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23. Given natural numbers $m$ and $n$. Prove the inequality
$$
\frac{1}{\sqrt[m]{n+1}}+\frac{1}{\sqrt[n]{m+1}} \geqslant 1
$$ | 72.23. $\sqrt[n]{m+1} \leqslant 1+m / n, \quad$ since $m+1 \leqslant(1+m / n)^{n}$ (use the binomial theorem of Newton). Similarly $\sqrt[m]{n+1} \leqslant 1+n / m$. Therefore,
$$
\frac{1}{\sqrt[n]{m+1}}+\frac{1}{\sqrt[m]{n+1}} \geqslant \frac{1}{1+m / n}+\frac{1}{1+n / m}=1
$$
which was to be proved. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,842 |
24. Given a finite sequence of integers. Under each of them, we write how many times it has occurred in this sequence. From the resulting sequence, we similarly construct a new sequence, and so on. Prove that at some step, we will obtain two identical sequences in a row.
## 10th grade | 72.24. It is clear that, starting from the third sequence, each number does not exceed the one under which it is written. Since a non-increasing sequence of natural numbers eventually becomes constant, this implies the required result - starting from some moment $t_{1}$, the first number in the sequence will be the sam... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,843 |
26. A closed spatial broken line is called regular if both the segments of the broken line and the angles between adjacent segments are equal. Prove that for any $N$, greater than 5, there exists a regular $N$-segment broken line that does not lie in one plane. | 72.26. Let's construct a broken line where all angles between segments are $90^{\circ}$. If a broken line with $N$ segments has already been constructed, in which there are three consecutive segments $A B, B C$ and $C D$ lying in the same plane, then a broken line with $N+2$ segments is constructed as follows: the segm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,844 |
27. A prime number $p$ is not equal to 3. Prove that the number $4 p^{2}+1$ can be represented as the sum of three squares of natural numbers. | 72.27. If $p=3 k+1$, then $4 p^{2}+1=(4 k+2)^{2}+(4 k+1)^{2}+(2 k)^{2}$. If $p=3 k+2$, then $4 p^{2}+1=(4 k+3)^{2}+(4 k+2)^{2}+(2 k+2)^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,845 |
30. In a volleyball tournament, several matches were played, after which each team had 10 wins and 10 losses. Prove that from the played matches, one can select several such that in these matches, each team will have exactly one loss and exactly one win.
## ELIMINATION ROUND | 72.30. Let's represent the matches of the tournament with a picture, where points will correspond to teams, and played matches will be assigned segments with arrows connecting the team points (the arrow is directed from the winning team to the losing team). Take any point-team $A$ and move out of it along some arrow $A... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,846 |
36. Each side of an equilateral triangle is divided into 30 equal parts. Lines drawn through the division points parallel to the sides of the triangle divide it into 900 small triangles. What is the maximum number of vertices of the partition, no two of which lie on the same drawn line or side? | 72.36. Answer. 21. Indeed, for each vertex, we can define three coordinates: the numbers of lines parallel to one of the sides of the triangle, counted from the vertex not belonging to the corresponding side (see Fig. 45). It is clear that the sum of the coordinates of any vertex is 30. Since the first (as well as the ... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,849 |
38 *. The city has the appearance of a square grid $100 \times 100$ with the side of each cell being 500 m. Along each side of each cell, one can move only in one direction. It is known that one can travel no more than 1 km through the city without violating the movement rules.
66
Prove that there will be no fewer tha... | 72.38. We will call an intersection a "source" if you can only leave from it, a "sink" if you cannot leave from it, and a "through" if you can enter and leave it. Then, no two intersections of the same type can be adjacent - for sinks and sources this is obvious, and if two throughs are adjacent, it means there is a 15... | 1300 | Combinatorics | proof | Yes | Yes | olympiads | false | 36,851 |
1. In three stores, there were 1973 textbooks. In the first three days, the first store sold respectively $1 / 47, 1 / 7$ and $1 / 2$ of its textbooks, the second store - $1 / 41, 1 / 5$ and $1 / 3$ of its textbooks, and the third store - $1 / 25, 1 / 20$ and $1 / 10$ of its textbooks. How many textbooks were in each s... | 73.1. Answer. In the first store, there were 658 textbooks, in the second - 615, and in the third - 700. | 658,615,700 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,852 |
10. Prove that $2^{10}+5^{12}$ is a composite number. | 73.10.
$$
\begin{gathered}
2^{10}+5^{12}=\left(2^{5}+5^{6}\right)^{2}-2^{6} 5^{6}=\left(2^{5}+5^{6}-2^{3} 5^{3}\right)\left(2^{5}+5^{6}+2^{3} 5^{3}\right)= \\
=14657 \cdot 16657
\end{gathered}
$$ | 14657\cdot16657 | Number Theory | proof | Yes | Yes | olympiads | false | 36,857 |
11. On each side of a parallelogram, a point was chosen such that the area of the quadrilateral formed by them is equal to half the area of the parallelogram. Prove that one of the diagonals of the quadrilateral is parallel to one of the sides of the parallelogram. | 73.11. Since the triangles remaining outside the constructed quadrilateral must also cover exactly half the area of the parallelogram, the parallelograms containing these triangles as their halves (see Fig. 48) must have a total area equal to the area of the parallelogram. However, then the quadrilateral formed by thei... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,858 |
12. The sides $a, b$ and $c$ of a certain triangle satisfy the equation $2\left(a^{8}+b^{8}+c^{8}\right)=\left(a^{4}+b^{4}+c^{4}\right)^{2}$. Prove that this triangle is a right triangle.
## 8th grade | 73.12. Factoring the expression $2\left(A^{8}+B^{8}+C^{8}\right)-\left(A^{4}+\right.$ $\left.+B^{4}+C^{4}\right)^{2}$, we get
$$
\left(A^{2}+B^{2}+C^{2}\right)\left(A^{2}+B^{2}-C^{2}\right)\left(A^{2}-B^{2}+C^{2}\right)\left(B^{2}+C^{2}-A^{2}\right) .
$$
Since the value of this expression is zero, then one of the fac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,859 |
13. $A D$ and $B E$ are the angle bisectors of triangle $A B C$. Prove that if $A C > B C$, then $A E > D E > B D$.
---
The translation maintains the original text's line breaks and formatting. | 73.13. Since $A E / E C = A B / B C > A B / A C = B D / D C$, point $E$ is farther from line $A B$ than $D$. Therefore, point $P = A B \cap \cap D E$ is located on line $A B$ such that $B$ lies between $A$ and $P$. This means $\angle A B E > \angle B D$. Similarly, we have $\angle E D A > \angle P A D$, i.e., $\angle D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,860 |
16. A triangle is cut into several convex polygons. Prove that among them, there is either a triangle, or there are two polygons with the same number of sides. | 73.16. Suppose there is no triangle, and there are a total of $n$ polygons. Then among them, there must be a polygon with no fewer than $n+3$ sides. Three of its sides can extend to the edge, while the remaining sides extend to the boundaries of the other $n-1$ polygons. However, two convex polygons obtained by cutting... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,863 |
17. Several natural numbers are written around a circle. Between every two adjacent numbers, their GCD is written. Then the old numbers are erased and the same operation is performed on the remaining numbers. Prove that after several steps all the numbers on the circle will become equal. | 73.17. If among the numbers on the circle there are different ones, then after performing the described operation, the sum of the numbers on the circle will decrease. This follows from the fact that the GCD of two numbers, written between them, does not exceed either of them and is equal to them only if these numbers c... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,864 |
18. Given several points, some of which are connected by segments, in such a way that one can travel from any point to any other point along these segments. Is it always possible to remove one of the points along with the segments emanating from it so that the remaining points are still connected to each other by paths... | 73.18. Yes, we can. First, let's mentally remove the maximum possible number of segments while ensuring that the connectivity property is preserved. After this, we will find a point from which exactly one segment extends - the existence of such a point is not difficult to prove by considering two points connected by a ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,865 |
26. Prove that if from an infinite arithmetic progression with the first term $A$ and difference $d$, not equal to zero, an infinite geometric progression can be extracted, then $A / d$ is a rational number. | 73.26. Let $A+k d, A+m d, A+n d$ be three consecutive terms of this geometric progression with common ratio $q$. Then we have $(A+m d) /(A+k d)=q$, i.e., if we denote $A / d$ by $t$, then:
$$
q=\frac{t+m}{t+k} ; \quad t=\frac{q k-m}{1-q}
$$
But since $(q k-m) /(1-q)=(q m-n) /(1-q)$ (the second expression is also equa... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,869 |
28. Prove that in a convex polyhedron there are two faces with the same number of sides. | 73.28. If a polyhedron has $N$ faces, then each can have from 3 to $N-1$ sides, i.e., no more than $N-3$ variants. Hence, some two faces must have the same number of sides. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,870 |
33. Given a polynomial $P(x)=a_{0} x^{n}+\ldots+a_{n-1} x+a_{n}$, all coefficients of which are integers. It is known that the equations $P(x)=1$, $P(x)=2$, and $P(x)=3$ have integer roots. Prove that the equation $P(x)=5$ cannot have two or more integer roots. | 73.33. Note. Use the fact that if a polynomial $P(x)$ has integer coefficients, then $P(a)-P(b)$ is divisible by $a-b$ for any integers $a$ and $b$.
| proof | Algebra | proof | Yes | Yes | olympiads | false | 36,872 |
36. Given a convex polygon with an area of 9. It is intersected by nine parallel lines, each a unit distance apart from each other. Prove that the sum of the lengths of the segments cut by the polygon on these lines does not exceed 10. | 73.36. Let the lengths of the cut segments be denoted by $a_{1}, a_{2}$, $\ldots, a_{10}$. Then we have two inequalities for the area:
$$
\frac{a_{1}+a_{2}}{2}+\frac{a_{2}+a_{3}}{2}+\ldots+\frac{a_{9}+a_{10}}{2} \leqslant 9 ; \quad 9\left(\frac{a_{1}+a_{10}}{2}\right) \leqslant 9
$$
which follow from comparing the ar... | a_{1}+a_{2}+\ldots+a_{10}\leqslant10 | Geometry | proof | Yes | Yes | olympiads | false | 36,873 |
1. Find all numbers $\overline{A B C}$ such that $\overline{A B C}=2(\overline{A B}+\overline{B C}+$ $+\overline{A C})$ | 74.1. Answer. $134,144,150,288,294$. | 134,144,150,288,294 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,877 |
2. Does there exist a convex polygon with exactly 1974 diagonals? | 74.2. Answer. No, it does not exist. If a convex polygon has $N$ vertices, then it has $N(N-3) / 2$ diagonals, and the number 1974 cannot be expressed as $N(N-3) / 2$. | No,itdoesnotexist | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,878 |
3. Several equilateral triangles were cut out of a sheet of cardboard. The numbers 1, 2, and 3 are written at the vertices of each. They were then stacked. Could it be that the sum of the numbers along each edge of the stack equals 55? | 74.3. No, it could not. The sum of all numbers should be equal to $6 k$, where $k$ is the number of triangles, and $55 \cdot 3=165$ is odd! | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,879 |
6. In the country, there are 1974 cities. From the capital, 101 air routes depart, and from the city of Distant - one air route. From all other cities, 20 air routes depart. Prove that it is possible to fly from the capital to Distant, possibly with layovers.
7th grade ${ }^{1}$ | 74.6. Let's consider all cities that can be reached from the capital (possibly with transfers). If Distant is not among them, then the sum of the number of air routes leaving from all these cities (including the capital itself) is odd. On the other hand, each air route is counted twice and therefore such a sum must be ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,882 |
13. On the plane, there are two circles, one outside the other. Is there a point on the plane, lying outside both circles, such that any line passing through this point intersects at least one of the circles? | 74.13. Yes, such a point exists - for example, point $M$ in Fig. 49.
Fig. 49
, no three of which lie on the same line, such that for any three points of this set, there is a fourth point in the set that forms a parallelogram with them. | 74.17. Consider the largest triangle $A B C$ with vertices at the given points. Suppose $D$ is a point from our set such that $A B C D$ is a parallelogram (see Fig. 50). All points of the given finite set lie inside the triangle $D E F$. Let another point $X$ from our set lie inside the triangle $A E C$ (other cases ar... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,890 |
18. There is a bacterium that divides into two at some point in time. One of the halves divides again into two, and so on. A total of 1000 bacteria have formed. Prove that there was a bacterium whose descendants contain no fewer than 334 and no more than 667 descendants.
## 9th grade | 74.18. Let $X_{1}$ be the number of offspring of the first bacterium, i.e., $X_{1}=1000$. One of the two "daughters" of this bacterium has a number of offspring not less than $X_{1} / 2$. Denote this number by $X_{2}$. Similarly, define the number $X_{3}$, which is not less than $X_{2} / 2$, and so on. Consider the fir... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,891 |
22. On the plane, there is a triangle $A B C$ and a circle $S$ with radius $R / 2$, where $R$ is the radius of the circumcircle of $A B C$. Prove that there exists a point $T$ such that the segments $T A, T B$, and $T C$ are bisected by the circle $S$. | 74.22. Since circle $S$ and the circumcircle of $A B C$ are homothetic with a homothety coefficient of $1 / 2$, there exists a center of homothety - this is the required point $T$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,892 |
23. On a circle with center at point $O$, there is a point $X$. On the diameter extending from point $X$, take point $Y$ such that $O$ lies between $X$ and $Y$. It is required to draw a chord $AB$ through point $Y$ so that $\angle A X B$ is minimized. | 74.23. Answer. The chord $A B$ should be drawn perpendicular to the line $O X$. | ThechordABshouldbedrawnperpendiculartothelineOX | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,893 |
24 *. In the Martian language, there are three words $A, B$, and $C$ such that the word $A A B B$ coincides with the word $C C$. Prove that there is a word $D$ such that each of the words $A, B$, and $C$ can be obtained by writing the word $D$ several times in a row.
## 10th grade | 74.24. Hint. Let $A$ consist of $m$ letters, and $B$ of $n$ letters. Prove that any letters in the word $C C$, which are in positions differing by $m$ or by $n$, coincide. Then the same is true for letters in positions differing by $m-n$ (if, for example, $m \geqslant n$) and so on.
Therefore, if $d=\mathrm{GCD}(m, n)... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,894 |
26. The rays $O S_{1}, O S_{2}, O S_{3}$, emanating from point $O$, intersect three parallel planes at points $A_{1}, B_{1}, C_{1} ; A_{2}, B_{2}, C_{2} ; A_{3}, B_{3}, C_{3}$, respectively. Let $V$ be the volume of the pyramid $O A_{1} B_{2} C_{3}$. The volumes of the pyramids $O A_{1} B_{1} C_{1}, O A_{2} B_{2} C_{2}... | 74.26. Note. Use the fact that if points $X, Y$ and $Z$ are taken on fixed rays $O S_{1}, O S_{2}$ and $O S_{3}$, then the volume of the pyramid $O X Y Z$ is equal to $k|O X\|O Y\| O Z|$, where $k$ is a constant that does not depend on the choice of points $X, Y, Z$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,896 |
27. Find the maximum of the expression
$$
\begin{aligned}
& x_{1}+x_{2}+x_{3}+x_{4}-x_{1} x_{2}-x_{1} x_{3}-x_{1} x_{4}-x_{2} x_{3}-x_{2} x_{4}-x_{3} x_{4}+ \\
& +x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}-x_{1} x_{2} x_{3} x_{4}
\end{aligned}
$$ | 74.27. The given expression can be rewritten as follows:
$$
1-\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)
$$
Since $\left(1-x_{1}\right)\left(1-x_{2}\right)\left(1-x_{3}\right)\left(1-x_{4}\right)$ is a non-negative number that can be equal to zero, the maximum value of the expres... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,897 |
30. The number $2^{n}$ is written on the board. Below it, two natural numbers are written, which sum up to the initial number. Then, under each of the sequentially obtained numbers, two natural numbers are written, which sum up to it, until a one is encountered. Prove that the sum of all the written numbers is not less... | 74.30. We will prove by induction that this sum is no less than $X \log X$ (the logarithm is taken to base 2), where $X$ is the original number written. Let the first step be the writing of $X=$ $=A+B$. Then, by the induction hypothesis, the sum of the numbers written under $A$ is no less than $A \log A$, and under $B$... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,898 |
31. Above a planet in the shape of a sphere, there are 37 point satellites flying. Prove that at any moment, there is a point on the surface of the planet from which no more than 17 satellites are visible. | 74.31. Let's pass a plane $P$ through a pair of satellites and the center of the planet and find two points on the planet that are the ends of its diameter perpendicular to the plane $P$. Then, no satellite can be observed simultaneously from both points, and the two selected satellites are also not visible from these ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,899 |
34. Natural numbers are written in the cells of a rectangular table. It is allowed to double simultaneously all the numbers in one column or subtract one from all the numbers in a row. Prove that using such operations, one can obtain a table consisting entirely of zeros. | 74.34. Let's ensure that all numbers in the first row become zeros. We will double all columns that contain the minimum number in the first row, and then by subtraction, we will achieve that some fields in the first row will be zeros. In this process, the sum of the numbers in the first row will decrease. Continuing si... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,901 |
35. The sides of a square are sequentially numbered with the numbers $1,2,3,4$. For an arbitrary point $A$ and side $k$, denote by $A_{k}$ the point symmetric to the projection of $A$ onto the line $k$ relative to the point $A$. Find all points $A$ such that each of the points $A_{1}, A_{12}, A_{123}, A_{1234}, A_{1234... | 74.35. Let's introduce coordinates in the square such that the origin is at the vertex belonging to sides 1 and 2. Then there is a unique point $A$ with the required property. This is the point with coordinates $(1 / 3 ; 1 / 3)$. Hint. Note that the operations $A \rightarrow A_{2}, A \rightarrow A_{4}$ do not affect on... | (1/3;1/3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,902 |
36. The sum of a hundred natural numbers less than 100 is 200. Prove that among them, several can be chosen whose sum is 100. | 74.36. Suppose this is not the case. Remove one of the numbers and renumber the rest. Then their sums of the form $X_{1}, X_{1}+X_{2}, \ldots$ $\ldots, X_{1}+X_{2}+\ldots+X_{99}$ give different non-zero remainders when divided by 100 - otherwise, by subtracting one sum from another, we would get several numbers whose s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,903 |
37. Find all natural numbers $k$ that have the property that there does not exist a $k$-gon such that the extension of any of its sides contains another side of this $k$-gon. Only polygons with non-parallel adjacent sides are considered. | 74.37. Answer. These are all even numbers greater than nine, and all odd numbers greater than 14.
Indeed, consider an arbitrary line on which lie at least two sides $AB$ and $CD$ of the given $k$-gon. Then there are at least four more lines on which lie the sides of the $k$-gon - specifically those sides whose one end... | These\\all\even\\greater\than\nine,\\all\odd\\greater\than\14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,904 |
5. Kolya and Vasya are writing a 20-digit number using only the digits $1,2,3,4,5$. Kolya writes the first digit, Vasya writes the second, and so on. Vasya wants to get a number divisible by nine. Will Kolya be able to prevent him? | 75.5. Answer. Yes, he will manage. First, Kolya writes a one, and then complements all the digits written by Vasya (of course, except the last one) to six. Then the total sum of the digits will be equal to $55+x$, where $x$ is the last digit written by Vasya. Since $x=1,2,3,4$ or 5, the sum of the digits is not divisib... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,907 |
9. In the record of a three-digit number, there are no zeros. Find the maximum value of the product of this number and the sum of the reciprocals of its digits. | 75.9. Answer. The maximum value is achieved when the number is 911; the value of the expression in this case is $911 \cdot 19 / 9$. | 911\cdot19/9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,911 |
14. Given five infinite geometric progressions, all members of which are integers. Prove that there exists a natural number that does not belong to any of these progressions. | 75.14. Let $\left\{A q^{n}\right\}$ be one of these progressions, $N$ - some natural number. Then on the segment $[0 ; N]$ there are no more than $\log _{q} N+1$ terms of this geometric progression. Therefore, five progressions cover no more than $5 \log _{a} N+5$ numbers not greater than $N$, where $a$ is the smallest... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,915 |
15. $F$ is the intersection point of the bisectors $A D$ and $C E$ of triangle $A B C$. It is known that points $B, D, E$ and $F$ lie on the same circle. Prove that the radius of this circle is not less than the radius of the inscribed circle of this triangle. | 75.15. It follows from the condition that $\angle B=60^{\circ}$. Therefore, the radius of the circle ($B D E F$) is not less than $B E / 2 = B E \sin \angle F B E$ (since $\angle F B E=30^{\circ}$). The latter number is nothing other than the length of the perpendicular dropped from point $F$ - the center of the inscri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,916 |
16. Prove that the points of intersection of the parabolas $Y=X^{2}+X-41$ and $X=Y^{2}+Y-40$ lie on the same circle. | 75.16. Adding the equations, we get $X^{2}+Y^{2}=81$. This equation must be satisfied by the coordinates of the points of intersection of the parabolas, which means that all of them lie on the circle $X^{2}+Y^{2}=$ $=81$. | X^{2}+Y^{2}=81 | Algebra | proof | Yes | Yes | olympiads | false | 36,917 |
18. In seven consecutive vertices of a regular 100-gon, chips of seven colors are placed. In one move, it is allowed to move any chip 10 fields clockwise to the $11-\mathrm{th}$, if it is free. It is required to collect the chips in the seven vertices following the initial ones. How many different arrangements of chips... | 75.18. Answer. There are seven possible arrangements. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,919 |
19. In a circle, two perpendicular diameters $A B$ and $C D$ are drawn. A point $X$ is taken on the arc $B D$, and $A X$ and $C X$ intersect $C D$ and $A B$ at points $E$ and $F$ respectively. Prove that if $C E / E D$ is rational, then $A F / F B$ is also rational. | 75.19. Since $X E$ is the bisector of angle $C D X$, then
$$
C E / E D=C X / D X=\operatorname{ctg} \alpha
$$
where $\alpha$ is the angle $D C X$. Similarly,
$$
A F / F B=A X / X B=\operatorname{ctg} \beta=\operatorname{ctg}\left(45^{\circ}-\alpha\right)
$$
where $\beta=\angle X A B$. If the first ratio is denoted ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,920 |
24. In a $100 \times 100$ square table, some cells are shaded. Each shaded cell is the only shaded cell either in its column or in its row. What is the maximum number of cells that can be shaded?
## 10th grade | 75.24. Answer: 198. Let's mark for each shaded cell the line in which it is the only one. Then among the marked lines - not all columns, otherwise there would be no more than 100 shaded cells. Similarly, not all rows are marked. Therefore, the number of marked lines is no more than 198. On the other hand, if we shade a... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,923 |
30. A convex $N$-gon is inscribed in a $1 \times 1$ square. Prove that there exist three consecutive vertices of this $N$-gon such that the area of the triangle formed by them is no more than $8 / N^{2}$.
## ELIMINATION ROUND
## 8th grade | 75.30. Let $\alpha_{i}$ be the angle between the $i$-th and $(i+1)$-th sides of a polygon, and $a_{i}$ be the length of its $i$-th side. The area of the triangle formed by three consecutive vertices is $a_{i} a_{i+1} \sin \alpha_{i} / 2$. The product of these areas is $(1 / 2)^{N}\left(a_{1} a_{2} \ldots a_{N}\right)^{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,926 |
35. In a group of 30 people, each person likes exactly $k$ people in the group. What is the minimum $k$ for which we can assert that there will definitely be two people who like each other? | 75.35. For $k=15$. Indeed, if we consider a graph with 30 vertices, each of which is connected by an edge, and place an arrow on edge $A B$ from $A$ to $B$ if person $B$ likes person $A$, then we should place $30 \cdot 15=$ $=450$ arrows, while the total number of edges is $30 \times$ $\times 29 / 2=435$. Therefore, on... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,930 |
38 *. The centers of all 64 squares on a chessboard are marked. Is it possible to cut the board into pieces with thirteen straight cuts so that each piece contains no more than one center of a square?
## $9-10$-th grades | 75.38. Let's mark 28 centers of boundary cells and connect them sequentially with segments, forming a contour of a $7 \times 7$ square. Any cut intersects two segments, and, consequently, 13 cuts will not cut more than 26 segments and will not be able to separate the centers from each other. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,931 |
40. Do there exist four pairwise non-intersecting spheres in space, not containing a given point $A$, such that any ray starting from point $A$ intersects at least one of these spheres? | 75.40. Yes, they do exist. Consider a tetrahedron with its center at point \( A \) and for each of its faces, construct a sphere that contains it but does not contain point \( A \). After this, using four homotheties, send these four spheres so far away from point \( A \) and to such different distances that the result... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,932 |
42 *. Does there exist a bijection $f: R \rightarrow R$ such that $f(x) +$ $+f^{-1}(x)=-x$ for any $x$? | 75.42. Let's denote \( f^{-1}(x) \) by \( y \). Then the condition can be rewritten as:
\[
f(f(y)) + f(y) + y = 0
\]
Assume \( f(0) = 0 \). Now, let's divide the line into segments of six types:
\[
\begin{aligned}
& A_{k} - \left[\varphi^{3 k} ; \varphi^{3 k+1}\left[; \quad B_{k} - \left[\varphi^{3 k+1} ; \varphi^{3... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,933 |
44 *. The sequence of integers $x_{0}, x_{1}, x_{2}, \ldots$ is such that $x_{0}=0$, and $\left|x_{n}\right|=\left|x_{n-1}+1\right|$ for each natural $n$. What is the smallest possible value of the expression $\mid \mathrm{x}_{1}+x_{2}+\ldots$ $\ldots+x_{1975} \mid$? | 75.44. Answer. 20. Since $x_{1}^{2}=1, x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \ldots$ $\ldots, x_{1976}^{2}=x_{1975}^{2}+2 x_{1975}+1$, then, adding all these equalities, we get
$$
\sum_{i=1}^{1975} x_{i}=\left(x_{1976}^{2}-1976\right) / 2
$$
Since the closest square to 1976 of an even number ( $x_{1976}$ is even!) is $44^{2... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,934 |
46 *. Each of the numbers $x_{0}, x_{1}, \ldots, x_{n}$ is equal to 0 or 1. Prove that
$$
\begin{aligned}
& x_{0}+\frac{x_{1}}{\sqrt{2}}+\frac{x_{2}}{(\sqrt{2})^{2}}+\ldots+\frac{x_{n}}{(\sqrt{2})^{n}} \leqslant \\
& \leqslant(1+\sqrt{2}) \sqrt{x_{0}+\frac{x_{1}}{2}+\ldots+\frac{x_{n}}{2^{n}}}
\end{aligned}
$$ | 75.46. Let's define the sequence $e_{0}, e_{1}, \ldots$ by the following rule:
$$
e_{0}=\sqrt{2}, \quad e_{n-1}-e_{n}=\frac{1}{\sqrt{2^{n}}}-(1+\sqrt{2})\left(\sqrt{S_{n}}-\sqrt{S_{n-1}}\right)
$$
where $S_{n}=1+1 / 2+1 / 4+\ldots+1 / 2^{n}$.
It is not difficult to prove by induction on $n$ that the inequality
$x_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,935 |
3. Villages $A, B$ and $C$ are located at the vertices of an equilateral triangle. In village $A$, there are 100 schoolchildren, in village $B-200$, in village $C-300$. Where should a school be built so that the total distance traveled by all schoolchildren is as small as possible? | 76.3. Answer. In village $C$. If it is built at any other point $X$, then when it is moved to $C$, the total distance traveled by the schoolchildren will change by $300|C X|-100(|A X|-$ $-|A C|)-200(|B X|-|B A|)$, i.e., it will obviously decrease, since $|A X|-|A C| \leqslant|C X|,|B X|-|B C| \leqslant|C X|$. | In\village\C | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,938 |
4. The strip is divided into 30 cells in one row. At the edge cells, there is a chip each. Two players take turns moving their chips one or two cells in any direction. Moving over the opponent's chip is not allowed. The player who cannot make a move loses. How should the starting player play to win? | 76.4. The beginner must move their token one space on the first move - there will be 27 free spaces between the tokens. After this, the player should respond to their opponent's moves as follows: when the opponent moves one space, the player should move their token two spaces, and when the opponent moves two spaces, th... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,939 |
5. On a grid paper, a square of $11 \times 11$ cells is drawn. It is required to mark the centers of some cells in such a way that the center of any other cell is located on a segment connecting some two marked points lying on the same vertical or on the same horizontal. What is the smallest number of cells that need t... | 76.5. Answer. 22 cells.
76,6. Let $a_{1}, a_{2}, \ldots, a_{m}$ be the sizes of the smaller square plots, $n$ be the side length of the larger square, and $S$ be the sum of the lengths of the internal fences. Then, obviously, $4 \sum a_{i}=2 S+4 n$, $\sum a_{i}^{2}=n^{2}$. From the second equality, it follows that $\s... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,940 |
13. $x_{1}, x_{2}, \ldots, x_{25}$ are some integers, and $y_{1}, y_{2}, \ldots$, $\mathrm{y}_{25}$ is a permutation of these numbers. Prove that the number $\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right) \ldots\left(x_{25}-y_{25}\right)$ is even. | 76.13. If the product is odd, then $x_{k}-y_{k}$ is odd for any $k$. But then $\left(x_{1}-y_{1}\right)+\left(x_{2}-y_{2}\right)+\ldots+\left(x_{25}-y_{25}\right)=$ $=0$ - is also odd. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,944 |
17. In quadrilateral $ABCD$, angle $C$ is the largest. $K$ is the intersection point of line $AD$ and the line passing through $C$ and parallel to $AB$; $M$ is the intersection point of line $AB$ and the line passing through $C$ and parallel to $AD$; $P$ is the intersection point of lines $BK$ and $MD$. Prove that the ... | 76.17. Adding the equalities $S(B C K)=S(M K A)$, $S(C D K)=$ $=S(M D K)$ and canceling the areas of the parts that appear on both sides of the equalities, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,948 |
18. In the set of natural numbers, three pairwise non-intersecting subsets were chosen. Prove that it is possible to choose numbers \(X\) and \(Y\), belonging to two different chosen subsets, such that the sum \(X+Y\) does not belong to the third subset.
## 9th grade | 76.18. Let these sets be denoted as $M_{1}, M_{2}$, and $M_{3}$. Suppose $A$ and $B$ are two consecutive natural numbers belonging to different sets (without loss of generality, we can assume that $A \in M_{1}, B \in M_{2}$, and also that $A$ is odd, and $B$ is even). Now, if we assume the opposite of the statement in ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,949 |
19. In triangle $A B C \quad A C=(A B+B C) / 2$. Prove that the radius of the inscribed circle of triangle $A B C$ is three times smaller than one of its heights. | 76.19. Let $a=|B C|, b=|A C|, c=|A B|$. Since $b=(a+c) / 2$, then $(a+b+c) / 2=3 b / 2$ and $r=2 S /(a+b+c)=$ $=2 S / 3 b=(2 S / b) / 3$, i.e., the radius of the inscribed circle is one third of the height dropped to $A C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,950 |
23. Tsar Bureaucrat has 12 deputies and forms commissions from them such that any two commissions have at least one common member, but must differ in composition. At some point, Bureaucrat formed 1000 commissions. Prove that he can form one more such commission while meeting all the described conditions. | 76.23. For a 12-element set of deputies, there are $2^{12} = 4096$ subsets, which can be divided into $2^{11}$ pairs of the form $(A, \bar{A})$, where $\bar{A}$ is the complement of set $A$. Since $2^{11} = 2048 > 1000 + 2$, there will be a pair $(A, \bar{A})$ where $A$ and $\bar{A}$ are non-empty sets that have not ye... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,952 |
24 *. Given triangle $ABC$ and the circumcircle around it. $K$ is the intersection point of the bisectors of the internal angle $B$ and the external angle $C; L$ is the intersection point of the bisectors of the internal angle $C$ and the external angle $B; M$ is the midpoint of segment $KL$. Prove that $M$ is the midp... | 76.24. Since $A K$ is the bisector of the external angle $A$, as well as $A L$, then $A$ lies on the segment $K L$. It is clear that this bisector passes through the midpoint of the arc $C A B$, since the bisector of the internal angle $A$ passes through the diametrically opposite point - the midpoint of the arc $C B$.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,953 |
26. Find the function $f$, defined on the set of all real numbers, if it is known that $f^{2}(x+y)=f^{2}(x)+f^{2}(y)$ for any $x$ and $y$. | 76.26. Let's denote $f^{2}(x)$ by $g(x)$. Let $x$ and $y$ be arbitrary numbers, and $a=y-x$. Then $g(y)=g(x)+g(a)$ and also $g(x)=g(y-a)=g(y)+g(-a)$. Since $g(a) \geqslant 0, g(-a) \geqslant 0$, we get that $g(y) \geqslant g(x) \geqslant g(y)$. Therefore, $g$ is a constant, and obviously, it is zero. Thus, $g(x)=0$ and... | f(x)=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,955 |
27. Find all real solutions of the equation
$$
\begin{aligned}
& \sqrt{a+b x}+\sqrt{b+c x}+\sqrt{c+a x}= \\
& =\sqrt{b-a x}+\sqrt{c-b x}+\sqrt{a-c x}
\end{aligned}
$$
if it is known that it has solutions. | 76.27. Answer. If the numbers $a, b, c$ are non-negative and at least one of them is not zero, then $x=0$. If $a=b=c=0$, then $x$ is any real number. Other options are not possible if it is known that the equation has a solution. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,956 |
29. A $5 \times 5$ table is filled with zeros and ones. It is known that the top-left and bottom-right corners contain ones, while the other two corners contain zeros. Prove that in the table, it is possible to select two different $2 \times 2$ squares (possibly intersecting) with the same arrangement of numbers. | 76.29. Suppose all 16 variants occur once. Let the number of ones in the central $3 \times 3$ square be $x$, and on the border (excluding the corners) be $-a$. Then, summing the number of ones in the left upper and right lower $4 \times 4$ squares, we get $2 x+a+2=16$, since each of these squares must contain eight one... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,957 |
32. Space is divided into five non-empty subsets. Prove that there is a line intersecting at least three of them. | 76.32. Instruction. First, we assume that we can choose five points - one from each subset - such that no four of them lie in the same plane. We connect these points pairwise with ten lines and assume that each of them is colored in only two colors. Consider a plane that is not parallel to any of these lines. Then, amo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,960 |
33. Solve the system in real numbers:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=x_{3}^{2} \\
x_{2}+x_{3}=x_{4}^{2} \\
x_{3}+x_{1}=x_{5}^{2} \\
x_{4}+x_{5}=x_{1}^{2} \\
x_{5}+x_{1}=x_{2}^{2}
\end{array}\right.
$$ | 76.33. Let's first consider the case when all unknowns are non-negative. If we denote by $X$ and $Y$ the largest and smallest numbers in the set of unknowns, then it is obvious that $X^{2} \leqslant 2 X, Y^{2} \geqslant 2 Y$, i.e., for non-zero $X$ and $Y$ we get $2 \leqslant Y \leqslant X \leqslant 2$, from which we s... | x_{1}=x_{2}=\ldots=x_{5}=2x_{1}=x_{2}=\ldots=x_{5}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,961 |
34. Two players play a game with a finite number of positions, and for any position, the sum of the number of positions that can be moved to from it and the number of positions that can move to it is the same. A player loses if they have no move. Prove that the number of positions in which the first player loses when t... | 76.34. Suppose the number of losing positions $X$ is greater than the number of winning positions $Y$. Then we will prove that there is a pair of losing positions $A$ and $B$ such that one can move from $A$ to $B$ in one move. Indeed, if this is not the case, then the number of positions adjacent to losing positions is... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,962 |
2. On a chessboard, 20 rooks are placed such that every square is under attack by at least one of them (the square on which a rook stands is also attacked by it). Prove that it is possible to remove 12 rooks so that the remaining eight rooks still cover the entire board. | 77.2. If we assume that there is a rook on each vertical, then all rooks except these eight can be removed. If, however, there is no rook on some vertical, it is clear that there must be a rook on each horizontal. Then a similar operation of removing rooks can be performed. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,965 |
3. A regular octagon is divided into four parts of equal area by two straight cuts. Prove that these lines are perpendicular to each other. | 77.3. If these lines intersect not at the center of the octagon, then by shifting them parallel so that both pass through the center, we obtain a new arrangement of lines such that opposite parts have the same area. However, it is clear that with such a shift, in one of the pairs of opposite parts, the area of one will... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,966 |
6. Two players take turns placing digits in a strip of 12 cells until a 12-digit number is formed. In this process, the digits 0 and 9 are not allowed. Prove that the second player can ensure that the resulting number is divisible by 77.
## 7th grade | 77.6. The second player's strategy is as follows: he mentally divides the 12 positions into two sets of six. For any move by the first player, which involves writing a digit $X$, the second player writes the digit $9-X$ in the corresponding position in the other set of six. The resulting number will have the form $A \c... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,969 |
8. The number 197719771977 is represented as
$$
a_{0}+a_{1} \cdot 10^{1}+a_{2} \cdot 10^{2}+\ldots+a_{11} \cdot 10^{11}
$$
where $a_{0}, a_{1}, \ldots, a_{11}$ are non-negative integers, the sum of which does not exceed 72. Find these numbers. | 77.8. Consider the record of the number 197719771977, for which the sum of the numbers $a_{i}$ is minimal. If one of the numbers $a_{i}$ is greater than 10, then it can be replaced by $a_{i}-10$, while adding one to $a_{i+1}$ (for $i<11$). From this, it follows that in this case we will have the decimal representation ... | 1,9,7,7,1,9,7,7,1,9,7,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,970 |
9. On the sides $A B, B C, C D$ and $D A$ of the square $A B C D$, points $K, L, M, N$ are taken respectively. Prove that $K L+L M+$ $+M N+N K \geqslant 2 A C$. | 77.9. Hint. Let $O$ be the center of the given square. Project the segments $K L, L M, M N$ and $N K$ onto the diagonals $A C$ and $B D$. Then the projections will coincide with the unions of the projections of the segments $O K, O L, O M$ and $O N$ onto the corresponding diagonals. It remains to use the fact that for ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,971 |
12. Point $O$ inside a convex polygon $P$ is such that any line passing through it divides $P$ into two parts of equal area. Prove that $O$ is the center of symmetry of the polygon $P$.
## 8th grade | 77.12. Instruction. First, prove that if $M$ is a vertex of a polygon, then there exists a vertex $N$ such that $O \in [M N]$, and then prove that for any such pair $M O = O N$.
| proof | Geometry | proof | Yes | Yes | olympiads | false | 36,972 |
14. $A_{1}, A_{2}, \ldots, A_{n}$ - subsets of the set of natural numbers. Prove that there exist such natural numbers $X$ and $Y$, that each of these subsets either contains both $X$ and $Y$, or does not contain either $X$ or $Y$. | 77.14. If $A$ is some subset of $N$, then we will denote its complement by $\bar{A}$. One of the sets $A_{1}$ and $\overline{A_{1}}$ is infinite. We will choose it and denote it by $C_{1}$. One of the sets $A_{2} \cap C_{1}$ and $\overline{\bar{A}}_{2} \cap C$ is infinite. We will call it $C_{2}$ and so on. As a result... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,973 |
16. We consider all possible broken lines that follow the sides of the cells and connect the two opposite corners of a square sheet of grid paper measuring $100 \times 100$ by the shortest path. What is the smallest number of such broken lines needed so that their union contains all the vertices of the cells? | 77.16. Answer. 101 broken lines. The fact that there are no fewer than 101 broken lines follows from the fact that if we consider all the vertices lying on the diagonal connecting two other opposite vertices of the square (there are 101 of them), then no broken line of the specified type can contain two vertices from t... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,975 |
17. Segments $A_{1}, A_{2}, \ldots, A_{1977}, B_{1}, B_{2}, \ldots, B_{1977}$ lie on the same line. It is known that each segment $A_{k}$ has a common point with each of the segments $B_{k-1}, B_{k+1}$. Moreover, segment $A_{1977}$ has a common point with $B_{1}$, and $A_{1}$ has a common point with $B_{1977}$. Prove t... | 77.17. Suppose that for no $k$ the segments $A_{k}$ and $B_{k}$ intersect. Let $A_{1}$ lie to the left of $B_{1}$. Since $B_{2}$ intersects with $A_{1}$, and $A_{2}$ intersects with $B_{1}$, it is easy to see that $B_{2}$ must lie to the left of $A_{2}$. Continuing this reasoning, we get that $A_{1977}$ lies to the lef... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,976 |
18. $\vec{a}_{1}, \vec{a}_{2}, \ldots, \vec{a}_{n}$ - vectors on a plane, no two of which are collinear. It is known that for any two different indices $i$ and $j$ among these vectors, there is a vector of the form $x \vec{a}_{i}+y \vec{a}_{j}$, where $x$ and $y$ are some negative numbers. Prove that $n-$ is odd.
## 9... | 77.18. Consider the unit circle centered at the origin $O$ and the points of its intersection with lines parallel to the given vectors. If point $A$ is such that $\overrightarrow{O A}$ is in the same direction as the corresponding vector, then we will paint point $A$ red, and if it is in the opposite direction, then we... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,977 |
23. The function $f$ is defined on [0; 1[ by the following rule:
$$
f(x)= \begin{cases}x+\frac{2-\sqrt{2}}{2}, & x \in\left[0 ; \frac{\sqrt{2}}{2}[\right. \\ x-\frac{\sqrt{2}}{2}, & x \in\left[\frac{\sqrt{2}}{2} ; 1[\right.\end{cases}
$$
Prove that for any interval ] $a ; b[\subset[0 ; 1[$, there exist a point $x$ fr... | 77.23. Let us choose a natural number $n$ such that
$$
] x-\frac{1}{n} ; \left.x+\frac{1}{n} \right\rvert\, \subset\right] a ; b[.
$$
Let $1-1 / \sqrt{2}$ be denoted by $t$. Then $f(x)=\{x+t\}$, where the curly braces denote the fractional part of a number. Thus,
$$
f(f(\ldots(x) \ldots)=\{x+n t\}
$$
where the func... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,978 |
2. A polygon is divided by diagonals into triangles, which are colored black and white such that any two triangles sharing a common side are colored differently. Prove that the number of black triangles does not exceed three times the number of white triangles. | 78.2. Since any black triangle borders at least one white triangle, their number does not exceed the total number of sides of white triangles, which is triple the number of white triangles. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,984 |
3. Is it true that 57599 is a prime number? | 78.3. No, incorrect. $57599=240^{2}-1=239 \cdot 241$. | 57599=239\cdot241 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,985 |
15. The route of the cycling relay is a hexagon, all angles of which are $120^{\circ}$, and the lengths of the sides are expressed in whole numbers of kilometers. Each stage of the relay is one of the sides of the hexagon, with the first, third, and fifth stages being ridden by women, and the second, fourth, and sixth ... | 78.15. Answer. Men are obviously wrong. Prove that the superiority of one sum of lengths over another must be divisible by three.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,994 | |
16. Does there exist an arrangement of integers in the cells of an infinite sheet of graph paper such that in any rectangle $1918 \times 1978$ the sum of the numbers is 60? | 78.16. Yes, such an arrangement exists. To describe it, let's introduce a standard coordinate system on the plane. Now, in cells where the sum of the coordinates is divisible by 1918, we will place ones, and in cells where the sum of the coordinates is divisible by 1978, we will place the number -1. If a cell needs to ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,995 |
17. On a plane, six circles are drawn, with the first touching the sixth and second; the second touching the first and third; the third touching the second and fourth, and so on. Prove that there exists a new circle that intersects all six given circles.
## 9th grade | 78.17. Construct a circle through the points of tangency of the following pairs of circles: the first and second, the third and fourth, the fifth and sixth. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,996 |
20. For positive numbers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$, it is known that the inequalities are satisfied:
$$
\sum_{i \leqslant j} a_{i} a_{j} \leqslant 1, \quad \sum_{i \leqslant j} b_{i} b_{j} \leqslant 1
$$
Prove that
$$
\sum_{i \leqslant j}\left(a_{i}-b_{i}\right)\left(a_{j}-b_{j}\right) \leqslant 1
$... | 78.20. Adding obvious equalities
$$
\left(a_{i}-b_{i}\right)\left(a_{j}-b_{j}\right)=a_{i} a_{j}+b_{i} b_{j}-\left(a_{i} b_{j}+a_{j} b_{i}\right)
$$
we get
$$
S=\sum_{i \leqslant j}\left(a_{i}-b_{i}\right)\left(a_{j}-b_{j}\right)=\sum_{i \leqslant j} a_{i} a_{j}+\sum_{i \leqslant j} b_{i} b_{j}-\sum_{i \leqslant j}\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,998 |
27. Given a convex hexagon, each of its major diagonals bisects its area. Prove that these diagonals intersect at one point.
## FINAL ROUND | 78.27. Suppose that in the hexagon $A B C D E F$ the diagonals $A D, B E$ and $C F$ do not intersect at one point (see Fig. 51). Then, since the equalities
$$
S(A B X)=S(X D E), \quad S(B C Z)=S(Z E F), \quad S(C D Y)=S(Y A F)
$$
must hold, the inequalities
$$
A X \cdot B X > D Y \cdot E Z, \quad F Z \cdot E Z > B X... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,001 |
30. At the vertices of a regular 100-gon, integers are placed. Every minute, each number is replaced by its difference with the number following it in a clockwise direction. Prove that after 5 minutes, the sum of the numbers at the vertices of our 100-gon will be divisible by five. | 78.30. If we denote six consecutive numbers by $a_{k}, a_{k+1}, \ldots, a_{k+5}$, then after 5 minutes, the number at the position of $a_{k}$ will be $a_{k}-5 a_{k+1}+10 a_{k+2}+10 a_{k+3}+5 a_{k+4}-a_{k+5}$. Summing all such expressions and discarding terms that are obviously divisible by five, we get $\sum_{k}\left(a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,003 |
31. On a line, there are 1978 segments, no two of which have common endpoints. Prove that these segments cannot be numbered such that for any $k$ from 1 to 1978, the $k$-th segment contains exactly $k$ endpoints of other segments. | 78.31. If segment $A B$ contains $x$ ends of segment $C D$, and segment $C D$ contains $y$ ends of segment $A B$, then, as can be easily obtained by simple enumeration, the numbers $x$ and $y$ have the same parity. From this, it follows that if we sum all the numbers $a_{k}$, where $a_{k}$ denotes the number of ends of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,004 |
$32 *$. The sequence $\left(a_{n}\right)$, all terms of which are equal to 0 or 1, is such that if $k<2^{n}$, then $a_{k}$ is not equal to $a_{k+2^{n}}$. Prove that it is non-periodic. | 78.32. Suppose this is not the case, and for convenience, let's assume that $a_{1}=1$. From the definition of the sequence, it follows that if the binary representation of the number $n$ contains an even number of ones, then $a_{n}=0$; otherwise, $a_{n}=1$. If the period has length $p$, then for sufficiently large $k$ ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,005 |
35 *. Given a convex polygon $M$ such that the lengths of all its sides and diagonals are integers. $K$ is a given square. Prove that there is a finite set of polygons congruent to $M$ such that their union contains $K$ and any point of the square not lying on the side of one of the polygons is covered by the same numb... | 78.35. Let us position the polygon so that one of its vertices $A_{1}$ is at the origin, and the side $A_{1} A_{2}$ lies along the x-axis (see Fig. 52). Since all the lengths of the diagonals and sides are rational, the cosines of the angles $A_{k} A_{1} A_{l}$ are also rational. Therefore, since $\cos (\alpha-\beta)=\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,007 |
6. $a, b, c$ are prime numbers, and $a+b$ and $a b$ are divisible by $c$. Prove that $a^{3}-b^{3}$ is divisible by $c$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 79.6. $\quad$ Since $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)=(a-b) \times$ $\times\left((a+b)^{2}-a b\right)$, then due to the fact that $a+b$ and $a b$ are divisible by $c$, we get that $(a+b)^{2}-a b$ is divisible by $c$, and therefore, the difference $a^{3}-b^{3}$ is also divisible by $c$. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,010 | |
10. What is the greatest number of natural numbers less than 50 that can be chosen so that any two are coprime? | 79.10. Answer. 16. Hint. Replace each number, except 1, with any of its prime divisors.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,012 |
12. Solve the system of equations:
$$
\left\{\begin{array}{l}
1+A+B=A B \\
2+A+C=A C \\
5+B+C=B C
\end{array}\right.
$$ | 79.12. Answer. $A=2, B=3, C=4$ or $A=0, B=-1, C=$ $=-2$. | A=2,B=3,C=4orA=0,B=-1,C=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,014 |
13. In triangle $A B C$, $A A_{1}$, $B B_{1}$, $C C_{1}$ are altitudes, and $A A_{0}$, $B B_{0}$, $C C_{0}$ are medians. Prove that the length of the broken line $A_{0} B_{1} C_{0} A_{1} B_{0} C_{1} A_{0}$ is equal to the perimeter of triangle $A B C$. | 79.13. Each segment of this broken line is the median in some right-angled triangle (for example, $A_{0} B_{1}$ is the median of triangle $B B_{1} C$), and therefore its length is half the length of the hypotenuse of the corresponding triangle ( $\left.A_{0} B_{1}=B C / 2\right)$. By adding all these equalities, we obt... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,015 |
16. Triangle $A B C$ is inscribed in a circle. $A_{1}$ is the midpoint of arc $B C$, $B_{1}$ is the midpoint of arc $A C$, $C_{1}$ is the midpoint of arc $A B$. The sides of triangle $A B C$ cut off smaller segments on segments $A_{1} B_{1}$, $B_{1} C_{1}$, $A_{1} C_{1}$ with midpoints $M_{1}$, $M_{2}$, $M_{3}$ respect... | 79.16. Let $P=A_{1} C_{1} \cap A B, \quad Q=A_{1} C_{1} \cap B C$. Then $\triangle B P Q$ is an isosceles triangle, from which it follows that the median $B M_{3}$ in it coincides with the bisector lying on the line $B B_{1}$. Further, $B M_{3} \perp A_{1} C_{1}$ and, therefore, the angle $\angle C_{1} M_{3} B_{1}=90^{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,018 |
17. Prove that there are infinitely many natural numbers that cannot be represented as the sum of the square of an integer and a prime number.
## 8th grade | 79.17. We can take numbers of the form $(3 k+2)^{2}$. Indeed,
$$
n^{2}+p=(3 k+2)^{2} \Rightarrow p=(3 k+2-n)(3 k+2+n)
$$
Such a factorization is only possible if the first factor is 1, but then $p=3 k+2+3 k+1=6 k+3$, which is impossible for $k>0$ since $p$ is a prime number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,019 |
18. Prove that for any natural $k>1$ the number $1010 \ldots 0101$ ($k+1$ ones and $k$ zeros) is composite. | 79.18. For odd $k>1$, the given number is obviously divisible by 101. For even $k$, after multiplying by 11, we get a number that is divisible by $A=111 \ldots 11(k+1$ ones). Since $A$ is coprime with 11, the original number is also divisible by $A$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,020 |
20. In triangle $A B C$, angle $B$ is equal to $60^{\circ}, A K$ and $C E$ are angle bisectors, and points $K$ and $E$ lie on the corresponding sides. Segments $A K$ and $C E$ intersect at point $O$. Prove that $O K=O E$. | 79.20. Direct calculation of angles shows that $\angle E O K=120^{\circ}$. Therefore, the quadrilateral $B E O K$ is cyclic. Since $B O$ is the bisector of angle $B$, then $\angle O B E=\angle O B K$, and thus the chords $O E$ and $O K$ are equal in length. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,021 |
22. Given $n$ pairwise coprime natural numbers $a_{1}, a_{2}, \ldots, a_{n}$, such that $1<a_{i}<(2 n-1)^{2}$. Prove that among them there is at least one prime. | 79.22. If all these numbers are composite, then each of them must have a prime divisor less than $2 n-1$. But there are no more than $n-1$ such prime numbers, and therefore, some two of them must coincide, which contradicts their mutual simplicity. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,023 |
23. In each cell of an $n \times m$ table, there is a one. It is allowed to take any $2 \times 2$ square and change the sign of all numbers in it. Can we achieve a chessboard pattern of signs in the table using such operations? The answer should depend on $m$ and $n$.
## 9th grade | 79.23. Answer. Only if $n$ and $m$ are both divisible by four. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,024 |
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