertices, can intersect? | The plane does not intersect at least one edge in each face.
## Solution
The answer is 66. We need to prove that this number is possible, and that a larger number is impossible. First, let's prove the latter. Each face of the polyhedron is intersected by the plane in no more than two edges, meaning that at least one ... | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,585 |
Yamenniko i.v.
The numbers 2, 3, 4, ..., 29, 30 are written on the board. For one ruble, you can mark any number. If a number is already marked, you can freely mark its divisors and numbers that are multiples of it. What is the minimum number of rubles needed to mark all the numbers on the board? | Let's mark the numbers $17, 19, 23$, and 29, spending four rubles. Then mark the number 2, spending another ruble. After this, we can freely mark all even numbers (since they are divisible by 2), and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$) the even number $2n$ is already m... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,586 |
Shapovalov A.V.
In the center of each cell of a rectangular grid $\$ M \$$ there is a point light bulb, initially all of them are off. In one move, it is allowed to draw any straight line that does not touch the light bulbs and turn on all the light bulbs on one side of this line, if all of them are off. Each move mus... | Instead of the original rectangle $\$ M$ b, we will consider the rectangle $\$ \mathrm{~N} \$$ with vertices at the corner lamps.
Estimates. Note that with each move, at least one of the four corner lamps is lit. Therefore, the number of moves is no more than 4. In part a), note also that we must light the central lam... | )3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,587 |
Evdokimov M.A.
16 cards with integers from 1 to 16 are laid face down in a $4 \times 4$ table so that cards with consecutive numbers are adjacent (touching by a side). What is the minimum number of cards that need to be flipped simultaneously to definitely determine the location of all numbers (regardless of how the c... | Evaluation. Let's number the cells as shown in Figure 1.
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 2 | 1 | 4 | 3 |
| 5 | 6 | 7 | 8 |
| 6 | 5 | 8 | 7 |
Fig. 1
Fig. 2
\left(x y+y z+x z-x^{2}-y^{2}-z^{2}\right)=2 b\left(x y+y z+x z-x^{2}-y^{2}-z^{2}\right)$ (see problem $\underline{61005}$ g). But the expression in parentheses is non-positive and turns to zero only when $x=y=z$. Therefore, the equation has a solution only when $b=0$. In this ca... | (0,,-),(,0,-)forb=0;nosolutionsforb\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,590 |
Points $A$ and $B$ move uniformly and with equal angular velocities along circles $O_{1}$ and $O_{2}$ respectively (clockwise). Prove that vertex $C$ of the equilateral triangle $A B C$ also moves uniformly along some circle. | Without loss of generality, we can assume that $A$ and $B$ move clockwise with unit angular velocity, i.e., at time $t$ the arguments of vectors $\overrightarrow{O_{1} A}, \overrightarrow{O_{2} B}$ are $t_{1}+t$ and $t_{2}+t$ respectively. All vectors on the plane can be associated with complex numbers. Then $\overrigh... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,591 |
$\left\{a_{1}, a_{2}, \ldots, a_{20}\right\}$ - a set of positive integers.
We construct a new set of numbers $\left\{b_{0}, b_{1}, b_{2}, \ldots\right\}$ according to the following rule:
$b_{0}$ - the number of elements in the original set that are greater than 0,
$b_{1}$ - the number of elements in the original se... | Let's consider the following construction. Take a set of cubes and build "towers" - columns of heights $a_{1}, a_{2}, \ldots, a_{n}$ (see the figure). We will count the number of cubes needed in two ways. Counting by columns, we get that the number of cubes is equal to the sum of the numbers in the first set. Another w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,598 |
Chekanov Yu.V.
Kolya has a segment of length $k$, and Lev has a segment of length $l$. First, Kolya divides his segment into three parts, and then Lev divides his segment into three parts. If two triangles can be formed from the resulting six segments, Lev wins; otherwise, Kolya wins. Depending on the ratio of $k/l$, w... | First case. If $k>l$, then Kolya wins: he can cut off a part from $k$ that will be greater than the sum of all the rest.
For example, $k$ can be cut into parts (fig.)
$$
l+\frac{2}{3}(k-l), \frac{1}{6}(k-l), \frac{1}{6}(k-l)
$$
Then the largest part $l+\frac{2}{3}(k-l)$ cannot be a side of any triangle: by the trian... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,599 |
$3+$
[ Proof by contradiction ]
In a chess tournament for the title of Master of Sports, 12 people participated, each playing one game against every other participant. 1 point is awarded for a win, 0.5 points for a draw, and 0 points for a loss. At the end of the tournament, the title of Master of Sports was awarded t... | Calculate the total points scored by players who received the title of Master of Sports.
## Solution
a) Let 7 participants win all games against the remaining five, and all games between themselves ended in a draw. Then each of them scored
$5+0.5 \cdot 6=8$ points, which is more than the required $0.7 \cdot 11=7.7$ ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,603 |
The corridor is covered by several carpet runners (possibly with overlaps). Prove that it is possible to remove some runners in such a way that the remaining runners still cover the corridor and the sum of their lengths does not exceed twice the length of the corridor.
# | Consider the minimal system of paths covering the entire corridor. Show that in each overlap, no more than two paths are involved.
## Solution
Let's rephrase the problem as follows. Suppose a segment of length 1 is covered by segments. We need to prove that some segments can be excluded from the cover so that the rem... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,604 |
[ Circles on a Sphere
On a spherical Sun, a finite number of circular spots have been discovered, each occupying less than half of the Sun's surface. These spots are assumed to be closed (i.e., the boundary of the spot belongs to it) and do not intersect with each other. Prove that there will be two diametrically oppo... | Consider the spot with the largest radius. Slightly increase the radius of the circle containing it so that it is again smaller than the radius of the Sun and this circle does not intersect with the other spots. Reflect the boundary of the circle relative to the center of the Sun. The resulting circle cannot be complet... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,608 |
On a sheet of paper, points $A, B, C, D$ are marked. A recognition device can perform two types of operations with absolute accuracy: a) measure the distance in centimeters between two given points; b) compare two given numbers. What is the minimum number of operations this device needs to perform to definitely determi... | To determine whether $A B C D$ is a rectangle, it is sufficient to check the equalities $A B=C D$, $B C=A D$, and $A C=B D$ - a total of 9 operations (3 operations for each equality: two measurements and one comparison). The rectangle $A B C D$ will be a square if $A B=B C$ - for this, one more, the 10th, operation of ... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,609 |
Can you place crosses and noughts on a sheet of squared paper so that on no horizontal, vertical, or diagonal line can you find three identical signs in a row?
# | Answer: It is possible. An example of such an arrangement is shown in the figure.
Send a comment | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,611 |
Folkpor
a) Two identical gears with 14 teeth each are placed one on top of the other so that the teeth align (so the projection onto a plane looks like a single gear). After this, four pairs of aligned teeth are cut out. Is it always possible to rotate these gears relative to each other so that the projection onto a p... | a) In the initial position, four pairs of cut-out teeth coincide (one tooth on the first gear and one on the second). There are 16 such pairs in total. Therefore, in one of the 13 remaining positions, no pair coincides.
b) Counterexample: number the teeth in order and cut out the teeth with numbers $1, 2, 5, 7$. As is... | ) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,612 |
Frankin B.R.
Given that for no side of the triangle can a triangle be formed from the height, bisector, and median drawn to it.
Prove that one of the angles of the triangle is greater than $135^{\circ}$. | From the condition, it follows that each median is not less than the sum of the bisector and the height drawn from the same vertex. If the angle between some median and the corresponding height is not greater than $60^{\circ}$, then the median is not greater than twice the height, and the sum of the bisector and the he... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,613 |
Kovaldzhi A.K.
Four grasshoppers sit at the vertices of a square. Every minute, one of them jumps to the point symmetric to it relative to another grasshopper. Prove that the grasshoppers cannot at some point end up at the vertices of a larger square.
# | $1^{\circ}$. Let's imagine that the square, with grasshoppers sitting at its vertices, is a square on a grid paper
(the size of the square is $1 \times 1$). Note that the grasshoppers always j... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,614 |
## [ Connectivity and Decomposition into Connected Components]
In a connected graph, the degrees of four vertices are 3, and the degrees of the remaining vertices are 4.
Prove that it is impossible to remove an edge such that the graph splits into two isomorphic connected components.
# | Let it succeed. If the removed edge connected vertices of the same degree, then in each obtained component there will be an odd number of odd vertices (one or three), which is impossible.
If, however, the removed edge connected vertices of degrees 3 and 4, then only in one component will there be a vertex of degree 2,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,616 |
[Sum of angles in a triangle. Theorem of the exterior angle.] Counting in two ways
In a square, 20 points were marked and connected by non-intersecting segments to each other and to the vertices of the square, such that the square was divided into triangles. How many triangles were formed? | We will consider the marked points and the vertices of the square as vertices, and the segments connecting them and the sides of the square as edges of a planar graph. For each piece into which this graph divides the plane, we will count the number of edges bounding it, and sum all the obtained numbers. Since each edge... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,617 |
[ $\quad\left[\begin{array}{l}\text { Factorization } \\ {\left[\begin{array}{l}\text { Induction (other) }\end{array}\right]}\end{array}\right]$
Prove that $2^{2 n-1}+3 n+4$ is divisible by 9 for any $n$.
# | $2^{2 n-1}+3 n+4=\left(2^{2 n-1}+1\right)+3 n+3=3\left(2^{2 n-2}-2^{2 n-3}+\ldots-2+1+n+1\right)$, a
$2^{2 n-2}-2^{2 n-3}+\ldots-2+1+n+1 \equiv(-1)^{2 n-2}-(-1)^{2 n-3}+\ldots-(-1)+1+n+1=3 n(\bmod 3)$, that is, it is divisible by 3.
Send a comment | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,619 |
In a hockey tournament, $n$ teams are participating. Each team meets every other team once, with the winning team being awarded 2 points, the team that draws - 1, and the losing team - 0 points. What is the maximum point difference that can exist between teams that finish in adjacent positions? | Evaluate the number of points scored by the first few teams and the last few teams.
## Solution
Estimation. We divide the teams into two groups - the first group includes the teams that took the first $k$ places, and the second group includes the teams that took the last $m=n-k$ places. The first $k$ teams scored $k(... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,621 |
Tereshin D.
On the sides $AB$ and $AD$ of the square $ABCD$, points $K$ and $N$ are taken, respectively. Given that $AK \cdot AN = 2 BK \cdot DN$. Segments $CK$ and $CN$ intersect the diagonal $BD$ at points $L$ and $M$. Prove that points $K, L, M, N$, and $A$ lie on the same circle. | Prove that $\angle B K C+\angle D N C=135^{\circ}$.
## Solution
Let the side of the square be 1. Denote $B K=a, D N=b, \quad \angle B K C=\alpha, \angle D N C=\beta$. Then $(1-a)(1-b)=$ $2 a b$, from which $1-a b=a+b$.
Since $\operatorname{tg} \alpha=1 / a, \operatorname{tg} \beta=1 / b$, then $\operatorname{tg}(\al... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,623 |
In a regular eighteen-sided polygon $A_{0} . . A_{17}$, the diagonals $A_{0} A_{p+3}, A_{p+1} A_{18-r}$, and $A_{1} A_{p+q+3}$ are drawn. Prove that the specified diagonals intersect at one point in any of the following cases:
a) $\{p, q, r\}=\{1,3,4\}$
b) $\{p, q, r\}=\{2,2,3\}$. | Apply Ceva's theorem (see problem 35216) to the hexagon $A_{0} A_{1} A_{p+1} A_{p+3} A_{p+q+3} A_{18-r}$.
a) We need to check the equality $\sin 10^{\circ} \sin 20^{\circ} \sin 70^{\circ}=\sin 10^{\circ} \sin 30^{\circ} \sin 40^{\circ}$. We have $\sin 20^{\circ} \sin 70^{\circ}=\sin 20^{\circ} \cos$ $20^{\circ}=1 / 2 ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,624 |
[ Products and factorials ]
Prove that for a prime $p(p-1)!\equiv-1(\bmod p)$.
# | All numbers from 2 to $p-2$ can be divided into pairs of mutually inverse numbers under multiplication, that is, for each $a$ in this interval, there exists a $b$ (different from $a$ according to problem $\underline{60718}$) such that $a b \equiv 1(\bmod p)$. Therefore, $(p-1)!\equiv p-$ $1(\bmod p)$ | (p-1)!\equivp-1(\bmodp) | Number Theory | proof | Yes | Yes | olympiads | false | 52,627 |
Using the result of problem 61403, prove the inequalities:
a) $\sqrt[n]{a_{1} \ldots a_{n}} \leq \frac{a_{1}+\ldots+a_{n}}{n}$ (Cauchy's inequality);
b) $\left(\frac{b_{1}+\ldots+b_{n}}{n}\right)^{b_{1}+\ldots+b_{n}} \leq b_{1}^{b_{1}} \ldots b_{n}^{b_{n}}$
c) $c_{1}^{b_{1}} \ldots c_{n}^{b_{n}} \leq c_{1} b_{1}+\ld... | a) The first method. From the Cauchy inequality for two numbers (see problem 20862), the Cauchy inequality for \( n = 2^k \) numbers can easily be derived by induction. Therefore, it is sufficient to prove that the Cauchy inequality for \( n \) numbers implies the Cauchy inequality for \( n-1 \) numbers.
Indeed, accor... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,629 |
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