problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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The cells of a $100 \times 100$ table are colored in 4 colors such that in each row and each column there are exactly 25 cells of each color.
Prove that there exist two rows and two columns, all four cells at their intersections being colored in different colors. | Assume the opposite: let among the four cells at the intersection of any two rows and any two columns, there are two cells of the same color.
We will call a horizontal (vertical) pair two cells of different colors lying in the same row (column). We will call a horizontal (vertical) match two cells of the same color ly... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,719 |
Tokaeva I.
Let $F_{1}, F_{2}, F_{3}, \ldots$ be a sequence of convex quadrilaterals, where $F_{k+1}$ (for $k=1,2,3, \ldots$) is obtained by cutting $F_{k}$ along a diagonal, flipping one of the parts, and gluing it back along the cut line to the other part. What is the maximum number of different quadrilaterals that t... | Let $ABCD$ be the original quadrilateral $F_{1}$. We can assume that each time the half of the quadrilateral containing side $CD$ is flipped, while side $AB$ remains stationary. In this process, the sum of angle $A$ and the opposite angle does not change. Additionally, the set of side lengths does not change. However, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,721 |
Kanel-Belov A.Ya.
Prove that in space there exists such an arrangement of 2001 convex polyhedra that no three of the polyhedra have common points, but every two touch each other (that is, have at least one boundary point, but do not have common interior points).
# | Let $N=2001$ and describe the construction of the arrangement of $N$ convex polyhedra, none of which intersect with any other two and each pair of which touch.
Consider an infinite circular cone with vertex $O$ at the origin and axis along the $O z$ axis. On the circle, which is the section of the cone by the plane $z... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,723 |
\{ दккубов A.
In an acute scalene triangle $ABC$, the altitude $AH$ is drawn. Points $B_{1}$ and $C_{1}$ are marked on sides $AC$ and $AB$ respectively, such that $HA$ is the bisector of angle $B_{1}HC_{1}$ and the quadrilateral $BC_{1}B_{1}C$ is cyclic. Prove that $B_{1}$ and $C_{1}$ are the feet of the altitudes of ... | Let $O$ be the center of the circumcircle of triangle $A B_{1} C_{1}$ (see figure). The quadrilateral $B C_{1} B_{1} B$ is inscribed, hence $B_{1} C_{1}$ and $B C$ are antiparallel. Therefore, triangle $A B_{1} C_{1}$ is homothetic to the triangle with vertices at point $A$ and the bases of the altitudes dropped from v... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,724 |
Shapovodov A.B.
In a rectangular box with a base of $m \times n$, where $m$ and $n$ are odd numbers, dominoes of size $2 \times 1$ are placed such that only one $1 \times 1$ square (a hole) remains uncovered in the corner of the box. If a domino tile is adjacent to the hole by its short side, it is allowed to slide al... | Let's number the rows and columns in order, starting from the edge, and call the cells at the intersections of rows with odd numbers "fields." In particular, all cells in the corners are fields. Note that the hole will only move across fields.
If a field $A$ is covered by a domino, then one of the narrow sides of the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,726 |
| Orthogonal (rectangular) projections | ] |
| :---: | :---: | :---: |
| | Auxiliary similar triangles | ] |
| | [ Properties of perpendicular bisectors of the sides of a triangle | |
| | Ordering by increasing (decreasing). | |
| | Inequalities for elements of a triangle (other). | ] |
| | Examples and countere... | a) Let in triangle $ABC \quad \angle A < \angle B \leq \angle C$. Then the perpendicular bisectors of sides $AC$ and $BC$ intersect side $AB$. The segments of these perpendicular bisectors lying inside the triangle have equal projections on lines perpendicular to $AB$, but they form different angles with these lines. T... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,727 |
Rubasnov I.S.
On a plane, $n (n>2)$ lines passing through a single point $O$ are marked in such a way that for any two of them, there is a marked line that bisects one of the pairs of vertical angles formed by these lines. Prove that the drawn lines divide the full angle into equal parts. | Let's call the $2n$ angles into which the marked lines divide the full angle elementary, and the two marked lines forming an elementary angle - adjacent.
Since there are no marked lines inside the elementary angles, for any two adjacent marked lines, there will be a line that bisects the second pair of angles formed b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,728 |
Rubanov I.S.
A set of 2003 positive numbers is such that for any two numbers $a$ and $b$ ($a > b$) in the set, at least one of the numbers $a + b$ or $a - b$ is also in the set. Prove that if these numbers are arranged in ascending order, the differences between consecutive numbers will be the same.
# | Let's number the numbers in the set in ascending order: $0a 2002+a 1=a 2003$, the set includes the difference $a 2002-a 2$. For the same reasons, the set includes differences $a 2002-a 3$, $a 2002-a 2001$.
There are 2000 such differences, all of which are distinct and less than $a 2001$ (since $a 2001=a 2003-a_{2}>a 2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,729 |
Tokarev S.I.
A set of five-digit numbers $\left\{N_{1}, N_{k}\right\}$ is such that any five-digit number, all digits of which are in non-decreasing order, coincides in at least one digit with at least one of the numbers $N_{1}, N_{k}$. Find the smallest possible value of $k$. | A set with the specified properties cannot consist of a single number. Indeed, for each $N=\overline{a b c d e}$, there is a number $G=\overline{\boldsymbol{g g 9 g g}}$ that differs from $N$ in all digits, where $g$ is a non-zero digit different from $a, b, c, d, e$. We will show that the numbers $N_{1}=13579$ and $N_... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,730 |
Kopoyiky G.A.
A cyclist is traveling on a circular road, moving in one direction. Each day he covers 71 km and stops to sleep by the roadside. There is an anomalous zone 71 km long on the road. If the cyclist stops in it for the night at a distance of y km from one of its boundaries, he wakes up in the opposite part o... | Let's call points on the track $A$ and $A^{\prime}$ symmetric if the length of the path from point $A$ to one boundary of the zone in one direction is equal to the length of the path from point $A^{\prime}$ to the other boundary in the opposite direction. For example, if a cyclist falls asleep at a point in the anomalo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,732 |
$\underline{\text { Problem } A .}$
A natural number is written on the board. If the number $x$ is written on the board, then you can add the number $2 x+1$ or ${ }^{x} / x+2$. At some point, it was found that the number 2008 is on the board. Prove that it was there from the beginning. | We can assume that no "extra" numbers were written on the board, meaning all numbers "participated" in obtaining the number 2008. Note that all the numbers written are positive. Suppose at some point the board has a rational number, in its irreducible form, written as
$x = p / q$. Then we can write the number $2x + 1 ... | 2008 | Number Theory | proof | Yes | Yes | olympiads | false | 54,734 |
Shapovalov A.V.
55 boxers participated in a tournament with a "loser leaves" system. The fights proceeded sequentially. It is known that in each match, the number of previous victories of the participants differed by no more than 1. What is the maximum number of fights the tournament winner could have conducted? | We will prove by induction that
a) if the winner has conducted no less than $n$ fights, then the number of participants is no less than $u_{n+2}$;
b) there exists a tournament with $u_{n+2}$ participants, the winner of which has conducted $n$ fights ( $u_{k}-$ Fibonacci numbers).
Base case $\left(n=1, u_{3}=2\right)... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,737 |
Franklin 5.P.
On a circular track, $2 n$ cyclists started simultaneously from the same point and rode with constant but different speeds (in the same direction). If after the start, two cyclists are again in the same point at the same time, we call this a meeting. By noon, every two cyclists have met at least once, an... | Let $S-$ be the length of the track, $v_{1}<v_{2}<\ldots<v_{2 n}-$ be the speeds of the cyclists, $u=\min \left\{v_{2}-v_{1}, v_{3}-v_{2}, \ldots, v_{2 n}-v_{2 n-1}\right\}$. Cyclists with numbers $i<j$ meet at intervals of time $\frac{S}{v_{j}-v_{i}}$. Clearly, the largest of these intervals is $s /$, and we will have... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,738 |
Loatnikov A.A.
Can all cells of an infinite grid plane be colored in white and black so that each vertical line and each horizontal line intersect a finite number of white cells, and each diagonal line intersects a finite number of black cells? | Let's introduce a coordinate system $O x y$ such that the vertical and horizontal grid lines have equations $x = n$ and $y = m$ (where $n$ and $m$ are integers). We will color black those and only those cells, all points of which satisfy one of the inequalities $|y| \geq x^{2}$ or $|x| \geq y^{2}$ (see the figure), and... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,739 |
$\underline{\text { Problem } B .}$
The vertices of a regular 45-gon are colored in three colors, with an equal number of vertices of each color. Prove that it is possible to choose three vertices of each color such that the three triangles formed by the chosen monochromatic vertices are equal. | Let the colors be blue, red, and yellow. Draw a black 45-gon on a transparent film and overlay it on the original. Let's call this position C. Circle 15 blue vertices on the film. By rotating the film each time by an angle of $360^{\circ} : 45 = 8^{\circ}$, align the vertices on the film with the vertices of the origin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,740 |
Mitrofanov I.V.
5000 cinephiles gathered for the congress, each having seen at least one film. They are divided into two types of sections: either discussing a film that all members of the section have seen, or each person tells about a film they have seen that no one else in the section has seen. Prove that all of th... | It is sufficient to distribute the participants of the congress into no more than 100 sections: later, the number of sections can be increased to 100 by dividing them into arbitrary parts.
If there is a film that has been seen by more than 100 people, we will allocate all of them to a separate section. If now there is... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,741 |
A.K.
On the plane, there is an open non-intersecting broken line with 31 segments (adjacent segments do not lie on the same straight line). Through each segment, a line containing this segment was drawn. As a result, 31 lines were obtained, some of which may have coincided. What is the smallest number of different lin... | Evaluation. Except for the ends, the broken line has 30 vertices, and each is the intersection of two lines. If there are no more than eight lines, then there are no more than $7 \cdot 8: 2=28$ intersection points - a contradiction.
Example - in the figure.
 initially all the cement is located at one warehouse $S$, and it needs to be distributed evenly among all warehouses. Then, to each warehouse, except for $S$, cement must be delivered in some trip, so there should be at least 2010 trips in total.
We will prove by induction on $n$ tha... | 2010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,744 |
уvin К.
The main audience of the company "Horns and Hooves" is a square hall with eight rows of eight seats each. 64 employees of the company took a test in this hall, which had six questions with two answer options for each. Could it be that among the answer sets of the employees, there were no identical sets, and th... | Let's denote the first answer to each question as zero, and the second as one. Then, each set of answers corresponds to a set of six zeros and ones. We will represent the audience as an $8 \times 8$ table, where sets of six digits are placed in the cells. We need to fill the table such that adjacent sets (horizontally ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,745 |
Shapovalov A.V.
Having defeated Koschei, Ivan demanded gold to ransom Vasilisa from the bandits. Koschei led him to a cave and said: "In the chest lie gold ingots. But you cannot simply take them: they are enchanted. Put one or several into your bag. Then I will put one or several from the bag back into the chest, but... | Ivan will act in such a way that each time Kashchey's move will be the only one: all other numbers either have already appeared in previous moves or are too large - Ivan does not have such a number of ingots at that moment. We will record the moves of the game as follows:
number of ingots moved (number of ingots Ivan ... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,746 |
Senderov V.A.
Do there exist natural numbers $a, b, c$, greater than $10^{10}$, such that their product is divisible by any of them, increased by 2012? | Let's choose some $t>10^{10}$ and set $a=b=2012 t, c=2012\left(t^{2}-1\right)$. $c$ is divisible by $2012(t+1)=a+2012=b+$ 2012 , and $a b=2012^{2} t^{2}$ is divisible by $2012 t^{2}=c+2012$.
## Answer
There exist. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,747 |
Zaslavsky A.A.
A square is cut into several (more than one) convex polygons with pairwise distinct numbers of sides.
Prove that among them there is a triangle.
# | Let a square be cut into $n$ polygons. Each of these polygons has no more than one side on each side of the square; with any other of the polygons in the partition, it shares no more than one side. Therefore, it can have no more than $4+(n-1)=n+3$ sides.
Thus, the number of sides of any polygon in the partition is no ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,748 |
Gladkova E.B.
Computers $1,2,3, \ldots, 100$ are connected in a ring (the first with the second, the second with the third, ..., the hundredth with the first). Hackers have prepared 100 viruses, numbered them, and at different times, in a random order, they launch each virus on the computer with the same number. If a ... | Let's prove several statements.
1) Every computer is infected by viruses at least twice.
Consider, for example, computer 1. If the first virus to infect it came from computer 100, then it will also be infected by virus 1. If the first virus to infect computer 1 is virus 1, then the virus that first infected computer ... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,750 |
On an infinite chessboard, a pawn is placed every three squares horizontally and vertically. Is it possible for a knight to tour the remaining part of the board, visiting each square exactly once?
# | ## Solution
Assume that the chips are placed on black squares (see the picture). Consider a sufficiently large square of size $(4N-3) \times (4N-3)$ (the value of N will be specified later), with chips placed at its corners, as well as a square of size $(4N+1) \times (4N+1)$, obtained by expanding the square $(4N-3) \... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,752 |
On the Island of Misfortune with a population of 96 people, the government decided to carry out five reforms. Each reform is opposed by exactly half of all citizens. A citizen will go to a rally if they are dissatisfied with more than half of all the reforms. What is the maximum number of people the government can expe... | Let $x$ be the number of people who came out to the rally. Consider the total number of "dissatisfactions." On one hand, each reform is opposed by exactly 48 residents, so the total number of dissatisfactions is $48 \cdot 5 = 240$. On the other hand, each person who came out to the rally is dissatisfied with at least t... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,754 |
$[\quad$ Regular Polygons [ Application of Trigonometric Formulas (Geometry).]
A regular ( $4 k+2$ )-gon is inscribed in a circle of radius $R$ with center $O$.
Prove that the sum of the lengths of the segments cut by the angle $A_{k} O A_{k+1}$ on the lines $A_{1} A_{2 k}, A_{2} A_{2 k-1}, \ldots, A_{k} A_{k+1}$, is... | For $k=3$ the solution to the problem is clear from the figure.
Indeed, $A_{3} A_{4}=O Q$, $K L=Q P$ and $M N=P A_{14}$, so $A_{3} A_{4}+K L+M N=O Q+Q P+P A_{14}=O A_{14}=R$. The proof is co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,758 |
[ Systems of nonlinear algebraic equations ]
Solve the systems:
a) $\left\{\begin{array}{l}x+3 y=4 y^{3} \\ y+3 z=4 z^{3} \\ z+3 x=4 x^{3}\end{array}\right.$
b) $\left\{\begin{array}{l}2 x+x^{2} y=y \\ 2 y+y^{2} z=z \\ 2 z+z^{2} x=x\end{array}\right.$
c) $\left\{\begin{array}{l}3\left(x+\frac{1}{x}\right)=4\left(y+... | a) If one of the unknowns (for example, $x$) is greater than 1, then $z=4 x^{3}-3 x=x^{3}+3\left(x^{3}-x\right)>1$; similarly, $y>1$. But then
$x^{3}+y^{3}+z^{3}>x+y+z$, which contradicts the equality obtained by adding all the equations. Therefore, each of the unknowns does not exceed 1. Similarly, it is proved that ... | (0,0,0),\(1,1,1),\(\cos\pi/14,-\cos5\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,760 |
[ Geometric progression $]$ $[$ Classical combinatorics (other). ]
A frog jumps from vertex to vertex of a hexagon $A B C D E F$, each time moving to one of the adjacent vertices.
a) In how many ways can it get from $A$ to $C$ in $n$ jumps?
b) The same question, but with the condition that it cannot jump to $D$?
##... | a) It is clear that after an even number of jumps, the frog can only be at vertices $A, C$ or $E$.
Let $a_{k}, c_{k}, e_{k}$ denote the number of paths of length $2k$ leading from $A$ to $A, C$, and $E$ respectively. By symmetry, $c_{k} = e_{k}$. It is easy to see that the following equalities hold: $c_{k+1} = a_{k} +... | )\frac{1}{3}(4^n-1)forevenn;cannotreachforoddn.\quadb)3^{n/2-1}forevenn;cannotreachforoddn.\quad)(3/4)^{k-1}forn=2k-1forn | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,761 |
$\left[\begin{array}{ll}{\left[\begin{array}{l}\text { Graph Theory (miscellaneous). }\end{array}\right]} \\ \text { [Examples and counterexamples. Constructions ] }\end{array}\right]$
Authors: Pakharev A., Skopenkov M.B., Ustinov A.V.
Each city in a certain country is assigned a unique number. There is a list in whi... | Consider a country of 12 cities connected by roads as shown in the figure.
Notice that the figure is symmetric with respect to each diameter passing through the midpoints of the small chords... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,764 |
Bakayev E.V.
On a circle, 10 points are marked and numbered clockwise: $A_{1}, A_{2}, \ldots, A_{10}$, and they can be paired symmetrically relative to the center of the circle. Initially, there is a grasshopper sitting in each marked point. Every minute, one of the grasshoppers jumps along the circle over its neighbo... | Ten grasshoppers divide a circle into 10 arcs. We will paint these arcs alternately in black and white (see the left figure). Initially, the total lengths of the black and white arcs are equal, since the arc symmetric to a black arc relative to the center is white, and vice versa.
.] Semi-invariants $\quad]$
Santa Claus had $n$ types of candies, with $k$ pieces of each type. He distributed all the candies randomly into $k$ gifts, each containing $n$ candies, and gave them to $k$ children. The children decided to restore fairness. Two children are willin... | Let's take a child $A$ with the smallest number of types. If they have $n$ types, then everything is fine. If not, then they have more than one candy of some type. This means that some child $B$ does not have this type at all. Then there will be a type that $B$ has but $A$ does not. Let $A$ and $B$ exchange these types... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,767 |
Two people are dividing a piece of cheese. First, the first person cuts the cheese into two pieces, then the second person cuts any of the pieces into two, and so on, until there are five pieces. Then the first person takes one piece, then the second person takes one of the remaining pieces, then the first person again... | Let the total amount of cheese be 50 (pounds). Players $A$ (first) and $B$ are playing. It is clear that in the division, both would benefit from a greedy algorithm (each time the largest piece is chosen). In this case, $B$ will get the second and fourth largest pieces by weight.
Let's denote their total weight by $R$... | 0.60.4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,768 |
$\underline{\text { Blasova } H}$.
$n$ boys and $n$ girls are standing in a circle. We will call a pair of a boy and a girl good if there are an equal number of boys and girls on one of the arcs between them (in particular, a boy and a girl standing next to each other form a good pair). It turns out that there is a gi... | Note that on any arc between members of a good pair, there are an equal number of girls and boys.
Let $D$ be a girl participating in 10 good pairs. Denote all the children in a clockwise order as $K_{1}, K_{2}, \ldots, K_{2 n}$ such that $K_{1}$ is $D$, and continue the numbering cyclically (for example, $K_{0}=K_{2 n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,769 |
[ Auxiliary area. The area helps to solve the task]
Authors: Grigalko A.V., Mitranov I.V.
a) There is an unlimited set of cards with the words "abc", "bca", "cab". From them, a word is formed according to the following rule. Any card is chosen as the initial word, and then at each step, a card can be attached to the ... | We will consider all possible pairs of letters in a word. Let's call pairs $a \ldots . b, b . . . c$ and $c...a$ good,
a) Let's consider how the number of pairs changes when adding one card. ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,770 |
Kanel-Belov A.Y.
From a regular octahedron with side 1, six corners - pyramids with a square base and edge $1 / 3$ - were cut off. The resulting polyhedron has faces that are squares and regular hexagons. Can copies of this polyhedron tile space? | Let's mark integer points in space where all coordinates have the same parity. Consider for each such point the set of points that are no further from it than from other marked points. The entire space will be divided into such sets (up to common boundaries). All these sets are obtained from each other by translation a... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,772 |
Shapovalov A.V.
Decided to check the court sage with a chess move. "Here are six boxes," said the Shah, - "with the numbers $1,2,3,4$, 5, 6 on the lids. Each box contains a gold coin that weighs exactly as many grams as the number written on it. You arrange the boxes as you wish in the cells of a $2 \times 3$ rectangl... | A wise man can arrange the boxes, for example, like this:
| 4 | 5 | 1 |
| :--- | :--- | :--- |
| 2 | 6 | 3 |
| | | 3 |
And he can point out to the king the boxes with the weights 2, 3, and 5 on their lids. If the king names the sum 10, it means he did not change anything. If he swapped some coins, the sum will be d... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,774 |
8,9,10 |
| :---: | :---: | :---: |
| | Geometric Interpretations in Algebra | |
| | $\underline{\text { Equations in Integers }}$ | |
| | Integer Lattices | |
Author: Kirillov A.A.
Given two coprime natural numbers $a$ and $b$. Consider the set $M$ of integers that can be represented in the form $a x + b y$, wh... | Let's formulate our problem in geometric terms. Each pair of integers $(x, y)$ will be called an "integer point" and will be represented as a red point if both its coordinates are non-negative, and as a blue point if at least one coordinate is negative. For each integer $n$, the equation $a x + b y = n$ defines a line ... | -b | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,775 |
9В In a rectangular table, the product of the sum of the numbers in any column and the sum of the numbers in any row equals the number at their intersection.
Prove that the sum of all numbers in the table is one, or all numbers are zero.
# | Let $x_{1}, \ldots, x_{n}$ be the sums of the numbers in the rows, and $y_{1}, \ldots, y_{m}$ be the sums of the numbers in the columns. At the intersection of the $i$-th row and the $j$-th column, the number is $x_{i} y_{j}$. Therefore, the sum of the numbers in the $i$-th row is $x_{i} y_{1} + x_{i} y_{2} + \ldots + ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,777 |
}
Given a natural number $N$. The following operation is performed with it: each digit of this number is written on a separate card (it is allowed to add or discard any number of cards with the digit 0), and then these cards are divided into two piles. In each of them, the cards are arranged in any order, and the two ... | Let's prove that even 12 operations are sufficient. Note that if from a set of cards $A$ we can obtain a set of cards $A_{1}$ in one operation, and from a set of cards $B$ we can obtain a set of cards $B_{1}$ in one operation, then from the union of sets of cards $A \backslash B$ we can obtain the set $A_{1} \backslash... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,778 |
$[\quad$ Product Rule The product of some 1986 natural numbers has exactly 1985 distinct prime divisors. Prove that either one of these numbers, or the product of several of them, is a square of a natural number.
# | Consider sets of initial numbers taken in all possible quantities: one by one, in pairs, in threes, and so on. There are $2^{1986}-1$ such sets. We will multiply the numbers in each such set and represent the resulting product as the product of the largest perfect square and several prime factors (for example, $2^{16} ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,779 |
[Pairing and grouping; bijections] Proof by contradiction
What is the smallest number of weights in a set that can be divided into 3, 4, and 5 equal-mass piles? | Answer: 9. First, let's prove that the set cannot contain fewer than nine weights. Suppose this is not the case, that is, there are no more than eight. Let $60 m$ be the total mass of all the weights in the set. First, note that the set cannot contain weights with a mass greater than 12 t (since the set can be divided ... | 9 | Number Theory | proof | Yes | Yes | olympiads | false | 54,780 |
Chebotarev A.S.
Given an infinite sequence of polynomials $P_{1}(x), P_{2}(x), \ldots$. Does there always exist a finite set of functions $f_{1}(x), f_{2}(x), \ldots, f_{N}(x)$, such that any of them can be written as compositions of these functions (for example, $P_{1}(x)=$ $\left.f_{2}\left(f_{1}\left(f_{2}(x)\right... | Let, for example, $f(x)=\pi+\operatorname{arctg} x, g(x)=x+\pi, h(x)$ be a function which on the interval $(-\pi / 2+\pi n, \pi / 2+\pi n)$ equals $P_{n}(\operatorname{tg} x)$ (see figure). Then
$P_{n}(x)=h(g(g(\ldots(g(f(x))) \ldots)))$ (function $g$ is used $n-1$ times).
## Answer
Always Send comment
Authors: Gal... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,781 |
Galochnik A.I.
Prove that if $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1$, then $x+y=0$. | Multiplying both sides of the given inequality by $x-\sqrt{x^{2}+1}$, we obtain that $-y-\sqrt{y^{2}+1}=x-\sqrt{x^{2}+1}$. Similarly, multiplying both sides of the given equality by $y-\sqrt{y^{2}+1}$, we arrive at the equality $-x-\sqrt{x^{2}+1}=y-\sqrt{y^{2}+1}$. Adding the obtained equalities, we arrive at the equal... | x+0 | Algebra | proof | Yes | Yes | olympiads | false | 54,782 |
Senderov B.A.
Find all pairs $(a, b)$ of natural numbers such that for any natural $n$, the number $a^{n}+b^{n}$ is a perfect $(n+1)$-th power. | Let there be a sequence of natural numbers $c_{n}$ such that $a^{n}+b^{n}=c_{n}^{n+1}$. Clearly, $c_{n} \leq a+b$, because otherwise $c_{n}^{n+1}>(a+b)^{n+1}>a^{n}+b^{n}$. Therefore, in the sequence $c_{n}$, at least one number $c \leq a+b$ repeats infinitely many times, meaning the equation
$a^{n}+b^{n}=c^{n+1}$ (wit... | (2,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,786 |
Geometry on graph paper
In each cell of a chessboard, there are two cockroaches. At a certain moment, each cockroach crawls to an adjacent (by side) cell, and the cockroaches that were in the same cell crawl to different cells. What is the maximum number of cells on the board that can remain free after this? | Evaluation. We will color 20 cells of the board as shown in the left figure. Each colored cell has the following property: no matter which two adjacent cells the cockroaches crawl into from it, the cockroaches will not be able to reach these cells from any other colored cell. Therefore, after all the cockroaches have m... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,787 |
M. Murraikin
On an island, there live 100 knights and 100 liars, each of whom has at least one friend. Knights always tell the truth, while liars always lie. One morning, each resident either said the phrase "All my friends are knights" or the phrase "All my friends are liars," and exactly 100 people said each phrase.... | Let's show that it is possible to find no less than 50 pairs of friends, one of whom is a knight, and the other is a liar. If the phrase "All my friends are liars" was said by no less than 50 knights, then each of them knows at least one liar, and the 50 required pairs are found. Otherwise, the phrase "All my friends a... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,788 |
Shapovevo A.B.
Two players take turns writing one natural number each (repetitions are allowed) on their half of the board so that the sum of all numbers on the board does not exceed 10000. Once the sum of all numbers on the board becomes equal to 10000, the game ends with the calculation of the sum of all digits on e... | The second player can guarantee a draw: it is enough for him to always write numbers with a digit sum of 1 (for example, just the number 1) - indeed, he makes no more moves than the first, and on each move, he writes a number with no greater digit sum. Moreover, for the same reason, the first player will lose if he wri... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,789 |
Bogdanov I.I.
In the cells of a $100 \times 100$ square, the numbers $1,2, \ldots, 10000$ are placed, each exactly once; at the same time, numbers differing by 1 are placed in cells that share a side. After this, the distances between the centers of each pair of cells, the numbers in which differ by exactly 5000, were... | We will number the rows (from bottom to top) and columns (from left to right) of the square with numbers from 1 to 100; we will denote a cell by a pair of the numbers of its row and column. We will call the distance between cells the distance between their centers. Cells will be called paired if the numbers in them dif... | 50\sqrt{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,790 |
The game board is a strip $1 \times N$. At the beginning of the game, there is one white checker on several of the leftmost squares, and one black checker on the same number of rightmost squares. White and Black take turns, with White starting. A move consists of moving one of their checkers in the direction of the opp... | In this game, White undoubtedly has an advantage, although they sometimes lose. For convenience, we will sometimes number the cells of the board from left to right: $1, 2, 3, . .(N-1), N$.
In point "a", when $N=3$, White will lose (this trivial case was often "overlooked"), while in all other cases, they will win by m... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,791 |
Frankin B.P.Dan gives a convex $n$-gon $A_{1} \ldots A_{n}$. Let $P_{i}(i=1, \ldots, n)$ be a point on its boundary such that the line $A_{i} P_{i}$ divides its area in half. It is known that all points $P_{i}$ do not coincide with the vertices and lie on $k$ sides of the $n$-gon. What is a) the smallest; b) the larges... | a) Since the segments $A_{i} P_{i}$ divide the area of the polygon in half, any two of them intersect. Suppose point $P_{i}$ lies on side $A_{j} A_{j+1}$. Then points $P_{j}$ and $P_{j+1}$ lie on different sides of $A_{i}$, meaning there will always be three points lying on different sides. On the other hand, if two ve... | )3;b)n-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,793 |
$\left[\begin{array}{l}\text { Arithmetic of residues (miscellaneous) }) \\ {[\underline{\text { Induction (miscellaneous) }}]}\end{array}\right.$
Prove that for $n>1$ the number $1^{1}+3^{3}+\ldots+\left(2^{n}-1\right)^{2^{n}-1}$ is divisible by $2^{n}$, but not by $2^{n+1}$. | Lemma 1. $k^{2^{n}} \equiv 1\left(\bmod 2^{n+2}\right)$ for each odd number $k$.
Proof. $k^{2^{n}}-1=(k-1)(k+1)\left(k^{2}+1\right)\left(k^{4}+1\right) \ldots\left(k^{2^{n-1}}+1\right)$ - a product of $n+1$ even factors. Additionally, one of the first two factors is divisible by 4.
Lemma 2. $\left(k+2^{n}\right)^{k} ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,794 |
Kosukhin O.n.
For $n=1,2,3$, we will call a number of type $n$ any number that is either equal to 0, or belongs to the infinite geometric progression
1, $(n+2),(n+2)^{2}, \ldots$, or is the sum of several different terms of this progression. Prove that any natural number can be represented as the sum of a number of t... | Let $a_{1}, a_{2}, a_{3}, \ldots$ be a non-decreasing sequence of natural numbers. For $k \geq 2$, denote by $S_{k} = a_{1} + \ldots + a_{k-1}$ the sum of the first $k-1$ terms of this sequence, and set $S_{1} = 0$.
We will prove that every natural number can be represented as the sum of distinct terms of the sequence... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,797 |
Dolnikov V.L.
In a certain city, the network of bus routes is arranged in such a way that every two routes have exactly one common stop, and each route has at least 4 stops. Prove that all stops can be distributed between two companies in such a way that on each route there will be stops of both companies. | Consider two arbitrary routes $l_{1}$ and $l_{2}$; let $A$ be their common stop. If stop $A$ is on all routes, we can assign it to one company and all other stops to another.
Suppose there exists a route $l_{3}$ that does not pass through stop $A$, and $B$ and $C$ are its common stops with $l_{1}$ and $l_{2}$, respect... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,798 |
Can the set of all natural numbers be partitioned into non-intersecting finite subsets $A_{1}, A_{2}$, $A_{3}, \ldots$ such that for any natural $k$ the sum of all numbers in the subset $A_{k}$ equals $k+$ $2013 ?$ | Suppose the desired partition exists. Let's call a set $A_{k}$ large if it contains more than one element. Suppose the number of large sets is finite. Then there exists a number $t>1$ such that each of the sets $A_{t}, A_{t+1}, A_{t+2}, \ldots$ consists of a single element. Thus, the union of the sets $A_{t}, A_{t+1}, ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,799 |
$\underline{\text { Gopovanov A.S. }}$The sum of the digits in the decimal representation of a natural number $n$ is 100, and the sum of the digits of the number $44 n$ is 800. What is the sum of the digits of the number $3 n$? | Note that $44 n$ is the sum of 4 instances of the number $n$ and 4 instances of the number $10 n$.
If we add these numbers digit by digit, then in each digit place, there will be the sum of the quadrupled digit from the same place in the number $n$ and the quadrupled digit from the next place. If no carries occur in t... | 300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,800 |
$\left[\begin{array}{l}{[\text { Theory of algorithms (other). }} \\ {[\quad \text { Estimation + example }}\end{array}\right]$
There are 2004 boxes on the table, each containing one ball. It is known that some of the balls are white, and their number is even. You are allowed to point to any two boxes and ask if there... | We will number the boxes (and thus the balls in them) from 1 to 2004 and denote a question by a pair of box numbers. We will call non-white balls black.
We will show that a white ball can be found in 2003 questions.
We will ask the questions $(1,2),(1,3),(1,2004)$. If all answers are positive, then the first ball is ... | 2003 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,801 |
There are 2004 boxes on the table, each containing one ball. It is known that some of the balls are white, and their quantity is even. You are allowed to point to any two boxes and ask if there is at least one white ball in them. What is the minimum number of questions needed to guarantee identifying any two boxes cont... | We will number the boxes (and the balls in them) from 1 to 2004 and denote the question by a pair of box numbers. We will call non-white balls black.
We will show that in 4005 questions, we can find two white balls.
We will ask the questions (1,2), (1,3), (1,2004), (2,3), (2,4), (2,2004).
If all answers are positive... | A-B=4(-b) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,802 |
Consider the sequence \(a_1, a_2, \ldots, a_{2000}\) of real numbers such that for any natural number \(n, 1 \leq n \leq 2000\), the following equality holds:
\[
a_1^3 + a_2^3 + \ldots + a_n^3 = \left(a_1 + a_2 + \ldots + a_n\right)^2
\]
Prove that all terms of this sequence are integers. | We will prove by induction on $n$ that for any $n$ the sum $S_{n}=a 1+\ldots+a_{n}$ can be represented as $\frac{1}{2} b_{n}\left(b_{n}+1\right)$, where $b_{n}$ is an integer. Base of induction: $a 13=a 12 \Rightarrow S 1=a 1=0$ or 1, i.e., $S 1=\frac{1}{2} \cdot 0 \cdot 1$ or $S 1=\frac{1}{2} \cdot 1 \cdot 2$. Inducti... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,803 |
Do there exist 100 rectangles such that no one of them can be covered by the other 99? | Choose rectangles one by one, each subsequent one being much thinner and smaller in area than the previous one.
## Solution
We will choose rectangles one by one. As the first rectangle, we take a square with a side length of 1. Let $s_{i}$ be the area of the i-th rectangle, $l_{i}$ be the length of its longer side (t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,804 |
Does there exist a figure with exactly two axes of symmetry but no center of symmetry?
# | If a figure has exactly two axes of symmetry, then they are mutually perpendicular. If a figure has two mutually perpendicular axes of symmetry, then it has a center of symmetry.
## Solution
First, let's prove that if a figure has exactly two axes of symmetry, then they are mutually perpendicular. Indeed, let $l_{1}$... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,805 |
Author: $\underline{\text { Xhitin } C}$.
On a circular road, several identical cars are parked. If all the gasoline from these cars were poured into one car, this car would be able to travel the entire circular road and return to its original position. Prove that at least one of these cars can drive around the entire... | The most straightforward proof - by induction on the number $n$ of cars - proceeds as follows. The case $n=1$ is obvious. Suppose the statement is proven for $n$ cars. Let there be $n+1$ cars. Then among them, there is a car $A$ that can, using only the fuel it has, reach the next car $B$ (this is easily proven by cont... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,808 |
Some of the numbers $a_{1}, a_{2}, \ldots a_{\mathrm{n}}$ are equal to +1, the others are equal to -1. Prove that
$$
\begin{gathered}
2 \sin \left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\ldots+\frac{a_{1} a_{2} \cdot \ldots \cdot a_{n}}{2^{n-1}}\right) \frac{\pi}{4}= \\
=a_{1} \sqrt{2+a_{2} \sqrt{2+a_... | Apply induction on $n$. For $n=1$, we obtain an obvious identity. The equality
$2\left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\ldots+\frac{a_{1} a_{2} \cdot \ldots \cdot a_{n} a_{n+1}}{2^{n}}\right) \frac{\pi}{4}=$
$=a_{1} \frac{\pi}{2}+a_{1}\left(a_{2}+\frac{_{2} a_{3}}{2}+\frac{a_{2} a_{3} \cdots a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,809 |
[ Theory of algorithms (miscellaneous). $\quad]$ [Pairings and groupings; bijections ]
100 numbers, including positive and negative ones, are written in a row. Underlined are, first, each positive number, second, each number whose sum with the next one is positive, and third, each number whose sum with the next two is... | Answer: No, it cannot. We will emphasize numbers as follows: 1) positive numbers with one line; 2) negative numbers, the sum of which with the next number is positive, with two lines; 3) negative numbers, for which the sum with the next number is non-positive, but the sum with the next two numbers is positive, with thr... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,811 |
On a grid paper, three points $A, B, C$ are chosen, located at the vertices of the cells. Prove that if triangle $A B C$ is acute, then there is at least one more vertex of a cell inside or on the sides of the triangle. | Let's construct a rectangle with sides along the lines of a grid paper, so that vertices $A, B, C$ lie on its sides. None of the vertices $A, B, C$ can be inside this rectangle, since otherwise the angle at this vertex would be obtuse. At least one of the points $A, B, C$ lies on the side of the rectangle, not at its v... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,812 |
$n$ numbers are written in a row, among which there are positive and negative numbers. Each positive number is underlined, as well as each number whose sum with several immediately following numbers is positive. Prove that the sum of all underlined numbers is positive.
# | Let $a_{1}, a_{2}, \ldots, a_{\mathrm{n}}$. Underline the number $a_{\mathrm{i}} k+1$ with lines, if $k$ is the smallest number for which $a_{\mathrm{i}}+a_{\mathrm{i}+1}+a_{\mathrm{i}}$ $+2+\ldots+a_{\mathrm{i}+\mathrm{k}}>0$ (a positive number $a_{1}$ is underlined with one line). Clearly, if the number $a_{\mathrm{i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,813 |
An infinite broken line $A_{0} A_{1} \ldots A_{n} \ldots$, all angles of which are right angles, starts at point $A_{0}$ with coordinates $x=0$, $y=1$ and encircles the origin $O$ in a clockwise direction. The first segment of the broken line has a length of 2 and is parallel to the bisector of the 4th coordinate angle... | Note that the coordinates of the vector $\overrightarrow{A_{i} A_{i+1}}$ are $\left( \pm \frac{l}{\sqrt{2}}, \pm \frac{l}{\sqrt{2}}\right)$, where $l-$ is the length of this vector, an integer.
Therefore, the coordinates of any point $A_{\mathrm{i}}$ have the form $\left(\frac{n}{\sqrt{2}}, 1+\frac{m}{\sqrt{2}}\right)... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,814 |
Given two circles, the length of each of which is 100 cm. On one of them, 100 points are marked, and on the other, several arcs are marked, the total length of which is less than 1 cm. Prove that these circles can be superimposed so that no marked point falls on a marked arc.
# | Combine these circles and place the painter at a fixed point on one of them. We will rotate this circle and instruct the painter to paint the point on the circle that he passes by every time any marked point lies on the marked arc. We need to prove that after a full rotation, part of the circle will remain unpainted. T... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,816 |
The pairwise distances between points $A_{1}, \ldots, A_{n}$ are greater than 2. Prove that any figure with an area less than $\pi$ can be translated by a vector of length no more than 1 so that it does not contain any of the points $A_{1}, \ldots, A_{n}$. | Let $\Phi$ be a given figure, $S_{1}, \ldots, S_{\mathrm{n}}$ be circles of radius 1 with centers at points $A_{1}, \ldots, A_{\mathrm{n}}$. Since the circles $S_{1}, \ldots, S_{\mathrm{n}}$ do not intersect each other, the figures $V_{\mathrm{i}}=\Phi \cap S_{\mathrm{i}}$ do not intersect each other either, and thus t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,817 |
In a circle of radius 16, 650 points are arranged. Prove that there exists a ring with an inner radius of 2 and an outer radius of 3, in which at least 10 of the given points lie.
# | First, note that a point $X$ belongs to a ring with center $O$ if and only if the point $O$ belongs to the same ring with center $X$. Therefore, it is sufficient to prove that if rings are constructed with centers at the given points, then one of the points of the considered circle will be covered by at least 10 rings.... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,818 |
Berroov S.l.
Inside a convex hundred-gon, $k$ points are chosen, $2 \leq k \leq 50$. Prove that it is possible to mark $2k$ vertices of the hundred-gon such that all the chosen points are inside the $2k$-gon formed by the marked vertices. | We will call the convex hull of a finite set of points the smallest convex polygon containing all these points. It can be proven that for any finite set of points, there exists a unique convex hull. Let $M=A_{1} A_{2} \ldots A_{n}$ be the convex hull of the selected $k$ points ($n \leq k$), and let point $O \in M$ be d... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,819 |
[ Smallest or Largest Angle ]
a) The lengths of the angle bisectors of a triangle do not exceed 1. Prove that its area does not exceed $1 / \sqrt{3}$.
b) On the sides $B C, C A$, and $A B$ of triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are taken. Prove that if the lengths of segments $A A_{1}$, $B B_{1}$, an... | a) Let $\alpha$ be the smallest angle of triangle $ABC$; $AD$ is the angle bisector. One of the sides $AB$ and $AC$ does not exceed $AD / \cos (\alpha / 2)$, since otherwise the segment $BC$ would not pass through point $D$. Let for definiteness $AB \leq AD / \cos (\alpha / 2) \leq AD / \cos 30^{\circ} \leq 2 / \sqrt{3... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,820 |
8,9
In the plane, there is a finite number of pairwise non-parallel lines, and through the intersection point of any two of them, there passes another one of the given lines. Prove that all these lines pass through one point.
# | Suppose not all lines pass through one point. Consider the intersection points of the lines and choose the smallest non-zero distance from these points to the given lines. Let the smallest distance be from point $A$ to line $l$. Through point $A$, at least three given lines pass. Let these lines intersect line $l$ at p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,821 |
On the plane, there are $n$ points, and the midpoints of all segments with endpoints at these points are marked. Prove that the number of distinct marked points is at least $2 n-3$.
# | Let $A$ and $B$ be the most distant points from each other. The midpoints of the segments connecting point $A$ (respectively point $B$) with the other points are all distinct and lie inside the circle of radius $A B / 2$ centered at $A$ (respectively $B$). The two circles have only one common point, so the number of di... | 2n-3 | Combinatorics | proof | Yes | Yes | olympiads | false | 54,822 |
[ Convex hull and supporting lines (planes).]
On the plane, there are $n \geq 4$ points, and no three of them lie on the same line. Prove that if for any three of them, there exists a fourth (also from the given points) such that they form the vertices of a parallelogram, then $n=4$.
# | Consider the convex hull of the given points. There are two possible cases.
1. The convex hull is a parallelogram $A B C D$. If point $M$ lies inside the parallelogram $A B C D$,
then the vertices of all three parallelograms with vertices $A, B$ and $M$ lie outside $A B C D$ (see figure). Therefore, in this case, the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,823 |
Geometry on graph paper ] Colorings
Nodes of an infinite graph paper are colored in three colors. Prove that there exists an isosceles right triangle with vertices of the same color. | Suppose there is no isosceles right triangle with legs parallel to the sides of the cells and vertices of the same color. For convenience, we can assume that the cells, not the nodes, are colored. We divide the sheet into squares with side length 4; then on the diagonal of each such square, there will be two cells of t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,824 |
An equilateral triangle is divided into $n^{2}$ identical equilateral triangles (see figure). Some of them are numbered with the numbers $1,2, \ldots, m$, such that triangles with consecutive numbers share a side. Prove that $m \leq n^{2}-n+1$.
# | Let's color the triangles as shown in the figure. Then the number of black triangles will be \(1+2+\ldots+n=n(n+1)/2\), and the number of white triangles will be \(1+2+\ldots+(n-1)=n(n-1)/2\). It is clear that two triangles with consecutive numbers are of different colors. Therefore, among the numbered triangles, there... | \leqn^{2}-n+1 | Combinatorics | proof | Yes | Yes | olympiads | false | 54,826 |
[ Auxiliary coloring (other) ]
A triangulation of a polygon is its division into triangles with the property that these triangles either share a side, share a vertex, or have no points in common (i.e., a vertex of one triangle cannot lie on the side of another). Prove that the triangles of a triangulation can be color... | We will prove this statement by induction on the number of triangles in the triangulation. For a single triangle, the required coloring exists. Now suppose that any triangulation consisting of fewer than $n$ triangles can be colored in the required manner, and we will prove that then any triangulation consisting of $n$... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,827 |
Some number of points are located on a plane such that any 3 of them can be enclosed in a circle of radius $r=1$. Prove that then all the points can be enclosed in a circle of radius 1.
# | Consider a circle containing all given points. We will decrease the radius of such a circle until it is no longer possible. Let $R$ be the radius of the resulting circle. At least two given points lie on the boundary of this circle. First, consider the case when exactly two points $A$ and $B$ lie on the boundary. Clear... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,829 |
[ Convex hull and supporting lines (planes).]
On the plane, there are several points, the pairwise distances between which do not exceed 1. Prove that these points can be covered by an equilateral triangle with side $\sqrt{3}$.
# | Consider three lines, each pair forming angles of $60^{\circ}$, and draw three pairs of supporting lines parallel to the chosen lines for a given set of points. The drawn supporting lines define two equilateral triangles, each of which covers the given points. We will prove that the side of one of them does not exceed ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,830 |
$\left[\begin{array}{l}{[\underline{\text { Induction in Geometry }}]} \\ {[\quad \underline{\text { Systems of Points }}]}\end{array}\right]$
Prove that if $n$ points do not lie on the same line, then among the lines connecting them, there are at least $n$ different ones. | We will prove this by induction on $n$. For $n=3$, the statement is obvious. Suppose we have proven it for $n-1$ points, and we will prove it for $n$ points. If on every line passing through two given points, there lies another given point, then all the given points lie on one line (see problem 20.13). Therefore, there... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,832 |
Font der Flasche D:
On an infinite strip of cells numbered by integers, several stones lie (possibly several in one cell). The following actions are allowed:
1. Remove one stone from cells $n-1$ and $n$ and place one stone in cell $n+1$;
2. Remove two stones from cell $n$ and place one stone in cells $n+1, n-2$.
Pro... | Let $a_{i}$ denote the number of stones in the cell numbered $i$.
Then the sequence $A=\left(a_{i}\right)$ defines a configuration - the arrangement of stones in the cells.
Let $\alpha$ be the root of the equation $x^{2}=x+1$, which is greater than 1.
We will call the weight of the configuration $A$ the number $w(A)... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,833 |
[ Invariants $]$
Given points $A_{1}, \ldots, A_{\text {n }}$. Consider a circle of radius $R$ containing some of them. We then construct a circle of radius $R$ centered at the center of mass of the points lying inside the first circle, and so on. Prove that this process will stop, i.e., the circles will begin to coin... | Let $S_{\mathrm{n}}$ be the circle constructed at the $n$-th step, and $O_{\mathrm{n}}$ be its center. Consider the quantity $F_{\mathrm{n}}=\sum\left(R^{2}-O_{\mathrm{n}} A_{\mathrm{i}}{ }^{2}\right)$, where the summation is only over the points that lie inside the circle $S_{\mathrm{n}}$. We will denote the points ly... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,837 |
|
Rabbits are sawing a log. They made 10 cuts. How many chunks did they get?
# | Into how many parts is a log divided by the first cut? How does the number of pieces change after each subsequent cut?
## Solution
The number of chunks is always one more than the number of cuts, since the first cut divides the log into two parts, and each subsequent cut adds one more chunk. Answer: 11 chunks.
## An... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,838 |
Rabbits are sawing a log again, but now both ends of the log are secured. Ten middle chunks have fallen, while the two at the ends remain secured. How many cuts did the rabbits make?
# | How many logs did the rabbits get?
## Solution
The rabbits got 12 logs - 10 fallen and 2 secured. Therefore, there were 11 cuts.
## Answer
11 cuts.
## Problem | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,841 |
[ Cutting (other).]
What is the maximum number of pieces into which a round pancake can be divided using three straight cuts?
# | Recall problem 11.
## Solution
If from three lines each pair intersects inside the pancake, it will result in 7 pieces (see Fig. 11.4). If, however, any two of these lines are parallel or intersect outside the pancake, there will be fewer pieces.
## Answer
7 pieces. | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,842 |
$\left.\begin{array}{l}{[\text { Decimal numeral system }} \\ {[\text { Case enumeration }}\end{array}\right]$
One three-digit number consists of different digits in ascending order, and in its name, all words start with the same letter. Another three-digit number, on the contrary, consists of the same digits, but in ... | Notice, "the numbers are equal" and "the numbers start with the same letter" are two completely different statements.
## Solution
These numbers are 147 and 111, respectively. The problem is solved by simply enumerating the options, of which there are not so many.
## Answer
147 and 111. | 147111 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,843 |
Write in a line the first 10 prime numbers. How to cross out 6 digits so that the largest possible number is obtained
# | Notice that the first 10 prime numbers form a 16-digit number.
## Solution
Here are the first ten prime numbers written in sequence: 2357111317192329
## Answer
The resulting number is 7317192329. | 7317192329 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,844 |
Pinocchio and Pierrot had a bicycle, on which they set off to the neighboring village. They took turns riding, but every time one was riding, the other was walking, not running. Despite this, they managed to arrive at the village almost twice as fast as if they had both walked. How did they manage to do this?
# | Try to organize a trip so that both Buratino and Piero ride a bicycle for exactly half the distance.
## Solution
Buratino rode the bicycle for half the distance, got off, and continued on foot. Piero walked the first half of the distance, then reached the bicycle, got on it, and rode it. This way, they saved time. Se... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,845 |
Kovaldzhi A.K.
Three hedgehogs were dividing three pieces of cheese weighing 5 g, 8 g, and 11 g. A fox came to help them. She can cut and eat 1 g of cheese from any two pieces at the same time. Can the fox leave the hedgehogs equal pieces of cheese? | For example, the fox first cuts off 1 g three times from the pieces weighing 5 g and 11 g. This will result in one piece weighing 2 g and two pieces weighing 8 g each. Now, there are six more times to cut and eat 1 g from the pieces weighing 8 g.
## Answer
Yes, she can. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,846 |
Chebotarev A.S.
A square napkin was folded in half, and the resulting rectangle was folded in half again (see figure). The resulting square was cut with scissors (in a straight line). Could the napkin have split a) into 2 pieces? b) into 3 pieces? c) into 4 pieces? d) into 5 pieces? If yes - draw such a cut, if no - w... | If the cut is "moved" within the same side, the number of pieces will not change. When the cut passes through a corner or moves to an adjacent side, the number of pieces may change.
## Solution
Notice that if the cut is "moved" within the same side, the number of pieces will not change. When the cut passes through a ... | all\possible | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,847 |
Can five rays be drawn from one point on a plane so that among the angles they form, there are exactly four acute angles?
Angles are considered not only between adjacent rays but also between any two rays.
# | An example of the required layout is shown in the figure:
## Answer
It can be done. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,848 |
In a company of $k$ people ($k>3$), each person has a piece of news known only to them. In one phone call, two people share all the news they know. Prove that in $2k-4$ calls, everyone can learn all the news.
# | Information can be transmitted in the following way. Consider four people in a company - let's call them A, B, C, and D. Initially, all members of the company except B, C, and D call A and tell him their news. This will require $\mathrm{k}-4$ calls. Then A and B talk to each other, as well as C and D. After that, A tal... | 2k-4 | Combinatorics | proof | Yes | Yes | olympiads | false | 54,849 |
In each cell of a rectangular table of size $M \times K$, a number is written. The sum of the numbers in each row and in each column is equal to 1.
Prove that $M=K$. | Let's find the sum of all numbers in the table in two ways. Since the table contains $M$ rows and the sum of numbers in each row is 1, the total sum is $M$. Similarly, calculating this same sum by columns, we get $K$ in the end. | K | Combinatorics | proof | Yes | Yes | olympiads | false | 54,850 |
10 guests came to visit and each left a pair of galoshes in the hallway. All pairs of galoshes are of different sizes. The guests began to leave one by one, putting on any pair of galoshes that they could fit into (i.e., each guest could put on a pair of galoshes not smaller than their own). At some point, it was disco... | If there are more than five guests left, then the guest with the smallest size of galoshes among the remaining guests can put on the largest of the remaining galoshes.
## Solution
Let's number the guests and their pairs of galoshes from 1 to 10 in ascending order of the size of the galoshes. Suppose there are 6 guest... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,851 |
Find the number of all Young diagrams with weight $s$, if
a) $s=4$
b) $s=5$;
c) $s=6$;
d) $s=7$.
For the definition of Young diagrams, see the reference manual. | List all the ways to break down the given weights into sums of natural addends, arranged in non-decreasing order.
a) $4=3+1=2+2=2+1+1=1+1+1+1$.
b) $5=4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1$.
c) $6=5+1=4+2=4+1+1=3+3=3+2+1=3+1+1+1=2+2+2=2+2+1+1=2+1+1+1+1=1+$ $1+1+1+1+1$
d) $7=6+1=5+2=5+1+1=4+3=4+2+1=4+1+1+1=3+3+1=3+2+... | )5;b)7;)11;)15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,853 |
$\left.\quad \begin{array}{ll}{\left[\begin{array}{l}\text { Principles of divisibility by 3 and 9 } \\ \text { [Examples and counterexamples. Constructions] }\end{array}\right]}\end{array}\right]$
Write the number 2013 several times in a row so that the resulting number is divisible by 9. | The number 201320132013 is divisible by 9, since the sum of its digits is equal to $(2+0+1+3) \cdot 3=18$.
## Answer
Given a sheet of graph paper. Each node of the grid is marked with some letter. What is the smallest number of different letters needed to mark these nodes so that on any segment (going along the sides... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,854 |
8,9
Find the equation of the plane passing through the point $M(-2 ; 0 ; 3)$ and parallel to the plane $2 x-y-3 z+5=0$. | The vector perpendicular to the desired plane has coordinates (2; -1; -3). Since the plane passes through the point $M(-2; 0; 3)$, its equation is
$$
2(x+2)-y-3(z-3)=0, \text { or } 2 x-y-3 z+13=0 \text {. }
$$
## Answer
$2 x-y-3 z+13=0$. | 2x-y-3z+13=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,855 |
Grandpa is twice as strong as Grandma, Grandma is three times stronger than Granddaughter, Granddaughter is four times stronger than Bug, Bug is five times stronger than Cat, and Cat is six times stronger than Mouse. Grandpa, Grandma, Granddaughter, Bug, and Cat together with Mouse can pull out the Turnip, but without ... | How many Mice does a Cat replace? And a Granddaughter?
## Solution
A Cat replaces 6 Mice. A Bug replaces $5 \bullet 6$ Mice. A Granddaughter replaces 4$\cdot$5$\cdot$6 Mice. A Grandma replaces $3 \cdot 4 \cdot 5 \cdot 6$ Mice. A Grandpa replaces $2 \cdot 3 \cdot 4 \cdot 5 \cdot 6$ Mice. In total, it requires:
$(2 \c... | 1237 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,856 |
Chuk and Gek were decorating the Christmas tree with their mother. To prevent them from fighting, their mother gave each of them the same number of branches and the same number of ornaments. Chuk tried to hang one ornament on each branch, but he was short of one branch. Gek tried to hang two ornaments on each branch, b... | Try to do as Chuk did - hang one toy on each branch.
## Solution
Let's try to do as Chuk did — hang one toy on each branch, then one toy will be left over. Now, let's take two toys — one that is left over, and another one from one of the branches. If we now hang these toys as the second ones on the branches that stil... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,857 |
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