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Knop K.A.
Kostya had a pile of 100 pebbles. Each move, he divided one of the piles into two smaller ones, until he ended up with
100 piles of one pebble each. Prove that
a) at some point, there were 30 piles that contained exactly 60 pebbles in total;
b) at some point, there were 20 piles that contained exactly 60... | We will call piles with 1, 2, 3, 4 pebbles a unit, a double, a triple, and a quadruple, respectively.
a) Wait until there are 70 piles. Among them, there will be 40 units (otherwise, there would be no fewer than $2 \cdot 31 + 39 = 101$ pebbles).
If we discard them, 30 piles will remain, containing 60 pebbles.
b) We w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,256 |
Mitrofanov I.v.
A strip of cells $1 \times 1000000$ is divided into 100 segments. Each cell contains an integer, and the numbers in the cells within the same segment are the same. A chip is placed in each cell. Then the following operation is performed: all chips are simultaneously moved, each by the number of cells t... | For each cell of the first segment, let's define its route - the sequence of cells that the chip standing on it passes through until it returns to the first segment. The "return" cell is not included in the route. The length of the route is the number of cells included in it. This is the number of operations that are c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,257 |
10,11 |
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In the school, $2 n$ subjects are studied. All students receive grades of 4 and 5. No two students have the same grades, and it cannot be said that one student performs better than another. Prove that the number of students in the school is no more than $C_{2 n}^{n}$.
(We consider th... | Let's slightly change the condition: suppose there are $2^{2 n}$ students in the school with all possible sets of fives and fours. We will select from them the largest group A of pairwise incomparable students (in the sense of the problem's condition). We will prove that this group consists exactly of all students who ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,259 |
Motion Problems ] [ Coordinate Method on the Plane ]
Cowboy Jimmy bet his friends that he could shoot through all four blades of the ventilator with one shot. (The ventilator is constructed as follows: on an axis rotating at a speed of 50 revolutions per second, four semicircles are arranged at equal distances from ea... | Let's assume Jimmy shoots parallel to the axis of the ventilator, slightly above it. Then the trajectory of the bullet is represented by a straight line with a positive slope on the plane $(x, t)$ (where $x$ is the coordinate along the axis, and $t$ is time). To understand which shot will be successful, let's consider ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,260 |
Kosukhin O.N.
On a rectangular sheet of paper, a circle is drawn, inside which Misha mentally selects $n$ points, and Kolya tries to guess them. In one attempt, Kolya indicates one point on the sheet (inside or outside the circle), and Misha tells Kolya the distance from it to the nearest unguessed point. If this dist... | Let there be $k \geq 1$ unsolved points $c_{k, 1}, c_{k, 2}, \ldots, c_{k, k}$ left on the sheet of paper. We will show how to solve one of them using $2k + 1$ attempts.
Draw a line segment $l$ on the sheet of paper that does not intersect the marked circle. On this segment, indicate $(k + 1)$ points
$a_{k, 1}, a_{k,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,261 |
Gerekо A.A.
In $n$-athlon competitions, $2^{n}$ people participate. For each athlete, their strength in each event of the program is known. The competitions proceed as follows: first, all athletes participate in the first event, and the best half of them advance to the next round. This half participates in the next ev... | a) Induction on $n$. The base case is obvious: one winner of a single competition between two participants is already half. Suppose there is an example of $n$ competitions for $2^{n}$ athletes with $2^{n-1}$ possible winners. Divide $2^{n+1}$ athletes into two equal groups $A$ and $A^{\prime}$. Assume that in some type... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,262 |
Shen A.H.
A regular triangle $ABC$ cut out of plywood was placed on the floor. Three nails were hammered into the floor (one next to each side of the triangle) in such a way that the triangle cannot be rotated without lifting it off the floor. The first nail divides side $AB$ in the ratio $1:3$, counting from vertex $... | $1^{0}$. Let's first solve the "one nail problem". Suppose one nail is hammered in, touching the triangle at point $M$ on side $A C$. We fix the center of the supposed rotation (point $O$). Can the triangle be rotated around point $O$ by a small angle, and if so, in which direction?
Suppose the triangle is positioned ... | \frac{5}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,263 |
Auxiliary coloring (miscellaneous). $\underline{\text { Induction (miscellaneous). }}$ Authors: Dolnikov V., Karasev D.V., Berlov S.l. In a country, there are 2000 cities, some pairs of which are connected by roads. It is known that through any city, no more than $N$ different non-self-intersecting cyclic routes of o... | Consider a graph with vertices in cities, the edges of which correspond to roads. From the condition, it follows that in this graph, through each vertex, there are no more than $N$ odd cycles.
We will prove by induction on the number of vertices that the vertices of such a graph can be colored in $N+2$ colors so that ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,265 |
Vilenkin A.N.
The grid of lines shown in the figure consists of concentric circles with radii $1,2,3,4, \ldots$ and center at point $O$, a straight line $l$ passing through point $O$, and all possible tangents to the circles parallel to $l$. The entire plane is divided into cells by these lines, which are colored in a... | Let's take the line $l$ as the $Ox$ axis, and the point $O$ as the origin of the rectangular coordinate system, as shown in Fig. 5. The unit of scale is already given in the condition. The grid of lines mentioned in the condition is composed of circles (circles $x^{2}+y^{2}=n^{2}$) and lines $y=m$, where $m$ and $n$ ar... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,266 |
Ivlev B.M.
Each of the nine lines divides the square into two quadrilaterals, the areas of which are in the ratio 2:3.
Prove that at least three of these nine lines pass through one point. | If a line intersects two adjacent sides of a square, it obviously divides the square into a triangle and a pentagon. However, by the condition, each of the 9 lines divides the square into quadrilaterals. Therefore, such a line intersects two opposite sides of the square, i.e., it divides the square into two trapezoids ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,267 |
Podlipsky O.K.
In a rectangular grid of $49 \times 69$, all $50 \cdot 70$ vertices of the cells are marked. Two players play the following game: each move, a player connects two points with a segment, with the condition that one point cannot be the endpoint of two drawn segments. Segments can share common points. Segm... | The first solution. Divide all marked points into pairs such that any segment with endpoints in points of the same pair is a horizontal segment of length 1.
We will describe a winning strategy for the first player.
Suppose the first player connects the points of some pair on the first move.
Next, if the second playe... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,268 |
Mussi 0. On a line, $n$ distinct blue points and $n$ distinct red points are marked. Prove that the sum of pairwise distances between points of the same color does not exceed the sum of pairwise distances between points of different colors.
# | Let's prove the statement of the problem in a more general assumption, when the considered points can coincide. We will prove by induction on the number $N$ of distinct points among the $2n$ marked ones. In the case $N=1$, the desired inequality is obviously satisfied. For $N$ distinct points, let $S_{1}^{N}$ denote th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,271 |
Bogdanov I.I.
2009 non-negative integers not exceeding 100 are arranged in a circle. It is allowed to add 1 to two adjacent numbers, and this operation can be performed with any two adjacent numbers no more than $k$ times. For what least $k$ can all the numbers be guaranteed to be made equal? | Let the numbers on the circle be denoted by $a_{1}, \ldots, a_{2009}$, and set $a_{n+2009} = a_{n} = a_{n-2009}$. Let $N = 100400$.
1. Set $a_{2} = a_{4} = \ldots = a_{2008} = 100$ and $a_{1} = a_{3} = \ldots = a_{2009} = 0$. Suppose we managed to make all the numbers equal for some value of $k$. Consider the sum $S =... | 100400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,272 |
[ Rectangles and Squares. Properties and Characteristics ] Problem 111876 Topics: [ Rectangles and Squares. Properties and Characteristics ]
On a plane, several rectangles with sides parallel to the coordinate axes are drawn. It is known that any two rectangles can be intersected by a vertical or horizontal line. Pro... | Lemma. Suppose in a family of rectangles, any two can be intersected by a vertical line. Then all of them can be intersected by a vertical line.
Proof. Consider the rectangle with the leftmost right boundary and the rectangle with the rightmost left boundary. By the condition, they can be intersected by a line. Then, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,273 |
For which $n$ can a regular $n$-gon be placed on a sheet of lined paper so that all its vertices lie on the lines?
(Lines are parallel straight lines, located at equal distances from each other.) | Suppose a regular $n$-gon can be placed on lined paper. Let $O$ be the center of this polygon. Rotate all lines by an angle of ${ }^{360} \%$ around point $O$ and draw on the same diagram both the original and the new (resulting from the rotation) parallel lines. Now each vertex of the $n$-gon lies at the intersection ... | 3,4,6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,274 |
$\underline{\text { Ionin Yu.I. }}$ y.
On $n$ cards laid out in a circle, numbers are written, each of which is either 1 or -1. What is the minimum number of questions needed to definitely determine the product of all $n$ numbers, if in one question it is allowed to find out the product of the numbers on a) any three ... | Let the numbers written on the cards be denoted by $a_{1}, a_{2}, \ldots, a_{n}$, and the number of questions we are looking for by $p$.
a) First, note that each of the given numbers must be included in at least one of the products whose values we determine by asking our questions. Otherwise, by changing the sign of t... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,277 |
Berpow S.l. Given natural numbers $p<k<n$. On an infinite grid, some cells are marked such that in any rectangle of size $(k+1) \times n$ (with $n$ cells horizontally and $k+1$ vertically), exactly $p$ cells are marked. Prove that there exists a rectangle of size $k \times (n+1)$ (with $n+1$ cells horizontally and $k$ ... | Consider two rectangles: a rectangle $(k+1) \times n$ and a rectangle $k \times (n+1)$, which share the same lower-left cell.
We will call the part of the rectangle $(k+1) \times n$ not covered by the rectangle $k \times (n+1)$ black (this is a strip $1 \times n$), and the part of the rectangle $k \times (n+1)$ not co... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,279 |
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On a line, there are 50 segments. Prove that at least one of the following statements is true:
- some 8 of these segments have a common point;
- some 8 of these segments are such that no two of them intersect. | Probably, one can cite many different theorems of which the statement of this problem is a particular case or a consequence. Here we will indicate only one of them.
Theorem. Let a system of segments be given on a line. Denote by $M$ the smallest number of points on the line such that each of the segments of the system... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,280 |
Senderov V.A.
For which natural numbers $n$ is the inequality
$$
\sin n \alpha+\sin n \beta+\sin n \gamma<0
$$
true for any angles $\alpha, \beta, \gamma$ of an acute triangle? | For any triangle $T$ with angles $\alpha, \beta, \gamma$, denote $f_{n}(T)=\sin n \alpha+\sin n \beta+\sin n \gamma$.
Lemma. Suppose $x+y+z=\pi k$, where $k \in \mathbb{Z}$. Then
$|\sin x| \leq|\sin y|+|\sin z|$.
When $y \neq 7$ t $\pi l, z \neq \boldsymbol{f} \pi l$, where $l \in \mathbb{Z}$, this inequality is str... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 54,281 |
All natural numbers, in decimal notation of which there are no more than $n$ digits, are divided into two sets as follows. The first set includes numbers with an odd sum of digits, and the second set includes numbers with an even sum of digits. Prove that for any natural number $k \leq n$, the sum of the $k$-th powers ... | Let's prove the given statement by induction on $n$. We agree that when considering numbers with no more than $n$ digits, we will prepend zeros to all numbers so that they all become $n$-digit numbers.
The validity of the statement for $n=2$ (then $k$ can only take the value $k=1$) is easy to verify:
| Numbers with o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,282 |
In space, there are 3 planes and a sphere. In how many different ways can a second sphere be placed in space so that it touches the three given planes and the first sphere? (In this problem, we are actually talking about the tangency of spheres, i.e., it is not assumed that the spheres can only touch externally - ed. n... | Answer: from 0 to 16 (depending on the arrangement of the planes and the sphere). Let $O$ and $R$ be the center and radius of the given sphere $S$. Suppose the sphere $S_{1}$ with center $O_{1}$ touches the given sphere and three given planes. We associate the sphere $S_{1}$ with the sphere $S_{1}'$ with center $O_{1}$... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,283 |
| [ | Minimum or maximum area (volume). |
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| $\left[\begin{array}{llll} \\ \text { L }\end{array}\right.$ | Convex hull and supporting lines (planes). |
| [ | Median of a pyramid (tetrahedron). |
| [ | Area and orthogonal projection |
| [ | Inequalities with areas |
| | Area. One figure lies inside an... | 
a) Let's show that in a regular octahedron \(ABCDEF\) (Fig. 11-7-1), it is impossible to select four vertices with the property mentioned in part a). Without loss of generality, we can assume... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,285 |
| Inversion helps solve the task | |
| 2054 | Induction in geometry | Class |
| [ | Four points lying on one circle | ] |
In this problem, we will consider sets of $n$ lines in general position, i.e., sets in which no two lines are parallel and no three lines pass through the same point.
To a set of two lines in gen... | a) Let $M_{ij}$ denote the intersection point of the lines $l_i$ and $l_j$, and $S_{ij}$ the circle corresponding to the remaining three lines. Then the point $A_1$ is the point of intersection of the circles $S_{15}$ and $S_{12}$, distinct from the point $M_{34}$. Repeating this reasoning for all points $A_i$, we obta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,288 |
[ Inversion helps solve the problem. Induction in geometry Four points lying on one circle Let points $M_{1}$ and $M_{2}$ be chosen on two intersecting lines $l_{1}$ and $l_{2}$, not coinciding with the intersection point $M$ of these lines. We will correspond to them a circle passing through $M_{1}, M_{2}$, and $M$.
... | a) Let $M_{ij}$ denote the intersection point of lines $l_i$ and $l_j$. Then the point $A_1$, corresponding to the triplet $l_2, l_3, l_4$, is the intersection point of the circumcircles of triangles $M_2 M_3 M_{23}$ and $M_3 M_4 M_{34}$. Reasoning similarly for points $A_2, A_3$, and $A_4$, we obtain that points $A_1,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,289 |
$2+$
John had a full basket of trempons. First, he met Anna and gave her half of his trempons and another half-trempon. Then he met Banna and gave her half of the remaining trempons and another half-trempon. After meeting Vanna and giving her half of the trempons and another half-trempon, the basket was empty. How man... | Notice that before meeting Vanna, John had one tremponch left, as half of this amount was half a tremponch. Before meeting Banna, he had 3 tremponchs, because half of this amount was one and a half tremponchs, that is, one and a half. Similarly, we get that initially there were 7 tremponchs.
## Answer
7 tremponchs. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,293 |
Let $M$ be a finite set of numbers. It is known that among any three of its elements, there are two whose sum belongs to $M$.
What is the maximum number of elements that can be in $M$? | Consider either the four largest or the four smallest numbers.
## Solution
Example of a set of 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that the set $M=\left\{a_{1}, a_{1}, \ldots, a_{n}\right\}$ of $n>7$ numbers does not have the required property. We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,299 |
8,9 |
Find the cosine of the angle at the base of an isosceles triangle if the point of intersection of its altitudes lies on the inscribed circle of the triangle.
# | Let $O$ be the center of the circle inscribed in triangle $ABC (AC = BC)$, $H$ be the point of intersection of the altitudes, $\angle A = \angle B = 2\alpha$, $K$ be the midpoint of $AB$. Then $OK = AK \tan \alpha$, $HK = AK \tan(90^\circ - 2\alpha) = AK \cot 2\alpha$.
Since $HK = 2OK$, then $2 \tan \alpha = \cot 2\al... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,300 |
## [ Arrangements and Partitions ] [ Gaussian Polynomials ] [ Generating Functions ] [ Induction (other) . ]
Let $f_{k, l}(x)$ be the generating function of the sequence $P_{k, l}(n)$ from problem $\underline{61525:} f_{k, l}(x)=P_{k, l}(0)+x P_{k, l}(1)+\ldots+$ $x^{k l} P_{k, l}(k l)$.
a) Prove the equalities: $... | a) Both equalities immediately follow from the definition and formulas a) and b) of problem $\underline{61525}$.
b) Verify that the Gaussian polynomials satisfy, for example, the first of the equalities from part a). | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,302 |
$\underline{\text { Akulichi I.F. }}$
Do there exist ten pairwise distinct natural numbers such that their arithmetic mean is
a) exactly six times;
b) exactly five times;
their greatest common divisor? | a) For example, $1, 2, \ldots, 9, 15$. The sum of these numbers is 60, the arithmetic mean is 6, and the GCD is 1.
b) Let the GCD of ten numbers $a_{1}<a_{2}<\ldots<a_{10}$ be $d$. Then $a_{1} \geq d, a_{2} \geq 2 d, \ldots, a_{10} \geq 10 d$. Therefore, the sum of these numbers is not less than $55 d$, and the arithm... | )Theyexist;b)theydonotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,304 |
Segrackyan $H$.
In the cells of an $n \times n$ board, numbers from 1 to $n^{2}$ are arbitrarily placed. Prove that there will be two such adjacent cells (having a common vertex or a common side) such that the numbers in them differ by at least
$n+1$. | Suppose that for any two adjacent cells, the numbers written in them differ by no more than $n$.
Consider the cells containing the numbers 1 and $n^{2}$. A chess king can move from the first cell to the second in no more than $n-1$ moves, so the difference $n^{2}-1$ is no greater than $n(n-1)$. Contradiction. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,307 |
Tokarev S.i.
An ant is crawling along the wire frame of a cube, never turning back.
Can it happen that it visited one vertex 25 times, and each of the others 20 times? | Mark the vertices of a cube with numbers 1 and -1 in a "chessboard" pattern. Suppose an ant has crawled along the edges of the cube in such a way that it visited one vertex 25 times, and each of the others 20 times. Let's calculate the sum of the numbers visited by the ant (each number appears in this sum several times... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,308 |
Kovaldzhi A.K.
Three grasshoppers sit on a straight line such that the two outer ones are 1 m away from the middle one. Every second, one of the grasshoppers jumps over another to a symmetrical point (if $A$ jumps over $B$ to point $A_{1}$, then $A B=B A_{1}$).
After some time, the grasshoppers ended up in the same p... | Let grasshoppers sit on the coordinate axis at points $-1,0,1$. Note that as a result of each jump, the grasshopper's coordinate remains an integer. In addition, the grasshopper always jumps an even distance. From this, it follows that if the grasshopper's coordinate was initially even, it will always remain even. Ther... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,309 |
5,6,7 |
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| | [ Divisibility of numbers. General properties ] | |
| | [ Decimal number system ] | |
| | Case analysis | |
Which digits can stand in place of the letters in the example $A B \cdot C=D E$, if different letters represent different digits and the digits are written from left t... | Let's consider the largest possible values of the number $DE$. The numbers 89 and 79 are prime, so they do not fit. Next, let's consider the number 78, and check its single-digit divisors to get the answer. Further checking can confirm that this answer is unique.
## Answer
$13 \cdot 6=78$. | 13\cdot6=78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,310 |
8,9,10,11 |
Author: S $\underline{\text { Saghafian M. }}$.
In the plane, five points are marked. Find the maximum possible number of similar triangles with vertices at these points. | Example. Vertices and the center of a square.
Evaluation. Let's describe all configurations of four points forming four similar triangles. Let $\$ \mathrm{~A} \$$, $\$ \mathrm{~B} \$$, \$C \$ \$ \$
1. Point $\$ \mathrm{~A} \$$ lies inside triangle $\$ B C D \$$. Let $\$ B \$$ be the largest angle of triangle $\$ B C ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,313 |
How many pounds of grain need to be ground to have exactly 100 pounds of flour left after paying for the work - 10% of the milling?
There are no losses during milling
# | Think about what part of the payment will be from the final revenue, not the initial revenue.
## Solution
The remaining 100 pounds constitute $90\%$ (that is, $9/10$) of the weight of the grain. Therefore, the weight of the grain was $100 \cdot \frac{10}{9} = 111 \frac{1}{9}$ pounds.
## Answer
$111 \frac{1}{9}$ pou... | =8,b=2;1976 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,317 |
The number 12 is written on the board. Every minute, the number is either multiplied or divided by 2 or 3, and the result is written on the board in place of the original number. Prove that the number that will be written on the board exactly one hour later will not be equal to 54.
# | After each operation, the parity of the total number of twos and threes in the prime factorization of the number on the board changes. At the beginning, this number is odd (equal to 3), since $12=2 \cdot 2 \cdot 3$. Therefore, after 60 operations (seconds), it should be odd, but in the factorization $54=2 \cdot 3 \cdot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,318 |
[ Chessboards and chess pieces ] [ Examples and counterexamples. Constructions ]
What is the maximum number of queens that can be placed on an $8 \times 8$ chessboard without attacking each other? | You can place no more than eight queens that do not attack each other, as each queen controls one row.
Let's show how you can place eight queens:
Send a comment | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,319 |
Can a knight's move cover all the squares of a chessboard, starting from square $a1$, ending at square $h8$, and visiting each square of the board exactly once?
# | Notice that after each odd move, the knight is on a white square, and after each even move, it is on a black square.
## Solution
No, it is not possible. To cover all the squares of the chessboard, 63 moves are required. After each odd move, the knight is on a white square, and after each even move, it is on a black s... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,320 |
A crazy cashier exchanges any two coins for any three of your choice, and any three coins for any two. Will Petya be able to exchange 100 coins worth 1 ruble each for 100 coins worth 1 forint each, giving the cashier exactly 2001 coins in the process? | If Petya exchanges two coins for three, the number of coins he has increases by one. Let him make $N$ such exchanges. He gives the cashier $2 N$ coins. To keep the total number of coins, Petya is forced to make the same number of exchanges of three coins for two. In doing so, he will give the cashier another $3 N$ coin... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,325 |
[ Methods for solving problems with parameters ] [ [Graphs and GMT on the coordinate plane $]$
Can the vertex of the parabola $y=4 x^{2}-4(a+1) x+a$ lie in the second coordinate quadrant for some value of $a$?
# | The x-coordinate of the vertex of the parabola is equal to $1 / 2(a+1)$, and the y-coordinate is equal to $(a+1)^{2}-2(a+1)^{2}+a=-a^{2}-a-1<0$. Therefore, the vertex of the parabola cannot be in the second quadrant.
## Answer
It cannot.
Send a comment | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,327 |
$\underline{\text { frankin } 5 . P .}$
All integers from 1 to 2010 are written in a circle in such an order that, when moving clockwise, the numbers alternately increase and decrease.
Prove that the difference between some two adjacent numbers is even. | Let all the differences between adjacent numbers be odd. Then even and odd numbers alternate around the circle. But this means that either each even number is greater than both adjacent odd numbers, or each even number is less than both adjacent odd numbers. In the first case, there will be no place for the number 2, a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,329 |
3[ Examples and counterexamples. Constructions]
In the cells of a $5 \times 5$ square table, the numbers 1 and -1 are arranged. It is known that the number of rows with a positive sum is greater than the number of rows with a negative sum.
What is the maximum number of columns in this table that can have a negative s... | Let's consider one of the possible examples (see the table). The sum of the numbers in each of the three upper rows is positive, while the sum of the numbers in each column is negative.
| 1 | 1 | 1 | -1 | -1 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | -1 | 1 | -1 | 1 |
| -1 | 1 | -1 | 1 | 1 |
| -1 | -1 | -1 | -1... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,330 |
A line intersects the graph of the function $y=x^{2}$ at points with abscissas $x_{1}$ and $x_{2}$, and the x-axis at a point with abscissa $x_{3}$. Prove that $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{1}{x_{3}}$. | The first method. The equation of the given line has the form $y=k\left(x-x_{3}\right)$. From the condition, it follows that $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}=k\left(x-x_{3}\right)$ or
$x^{2}-k x+k x_{3}=0$. By Vieta's theorem, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=\fr... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,331 |
Bakayev E.V.
About the group of five people, it is known that:
Alyosha is 1 year older than Alekseev,
Borya is 2 years older than Borisov,
Vasya is 3 years older than Vasiliev,
Grisha is 4 years older than Grigoryev, and there is also Dima and Dmitriev in this group.
Who is older and by how many years: Dima or Dm... | The sum of the ages of Alyosha, Borya, Vasya, Grisha, and Dima is equal to the sum of the ages of Alexeev, Borisov, Vasilyev, Grigoryev, and Dmitriev. Therefore, Dmitriev is older than Dima by \(1+2+3+4=10\) years.
## Answer
\[
\begin{aligned} &\({[\quad\) The Pigeonhole Principle (other). \(] } \\ &\) Problem \(\und... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,332 |
Let $\mathrm{f}(\mathrm{x})$ be some polynomial, for which it is known that the equation $\mathrm{f}(\mathrm{x})=\mathrm{x}$ has no roots. Prove that then the equation $f(f(x))=x$ also has no roots.
# | From the condition, either $f(x) > x$ for any $x$, or $f(x) < x$. Let $f(x) = y$. Then $f(f(x)) = f(y) > y = f(x) > x$. Thus, for any $x$, $f(f(x)) - x > 0$, i.e., the equation $f(f(x)) = x$ has no roots. Similarly, we show that the equation $f(f(x)) = x$ has no roots in the case when for any $x$ the inequality $f(x) <... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,333 |
The numbers $1,2, \ldots, 9$ are divided into three groups. Prove that the product of the numbers in one of the groups is not less than 72.
# | The product of the numbers in all groups is $9!=362880$, while $71^{3}=357911<9$!.
保留源文本的换行和格式，翻译结果如下：
The product of the numbers in all groups is $9!=362880$, while $71^{3}=357911<9$!. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,335 |
Around a round table, 25 boys and 25 girls are sitting. Prove that someone sitting at the table has two boys as neighbors.
# | Let's remove every second person from the table and seat them at another table in the same order. Now there are 25 people at each table, so the number of boys and girls is not equal - there is a table where there are more boys than girls. At this table, two boys should sit next to each other (if each boy is seated betw... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,336 |
Invariants
$\left[\begin{array}{ll}\text { Invariants } \\ {\left[\begin{array}{l}\text { Evenness and Oddness }\end{array}\right]}\end{array}\right]$
In an $8 \times 8$ table, one of the cells is colored black, and all the others are white. Prove that it is impossible to make all cells white by recoloring rows and c... | When repainting a row or column, the parity of the number of black cells in the table does not change, and initially, their quantity is odd. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,337 |
Is it true that two graphs are isomorphic if
a) they both have 10 vertices, each with a degree of 9?
b) they both have 8 vertices, each with a degree of 3?
c) they are connected, acyclic, and contain 6 edges? | a) In such a graph, each vertex is connected to all others.
b) See counterexample:
c) See counterexample:
=1 / 4-(1 / 2-a)^{2}$.
## Solution
Since $a+b=1$, then $b=1-a$, and thus
$a b=a(1-a)=1 / 4-(1 / 2-a)^{2} \leq 1 / 4$,
since the square of any expression is non-negative.
## Answer
$1 / 4$. | \frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 54,339 |
$\left[\begin{array}{ll}{[\text { Algebraic inequalities (miscellaneous). }]}\end{array}\right]$
Prove that for any $x, y, z$ the inequality holds: $x^{4}+y^{4}+z^{2}+1 \geq 2 x\left(x y^{2}-x+z+1\right)$. | $x^{4}+y^{4}+z^{2}+1-2 x\left(x y^{2}-x+z+1\right)=x^{4}-2 x^{2} y^{2}+y^{4}+z^{2}-2 x z+x^{2}+x^{2}-2 x+1=\left(x^{2}-y^{2}\right)^{2}+(z-x)^{2}+(x-1)^{2} \geq 0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,340 |
$[\underline{\text { Algebraic Inequalities }}[\underline{\text { proof }})]$
Prove that for $n \geq 3$ the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{3}{5}$ holds. | Induction by p. Base: $1 / 4+1 / 5+1 / 6=37 / 60>3 / 5$. The induction step is made as in problem $\underline{60304}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,341 |
$x \geq-1, n-$ natural number. Prove that $(1+x)^{n} \geq 1+n x$.
# | We will prove the inequality by induction on $n$.
Base case. When $n=1$, the inequality becomes an equality.
Inductive step. Suppose it has already been proven that $(1+x)^{n} \geq 1+n x$. Then $(1+x)^{n+1} \geq(1+n x)(1+x)=1+n x+x+n x^{2} \geq 1$ $+(n+1) x$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,342 |
## [ Algebraic inequalities (other).] Induction (other). $\quad]$
For which natural numbers $n$ does the inequality $2^{n} \geq n^{3}$ hold?
# | $2^{9}=526n^{3}$ for $n>9$. The base case ($n=10$) is obvious.
Induction step. $2^{n+1}=2 \cdot 2^{n}>2 n^{3}>(n+1)^{3}$. Indeed, $\frac{(n+1)^{3}}{n^{3}}=\left(1+\frac{1}{n}\right)^{3}=1+\frac{3}{n}+\frac{3}{n^{2}}+\frac{1}{n^{3}}<2$ for $n>9$.
## Answer
For $n \geq 10$. | n\geq10 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 54,343 |
3 [Completing the Square. Sums of Squares]
Prove that for any $x$ the inequality $x^{4}-x^{3}+3 x^{2}-2 x+2 \geq 0$ holds.
# | $x^{4}-x^{3}+3 x^{2}-2 x+2=\left(x^{2}-x / 2\right)^{2}+7 / 4 x^{2}+(x-1)^{2}+1 \geq 0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$x^{4}-x^{3}+3 x^{2}-2 x+2=\left(x^{2}-x / 2\right)^{2}+7 / 4 x^{2}+(x-1)^{2}+1 \geq 0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,344 |
[ Algebraic inequalities (miscellaneous).] [ Case analysis $]$
$x, y>0$. Let $S$ denote the smallest of the numbers $x, 1 / y, y+1 / x$. What is the maximum value that $S$ can take? | If $x \leq \sqrt{2}$, then $S \leq \sqrt{2}$. If $y \geq \frac{1}{\sqrt{2}}$, then $S \leq y \leq \sqrt{2}$. If $x>\sqrt{2}, y<\frac{1}{\sqrt{2}}$, then $S \leq y+\frac{1}{x}<\frac{2}{\sqrt{2}}=\sqrt{2}$. Therefore, $S \leq \sqrt{2}$. Equality is achieved when $x=\frac{1}{y}=\sqrt{2}$.
## Answer
$\sqrt{2}$. | \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 54,345 |
3 [Central symmetry helps solve the task]
What is the maximum number of pawns that can be placed on a chessboard (no more than one pawn per square), if:
1) a pawn cannot be placed on the e4 square;
2) no two pawns can stand on squares that are symmetric relative to the e4 square?
# | Evaluation. All fields of the board except for the a-file, the 8th rank, and the e4 square can be divided into pairs symmetrical relative to e4. Such pairs form 24. According to the condition, no more than one pawn can be placed on the fields of each pair. In addition, no more than one pawn can be placed on the fields ... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,346 |
What is the minimum number of sportlotto cards (6 out of 49) you need to buy to ensure that at least one number is guessed correctly in at least one of them?
# | Let's fill eight cards as follows: in the first one, we will strike out numbers from 1 to 6, in the second one - from 7 to 12, and so on, in the last one - from 43 to 48. The number 49 will remain unstruck in any card. Therefore, at least five of the winning numbers will be struck out.
We will prove that seven cards m... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,347 |
In the class, each boy is friends with exactly two girls, and each girl is friends with exactly three boys. It is also known that there are 31 pioneers and 19 desks in the class. How many people are in this class?
# | The number of "friendly connections" in the class is three times the number of girls and twice the number of boys. This means that the number of boys to the number of girls is in the ratio of $3: 2$, and the total number of students is divisible by 5. On the other hand, there are no fewer than 31 and no more than 38 st... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,348 |
Around a round table, there were a) 15; b) 20 people sitting. They want to reshuffle their seats so that those who previously sat next to each other now sit two people apart. Is this possible?
# | a) Let's number all the seats around the table and all the people sitting in them according to the seats they occupy. Suppose we have managed to seat everyone as required. Without loss of generality, we can assume that person 1 remains in their place. If this is not the case, we can achieve this by rotating the table; ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,349 |
At the Lomonosov tournament at the MINIMO institute, there were competitions in mathematics, physics, chemistry, biology, and ballroom dancing. When the tournament ended, it turned out that an odd number of schoolchildren participated in each competition, and each schoolchild participated in an odd number of competitio... | Let's consider the sum of the number of participants in all five competitions. This sum is odd, as an odd number of schoolchildren participated in each competition.
On the other hand, this number can also be obtained by summing the number of competitions each schoolchild participated in. Since the sum is odd and all t... | Odd | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,351 |
In the plane, 101 points are marked, not all of them lie on the same line. Through each pair of marked points, a line is drawn with a red pencil. Prove that there exists a point in the plane through which at least 11 red lines pass. | Let through one of the marked points $A$ pass no more than ten red lines. On these lines, there are, excluding $A$, 100 marked points. Therefore, on one of these lines $l$ there are at least ten of them. Together with $A$, the line $l$ contains at least 11 marked points. Consider a point $B$ that does not lie on $l$. I... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,353 |
In Moscow, there are 2000 rock climbers, in St. Petersburg and Krasnoyarsk - 500 each, in Yekaterinburg - 200, and the remaining 100 are scattered throughout Russia. Where should the Russian Rock Climbing Championship be held to minimize the transportation costs for the participants
# | Match each climber not living in Moscow with one Muscovite. Each such pair (there are 1300 of them) is indifferent to where to host the championship, on the segment between Moscow and the corresponding city. But there are still 700 climbers from Moscow left, for whom it is, of course, more beneficial to host the champi... | Moscow | Other | math-word-problem | Yes | Yes | olympiads | false | 54,354 |
Kostrikina I.A.
At a round table, pastries are placed at equal intervals. Igor walks around the table and eats every third pastry he encounters (each pastry can be encountered several times). When there were no pastries left on the table, he noticed that the last pastry he took was the first one he encountered, and he... | Since Igor passed an integer number of circles, he met the first pastry at the moment of approaching the table. In addition, the sequence of eating pastries will not change if we remove the requirement of "equal intervals". If there were four pastries, Igor would have walked exactly five circles (see the figure).
. Upon reaching a new cell, he either changes to its color or repaints it to his own color. A white chameleon-painter is placed on a black $8 \times 8$ board. Can he paint it in a checkerboard pattern... | Consider the moment when the last cell was repainted.
## Solution
Suppose that it was possible to repaint in a checkerboard pattern. Consider the last repainted cell. Suppose it became black. Then all its neighbors are white. The painter came to it being white, so he could not have repainted it to black.
## Answer
... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,356 |
$\begin{array}{ll}{\left[\begin{array}{l}\text { Game Theory (other) }\end{array}\right]} \\ {\left[\begin{array}{l}\text { Pairings and Groupings; Bijections }\end{array}\right]}\end{array}$
A token is placed on a chessboard. Two players take turns moving the token to an adjacent square. It is forbidden to place the ... | Match the second move with some move of the first.
## Solution
We will divide the board into dominoes $1 \times 2$. The strategy of the first player: if before the first player's move, the token is in one of the cells belonging to some domino, then by his move, the first player moves it to the other cell of the same ... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,357 |
Soldiers are lined up in two rows of $n$ people each, so that each soldier in the first row is not taller than the soldier standing behind him in the second row. The soldiers in the rows are arranged by height. Prove that after this, each soldier in the first row will also be not taller than the soldier standing behind... | Let $a_{1}, a_{2}, \ldots, a_{n}$ denote the heights of the soldiers in the first row in descending order, and $b_{1}, b_{2}, \ldots, b_{n}$ denote the heights of the soldiers in the second row in descending order (we use the same notation for the soldiers themselves). Suppose the statement of the problem is false: $a_... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,358 |
3 [ Completing the square. Sums of squares ]
Find all real solutions of the equation with 4 unknowns: $x^{2}+y^{2}+z^{2}+t^{2}=x(y+z+t)$. | Move all terms to the left side and represent the expression on the left as the sum of several squares.
## Solution
$1 / 4 x^{2}+(1 / 2 x-y)^{2}+(1 / 2 x-z)^{2}+(1 / 2 x-t)^{2}=0$. From this, it is clear that all terms are equal to zero. We get: $x=1 / 2 x-y=1 / 2 x-z=1 / 2 x-t=0$, which means $x=y=z=t=0$.
## Answer... | (0,0,0,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,360 |
Two geniuses were told natural numbers and were informed that these numbers differ by 1. After that, they take turns asking each other the same question: "Do you know my number?". Prove that sooner or later one of them will answer positively.
# | If one of the geniuses is told the number 1, then he knows that the second one was told the number 2.
## Solution
If the number of one of the geniuses is $m$, then he knows that the number of the other genius is either $m+1$ or $m-1$; he only needs to determine which of these two possibilities is the case. When geniu... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,361 |
The cells of a $7 \times 7$ board are colored in a checkerboard pattern such that the corners are colored black. It is allowed to repaint any two adjacent cells in the opposite color. Is it possible to repaint the entire board white using such operations?
# | When repainting, the parity of the number of cells of each color does not change.
## Solution
Notice that when repainting two cells, the number of white (or black) cells either increases by 2, decreases by 2, or remains unchanged. In any case, the parity of the number of cells of each color does not change. Initially... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,362 |
On the plane, several lines are drawn (no less than two), no two of which are parallel and no three pass through the same point. Prove that among the parts into which these lines divide the plane, there will be at least one angle.
# | Consider the convex hull of the set of intersection points of lines.
## Solution
Consider the set of intersection points $\mathrm{P}$ of pairs of lines. Take the convex hull of the set P, which is the smallest convex polygon M containing this set. The vertices of the polygon M are some of the points in the set P. Con... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,363 |
There are seven glasses on the table - all upside down. In one move, you can flip any four glasses.
Can you achieve, in several moves, that all the glasses are standing upright?
# | How does the parity of the number of glasses standing upside down change?
## Solution
The first way. Suppose at some point we flipped 4 glasses, of which $k$ glasses were standing upside down, and $4-k$ - correctly ( $k$ can take values from 0 to 4). After flipping, out of these four glasses, $k$ will be standing cor... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,364 |
[Pairings and groupings; bijections ] [ Products and factorials $\quad$]
A natural number $A$ has exactly 100 different divisors (including 1 and $A$). Find their product. | Each divisor $d$ of the number $A$ corresponds to another divisor $A / d$.
## Solution
Notice that if the number $d$ is a divisor of the number $A$, then the number $A / d$ is also a divisor of the number $A$. Since $A$ has 100 divisors, they are divided into 50 such pairs.
## Answer
$A^{50}$. | A^{50} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,365 |
Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 10 games, the second - 21. How many games did the third player play? | The first player plays the rarest every second game.
## Solution
According to the condition, the second player played 21 games, so there were at least 21 games in total. Out of every two consecutive games, the first player must participate in at least one, so the number of games was no more than $2 \cdot 10 + 1 = 21$... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,366 |
On the plane, 2000 points are marked. Can a line be drawn such that 1000 points lie on each side of it? | A straight line can be moved continuously so that it remains parallel to itself, while points will gradually transition from one part of the plane to another.
## Solution
Consider all lines connecting pairs of given points. Take some line \( l \) that is not perpendicular to any of these lines. Introduce a coordinate... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,367 |
[ [Properties of numbers. General properties] ] |
| :---: |
It is known about seven natural numbers that the sum of any six of them is divisible by 5. Prove that each of these numbers is divisible by 5. | Prove that the sum of all seven numbers is divisible by 5.
## Solution
Let the given numbers be $a, b, c, d, e, f, g$, and $S$ be their sum. According to the condition, the numbers $S-a, S-b, S-c, S-d, S-e, S-f, S-g$ are divisible by 5. Therefore, their sum,
$7S - S = 6S$ is divisible by 5. But then $S$ is also divi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,368 |
[ The Pigeonhole Principle (continued). ] [ Tables and Tournaments (continued). $\quad]$
On each cell of a $9 \times 9$ board, there is a beetle. At a whistle, each beetle moves to one of the diagonally adjacent cells. As a result, some cells may end up with more than one beetle, while some cells may remain unoccupie... | Paint the vertical lines of the board in black and white, alternating. Then, from a black cell, the beetle moves to a white one, and from a white one - to a black one.
## Solution
We will paint the vertical lines of the board in black and white, alternating. As a result, $5 \times 9$ $=45$ cells will be painted black... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,370 |
3 [ Examples and counterexamples. Constructions]
Can ten numbers be written in a row so that the sum of any five consecutive numbers is positive, while the sum of any seven consecutive numbers is negative
# | From the condition, it follows that the sum of any two consecutive numbers (except, possibly, the fifth and sixth) must be negative.
## Solution
The condition of the problem is satisfied, for example, by the numbers 20, -30, 20, -30, 24, 24, -30, 20, -30, 20. | 20,-30,20,-30,24,24,-30,20,-30,20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,371 |
In the company, every two people have exactly five common acquaintances. Prove that the number of pairs of acquaintances is divisible by 3.
# | Express the number of triples of people who know each other pairwise through the number of pairs of acquaintances.
## Solution
Let $P$ denote the number of pairs of acquaintances (i.e., the number of edges in the corresponding graph), and $T$ the number of triangles in this graph. By the condition, each edge is part ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,372 |
Prove that the system of inequalities $|x|<|y-z|,|y|<|z-x|,|z|<|x-y|$ has no solutions.
# | Consider two possibilities - when the numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ have the same sign, and when among these numbers there are numbers of different signs.
## Solution
Consider two possibilities. If all numbers have the same sign, then choose the one with the largest absolute value from the numbers x, y... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 54,374 |
8,9
After the third transport ring was put into operation, it was planned to install exactly 1998 traffic lights on it. Every minute, they all simultaneously change color according to the following rule: Each traffic light changes color depending on the color of its two neighbors (to the right and left), and 1) if the... | Consider the situation preceding the moment when all traffic lights switched to yellow.
## Solution
No, this is incorrect. For the sake of contradiction, let's assume that at some point in time, all traffic lights first turned yellow. Consider the situation one minute earlier. Clearly, some traffic light is showing a... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 54,376 |
From the book, 25 pages were torn out. Can the sum of 50 numbers, which are the page numbers (on both sides) of these pages, be equal to 2001?
# | The sum of two numbers written on both sides of one page gives a remainder of 3 when divided by 4.
## Solution
Consider one of the pages. On one side, an odd number $2 m-1$ is written, and on the other side, the next even number $2 m$. The sum of these two numbers is $4 m-1$, which gives a remainder of 3 when divided... | Itcannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,377 |
Can a plane be covered with parquet made of rectangles in such a way that all these rectangles can be cut with one straight cut?
# | First, cut the square into many rectangles so that some line parallel to the diagonal of the square intersects each rectangle.
## Solution
One of the tiling options is shown in the picture (the line cuts off the upper left corners of the blue squares and the lower right corners of the green squares).
.]
What is the minimum number of shots in the game "Battleship" on a 7*7 board needed to definitely hit a four-deck ship (a four-deck ship consists of four cells arranged in a row)?
# | If on a $7 * 7$ board, n non-overlapping four-deck ships can fit, this means that (n-1) shots would not be enough.
## Solution
In the first picture, an example sequence of 12 shots is provided, with which any four-deck ship would be hit. In the second picture, an example of the placement of twelve non-overlapping fou... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,379 |
Petya takes black and red cards out of a bag and stacks them in two piles. It is forbidden to place a card on another card of the same color. The tenth and eleventh cards laid out by Petya are red, and the twenty-fifth is black. What color is the twenty-sixth card laid out? | Show that after a card with an odd number was placed, two cards of the same color ended up on top.
## Solution
Notice that the positions where two cards of the same color are on top and the positions where two cards of different colors are on top alternate. Since the 10th and 11th cards are red, after the 11th card w... | Red | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,380 |
10,11 |
Can 7 different lines be drawn through a point in space such that for any two of them, there is a third one that is perpendicular to both?
# | Construct an example based on the example of three mutually perpendicular lines.
## Solution
Place three pairs of mutually perpendicular lines in some plane P and another line $m$ perpendicular to this plane. If any two lines from this set of seven lie in plane P, then they are perpendicular to line $m$, and if one o... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,381 |
On the test, the teacher gave five problems and graded the test with a score equal to the number of problems solved. All students, except for Petya, solved the same number of problems, while Petya solved one more. The first problem was solved by 9 people, the second by 7 people, the third by 5 people, the fourth by 3 p... | Suppose Petya scored no less than 4, then the others solved no less than 3 problems each, and the total number of problems solved by all students is no less than $3 \cdot 9=27$ (it is clear from the condition that the number of students is no less than 9). However, on the other hand, this number is equal to $9+7+5+3+1=... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,382 |
Two players are playing double chess: all pieces move as usual, but each makes two chess moves in a row. Prove that the first can at least force a draw.
# | $\quad$ -
Assume that the second player has a winning strategy and show that the first can apply this
strategy.
## Solution
Assume the opposite, i.e., the second player has a winning strategy. This means that the second player has a rule P for responding to any possible moves of the first such that if the second fo... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,383 |
From the numbers $1,2, \ldots, 49,50$, 26 numbers were chosen. Is it necessarily true that among them there will be two numbers that differ from each other by 1? | Divide all numbers into pairs of adjacent numbers.
## Solution
Divide all numbers into 25 pairs of adjacent numbers: 1-2, 3-4, ..., 49-50. If no more than one number was chosen from each pair, then no more than 25 numbers would have been chosen in total. However, 26 numbers were chosen according to the condition. Thi... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,384 |
There is a strip $1 \times 99$, divided into 99 cells $1 \times 1$, which are alternately painted in black and white. It is allowed to repaint simultaneously all cells of any rectangular cell $1 \times k$. What is the minimum number of repaintings required to make the entire strip monochromatic?
# | Each repainting destroys no more than two borders between cells of different colors.
## Solution
Estimate. Let's call a bridge an segment separating two cells of different colors. Initially, we have 98 bridges. Each repainting changes the number of bridges by no more than 2. Therefore, 48 repaintings are not enough.
... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,386 |
Prove that no convex polygon can be cut into 100 different equilateral triangles.
# | Assume the opposite and consider the smallest triangle adjacent to the boundary of the polygon.
## Solution
Suppose that a convex polygon is divided into 100 different equilateral triangles. Consider the smallest of all triangles adjacent to the boundary of the polygon. Let this triangle be $A B C$, with side $A B$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,387 |
[ Area of a triangle (using height and base).
[ Area of a triangle (using semiperimeter and radius of inscribed or excircle).
The radius of the inscribed circle of a triangle is 1. Prove that the smallest height of this triangle does not exceed 3.
# | Write the expression for the area of a triangle in terms of the radius of the inscribed circle and the sides, as well as in terms of the heights and sides.
## Solution
Let \( r = 1 \) be the radius of the circle inscribed in the given triangle, \( a, b, c \) be the lengths of its sides, with \( a \) being the largest... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,388 |
There is a 1999×2001 table. It is known that the product of the numbers in each row is negative.
Prove that there will be a column, the product of the numbers in which is also negative.
# | According to the condition, in each row there is an odd number of negative numbers (and there are no zeros). Since the number of rows is odd, there is an odd number of negative numbers in the entire table. Therefore, in at least one of the columns (more precisely, in an odd number of columns), there is also an odd numb... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,389 |
Find the maximum value that the expression $a e k-a f h+b f g-b d k+c d h-c e g$ can take if each of the numbers $a, b, c, d, e, f, g, h, k$ is equal to $\pm 1$.
# | Evaluation. The product of all addends equals - $(a b c d e f g h k)^{2}$, which is negative. Therefore, the number of minus ones among the addends is odd and thus not equal to zero. Consequently, the sum does not exceed 4.
Example. Let $a=c=d=e=g=h=k=1, b=f=-1$. Then the given expression equals 4.
## Answer
4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,390 |
[ Trapezoids (miscellaneous). $[$ Application of trigonometric formulas (geometry).]
In trapezoid $A B C D$, angles $A$ and $D$ are right angles, $A B=1, C D=4, A D=5$. A point $M$ is taken on side $A D$ such that $\angle C M D=$ $2 \angle B M A$.
In what ratio does point $M$ divide side $A D$? | Let $\angle B M A=\alpha$. Then $\angle C M D=2 \alpha, A M=A B \operatorname{ctg} \alpha=\operatorname{ctg} \alpha, M D=C D \operatorname{ctg} 2 \alpha=4 \operatorname{ctg} 2 \alpha=2\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{ctg} \alpha$. Since $A M+M D=A D$, then $\operatorname{ctg} \alpha+2\left(1-\o... | 2:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,391 |
Through a point on a plane, 10 lines were drawn, after which the plane was cut along these lines into angles. Prove that at least one of these angles is less than $20^{\circ}$. | Assume that each of the obtained angles is not less than $20^{\circ}$.
## Solution
Ten lines drawn through one point divide the plane into 20 angles. If all of them are not less than $20^{\circ}$, then their sum is not less than
$20 \cdot 20^{\circ}=400^{\circ}>360^{\circ}$. Contradiction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,392 |
Given a chessboard. It is allowed to repaint all the cells of any row or column to another color at once.
Can it result in a board with exactly one black cell? | When repainting a row or column containing $k$ black and $8-k$ white cells, it will result in $8-k$ black and $k$ white cells. Therefore, the number of black cells will change by $(8-k)-k=8-2k$, that is, by an even number. Since the parity of the number of black cells is preserved, we cannot achieve one black cell from... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,393 |
Given natural numbers $x_{1}, \ldots, x_{n}$. Prove that the number $\left(1+x_{1}^{2}\right) \ldots\left(1+x_{n}^{2}\right)$ can be represented as the sum of squares of two integers. | The statement immediately follows from the fact that the product of two sums of two squares is a sum of two squares (see problem $\underline{61078})$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,394 |
Prove that a square can be cut into $n$ squares for any $n$, starting from six.
# | If a square admits a partition into $n$ squares, then it admits a partition into $n+3$ squares (it is sufficient to cut one of the squares into four). Let's divide all natural numbers into three arithmetic progressions $n=$ $3 k, n=3 k+1, n=3 k+2$, and in each of them find the minimum $n$ for which the problem has a so... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,395 |
[ Ordering in ascending (descending) order. ] [ Classical combinatorics (other). $\quad]$
a) A traveler stopped at an inn, and the owner agreed to accept rings from a golden chain the traveler wore on his wrist as payment for lodging. However, he set a condition that the payment should be daily: each day the owner sh... | a) It is enough to cut two rings so that pieces of three and six rings are separated. On the third day, the traveler gives the piece of three rings and receives two rings as change, and on the sixth day, the piece of six rings and receives five rings as change.
b) Arrange the resulting pieces of the chain (not countin... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,396 |
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