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3 [Systems of segments, lines, and circles]
How many parts do $n$ lines in general position divide the plane into, that is, such that no two are parallel and no three pass through the same point?
# | We will prove by induction that the number of parts is $1+1 / 2 n(n+1)$. The base case ($n=1$) is obvious.
Inductive step. Let $n>1$. By the inductive hypothesis, before the $n$-th line is drawn, there were $1+1 / 2(n-1) n$ parts. The new line is divided into $n$ segments by the points of intersection with the old lin... | 1+\frac{1}{2}n(n+1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,397 |
\[
\left[\begin{array}{l}
\text { Pascal's Triangle and the Binomial Theorem } \\
{[\underline{\text { Induction (other) }}]}
\end{array}\right.
\]
\[
\begin{aligned}
& \frac{1}{1} \\
& \frac{1}{2} \quad \frac{1}{2} \\
& \frac{1}{3} \quad \frac{1}{6} \quad \frac{1}{3} \\
& \begin{array}{llll}
\frac{1}{4} & \frac{1}{12... | The denominators of the numbers located in the rows of the harmonic triangle are proportional to the elements of Pascal's triangle, with the boundary members serving as the coefficients of proportionality. Where Pascal's triangle has the number $C_{n}^{k}$, Leibniz's triangle contains $\frac{1}{(n+1) C_{n}^{k}}$. The r... | \frac{1}{(n+1)C_{n}^{k}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,398 |
[ $\quad$ Monotonicity and Boundedness $]$ [ Mixed Equations and Systems of Equations]
Solve the system of equations:
$x^{2}+4 \sin ^{2} y-4=0$,
$\cos x-2 \cos ^{2} y-1=0$. | $2 \cos ^{2} y=\cos x-1 \leq 0$, from which $\cos ^{2} y=0, \sin ^{2} y=1$. Now from the first equation we get $x=0$.
## Answer
$x=0, y=\pi / 2+k \pi$.
## Problem | 0,\pi/2+k\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,399 |
3 [Problems on Percentages and Ratios]
Mom gave Vasya money for 30 pencils. It turned out that the pencil factory was running a promotional campaign in the store: in exchange for a receipt for a set of 20 pencils, they return $25 \%$ of the set's cost, and for a receipt for a set of 5 pencils, they return $10 \%$. Wha... | Note that $25 \%$ of the cost of 20 pencils is the cost of 5 pencils, and $10 \%$ of the cost of 5 pencils is half the cost of a pencil. It is clear that to get the maximum discount, Vasya should act as follows: as long as he has enough money, he should buy a set of 20 pencils and immediately exchange the receipt; if h... | 36 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,400 |
Kanel-Belov A.Y.
Two numbers are written on the board in the laboratory. Every day, the senior researcher Petya erases both numbers on the board and writes down their arithmetic mean and harmonic mean instead. On the morning of the first day, the numbers 1 and 2 were written on the board. Find the product of the numbe... | The product of the numbers on the board does not change. Indeed, $\frac{a+b}{2} \cdot \frac{2 a b}{a+b}=a b$. Therefore, the desired product is 2.
## Answer
2.
Send a comment | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,401 |
On each kilometer of the highway between the villages of Yolkiino and Palkino, there is a post with a sign. On one side of the sign, it shows how many kilometers are left to Yolkiino, and on the other side, how many kilometers are left to Palkino. Borya noticed that on each post, the sum of all the digits is 13. What i... | Let the distance from Yolkiino to Palkiino be $n$ kilometers. Clearly, $n \geq 10$. Moreover, $n \leq 49$ (otherwise, the sum of the digits on the 49th milestone would be greater than 13).
On the tenth milestone from Yolkiino, one side reads 10, and the other side reads $n-10$, which does not exceed 39. The sum of its... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,402 |
3 [ Decimal numeral system $]
Which numbers are more numerous among all numbers from 100 to 999: those for which the middle digit is greater than both extreme digits, or those for which the middle digit is less than both extreme digits? | Note that if the middle digit of a number $a$ is greater than both of the outer digits, then for the number 999 - $a$, the middle digit is less than both of the outer digits. Therefore, among the numbers from 100 to $999-100=899$, there is an equal number of numbers with the largest middle digit and with the smallest m... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,403 |
Galochkin A.i. Prove that in any convex polygon there are no more than 35 angles less than $170^{\circ}$.
# | Suppose this is not the case. Then, in some convex $n$-gon, there are at least 36 angles less than $170^{\circ}$ (the remaining $n-36$ angles do not exceed $180^{\circ}$). The sum of all angles in a convex $n$-gon is $180^{\circ}(n-2)$.
Therefore,
$$
180^{\circ}(n-2)<170^{\circ} \cdot 36+180^{\circ}(n-36)
$$
Problem | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,405 |
$\left[\begin{array}{l}\text { Method of Coordinates on the Plane } \\ \text { [Trigonometric Ratios in a Right Triangle }]\end{array}\right]$
Prove that any line not parallel to the y-axis has an equation of the form $y=k x+l$. The number $k$ is called the slope of the line. The slope of the line, up to its sign, is ... | Let the line be given by the equation $a x + b y + c = 0$. If $b \neq 0$, then this equation can be written in the form: $y = -\frac{a}{b} x - \frac{c}{b}$. Let $-\frac{a}{b} = k, -\frac{c}{b} = l$. We obtain the equation $y = k x + l$.
If the coordinates of two different points on the line are equal, then the line is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,406 |
Agakhanov N.X.
Prove that the numbers from 1 to 16 can be written in a row, but cannot be written in a circle such that the sum of any two adjacent numbers is a square of a natural number.
# | If next to 16 stands the number $x$, then $16+1 \leq 16+x=a^{2} \leq 16+15$, from which $a^{2}=25$ and $x=9$. Therefore, 16 cannot have more than one neighbor, and the arrangement of numbers in a circle that satisfies the condition is impossible. An example of an arrangement in a row: $16,9,7,2,14,11,5,4,12,13,3,6,10,1... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,409 |
Izmeystev I.v.
In a deck of 52 cards, there are 13 cards of each suit. Vanya draws one card at a time from the deck. The drawn cards are not returned to the deck. Each time before drawing a card, Vanya guesses a suit. Prove that if Vanya always guesses a suit that has no fewer cards remaining in the deck than any othe... | Let's call the characteristic of a deck the number of cards of the suit that has the most cards left in the deck. With each move, the characteristic either remains unchanged or decreases by 1. In the latter case, it is obvious that a card of the guessed suit is taken. It remains to note that at the beginning of the gam... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,410 |
Can the circles on a five-pointed star (see figure) be filled with 4 ones, 3 twos, and 3 threes such that the sums of the four numbers on each of the five lines are equal?
# | Suppose the numbers have been arranged in the required manner. Let $S$ be the sum of the numbers on each line, then the sum of the numbers on all five lines is $5S$. Since each circle lies at the intersection of two lines, in this calculation, the number written in each circle is counted twice. Therefore, the found sum... | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,413 |
Consider $C$.
In 25 boxes, there are balls of several colors. It is known that for any $k(1 \leq k \leq 25)$, in any $k$ boxes, there are balls of exactly $k+1$ different colors. Prove that balls of one of the colors lie in all the boxes. | Let the boxes be denoted as $B_{1}, \ldots, B_{25}$. According to the problem, the total number of colors is 26. If we consider all the boxes except $B_{i}$, the total number of colors in them is 25. Therefore, there is a color that is present only in the box $B_{i}$; let's call it $C_{i}$. Since the total number of co... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,414 |
Raskina I.V.
In a singing competition, a Rooster, a Crow, and a Cuckoo participated. Each member of the jury voted for one of the three performers. The Woodpecker calculated that there were 59 judges in total, and that the Rooster and the Crow received a total of 15 votes, the Crow and the Cuckoo received 18 votes, an... | The number of votes cast for the Rooster and the Raven cannot be more than $15+13=28$. Similarly, the total number of votes for the Raven and the Cuckoo cannot exceed $18+13=31$, and for the Cuckoo and the Rooster, it cannot exceed $20+13=33$ votes. By adding these three results, we estimate the upper bound of twice th... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,415 |
Shatov A.V.
On an $8 \times 8$ chessboard, a cube (the bottom face coincides with one of the cells of the board) is placed. It was rolled across the board, flipping over its edges, such that the cube visited all the cells (some cells, possibly, more than once). Could it happen that one of its faces never lay on the bo... | Place the cube on square a1 and roll it along the route a1 - a2 - b2 - b1. In doing so, it will end up in the adjacent square b1 and will again be standing on the bottom face (the one it initially stood on), while the face that was initially on top never touched the board. Each time we move the cube to an adjacent squa... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,416 |
8
Petya can mark points on any segment that divide this segment in half or in the ratio $n:(n+1)$, where $n$ is any natural number. Petya claims that this is enough to mark a point on any segment that divides it in any given rational ratio. Is he right? | Rational ratio is the ratio of integers. To divide a segment in the ratio $k: l$, it is enough to divide it into $m=k+l$ equal parts. We will show how to do this by induction on $m$. Base: $m=1$. Inductive step. Suppose we can divide a segment into any number of parts less than $m$. If $m=2 n$, we divide the segment in... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 54,417 |
Borodin P.A.
Is it true that any 100 cards, each with a single digit 1, 2, or 3, each digit appearing no more than 50 times, can be arranged in a single row so that there are no fragments 11, 22, 33, 123, and 321? | Let there be $x$ cards with the digit 1, $y$ cards with the digit 2, and $z$ cards with the digit 3. Then $x+y+z=100$, and since
$$
\frac{x+y-z}{2}+\frac{z+y-x}{2}+\frac{x+z-y}{2}=\frac{x+y+z}{2}=50
$$
the desired sequence can be formed from $\frac{x+y-z}{2}=50-z$ fragments of 21, then from $\frac{z+y-x}{2}=50-x$ fra... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,420 |
Do there exist natural numbers $a, b, c, d$ such that $a^{3}+b^{3}+c^{3}+d^{3}=100^{100}$? | For example $\left(100^{33}\right)^{3}+\left(2 \cdot 100^{33}\right)^{3}+\left(3 \cdot 100^{33}\right)^{3}+\left(4 \cdot 100^{33}\right)^{3}=\left(1^{3}+2^{3}+3^{3}+4^{3}\right) \cdot 100^{99}=100 \cdot 100^{99}=100^{100}$.
## Answer
There are. | 100^{100} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,421 |
Shapovalov A.V.
Through the origin, lines (including the coordinate axes) are drawn that divide the coordinate plane into angles of $1^{\circ}$.
Find the sum of the abscissas of the points of intersection of these lines with the line $y=100-x$.
# | The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the origin, 180 lines are drawn, the line $y=100-x$ intersects 179 of them. For each point on the line $y=100-x$, the sum of the coordinates is 100, therefore, the total sum of the abscissa... | 8950 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,422 |
Coordinate Method ] Complex [ The ratio of areas of triangles with a common base or common height] Class
In quadrilateral $A B C D$, find a point $E$ such that the ratio of the areas of triangles $E A B$ and $E C D$ is $1: 2$, and the ratio of the areas of triangles $E A D$ and $E B C$ is $3: 4$, given the coordinate... | Since
$$
\overrightarrow{A B}=\overrightarrow{(-2-(-2): 3-(-4))}=\overrightarrow{(0: r)} \overrightarrow{D C}=\overrightarrow{(+-4: 6-(-1))}=\overrightarrow{(0: r)}=\overrightarrow{A B}
$$
then the quadrilateral $A B C D$ is a parallelogram, with sides $A B$ and $C D$ parallel to the y-axis. Let $E$ be the desired po... | E(0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,423 |
$\underline{\text { Avilov N.I. }}$.
In the vertices of a 33-gon, integers from 1 to 33 were written in some order. Then, on each side, the sum of the numbers at its ends was written.
Can the sides contain 33 consecutive integers (in some order)? | Let the numbers at the vertices go in the following order: $1,18,2,19,3,20, \ldots, 16,33,17$. It is not hard to verify that the sums of two adjacent numbers will increase in order from 19 to 50. And the sum of the first and last is 18.
## Answer
They can. | Theycan | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,424 |
Gálnierin $\Gamma$.
In each cell of a secret $n \times n$ table, one of the digits from 1 to 9 is written. From these, $n$-digit numbers are formed, written from left to right in the rows and from top to bottom in the columns. Petya wants to write an $n$-digit number without zeros in its representation, such that neit... | If less than $n$ cells have been checked, then in some row of the checked cells there is none, and any numbers can be there.
Suppose Petya has checked $n$ cells along the diagonal, at the intersection of rows and columns with the same numbers. Then it is enough for Petya to present a palindrome number, in which the $i... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,425 |
Orinov E.n.
100 pirates played cards for gold sand, and then each counted how much they won or lost in total. Each loser has enough gold to pay off their debt. In one operation, a pirate can either distribute gold equally to everyone or collect an equal amount of gold from everyone. Prove that it is possible to achiev... | Let $Z$ denote the sum of winnings (which is also the sum of losses). Suppose initially all the sand of each pirate is in their right pocket. We will ask each loser to give 0.01 of their loss to everyone (including themselves). The sand received in this process is placed in their left pocket. As a result, each pirate w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,426 |
Bakayev E.V.
In one operation, you can either add 9 to a number or erase the digit 1 in it at any position.
Can you get the number $A+1$ from any natural number $A$ using such operations?
(If a 1 is erased at the beginning of the number, and it is immediately followed by zeros, these zeros are also erased.) | Let's prepend 8 ones to the number $A+1$. The resulting number $B$, when divided by 9, gives the same remainder as $A$, so it can be obtained from $A$ by adding nines. By then erasing the 8 ones, we get the number $A+1$.
## Answer
From any. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,427 |
[ Processes and Operations ]
Authors: Martynov P., Martynova N.
Paddy is stored in the warehouses of two stores: on the first warehouse, there is 16 tons more than on the second. Every night
at midnight, the owner of each store steals a quarter of the paddy available on the competitor's warehouse
and moves it to their... | Let at some point before midnight, the first warehouse has $d$ tons of millet more than the second. At the moment of the theft, this difference decreases by $d / 4$, and at the moment when the stolen grain reaches the "foreign" warehouse, it decreases by another $d / 4$. Thus, every night the difference is halved. Sinc... | 2^{-6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,429 |
The number 61 is written on the board. Every minute, the number is erased from the board and the product of its digits, increased by 13, is written in its place. After the first minute, the number 19 is written on the board ( $6 \cdot 1+13=19$). What number will be on the board after an hour? | Let's consider the numbers that will be written on the board over the first few minutes:
| After one minute | $6 \cdot 1+\mathbf{1 3}=\mathbf{1 9}$ |
| :---: | :---: |
| After two minutes | $1 \cdot 9+13=\mathbf{2 2}$ |
| After three minutes | $2 \cdot 2+13=\mathbf{1 7}$ |
| After four minutes | $1 \cdot 7+13=\mathbf{... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,432 |
Folklore
Plot on the coordinate plane the set of all points whose coordinates $x$ and $y$ satisfy the inequality $\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}} \geq x y$. | The desired set decomposes into two: $A=\{(x, y) \mid |x| \leq 1, |y| \leq 1, xy \leq 0\}$ and $B=\left\{(x, y) \mid |x| \leq 1, |y| \leq 1, (1-x^2)(1-y^2) \geq x^2 y^2\right\}$
Set $A$ represents the union of two squares with side length 1, located in the II and IV coordinate quadrants.
The last inequality in the de... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 54,433 |
1000 apples are distributed among several baskets. You can remove baskets and take apples out of the baskets. Prove that it is possible to achieve an equal number of apples in all the remaining baskets and the total number of remaining apples is no less than 100.
# | Reasoning by contradiction, evaluate the number of baskets that initially contained no less than one apple, no less than two apples, etc.
## Solution
Assume the opposite. Then initially there were fewer than 100 baskets with at least one apple, otherwise we would have taken all the apples except one from each basket,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,434 |
$\left[\begin{array}{l}\text { Coordinate Method on the Plane }] \\ {[\quad \underline{\text { Circles (Miscellaneous). }}]}\end{array}\right.$
Find the radius and coordinates of the center of the circle given by the equation
a) $(x-3)^{2}+(y+2)^{2}=16$;
b) $x^{2}+y^{2}-2(x-3 y)-15=0$;
c) $x^{2}+y^{2}=x+y+\frac{1}{... | a) A circle with radius $R$ and center at point $A(a ; b)$ has the equation
$$
(x-a)^{2}+(y-b)^{2}=R^{2}
$$
In this case, $a=3, b=-2, R=4$.
b)
$$
\begin{aligned}
x^{2}+y^{2}-2(x-3 y)-15= & 0 \Leftrightarrow x^{2}-2 x+1+y^{2}+6 y+9-1-9-15=0 \Leftrightarrow \\
& \Leftrightarrow(x-1)^{2}+(y+3)^{2}=25
\end{aligned}
$$
... | (3,-2),R=4;(1,-3),R=5;(\frac{1}{2},\frac{1}{2}),R=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,435 |
\left.\begin{array}{l}{[\text { Coordinate method on the plane }} \\ {[\text { Pythagorean Theorem (direct and inverse) }}\end{array}\right]
Given points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Prove that
$$
A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}
$$ | Apply the Pythagorean theorem.
## Solution
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be the given points, where $x_{1} \neq x_{2}$ and $y_{1} \neq y_{2}$. Let the line passing through point $A$ parallel to the $O X$ axis and the line passing through point $B$ parallel to the $O Y$ axis interse... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,436 |
[Coordinate method on the plane]
Prove that the equation of the line passing through the points $M_{0}\left(x_{0} ; y_{0}\right)$ and $M_{1}\left(x_{1} ; y_{1}\right)\left(x_{1} \neq x_{0}, y_{1} \neq y_{0}\right)$ has the form
$$
\frac{y-y_{0}}{y_{1}-y_{0}}=\frac{x-x_{0}}{x_{1}-x_{0}}
$$ | Substitute the coordinates of the point $M_{1}\left(x_{1} ; y_{1}\right)$ into the equation of the line with slope $k$, passing through the point $M_{0}\left(x_{0} ; y_{0}\right)$, and find $k$.
## Solution
The equation of a line with slope $k$, passing through the point $M_{0}\left(x_{0} ; y_{0}\right)$, is $y-y_{0}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,437 |
Prove that the line $3 x-4 y+25=0$ is tangent to the circle $x^{2}+y^{2}=25$ and find the coordinates of the point of tangency.
## Answer
$(-3 ; 4)$.
Let's find the coordinates of all common points of the given line and circle. For this, we will solve the system of equations
$$
\left\{\begin{array}{l}
3 x-4 y+25=0 ... | Since the circle touches the coordinate axes and passes through a point located in the first coordinate quadrant, the center of the circle lies on the line $y=x$. Therefore, the abscissa and ordinate of the center of the circle are equal to its radius. Thus, the equation of the circle has the form
$$
(x-R)^{2}+(y-R)^{... | (x-5)^{2}+(y-5)^{2}=25or(x-1)^{2}+(y-1)^{2}=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,438 |
Find the coordinates of the points of intersection of the circles
$$
(x-2)^{2}+(y-10)^{2}=50 \text { and } x^{2}+y^{2}+2(x-y)-18=0 .
$$ | Let's find the coordinates of the intersection points $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)$ of the given circles by solving the system of equations
$$
\left\{\begin{array}{l}
(x-2)^{2}+(y-10)^{2}=50 \\
x^{2}+y^{2}+2(x-y)-18=0
\end{array}\right.
$$
We will get: $x_{1}=3, y_{1}=3, x_{2}=-3, y_... | (3;3),(-3;5) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,439 |
[ $\underline{\text { Invariants }}$ ]
On the board, the number $8^{n}$ is written. The sum of its digits is calculated, then the sum of the digits of the resulting number is calculated again, and so on, until a single-digit number is obtained. What is this number if $n=1989$? | Use the fact that the sum of the digits has the same remainder when divided by 9 as the number itself. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,442 | |
[ Dirichlet's Principle (continued) ]
Numbers are placed on all edges of a cube. On each face (square), the sum of the four numbers located on its edges (sides of the square) is written. Arrange the numbers 1 and -1 on the edges so that all the numbers on the faces are different.
# | Suppose we managed to do this. Then on each of the six faces of the cube, one of the five numbers: $-4, -2, 0, 2$ or 4 is placed. But there are six faces, so on two of the faces, the same number must be placed; a contradiction. Therefore, it is impossible to arrange the numbers this way.
## Answer
This cannot be done... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,443 |
Each point on the plane with integer coordinates is colored in one of p colors. Prove that there exists a rectangle with vertices of the same color. | Consider integer points located in an infinite horizontal strip. In this strip, there will be two identically colored vertical rows of points.
## Solution
Consider a strip of $\mathrm{n}+1$ consecutive horizontal rows of points. Consider the vertical rows of this strip, consisting of $\mathrm{n}+1$ points. There are ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,444 |
[ Dirichlet's Principle (continued).]
There are 101 natural numbers, and the sum of these numbers is 200. Prove that it is always possible to select several of these numbers whose sum is 100. | The following interpretation of the condition can help solve the problem. Let's consider a circle of length 200, divided into 101 arcs, the lengths of which are the given numbers.
## Solution
Let the given numbers be $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{101}$. Take a circle of length 200. Mark 200 poi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,445 |
On a square field $10 * 10$, nine $1 * 1$ cells are overgrown with weeds. After this, the weeds can spread to a cell if it has at least two neighboring cells already overgrown with weeds. Prove that nevertheless, the weeds will not be able to spread to all cells.
# | How does the perimeter of the area overgrown with weeds change?
## Solution
Consider the boundary of the area overgrown with weeds (i.e., all segments of length 1 between nodes, on one side of which there is weed, and on the other side there is not). Initially, the length of the boundary was no more than $9 * 4=36$, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,446 |
[ Dirichlet's Principle (continued).]
In an $n * n$ table, some $2 n$ cells are marked. Prove that there will be a parallelogram with vertices at the centers of the marked cells.
# | Prove that there exists a parallelogram with a pair of sides parallel to a row of the table.
## Solution
We will prove that there exists a parallelogram with a pair of sides parallel to a row of the table.
Consider in each row the leftmost marked cell (if there are marked cells in that row). Let the leftmost marked ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,447 |
In a convex polygon on the plane, there are no fewer than $m^2 + 1$ points with integer coordinates.
Prove that there are $m + 1$ points with integer coordinates that lie on the same line. | Paint the integer points of the plane in $\mathrm{m}^{2}$ colors. There will be two points of the same color.
## Solution
By the Pigeonhole Principle, among $\mathrm{m}^{2}+1$ points with integer coordinates, there will be two points $(\mathrm{k}, \mathrm{l})$ and $\left(\mathrm{k}_{1}, \mathrm{l}_{1}\right)$ such th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,448 |
Gribalko A.V.
a) A convex pentagon is divided into three triangles by non-intersecting diagonals. Can the points of intersection of the medians of these triangles lie on one straight line?
b) The same question for a non-convex pentagon. | Clearly, exactly two diagonals have been drawn, and they both emanate from the same vertex (let's say from $A$). Then the indicated points of intersection of the medians are obtained by a homothety with center $\$ 4 \$$ and coefficient $\$ 2 / 3 \$$ from the midpoints of sides $BC$, $CD$, and $\$ 2$.
a) These midpoint... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,450 |
Kuuyggin A.K.
Among any five nodes of a regular square grid, there will always be two nodes such that the midpoint of the segment between them is also a node of the grid. What is the minimum number of nodes of a regular hexagonal grid that must be taken so that among them there will always be two nodes such that the m... | 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 of the lemma. Introduce the o... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,451 |
[ Examples and counterexamples. Constructions ]
There are 200 cards of size $1 \times 2$, on each of which the numbers +1 and -1 are written. Can these cards be used to fill a sheet of grid paper of size $4 \times 100$ in such a way that the products of the numbers in each column and each row of the resulting table ar... | Answer: Yes. The cards can be arranged as follows: 50 rows of $4 \times 2$ rectangles, filled with cards such that the first column contains +1, and the second column contains -1. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,452 |
8,9
Given a point $M(x ; y)$. Find the coordinates of the point symmetric to point $M$ with respect to a) the origin; b) the point $K(a ; b)$.
# | Let the point $M^{\prime}\left(x^{\prime}\right.$; $\left.y^{\prime}\right)$ be symmetric to the point $M(x ; y)$ with respect to the origin, i.e., the point $O(0 ; 0)$. Since $O$ is the midpoint of the segment $M M^{\prime}$, and the coordinates of the midpoint of a segment are the arithmetic means of the correspondin... | )(-x;-y);b)(2-x;2b-y) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,453 |
Given points $A(-2 ; 0), B(1 ; 6), C(5 ; 4)$ and $D(2 ;-2)$. Prove that quadrilateral $A B C D$ is a rectangle.
# | Prove that $\overrightarrow{A B}=\overrightarrow{D C}$ and $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$.
## Solution
Quadrilateral $A B C D$ is a rectangle if $A B \| D C, A B=D C$ and $A B \perp A D$. To do this, it is sufficient to prove that $\overrightarrow{A B}=\overrightarrow{D C}$ and $\overrightarrow{A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,454 |
At the vertices of a hexagon, the numbers $12,1,10,6,8,3$ (in that order) are written. In one move, it is allowed to choose two adjacent vertices and simultaneously add one to the numbers in these vertices or simultaneously subtract one from them. Can we end up with six numbers in the following order:
a) $14,6,13,4,5,... | a) Yes, for example: $(12,1,10,6,8,3)$ - $(12,1,10,6,7,2)$ - $(14,3,10,6,7,2)$ - $(14,3,10,4,5,2)$ - $(14,6,13,4,5,2)$.
b) No, because the parity of the sum of all numbers cannot change, and at the beginning, the sum was 40, an even number. Therefore, it cannot become $6+17+14+3+15+2$ = 57, an odd number.
## Answer
... | )yes;b)no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,455 |
With the help of a magic ATM, you can exchange any banknote for any finite number of banknotes of lesser value. Having received 1000 francs in one bill, will you be able to pay the rent every month? (This takes place in Switzerland, where the rent is constant, and life is infinite.)
# | If one can make do with a certain coin, then a coin of lesser value will suffice as well. By descending, we get that even a cent will suffice, which is incorrect.
## Answer
No. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,456 |
[ $\left.\begin{array}{l}\text { Coordinate method on the plane } \\ \text { [Pythagorean Theorem (direct and inverse) }\end{array}\right]$
Prove that a circle of radius $R$ with center at point $A(a ; b)$ has an equation of the form
$$
(x-a)^{2}+(y-b)^{2}=R^{2}
$$
# | Recall the definition of a circle and apply the formula for the distance between two points on a coordinate plane.
## Solution
Let the point $M(x ; y)$ belong to a circle of radius $R$ with center $A(a ; b)$. Then the point $M$ is at a distance $R$ from the point $A$, i.e., $M A=R$. By the formula for the distance be... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,457 |
[Coordinate method on the plane]
Find the equation of the line passing through the point $M(-1 ; 4)$ and perpendicular to the line $x-2 y+4=0$.
# | Apply the condition of perpendicularity of two lines ( $\left.k_{1} \cdot k_{2}=-1\right)$.
## Solution
The equation of the given line can be written as $y=\frac{1}{2} x+2$. Then its slope $k_{1}=\frac{1}{2}$. If $k_{2}$ is the slope of the line perpendicular to the given one, then $k_{1} \cdot k_{2}=-1$. Therefore,
... | 2x+y-2=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,459 |
[Coordinate method on the plane]
Given points $A(6 ; 1), B(-5 ;-4), C(-2 ; 5)$. Form the equation of the line on which the height of triangle $A B C$, drawn from vertex $A$, lies.
# | Apply the condition of perpendicularity of two lines ( $k_{1} \cdot k_{2}=-1$ ).
## Answer
$$
x+3 y-9=0
$$
Let's find the equation of line BC through two points:
$$
\frac{y-(-4)}{5-(-4)}=\frac{x-(-5)}{-2-(-5)}, \text { or } y=3 x+11
$$
Then its slope $k_{1}=3$. If $k_{2}$ is the slope of the line perpendicular to ... | x+3y-9=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,460 |
8,9 Find the distance between the parallel lines $y=-3 x+5$ and $y=-3 x-4$.
# | ## First method.
Since the coordinates of point $A(0 ; 5)$ satisfy the equation $y=-3 x+5$, this point lies on the first line. Since the lines are parallel, the distance between them is equal to the distance from point $A$ to the second line. Let's write the equation of this line in general form $(y+3 x+4=0)$ and use ... | \frac{9}{\sqrt{10}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,461 |
[ Coordinate method on the plane ] [Triangle inequality (other).]
Point $M$ lies on the line $3 x-4 y+34=0$, and point $N-$ lies on the circle $x^{2}+y^{2}-8 x+2 y-8=0$. Find the minimum distance between points $M$ and $N$. | Notice that
$$
x^{2}+y^{2}-8 x+2 y-8=0 \Leftrightarrow x^{2}-8 x+16+y^{2}+2 y+1=25 \Leftrightarrow(x-4)^{2}+(y+1)^{2}=5^{2}
$$
This means the center of the circle is the point $Q(4, -1)$, and the radius is 5.
Let $d$ be the distance from the point $Q$ to the line $3 x-4 y+34=0$. Then
$$
d=\frac{|3 \cdot 4-4 \cdot(-... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,462 |
A circle is divided into six sectors, each containing a chip. It is allowed to move any two chips to their adjacent sectors in one move.
Can all the chips be collected in one sector using such operations? | Let's number the sectors around the circle from 1 to 6 and for any arrangement of chips, consider the sum $S$ of the sector numbers where the given six chips are placed (taking into account multiplicity). Clearly, when a chip is moved to an adjacent sector, the corresponding term in the sum $S$ changes parity. Therefor... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,463 |
[ Invariants $]$
On the board, the numbers $1,2, \ldots, 20$ are written. It is allowed to erase any two numbers $a$ and $b$ and replace them with the number $a b + a + b$.
What number can remain on the board after 19 such operations? | Note that $a b+a+b+1=(a+1)(b+1)$. This means that if all numbers are increased by 1, the product of the new numbers under the specified operation does not change. At the beginning (and therefore at the end), it is equal to $21!$.
## Answer
$21!-1$. | 21!-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,464 |
[ Invariants $]
In the countries of Dillia and Dallia, the monetary units are dillers and dallers, respectively, with 1 diller in Dillia being exchangeable for 10 dallers, and 1 daller in Dallia being exchangeable for 10 dillers. A beginning financier has 1 diller and can freely move between the two countries and exch... | Invariant - the remainder of the difference between the number of dealers and the number of dallers modulo 11 for the financier. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,466 |
There is a pile of 1001 stones on the table. A move consists of taking a stone from any pile that contains more than one stone, and then dividing one of the piles into two. Is it possible to leave only piles consisting of three stones on the table after several moves? | Cannot. Consider the quantity $s$, equal to the sum of the number of stones and the number of piles. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,467 |
Given some triple of numbers. With any two of them, it is allowed to perform the following: if these numbers are equal to $a$ and $b$, then they can be replaced by $\frac{a+b}{\sqrt{2}}$ and $\frac{a-b}{\sqrt{2}}$. Is it possible to obtain the triple $(1, \sqrt{2}, 1+\sqrt{2})$ from the triple $\left(2, \sqrt{2}, \frac... | Consider the sum of the squares of the numbers in the triple.
## Solution
Note that $\left(\frac{a+b}{\sqrt{2}}\right)^{2}+\left(\frac{a-b}{\sqrt{2}}\right)^{2}=a^{2}+b^{2}$. Therefore, the sum of the squares of the numbers in the triple does not change. However,
$$
2^{2}+(\sqrt{2})^{2}+\left(\frac{1}{\sqrt{2}}\righ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,468 |
$3+$ [Tables and Tournaments (other) ]
In a $3 \times n$ rectangle, there are chips of three colors, with $n$ chips of each color.
Prove that the chips can be rearranged in each row so that each column contains chips of all colors. | We will prove this by induction on $n$. The base case ($n=1$) is obvious.
Induction step. Suppose the chip in the top cell of the last column is of color $a$. We will consider two cases.
1) The third row contains chips of both remaining colors. Note that in the second row, there is a chip of a color different from $a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,469 |
Upon the completion of a ballroom dance competition, in which 7 boys and 8 girls participated, each of them stated the number of their partners: $3,3,3,3,3,5,6,6,6,6,6,6,6,6,6$. Did anyone of them make a mistake? | The sum of the numbers called out by all the boys should equal the sum of the numbers called out by all the girls.
Indeed, both these sums should equal the number of pairs formed during the competition.
Since the sum of all the numbers called out is 74, each of the sums of the numbers called out by the boys and girls... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,471 |
8,9
On the plane, there is a non-self-intersecting closed broken line, no three vertices of which lie on the same
line. We will call a pair of non-adjacent segments special if the extension of one of them intersects the other segment.
Prove that the number of special pairs is even. | The following consideration can be used: a broken line enters a certain area as many times as it exits.
## Solution
Consider two adjacent segments of the broken line $A B$ and $B C$. Extend these segments beyond vertex $B$ to form rays $B M$ and $B N$, which form angle $M B N$ (points $M$ and $N$ are chosen sufficien... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,472 |
Evenness and Oddness of Point Systems
On each of 15 planets, the distances between which are all different, there is an astronomer who observes the nearest planet to them. Prove that some planet is not observed by anyone.
# | Choose a pair of planets that are closest to each other.
## Solution
Let's take two planets, the distance between which is the smallest among all pairwise distances. It is clear that astronomers on these two planets are looking at each other. Consider the remaining 13 planets. If at least one of the astronomers on th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,473 |
A castle has the shape of a rectangle measuring $7 \times 9$ cells. Each cell, except the central one, is a room in the castle, and the central cell contains a pool. In each wall (side of the cell) separating two adjacent rooms, there is a door. Is it possible to walk through all the rooms without leaving the castle an... | Use a chessboard coloring.
## Solution
We will color the cells of the $7 \times 9$ square in a checkerboard pattern, so that the corner cells are black. In this case, there will be one more black cell than white. It is easy to check that the central cell will be white. Let's assign the rooms of the castle black and w... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,474 |
Several numbers are written on the board. In one move, it is allowed to take any two of them $a$ and $b$, both not equal to zero, and replace them with the numbers $a-b / 2$ and $b+\frac{a}{2}$. Can the original numbers be obtained on the board after several such moves?
# | What happens to the sum of the squares of the numbers written on the board?
## Solution
Consider the sum of the squares of the numbers written on the board, and show that this sum increases. This will justify the negative answer to the question of the problem.
Indeed, $(a - b/2)^2 + (b + a/2)^2 = 5/4 (a^2 + b^2) > a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,475 |
The numbers $1, 1/2, \ldots, 1/n$ are written on a board. It is allowed to erase any two numbers $a$ and $b$ and replace them with the number $ab + a + b$.
What number will remain after $n-1$ such operations? | During this operation, the product of the numbers increased by 1 is preserved.
## Solution
Let at some point the numbers $a, b, \ldots, z$ be written on the board. Consider the product $P=(a+1)(b+1) \ldots(z+1)$. After replacing the pair of numbers $a, b$ with $ab+a+b$, the term $(a+1)(b+1)$ in the product will be re... | n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,477 |
$\left.\frac{[\text { Counting in two ways }]}{[\text { Trees }}\right]$
Tsar Guidon had 5 sons. Among his descendants, 100 each had exactly 3 sons, and the rest died childless.
How many descendants did Tsar Guidon have
# | When counting the descendants, don't forget about the sons of Gvidon.
## Solution
Every descendant of King Gvidon is either a son of one of his descendants or a son of Gvidon himself. According to the condition, all the descendants of Gvidon had a total of 300 sons. And Gvidon himself had 5 sons, so the total number ... | 305 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,478 |
[ Cutting into parts with special properties ]
The vertices of a convex polygon are colored in three colors such that each color is present and no two adjacent vertices are colored the same. Prove that the polygon can be divided by diagonals into triangles such that each triangle has vertices of three different colors... | Use induction on the number of vertices of the polygon.
## Solution
Let the colors be denoted by the numbers $1, 2, 3$. We will prove the statement by induction on the number $n$ of vertices of the polygon. The base case ($n=3$) is trivial.
Inductive step. Suppose $n>3$. Choose two vertices $A$ and $B$ of the same c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,479 |
] Parity ] [Chessboards and chess pieces.
On a chessboard, 8 rooks are placed so that they do not attack each other.
Prove that an even number of rooks are located on the black squares. | Enter the coordinates on the chessboard and calculate the sum of the coordinates of all the rooks.
## Solution
The first way. Let's number all the columns, starting from the leftmost one, and all the rows, starting from the bottommost one, with numbers from 1 to 8. Thus, each rook is assigned a pair of "coordinates."... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,480 |
Tokarev S.i.
Prove that every natural number is the difference of two natural numbers that have the same number of prime divisors.
(Each prime divisor is counted once, for example, the number 12 has two prime divisors: 2 and 3.)
# | If the number $n$ is even, that is, $n=2 m$, then the required numbers will be $4 m$ and $2 m$. Let $n$ be odd, $p_{1}, \ldots, p_{s}$ be its prime divisors, and $p$ be the smallest odd prime number not included in the set $\left\{p_{1}, \ldots, p_{s}\right\}$. Then the required numbers will be $p n$ and $(p-1) n$, sin... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,481 |
Rubanov I.S. In the country there are 2000 cities. Each city is connected by non-stop two-way air routes to some other cities, and for each city, the number of outgoing air routes is a power of two (i.e., $1, 2, 4, 8, \ldots$). For each city $A$, a statistician calculated the number of routes with no more than one tran... | Let's call a non-stop flight from one city to another a short route, and a flight from one city to another with one stopover a long route.
We will renumber the cities and denote by $2^{n_{i}}(i=1, \ldots, 2000)$ the number of flights departing from the $i$-th city.
We will count the short routes at their final destin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,483 |
Rubanov I.S.
Can the numbers $1,2, \ldots, 60$ be arranged in a circle in such an order that the sum of any two numbers, with one number between them, is divisible by 2, the sum of any two numbers, with two numbers between them, is divisible by 3, and the sum of any two numbers, with six numbers between them, is divis... | Let us assume that we have arranged the numbers in the required manner. Take the number 7. According to the condition, the 7th, 14th, 21st, 28th, 35th, 42nd, 49th, and 56th numbers from it clockwise are multiples of 7. We have already counted 9 different numbers that are multiples of 7. But among the numbers from 1 to ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,484 |
Rubanovo I.S.
There is a set of weights with the following properties:
a. It contains 5 weights, all of different weights.
b. For any two weights, there are two other weights with the same total weight.
What is the smallest number of weights that can be in this set? | Let $A$ be one of the lightest weights, and $B$ be one of the weights that follow $A$ in weight. Clearly, the pair of weights $\{A, B\}$ can only be balanced by an identical pair.
Therefore, there are at least two weights $A$ and two weights $B$. The pair $\{A, A\}$ can also only be balanced by an identical pair. Ther... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,487 |
Slobodnik S.G
Given positive numbers $a_{1}, a_{2}, \ldots, a_{n}$. It is known that $a_{1}+a_{2}+\ldots+a_{n} \leq 1 / 2$. Prove that $\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots(1$ $\left.+a_{n}\right)<2$.
# | Expanding the parentheses on the left side, we get the sum $1+\left(a_{1}+\ldots+a_{n}\right)+\left(a_{1} a_{2}+\ldots+a_{n-1} a_{n}\right)+\left(a_{1} a_{2} a_{3}+\ldots+a_{n-2} a_{n-}\right.$ $\left.{ }_{1} a_{n}\right)+\ldots+a_{1} a_{2} \ldots a_{n}$. The sum of the numbers in the second parenthesis does not exceed... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,489 |
Raskina I.V.
A miser knight keeps gold coins in 77 chests. Once, while counting them, he noticed that if he opens any two chests, he can redistribute the coins lying in them equally between these two chests. Then he noticed that if he opens any 3, or any 4, ..., or any 76 chests, he can also redistribute the coins lyi... | Let's divide the chests into 11 groups, with 7 chests in each group. The total number of coins in each group of chests must be divisible by 7, which means the total number of coins in all 77 chests is also divisible by 7.
Now, let's divide the chests into 7 groups, with 11 chests in each group. The number of coins in ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,491 |
$\underline{\text { Induction (miscellaneous). }}$ Graph theory (miscellaneous).
Authors: Bogdanov et al.
In a certain country, there are 100 cities (consider the cities as points on a plane). In a directory, for each pair of cities, there is a record of the distance between them (a total of 4950 records).
a) One re... | a) Let 98 points lie on a straight line $l$, and two points $A$ and $B$ lie outside it. If the distance between $A$ and $B$ is unknown, it cannot be restored: by replacing point $B$ with $B^{\prime}$, which is symmetric to $B$ relative to $l$, the other distances will not change.
b) By induction, we will show that for... | )Notalways;b)k=96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,494 |
Zaslavsky A.A.
Two ants each crawled along their own closed path on a $7 \times 7$ board. Each ant only crawled along the sides of the cells and visited each of the 64 vertices of the cells exactly once. What is the minimum possible number of such sides that both the first and the second ants crawled along? | An example where the number of "common" sides equals 16 is shown in the figure (one path is black, the other is red).
Estimate. Each ant visited 64 different sides. In total, there are 7$\cd... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,496 |
Berrov S.l.
There is a pile of more than $n^{2}$ stones on the table. Petya and Vasya take turns taking stones from the pile, with Petya going first. In one move, one can take any prime number of stones less than $n$, or any multiple of $n$ stones, or one stone. Prove that Petya can act in such a way as to take the la... | Assume the opposite: Vasya has a strategy that allows him to always take the last stone. Let $d$ be the initial number of stones, and $r$ be the remainder of the division of $d$ by $n$. ($r \neq 0$, otherwise Petya can immediately take all the stones).
After his first move, Petya can, by taking a multiple of $n$ stone... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,497 |
$\underline{\text { Folklore }}$
In each cell of a $10 \times 10$ table, a number is written. In each row, the largest number (or one of the largest, if there are several) is underlined, and in each column, the smallest number (or one of the smallest) is underlined. It turns out that all underlined numbers are underli... | Consider two arbitrary underlined numbers $A$ and $B$. From the condition, it follows that they are located in different rows and different columns. Let the number at the intersection of the row containing number $A$ and the column containing number $B$ be $C$, and the number at the intersection of the row containing n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,498 |
Klepitsyn V.A.
Cells of a $5 \times 5$ board are painted in a checkerboard pattern (corner cells are black). A figure, a mini-bishop, moves along the black cells of this board, leaving a trail on each cell it visits and not returning to that cell again. The mini-bishop can move either to free (diagonally) adjacent cel... | Let's first provide an example of the mini-elephant's route that ensures it visits twelve cells (see the figure, numbers from 1 to 12 show the order of cell visits).
| 1 | | 4 | | 6 |
| :--- | :--- | :--- | :--- | :--- |
| | 2 | | 5 | |
| 3 | | 7 | | 9 |
| | 11 | | 8 | |
| 12 | | 10 | | |
We will prove t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,499 |
$\underline{\text { Folklore }}$
Consecutive natural numbers 2 and 3 are divisible by consecutive odd numbers 1 and 3, respectively; numbers 8, 9, and 10 are divisible by 1, 3, and 5, respectively. Will there be 11 consecutive natural numbers that are divisible by $1, 3, 5, 7, 9, 11, 13, 15, 17, 19$, and 21, respectiv... | Consider the number $A=1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19 \cdot 21$. Then the numbers $1 / 2(A+1), 1 / 2(A+3), 1 / 2(A+5), \ldots, 1 / 2(A+19)$, $1 / 2(A+21)$ are consecutive natural numbers and are divisible by $1,3,5, \ldots, 19$ and 21 respectively.
## Answer
They will b... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,502 |
Shapovalov A.V.
In school, all students are either excellent, good, or poor performers. 99 students stood in a circle. Each of them has at least one poor performer among their three left neighbors, at least one excellent performer among their five right neighbors, and at least one good performer among their four neigh... | Note that the first two conditions can be more simply formulated as follows: among any three standing in a row, there is a C-student, and among any five standing in a row, there is an A-student. In addition, next to each B-student or one person away from him, there must be another B-student (let's call such two B-stude... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,503 |
Four spotlights are located at four given points on a plane, each of which can illuminate a right angle. The sides of these angles can be directed north, south, west, or east. Prove that these spotlights can be directed in such a way that they illuminate the entire plane.
# | Among the given points, let's choose the two most northern ones (if there are several most northern points, then we choose any two of them). We will direct one side of each right angle at these points southward, and the other sides toward each other. As a result, the entire half-plane south of these two spotlights will... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,504 |
Shapovalov A.V.
The dragon imprisoned the knight and gave him 100 different coins, half of which are magical (which ones exactly - only the dragon knows). Every day, the knight sorts all the coins into two piles (not necessarily equal). If the piles end up with an equal number of magical coins or an equal number of or... | b) First, we will divide the coins into two piles: the first pile with 49 coins, and the second with 51. Then, each day we will move one coin from the first pile to the second. On the 25th day, the first pile will have 25 coins, and the second pile will have 75. Therefore, on the 25th day, the first pile will have no m... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,506 |
In a certain country, every two cities are connected either by an airline or by a railway. Prove that
a) one can choose a mode of transport so that from each city one can reach any other city using only this mode of transport
b) from some city, by choosing one mode of transport, one can reach any other city with no m... | a) First method. Suppose from some city $A$ it is impossible to reach some city $B$ by railway. Consider the set $M$ of all cities that can be reached from city $A$ by railway. The set of cities not in $M$ is denoted by $N$. The set $N$ is non-empty, as it contains city $B$. It is clear that from the cities in set $M$,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,508 |
In one country, there are 100 cities, and each is connected to every other city by a road with one-way traffic. Prove that it is possible to change the direction of traffic on no more than one road so that it is possible to travel from any city to any other city.
# | We will prove by induction that the statement is true for any number of cities $n$ greater than 2.
Base case - for three cities - is obvious.
Induction step. Remove a city $A$ that has both incoming and outgoing roads (such a city exists because there are no more than one city from which roads only exit; similarly fo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,509 |
In a certain state, there are 101 cities.
a) Each city is connected to each of the others by a one-way road, and 50 roads enter each city and 50 roads leave each city. Prove that from each city, you can reach any other city by traveling no more than two roads.
b) Some cities are connected by one-way roads, and 40 roa... | Consider cities $A$ and $B$. Let the road connecting them lead from $B$ to $A$.
a) Consider 50 cities that have roads leading into $A$, and 50 cities that have roads leading out of $B$. Since $50+50>99$, there is a city $C$ that belongs to both sets. Therefore, it is possible to travel along the route $ACB$.
b) Let t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,510 |
6,7
The sum of positive numbers $x_{1}, x_{2}, \ldots, x_{n}$ is $1 / 2$. Prove that $\frac{1-x_{1}}{1+x_{1}} \cdot \frac{1-x_{2}}{1+x_{2}} \cdot \ldots \cdot \frac{1-x_{n}}{1+x_{n}} \geq \frac{1}{3}$. | We will prove a more general statement by induction: if the sum of positive numbers $x_{1}, x_{2}, \ldots, x_{n}$ is $s<1$, then
$$
\frac{1-x_{1}}{1+x_{1}} \cdot \frac{1-x_{2}}{1+x_{2}} \cdot \ldots \cdot \frac{1-x_{n}}{1+x_{n}} \geq \frac{1-s}{1+s}
$$
Base case. For $n=1$, the statement is obvious. Let's check it fo... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,511 |
$[$ Arithmetic of residues (miscellaneous).]
Solve the equation $x^{2}+y^{2}+z^{2}=2 x y z$ in integers. | If there is a non-zero solution to this equation in integers, then there is also a solution in natural numbers. Let's prove that there are no solutions in natural numbers.
Suppose this is not the case. Then consider the smallest natural number \(a\), for which there exist such natural numbers \(b, c, n\) that \(a^{2} ... | (0,0,0) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,512 |
Baron Munchausen claimed to Georg Cantor that he could list all natural numbers except one in such a way that only a finite number of them would be greater than their position. Is the baron boasting? | Let $a_{1}, a_{2}, a_{3}, \ldots$ be all natural numbers listed in a row without the number one. We will construct a new sequence $\left(b_{n}\right)$ of natural numbers defined as follows: $b_{1}=1, b_{n+1}=a_{b_{n}}$ for $n \geq 2$.
We will prove that all terms of the sequence $(b_{n})$ are distinct. Indeed, suppose... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,513 |
[Examples and counterexamples. Constructions]
A snail woke up, crawled from the mushroom to the spring, and fell asleep. The journey took six hours. The snail moved sometimes faster, sometimes slower, and stopped. Several scientists observed the snail. It is known that:
1) At every moment of the journey, the snail wa... | Example. Let's divide the entire time of the snail's movement into 30 intervals of 12 minutes each. Suppose the snail crawls 1 m in each of the 10 intervals numbered $1, 6, 7, 12, 13, 18, 19, 24, 25$, and 30, and rests the rest of the time. Scientists observe it in intervals 1-5 (from 1st to 5th - exactly one hour), 2-... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,515 |
Fomin S.B.
Two people toss a coin: one tossed it 10 times, the other - 11 times.
What is the probability that the second one got heads more times than the first one? | The events "the second one got more heads than the first" and "the second one got more tails than the first" are obviously equally probable. But they also complement each other (since the second one tossed the coin exactly one more time than the first, either heads or tails must be more, but not both at the same time).... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,516 |
[ Degree of a vertex ] $[\underline{\text { Induction }}($ other) $)]$
On a plane, several points are drawn, and some pairs of points are connected by segments. It is known that no more than $k$ segments come out of each point. Prove that the points can be colored in $k+1$ colors such that any two points connected by ... | Use induction on the number of points.
## Solution
Induction on the number of points. Base: for one point, the statement is obvious.
Induction step. Suppose there are $n+1$ points drawn on the plane, some pairs of which are connected by segments. Consider one of the points - A. For a moment, forget about it and the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,517 |
9,10 | |
Prove that for no integer $k$ the number $k^{2}+k+1$ is divisible by 101. | Prove by contradiction; we can use Fermat's little theorem.
## Solution
Suppose that for some $k$ the number $k^{2}+k+1$ is divisible by 101. Then $(k-1)\left(k^{2}+k+1\right)=k^{3}-1$ is also divisible by 101, that is, $k^{3} \equiv 1(\bmod 101)$. Therefore,
$k^{100}(k-1)=k^{101}-\left(k^{3}\right)^{33} k \equiv k-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,518 |
|
A city in the shape of a triangle is divided into 16 triangular blocks, and at the intersection of any two streets, there is a square (there are a total of 15 squares in the city). A tourist started touring the city from a certain square and ended the tour at a different square, visiting each square exactly once. Pr... | Paint the blocks in black and white in a checkerboard pattern. If a tourist has passed through two streets of one block, then he has made a turn of $120^{0}$.
## Solution
There are 15 areas in total, so the tourist has passed through 14 streets connecting pairs of adjacent areas. We will paint the blocks in black and... | 4 | Combinatorics | proof | Yes | Yes | olympiads | false | 54,519 |
\begin{aligned} & {[\text { Pythagorean Theorem (direct and inverse) }} \\ & {[\quad \underline{\text { Coordinate Method }}]}\end{aligned}
Given a triangle $ABC$ and a point $H$ on the line $AB$. Prove that $CH$ is the altitude of triangle $ABC$ if and only if $AC^2 - BC^2 = AH^2 - BH^2$. | Necessity.} Let $CH$ be the height of triangle $ABC$ (Fig.1). If point $H$ coincides with vertices $A$ or $B$, the statement is obvious. Suppose point $H$ does not coincide with either $A$ or $B$. Then, by the Pythagorean theorem from right triangles $ACH$ and $BCH$, we find that
$$
AC^2 - AH^2 = CH^2, \quad BC^2 - BH... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,520 |
Dirichlet's Principle (angles and lengths). Circle, sector, segment, etc.
On a plane, there are 25 points, and among any three of them, there are two at a distance less than 1. Prove that there exists a circle of radius 1 containing at least 13 of these points. | Let $A$ be one of the given points. If all other points lie within the circle $S_{1}$ of radius 1 centered at $A$, there is nothing more to prove. Now let $B$ be a given point lying outside the circle $S_{1}$, i.e., $A B>1$. Consider
the circle $S_{2}$ of radius 1 centered at $B$. Among the points $A, B$, and $C$, wher... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,522 |
8,9
We will call a convex heptagon special if three of its diagonals intersect at one point. Prove that by slightly moving one of the vertices of a special heptagon, one can obtain a non-special heptagon.
# | Let $P$ be the intersection point of the diagonals $A_{1} A_{4}$ and $A_{2} A_{5}$ of the convex heptagon $A_{1} \ldots A_{7}$. One of the diagonals $A_{3} A_{7}$ and $A_{3} A_{6}$, for definiteness the diagonal $A_{3} A_{6}$, does not pass through the point $P$. The number of intersection points of the diagonals of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,523 |
10,11 |
In a rectangle of area 1, five figures of area $1 / 2$ each are placed. Prove that there are
a) two figures, the area of the common part of which is not less than $3 / 20$;
b) two figures, the area of the common part of which is not less than $1 / 5$;
c) three figures, the area of the common part of which i... | We will use the notations of problem 58106.
a) According to problem $\underline{58106}$ b) $1 \geq 5 \cdot 0.5 - M_{2}$, which means $M_{2} \geq 1.5$. $M_{2}$ is the sum of the areas of the pairwise intersections of five figures, so the area of the largest of these intersections is not less than 0.15.
Remark. The same... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,525 |
Positive numbers $a, b, c, x, y$, such that
$x^{2}+x y+y^{2}=a^{2}$,
$y^{2}+y z+z^{2}=b^{2}$,
$x^{2}+x z+z^{2}=c^{2}$.
Express the value of $x y+y z+x z$ in terms of $a, b$ and $c$. | The equation $x^{2}+x y+y^{2}=a^{2}$ can be interpreted as the Law of Cosines in a triangle with sides $x, y, a$ and an angle of $120^{\circ}$.
## Solution
Draw three rays from point $O$ with angles of $120^{\circ}$ between them, and mark segments $O A=z, O B=x, O C=y$ on them. According to the Law of Cosines, the si... | xy+yz+x4\sqrt{\frac{p(p-)(p-b)(p-)}{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,528 |
10,11 [ Investigation of a quadratic trinomial ]
Plot on the phase plane $\operatorname{Opq}$ the sets of points $(p, q)$ for which the equation $x^{3}+p x+q=0$ has
a) one root
b) two roots
c) three distinct roots;
d) three coincident roots. | If $a$ is a root of our equation, then $x^{3}+p x+q=(x-a)\left(x^{2}+a x+b\right)$. In this case, $p=b-a^{2}, q=-a b$. Then $4 p^{3}+27 q^{2}=4 b^{3}-12 a^{2} b^{2}+12 a^{4} b-4 a^{6}+27 a^{2} b^{2}=4 b^{3}-a^{2} b^{2}+16 a^{2} b^{2}-4 a^{4} b+16 a^{4} b-4 a^{6}=\left(4 b-a^{2}\right)\left(b^{2}+4 a^{2} b+\right.$ $\le... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,529 |
[ Methods for solving problems with parameters ] [ Phase plane of coefficients ]
Plot on the phase plane $O p q$ the set of points $(p, q)$ for which the equation $x^{3}+p x+q=0$ has three distinct roots belonging to the interval $(-2,4)$. | See solutions of problems $\underline{61272}, \underline{61273}$.
## Answer
The set of points defined by the inequalities $4 p^{3}+27 q^{2}<0,-4 p-64<q<2 p+8$.[^0]
How many roots does the equation $8 x\left(1-2 x^{2}\right)\left(8 x^{4}-8 x^{2}+1\right)=1$ have on the interval $[0,1]$?
## Solution
Notice that $8 x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,530 |
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