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 |
|---|---|---|---|---|---|---|---|---|---|
17. In a certain state, every two cities are connected by a road. On each road, movement is allowed only in one direction. Prove that there exists a city from which you can travel around the entire state, visiting each city exactly once. | 65.17. We will prove this by induction on the number of cities. The base case is obvious. Now, if we separate some city $A$ from the given $n$ cities, then the remaining $n-1$ cities can be toured in a route of the form $B_{1} \rightarrow B_{2} \rightarrow \ldots \rightarrow B_{n-1}$. If there is a road $A \rightarrow ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,682 |
18. Find all octets of prime numbers such that the sum of the squares of the numbers in the octet is 992 less than their quadrupled product.
## 9th grade | 65.18. Answer. There is only one such set, consisting of eight twos. | 8\times2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,683 |
23. Let $A_{1}, A_{2}, \ldots, A_{n}$ and $B_{1}, B_{2}, \ldots, B_{n}$ be two permutations of the numbers $1,2,3, \ldots, n$. Prove that for even $n$, some two numbers in the set $A_{1}+B_{1}, A_{2}+B_{2}, \ldots, A_{n}+B_{n}$ give the same remainder when divided by $n$. | 65.23. If all remainders are distinct, then their sum is $0+1+\ldots+(N-1)=N(N-1) / 2$, i.e., the sum is not divisible by $N$. On the other hand, the sum of such numbers is $2(1+2+\ldots+N)=N(N+1)$ and is divisible by $N$. We obtain a contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,685 |
24*. $N$ circles on the plane occupy an area of 1. Prove that from them, one can select several non-overlapping circles, the sum of whose areas is greater than $1 / 9$.
## $10-11$-th grades | 65.24. We will increase the radii of all circles by three times, keeping the original centers. After this, if any two of the original circles intersect, we will remove the circle with the smaller radius. In this case, the following property will be preserved: the tripled circles will still cover the set that was covere... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,686 |
28.
Variant for 10th grade. On the side $AB$ of triangle $ABC$, an arbitrary point $M_{1}$ is taken. From the center $A$ with radius $AM_{1}$, an arc $M_{1} M_{2}$ is drawn inside triangle $ABC$ to side $AC (M_{2} \in AC)$, then from the center $C$ with radius $CM_{2}$, an arc $M_{2} M_{3}$ is drawn inside $ABC$ to s... | 65.28. We will conduct the proof by induction on $n$. The base case $n=1$ is obvious. The transition:
$$
\begin{gathered}
(A+B)^{n+1} \leqslant (A+B) 2^{n-1}\left(A^{n}+B^{n}\right)= \\
=2^{n-1}\left(A^{n+1}+B^{n+1}+A B^{n}+B A^{n}\right)
\end{gathered}
$$
Since $A B^{n}+B A^{n} \leqslant A^{n+1}+B^{n+1} \quad$ (beca... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,689 |
29. Given a cube $12 \times 12 \times 12$, which is cut by planes parallel to its faces into unit cubes. How many parts will the cube be divided into if a section in the form of a regular hexagon is made? | 65.29. Indication. Introduce a Cartesian coordinate system with the origin at one of the cube's vertices. Then the plane $x+y+z=18$ intersects the cube with a corner at point $(a, b, c)$ if and only if $15<a+b+c<18$, i.e., $a+b+c=16$ or 17 (the corner of the cube that is closest to the origin is meant). It remains to f... | 216 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,690 |
31 *. In the cells of an $n \times n$ table, non-negative integers are placed, and if a zero is in some cell, then the sum of all numbers in the same row or column with this cell is at least $n$. Prove that the sum of the numbers in the table is at least $n / 2$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 65.31. Consider a string with the minimum sum $S$ of numbers in it. Then it contains at least $n-S$ zeros and in the corresponding columns (there are $n-k$ of them; $k \leqslant S$) the sum of numbers is at least $n-S$ in each. In the remaining $k$ columns, the sum is at least not less than $S$. Therefore, the sum of a... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 36,691 | |
33. Among all polynomials of the form $x^{2}+a x+b$, find the one for which the maximum modulus on the interval $[-1 ; 1]$ is minimal. | 65.33. Answer. $x^{2}-1 / 2$. | x^{2}-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,692 |
35. On a square grid, there is a figure with an area less than 1. Prove that this figure can be translated so that no grid node lies inside it. | 65.35. Suppose this is not the case, and we number all the lattice nodes located at a distance of no more than 2 from the given figure $F$. Consider a unit square $K$, the lower left corner of which is the origin $O$. For each point $M$ in this square, consider the vector $\overrightarrow{O M}$ and subject the given fi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,693 |
3. All integers from 1 to 1966 are written on the board. It is allowed to erase any two numbers and replace them with their difference. Prove that it is impossible to achieve, by repeatedly performing this operation, that only zeros remain on the board. | 66.3. Let's track the sum of the numbers. Initially, it equals $1965 \times$ $\times 983$ and is odd. Note that replacing a pair of numbers $A$ and $B$ with their difference $A-B$ does not change the parity of the sum of all numbers. Therefore, the sum of all numbers on the board is always odd, and thus, all zeros cann... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,697 |
4. Black paint was splattered onto a white plane. Prove that there will be three points of the same color lying on one straight line, such that one of the points lies exactly midway between the other two. | 66.4. Consider two points of the same (say, white) color lying on the same line (the real axis), and for convenience, let's assume that they correspond to the numbers 0 and 1. If the points 2 or -1 are white, then the required triplet is found; the same applies to the point $1 / 2$. However, if all three of these point... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,698 |
7. Prove that the radius of a circle is equal to the difference in lengths of two chords, one of which subtends an arc of $1 / 10$ of the circumference, and the other subtends an arc of $3 / 10$ of the circumference. | 66.7. Consider six consecutive points of division of a circle into 10 equal arcs: $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, and $A_{6}$. Then the line $A_{2} A_{5}$ is parallel to the diameter $A_{1} A_{6}$ and the line $A_{3} A_{4}$, and the line $A_{3} A_{6}$ is parallel to the line $A_{4} A_{5}$. Denote the intersection ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,700 |
11. $N$ points on the plane are arranged in such a way that any triangle with vertices at these points has an area less than 1. Prove that all these points can be enclosed in a triangle of area 4 *.
## 8th grade | 66.11. Consider a triangle $ABC$ of maximum area with vertices at the given points. It is clear that each of the given $N$ points lies within triangle $A^{\prime} B^{\prime} C^{\prime}$ (see Fig. 31), where $A^{\prime} B^{\prime} \parallel A B, B^{\prime} C^{\prime} \parallel B C, A^{\prime} C^{\prime} \parallel A C$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,703 |
15. Prove that the sum of all divisors of the number $n^{2}$ is odd. | 66.15. Hint. It is enough to check this for the case of odd $n$. Then all divisors are divided into pairs $\left(d, n^{2} / d\right)$ with an even sum, except for the "lonely" odd $n$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,704 |
17. The numbers $x_{1}, x_{2}, \ldots$ are constructed according to the following rule: $x_{1}=2$; $x_{2}=\left(x_{1}^{5}+1\right) / 5 x_{1} ; x_{3}=\left(x_{2}^{5}+1\right) / 5 x_{2}, \ldots$ Prove that no matter how long we continue this construction, all the resulting numbers will be no less than $1 / 5$ and no more... | 66.17. Since $x^{4}+1>x$ for any real $x$, then $\left(x^{4}+1\right) / 5 x \geqslant 1 / 5$ for positive $x$. Since the function $x^{4}-10 x+1$ is convex, it attains its maximum value on the interval $[1 / 5 ; 2]$ at its endpoints. At the points $1 / 5$ and 2, its values are negative. Therefore, if $1 / 5 \leqslant x ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,705 |
20. Prove that it is possible to color the plane using nine colors in such a way that the distance between any two points of the same color is different from 1966 m. | 66.20. Instruction. Tile the plane with regular hexagons with a side length of 982 m. After that, color each hexagon in its own color so that any two hexagons of the same color are separated from each other by at least two other hexagons. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,707 |
21. $P$ and $Q$ are prime numbers; $Q^{3}-1$ is divisible by $P$; $P-1$ is divisible by $Q$. Prove that $P=1+Q+Q^{2}$. | 66.21. $Q^{3}-1=(Q-1)\left(Q^{2}+Q+1\right)$ is divisible by $P$. Since $P>Q$, then $Q^{2}+Q+1$ is divisible by $P=k Q+1$. Let $(k Q+1) m=Q^{2}+Q+1$, i.e., $m=Q^{2}+Q-k m Q+1$. Then either $m=1$, or $m \geqslant Q+1$. But in the second case $Q^{2}+Q+$ $+1 \geqslant(Q+1)^{2}$ - absurd. Therefore, $m=1$. | 1+Q+Q^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 36,708 |
22. On the sides of triangle $A B C$, right-angled isosceles triangles $A B D, B C E$, and $A C F$ are constructed outwardly. Prove that the segments $D E$ and $B F$ are equal and mutually perpendicular. | 66.22. Consider point $M$ as the midpoint of segment $A B$. It is not difficult to prove that segments $M F$ and $M E$ are equal in length and perpendicular. This follows from the fact that triangles $C Y A$ and $C B U$ are obtained from each other by a $90^{\circ}$ rotation around point $C$ (here $Y$ and $U$ are the v... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,709 |
23 *. There are $k$ colors. In how many ways can the sides of a given regular $n$-gon be painted so that adjacent sides are painted in different colors (the polygon cannot be rotated)?
## $10-11$ grades | 66.23. Let's denote this number of ways by $x_{n}$. Then, as we can see, $x_{1}=0, x_{2}=k(k-1)$ and for any $n$ the relation $x_{n}=(k-2) x_{n-1}+(k-1) x_{n-2}$ holds. Indeed, let's paint the sides from the first to the $(n-2)$-th. The number of such colorings, in which the color of the $(n-2)$-th side does not match ... | x_{n}=(k-1)^{n}+(k-1)(-1)^{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,710 |
30. $m$ and $n$ are natural numbers, and $m$ is odd. Prove that the numbers $2^{n}+1$ and $2^{m}-1$ are coprime. | 66.30. Suppose there exists a prime number $p$ such that $2^{n}+1$ and $2^{m}-1$ are divisible by $p$, i.e., $2^{n}$ and $2^{m}$ have remainders of -1 and 1, respectively, when divided by $p$. Let's find the GCD $(n, m)=d$. Since there exist natural numbers $a$ and $b$ such that $a n - b m = d$, then $2^{d}$ has a rema... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,712 |
33. A square tiling is a partition of a square into a finite number of rectangles whose sides are parallel to the sides of the square. A tiling is called primitive if it is not a subdivision of a larger tiling. For which $n$ does there exist a primitive tiling of a square into $n$ rectangles? | 66.33. Answer. For $n=5$ and for $n \geqslant 7$. In Fig. 32, a, the division into an arbitrary odd number of rectangles greater than 4 is illustrated - an example for $n=9$. In Fig. 32, b, the left half of the square is divided according to Fig. 32, a, while the division of the right half into four rectangles is fixed... | For\n=5\\for\n\\geqslant\7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,713 |
1. The capacities of cubic vessels are in the ratio $1: 8: 27$, and the volumes of the liquid poured into them are in the ratio $1: 2: 3$. After this, a certain amount of liquid was transferred from the first to the second vessel, and then from the second to the third, so that the level of liquid in all three vessels b... | 67.1. Answer. 350,700 and 1,050 liters. | 350,700,1050 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,714 |
4. Each of the eight given distinct natural numbers
is less than 16. Prove that among their pairwise differences there are at least three identical ones. | 67.4. There are 14 possible differences in total - from 1 to \(14=15\) - I. For the given eight numbers, there are 28 possible pairwise differences. It is clear that the number 14 can be represented as a difference of natural numbers less than 16 in only one way (see above). Therefore, the other 13 numbers correspond t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,716 |
7. Prove that $\left(1+x+x^{2}+\ldots+x^{100}\right)\left(1+x^{102}\right)-102 x^{101} \geqslant$ $\geqslant 0$. | 67.7. Expanding the brackets, we get the sum of all even powers of \( x \) from 1 to \( x^{202} \). It remains to write inequalities of the form \( 1 + x^{202} \geqslant 2 x^{101} ; x^{2} + x^{200} \geqslant 2 x^{101} \) and so on, which are equivalent to the obvious inequalities \( \left(1 - x^{101}\right)^{2} \geqsla... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,718 |
8. In quadrilateral $A B C D$, $M$ is the midpoint of $A B$, and $N$ is the midpoint of $C D$. Lines $A D$ and $B C$ intersect $M N$ at points $P$ and $Q$ respectively. Prove that if $\angle B Q M=\angle A P M$, then $B C=$ $=A D$. | 67.8. Reflect vertices $B$ and $C$ with respect to the line $M N$ and obtain points $B^{\prime}$ and $C^{\prime}$ respectively. The lines $A D$ and $B^{\prime} C$ should be parallel. But since both lines $A B^{\prime}$ and $D C^{\prime}$ are parallel to $M N$, we get that $A B^{\prime} C^{\prime} D$ is a parallelogram ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,719 |
12. $X$ and $Y$ are the roots of the equation $t^{2}-c t-c=0$. Prove that the inequality $X^{3}+Y^{3}+(X Y)^{3} \geqslant 0$ holds. | 67.12. Since $X+Y=c, X Y=-c$, a short calculation shows that $X^{3}+Y^{3}+(X Y)^{3}=2 c^{2} \geqslant 0$. | 2^{2}\geqslant0 | Inequalities | proof | Yes | Yes | olympiads | false | 36,721 |
13. Two circles touch internally at point $A$. Through point $B$ of the inner circle, different from $A$, a tangent to this circle is drawn, intersecting the outer circle at points $C$ and $D$. Prove that $A B$ is the bisector of angle $C A D$. | 67.13. Let's draw a common tangent to both circles at point $A$. The intersection point of this tangent and the line $C D$ will be denoted as $E$. It is clear that $\angle E D A = \angle C A E$, $\angle E C A = \angle D A E$, since triangles $C E A$ and $D E A$ are similar. Triangle $B E A$ is isosceles, so $2 \angle B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,722 |
14. Prove that $2^{3100}+1$ is divisible by $3^{101}$. | 67.14. It is sufficient to prove that $2^{3 k}+1$ is divisible by $3^{k+1}$. The proof is conducted by induction. The base case $k=1$ is obvious. The inductive step is carried out using the identity $2^{3 x}+1=\left(2^{x}+1\right)\left(2^{2 x}-2^{x}+1\right)$; the second factor on the right-hand side is divisible by th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,723 |
16. In a certain group of people, each person has one enemy and one friend. Prove that these people can be divided into two groups such that in each group there are neither enemies nor friends. | 67.16. First, let's divide everyone into two companies arbitrarily and consider the following quantity \( S \): for each person, determine the number of their friends and enemies in the same company they are in, and then sum all these numbers. Now, consider some person \( A \) for whom the corresponding term is greater... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,724 |
17. The numbers $a_{1}, a_{2}, \ldots, a_{100}$ are such that:
$$
\begin{aligned}
& a_{1}-2 a_{2}+a_{3} \leqslant 0 \\
& a_{2}-2 a_{3}+a_{4} \leqslant 0 \\
& \cdots \cdots \\
& a_{98}-2 a_{99}+a_{100} \leqslant 0
\end{aligned}
$$
and at the same time $a_{1}=a_{100} \geqslant 0$. Prove that all these numbers are non-n... | 67.17. If the set contains a negative number, then the minimum among the numbers in the set - let it be $a_{i}$ - is not at the edge. But since $a_{i}>\left(a_{i+1}+a_{i-1}\right) / 2$, both numbers $a_{i+1}$ and $a_{i-1}$ are equal to $a_{i}$. Continuing to reason similarly, we get that all numbers are equal and negat... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,725 |
18. Given consecutive odd numbers $p$ and $q$. Prove that $p^{p}+q^{q}$ is divisible by $p+q$.
翻译结果如上，保留了原文的换行和格式。 | 67.18 .
$$
p^{p}+q^{q} \equiv(-q)^{p}+q^{q}=q^{2 k-1}\left(q^{2}-1\right)(\bmod p+q)
$$
and if $q=2 k+1, p=2 k-1$, and this is equal to
$$
q^{2 k-1}(q-1)(q+1)=q^{2 k-1} 2 k(2 k+2)
$$
which is divisible by $4 k=p+q$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,726 |
20. Given a sequence of $k$ numbers. It is allowed to replace any number with the sum of the numbers to the right of it. Prove that if this operation is performed enough times, some sequence will repeat twice in a row. | 67.20. Let $t_{2}$ be the moment when a similar operation is performed for the first time on the second number from the right. It is clear that such an operation should not be performed again, otherwise the sequence will repeat. If the operation is never performed on the second number from the right, we set $t_{2}=0$. ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,727 |
22. $a_{1} \geqslant a_{2} \geqslant \ldots \geqslant a_{n} \geqslant 0 ; B$ - the largest of the numbers $\left|b_{1}\right|$, $\left|b_{1}+b_{2}\right|, \ldots,\left|b_{1}+b_{2}+\ldots+b_{n}\right|$. Prove that $\mid a_{1} b_{1}+a_{2} b_{2}+$ $+\ldots+a_{n} b_{n} \mid \leqslant B a_{1}$. | 67.22. Consider the expression
$$
S=\sum_{k=1}^{n-1}\left(a_{k}-a_{k+1}\right)\left(b_{1}+\ldots+b_{k}\right)
$$
By expanding the brackets, it is not hard to notice that it equals $a_{1} b_{1}+$ $+a_{2} b_{2}+\ldots+a_{n} b_{n}$. On the other hand, since $a_{k}-a_{k+1}$ is non-negative, and any expression of the form... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,728 |
29. Prove that on a plane there exists a finite set of points such that for each point in this set, there are at least one hundred points in the set that are equidistant from it.
## ELIMINATION ROUND | 67.29. We will prove that for any $k$ there exists a system of points such that for any point in the system there are at least $k$ points of the system that are at a unit distance from it. Induction on $k$. The base case $k=1$ does not require proof. To prove the transition from $k=n-1$ to $k=n$, consider a system of p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,731 |
31. Two circles intersect at point $A$; $AB$ and $AC$ are their chords, which are also tangents to another circle at point $A$. Denote the second intersection point of the circles by $D$. Prove that $|AB|^2 / |AC|^2 = |BD| / |CD|$. | 67.31. Let the centers of the first and second circles be at points \(O_{1}\) and \(O_{2}\) respectively. Then, as is not difficult to see, triangles \(O_{1} B D\) and \(O_{2} A D\) are similar (as are triangles \(O_{2} C D\) and \(O_{1} A D\)). This gives us the equalities
\[
\frac{R}{A D}=\frac{r}{B D} \quad \text {... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,733 |
32. Two polynomials with real coefficients take integer values at the same points. Prove that either their sum or their difference is a constant. | 67.32. Let us denote these polynomials by $F$ and $G$, and we can assume that $F$ is increasing for $x > x_{0}$. Let $\varepsilon = 1$ or -1 such that $\varepsilon G$ is also increasing for $x > x_{0}$. Consider $F - \varepsilon G$ and assume that this is not a constant. Let $F(x_{1}) = a, G(x_{1}) = b$, where $x_{1} >... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,734 |
35 *. In a square table of size $N \times N$, non-negative numbers are written such that the sum in any row, as well as the sum in any column, is equal to 1. Prove that in this table, it is possible to select $N$ positive numbers, no two of which will be in the same row or in the same column. | 67.35. Note. Let's call the columns of the table "boys" and the rows "girls" and say that a boy is acquainted with a girl if a positive number is placed at the intersection of the corresponding lines. Then the result of the problem reduces to the well-known "marriage lemma": there are $N$ boys and $N$ girls, and it is ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,736 |
5. Several teams participated in a volleyball tournament. Team $A$ is considered stronger than team $B$ if either $A$ won against $B$, or there exists a team $C$ such that $A$ won against $C$, and $C$ won against $B$. Prove that if team $T$ is the tournament winner, then it is stronger than all other teams. | 68.5. Suppose team $T$ has won against $N$ teams. If some team $X$ is not among them and has not lost to any of them, then it has scored no less than $N+1$ points, which contradicts the fact that $T$ is the tournament winner. Therefore, $X$ has lost either to $T$ itself or to one of the teams that $T$ has won against. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,739 |
8. Find five numbers, the pairwise sums of which are 0, $2, 4, 5, 7, 9, 10, 12, 14, 17$. | 68.8. Answer. These are the numbers: $-1 ; 1 ; 3 ; 6 ; 11$. | -1,1,3,6,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,741 |
12. In a circle of radius 3, several circles are placed arbitrarily, the sum of whose radii is 25. Prove that there exists a line that intersects at least nine of these circles.
## 8th grade | 68.12. Let's project all circles onto the diameter of a large circle of arbitrary diameter. The sum of the lengths of the projections is obviously equal to the sum of the diameters of the circles, i.e., 50. Since the length of the large diameter is 6, if each of its points is covered by projections no more than eight t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,742 |
13. In parallelogram $A B C D$, diagonal $A C$ is greater than diagonal $B D$. Point $M$ on diagonal $A C$ is such that a circle can be circumscribed around quadrilateral $B C D M$. Prove that $B D$ is a common tangent to the circles circumscribed around triangles $A B M$ and $A D M$. | 68.13. We will prove that $B D$ is tangent to the circumcircle of triangle $A B M$. For this, it is sufficient to check that angle $A B D$ is equal to the angle subtended by the arc $A M B$ on the specified circle. It is clear, however, that this angle is equal to angle $B M C$, since they both complement angle $A M B$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,743 |
14. $A$ - an odd number, $X$ and $Y$ - roots of the equation $t^{2} + A t - 1 = 0$. Prove that $X^{4} + Y^{4}$ and $X^{5} + Y^{5}$ are integers and are coprime. | 68.14. Using Vieta's theorem, it can be calculated that $\quad S=X^{4}+Y^{4}=A^{4}+4 A^{2}+2, \quad$ and $\quad T=X^{5}+Y^{5}=-A^{5}-$
$-4 A^{3}-3 A^{2} . \quad$ Since $\quad A S+T=-A, \quad$ then $\quad$ GCD $(S, T)=$ GCD $(A, S)=$ GCD $(A, 2)=1$, since $A$ is odd. | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,744 |
15. A regular triangle is reflected symmetrically relative to one of its sides. The new triangle is again reflected symmetrically relative to one of its sides. This is repeated several times. It turns out that the triangle obtained in the end coincides with the original one. Prove that an even number of reflections wer... | 68.15. Let the vertices of the triangle be $A, B, C$ and color the plane as shown in Fig. 35, - in black and white colors. Note that with each flip, the color of the plane on which our triangle $A B C$ lies changes from black to white or vice versa. Since in the end the color turned out to be the same as before, it is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,745 |
22. In an acute-angled triangle $A B C$, two circles are placed, touching sides $A C$ and $B C$ and sides $A B$ and $B C$ respectively, as well as each other. Prove that the sum of their radii is greater than the radius of the circle inscribed in triangle $A B C$. | 68.22. Draw segments $A D$ and $A E$ from vertex $A$, tangent to the first and second circles respectively, so that $D$ and $E$ lie on $B C$. Then, obviously, triangles $A C D$ and $A E B$ cover triangle $A B C$ and, therefore, the sum of the radii of these circles is
$$
\frac{2 S(A C D)}{p(A C D)}+\frac{2 S(A B E)}{p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,748 |
24 *. Prove that an equilateral triangle cannot be cut into several pairwise unequal equilateral triangles.
## $10-$ th grade | 68.24. Suppose this is possible. Then let's classify the ends of the internal sides of the triangles in the partition into three types: points lying on the side of the large triangle (in which four internal sides meet - see Fig. $36, a$); points inside the triangle, in which six small triangles of the partition meet, i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,749 |
31. In a circle, two chords $A B$ and $A C$ are drawn. From the midpoint $M$ of the arc $B A C$, a perpendicular is dropped to the longer of the chords. Prove that its foot divides the broken line $B A C$ in half. | 68.31. Suppose that $A B>B C$, and let the foot of the perpendicular be denoted by $H$. Now, on the extension of the segment $A B$, mark a point $C^{\prime}$ such that $B C=B C^{\prime}$. It is not hard to see, by calculating the angles, that $B M$ is the bisector of angle $C B C^{\prime}$. Since triangle $B C C^{\prim... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,752 |
33. On a plane, there are $M$ points, and not all of them lie on the same line. Prove that one can find at least $(M-1) \times$ $\times(M-2) / 2$ triangles with vertices at these points. | 68.33. Hint. Induction on $M$. Consider $M+1$ points; take the smallest polygon containing all of them (this is the so-called "convex hull" of the given set of points), and remove one of its vertices $A$
, inside which lie the points \( C, D \) and \( O \), where \( A, B, C, D \) are the projections of the tetrahedron's vertices, and \( O \) is the projection of a given point. By considering all possible cases, it is not diffic... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,755 |
37 *. Given a convex polyhedron, at each vertex of which three faces meet. Each face is painted in one of four colors, and faces sharing an edge have different colors. Prove that the number of faces of the first color with an odd number of sides has the same parity as the number of faces of the second color with an odd... | 68.37. Let's mentally combine the faces of the first and second colors into a single common fifth color. Considering the areas of the fifth color on the surface of the polyhedron, as we move along the boundary of such an area, we see that the colors of the adjacent areas (third and fourth colors) alternate, and therefo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,757 |
1. On a chessboard, eight rooks are placed such that no two of them attack each other. Prove that an even number of rooks stand on black squares. | 69.1. A square on the board is white if and only if the sum of its column and row numbers is odd. Since the sum of all row and column numbers for the given eight rooks is $2(1+\ldots+8)=72$ - even, the number of squares contributing an odd value to this sum is even. Therefore, the number of black squares occupied by ro... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,759 |
6. Two brilliant mathematicians were each told a natural number and were informed that these numbers differ by one. After that, they take turns asking each other the same question: "Do you know my number?". Prove that sooner or later one of them will answer affirmatively.
## 7th grade | 69.6. Suppose that $N$ similar questions have been asked. We will prove that then each mathematician can be sure that his partner's number is not less than $N$. Induction on $N$. The base case $N=1$ is obvious. After $N-1$ such questions have been asked, each of them is sure that his partner's number is not less than $... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,762 |
9. Prove the identity
\[
\begin{gathered}
\frac{2}{x^{2}-1}+\frac{4}{x^{2}-4}+\frac{6}{x^{2}-9}+\ldots+\frac{20}{x^{2}-100}= \\
=11\left[\frac{1}{(x-1)(x+10)}+\frac{1}{(x-2)(x+9)}+\ldots+\frac{1}{(x-10)(x+1)}\right]
\end{gathered}
\]
$10^{*}$. In the center of a field shaped as a square, there is a wolf, and at the v... | 69.9. The left expression can be rewritten as:
$$
\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+\ldots+\frac{1}{x-10}-\frac{1}{x+10}
$$
By combining the terms of this sum differently, we get
$$
\begin{aligned}
& {\left[\frac{1}{x-1}-\frac{1}{x+10}\right]+\left[\frac{1}{x-2}-\frac{1}{x+9}\right]+\ldots+\lef... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,764 |
13. On the base $AD$ of the trapezoid $ABCD$, there is a point $E$ such that the perimeters of triangles $ABE$, $BCE$, and $CDE$ are equal. Prove that $BC = AD / 2$. | 69.13. Let $X$ be a point on $[A D]$ such that $B X \| C D$. Then, obviously, $p(A B X)=p(B C X)$, where $p(T)$ denotes the perimeter of triangle $T$. But since the same is true for triangles $A B E$ and $B C E$, we have $p(A B X)-p(B C X)=X E \pm|C E-C X|$, and by the triangle inequality, we get that this difference i... | BC=AD/2 | Geometry | proof | Yes | Yes | olympiads | false | 36,766 |
18. Given a finite sequence of zeros and ones, possessing two properties:
1) if in some arbitrary place in the sequence you select five digits in a row and in any other place also select five digits in a row, then these quintets will be different (they can overlap)
2) if you add any digit to the right of the sequence, ... | 69.18. Somewhere in the sequence, the quintets $a b c d 0$ and $a b c d 1$ should appear, where $a b c d$ are the last four digits of the sequence. This means that the quartet $a b c d$ appears three times, but before it can only be 0 and 1. To avoid repeating a quintet of digits, nothing should stand before one of the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,768 |
20. A tram ticket is called lucky in the Leningrad style if the sum of its first three digits equals the sum of the last three digits. A tram ticket is called lucky in the Moscow style if the sum of its digits in even positions equals the sum of its digits in odd positions. How many tickets are lucky both in the Lening... | 69.20. There are 6700 such tickets. Note. The second digit of the number should match the fifth.
保留源文本的换行和格式，翻译结果如下：
69.20. There are 6700 such tickets.
Note. The second digit of the number should match the fifth. | 6700 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,769 |
21. $m, n$ - natural numbers, and $\frac{m}{n}<\sqrt{2}$. Prove that $\frac{m}{n}<\sqrt{2}\left(1-\frac{1}{4 n^{2}}\right)$. | 69.21. Indeed,
$$
\sqrt{2}-\frac{m}{n}=\frac{\left(2-m^{2} / n^{2}\right)}{(\sqrt{2}+m / n)}>\frac{\left(2 n^{2}-m^{2}\right)}{2 \sqrt{2} n} \geqslant \frac{1}{2 \sqrt{2} n}
$$
which is what we needed to prove. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,770 |
23. $K>1$ - a natural number. The sequence $\left(x_{n}\right)$ is constructed as follows: $x_{1}=1, x_{2}=K, x_{n}=K x_{n-1}-x_{n-2}$ for $n>2$. Prove that for any natural $n$ there exists such $m>n$ that $x_{m}$ is divisible by $x_{n}$. | 69.23. If $x_{n-1} \equiv -a\left(\bmod x_{n}\right)$, then $x_{n+l} \equiv a x_{l}\left(\bmod x_{n}\right)$. Therefore, $x_{2 n}$ is divisible by $x_{n}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,771 |
24*. A group of 60 vacationers arrived at a resort with four meals a day for a 15-day stay. The dining table has 61 seats. The director of the resort always sits in one seat. The director wants to meet each vacationer personally and introduce them to each other. To do this, he wants to seat the vacationers in a differe... | 69.24. Let's number all the vacationers from 0 to 59 and on the first day, seat them in the following order (counting from the director's place clockwise):
\[
\begin{aligned}
& 0,0-1,0-1+2,0-1+2-3, \ldots \\
& \ldots, 0-1+2-\ldots+58-59(\bmod 60)
\end{aligned}
\]
where $(\bmod 60)$ means the remainder modulo 60 (i.e.... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,772 |
25. In the tetrahedron $ABCD$, $AB$ is perpendicular to $CD$, and $O$ is an arbitrary point in space. Prove that the sum of the squares of the distances from point $O$ to the midpoints of edges $AC$ and $BD$ is equal to the sum of the squares of the distances from point $O$ to the midpoints of $AD$ and $BC$. | 69.25. Let's denote the vectors $\overrightarrow{A B}, \overrightarrow{A C}$, and $\overrightarrow{A D}$ as $\vec{b}, \vec{c}$, and $\vec{d}$ respectively. The condition of perpendicularity between $\overrightarrow{A B}$ and $\overrightarrow{C D}$ gives the equality $\overrightarrow{b c}=\vec{b} \vec{d}$ for scalar pro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,773 |
29. Construct such a sequence of natural numbers that among the differences between its terms, all natural numbers occur exactly once. | 69.29. We will sequentially construct a sequence, obtaining a set of $2 n$ numbers on the $n$-th step such that among the differences between its members, the numbers 1, 2, ..., $n$, as well as some other natural numbers, will appear exactly once without repetition. On the first step, consider the set 1, 2. On the $(n+... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,775 |
30 *. $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$ are the altitudes of an acute-angled triangle $A B C$ with the lengths of the corresponding sides $a, b$ and $c$. $A_{1}$ and $A_{2}$ are the projections of point $A^{\prime}$ onto sides $A B$ and $A C$ respectively; $B_{1}$ and $B_{2}$ are the projections of point ... | 69.30. Expressing the areas of the triangles in terms of $a, b, c$ and the angles $\boldsymbol{\alpha}, \boldsymbol{\beta}$, and $\gamma$ of the triangle ($\angle B A C, \angle A B C$, and $\angle A C B$), we obtain that the left side of the equality transforms into the expression
$\frac{1}{2} a b c \sin \alpha \sin \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,776 |
34. A plane is painted in three colors. Prove that on it, you can find a triangle of area 1, all vertices of which are painted the same color. | 69.34. Suppose that such a triangle does not exist. Consider two points $A$ and $B$ of the same (say, first) color, located at a distance $d$ from each other. Draw lines $L_{1}$ and $L_{2}$, parallel to $A B$, at a distance of $2 / d$ from it. Clearly, there are no points of the first color on these lines. We will prov... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,779 |
37. $F_{1}, F_{2}, \ldots, F_{n}$ - polynomials with integer coefficients. Prove that for some integer $a$ all numbers $F_{1}(a)$, $F_{2}(a), \ldots, F_{n}(a)$ are composite.
38 * . $x_{0}<x_{1}<\ldots<x_{n}$ - natural numbers. Prove the inequality
$$
\frac{\sqrt{x_{1}-x_{0}}}{x_{1}}+\ldots+\frac{\sqrt{x_{n}-x_{n-1}}... | 69.37. Let $x_{0}$ be an integer such that $y_{1}=\left|F_{1}\left(x_{0}\right)\right|>$ $>1, \ldots, y_{n}=\left|F_{n}\left(x_{0}\right)\right|>1$. Consider then the number $a=x_{0}+$ $+k y_{1} y_{2} \ldots y_{n}$. By the standard property of polynomials with integer coefficients, $F(p)-F(q)$ is divisible by $p-q$ for... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,780 |
3. In a certain village, there are 1970 residents. From time to time, they exchange one 10-kopeck coin for two 5-kopeck coins or vice versa. Can it happen that during a certain week, each of them gave away exactly 10 coins in such exchanges? | 70.3. No, this is impossible. Indeed, in each exchange, exactly three coins are involved, which means the number of coins that participated in the exchanges must be divisible by three. If each resident gave away exactly 10 coins, then a total of 19700 coins would have participated in the exchanges, and this number is n... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,783 |
5. In a certain state, any two cities are connected by either an air or a water route. Prove that from any city to any other, one can travel by water or from any city to any other, one can travel by air. | 70.5. Induction on the number of cities n. The base case is obvious. Assume there are $n+1$ cities. Choose a mode of transport that connects all cities except for some distinguished city $A$. If $A$ is connected to at least one of the other cities by the same mode of transport, then all $n+1$ cities are connected by th... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,784 |
6. In a volleyball tournament, 12 teams participated. No team scored seven points. Prove that there will be three teams $A$, $B$, and $C$ such that $A$ won against $B$, $B$ won against $C$, and $C$ won against $A$.
## 7th grade | 70.6. Suppose this is not the case and for any teams $A, B$ and $C$ it is known that if $A$ won against $B$, and $B$ won against $C$, then $A$ won against $C$. But then it is clear that the team $X$, which scored the most points, won against all teams. Further, the team $Y$, which came in second place, won against all ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,785 |
9. Given an isosceles triangle with the angle at the vertex $20^{\circ}$. Prove that:
a) the lateral side is greater than twice the base;
b) the lateral side is less than three times the base. | 70.9. On the side $A B$, we lay off a segment $B E$ equal to the base $B C$ - see Fig. 38, $a$. Then $\angle C E B = \angle E C B = 50^{\circ}$,
 $y^{3}+y^{2}+5 y+1=0$. But since $2 y-1=x^{2} \geqslant 0$, $y$ is positive. However, the equation (*) clearly has no positive solutions. Therefore, $y=1$ is the only option. Accordingly, $x=1$ o... | 1or-1,\,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,788 |
13. Two circles are inscribed in angle $ABC$, one of which touches side $AB$ at point $A$, and the other touches side $BC$ at point $C$. Prove that these circles intercept equal segments on line $AC$. | 70.13. Let's denote the other points of tangency lying on the sides $A B$ and $B C$ as $D$ and $E$ respectively, and the second points of intersection of the first and second circles with the segment $A C$ as $X$ and $Y$ respectively. Then, by the property of secants, $C X \cdot C A = C E^2$, $A Y \cdot A C = A D^2$. B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,789 |
16. In a $5 \times 5$ square, 16 cells are shaded. Prove that it is possible to choose a $2 \times 2$ square in which at least three cells are shaded. | 70.16. Suppose this is not the case. Consider the rightmost column and the bottom row of the table. In them, at least eight cells must be shaded (but not all nine). From this, we can see that the cells marked with asterisks in Fig. 39 must not be shaded. Then, in the top-left $3 \times 3$ square, eight cells are shaded... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,792 |
18. On a plane, there are 35 points, no three of which lie on the same line. Some of them are connected by segments - there are 100 in total. Prove that some two of these segments intersect.
## 9th grade | 70.18. Instruction. Prove that for any picture of this kind, the inequality $X \leqslant 3 Y-6$ holds, where $X$ is the number of segments, and $Y$ is the number of points. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,793 |
26. Prove the inequality $\sin 2+\cos 2+2(\sin 1-\cos 1) \geqslant$ $\geqslant 1$. | 70.26. Let $x=\sin 1, y=\cos 1$. Then we need to prove the inequality
$$
2 x y+1-2 x^{2}+2(x-y) \geqslant 1
$$
Rewrite it as
$$
(x-1)(y-x) \geqslant 0
$$
This is true because $x<1$, and $y<x$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,797 |
29. Find a function $f(x)$ such that for any real $x$, except 0 and 1, $f(1 / x) + f(1 - x) = x$. | 70.29. Solving the system of equations:
$$
\left\{\begin{array}{l}
f(x)+f\left(1-\frac{1}{x}\right)=\frac{1}{x} \\
f\left(1-\frac{1}{x}\right)+f\left(\frac{1}{1-x}\right)=\frac{x}{x-1} \\
f\left(\frac{1}{1-x}\right)+f(x)=1-x
\end{array}\right.
$$
we get $f(x)=\frac{x^{3}-x^{2}+1}{2 x(1-x)}$. | \frac{x^{3}-x^{2}+1}{2x(1-x)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,798 |
30 *. A square $N \times N$ is divided into three vertical strips. The width of the first strip from the left is $-k$ cells, and the third is $-m$ cells, with the numbers $N-k$ and $N-m$ being coprime. In the first row, the numbers from 1 to $N$ are written in order from left to right. In the second row, the numbers ar... | 70.30. It is sufficient to prove that none of the numbers will end up in their previous positions in the rows numbered $2,3, \ldots, N$. Clearly, each number, when moving to the next row, increases its position by one of the numbers $m-N, N-k, m-k$. Suppose that a given number $A$ has been in the left group $A$ times, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,799 |
32. The numbers $t_{1}, t_{2}, \ldots, t_{n}$ are positive and their product is 1. Prove that there is an index $k<n$ such that $t_{k}\left(t_{k+1}+1\right) \geqslant 2$ (where $t_{n+1}$ denotes the number $t_{1}$). | 70.32. Suppose that $t_{k}\left(t_{k+1}+1\right)<2$ for any $k$. By multiplying all these inequalities, we get that
$$
\left(t_{1}+1\right)\left(t_{2}+1\right) \ldots\left(t_{n}+1\right)<2^{n}
$$
But since $1+t \geqslant 2 \sqrt{t}$ for any positive $t$, we have $\left(t_{1}+1\right)\left(t_{2}+1\right) \ldots\left(t... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,800 |
35 *. $D$ and $C$ are arbitrary points on a circle that touches the segment $A B$ at its midpoint. $A D$ and $B C$ intersect the circle at points $X$ and $Y$ respectively, and $C X$ and $D Y$ intersect $A B$ at points $M$ and $N$ respectively. Prove that $A M=B N$. | 70.35. Consider the circle passing through points $A, D$ and $N$ and intersecting the given circle a second time at point $E$. The line $N E$ intersects
Fig. 40 the original circle again a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,802 |
3. Kolya, Zhenya, and Nadya take several exams, where they receive a certain whole number of points depending on the place they occupy - first, second, or third, with the person who occupies a higher place receiving more points. After all the exams, Kolya scored 22 points, while Zhenya and Nadya each scored 9 points. Z... | 71.3. Instruction. A total of 40 points were scored. This means $40 = S n$, where $S$ is the sum of points for one exam, and $n$ is the number of exams. Given that $S \geqslant 6$, thus $S = 8, 10$ or 20. Since Zhena was first in algebra, there were more than two exams; otherwise, 22 would be less than $9 \cdot 2 = 18$... | Kolyawasinphysics | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,807 |
6. Can the entire plane be covered with squares, among which only two are identical?
## 7th grade | 71.6. Answer. Yes, it is possible - see fig. 41. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,809 |
9. Inside the square $A B C D$, a point $K$ is chosen. Perpendiculars are drawn through the vertices $A, B, C$ and $D$ to the lines $B K, C K$, $D K$ and $A K$ respectively. Prove that all of them intersect at one point.
60 | 71.9. Consider a $90^{\circ}$ rotation that maps a square onto itself. Then the lines $A K, B K, C K$, and $D K$ will precisely transition into the dropped perpendiculars, and the point $K$ will become the common point of these perpendiculars. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,812 |
12. The bottom of a rectangular box was tiled with tiles of sizes $2 \times 2$ and $1 \times 4$. They were poured out of the box, and one of the $2 \times 2$ tiles was lost. It had to be replaced with a spare $1 \times 4$ tile. Prove that it is now impossible to tile the bottom of the box.
## 8th grade | 71.12. Considering the bottom of the box divided into unit squares, let's mark some of them as shown in Fig. 42. Then each $2 \times 2$ tile covers exactly one marked square, while each $1 \times 4$ tile covers zero or two squares, i.e., the parity of the number of $2 \times 2$ tiles must match the parity of the number... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,813 |
14. Given the numbers $5^{1971}$ and $2^{1971}$. They are written consecutively. What is the number of digits in the resulting number? | 71.14. Answer. 1972 digits, since the product of these numbers is equal to $10^{1971}$. | 1972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,814 |
15. Around a circle, 100 integers are written, the sum of which is 1. A chain is defined as several consecutive numbers. Find the number of chains whose sum of numbers is positive. | 71.15. Answer. 4951. Use the fact that from two chains, one of which complements the other, the sum of the numbers is positive in exactly one of them. | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,815 |
16. There are two piles of 100 matches each. Two players play the following game: the first player discards one of the two piles and divides the other into two not necessarily equal piles. Then the second player performs the same operation, and so on. Can the first player win if the loser is the one who cannot split a ... | 71.16. Yes, the first player can win, and to do this, he needs to divide the pile with an even number of matches into two piles, each containing an odd number of matches every time. This strategy obviously leads him to victory. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,816 |
18. $A B C$ is a triangle, $B D$ is its bisector. The length of side $A B$ is 15, and side $B C$ is 10. Prove that the length of $B D$ does not exceed 12.
## 9th grade | 71.18. Consider a point $E$ on side $B C$ such that $B E = E D$. Then the lines $A B$ and $D E$ are parallel, triangles $A B C$ and $D E C$ are similar, and thus $B E = E D = 6$. Therefore, by the triangle inequality, $B D < B E + E D = 12$. | BD<12 | Geometry | proof | Yes | Yes | olympiads | false | 36,817 |
19. Solve the system of equations:
$$
\left\{\begin{array}{l}
x_{0}=x_{0}^{2}+x_{1}^{2}+\ldots+x_{100}^{2} ; \\
x_{1}=2\left(x_{0} x_{1}+x_{1} x_{2}+\ldots+x_{99} x_{100}\right) ; \\
x_{2}=2\left(x_{0} x_{2}+x_{1} x_{3}+\ldots+x_{98} x_{100}\right) \\
x_{3}=2\left(x_{0} x_{3}+x_{1} x_{4}+\ldots+x_{97} x_{100}\right) ;... | 71.19. Answer. There are a total of 202 solutions. The main series: $x_{0}=0, x_{k}= \pm 1 / 2, x_{j}=0$ for $j \neq k$ (200 solutions). Another solution: all $x_{i}$ are equal to zero, and the last solution is: $x_{0}=$ $=1, x_{i}=0$ for $i>0$. | 202 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,818 |
21. The number $\overline{A B C}$ is prime. Prove that the number $B^{2}-4 A C$ cannot be a perfect square. | 71.21. Suppose that $B^{2}-4 A C=n^{2}$. Then we have $4 A\left(A x^{2}+\right.$ $+B x+C)=(2 A x+B+n)(2 A x+B-n)$. From this, substituting $x=10$, we get that $20 A+B+n$ or $20 A+B-n$ is divisible by the prime number $\overline{A B C}$. But then, obviously, $20 A+B+n \geqslant 100 A+$ $+10 B+C$, which is impossible, si... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,819 |
24. Can the plane be tiled with pairwise distinct triangles with rational side lengths?
## $10-$ th grade | 71.24. Yes, it is possible. Consider the partition of the rectangle $12 \times 16$, as shown in Fig. $43, a$, and then cover the plane with similar rectangles - see Fig. $43, \sigma$, each of which we will partition similarly. Check that all the resulting triangles are different.
a)
^{200}+\left(\frac{1}{2}\right)^{199} \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{198} \cdot \frac{1}{3}+\ldots+\frac{1}{2} \cdot \frac{1}{200}+\frac{1}{201}<\frac{1}{90}$. | 71.28. Let $x=1 / 2^{200}+\ldots+1 / 201$. Then
$$
2: \preccurlyeq x+\sum_{k=1}^{200} \frac{1}{2^{200-k}}\left(\frac{1}{k}-\frac{1}{k+1}\right)+\frac{2}{201}
$$
Therefore,
$$
\begin{gathered}
x \leqslant \frac{2}{201}+\sum_{k=1}^{200} \frac{1}{2^{200-k}} \frac{1}{k(k+1)}=\frac{2}{201}+\sum_{k=1}^{180}+\sum_{k=181}^{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,823 |
30*. On a plane, there are $N$ pairwise non-parallel lines and a point $P$. Point $P$ is projected onto all the given lines, the obtained projections are again projected onto all the lines, and so on. Prove that all points obtained in this way can be covered by one circle.
## ELIMINATION ROUND | 71.30. Enclose all intersection points of the lines and point $P$ in a circle of radius $R$. Now consider $t<1$ such that the cosines of all angles between the given lines are less than $t$. Then the points specified in the condition cannot appear outside a circle with the same center $O$ and radius $\frac{2 R}{1-t}$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,824 |
31. Real numbers $a_{1}, a_{2}, \ldots, a_{n}$ are positive. Prove that the equation
$$
x^{n}+a_{1} x^{n-1}-a_{2} x^{n-2}-\ldots-a_{n}=0
$$
has no more than one positive root. | 71.31. Let's rewrite the equation as: $x^{n}+a_{1} x^{n-1}=a_{2} x^{n-2}+$ $+\ldots+a_{n}$ and divide it by $x^{n-1}$. We get $x+a_{1}=a_{2} / x+$ $+\ldots+a_{n} / x^{n}$. Since on the left we have a monotonically increasing function, and on the right a monotonically decreasing function on the positive semi-axis, the e... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,825 |
36 *. The edges of a complete graph with $2 n+1$ vertices are colored in three colors. Prove that one can choose one of the colors and $n+1$ vertices of the graph such that from each of them to any other chosen vertex one can reach by edges of the specified color. | 71.36. Induction on $n$. Choose a set $X$ of $n$ vertices connected by edges, say, of the first color. Now remove one of them and from the remaining vertices select another $n$-set $Y$, connected by one color.
1st case. This color is the first. Then the two selected sets of $n$ vertices should not intersect, and one r... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,826 |
37. Prove that the equation $X^{3}+Y^{3}+Z^{3}=2$ has infinitely many solutions in integers. | 71.37. For any integer $t$ the triple $X=1+6 t^{3}, Y=1-6 t^{3}$, $Z=-6 t^{2}$ is a solution to our equation. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,827 |
6. Is it possible to place the numbers from 1 to 12 on the edges of a cube so that the sums at all vertices (the numbers on the edges emanating from the given vertex are summed) are equal?
## 7th grade | 72.6. No, it is not possible. Let's sum all such sums at the vertices, if the arrangement is possible. On the one hand, the result should be of the form $8 k$, and on the other hand, each number on an edge enters this sum twice and it should be equal to $12 \cdot 11$ | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,833 |
7. Can two straight cuts passing through two vertices of a triangle divide it into four parts such that three of them are equal-area triangles? | 72.7. No, it cannot. For the equality $S(A O B)=$ $=S(B O D)=S(A O E)$ (see Fig. 44) to hold, it is necessary that $A O=$ $=O D, B O=O E$. But this would mean that $O$ lies on two medians of triangle $A B C$, which is impossible. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,834 |
12. Prove that if an even number can be represented as the sum of the squares of two integers, then its half can also be represented in such a form.
## 8th grade | 72.12. Since $2 n=x^{2}+y^{2}$, where $n, x$ and $y$ are integers, then $x$ and $y$ have the same parity. Therefore, the equality $n=((x+y) / 2)^{2}+((x-y) / 2)^{2}$ gives us the required representation. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,836 |
13. $K$ and $M$ are the points of intersection of two circles. From point $K$, two rays are drawn, one intersecting the first circle at point $A$ and the second circle at point $B$, the other intersecting the first circle at point $C$ and the second circle at point $D$. Prove that the measures of angles $MAB$ and $MCD$... | 72.13. Since quadrilateral $A K C M$ is cyclic, $\angle K A M=180^{\circ}-\angle M C D$. But since
Fig. 44 $\angle M A B=180^{\circ}-\angle K A M$, then $\angle M A B=$ $=\angle M C D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,837 |
16. On a plane, several circles are drawn, some of which are connected by segments. Prove that integers can be placed in these circles so that two circles are connected by a segment if and only if the numbers in them are coprime. | 72.16. Let's mentally draw curves that connect exactly those of the circles that are not connected by line segments. On each curve, we will write a prime number, choosing all these numbers to be different. After that, we will write in each circle a number equal to the product of all prime numbers written on the curves ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,838 |
17 *. A regular triangle with a side of 32 is given. From its vertex, a regular triangle with a side of 1 is cut off. The remaining part is divided into regular triangles. Prove that there are no fewer than 15 of them. | 72.17. Let's define the function $f(x)$ by the formula
$$
f(x)= \begin{cases}2 \log _{\varphi}(1-x)+2, & 0 \leqslant x \leqslant 2-\varphi \\ 0, & x>2-\varphi\end{cases}
$$
where $\varphi=(1+\sqrt{5}) / 2$ is the "golden ratio". Then the inequality holds: when cutting a triangle with side $n$ and removing the upper t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,839 |
18. Natural numbers \( m, n, a, b, k, l \) are such that
\[
\frac{m}{n} < \frac{a}{b} < \frac{k}{l}, \quad |m l - k n| = 1
\]
Prove that \( b \geqslant n + l \).
## 9th grade | 72.18 .
$$
\begin{aligned}
b & =b(k n-m l)=b k n-b m l=b k n-a n l+a n l-b m l= \\
& =n(b k-a l)+l(a n-b m) \geqslant n \cdot 1+l \cdot 1=n+l
\end{aligned}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,840 |
20. Quadrilateral $ABCD$ is inscribed in a circle. Prove that if the point of intersection of the tangents to the circle at vertices $A$ and $C$ lies on the line $BD$, then $AB \cdot CD = BC \cdot AD$. | 72.20. Since $\quad S(A D B) / S(C D B)=A D \sin \angle A D B / C D \times$ $\times \sin \angle C D B$, and on the other hand, $\quad S(A D B) / S(C D B)=$ $=A B \sin \angle A B D / C B \sin \angle C B D$, it is sufficient to prove that $\sin \angle A D B / \sin \angle A B D=\sin \angle C D B / \sin \angle C B D$. Appl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,841 |
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