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13. In some cells of a $50 \times 50$ square table, numbers +1 and -1 are placed in such a way that the absolute value of the sum of all numbers in the table does not exceed 100. Prove that in some $25 \times 25$ square, the absolute value of the sum of the numbers does not exceed 25.
## 8th grade | 90.13. Let's divide the square into four quarters. If the sums of the numbers in them have the same sign, then it is obvious that one of them in absolute value is not greater than 25. If there are two quarters, the sums of the numbers in which (let's say, $a$ and $b$) have different signs, then we will connect these tw... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,337 |
17. In the country of Faraway, there are 101 cities; the cities are connected by one-way roads such that any two cities are connected by no more than one road. It is also known that exactly 40 roads lead out of each city and exactly 40 roads lead into each city. Prove that from any city to any other, one can travel by ... | 90.17. Consider two arbitrary cities $A$ and $B$. Suppose there is no road $A \rightarrow B$, and also, there is no city $C$ such that there are roads $A \rightarrow C$ and $C \rightarrow B$. Let's find 40 cities $A_{1}$, $A_{2}, \ldots, A_{40}$, to which roads lead from $A$, and another 40 cities $B_{1}, B_{2}, \ldots... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,340 |
18. Among 103 coins, two are counterfeit and differ in weight from the genuine ones. It is known that all genuine coins weigh the same, as do the two counterfeit coins. Determine in three weighings on a two-pan balance without weights which is heavier: the genuine or the counterfeit coin. | 90.18. Let's divide the coins into three piles of 34 coins each and compare the weight of the first and second piles, and then the second and third. It is clear that all of them cannot be equal. Suppose that the weights of the piles are different in the first weighing, but equal in the second, and for definiteness, let... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,341 |
19. On the Island of Logic, every person is either a liar, who always tells lies, or a knight, who always tells the truth. Each islander said the following two phrases:
all my acquaintances are acquainted with each other;
among my acquaintances, there are no fewer liars than knights.
Prove that on the island, there ... | 90.19. We will select several liars (let's call them main liars) such that no two main liars are acquainted with each other, and every liar is acquainted with at least one main liar. This can be done as follows: consider any liar and declare them a main liar. Then, if there is at least one liar who is not acquainted wi... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,342 |
20. How many pairs of natural numbers ( $m, n$ ) exist such that $m, n \leqslant 1000$ and
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} ?
$$
## 9th grade | 90.20. Answer. 1706. Hint. The pair ( $m, n$ ) satisfies the inequalities from the condition if and only if the line $y=\sqrt{2 x}$ intersects a cell on the coordinate grid plane, the lower left corner of which has coordinates $(m, n)$. | 1706 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,343 |
23. Prove that in any arithmetic progression, the terms of which are natural numbers, there are two terms with the same sum of digits. | 90.23. Let $A$ be any term of an arithmetic progression, and $d$ be its common difference. Then all terms of the progression of the form $A+10^{k} d$, where $10^{k}>A$, have the same sum of digits. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,344 |
24. In a convex quadrilateral $A B C D$, angle $B$ is a right angle, and the length of diagonal $A C$, which is the bisector of angle $A$, is equal to the length of side $A D$. In triangle $A D C$, the altitude $D H$ is drawn. Prove that line $B H$ bisects segment $C D$. | 90.24. The equality of triangles $A H D$ and $A B C$ gives us the equality $A B=A H$. From this, the isosceles triangles $A B H$ and $A C D$ are similar, and therefore $\angle H C D=\angle B H A=\angle C H K$, i.e., $C K=$ $=H K$. Since $\angle K D H=90^{\circ}-\angle H C D=90^{\circ}-\angle C H K=\angle K H D$, then $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,345 |
32. Can a plane be covered without overlap using squares with side lengths $1,2,4,8,16, \ldots$, if each square is allowed to be used no more than: a) 10 times; b) 1 time?
## 10th grade
(Physics and Mathematics Schools) | 90.32. Answer. Variant $a$: can; variant b: cannot. Hint. Variant b: prove that each vertex of any square lies on the side of a larger square. Now consider 264
squares with the minimum side length involved in the "tiling". Then for one of the four squares adjacent to it, the specified property is not satisfied. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,348 |
35. Given a polynomial $F$ with integer coefficients such that $F(2)$ is divisible by five, and $F(5)$ is divisible by two. Prove that $F(7)$ is divisible by 10. | 90.35. Hint. Using the fact that $F(n)-F(m)$ is divisible by $n-m$, prove that $F(7)$ is divisible by both two and five. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,349 |
43. Natural numbers $a$ and $b$ are such that $a^{2}+a b+1$ is divisible by $b^{2}+b a+1$. Prove that $a=b$.
10 Order No. 64
145 | 90.43. From the identity $b\left(a^{2}+a b+1\right)-a\left(b^{2}+b a+1\right)=b-a$ it follows that $a-b$ is divisible by $b^{2}+b a+1$, which is only possible when $a-b=0$. | b | Number Theory | proof | Yes | Yes | olympiads | false | 37,351 |
45. On the side $B C$ of the square $A B C D$, an arbitrary point $P$ is taken, and a circle is drawn through $A, B$ and $P$, intersecting the diagonal $B D$ again at point $Q$. A circle is drawn through $C, P$ and $Q$, which intersects $B D$ again at point $R$. Prove that the points $A, R$ and $P$ lie on the same line... | 90.45. Consider the point $R^{\prime}=A P \cap B D$ and prove that the quadrilateral $C P R^{\prime} Q$ is cyclic. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,352 |
47. The vertices of the inscribed quadrilateral $ABCD$ are located at the nodes of a grid on a sheet of graph paper (with a cell side length of 1). It is known that $ABCD$ is not a trapezoid. Prove that $|AC \cdot AD - BC \cdot BD| \geqslant 1$. | 90.47. Let's use a well-known and easily provable fact: the area of any triangle with vertices at grid nodes is a number of the form $n / 2$, where $n$ is a natural number. Then $A C \cdot A D \sin \angle D A C$ is equal to $B C \cdot B D \sin \angle D B C$ - an integer. Therefore, $|A C \cdot A D - B C \cdot B D| = m ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,353 |
49*. The hostess baked a pie for the guests. At the table, there can be either $p$ people or $q$, where $p$ and $q$ are coprime. Into what minimum number of pieces (not necessarily equal) should the pie be cut in advance so that in any case it can be distributed equally? | 90.49. Answer: $p+q-1$. Constructing an example is left to the reader, while we will prove that it is impossible to cut the cake into fewer pieces. Suppose all $p+q$ guests have gathered in the room (we will assume that among the guests who were supposed to come in the number of $p$ people, there are no guests who were... | p+q-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,355 |
50*. Twenty numbers are arranged in a circle. It is allowed to replace a triplet of consecutive numbers $x, y, z$ with the triplet $x+y, -y, z+y$ (in that exact order). Is it possible to use these operations to transform the set $\{1,2,3, \ldots, 9,10, -1,-2,-3, \ldots,-9,-10\}$ into the set $\{10,9,8, \ldots, 2,1, -10... | 90.50. Answer. Yes, it is possible. Hint. Consider a set of 20 integers $x_{1}, x_{2}, \ldots, x_{20}$ such that the differences $x_{2}-x_{1}, x_{3}-x_{2}, \ldots, x_{1}-x_{20}$ give exactly the set of numbers standing on the circle; determine how the transformation mentioned in the problem affects the set $\left\{x_{k... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,356 |
56*. On a shelf, a 100-volume collection of L. N. Tolstoy's works stands in a random order. It is allowed to take any two volumes with numbers of different parity and swap their places. What is the minimum number of such swaps required to always arrange the volumes in order? | 90.56. Answer. 124 permutations. Hint. Break down the entire permutation of volumes into cycles. We will call the operation of exchanging volumes from different cycles merging, and the operation of exchanging volumes from the same cycle - breaking. Prove that by applying no more than 25 merging operations and no more t... | 124 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,357 |
57*. Given a polynomial $F(x)$ with integer coefficients, and it is known that $F(n)$ is divisible by one of the integers $a_{1}$, $a_{2}, \ldots, a_{m}$ for any integer $n$. Prove that one can choose one of these numbers such that for any integer $n$, $F(n)$ will be divisible by it. | 90.57. Let's prove this by contradiction. Suppose there are numbers $x_{1}$, $x_{2}, \ldots, x_{m}$ such that $F\left(x_{k}\right)$ is not divisible by $a_{k}$ for any $k=1$, $2, \ldots, m$. This means that there exist numbers $d_{k}=p_{k}^{\alpha^{k}}$, where $p_{k}$ are prime numbers such that $a_{k}$ is divisible by... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,358 |
60. A segment contains several smaller segments that cover it. From each of them, half was discarded: the left or the right. Prove that the remaining halves cover at least one third of the length of the original segment. | 90.60. Instruction. Let's sequentially exclude segments that are entirely contained within others. Now, let's subject each segment, again sequentially, to a homothety with its center at the midpoint and with the minimum possible coefficient—such that the segments still cover the original. Then each segment will have co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,359 |
62. For a $100 \times 100$ checkerboard, the top edge was glued to the bottom edge, and the right edge to the left edge, after which the board took the shape of a donut. Can 50 rooks - red, blue, and green - be placed in its cells such that each red rook attacks at least two blue rooks, each blue rook attacks at least ... | 90.62. Answer. No, it cannot. Hint: Prove that the number of red, blue, and green rooks in such an arrangement must be equal. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,361 |
64*. In the kingdom of Olympia, there are 100 cities, and each pair of them is connected by exactly one one-way road. It turned out that not from every city can one travel to any other city while adhering to the traffic rules. Prove that the king can choose one of the cities and, by changing the directions of all roads... | 90.64. Instruction: Prove this by induction on the number of cities $n$, starting with $n=7$. For the inductive step, use the lemma: there exists a city from which at least two roads lead out and into which at least two roads lead.
Please note that the proof of the base case in this instance is arguably the most chall... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,363 |
1. In a heavy rocketry club, 40 schoolchildren are enrolled. Each of them has bolts, nuts, and nails. It is known that the number of club members who have a different number of nails than bolts is exactly 15 people. The number of those who have the same number of nuts as nails is 10. Prove that there are at least 15 cl... | 91.1. It is clear that the number of club members who have the same number of nails as bolts, but a different number of screws than nails, is no less than $40-15-10=15$ people. On the other hand, for each such club member, the number of screws is not equal to the number of bolts. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,366 |
2. On the black market in the village of Perestroyno, if you try hard enough, you can exchange any two food coupons for three others and vice versa. Can the cooperator Vasya exchange 100 butter coupons for 100 sausage coupons, giving away exactly 1991 coupons in the process? | 91.2. Answer. No, it cannot. Hint. The number of exchanges of type three - for two must be equal to the number of exchanges two - for three. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,367 |
5. There are three identical meter-long rods: red, blue, and white. Kolya breaks the first rod into three pieces, after which Vasya does the same with the second rod. Finally, Kolya breaks the third rod into three pieces as well. Can Kolya act in such a way that regardless of how Vasya breaks the rod, it is possible to... | 91.5. Answer. Yes, he can. He needs to break both rods (the first and the third) in the same way: each into parts with lengths of 50 cm, 25 cm, and 25 cm. | Yes,hecan.Heneedstobreakbothrods(thefirstthethird)inthesame:eachintopartswithlengthsof50,25,25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,369 |
6. Nine teams played a single round-robin volleyball tournament. Is the following property necessarily true: there exist two teams $A$ and $B$ such that any other team lost to either team $A$ or team $B$ in the tournament?
## 7th grade | 91.6. No, not necessarily. Consider a tournament where teams $A_{1}, A_{2}, A_{3}$ each earned one point in their matches against each other, just like the teams in the triplets $B_{1}, B_{2}, B_{3}$ and $C_{1}, C_{2}, C_{3}$. Then all teams $A_{i}$ won against all teams $B_{j}$, all teams $B_{j}$ won against all teams... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,370 |
12. In a certain state, every two cities are connected by exactly one road: either a highway or a railway. Prove that one can choose a single mode of transport - car or train - such that from any city to any other city, one can travel by passing through no more than two other cities and using only this mode of transpor... | 91.12. Suppose that it is impossible to travel from $A$ to $B$ by railway with no more than two transfers, and it is impossible to travel from $C$ to $D$ by highway, stopping in no more than two other cities along the way (some of the four cities mentioned may coincide). Let us then consider only the roads connecting c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,374 |
17. From the vertex $A$ of triangle $A B C$, perpendiculars $A X$ and $A Y$ are dropped to the bisectors of the external angles $B$ and $C$. Prove that the length of the segment $X Y$ is equal to the semiperimeter of triangle $A B C$. | 91.17. Instruction. Extend the segments $A X$ and $A Y$ until they intersect the line $B C$ at points $K$ and $L$. Prove that the segment $X Y$ is the midline of triangle $A K L$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,377 |
20. Given a set $A$ of 10 distinct real numbers, one can construct a set $A(5)$ consisting of all possible sums of five different numbers from the set $A$. Do there exist two different sets $A$ and $B$ such that the sets $A(5)$ and $B(5)$ are identical?
## 9th grade | 91.20. Answer. Yes, such sets exist. For example, the sets $A=\{1,2, \ldots, 9,-45\}$ and $B=\{-1,-2, \ldots,-9, 45\}$ are suitable. For any number $x$ from the set $A(5)$, it is clear that there are five numbers in the set $B$ whose sum is equal to $-x$. Then the sum of the other five numbers from the set $B$ is equal... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,380 |
22. In an acute-angled triangle $ABC$, $AB > BC$. Points $X$ and $Y$ are taken on sides $AB$ and $BC$ respectively such that $AX = BY$. Prove that $XY \geqslant AC / 2$.
150 | 91.22. Let's conduct an additional construction - consider a line parallel to $BC$, passing through point $A$. Mark a point $Z$ on it such that $AZ = BX$. It is easy to see that $YZ \geqslant AC$ and by the triangle inequality, $XZ + XY \geqslant YZ \geqslant AC$. Since triangles $AXZ$ and $BXY$ are equal, then $XY = \... | 2XY\geqslantAC | Geometry | proof | Yes | Yes | olympiads | false | 37,381 |
23. A triangle has integer side lengths $x, y$, and $z$, and it is known that the length of one of its altitudes is equal to the sum of the lengths of the other two altitudes. Prove that $x^{2}+y^{2}+z^{2}$ is a perfect square. | 91.23. Hint. Prove that $x^{2}+y^{2}+z^{2}=(x+y-z)^{2}$, where $z$ is the smallest side of the triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,382 |
24. The sequence of natural numbers $\left(a_{n}\right)$ is constructed according to the following rule. Each of its members with an even index $a_{2 n}$ is obtained from $a_{2 n-1}$ by subtracting one of its digits, and each member with an odd index $a_{2 n+1}$ is obtained from $a_{2 n}$ by adding one of the digits of... | 91.24. Instruction: Check that if $a_{1} \leqslant 10^{p}$, no term of the sequence can be greater than $10^{p}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,383 |
26. At the symposium, each delegate is acquainted with at least one of the other participants, but for any two delegates, there is a third who is not acquainted with either of them. Prove that all delegates can be divided into three groups such that each participant of the symposium is acquainted with at least one pers... | 91.26. Consider a friendship graph and remove all unnecessary edges, i.e., leave only as many as needed to ensure that each connected component remains connected. After this, there will be a vertex \(X_{0}\) connected to exactly one vertex \(Y\), and we can consider the vertices \(X_{0}, X_{1}, \ldots, X_{k}\) that are... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,384 |
29. Two circles with centers $O_{1}$ and $O_{2}$ intersect at points $A$ and $B$. The circle ($O_{1} B O_{2}$) intersects the second circle again at point $P$. Prove that points $O_{1}, A$, and $P$ lie on the same line. | 91.29. There are several ways of mutual arrangement of circles and points on them. Consider the way indicated in Fig. 67. Suppose that the line $O_{1} A$ intersects the second circle at some point $P^{\prime}$. Then $\angle A P^{\prime} O_{2}=\angle P^{\prime} A O_{2}=180^{\circ}-$ $-\angle O_{1} A O_{2}=180^{\circ}-\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,385 |
31. A continuous and strictly monotonically increasing function $f$ is such that $f(0)=0$ and $f(1)=1$. Prove that
$$
\begin{gathered}
f\left(\frac{1}{10}\right)+f\left(\frac{2}{10}\right)+\ldots+f\left(\frac{9}{10}\right)+f^{-1}\left(\frac{1}{10}\right)+f^{-1}\left(\frac{2}{10}\right)+\ldots \\
\ldots+f^{-1}\left(\fr... | 91.31. Instruction. Evaluate the two halves of the specified sum through the areas under the subgraph and above the supergraph of the function $f(x)$ within the square $[0 ; 1] \times [0 ; 1]$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,387 |
32. The computer "KPK-1991" can perform the following two operations on natural numbers:
a) squaring;
b) an $n$-digit $(n>3)$ number $X$ transitions to the number $A+B$, where $A$ is the number formed by the last three digits of $X$, and $B$ is the number formed by the first $n-3$ digits of $X$.
Can the computer obt... | 91.32. Answer. No, it cannot. Hint. Track the remainder when dividing by 37. | No,itcannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,388 |
34*. The cells of an $N \times N$ board are painted in red, blue, and green, such that next to any red cell there is a blue one, next to any blue one there is a green one, and next to any green one there is a red cell (i.e., the corresponding cells share a side). Prove that for the number of red cells $k$ the following... | 91.34. Note. If we denote by $A, B$ and $C$ the quantities of red, blue, and green cells respectively, then the inequalities $A \leqslant 3 B, B \leqslant 3 C, C \leqslant 3 A$ are obvious. Now prove the inequalities $A \leqslant B+4 C, B \leqslant C+4 A, C \leqslant A+4 B$. Use the following construction: for each red... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,389 |
38. The sequence of natural numbers $\left(a_{n}\right)$ is constructed according to the following rule. Each of its members with an even index $a_{2 n}$ is obtained from $a_{2 n-1}$ by adding to it some digit of the number $a_{2 n-1}$, and each member with an odd index $a_{2 n+1}$ is obtained from $a_{2 n}$ by subtrac... | 91.38. Note: Check that if $a_{1} \leqslant 10^{p}-10$ for any $n$ we have $a_{n}<10^{p}$, and if $a_{1} \leqslant 4 \cdot 10^{p}$ for any $n$ we have $a_{n} \leqslant 4 \cdot 10^{p}+8$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,390 |
39. Do there exist four different numbers such that any two of them - $x$ and $y$ - satisfy the relation
$$
x^{10}+x^{9} y+x^{8} y^{2} \ldots+x y^{9}+y^{10}=1 ?
$$ | 91.39. Answer. No, it does not exist. Multiplying by $x-y$, we get that $x^{11}-y^{11}=x-y$, i.e., $x^{11}-x=y^{11}-y$. From this, the polynomial $f(t)=t^{11}-t+C$ must have four distinct roots, which is impossible, as $f^{\prime \prime}(t)$ has only one root. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,391 |
41*. On the planet Tranai, there are 1991 cities, each of which is connected by a road. Every day, the Ministry of Construction closes three roads for repair, after which the Ministry of Transport introduces one-way traffic on one of the remaining open roads. Roads where one-way traffic has already been introduced are ... | 91.41. Instruction. In the first 500 days, the Ministry of Transport should create a closed cycle of length no less than 400. Then, the number of cities from which there are no more than 300 unopened roads to the cities in the cycle should not exceed 15. Next, within 30 days, these cities should be connected by two roa... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,393 |
43. Two circles of the same radius intersect at points $A$ and $B$. An arbitrary line passing through point $B$ intersects the circles at points $X$ and $Y$ as well. Find the geometric locus of the midpoints of segments $X Y$. | 91.43. Answer. This is a circle with diameter $A B$. | This\is\\circle\with\diameter\AB | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,394 |
46. Does there exist a function $f: N \rightarrow N$ such that $f(f(f(\ldots f(x)) \ldots)=x+1$ for any natural $x$ (f applied $f(x)$ times)? | 91.46. Answer. Such a function does not exist. Hint. Consider the sequence $a_{1}=1, \quad a_{2}=f\left(a_{1}\right), \ldots, \quad a_{n}=$ $=f\left(a_{n-1}\right), \ldots$ and prove that each natural number appears exactly once among its terms and that the sequence is monotonic. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,397 |
47*. A sequence of 26 non-zero digits is written out. Prove that this sequence can be divided into several parts so that the sum of the numbers formed by the digits of each part is divisible by 13. | 91.47. Let us use the following lemma, the proof of which is left to the reader as an exercise in applying induction.
Lemma. Let $S$ be a set of $k \leqslant 13$ integers not divisible by 13; $A(S)$ be the set of remainders when all possible sums of several numbers from $S$ are divided by 13. Then the set $A(S)$ has a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,398 |
49*. A shuffle of a deck of $n$ cards is defined as the following operation: the deck is divided into some (arbitrary) number of parts, which are then laid out in reverse order without changing the order of the cards within each part. Prove that a deck of 1000 cards can be moved from any position to any other position ... | 91.49. Note. Shuffling can be perceived as dividing the deck into several parts, each of which is flipped, remaining in the same place, after which the entire deck is flipped. Excluding the flipping of the deck, we will only perform the flipping of the parts into which the deck is divided - let's call this operation a ... | 56 | Combinatorics | proof | Yes | Yes | olympiads | false | 37,399 |
51. $A B$ is a chord of a circle, dividing it into two segments. $M$ and $N$ are the midpoints of the arcs into which points $A$ and $B$ divide the circle. When the circle is rotated around point $A$ by a certain angle, point $B$ moves to point $B^{\prime}$, and point $M$ moves to point $M^{\prime}$. Prove that the seg... | 91.51. Let the midpoint of segment $B B^{\prime}$ be denoted by $D$. Since $\angle A N M^{\prime} = \angle A B B^{\prime}$, the lines $M M^{\prime}$ and $B B^{\prime}$ intersect at a point $N^{\prime}$ lying on our circle. A short calculation shows that $N M / M M^{\prime} = N B / B D$, and the equality of angles $N M ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,400 |
52. Real numbers $x_{1}, x_{2}, \ldots, x_{n}$ lie in the interval $[-1 ; 1]$, and the sum of their cubes is zero. Prove that the sum $x_{1}+x_{2}+\ldots+x_{n}$ does not exceed $n / 3$. | 91.52. Note. Use the fact that for any $x \geqslant-1$ we have
$$
4 x^{3}-3 x+1=(x+1)(2 x-1)^{2} \geqslant 0
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,401 |
55*. We will call a finite sequence $a_{1}, a_{2}, \ldots, a_{n}$ $p$-balanced if all sums of the form $a_{k}+a_{k+p}+a_{k+2 p}+\ldots$ 154
$(k=1,2, \ldots, p)$ are equal to each other. Prove that if a 50-term sequence is $p$-balanced for $p=3,5$, $7,11,13,17$, then all its terms are zero. | 91.55. Let $z$ be a complex root of unity of degree $p$, i.e., $z^{p}=1$. Since $0=z^{p}-1=(z-1)\left(z^{p-1}+z^{p-2}+\ldots+z+1\right)$, we obtain that $z^{p-1}+z^{p-2}+\ldots+z+1=0$ for $z \neq 1$. Now consider the polynomial $f(t)=a_{1}+a_{2} t+a_{3} t^{2}+\ldots+a_{n} t^{n-1}$. If the sequence $\left(a_{k}\right)$ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,403 |
56*. Prove that the number $512^{3}+675^{3}+720^{3}$ is composite. | 91.56. Let $x=512, y=675, z=720$. Then $2 z^{2}=3 x y$ and
$$
\begin{gathered}
x^{3}+y^{3}+z^{3}=x^{3}+y^{3}-z^{3}+3 x y z= \\
=(x+y-z)\left(x^{2}+y^{2}+z^{2}-x y+x z+y z\right)
\end{gathered}
$$
is divisible by $x+y-z=467$. More precisely, this number is equal to $229 \cdot 467 \cdot 7621$. | 229\cdot467\cdot7621 | Number Theory | proof | Yes | Yes | olympiads | false | 37,404 |
1. In a tic-tac-toe tournament, one point is awarded for a win, zero points for a draw, and one point is deducted for a loss. Several schoolchildren played a tic-tac-toe tournament where each played exactly one game with every other participant. One of the participants scored seven points, while another scored 20 point... | 92.1. Suppose there were no draws. Then the team that scored 20 points must have played an even number of games (since the number of games it won exceeds the number of games it lost by 20), while the team that scored seven points must have played an odd number of games. This leads to a contradiction, since any two team... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,406 |
3. Given a natural number $N$. Prove that the numbers $N(N-1)$ and $(N+1)^{2}$ have different digit sums. | 92.3. The difference between two given numbers is $3 N+1$. However, one of these numbers is divisible by three and the sum of its digits is divisible by three. Therefore, the other number is not divisible by three and the sum of its digits cannot be the same as that of the first number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,408 |
6. Three counterfeiters printed many different banknotes - each for 100 rubles. It is known that each of them can pay any other any amount from 1 to 25 rubles (possibly with change). Prove that together they can pay any amount from 100 to 200 rubles.
Note. The banknotes are counterfeit and can be of any denomination. ... | 92.6. Since the first counterfeiter can pay the second exactly 25 rubles (possibly with change), one of them must have several bills totaling no less than 25 and no more than 50 rubles. Indeed, if the first gave the second $x$ rubles and received $y$ rubles in change, then $x-y=25$, and $x, y \leqslant 100$. Clearly, i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,410 |
10. On a grid paper, 100 lattice points are marked. Prove that there exist two of them - $A$ and $B$ - such that the rectangle $A X B Y$ with sides parallel to the grid lines contains at least 20 marked points (including the points lying on the sides of the rectangle). | 92.10. Enclose all marked nodes in the smallest possible rectangle P with sides running along the grid lines. Then, on each side of this rectangle, there is one of the marked nodes - let's denote them in the order of traversing the perimeter of the rectangle as \(A, B, C\) and \(D\). Then the five rectangles defined by... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,413 |
11. Solve the system of equations:
$$
\left\{\begin{array}{l}
a^{2}+3 a+1=(b+c) / 2 \\
b^{2}+3 b+1=(a+c) / 2 \\
c^{2}+3 c+1=(a+b) / 2
\end{array}\right.
$$ | 92.11. Adding all the equations, we get $(a+1)^{2}+(b+1)^{2}+$ $+(c+1)^{2}=0$. From this, $a=b=c=-1$. This is the only solution to the system. | =b==-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,414 |
12. All coins in Fedya's collection have a diameter no larger than $10 \mathrm{~cm}$. He stores them in a flat box measuring $30 \times 70$ cm. He was given a coin with a diameter of $25 \mathrm{~cm}$. Prove that now he will be able to fit all his coins in a flat box measuring $55 \times 55 \mathrm{cm}$. | 92.12. Let's cover a rectangle $30 \times 70$ with two overlapping rectangles $30 \times 55$ and $30 \times 25-$ see Fig. 68. It is clear that each coin lies entirely within one of these rectangles. By placing them inside a square $55 \times 55$, we see that a new coin from Fedya's collection can fit exactly into the u... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,415 |
18. Given natural numbers $k$ and $n$, the difference of which is greater than 1. It is known that $4kn + 1$ is divisible by $k + n$. Prove that the numbers $2n - 1$ and $2k + 1$ have a common divisor greater than 1. | 92.18. Since $4 k(k+n)$ and $4 n(k+n)$ are divisible by $k+n$, we obtain that $4 n^{2}-1$ and $4 k^{2}-1$, which are equal to $(2 n-1)(2 n+1)$ and $(2 k-1)(2 k+1)$, are divisible by $k+n$. Suppose that the numbers $2 n-1$ and $2 k+1$ are coprime. It is also true that the pairs of numbers $(2 k-1,2 n-1)$ and $(2 k+1,2 n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,418 |
21. Given two different quadratic trinomials $f(x)$ and $g(x)$, whose leading coefficients are equal to one. It is known that $f(19)+f(92)=g(19)+g(92)$. For which $x$ is the equality $f(x)=g(x)$ satisfied? | 92.21. Answer. Only at $x=55 \frac{1}{2}$. | 55\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,419 |
23. All 120 faces of twenty identical cubes with edge length 1 were painted black and white, with 60 faces of each color. Prove that the cubes can be placed on a table such that their bottom faces form a border of a $6 \times 6$ square, and among the faces that are visible, the number of black faces is equal to the num... | 92.23. Instruction. Let's first lay out the cubes in any manner. Suppose that $a$ black and $b$ white faces are visible. Prove that the cubes can be rotated such that the difference $a-b$ changes by two each time or does not change at all. After this, it only remains to transform the arrangement using similar
274
rota... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,420 |
24. Similar right triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ (angles at vertices $B$ and $B^{\prime}$ are right angles; corresponding vertices are denoted by the same letters; vertices are listed in a clockwise order) are placed on the plane such that $A = C^{\prime}$, and point $A^{\prime}$ lies on the r... | 92.24. Reflect point $C^{\prime}$ with respect to the line $A^{\prime} B^{\prime}$ and denote the result by $D$. Then, obviously, the quadrilateral $C A^{\prime} D C^{\prime}$ is cyclic. Therefore, the center of the circle circumscribed around triangle $A^{\prime} A C$ coincides with the center of the circle circumscri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,421 |
27. Given two ten-digit numbers $A$ and $B$, in the notation of which only 1 and 2 are used. Prove that the number of 20-digit numbers from which the number $A$ can be obtained by erasing digits is the same as the number of 20-digit numbers from which the number $B$ can be obtained.
## 10th grade | 92.27. We will prove by induction on $n$ the following fact: the number of $n$-digit numbers from which a given $k$-digit number $X$, written with ones and twos, can be obtained by erasing some digits, depends only on the numbers $k$ and $n$, but not on the specific number $X$—let's denote this number by $F(k, n)$. The... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,422 |
30. On each of $N \geqslant 3$ cards, a digit is written. Arranging these cards in a row in all possible ways, we obtain $N!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(N-1) N$ natural numbers. Can their product be a number whose decimal representation consists of only ones? | 92.30. Note: All digits must be odd, and therefore all $N!$ numbers must have the form $4k+1$ or $4k-1$. Prove that the numbers of the form $4k-1$ are even in quantity. Note. The number $111...11$ has the form $4k-1$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,423 |
31. A circle inscribed in an angle with vertex $O$ touches its sides at points $A$ and $B$. Ray $O X$ intersects this circle at two points $C$ and $D$ such that $O C=C D=1$. If $M$ is the point of intersection of ray $O X$ and segment $A B$, then what is the length of segment $O M$? | 92.31. Answer. 4/3. Hint: Prove that $|O D||C M|=$ $=|O C||M D|$. | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,424 |
32. A pile contains 1992 stones. Two players play the following game: on each turn, they can take from the pile any number of stones that is a divisor of the number of stones the opponent took on their previous turn. The first player can take any number of stones on their first move, but not all at once. The player who... | 92.32. The first player wins. With his first move, he takes eight stones, and 1984 stones remain in the pile, which is $1984=64 \cdot 31$. After this, the first player repeats all the moves of the second player. Since each move involves taking one, two, four, or eight stones from the pile, and since 1984 is divisible b... | The\first\player\wins | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,425 |
33. On a plane, there is a regular $n$-sided polygon with center $O$. The vectors $\overrightarrow{v_{1}}, \overrightarrow{v_{2}}, \ldots, \overrightarrow{v_{n}}$ lead from $O$ to the vertices of the polygon, and they are numbered in order, starting from some vertex. Prove that if $a_{1}<a_{2}<\ldots<a_{n}$ are some re... | 92.33. It is not difficult to prove that $\vec{v}_{1}+\vec{v}_{2}+\ldots+\vec{v}_{n}=0$. Transform the expression given in the condition as follows:
$$
\begin{gathered}
\left(a_{2}-a_{1}\right)\left(\overrightarrow{v_{2}}+\ldots+\overrightarrow{v_{n}}\right)+\left(a_{3}-a_{2}\right)\left(\overrightarrow{v_{3}}+\ldots+... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,426 |
37. A certain number of black and an equal number of white checkers are arranged in a circle. Let $A$ be the number of those triples of consecutive checkers in which the white color predominates, and $B$ be the number of triples of consecutive checkers in which the black color predominates. Prove that $A \leqslant 3 B$... | 92.37. Let $N$ be the number of black checkers. It is clear that the number of black checkers does not exceed $(3 B + A) / 3$, since in each black triplet there are no more than three black checkers, and in each white one, no more than one, and each checker is counted in three triplets. Therefore, $A + B = 2 N \leqslan... | A\leqslant3B | Combinatorics | proof | Yes | Yes | olympiads | false | 37,428 |
38. In an acute scalene triangle $ABC$, altitudes $AD$, $BE$, and $CF$ are drawn. Points $X$, $Y$, and $Z$ are such that $D$, $E$, and $F$ are the midpoints of segments $BX$, $CY$, and $AZ$ respectively. Prove that the centers of the circumcircles of triangles $ACX$, $ABY$, and $BCZ$ are the vertices of a triangle cong... | 92.38. Note. Check that the centers of the three circles are obtained by reflecting the center of the circle circumscribed around triangle $A B C$ with respect to its sides. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,429 |
43. For a natural number $\mathrm{K}$, 54 or 77 is added to it every second. Prove that after some time, a number will be obtained whose last two digits are the same. | 92.43. Arrange all possible remainders when divided by 100 in a circle, in the following order:
$$
0,77,77+77 \equiv 54,77+77+77, \ldots, 77 \cdot 99 \equiv 23
$$
Then adding 77 or 54 to a number means moving to the next number clockwise or skipping one. Clearly, moving in this way, we cannot jump over the pair $0,77... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,431 |
44. There are $n k$ stones, which are somehow distributed into $n$ piles. It is allowed to double any pile by moving stones from other piles into it in any way. For which $k$ can such operations always (for any $n$) ensure that all remaining piles have an equal number of stones? | 92.44. Answer. Only for $k$ being a power of two. Arrange $2k$ stones into two piles, one containing 1 stone and the other containing $2k-1$ stones. Suppose we can obtain two identical piles of $k$ stones each (or one pile of $2k$ stones, which is the same). What does the operation described in the condition look like?... | k | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,432 |
46. A regular $n$-gon $M$ is rotated around its center by an angle of $\pi / n$ to obtain the $n$-gon $M^{\prime}$. Into what minimum number of convex polygons can the figure $M \cup M^{\prime}$ be cut? | 92.46. Answer. $n+1$. The example is obvious. We will prove that it is impossible to cut it into a smaller number of polygons. For this, consider all the external vertices of the figure $M \cup M^{\prime}$ and, numbering the polygons into which it is cut, we will consider that these vertices are painted in the correspo... | n+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,434 |
47. The sequence $\left(F_{n}\right)$ is defined as follows: $F_{1}=1$, $F_{2}=2, F_{n+2}=F_{n+1}+F_{n}$ for any $n>0$. Prove that for any natural number $n$ the inequality
$$
\sqrt[n]{F_{n+1}}>1+\frac{1}{\sqrt[n]{F_{n}}}
$$
holds. | 92.47. Consider the numbers $F_{1} / F_{2}, F_{2} / F_{3}, \ldots, F_{n} / F_{n+1}$ and write the inequality for the means
$$
\frac{1}{n}\left(\frac{F_{1}}{F_{2}}+\frac{F_{2}}{F_{3}}+\ldots+\frac{F_{n}}{F_{n+1}}\right) \geqslant \sqrt[n]{\frac{1}{F_{n+1}}}
$$
Since for any $k$ we have $F_{k} / F_{k+1}=1-F_{k-1} / F_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,435 |
48*. The plane is painted in 1992 colors, and there is a certain triangle $T$ on it. Prove that there exists a triangle equal to $T$ on the plane, such that on any pair of its sides there are internal points of the same color. | 92.48. Placing the triangle \( T \) on the plane, we describe a circle around it with center \( O \) and consider all triangles obtained from \( T \) by rotations around \( O \) by a small angle. For each of them, we can draw a table of size \( 3 \times 1992 \), at the intersection of the \( k \)-th column and the \( j... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,436 |
49*. On the board, there are 128 ones. In one move, you can replace a pair of numbers $a$ and $b$ with the number $a b + 1$. Let $A$ be the maximum number that can appear on the board after 127 such operations. What is its last digit?
## 11th grade | 92.49. Hint. Prove that to obtain the maximum number, you can act according to the following scheme: at each moment, perform the operation on the two smallest numbers.
By now, carrying out the calculations in accordance with the derived algorithm, we get that the last digit of the maximum number is two. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,437 |
51. Find all triples of natural numbers such that the product of any two numbers in the triple, increased by 1, is divisible by twice the third number. | 92.51. There is exactly one such triple - $(1,1,1)$. Hint. All three numbers in the triple - let's denote them $a, b$ and $c$ - must be pairwise coprime and odd. In this case, the expression $a b+b c+$ $+c a+1$ is divisible by each of the numbers $a, b$ and $c$, and therefore, it is divisible by their product. | (1,1,1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,438 |
54. It is known that for any four lines in general position on a plane (i.e., such that no two are parallel and no three intersect at the same point), the circles circumscribed around the four triangles formed by their intersections pass through one point. Does there exist a set of 45 lines in general position such tha... | 92.54. Yes, this is possible. Consider two lines $L_{1}$ and $L_{2}$ intersecting at point $O$, and an arbitrary point $P$ not lying on them. Now, draw 43 different circles through points $O$ and $P$ such that they intersect each of the two given lines at two points, one of which, of course, is point $O$. Let the $i$-t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,439 |
56*. Let $L\left(a_{1}, a_{2}, \ldots, a_{k}\right)$ be the number of ways to place $a_{1}$ ones, $a_{2}$ twos, $\ldots, a_{k}$ numbers $k$ in the cells of an $m \times n$ table so that in each row the numbers increase from left to right in a non-strictly monotonic manner, and in each column the numbers increase from t... | 92.56. Let's check that
$$
L\left(a_{1}, \ldots, a_{p}, a_{p+1}, \ldots, a_{k}\right)=L\left(a_{1}, \ldots, a_{p+1}, a_{p}, \ldots, a_{k}\right)
$$
We will call a pattern a set of cells in an $m \times n$ table such that its intersection with any column and any row consists of several consecutive cells, and such that... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,441 |
57*. Several unit circles are placed on a plane. Is it true that it is always possible to mark several points so that there is exactly one marked point inside each circle? | 92.57. Answer. No, this is incorrect. Hint. Consider a set of circles, the centers of which are all the lattice nodes with a step of $1 / 1000$ within some specific $10 \times 10$ square. | No | Combinatorics | proof | Yes | Yes | olympiads | false | 37,442 |
2. The numbers $1,2,3, \ldots, 100$ are written in ascending order. It is allowed to erase several consecutive numbers and write down the number equal to their count (for example, erase the numbers $13,14,15$ and write the number 3 instead). Is it possible to achieve that after several such operations, the numbers 50,5... | 93.2. The given situation could not have arisen as a result of recording the number 51, since we had only 50 numbers exceeding 50. This means that the number 51 was on the board from the beginning, but then it turns out that all the numbers to the right of 51 disappeared without a trace, which is impossible. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,443 |
3. At the table, there are several boys and five girls, and on the table, on a plate, there are 30 buns. Each girl gave a bun (from the plate) to each boy she knew, and then each boy gave a bun (from the plate) to each girl he did not know. After this, it turned out that all the buns were given out. How many boys were ... | 93.3. Let the number of boys be $n$. Then there are $5 n$ different boy-girl pairs. Each such pair corresponds to
one passed bun: if the boy and girl are acquainted, the bun was passed by the girl, and if they are not acquainted - by the boy. Hence $5 n=30$. Answer. $n=6$. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,444 |
6. On a plane, there are 1993 triangles, and it is known that each of them contains at least four vertices of other triangles. Prove that some three triangles have a common point.
11 Order No. 64
## 7th grade | 93.6. Suppose that no three triangles have a common point. Then a point located inside a triangle can be a vertex of no more than one triangle, and a vertex of any triangle belongs to no more than one other triangle. How many vertices are there in total? On the one hand, no more than $1993 \cdot 3$, since each triangle... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,446 |
11. What is the greatest value that the expression
$$
a e k-a f h+b f g-b d k+c d h-c e g
$$
can take if each of the numbers $a, \ldots, k$ is equal to 1 or -1? | 93.11. Each of the six terms of the given sum is equal to 1 or -1, so this sum is even. It cannot take the value six, since in this case the terms $a e k, b f g, c d h$ must be equal to 1 and, therefore, their product is 1, and the terms $a f h, b d k, c e g$ must be equal to -1, and their product is -1. However, these... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,449 |
12. Let's call a person unsociable if they have fewer than 10 acquaintances. Let's call a person an eccentric if all of their acquaintances are unsociable. Prove that the number of eccentrics is no more than the number of unsociable people. | 93.12. Let's consider introverted eccentrics as normal people. We will count the number of acquaintances between the remaining introverts - let there be $m$ of them - and the remaining eccentrics - let there be $k$ of them. On the one hand, this number is no less than $10 k$, since each sociable eccentric is acquainted... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,450 |
13. In a rectangular table, some cells are marked: a star is drawn in them. It is known that for any marked cell, the number of stars in its column coincides with the number of stars in its row. Prove that the number of rows in the table that contain at least one star is equal to the number of columns in the table that... | 93.13. We will place in each cell containing a star the number that is the reciprocal of the number of stars in the column containing that cell. Then the sum of the numbers in any column containing a star is equal to 1. Similarly, the sum of the numbers in any row containing a star is also equal to 1. Therefore, the nu... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,451 |
16. Point $M$ is taken on side $A C$ of an equilateral triangle $A B C$, and on the extension of side $B C$ beyond vertex $C$, point $N$ is marked such that $B M=M N$. Prove that $A M=C N$.
The point $M$ is taken on the side $A C$ of an equilateral triangle $A B C$, and on the extension of the side $B C$ beyond the ve... | 93.16. Let's take a point $K$ on side $A B$ such that $K M \| B C$. It is not hard to see that triangles $M C N$ and $M K B$ are equal by side and three angles. Then $C N=K M=A M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,453 |
17. Natural numbers $x$ and $y$ are such that the sum of the fractions
$$
\frac{x^{2}-1}{y+1}+\frac{y^{2}-1}{x+1}
$$
162
is an integer. Prove that each of the two given fractions is an integer. | 93.17. Let $u$ be the first, and $v$ be the second of these fractions. Their sum and product are integers, so $u$ and $v$ are roots of a quadratic equation with integer coefficients, say, $x^{2}+$ $+m x+n=0$. Since $u$ and $v$ are rational roots, the square root of the discriminant of this equation $m^{2}-4 n$ is an in... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,454 |
19. Point $M$ is the midpoint of side $B C$ of a convex quadrilateral $A B C D$. It is known that the measure of angle $A M D$ is $120^{\circ}$. Prove the inequality $A B+\frac{B C}{2}+C D \geqslant D A$. | 93.19. Let $B^{\prime}$ be the point symmetric to $B$ with respect to $A M$, and $C^{\prime}$ be the point symmetric to $C$ with respect to $M D$. The angle $B^{\prime} M C^{\prime}$ is $60^{\circ}$, and the triangle $B^{\prime} M C^{\prime}$ is equilateral (see Fig. 70). The length of the broken line $A B^{\prime} C^{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,455 |
20. A chip is placed on one of the squares of a $7 \times 7$ board. It is allowed to sequentially place new chips on empty squares, but in such a way that each square on which the next chip is placed shares a side with no more than one already occupied square. What is the maximum number of chips that can be on the boar... | 93.20. Answer. 36 chips. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,456 |
22. A quadrilateral $A B C D$ is inscribed in a circle. Lines $A B$ and $C D$ intersect at point $M$, and lines $B C$ and $A D$ intersect at point $N$. It is known that $B M=D N$. Prove that $C M=C N$. | 93.22. By the Law of Sines, we have $B M / \sin \angle B C M = C M : : \sin \angle C B M$ and $D N / \sin \angle D C N = C N / \sin \angle C D N$. However, according to the condition, all these expressions are obviously equal, and taking into account that $\sin \angle C B M = \sin \angle A B C = \sin \angle A D C = \si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,458 |
24. Natural numbers $a, b$ and $c$ are such that $\frac{a b}{a-b}=c$. It is also known that the numbers $a, b$ and $c$ do not have a common natural divisor greater than 1. Prove that $a-b$ is a perfect square. | 93.24. Suppose that some prime number $p$ enters the factorization of the number $a-b$ in an odd power $2k-1$. Then, since $ab$ is divisible by $a-b$, one of the numbers $a$ and $b$ (say, $a$) is divisible by $p^k$. But then, obviously, the number $b$, equal to $a-(a-b)$, is also divisible by $p^k$, and thus the quotie... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,459 |
25. On a plane, several points are given. It is allowed to replace a pair of given points $A, B$ with a pair of new points $C, D$, if segments $A C$ and $B D$ have the same length, are perpendicular to $A B$, and points $C$ and $D$ lie on opposite sides of the line $A B$. Prove that after a series of such operations, w... | 93.25. Hint. Consider the quantity $S=\Sigma\left|O M_{i}\right|^{2}$, where $O$ is the origin, and $M_{i}$ are the given points, and trace the behavior of the quantity $S$ when the described operation is performed. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,460 |
26. Prove that for any positive numbers $x, y$ and $z$, the inequality
$$
\sqrt[2]{x+\sqrt[3]{y+\sqrt[4]{z}}} \geqslant \sqrt[32]{x y z}
$$
holds. | 93.26. Suppose the inequality is not satisfied. Then, obviously,
$$
\sqrt{x}<\sqrt[32]{x y z}, \quad \sqrt[6]{y}<\sqrt[32]{x y z}, \quad \sqrt[24]{z}<\sqrt[32]{x y z}
$$
These inequalities can be rewritten as:
$$
x^{16}<x y z, \quad y^{16}<(x y z)^{3}, \quad z^{16}<(x y z)^{12}
$$
Multiplying them, we get $x^{16} y... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,461 |
27*. Two regular 10-sided polygons are given, with a natural number written at each vertex of each polygon.
It is known that the sum of the numbers on each 10-sided polygon is 99. Prove that on both polygons, it is possible to mark several consecutive vertices (but not all) such that the sums of the marked numbers are... | 93.27. Consider a circle divided into 99 equal arcs. Among the ends of these arcs, mark 10 points that divide the circle into 10 arcs, the lengths of which correspond to the ten numbers on the first decagon (the order is also preserved). Similarly, construct a second circle modeling the second decagon. Now, superimpose... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,462 |
33. In a $17 \times 17$ square table, 80 cells are colored black, and the rest are white. It is allowed to color a row or a column black if the majority of the cells in that line are black. Prove that it is impossible to make the entire table black using such operations. | 93.33. Let's assume that this is possible. Consider the first moment when nine lines of one type (for example, nine rows) have been repainted. Suppose this happens after $k+9$ operations ($k$, obviously, no more than 8). Then it would only remain to perform no more than $17-k$ operations to repaint the columns. Thus, t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,465 |
34*. On the set $M$ of natural numbers from 1 to 1993, an operation * is defined which assigns to any two numbers $a$ and $b$ from the set $M$ some number $a * b$, also belonging to $M$. It is known that for any numbers $a$ and $b$ from $M$, the equality $(a * b) * a=b$ holds. Prove that there exists a number $a$ such ... | 93.34. Consider all possible ordered pairs $(a, b)$ of elements of the set $M$ and, assuming the opposite of the statement of the problem, indicate a natural way to partition this set of pairs into non-intersecting triples. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,466 |
36. Prove that no number of the form $3000 \ldots 001$ is a perfect square. | 93.36. Suppose that $300 \ldots 001=k^{2}$. Then $k^{2}-1=3 \cdot 10^{n}$ or $(k-1)(k+1)=3 \cdot 2^{n} 5^{n}$. Clearly, exactly one of the numbers $k-1, k+1$ can be divisible by 5, and therefore one of these numbers is divisible by $5^{n}$. From this, we obtain the inequality $5^{n} \leqslant 3 \cdot 2^{n}+2$, which is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,467 |
39. In a country with $n$ cities, some of which are connected to each other by roads, and any two cities are connected by no more than one road. Let's call the degree of a city the number of roads that lead out of it. Let $2 \leqslant k \leqslant n$ be some natural number. Prove that in the country there will definitel... | 93.39. Suppose this is not the case, and we order the degrees of all cities in ascending order: $d_{1} \leqslant d_{2} \leqslant \ldots \leqslant d_{n}$. Then the following chain of inequalities should hold:
$$
d_{k}-d_{1} \geqslant k-1, \quad d_{k+1}-d_{2} \geqslant k-1, \ldots, d_{n}-d_{n-k+1} \geqslant k-1
$$
We n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,469 |
40*. Differentiable functions $f$ and $g$, defined on the interval $[0 ; 1]$, are such that the function $f(0)=f(1)=1$ and the function $19 f^{\prime} g+93 f g^{\prime}$ is non-negative. Prove that $g(1) \geqslant g(0)$. | 93.40. Multiply the given inequality by $f^{18} g^{92}$ and we get that
$$
\left(f^{19} g^{93}\right)^{\prime}=19 f^{18} f^{\prime} g^{93}+93 g^{92} g^{\prime} f^{19} \geqslant 0
$$
Therefore, the function $f^{19} g^{93}$ is non-decreasing on the interval $[0 ; 1]$ and thus
$$
f^{19}(1) g^{93}(1) \geqslant f^{19}(0)... | (1)\geqslant(0) | Calculus | proof | Yes | Yes | olympiads | false | 37,470 |
41*. Let $n$ be an arbitrary natural number, and all natural numbers from $n$ to $3n-1$ are written on the board. It is allowed to erase any two numbers $a$ and $b (a \leqslant b)$ from the board and write the number $a / 2$ in their place. Prove that when after a series of such operations only one number remains on th... | 93.41. Hint. Consider the sum of the reciprocals of the values that are currently written on the board.
Consider the sum of the reciprocals of the values that are currently written on the board. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,471 |
42. The sequence ( $a_{k}$ ) consists of positive numbers and is such that $\left(a_{k+1}+k\right) a_{k}=1$ for any $k$. Prove that all its terms are irrational. | 93.42. Suppose that one of the members of the sequence is equal to $a_{k}=p / q$, where $p$ and $q$ are natural numbers. Then, as is easy to verify, the next member is equal to $(q-k p) / p$, and it is clear that the sum of the denominator and the numerator of the fraction has decreased. After a few steps, when either ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,472 |
43. Points $D, E$, and $F$ are chosen on the sides $A C, A B$, and $B C$ of isosceles triangle $A B C \quad(A B=B C)$ such that $D E = D F$ and $A E + F C = A C$. Prove that the angles $B A C$ and $F D E$ are equal. | 93.43. Consider on the base $AC$ a point $D'$ such that $AD' = FC$, $D'C = AE$. Then triangles $AED'$ and $FC D'$ are equal by two sides and the included angle, and therefore $ED' = D'F$.
From this, it follows that $D = D'$. Further, we obtain that the angle $\angle FDE$ complements the sum of angles $\angle ADE = \an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,473 |
49*. Inside a convex quadrilateral, four points are marked. Prove that there is a point on the perimeter of the quadrilateral such that the sum of the distances from this point to the vertices is greater than the sum of the distances to the marked points.
## 10th grade | 93.49. The four marked points can lie at the vertices of a convex quadrilateral or form a triangle with one point inside (degenerate cases are allowed). For brevity, we will only consider the first case (see Fig. 71.). First, let us introduce the following notation: for an arbitrary point $Z$ on the contour of the quad... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,477 |
51. Is it possible to place non-zero digits in the cells of a $10 \times 10$ table so that all 10-digit numbers formed in the rows are greater than the 10-digit number on the main diagonal, which, in turn, is greater than all 10-digit numbers in the columns of the table? | 93.51. Answer. No, it cannot. In the proof, it is sufficient to use the fact that any string-number is greater than any column-number. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,478 |
52. On the plane, a square $A B C D$ is given. Find the minimum of the ratio $\frac{O A+O C}{O B+O D}$, where $O$ is an arbitrary point on the plane. | 93.52. We will prove that $\sqrt{2(OA + OC)} \geqslant OB + OD$. Squaring the inequality and eliminating the denominator, since the equality $OA^2 + OC^2 = OB^2 + OD^2$ is a well-known fact, we get
$$
2[OA^2 + OC^2 + 2OA \cdot OC] \geqslant [OB^2 + OD^2 + 2OB \cdot OD]
$$
or
$$
OB^2 + OD^2 + 4OA \cdot OC \geqslant 2... | \frac{1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,479 |
54. On a piece of paper, 1993 points are marked, and some of them are connected by non-intersecting segments, in such a way that it is impossible to select several of these segments to form a closed broken line. Two players take turns placing chips on the marked points, with each subsequent chip covering a point adjace... | 93.54. Suppose that for any initial move of the first player to some point $A$, the second player has a winning response to point $B$. Consider the case where the first player moves to point $B$ on their first move. If the winning response of the second player is to a point $C$, different from $A$, then the sequence (1... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,481 |
55. The numbers $a_{1}, a_{2}, \ldots, a_{n}$ lie in the interval $[-1 ; 1]$. Prove the inequality
$$
\sum_{i=1}^{n} \frac{1}{1+a_{i} a_{i+1}} \geqslant \sum_{i=1}^{n} \frac{1}{1+a_{i}^{2}}\left(a_{n+1} \equiv a_{1}\right)
$$ | 93.55. For any positive numbers $x, y$ we have
$$
\frac{2}{1+x y} \geqslant \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}
$$
- this is proven by bringing to a common denominator. It remains only to add up all inequalities of the form
$$
\frac{2}{1+a_{i} a_{i+1}} \geqslant \frac{1}{1+a_{i}^{2}}+\frac{1}{1+a_{i+1}^{2}}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,482 |
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