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
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Evochkimov M.A.
A watermelon has the shape of a sphere with a diameter of 20 cm. Vasya made three mutually perpendicular planar cuts of depth $h$ (a cut is a segment of a circle, $h$ is the height of the segment, the planes of the cuts are pairwise perpendicular). Is it necessary that the watermelon will be divided in... | If a watermelon splits into parts, then its surface also splits into parts. Therefore, it is enough to draw three arcs of the required size on a sphere so that it does not split into parts. The points of intersection of pairs of cutting planes with the sphere will be called nodes, and we will have six of them. Obviousl... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,183 |
Ionin Yu. I.
The sum of $n$ positive numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ is 1.
Let $S$ be the maximum of the numbers $\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \ldots, \frac{x_{n}}{1+x_{1}+x_{2}+\ldots+x_{n}}$.
Find the smallest possible value of $S$. For which values of $x_{1}, x_{2}, \ldots, x_{... | Suppose that the positive numbers $a_{1}, a_{2}, \ldots, a_{n}$ are such that $a_{1}+a_{2}+\ldots+a_{n}=1$ and $\frac{a_{1}}{1+a_{1}}=\frac{a_{2}}{1+a_{1}+a_{2}}=\ldots=\frac{a_{n}}{1+a_{1}+a_{2}+\ldots+a_{n}}$. (We will see below that such a set of $a_{1}, a_{2}, \ldots, a_{n}$ indeed exists.) We will show that for $x... | 1-\frac{1}{\sqrt[n]{2}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 55,184 |
Shapovalov A.V. The host has a deck of 52 cards. The audience wants to know the order in which the cards are arranged (without specifying whether it is from top to bottom or bottom to top). They are allowed to ask the host questions of the form "How many cards lie between such-and-such and such-and-such cards?" One of... | Algorithm. Let's show how to establish the order of cards in 34 questions. We will divide all the cards in the deck (numbered from bottom to top), except for the 19th, into triples:
| The cards | 1 | 2 | 3 | $\ldots$ | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | $\ldots$ | 48 | 49 | 50 | 51 | 52 |
| :---: | :---: | :---: ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,187 |
Dolnikov V.L.
Given two convex polygons. It is known that the distance between any two vertices of the first is no more than 1, the distance between any two vertices of the second is also no more than 1, and the distance between any two vertices of different polygons is greater than $1 / \sqrt{2}$. Prove that the poly... | Let $F_{1}$ and $F_{2}$ denote the given polygons. Suppose they have a common interior point. There are two cases. 1) One polygon is contained within the other, say, $F_{1}$ lies inside $F_{2}$. Let $A$ be one of the vertices of $F_{1}$. Then, as it is easy to see, there will be three vertices $P, Q, R$ of the polygon ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,188 |
[Examples and Counterexamples. Constructions] [ Case Analysis $\quad$]
Authors: Volchenkov S.G., Mednikov L.E.
Let $S(x)$ denote the sum of the digits of the number $x$. Can three such natural numbers $a, b$, and $c$ be found such that $S(a+b) > 50$? | For example, the numbers $a=5555554445, b=5554445555, c=4445555555$ will work. Let's verify this:
$S(a+b)=S(11110000000)=50$. How can we find such numbers? Note that $S(2(a+b+c))=S((a+b)+(a+c)+(b+c)) \leq S(a+b)+S(a+c)+S(b+c) \leq 12$, i.e., the number $n=2(a+b+c)$, when divided by 2, should significantly increase its... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,189 |
Let $2 S$ be the total weight of some set of weights. We call a natural number $k$ average if we can choose $k$ weights from the set such that their total weight is $S$. What is the maximum number of average numbers that a set of 100 weights can have? | Note that if the number $m$ is average, then the number $100-m$ is also average. Therefore, if the number 1 is not average, then the number 99 is also not average, and the number of average numbers is no more than 97 (the number 100 is also not average). If, however, the number 1 is average, then the weight of one of t... | 97 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,191 |
$\underline{\text { Kanel-Belov A.Y. }}$
There is one red and $k(k>1)$ blue cells, as well as a deck of $2 n$ cards, numbered from 1 to $2 n$. Initially, the entire deck lies in a random order in the red cell. From any cell, you can take
the top card and move it either to an empty cell or on top of a card with a numbe... | Let's construct an example showing that when $n \geq k$ it is impossible. Suppose the cards (from top to bottom) are initially arranged as follows: first all the odd-numbered cards (in any order), followed by the even-numbered cards, with the top one being card $2n$. Then the first $k$ moves are uniquely determined - t... | k-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,192 |
Dolnikov V.L.
Given a tree with $n$ vertices, $n \geq 2$. In its vertices, numbers $x_{1}, x_{2}, x_{n}$ are placed, and on each edge, the product of the numbers at the ends of this edge is written. Let $S$ denote the sum of the numbers on all edges. Prove that $\sqrt{n-1}\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\rig... | Consider an edge $l$ connecting vertices with numbers $x_{i}$ and $x_{j}$. Denote by $k_{i}(l)$ the number of vertices from which it is impossible to reach vertex $x_{i}$ upon removing edge $l$.
It is clear that $1 \leq k_{i}(l), k_{j}(l) \leq n-1, k_{i}(l)+k_{j}(l)=n$. Moreover, for each $i$ the sum $\sum_{l} k_{i}(l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 55,193 |
Golovanov A.S.
Sasha wrote a non-zero digit on the board and keeps appending a non-zero digit to the right, until he writes a million digits. Prove that a perfect square was written on the board no more than 100 times. | Let's consider separately the numbers with an odd and even number of digits. Let $x_{1}^{2}, x_{2}^{2}, . .-$ be the squares encountered on the board with an even number of digits, and their representations contain $2 n_{1}, 2 n_{2}, . .\left(n_{1}<n_{2}<\ldots\right)$ digits respectively. From the inequality $x_{k}^{2... | 100 | Number Theory | proof | Yes | Yes | olympiads | false | 55,196 |
Determine the smallest real number $M$ such that the inequality $\mid a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-\right.$ $\left.a^{2}\right) \mid \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}$ holds for any real numbers $a, b, c$.
# | Author: Kanel-Belov A.Y.
The left side turns to zero when $a=b$, which means it is divisible by $a-b$. Similarly, it is divisible by $a-c$ and $b-c$. Now, from symmetry, it is clear that $a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)=\alpha(a-b)(a-c)(b-c)(a+b+c)$. By comparing the... | \frac{9\sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 55,197 |
Dolnikov V.L.
In a kingdom of $N$ cities, some pairs of which are connected by non-intersecting two-way roads (cities from such a pair are called neighboring). It is known that from any city, one can travel to any other city, but it is impossible to leave a city and return to the starting city by traveling along diffe... | We apply induction. The statement is obvious for $N=1$ and $N=2$.
Induction step. Let $\Gamma_{1}$ be the set of cities from which only one road leads out, and $\Gamma$ be the set of all other cities. Starting movement along different roads from some city, according to the condition, we cannot visit the same city twic... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,198 |
Romaschenko A.
In each cell of a $1000 \times 1000$ table, there is a zero or a one. Prove that it is possible to either cross out 990 rows so that each column has at least one uncrossed one, or cross out 990 columns so that each row has at least one uncrossed zero. | We will one by one select certain rows - columns and rows. A bad column intersects with all selected rows by zeros, a bad row - with all selected columns by ones. Initially, nothing is selected, and all rows are bad. We can assume that there are no less ones than zeros in the table. Then we choose a row where ones are ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,199 |
Let $P$ be a parabola.
On the coordinate plane, $n$ parabolas are drawn, which are graphs of quadratic trinomials; no two of them touch. They divide the plane into several regions, one of which is located above all the parabolas. Prove that the boundary of this region has no more than $2(n-1)$ vertices (i.e., points o... | Induction on $n$.
Base. For $n=1$, the statement is obvious.
Induction step. Let $f_{1}(x), f_{2}(x), \ldots, f_{n}(x)$ be the given quadratic trinomials $(n \geq 2)$, and let $f_{n}(x)$ be the trinomial with the smallest leading coefficient (if there are several, any of them). Denote the boundary of our region by $T... | 2(n-1) | Geometry | proof | Yes | Yes | olympiads | false | 55,200 |
Can a knight tour a board of size $4 \times N$, visiting each square exactly once, and return to the starting square? | We will color a $4 \times N$ board in 4 colors such that if a knight stands on a field of color 1 (respectively 2), then the next move it will stand on a field of color 3 (respectively 4).
Since the number of fields of colors 1 and 2 is the same as the number of fields of colors 3 and 4, if there is a knight's tour of... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,201 |
10,11 On a rectangular table, there are equal cardboard squares of $\mathrm{k}$ different colors with sides parallel to the sides of the table. If any $\mathrm{k}$ squares of different colors are considered, then some two of them can be nailed to the table with one nail. Prove that all squares of some color can be nail... | Conduct induction on the number of colors.
## Solution
We will prove the statement of the problem by induction on the number of colors \( n \). Base case for \( n=2 \). Consider the leftmost square \( K \). If it is of the first color, then all squares of the second color have a common point with it, hence each squar... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,202 |
[ $\quad \underline{\text { Radical Axis }}$
On a plane, there are two non-concentric circles $S_{1}$ and $S_{2}$. Prove that the geometric locus of points for which the power with respect to $S_{1}$ is equal to the power with respect to $S_{2}$ is a straight line.
# | Let $R_{1}$ and $R_{2}$ be the radii of the circles. Consider a coordinate system in which the centers of the circles have coordinates $(-a, 0)$ and $(a, 0)$. According to problem 3.52, the power of a point with coordinates $(x, y)$ relative to the given circles are $(x+a)^{2}+y^{2}-R_{1}^{2}$ and $(x-a)^{2}+y^{2}-R_{2... | (R_{1}^{2}-R_{2}^{2})/4 | Geometry | proof | Yes | Yes | olympiads | false | 55,203 |
Hoomoov A.
a) Prove that in any polygon, except for a triangle, there is at least one diagonal that lies entirely inside it.
b) Determine the smallest number of such diagonals that an $n$-gon can have. | a) If the polygon is convex, the statement is obvious. Suppose now that the internal angle of the polygon at vertex $A$ is greater than $180^{\circ}$. The visible part of the side is seen from point $A$ at an angle less than $180^{\circ}$, so from point $A$, parts of at least two sides are visible. Therefore, there exi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,205 |
In a certain state, golden and platinum sand are valued. Gold can be exchanged for platinum, and platinum for gold, according to a rate determined by natural numbers $g$ and $p$ such that: $x$ grams of gold sand are equivalent to $y$ grams of platinum sand if $x p = y g$ (the numbers $x$ and $y$ can be non-integers). C... | Let's prove that if initially the banker has 1 kg of each type of sand, and $g=p=k$, then at the end, the amount of at least one type of sand will not exceed $2-\frac{1}{k}$ kg. To do this, it is sufficient to prove that if initially the banker has $\frac{k}{2k-1}$ kg of each type of sand, then at the end, he cannot ob... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,206 |
Let $H$.
A line consists of $N$ boys of pairwise distinct heights. It was divided into the smallest possible number of groups of consecutive boys, in each of which the boys stand in increasing order of height from left to right (groups of one person are possible). Then, in each group, the boys were rearranged in decre... | Let's choose any number $h$ and call all the guys shorter than $h$ dwarfs, and the rest - giants. The place between neighbors, the left one of which is a dwarf and the right one - a giant, is called a seam. The weight of a seam is the number of dwarfs to the left and giants to the right of it (not necessarily in a row)... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,207 |
Petrov $\Phi$.
On a line, there are a finite number of frogs sitting at different integer points. In one move, exactly one frog jumps 1 to the right, and they must still be at different points. We calculated the number of ways the frogs can make $n$ moves (for some initial arrangement of frogs). Prove that if we allow... | Let's call a template a sequence of $n$ words, each of which is "Left" or "Right," where the $k$-th word indicates the direction in which frogs can jump on the $k$-th move. We will prove a stronger statement: the number of ways to make $n$ moves from a given position without frogs jumping onto each other does not depen... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,208 |
Zaslavsky A.A.
In a certain state, there are 32 cities, each of which is connected by a road with one-way traffic. The Minister of Communications, a secret villain, decided to organize traffic in such a way that, after leaving any city, it would be impossible to return to it. To achieve this, he can change the directi... | Let's prove the general formula. Suppose in the state there are $\$ 2 \wedge n$ \$ cities. Then the minister can achieve the desired result in no more than $\$ 2 \wedge\{n-2\}(2 \wedge n-n-1) \$$ days.
Lemma. Suppose in the state there are $\$ 2 \mathrm{k} \$$ cities, each two of which are connected by a one-way road.... | 208 | Combinatorics | proof | Yes | Yes | olympiads | false | 55,212 |
Arzhenev $A$.
Petya and Vasya take turns writing fractions of the form $\$ 1 / n \$$ on the board, where $\$ n \$$ is a natural number, and Petya starts. Petya writes only one fraction per turn, while Vasya writes one fraction on his first turn, two fractions on his second turn, and so on, increasing the number of fra... | Lemma. Any number $\$ 2 \$$ can be represented as a sum of $\$ \mathrm{k} \$$ fractions of the form $\$ \frac{1}{\mathrm{n}} \$$ in a finite number of ways (if it can be represented at all).
Proof of the lemma. We will prove by induction on $\$ \mathrm{k} \$$.
For $\$ \mathrm{k}=1 \$$, the statement is obvious. Suppos... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,215 |
Bourbaki n.
Students from one school often gather in groups and go to an ice cream cafe. After such a visit, they quarrel so much that no two of them will eat ice cream together again. By the end of the year, it turned out that they could only go to the ice cream cafe alone in the future. Prove that if the number of v... | Our task can be formulated as follows.
Given p different elements and there is a system of m sets composed of these elements, and:
1) no set in our system contains all the elements at once;
2) any two of the given p elements occur in one of the sets of the system;
3) if two elements occur in one of the sets, then the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,217 |
On the plane, it is impossible to arrange seven lines and seven points such that three lines pass through each point and three points lie on each line. Prove this.
# | Suppose such an arrangement of seven points and seven lines exists. First, we will prove that any two of these points lie on one of the given lines. Indeed, if $A$ is one of these points, then three lines pass through $A$, and on each of these lines lie two of the given points (excluding $A$); thus, $A$ and any of the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,218 |
Beikkreve v.P.
The sum of the tangents of angles measuring $1^{\circ}, 5^{\circ}, 9^{\circ}, 13^{\circ}, \ldots, 173^{\circ}, 177^{\circ}$ is 45. Prove this. | Author: Berkokoayko s.T.
Let's use the well-known identity $\operatorname{tg} \alpha+\operatorname{tg}\left(\alpha+60^{\circ}\right)+\operatorname{tg}\left(\alpha-60^{\circ}\right)=3 \operatorname{tg} 3 \alpha$ (which can be verified by expressing both sides in terms of $\operatorname{tg} \alpha$). From this, it follo... | 45 | Algebra | proof | Yes | Yes | olympiads | false | 55,219 |
$$
\begin{aligned}
& \text { [ Numerical tables and their properties ] } \\
& \text { Pigeonhole Principle ] }
\end{aligned}
$$
A bus route contains 14 stops (including the two terminal stops). The bus can carry no more than 25 passengers at any time. Prove that during the bus trip from one end to the other,
a) there... | a) We will only consider passengers who are in the bus when it travels from the 7th stop to the 8th. These will be exactly the passengers traveling from stop number $i \leq 7$ to stop number $j \geq 8$. Consider a $7 \times 7$ square, where the rows are numbered $1, \ldots, 7$, and the columns are numbered $8, \ldots, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,222 |
9,10 Given a sequence of integers constructed as follows: $a_{1}$ is an arbitrary three-digit number, $a_{2}$ is the sum of the squares of its digits, $a_{3}$ is the sum of the squares of the digits of $a_{2}$, and so on. Prove that in the sequence $a_{1}, a_{2}, a_{3}, \ldots$ there will inevitably be either 1 or 4. | First, note that $a_{2} \leq 9^{2} \cdot 3=243$, and therefore, $a_{3} \leq 2^{2}+9^{2} \cdot 2=166$. If $100 \leq a_{3} \leq 166$, then $a_{4} \leq 1+6^{2}+9^{2}=118$, and if $100 \leq a_{4} \leq 166$, then $a_{5} \leq 2+64<100$. Therefore, it is sufficient to verify the required statement only for numbers not exceedi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,223 |
In a tournament, 25 chess players are going to participate. They all play at different levels, and the stronger player always wins when they meet.
What is the minimum number of games required to determine the two strongest players? | Complexity: 5 Classes: 10,11
Example. Let's organize a tournament according to the Olympic system in five rounds: first, 12 pairs will play, then 6, then 3. Four people will remain, two pairs will play, and then the winners will play against each other. Thus, the strongest will be determined. He played no more than fi... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,224 |
The quadratic trinomial $a x^{2}+b x+c$ is a perfect square for all integer $x$. Prove that then $a x^{2}+b x+c=(d x+e)^{2}$. | Let $f(x)=a x^{2}+b x+c$. Then $\sqrt{f(x+1)}-\sqrt{f(x)}=\frac{f(x+1)-f(x)}{\sqrt{f(x+1)}+\sqrt{f(x)}}=\frac{a(2 x+1)+b}{\sqrt{f(x+1)}+\sqrt{f(x)}}$, so $\lim _{x \rightarrow \infty}(\sqrt{f(x+1)}-\sqrt{f(x)})=\sqrt{a}$. For integer $x$, the number $\sqrt{f(x+1)}-\sqrt{f(x)}$ is an integer, so $\sqrt{a}=d$ is an integ... | ^{2}++=(+e)^{2} | Algebra | proof | Yes | Yes | olympiads | false | 55,225 |
In a rectangle with an area of 5 square units, nine rectangles, each with an area of one square unit, are placed. Prove that the area of the common part of some two rectangles is greater than or equal to 1/9.
# | Suppose the area of the common part of any two rectangles is less than $1 / 9$. We will show that then they occupy an area greater than 5. Number the rectangles arbitrarily. The first rectangle occupies an area of 1. Add the second rectangle. The area of the common part of the first and second rectangle is less than 1/... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,226 |
[ Convex hull and supporting lines (planes).]
On a plane, a finite set of points is given. Prove that there is a point in it that has no more than three nearest points to it from the same set.
# | Suppose we have constructed a set $M$ of points on the plane, in which each point has at least four nearest neighbors. Let $r$ be the smallest distance between its points. Consider the set $L \subset M$ of all points whose distance to their nearest neighbors is $r$; in the set $L$, obviously, each point will also have ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,228 |
Gaykov S. 5.
In space, there are $2 n$ points, no four of which lie in the same plane. $n^{2}+1$ segments are drawn with endpoints at these points. Prove that the drawn segments form
a) at least one triangle;
b) at least $n$ triangles. | a) Let's choose a point from which the largest number of segments emanate. Denote it by $A_{1}$, the ends of the segments emanating from it by $B_{1}, \ldots, B_{k}$, and the remaining points by $A_{2}, \ldots, A_{2 n-k}$. If there are no triangles, then there are no segments between the points $B_{1}, \ldots, B_{k}$, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,229 |
$:$ Perlin $A$.
In the city $N$, each resident has acquaintances that constitute no less than 30% of the city's population. A resident goes to the election if at least one of their acquaintances is running. Prove that it is possible to conduct the mayoral election in city $N$ with two candidates in such a way that at ... | Solution I is based on the following lemma. Lemma. Let $S$ be an arbitrary non-empty set of residents. Then in city $N$ there will be a resident who is acquainted with not less than $30\%$ of the residents from $S$. Denote by $|X|$ the number of residents in the set $X$. We estimate the total number of (ordered) pairs ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,233 |
9,10,11 |
| :---: | :---: | :---: |
| | Convex Polygons | |
| | Graph Traversal | |
Authors: Misnik S., von der Flaass D.
Inside a circle, points $A_{1}, A_{2}, \ldots, A_{n}$ are located, and on its boundary - points $B_{1}, B_{2}, \ldots, B_{n}$, such that the segments $A_{1} B_{1}$, $A_{2} B_{2}, \ldots, A_{n}... | Induction on $n$. The base case $(n \leq 2)$ is obvious.
Inductive step. Let $n \geq 3$. Without loss of generality, we can assume that the polygon $M$ with vertices $A_{1}$, $A_{2}, \ldots, A_{k}$ is the convex hull of the set of points $A_{1}, A_{2}, \ldots, A_{n}$. Consider segments of the form $A_{m} B_{m} (1 \leq... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,234 |
Karpov D.V.
In the microcircuit, there are 2000 contacts, and initially, any two contacts are connected by a separate wire. Vasya and Petya, the troublemakers, take turns cutting the wires, with Vasya (who starts) cutting one wire per turn, and Petya cutting either two or three wires. The troublemaker who cuts the las... | Send a comment
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,236 | |
Dopnnikov V.L.
In a certain group of 12 people, among any nine there are five who are pairwise acquainted. Prove that in this group there are six who are pairwise acquainted.
# | Let's take a graph on 12 vertices, which correspond to people, and two of its vertices are connected if the people are not acquainted.
If this graph has no cycles of odd length, then its vertices can be partitioned into two parts, in each of which the vertices will not be connected (see the lemma to problem 110038), a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,237 |
Tokarev S.i.
The participants of the test olympiad were offered $n$ questions. The jury determines the difficulty of each question: a positive integer number of points awarded to participants for a correct answer to the question. For an incorrect answer, 0 points are awarded, and all points earned by a participant are... | Let's prove that such a distribution of points can exist with $n$ participants. The example is obvious - let the $k$-th participant answer only the $k$-th question. Then, by assigning the cost of questions $a_{1}, a_{2}, \ldots, a_{n}$, where $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\} = \{1,2, \ldots, n\}$, the jury w... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,239 |
D.V.
In the country, there are several cities, some pairs of which are connected by roads, and between any two cities, there exists a unique non-self-intersecting path along the roads. It is known that there are exactly 100 cities in the country from which only one road leads out. Prove that it is possible to build 50... | Consider a graph where the vertices correspond to cities and the edges to roads. In this graph, there is a unique path between any two vertices, hence there are no cycles (it is a tree). According to the problem, this graph has 100 vertices from which exactly one edge emerges (hanging vertices) - let these vertices be ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,240 |
Authors: Dolnikov V.L., Bogdanov I.I.
On a plane, a finite number of red and blue lines are taken, among which there are no parallel lines, such that through each intersection point of monochromatic lines, a line of the other color passes. Prove that all lines pass through one point. | Assume the opposite. Note that through each intersection point of two lines, there passes a red line. Consider a blue line $l$; let $A, B$ be the two points of intersection of $l$ with red lines that are farthest apart, and let $m$ and $n$ be the red lines passing through $A$ and $B$, respectively. Let $C$ be the inter... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,241 |
Bverovo S.L.
In the president's office, there are 2004 telephones, any two of which are connected by a wire of one of four colors. It is known that wires of all four colors are present. Is it always possible to select several telephones such that among the wires connecting them, wires of exactly three colors are prese... | Let's construct a graph where the vertices correspond to phones, and the edges correspond to wires. Consider the smallest such set of vertices in this graph such that among the edges connecting these vertices, there are edges of all four colors. Remove an arbitrary vertex from this set. Since the set was the smallest, ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,242 |
Galochkina A.I.
For angles $\alpha, \beta, \gamma$ the equality $\sin \alpha+\sin \beta+\sin \gamma \geq 2$ holds. Prove that $\cos \alpha+\cos \beta+\cos \gamma \leq \sqrt{5}$. | Suppose the opposite: $\cos \alpha+\cos \beta+\cos \gamma>\sqrt{5}$. Then for the vectors $\vec{a}=(\sin \alpha, \cos \alpha), \vec{b}=(\sin \beta, \cos \beta)$ and $\vec{c}=$ $(\sin \gamma, \cos \gamma)$ we have: $3<\sqrt{(\sin \alpha+\sin \beta+\sin \gamma)^{2}+(\cos \alpha+\cos \beta+\cos \gamma)^{2}}=|\vec{a}+\vec{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 55,243 |
Gribamko A.V.
On a $10 \times 10$ checkerboard, there is a bacterium in one of the cells. In one move, the bacterium moves to an adjacent cell and splits into two (both remain in the same cell). Then, one of the bacteria sitting on the board moves to an adjacent cell and splits into two, and so on. Can it happen that ... | We will color the cells of the board in white and black in a checkerboard pattern. Consider the difference between the number of bacteria on white cells and the number of bacteria on black cells. When a move is made from a black cell to a white one, this difference increases by 3, and when a move is made from a white c... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,246 |
The number 76 has an interesting property: the last two digits of the number $76^{2}=5776$ are 76 again.
a) Are there any other two-digit numbers with this property?
b) Find all such three-digit numbers $A$ such that the last three digits of the number $A^{2}$ form the number $A$.
c) Does there exist an infinite seq... | a) If $A^{2}$ ends with the same two digits as $A$, then $A^{2}-A=A(A-1)$ is divisible by $100=25 \cdot 4$. Since the numbers $A$ and $A-1$ are coprime, one of them must be divisible by 25, and the other by 4. Let's try whether each of the numbers 25, 50, and 75 can serve as $A$ or $A-1$ (both these numbers are two-dig... | )25,76;b)376,625;)exists | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,247 |
Limakov L.G.
a) The bottom of a rectangular box was paved with tiles of sizes $2 \times 2$ and $1 \times 4$. The tiles were spilled out of the box, and one $2 \times 2$ tile was lost. Instead of it, a $1 \times 4$ tile was found. Prove that it is now impossible to pave the bottom of the box with the tiles.
b) Will th... | a) Let's divide the bottom of the box into $1 \times 1$ squares and mark those squares that are in odd positions in odd rows (see the figure).
Notice that each $1 \times 4$ tile, covering fo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,248 |
Janovskaya C.A.
In a certain set, an operation * is defined which, for any two elements $a$ and $b$ of this set, computes some element $a^{*} b$ of this set. It is known that: $1^{\circ}$. For any three elements $a, b$, and $c$, $a^{*}\left(b^{*} c\right)=b *\left(c^{*} a\right)$.
$2^{\circ}$. If $a^{*} b=a^{*} c$, t... | From conditions $1^{\circ}$ and $2^{\circ}$ follows commutativity: substituting $a$ for $b$ in $1^{\circ}$, we get that for any $a$ and $c$
$$
a^{*}\left(a^{*} c\right)=a^{*}\left(c^{*} a\right)
$$
and from this, according to $2^{\circ}$, it follows that $a^{*} c=c^{*} a$.
From condition $1^{\circ}$ and commutativit... | proof | Algebra | proof | Yes | Yes | olympiads | false | 55,250 |
[ [Geometric progression $]$
$[$ Principle of the extreme (other).]
There are $4 n$ positive numbers such that from any four pairwise distinct numbers, a geometric progression can be formed. Prove that among these numbers, there will be $n$ identical ones. | Let's show that among the given numbers, there cannot be more than four pairwise distinct numbers. Combine equal numbers into groups, select one number from each group, and arrange the selected numbers in descending order: $a>b>c>d>e>\ldots$. The numbers $a, b, c, d$ form a geometric progression by the condition. But $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,251 |
Given a triangle $A B C$. Let $A_{1}, B_{1}, C_{1}$ be the points of intersection of the lines $A S, B S, C S$ with the sides $B C, C A, A B$ of the triangle, respectively, where $S$ is an arbitrary internal point of the triangle $A B C$. Prove that, in at least one of the quadrilaterals $A B_{1} S C_{1}, C_{1} S A_{1}... | Suppose that in each of the quadrilaterals $A B_{1} S C_{1}, C_{1} S A_{1} B, A_{1} S B_{1} C$ at least one of the angles at each pair of vertices $C_{1}$ and $B_{1}, C_{1}$ and $A_{1}, A_{1}$ and $B_{1}$ is acute. For example, in the quadrilateral $A B_{1} S C_{1}$, the angle at vertex $C_{1}$ is acute. Then in the qu... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,253 |
[ The principle of the extreme (miscellaneous).]
If given a sequence of 15 numbers
$$
a_{1}, a_{2}, \ldots, a_{15},(1)
$$
then one can write a second sequence
$$
b_{1}, b_{2}, \ldots, b_{15},(2)
$$
where $b_{\mathrm{i}}(i=1,2,3, \ldots, 15)$ is equal to the number of numbers in sequence (1) that are less than $a_{... | Answer: No, it does not exist. Suppose the required sequence of numbers $a_{\mathrm{i}}$ exists. When the numbers $a_{\mathrm{i}}$ are permuted, the numbers $b_{\mathrm{i}}$ are permuted in the same way. Moreover, if the numbers $a_{\mathrm{i}}$ are arranged in ascending order, then the numbers $b_{\mathrm{i}}$ are als... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,254 |
The numbers $[a],[2 a], \ldots,[N a]$ are distinct from each other, and the numbers $\left[\frac{1}{a}\right],\left[\frac{2}{a}\right], \ldots,\left[\frac{M}{a}\right]$ are also distinct from each other. Find all such $a$. | Answer: $\frac{N-1}{N} \leq|a| \leq \frac{M}{M-1}$. The numbers $[x]$ and $[y]$ are different if and only if the numbers $[-x]$ and $[-y]$ are different. Therefore, it is sufficient to consider the case when $a>0$. If $a<\frac{N-1}{N}$, then among the numbers $[a],[2 a], \ldots,[N a]$ there are coinciding ones, since t... | \frac{N-1}{N}\leq||\leq\frac{M}{M-1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,255 |
A segment of length $3^{\mathrm{n}}$ is divided into three equal parts. The first and third of these are called marked. Each of the marked segments is then divided into three parts, of which the first and third are again called marked, and so on until segments of length 1 are obtained. The ends of all marked segments a... | Let's call a segment, both ends of which are marked points, a pseudo-marked segment. We will prove the statement of the problem by induction on $n$. For $n=1$ the statement is obvious. Suppose that for $n=m$ the statement of the problem is true, then we will prove it for $n=m+1$. Notice that in a segment of length $3^{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,257 |
10,11
The game board has the shape of a rhombus with an angle of $60^{\circ}$. Each side of the rhombus is divided into nine parts. Lines are drawn through the division points, parallel to the sides and the smaller diagonal of the rhombus, dividing the board into triangular cells. If a chip is placed on a certain cell... | Let's replace the board with an equivalent $9 \times 9$ square board, where diagonals of the same direction are drawn in all cells (see Fig. 1).
Six chips are sufficient to cover all cells (see Fig. 1).
=2 \cdot 2520>4500$. Therefore, there exists an $n=1,2, \ldots, 9$, such that among our numbers, there are several equal to $n$, with a sum of at least 100 (otherwise, the sum of all numbers does not exceed $100 \cdot(1+\ldots+9)=4500$). Let's mark the minimum... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,259 |
$\left[\begin{array}{l}\text { Recurrence Relations }] \\ {[\quad \underline{\text { Induction (other) }}]}\end{array}\right]$
Given an arbitrary set of +1 and -1 of length $2^{\mathrm{k}}$. From it, a new set is obtained according to the following rule: each number is multiplied by the one following it; the last $2^{... | Let's consider the first few steps of the investigated process. The initial set:
$$
a_{1}, a_{2}, a_{3}, a_{4}, \ldots, a_{2^{\mathrm{k}}}
$$
The set obtained after the first step:
$$
a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4}, \ldots, a_{2} \mathrm{k} a_{1}
$$
The set obtained after the second step:
$$
a_{1} a_{3}, a_... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,261 |
8,9
What is the maximum number of numbers that can be chosen from the set $1,2, \ldots, 1963$, so that the sum of no two numbers is divisible by their difference? | Let's first show that it is possible to select 655 numbers from the given set in the required manner. Indeed, let's take all numbers of the form $3k+1; k=0,1, \ldots, 654$. The difference between any two such numbers is obviously divisible by 3, while the sum of any two of them gives a remainder of 2 when divided by 3.... | 655 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,262 |
Given a sequence ..., $a_{- \text{p }}, \ldots, a_{-1}, a_{0}, a_{1}, \ldots, a_{\text{p }}, \ldots$ infinite in both directions, and each of its terms is equal to $\frac{1}{4}$ of the sum of its two neighbors. Prove that if any two of its terms are equal, then there are an infinite number of pairs of equal terms in it... | Let $a_{i}=a_{j}$ for some $i<j$. We will then prove that $a_{i+k}=a_{j-k}$ for all $k$ - from this the statement of the problem will follow. We will first prove this for $1 \leq k \leq j-i-1$. Let $k$ in this interval be such that the value $\left|a_{i+k}-a_{j-k}\right|$ is maximal. Then $4\left|a_{i+k}-a_{j-k}\right|... | proof | Algebra | proof | Yes | Yes | olympiads | false | 55,263 |
Induction (other). $\quad]$
[Integer and fractional parts. Archimedes' Principle]
Given a sequence of positive integers $X_{1}, X_{2} \ldots X_{n}$, all elements of which do not exceed some number $M$. It is known that for all $k>2$, $X_{\mathrm{k}}=\left|X_{\mathrm{k}-1}-X_{\mathrm{k}-2}\right|$. What can be the max... | Let's denote by $\lceil x\rceil$ the smallest integer not less than $x$, and by $\lfloor x\rfloor$ the largest integer not exceeding $x$. We will prove that the maximum length $L(M)$ of such a sequence is
$$
L(M)= \begin{cases}2, & \text { if } M=1 \\ \left\lceil\frac{3 M}{2}\right\rceil+1, & \text { if } M \neq 1\end... | L(M)=\begin{cases}2,& | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,265 |
Prove that $4^{m}-4^{n}$ is divisible by $3^{k+1}$ if and only if $m-n$ is divisible by $3^{k}$.
# | Let $m>n$. The number $4^{m}-4^{n}=4^{n}\left(4^{m-n}-1\right)$ is divisible by $3^{k+1}$ if and only if $4^{m-n}-1$ is divisible by $3^{k+1}$. We will prove by induction on $k$ that the smallest natural number $a$ for which $4^{a}-1$ is divisible by $3^{k+1}$ is $3^{k}$; and that $4^{a}-1$ is not divisible by $3^{k+2}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,268 |
$$
\left[\begin{array}{l}
[\text { Theory of algorithms (read) })] \\
{[\underline{\text { Induction (read) }}]}
\end{array}\right.
$$
Let's define the "complexity" of a number as the smallest length of a numerical sequence (if such a sequence exists) that starts with zero and ends with this number, where each subsequ... | Answer: The maximum "complexity" is achieved by the number $\left(\left(2^{51}+1\right) / 3\right) / 2^{50}$.
We will prove that among numbers of the form $m / 2^{\mathrm{k}}$, where $m=1,3,5, \ldots, 2^{\mathrm{k}}-1$, the maximum "complexity" (equal to $2 k$) is achieved by the number for which $m=m_{0}(k)=\left(2^{... | ((2^{51}+1)/3)/2^{50} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,269 |
Gorsky E.A.
On the graph of a polynomial with integer coefficients, two points with integer coordinates are marked.
Prove that if the distance between them is an integer, then the segment connecting them is parallel to the x-axis.
# | Let points $A(a, f(a))$ and $B(b, f(b))$ be marked. According to the Bézout's theorem for integer polynomials (see the solution of problem 35562$) f(b)-f(a)=k(b-a)$, where the number $k$ is an integer. The square of the distance between points $A$ and $B$ is $(b-a)^{2} + (f(b)-f(a))^{2} = (b-a)^{2}(1+k^{2})$. By the co... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,270 |
To some natural number, two two-digit numbers were appended sequentially to the right. The resulting number turned out to be equal to the cube of the sum of the three original numbers. Find all possible triples of the original numbers.
# | Let $a$ denote the first natural number, and $b$ and $c$ the two-digit numbers written after it. Let $x = a + b + c$. According to the condition,
$10^{4} a + 100 b + c = x^{3}$.
If $x \geq 100$, then $x^{3} \geq 10^{4} x = 10^{4}(a + b + c) > 10^{4} a + 100 b + c$, which means the equation has no solutions.
Therefor... | 9,11,25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,271 |
4 [ Scalar product ]
Find the equation of the plane containing the line $\frac{x-1}{2}=-\frac{y}{3}=3-z$ and parallel to the line of intersection of the planes $4 x+5 z-3=0$ and $2 x+y+2 z=0$.
# | Vector $\overrightarrow{m_{1}}=(2 ;-3 ;-1)$ is the direction vector of the first line. Let $x=t$ and solve the system of equations for $y$ and $z$
$$
\left\{\begin{array}{l}
5 z=-4 t+3 \\
y+2 z=-2 t
\end{array}\right.
$$
We get
$$
z=-\frac{4}{5} t+\frac{3}{5}, y=-2 z-2 t=\frac{8}{5} t-\frac{6}{5}-2 t=-\frac{2}{5} t-... | 2x-y+7z-23=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,272 |
[ Coordinate method in space ] [ Parametric equations of a line ] Through the point $M(-2 ; 0 ; 3)$, draw a line that intersects the lines
$$
\left\{\begin{array} { l }
{ x = 2 - t } \\
{ y = 3 } \\
{ z = - 2 + t }
\end{array} \left\{\begin{array}{l}
2 x-2 y-z-4=0 \\
x+3 y+2 z+1=0
\end{array}\right.\right.
$$ | Through the point $M(-2 ; 0 ; 3)$ and the first line, we will construct the plane $\alpha$. For this, we will take any point on this line, for example, $A(2 ; 3 ;-2)$, find the vector $\overrightarrow{n_{1}}=(a ; b ; c)$, perpendicular to the direction vector $\overrightarrow{m_{1}}=(-1; 0 ; 1)$ and the vector $\overri... | -2+13,-3,3-12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,273 |
[ Coordinate method in space ] [ Parametric equations of a line]
Given points $A(1 ; 0 ; 1), B(-2 ; 2 ; 1), C(2 ; 0 ; 3)$ and $D(0 ; 4 ;-2)$. Formulate the parametric equations of the line passing through the origin and intersecting the lines $A B$ and $C D$. | Through the point $O(0 ; 0 ; 0)$ and the line $A B$, we draw the plane $\alpha$. Let $\overrightarrow{n_{\mathbf{I}}}=(a ; b ; c)$ be a non-zero vector perpendicular to the plane $\alpha$. Then $\overrightarrow{n_{\mathbf{1}}} \cdot \overrightarrow{O A}=0$ and $\overrightarrow{n_{\mathbf{1}}} \cdot \overrightarrow{O B}... | 8,2,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,275 |
$\underline{\text { Folklore }}$
$a_{1}, a_{2}, \ldots, a_{101}$ - a permutation of the numbers $2,3, \ldots, 102$ such that $a_{k}$ is divisible by $k$ for each $k$. Find all such permutations. | Let's add $a_{102}=1$. We have obtained a permutation on the set $\{1,2, \ldots, 102\}$. It decomposes into cycles. But the smallest number in a non-trivial (containing more than one number) cycle stands at a position with a number greater than the number itself. According to the condition, such a number can only be 1,... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,276 |
$\underline{\text { Kisil V.V. }}$.
Prove for each natural number $n>1$ the equality: $\left[n^{1 / 2}\right]+\left[n^{1 / 3}\right]+\ldots+\left[n^{1 / n}\right]=\left[\log _{2} n\right]+\left[\log _{3} n\right]+\ldots+$ $\left[\log _{n} n\right]$. | Add $n$ to both parts. We will prove that in the resulting equality, both sides are equal to the number of pairs $(k, m)$ of natural numbers satisfying the inequality $k^{m} \leq n$.
Indeed, the integer part of a number $a \geq 1$ is the number of natural numbers less than or equal to $a$.
Therefore,
$\left[n^{1 / m... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,277 |
Folklore
On a strip of paper, 60 characters: "x" and "0" are written in a row. This strip is cut into pieces with symmetrical arrangements of characters. For example:
$0, \times \times, 0 \times \times \times 0, \times 0 \times, \ldots$.
a) Prove that there exists a way to cut the strip such that there are no more t... | a) Let's divide the strip into 12 strips of length 5. It is not difficult to check by enumeration that any such strip can be divided into two symmetric pieces.
b) It is sufficient to construct a strip in which there are no symmetric pieces longer than 4. For example, a strip consisting of 10 pieces of the form
$\time... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,278 |
Prove that there are infinitely many pairs of consecutive natural numbers, each of whose prime factorization contains every prime factor with a power of at least two. Examples of such pairs of numbers: (8, 9), $(288,289)$.
# | Let's call a number good if its factorization contains any prime factor with an exponent of at least 2. It is clear that the product of good numbers is a good number. Let's construct an infinite sequence $\{a_n\}$ by setting $a_1=8, a_{n+1}=4a_n(a_n+1)$. We will prove by induction that $a_n$ and $a_n+1$ are good number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,280 |
Fomin S.V.
Two chess players are playing chess with clocks (after making a move, a player stops their own clock and starts the opponent's clock). It is known that after both have made 40 moves each, the clocks of both players show the same time: 2 hours 30 minutes.
a) Prove that at some point during the game, the clo... | a) Let the modulus of the difference in the clock readings at any moment be less than $t={ }^{37} / 20$ minutes. Then the time spent by the first chess player on the first move is less than $t$, and by each player on any subsequent move - less than $2 t$. Therefore, the total duration of the game is less than $79 \cdot... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 55,281 |
Kushnirenko A.G.
On a plane, there are 20 points, no three of which lie on the same line, including 10 blue and 10 red points.
Prove that it is possible to draw a line such that on each side of it there are five blue and five red points. | Let $R(\varphi)$ denote the set of lines with an angle of inclination $\varphi$ (measured counterclockwise from the horizontal direction), on either side of which there is the same number (5 or 4) of red points. Similarly, for blue points, we define the set $B(\varphi)$. Each of the sets $R(\varphi)$ and $B(\varphi)$ f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,283 |
Ivanov C.
Given an $n \times n$ table filled with numbers according to the following rule: in the cell located in the $i$-th row and $j$-th column of the table, the number $\frac{1}{i+j-1}$ is written. In the table, $n$ numbers are crossed out in such a way that no two crossed-out numbers are in the same column or the... | The sum of the numbers in the top-left and bottom-right vertices of any rectangle is greater than the sum of the numbers in the other two vertices.
\(\frac{1}{i+j-1}+\frac{1}{(i+p)+(j+q)-1} \geq \frac{1}{(i+p)+j-1}+\frac{1}{i+(j+q)-1}\). Indeed, this inequality has the form
\(\frac{1}{a}+\frac{1}{a+p+q} \geq \frac{1}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,284 |
Markelov S.V.
Are the following statements true:
a) If a polygon can be divided by a broken line into two equal polygons, then it can be divided by a segment into two equal polygons.
b) If a convex polygon can be divided by a broken line into two equal polygons, then it can be divided by a segment into two equal pol... | a) Here is an example:
We will prove that the hexagon in the figure cannot be divided by a segment into two equal polygons. Indeed, such a cut must pass through vertex $B$ (otherwise, one po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,287 |
[alpererin.A.
a) Two people are performing a card trick. The first person takes five cards from a deck of 52 cards (previously shuffled by one of the spectators), looks at them, and then lays them out in a row from left to right, placing one card face down and the others face up. The second participant of the trick th... | At least two of the five cards have the same suit.
## Solution
a) The participants agree in advance on numbering all the cards in the deck from 1 to 52, and numbering all possible permutations of the letters $a, b, c, d, e$ from 1 to 120. After receiving five cards, the first person flips any one of them and finds th... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 55,288 |
Auto: Pooizvoooov B.B.
$2 n$ radii divide the circle into $2 n$ equal sectors: $n$ blue and $n$ red, alternating in some arbitrary order. In the blue sectors, starting from a certain one, t... | Let's go around the sectors counterclockwise. Somewhere immediately after the red sector $K$ stands the blue sector $S$. We can assume that $S$ contains the number 1 (we can shift all the numbers in the sectors by the necessary number of times in a "circle" $1 \rightarrow 2 \rightarrow \ldots \rightarrow n \rightarrow ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,289 |
Shapovalov A.B.
Thieves Hapok and Glazok are dividing a pile of 100 coins. Hapok grabs a handful of coins from the pile, and Glazok, looking at the handful, decides who of the two will get it. This continues until one of them gets nine handfuls, after which the other takes all the remaining coins (the division may als... | Here is Hapka's strategy, which will ensure him no less than 46 coins: take six coins each time (as long as it is possible). The pile contains 16 such full handfuls and four more coins. With this strategy, Hapka (regardless of how Glazok acts) has only two possibilities:
a) at some point, Glazok will have nine full ha... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,290 |
Avor: Shapovalov A.v.
A deck of 52 cards is laid out in a $13 \times 4$ rectangle. It is known that if two cards lie next to each other vertically or horizontally, then they are of the same suit or the same rank. Prove that in each horizontal row (of 13 cards) all cards are of the same suit. | Let's call a rectangle composed of cards homogeneous if it contains an equal number of cards of each suit in terms of parity. For example, a $4 \times 13$ rectangle is homogeneous: it contains 13 cards of each suit. We will prove that any $2 \times 2$ square is homogeneous. Let's consider three cases.
1) Some card in ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,292 |
8,9 The area of a triangle is $6 \sqrt{6}$, its perimeter is 18, and the distance from the center of the inscribed circle to one of the vertices is $\frac{2 \sqrt{42}}{3}$. Find the smallest side of the triangle. | If the inscribed circle touches the side $A C$ of triangle $A B C$ at point $M$, and $p$ is the semiperimeter of the triangle, then $A M=p-B C$. Use this equality, and then apply the Law of Cosines.
## Solution
Let $O$ be the center of the circle inscribed in the given triangle $A B C$, $r$ its radius, $S=6 \sqrt{6}$... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,293 |
Kai was given a whole box of figurines in the shape of a "pedestal" (see the picture).
a) Can he tile an $8 \times 8$ chessboard with them?
b) And a $10 \times 10$ board?
 | a) Kai can tile an $8 \times 8$ board, for example, as shown in the figure:
b) However, he will not be able to tile a $10 \times 10$ board. Let's color this board in a checkerboard pattern.... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,294 |
Petya colored one cell of the rectangle. Sasha can color other cells of this rectangle according to the following rule: any cell can be colored if it has an odd number of colored neighbors (by side). Will Sasha be able to color all the cells of the rectangle (regardless of which cell Petya chose), if the size of the re... | a) First, we will color a row of 9 cells, containing the initially colored cell (see the first figure). Then we will paint every other column, starting the coloring from the painted row (see the second figure). After this, coloring the remaining cells of the board is quite simple.
Yes;b)No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,295 |
Participants in a chess tournament played one game against each other. For each participant A, the number of points scored (1 point for a win, $-1 / 2$ point for a draw, 0 points for a loss) and the strength coefficient were calculated using the formula: the sum of the points of the participants A defeated, minus the s... | Let the total points scored by the $i$-th participant be denoted by $S_{i}$, and their strength coefficient by $F_{i}$.
a) The first method. It is sufficient to prove that $\sum_{i} F_{i} \leq 0$. Let $S$ be the average number of points scored (this number of points will be for a player who finished all games in a dra... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,296 |
[ Weighings [Pairing and Grouping; Bijections]
An antiques dealer has acquired 99 old coins that look identical. He was told that exactly one of the coins is counterfeit - lighter than the genuine ones (and the genuine ones weigh the same). How, using a balance scale without weights, can the counterfeit coin be identi... | First, we will place 13 coins on each pan of the scales, then (if the scales are in balance) remove them and place 11 from those not yet used, then 9, 7, 5, 3, and 1 until one of the pans tips.
If this does not happen, then after the seventh weighing (when there will be only one coin on each pan), only one coin will r... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,297 |
On a circular railway, there are $n$ stations. Sometimes the station masters communicate with each other by radio. At any given time, only two people are in a communication session. Over the course of a day, exactly one radio session occurred between every two stations. For each station (considering only its sessions),... | The order in which four stations can communicate by radio: $A B, C D, B D, A C, A D, B C$.
Let's prove that five stations cannot communicate in the manner specified in the problem. Suppose the stations are located at the vertices of a pentagon $A B C D E$. Note that the first communication can only occur between two a... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,300 |
$\begin{aligned} & {\left[\begin{array}{l}\text { Method of coordinates in space } \\ \text { [Distance between two points. Equation of a sphere }]\end{array}\right]}\end{aligned}$
Find the distance from the point $M 0(x 0 ; y 0 ; z 0)$ to the plane $A x+B y+C z+D=0$. | Through the given point $M 0$ we draw a line perpendicular to the given plane. As the direction vector of this line, we take the vector $\vec{m}=(A ; B ; C)$. Then the parametric equations of the line are
$$
\left\{\begin{array}{l}
x-x_{0}=A t \\
y-y_{0}=B t \\
z-z_{0}=C t
\end{array}\right.
$$
Substituting $x, y$ an... | \rho=\frac{|Ax_{0}+By_{0}+Cz_{0}+D|}{\sqrt{A^{2}+B^{2}+C^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,301 |
[ Trigonometric Identities. ]
Show that $\sin 36^{\circ}=1 / 4 \sqrt{10-2 \sqrt{5}}$.
# | If the center of a regular decagon is connected to its vertices, it will be divided into 10 isosceles triangles with a vertex angle of $36^{\circ}$ and base angles of $72^{\circ}$. Consider one of these triangles (see figure) $\angle A O B=36^{\circ}, \angle O A B=\angle O B A=72^{\circ}, A B=a, A O=B O=b$. Draw the an... | \sin36=\frac{1}{4}\sqrt{10-2\sqrt{5}} | Algebra | proof | Yes | Yes | olympiads | false | 55,302 |
Solve the equation $\left(x^{2}-x+1\right)^{4}-10 x^{2}\left(x^{2}-x+1\right)^{2}+9 x^{4}=0$. | Let $y=\left(x^{2}-x+1\right)^{2}$, then $y^{2}-10 x^{2} y+9 x^{4}=0$. Solving this equation for $y$, we get: $y_{1}=9 x^{2}, y_{2}=$ $x^{2}$.
Thus, the given equation has been reduced to the following two: $\left(x^{2}-x+1\right)^{2}=9 x^{2}$ and $\left(x^{2}-x+1\right)^{2}=x^{2}$, which in turn lead to four quadrati... | -1,1,2-\sqrt{3},2+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,303 |
$\underline{\text { Fon-err-Flaass problem: }}$
A square board is divided by a grid of horizontal and vertical lines into $n^{2}$ unit squares. For what largest $n$ can one mark $n$ cells so that every rectangle of area at least $n$ with sides along the grid lines contains at least one marked cell? | Evaluation. It is clear that if $n$ cells are marked in such a way that the condition of the problem is satisfied, then in each row and each column there is exactly one marked cell. Assuming that $n \geq 3$ (it is obvious that $n=2$ is not the largest), let's take row $A$, in which the leftmost cell is marked, row $B$,... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,305 |
$:$ Sadikov $P$.
On a coordinate plane, there are four chips, the centers of which have integer coordinates. It is allowed to move any chip by a vector connecting the centers of any two of the remaining chips. Prove that by several such moves, any two predetermined chips can be made to coincide. | Lemma. If three chips lie on the same straight line and have integer coordinates, then any two of them can be combined.
Let the smallest distance among the three pairwise distances be between chips $A$ and $B$. Then, move the third chip $C$ several times by the vector $\overrightarrow{A B}$ or $\overrightarrow{B A}$ s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,307 |
Sonkin $M$.
Solve the equation $\left(x^{2}-y^{2}\right)^{2}=1+16 y$ in integers. | Clearly, $y \geq 0$, so the left side is not less than $(2 y-1)^{2}$, since the absolute difference between $y^{2}$ and any perfect square (if $y \geq 0$ and the squares are different) is not less than $|2 y-1|$. Therefore, $(2 y-1)^{2} \leq 1+16 y$, from which $y \leq 5$. Thus, the right side can take the values $1, 1... | (\1,0),(\4,3),(\4,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,309 |
Maiinniova E.
Let $P(x)$ be a quadratic trinomial with non-negative coefficients.
Prove that for any real numbers $x$ and $y$, the inequality $(P(x y))^{2} \leq P\left(x^{2}\right) P\left(y^{2}\right)$ holds. | Let $P(x)=a x^{2}+b x+c$. Then
$$
\begin{gathered}
(P(x y))^{2}-P\left(x^{2}\right) P\left(y^{2}\right)=\left(a x^{2} y^{2}+b x y+c\right)^{2}-\left(a x^{4}+b x^{2}+c\right)\left(a y^{4}+b y^{2}+c\right)= \\
=a^{2} x^{4} y^{4}+b^{2} x^{2} y^{2}+c^{2}+2 a b x^{3} y^{3}+2 a c x^{2} y^{2}+2 b c x y-a^{2} x^{4} y^{4}-b^{2... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 55,310 |
Auto: Shapovev A.V.
In a class of 33 students, each student was asked how many namesakes and how many namesakes by surname (including relatives) they have in the class. It turned out that among the numbers mentioned, all integers from 0 to 10 inclusive were encountered. Prove that there are two students in the class wi... | Let's combine students into groups by surnames and into groups by first names (groups consisting of a single person are possible - for example, a student without any namesakes). Each will be in two groups - by surname and by first name. From the condition of the problem, it follows that there are exactly eleven groups ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,311 |
Berrovv s.l.
Given a natural number $n>3$ and positive numbers $x_{1}, x_{2}, \ldots, x_{n}$, the product of which is 1.
Prove the inequality $\frac{1}{1+x_{1}+x_{1} x_{2}}+\frac{1}{1+x_{2}+x_{2} x_{3}}+\ldots+\frac{1}{1+x_{n}+x_{n} x_{1}}>1$ | $$
\begin{aligned}
& \frac{1}{1+x_{1}+x_{1} x_{2}}+\frac{1}{1+x_{2}+x_{2} x_{3}}+\ldots+\frac{1}{1+x_{n}+x_{n} x_{1}}> \\
& \quad>\frac{1}{1+x_{1}+x_{1} x_{2}+x_{1} x_{2} x_{3}+\ldots+x_{1} x_{2} \ldots x_{n-1}}+\frac{1}{1+x_{2}+x_{2} x_{3}+\ldots+x_{2} x_{3} \ldots x_{n}}+ \\
& \ldots+\frac{1}{1+x_{n}+x_{n} x_{1}+\ldo... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 55,313 |
Rubanov I.S.
What is the maximum finite number of roots that the equation
$$
\left|x-a_{1}\right|+. .+|x-a 50|=\left|x-b_{1}\right|+. .+|x-b 50|
$$
can have, where $a_{1}, a_{2}, \ldots, a_{50}, b_{1}, b_{2}, \ldots, b_{50}$ are distinct numbers? | Let $f(x)=\left|x-a_{1}\right|+\ldots+|x-a 50|-\left|x-b_{1}\right|-. .-|x-b 50|$ and rewrite the original equation in the form $f(x)=0$.
Let $c_{1}<c_{2}<. .<c 100$ be all the numbers from the set $\left\{a_{1}, \ldots, a 50, b_{1}, . ., b 50\right\}$, ordered in ascending order. On each of the 101 intervals $\left[-... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,314 |
Kulikov E.Yu.
In a $2 \times n$ table, positive numbers are arranged such that in each of the $n$ columns, the sum of the two numbers is 1. Prove that it is possible to cross out one number in each column so that in each row, the sum of the remaining numbers does not exceed ${ }^{(n+1) / 4}$. | Let the numbers $a_{1}, a_{2}, \ldots, a_{n}$ be in the top row. By rearranging the columns, we can assume that $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$.
Then in the bottom row, we have
$b_{1}=1-a_{1}, b_{2}=1-a_{2}, \ldots, b_{n}=1-a_{n};$ clearly, $b_{1} \geq b_{2} \geq \ldots \geq b_{n}$. If $a_{1}+a_{2}+\ldots+a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,315 |
G. Gooovanov A.C.
Sequences of positive numbers $\left(x_{n}\right)$ and $\left(y_{n}\right)$ satisfy the conditions
$x_{n+2}=x_{n}+x_{n+1}^{2}, y_{n+2}=y_{n}^{2}+y_{n+1}$ for all natural $n$. Prove that if all numbers $x_{1}, x_{2}, y_{1}, y_{2}$ are greater than 1, then $x_{n}>y_{n}$ for some natural $n$. | Obviously, starting from the second term, our sequences increase:
$x_{n+2}>x_{n+1}^{2}>x_{n+1}, y_{n+2}>y_{n+1}$. Since $x_{3}>1+1^{2}=2$,
$y_{3}>1^{2}+1=2$, all terms of each sequence, starting from the third, are greater than 2. Similarly, for $n>3$ we get $x_{n}>3, y_{n}>3$.
Now notice that $x_{n+2}>x_{n+1}^{2}>x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 55,316 |
Rubanov I.S.
There are $n$ matches on the table ($n>1$). Two players take them off the table in turns. On the first move, a player can take any number of matches from 1 to $n-1$, and then each time no more matches can be taken than the partner took on the previous move. The one who takes the last match wins. Find all ... | If $n$ is odd, the first player wins by taking one matchstick on the first move: from then on, both players are forced to take one matchstick at a time, and the last move will be the first player's.
If $n$ is even, the player who first takes an odd number of matchsticks loses, as they leave an odd number of matchstick... | n\neq2^{k}(k\in{N}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,317 |
Kanel-Belov A.Y.
Let's call a combination of digits several digits written in a row. In the country of Robotland, some combinations of digits are declared prohibited. It is known that the number of prohibited combinations is finite and there exists an infinite decimal fraction that does not contain prohibited combinat... | The number of forbidden digit combinations is finite, so there is a number $N$ such that all forbidden digit combinations are no longer than $N$ characters. In an infinite decimal fraction, one can find two identical segments of length $N$. Let the segments $a_{k} \ldots a_{k+N-1}$ and $a_{m} \ldots a_{m+N-1}$ of the a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,319 |
Fomin $A$.
Given a set consisting of 1997 numbers such that if each number in the set is replaced by the sum of the others, the resulting set is the same.
Prove that the product of the numbers in the set is 0. | Let the sum of the numbers in the set be $M$, then the number $a$ from the set is replaced by the number $b=M-a$. Summing these equations for all $a$:
$b_{1}+\ldots+b_{1997}=1997 M-\left(a_{1}+\ldots+a_{1997}\right)$, from which $M=0$, since $b_{1}+\ldots+b_{1997}=a_{1}+\ldots+a_{1997}=M$. Therefore, for any $a$ the n... | proof | Algebra | proof | Yes | Yes | olympiads | false | 55,320 |
Maiinnikova E.
At the enterprise, 50,000 people are employed. For each of them, the sum of the number of their immediate supervisors and their immediate subordinates is 7. On Monday, each employee of the enterprise issues an order and gives a copy of this order to each of their immediate subordinates (if any). Then, e... | If on an enterprise to the top executives, then each employee should eventually see at least one of the $k$ orders issued by these executives on Monday. On Monday, no more than $7 k$ employees saw them, on Tuesday - no more than $7 k \cdot 6$, on Wednesday - no more than $7 k \cdot 36$ employees. All those who saw thes... | 97 | Combinatorics | proof | Yes | Yes | olympiads | false | 55,321 |
Polsky o.k.
At the ends of a grid strip measuring $1 \times 101$ cells, two chips are placed: the first player's chip on the left and the second player's chip on the right. On a turn, it is allowed to move one's chip towards the opposite end of the strip by $1, 2, 3,$ or $4$ cells. It is allowed to jump over the oppon... | The first one wins. First, he needs to make moves of 4 cells until he reaches the 45th cell. Now it's the second's turn. If he also always makes moves of 4 cells, his next move will take him to cell 57. Then the first should move to cell 48 on his next turn and after the second's response, he should move so that there ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,322 |
$:$ enenorovo B.A.
Do there exist real numbers $a, b$, and $c$ such that for all real $x$ and $y$, the inequality
$$
|x+a|+|x+y+b|+|y+c|>|x|+|x+y|+|y| ?
$$
holds? | Suppose such numbers $a, b$, and $c$ exist. Choose $x>0$ and $y>0$ such that $x+a \geq 0, x+y+b \geq 0, y+c \geq 0$. Then the difference between the left and right sides is $a+b+c$. If we take $x0$, on the other hand, $a+b+c<0$. Contradiction.
## Answer
No.
Send a comment | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 55,323 |
The labyrinth is an $8 \times 8$ square, with each $1 \times 1$ cell containing one of four arrows (up, down, right, left). The top side of the top-right cell is the exit from the labyrinth. A token is placed in the bottom-left cell and moves one cell in the direction indicated by the arrow on each move. After each mov... | Suppose the chip never leaves the maze. Then the chip will land on cell number 1 (see fig.) a finite number of times (less than four), because otherwise, when the arrow points to the exit, the chip will leave the maze.
![](https://cdn.mathpix.com/cropped/2024_05_06_49e49f22191b45b40a57g-46.jpg?height=260&width=274&top... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 55,324 |
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