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Problem 4. 13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." The latter said to their right neighbor... | # Answer: 7.
Solution. It is clear that there were both boys and girls at the table. Let's see how the children were seated. After a group of boys sitting next to each other comes a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so t... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Small Island and Big Island have a rectangular shape and are divided into rectangular counties. In each county, a road is laid along one of the diagonals. On each island, these roads form a closed path that does not pass through any point more than once. Here is how Small Island is organized, with a total of... | Answer. Figure 2 provides an example for 9 counties.
Comment. We will show that examples do not exist for 7 counties (or fewer), while at the same time pointing out a property characteristic of all such examples.
All roads can be divided into two types: some roads connect the top-left corner of a county with the bott... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In the square, some cells are shaded as shown in the figure. It is allowed to fold the square along any grid line and then unfold it back. Cells that coincide with shaded cells when folded are also shaded. Can the entire square be shaded:
a) in 5 or fewer;
b) in 4 or fewer;
.
Solution. For example, it is possible to paint the entire lower half of the board with two vertical bends, after which the upper half can be painted with one horizontal bend - see the figure. (There are other solutions as well.)
Comment. It is impossible to paint all cells ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A bag of sunflower seeds was passed around a table. The first person took 1 seed, the second took 2, the third took 3, and so on: each subsequent person took one more seed than the previous one. It is known that in the second round, the total number of seeds taken was 100 more than in the first round. How ma... | Answer: 10 people.
Solution: Let there be $n$ people sitting at the table. Then on the second round, the first person took the $n+1$-th sunflower seed, the second person took the $n+2$-th - and generally, each person took $n$ more seeds than on the first round. Altogether, on the second round, they took $n \cdot n=n^{... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Three square paths with a common center are 1 m apart from each other (see figure). Three ants start simultaneously from the lower left corners of the paths and run at the same speed: Mu and Ra counterclockwise, and Wei clockwise. When Mu reaches the lower right corner of the largest path, the other two, who... | Answer: 4 m, 6 m, 8 m.
Solution: The lengths of the sides of two adjacent paths differ by 2 m (Fig. 3). Therefore, at the moment when Mu reached the corner, Ra had run 2 m along the right side of the path and was at a distance of $2+1=3$ m from the "lower" side of the outer path. Since $\mathrm{Pa}$ is halfway between... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Fox Alice and Cat Basil have grown 20 fake banknotes on a tree and are now filling in seven-digit numbers on them. Each banknote has 7 empty cells for digits. Basil calls out one digit at a time, either "1" or "2" (he doesn't know any others), and Alice writes the called digit in any free cell of any banknot... | Answer: 2.
Solution: Basil can always get two banknotes: he knows the place where the last digit should be written and names it so that it differs from the digit in the same place on some other banknote. Then the numbers on these two banknotes will be different, and the cat can take them.
We will show how Alice can e... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. On the surface of a planet shaped like a donut, two snails crawled, leaving trails behind: one along the outer equator, and the other along a spiral line (see figure). Into how many parts did the snails' trails divide the surface of the planet? (It is sufficient to write the answer.)
 and \(BC\) of square \(ABCD\) with side length 10, points \(K\) and \(L\) are marked such that \(AK = CL = 3\). On segment \(KL\), a point \(P\) is chosen, and on the extension of segment \(AB\) beyond point \(B\), a point \(Q\) is chosen such that \(AP = PQ = QL\) (see figure).
a) Prove... | Solution. Drop a perpendicular $PH$ from point $P$ to side $AB$. Since triangle $KBL$ is isosceles and right-angled, $\angle PKH = 45^\circ$. Therefore, triangle $KPH$ is also isosceles and $KH = HP$.
, while at the same time pointing out a property characteristic of all such examples.
All roads can be divided into two types: some roads connect the top-left corner of a county with the bott... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In the square, some cells are shaded as shown in the figure. It is allowed to fold the square along any grid line and then unfold it back. Cells that coincide with shaded cells when folded are also shaded. Can the entire square be shaded:
a) in 5 or fewer;
b) in 4 or fewer;
.
Solution. For example, it is possible to paint the entire lower half of the board with two vertical bends, after which the upper half can be painted with one horizontal bend - see the figure. (There are other solutions as well.)
Comment. It is impossible to paint all cells ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The numbers $2,3,4, \ldots, 29,30$ are written on the board. For one ruble, you can mark any number. If a number is already marked, you can freely mark its divisors and numbers that are multiples of it. What is the minimum number of rubles needed to mark all the numbers on the board? [6 points]
(I.V. Yashch... | Answer. For 5 rubles.
Solution. Let's mark the numbers $17, 19, 23$, and 29, spending four rubles. Then mark the number 2, spending another ruble. After this, we can freely mark all even numbers (since they are divisible by 2), and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$),... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A rectangular sheet of paper was folded, aligning a vertex with the midpoint of the opposite shorter side (Fig. 12). It turned out that triangles I and II are equal. Find the longer side of the rectangle if the shorter side is 8.
[6 points] (A. V. Khachatryan) | Answer: 12.
Fig. 12
Solution. Let's mark the equal segments (Fig. 13 - here we used the fact that in congruent triangles, sides opposite equal angles are equal). We see that the length of t... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Pete liked the puzzle, he decided to glue it together and hang it on the wall. In one minute, he glued together two pieces (initial or previously glued). As a result, the entire puzzle was joined into one complete picture in 2 hours. How long would it have taken to assemble the picture if Pete had glued togethe... | Answer: In one hour.
Solution 1. Each gluing reduces the number of pieces on the table by 1. Since after 120 gluings one piece (the complete puzzle) was obtained, there were 121 pieces at the beginning. Now, if three pieces are glued together per minute (i.e., the number of pieces is reduced by 2), one piece will rema... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 5. Replace in the equation
$$
\text { PIE = SLICE + SLICE + SLICE + ... + SLICE }
$$
identical letters with identical digits, and different letters with different digits, so that the equation is true, and the number of "slices of pie" is the largest possible. | Answer. The maximum number of "pieces" is seven, for example: ПИРОГ $=95207$, КУСОК $=13601$.
Solution. An example for seven "pieces" is given above. We will show that there cannot be more than seven "pieces". For this, it is convenient to rewrite the condition as a multiplication example: ПИРОГ $=$ КУСОК $\cdot n$, w... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. Let param1. What is the largest possible value of param2?
It is given that param1. Find the largest possible value of param2.
| param 1 | param 2 | Answer |
| :---: | :---: | :---: |
| $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=3$ | $6 \sin x$ | 4 |
| $\frac{25 \sin ^{2} x-37+40 \cos x}{... | # Solution
Notice that $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=\frac{6-(3 \sin x-2)^{2}}{2+(3 \cos x+\sqrt{5})^{2}}$. The obtained expression can equal 3 only if $3 \sin x-2=0$ and $3 \cos x+\sqrt{5}=0$. Therefore, the expression $6 \sin x$ can only take the value 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. What is the smallest number of different integers that need to be taken so that among them one can choose both a geometric and an arithmetic progression of length $5$?
(M.A. Evdokimov) | Solution. Let's provide an example of six integers that satisfy the condition: $-8, -2, 1, 4, 10, 16$. The numbers $1, -2, 4, -8, 16$ form a geometric progression, while the numbers $-8, -2, 4, 10, 16$ form an arithmetic progression.
We will show that no five distinct integers satisfy the condition of the problem. Sup... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. On an island, there live chameleons of five colors. When one chameleon bites another, the color of the bitten chameleon changes according to some rule, and the new color depends only on the color of the biter and the color of the bitten. It is known that 2023 red chameleons can agree on a sequence of bites, after wh... | # Answer. For $k=5$. Solution.
First, let's provide an example of rules under which the described recoloring would require at least 5 red chameleons. Let's number the colors so that red is the first color and blue is the last. Then, let the rules be as follows: if a chameleon of color $k1$ can reduce the number of its... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (B. Novikov) On each cell of an $8 \times 8$ board, a guard is placed. Each guard can look in one of four directions (along the lines of the board) and watch all the guards on the line of his sight. For what largest $k$ can the guards' gazes be directed so that each guard is watched by at least $k$ other guards? | Solution: Answer: 5. We will prove that $k \leq 5$. For this, assume that $k \geqslant 6$. Consider the guards standing at the corners of the board. Each of them is watched by at least 6 guards, and these guards must stand at the edge of the board. Moreover, if a guard sees one of the corner guards, they do not see the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the largest natural number $n$ with the following property: for any odd prime $p$ less than $n$, the difference $n-p$ is also a prime number.
(I. Akulich) | Answer: 10
Solution: Suppose that $n>10$. Notice that the numbers $n-3, n-5, n-7$ are all greater than three, and one of them is divisible by three, and consequently composite. Contradiction. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Among any five nodes of a regular grid paper, there will definitely be two nodes, the midpoint of the segment between which is also a node of the grid paper. What is the minimum number of nodes of a grid made of regular hexagons that need to be taken so that among them, there will definitely be two nodes, the midpoi... | # Answer. 9
Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof. Introduce the origin at one of the grid nodes and denote by $\vec{a}$ and $\vec{b}$ the radius vectors to the two nearest nodes (see picture). Th... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. A polyhedron with vertices at the midpoints of the edges of a certain cube is called a cuboctahedron. When the cuboctahedron is intersected by a plane, a regular polygon is obtained. What is the maximum number of sides this polygon can have?
(M. A. Evdokimov) | Solution.
Let the edge of the original cube, from which the cuboctahedron is obtained, be 1. Consider the sections of the cuboctahedron by a plane parallel to the base of the cube at a distan... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (I. Akulich) Find the largest natural $n$ with the following property: for any odd prime $p$ less than $n$, the difference $n-p$ is also a prime number. | Solution: Answer: 10. Indeed, $10=3+7=5+5$. We will prove that numbers greater than 10 do not work. Let $n$ be a number greater than 10. Note that the numbers $3, 5, 7$ give different remainders when divided by 3. Then the numbers $n-3, n-5, n-7$ give different remainders when divided by 3, meaning one of them is divis... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. What is the smallest number of triangular pyramids (tetrahedrons) into which a cube can be divided? | 14. It is easy to see that the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ can be divided into five tetrahedra: if we cut off the tetrahedra $B A C B_{1}$ and $D A C D_{1}$, as well as the tetrahedra $A_{1} B_{1} D_{1} A$ and $C_{1} B_{1} D_{1} C$, we will have one more (fifth) tetrahedron $A C B_{1} D_{1}$ left (Fig. 51; t... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
49. Let $h_{a}=\beta_{b}=m_{c}$, where $h_{a}=A P, \beta_{b}=B L$ and $m_{c}=C F-$ are the altitude, bisector, and median drawn from three different vertices of some triangle $A B C$. What value can the ratio of the greatest side of triangle $A B C$ to its smallest side have? | 49. First of all, note that if for triangle \(ABC\) the inequality \(m_{a}b = CA\) holds. This immediately follows from the known formula \(^{2}\) \(m_{a} = \frac{1}{2} \sqrt{2 b^{2} + 2 c^{2} - a^{2}}\), but it can also be proven without calculations. Translate triangle \(ABC\) parallel to vector \(\overline{ED}\) to ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In space, there are 4 points not lying in the same plane. How many planes can be drawn equidistant from these points? | 1. Since the four points do not lie in the same plane, the plane equidistant from these points cannot be located on the same side of all of them. Therefore, there are only two possible cases: 1) three points lie on one side of the considered plane, and the fourth point lies on the other side, and 2) two points lie on e... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. How many spheres exist that touch all the faces of the given triangular pyramid $T$?
| 3. The task is to determine how many points (centers of the sought spheres) are equidistant from the four faces of the pyramid.
The geometric locus of points equidistant from the faces of a given dihedral angle is a plane passing through the edge of the dihedral angle and bisecting this angle — the bisector plane of t... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
69. What is more likely: to win 3 games out of 4 or 5 games out of 8 against an equally matched opponent? | 69. The total number of possible outcomes of a sequence of four matches we get by combining a win or loss in the first match with a win or loss in the second, third, and fourth matches. This number is equal to $2^{4}=16$. All these outcomes are equally probable, as the opponents are of equal strength, and for each sepa... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
103*. The bus network of the city, consisting of several (more than two) routes, is organized in such a way that:
$1^{\circ}$ each route has at least three stops,
$2^{\circ}$ from any stop to any other stop, one can travel without transferring, and
$3^{\circ}$ for each pair of routes, there is one (and only one) sto... | 103. a) Let \( n \) be the number of stops on one of the bus routes in the city. We need to prove that in this case, each route will have exactly \( n \) stops and that exactly 2 routes will pass through each stop.
Let's denote the route with \( n \) stops by \( a \), and the stops on this route by \( A_{1}, A_{2}, \l... | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
4.11. The centers of three spheres, with radii of 3, 4, and 6, are located at the vertices of an equilateral triangle with a side length of 11. How many planes exist that are tangent to all three spheres simultaneously?
## § 3. Two intersecting circles lie on the same sphere | 4.11. Consider a plane tangent to all three given spheres, and draw a plane through the center of the sphere of radius 3 parallel to it. The resulting plane is tangent to the spheres of radii $4 \pm 3$ and $6 \pm 3$, concentric with the spheres of radii 4 and 6. When the signs of the number 3 are the same, the tangency... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.19. Chord $A B$ of a sphere with radius 1 has a length of 1 and is positioned at an angle of $60^{\circ}$ to the diameter $C D$ of this sphere. It is known that $A C=\sqrt{2}$ and $A C<B C$. Find the length of the segment $B D$. | 4.19. Let $O$ be the center of the sphere. Take a point $E$ such that $\overrightarrow{C E}=\overrightarrow{A B}$. Since $\angle O C E=60^{\circ}$ and $C E=1=O C$, then $O E=1$. Point $O$ is equidistant from all vertices of the parallelogram $A B E C$, so $A B E C$ is a rectangle and the projection $O_{1}$ of point $O$... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.35. a) Prove that the sum of the cosines of the dihedral angles of a regular tetrahedron is 2.
b) The sum of the plane angles of a trihedral angle is \(180^{\circ}\). Find the sum of the cosines of its dihedral angles.
## § 5. Orthocentric Tetrahedron
Definition. A tetrahedron is called orthocentric if all its alt... | 6.35. a) Let $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$ and $\mathbf{e}_{4}$ be unit vectors, non-perpendicular to the faces and directed outward. Since the areas of all faces are equal, then $\mathbf{e}_{1}+\mathbf{e}_{2}+\mathbf{e}_{3}+\mathbf{e}_{4}=0$ (see problem 7.19). Therefore, $0=\left|e_{1}+\mathbf{e}_{... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
7.37. What is the maximum number of planes of symmetry that a spatial figure consisting of three pairwise non-parallel lines can have?
Symmetry with respect to a line $l$ is a transformation of space that maps a point $X$ to a point $X^{\prime}$ such that the line $l$ passes through the midpoint of the segment $X X^{\... | 7.37. Let $P$ be a plane of symmetry of a figure consisting of three pairwise non-parallel lines. There are only two possible cases: 1) each given line is symmetric relative to $P$; 2) one line is symmetric relative to $P$, and the other two lines are symmetric to each other.
In the first case, either one line is perp... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.26. a) How many pairwise non-equal spatial quadrilaterals exist with one and the same set of side vectors?
b) Prove that the volumes of all tetrahedra defined by these spatial quadrilaterals are equal. | 8.26. a) We fix one of the vectors of the sides. It can be followed by any of the three remaining vectors, and then by any of the two remaining ones. Therefore, there are exactly 6 different quadrilaterals.
b) Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and $\mathbf{d}$ be the given vectors of the sides. Consider the pa... | 6 | Geometry | proof | Yes | Yes | olympiads | false |
9.9. a) Prove that it is possible to choose 8 vertices of a dodecahedron such that they are the vertices of a cube. How many ways can this be done?
b) Prove that it is possible to choose 4 vertices of a dodecahedron such that they are the vertices of a regular tetrahedron. | 9.9. a) From the solution to problem 9.2, it is clear that there exists a cube whose vertices are located at the vertices of the dodecahedron. Moreover, one edge of the cube is located on each face of the dodecahedron. It is also clear that choosing any of the five diagonals of a certain face of the dodecahedron as an ... | 5 | Geometry | proof | Yes | Yes | olympiads | false |
15.22. What is the smallest number of tetrahedra into which a cube can be cut? | 15.22. If a tetrahedron \(A^{\prime} B C^{\prime} D\) is cut out from the cube \(A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}\), the remaining part of the cube splits into 4 tetrahedra, i.e., the cube can be cut into 5 tetrahedra.
We will prove that the cube cannot be cut into fewer than 5 tetrahedra. The face ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15.32. A plane intersects the lower base of a cylinder along its diameter and has only one common point with the upper base. Prove that the area of the cut-off part of the lateral surface of the cylinder is equal to the area of its axial section. | 15.32. Let 0 be the center of the lower base of the cylinder; $AB$ - the diameter along which the plane intersects the base; $\alpha$ - the angle between the base and the intersecting plane; $r$ - the radius of the cylinder. Consider an arbitrary generatrix $XY$ of the cylinder, having a common point $Z$ with the inter... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
16.16. Given three pairwise tangent spheres $\Sigma_{1}$, $\Sigma_{2}$, $\Sigma_{3}$ and a set of spheres $S_{1}, S_{2}, \ldots, S_{n}$, such that each sphere $S_{i}$ is tangent to all three spheres $\Sigma_{1}$, $\Sigma_{2}$, $\Sigma_{3}$, as well as to the spheres $S_{i-1}$ and $S_{i+1}$ (it is meant that $S_{0}=S_{n... | 16.16. Consider the inversion with the center at the point of tangency of spheres $\Sigma_{1}$ and $\Sigma_{2}$. In this case, they transform into a pair of parallel planes, and the images of all other spheres touch these planes, and therefore their radii are equal. Thus, in a section by a plane equidistant from these ... | 6 | Geometry | proof | Yes | Yes | olympiads | false |
61. Determine the remainder of the division by 7 of the number
$$
3^{100}
$$
The 10-arithmetic and 7-arithmetic we have discussed are special cases of residue arithmetic modulo $m$, or $m$-arithmetic. Let $m$ be any positive integer. The elements of $m$-arithmetic are the numbers $0,1,2, \ldots, m-1$. Addition and mu... | 61. Let's compute $3^{100}$ in 7-arithmetic:
$$
3^{2}=2,3^{8}=6,3^{4}=4,3^{5}=5,3^{6}=1
$$
Therefore,
$$
3^{100}=3^{6 \cdot 16+4}=\left(3^{6}\right)^{16} \cdot 3^{4}=3^{4} \Rightarrow 4
$$
$14^{*}$ | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
117. Let three elements $x, y, z$ of the circular sequence $\breve{\Phi}_{m}$ be arranged such that the distance between $x$ and $y$ is equal to the distance between $y$ and $z$ (Fig. 54). Prove that if $x=y=0$, then $z=0$. | 117. The number $y=0$ is at an equal distance from $x$ and $z$. Therefore, based on problem 116, either $x+z=0$ or $x-z=0$. Since $x=0$, in both cases $z=0$, | 0 | Number Theory | proof | Yes | Yes | olympiads | false |
120. Prove that if a circular sequence without repetitions $\Phi_{m}$ contains zero and consists of an odd number of elements, then the number of its elements is three. | 120. If the circular sequence $\breve{\Phi}_{m}$ contains zero and consists of an odd number of terms, then there will be a pair of adjacent elements $x, y$ located at the same distance from zero (Fig. 143). According to problem 116, either
 A. P. Kiselev, Algebra, Part 2, 23rd edition, Uchpedgiz, M., 1946, p. 78.
18 Law. Zzya. E. Dynkin and V. Uspenskii
moment, the chip is located at point $a$ (Fig. 156). After each unit of time, it moves to... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Three circles with radii 1, 2, and 3 touch each other externally. Find the radius of the circle passing through the points of tangency of these circles. | 2. Note that the circle passing through the points of tangency is inscribed in a triangle with vertices at the centers of the circles. By equating the expressions for the area of the triangle obtained using Heron's formula and as the product of the semiperimeter and the radius of the inscribed circle, we find \( r=1 \)... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. In a trapezoid, the diagonals are equal to 3 and 5, and the segment connecting the midpoints of the bases is equal to 2. Find the area of the trapezoid. | 7. Let (Fig. 2) $ABCD$ be the given trapezoid ($BCAD$). Draw a line through $C$ parallel to $BD$, and denote by $K$ the point of intersection of this line with the line $AD$. The area of $\triangle ACK$ is equal to the area of the trapezoid, the sides $AC$ and $CK$ are equal to the diagonals of the trapezoid, and the m... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. A circle of radius 1 is inscribed in triangle $A B C$, where $\cos \widehat{A B C}=0.8$. This circle touches the midline of triangle $A B C$, parallel to side $A C$. Find the length of side $A C$. | 9. It follows from the condition that the height to side $AC$ is equal to two diameters of the inscribed circle, i.e., 4. If $M, N$ and $K$ are the points of tangency with $AB, BC$ and $CA$, then
$$
|BM|=|BN|=r \operatorname{ctg} \frac{\widehat{ABC}}{2}=1 \cdot \sqrt{\frac{1+0.8}{1-0.8}}=\sqrt{9}=3
$$
If $|MA|=|AK|=x... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
35*. What is the greatest number of rays in space that form pairwise obtuse angles? | 35. It is easy to see that it is possible to draw four rays, each pair of which forms obtuse angles: for example, it is sufficient to connect the center $O$ of a regular tetrahedron $ABCD$ with all its vertices (it is clear, for example, that $\angle AOB > \angle AO_1B = 90^\circ$, where $O_1$ is the center of the face... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
52. Given a triangle $T$.
a) Place a centrally symmetric polygon $m$ of the largest possible area inside $T$.
What is the area of $m$ if the area of $T$ is 1?
b) Enclose $T$ in a convex centrally symmetric polygon $M$ of the smallest possible area.
What is the area of $M$ if the area of $T$ is 1? | 52. a) If $O$ is the center of the sought centrally symmetric polygon $m$, then by symmetrically reflecting the given $\triangle ABC \equiv T$ together with $m$ (which will transform into itself), we can see that $m$ is also inscribed in $\triangle A_{1} B_{1} C_{1} \equiv T_{1}$, which is symmetric to $\triangle ABC$ ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
90. On a plane, there are 4 points $A_{1}, A_{2}, A_{3}, A_{4}$, the distance between any two of which is not less than 1. What is the maximum possible number of line segments $A_{i} A_{j}$ of length 1 connecting these points pairwise? | 90. The number $v_{2}(4)<6$, otherwise the number of diameters of a system of four points on a plane could equal the "complete" number of six segments connecting our points, and by the result of problem 87 a) this is not the case. However, the number of segments equal to 1, connecting the points of our system of four p... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. a) Points $A$ and $B$ move uniformly and with equal angular velocities along circles centered at $O_{1}$ and $O_{2}$, respectively (clockwise). Prove that vertex $C$ of the equilateral triangle $A B C$ also moves uniformly along some circle.
b) The distances from a fixed point $P$ on the plane to two vertices $A, B... | 6. a) Let $\overrightarrow{\mathrm{O}_{1} \mathrm{O}_{3}}$ be the vector obtained by rotating $\overrightarrow{O_{1} O_{2}}$ by $60^{\circ}$ (in the same direction as the rotation that transforms $\overrightarrow{A B}$ into $\overrightarrow{A C}$). Points $A^{\prime}$ and $B^{\prime}$ are the images of $A$ and $B$ unde... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. Kolya and Petya are dividing $2 n+1$ nuts, $n \geqslant 2$, with each wanting to get as many as possible. There are three ways of dividing (each goes through three stages).
1st stage: Petya divides all the nuts into two parts, each containing no fewer than two nuts.
2nd stage: Kolya divides each part again into t... | 10. Answer: the most profitable way for Kolya is the first one; with the second and third, he will get one nut less with correct play. (In general, as we will see, the dispute in this division is over one nut.)
No matter what piles (of $a$ and $b$ nuts, $a<b$) form after Petya's first move, Kolya can split the larger ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
21. Let's take any 1962-digit number divisible by 9. The sum of its digits will be denoted by $a$, the sum of the digits of $a$ by $b$, and the sum of the digits of $b$ by $c$. What is $c$?[^0] | 21. Answer. $c=9$. The sum of the digits of any number gives the same remainder when divided by 9 as the number itself (PZ). On the other
Fig. 30 hand, $a \leqslant 1962.9<19999$, so $b \l... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
23. What is the maximum area that a triangle with sides \(a, b, c\) can have, given the following constraints:
\[
0 \leqslant a \leqslant 1 \leqslant b \leqslant 2 \leqslant c \leqslant 3 \text { ? }
\] | 23. Answer: 1. Among triangles with two sides $a, b$ that satisfy the conditions $0<a \leqslant 1,1 \leqslant b \leqslant 2$, the one with the largest area is the right triangle with legs $a=1, b=2$ (indeed, $s \leqslant a b / 2 \leqslant 1$, since the height dropped to side $b$ is no more than $a$). The third side of ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
43. For each of the numbers from 1 to 1000000000, the sum of its digits is calculated, and for each of the resulting billion numbers, the sum of its digits is calculated again, and so on, until a billion single-digit numbers are obtained. Which number will there be more of: 1 or 2? | 43. Answer: the number of ones will be one more than the number of twos.
Any number gives the same remainder when divided by 9 as the sum of its digits (DS). Therefore, in our problem, ones come from numbers that give a remainder of 1 when divided by 9, i.e., from the numbers 1, 10, 19, 28, ..., 999999991, 1000000000,... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
46. Solve the equation in integers
$$
\sqrt{\sqrt{x+\sqrt{x+\sqrt{x+\ldots+\sqrt{x}}}}}=y
$$ | 46. Answer: the only solution is $x=y=0$.
Let $\boldsymbol{x}$ and $\boldsymbol{y}$ be integers satisfying the condition. After a series of squaring, we are convinced that $\sqrt{x+\sqrt{x}}=\boldsymbol{m}$ and $\sqrt{x}=k$ are integers, and
$$
m^{2}=k(k+1)
$$
If $k>0$, then it must be that $k^{2}<m^{2}<(k+1)^{2}$, ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
53. What is the smallest number of non-overlapping tetrahedra into which a cube can be divided? | 53. It is easy to see that a cube can be divided into 5 tetrahedra. In Fig. 40, these are the tetrahedra $A A^{\prime} B^{\prime} D^{\prime}, A B^{\prime} B C, A C D D^{\prime}, B^{\prime} C^{\prime} D^{\prime} C$, and $A C D^{\prime} B^{\prime}$.
Let us now prove that it is impossible to divide the cube into fewer te... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
92. Three consecutive vertices of a rhombus lie on the sides $A B, B C, C D$ of a given square with side 1. Find the area of the figure filled by the fourth vertices of such rhombi. | 92. Answer: $S=1$.
Let $K, L, N$ be the vertices of the rhombus on the sides $A B, B C$, and $A D$ of the square (Fig. $52, a$). Note that the length $K B$ is equal to the distance from point $M$ to the line $A D$. Therefore, if we fix point $K$, the possible positions of point $M$ fill a certain segment $M_{1} M_{2}$... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
99. In a regular $n$-gon ( $n>5$ ), the difference between the largest and smallest diagonals is equal to the side. Find $n$. | 99. Answer: $n=9$.
Let $a_{n}$ be the side length, and $D_{n}$ and $d_{n}$ be the lengths of the largest and smallest diagonals of a regular $n$-gon. For $n=4$ and $n=5$, all diagonals are equal. For $n=6$ and $n=7$, $D_{n}-a_{n}=2 A K=D_{8}-d_{8}$. For $n=9$ (Fig. 55,6), similarly, we get $\angle A B K=30^{\circ}$, s... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
123. In a certain state, the airline system is arranged in such a way that any city is connected by air routes to no more than three other cities, and from any city to any other, one can travel with no more than one layover.
What is the maximum number of cities that can be in this state? | 123. Answer: 10 cities.
From any city $A$, one can reach no more than three cities, and from each of them no more than two (excluding $A$). Thus, the total number of cities is no more than $1+3+3 \cdot 2=$ $=10$.
The example in Fig. 60 shows that the required system of air routes in a state with ten cities exists.
$... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
131. How many sides in a convex polygon can be equal in length to the longest diagonal? | 131. Answer: no more than two. An example of a polygon with two sides equal to the largest diagonal is shown in Fig. 61.
Suppose there are more than two such sides. We select two of them, $A B$ and $C D$, which do not share any vertices (this is possible, [^3] since a polygon with diagonals is not a triangle). Then at... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
190. Among the numbers of the form $36^{k}-5^{l}$, where $k$ and $l$ are natural numbers, find the smallest in absolute value. Prove that the found number is indeed the smallest. | 190. Answer: $11=36-5^{2}$.
The last digit of the number $36^{h}=6^{2 k}$ is 6, the last digit of the number $5^{t}$ is 5. Therefore, the number $\left|6^{2 k}-5^{t}\right|$ ends in either 1 (if $6^{2 k}>5^{t}$) or 9 (if $6^{2 k}<5^{t}$).
The equation $6^{2 k}-5^{t}=1$ is impossible, because then it would be $5^{l}=\... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
238. Several black and white chips are arranged in a circle. Two players take turns performing the following operation: the first player removes all black chips that have a white neighbor (at least on one side), and the second player then removes all white chips that have a black neighbor. They continue doing this unti... | 238. Answers: a) yes, b) 8 moves (each player - per move).
Fig. 102 shows an example of arranging 41 chips; next to each chip is its rank-number, indicating how many moves before the end it will be removed. The construction of such an arrangement is convenient "from the end"; to the remaining last black chip of rank 0... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
306. We will say that a number has the property $\mathrm{P}(k)$ if it can be factored into the product of $k$ consecutive natural numbers, all greater than 1.
a) Find $k$ such that some number $N$ simultaneously has the properties $\mathrm{P}(k)$ and $\mathrm{P}(k+2)$.
b) Prove that there are no numbers that simultane... | 306. a) $k=3 ; 720=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6=8 \cdot 9 \cdot 10$.
b) If $m(m+1)=n(n+1)(n+2)(n+3)$, then $m^{2}+m+1=\left(n^{2}+3 n+1\right)^{2}$, which is impossible. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
325. a) Find the smallest possible value of the polynomial
$$
P(x, y)=4+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}
$$
b) ${ }^{*}$ Prove that this polynomial cannot be represented as a sum of squares of polynomials in the variables $x, y$. | 325. a) Answer: 3 at $x=y=1$.
From the inequality of the arithmetic mean, it follows that $1+x^{2} y^{4}+x^{4} y^{2} \geqslant 3 x^{2} y^{2}$.
b) Let $P(x, y)=g_{1}^{2}(x, y)+g_{2}^{2}(x, y)+\ldots+g_{n}^{2}(x, y)$, where $g_{i}(x, y), i=1,2, \ldots, n-$ are polynomials. Since $P(x, 0)=$ $=P(0, y)=4$, the polynomials... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
389. The sequence $x_{n}$ is defined recursively: $\quad x_{1}=1 ; x_{2}=1 ; x_{n+2}=x_{n+1}^{2}-\frac{1}{2} x_{n}, \quad$ if $n \geqslant 1$. Prove that the sequence $x_{n}$ has a limit, and find it.
390 *. In the white cells of a chessboard of size $1983 \times 1984$, numbers 1 or -1 are written such that for any bl... | 389. Answer: $\lim _{n \rightarrow \infty} x_{n}=0\left(\left|x_{n+7}\right| \leqslant \frac{1}{4}\left(\frac{5}{6}\right)^{n}\right.$ for $\left.n \geqslant 1\right)$. | 0 | Algebra | proof | Yes | Yes | olympiads | false |
432. Milk is poured into 30 glasses. A boy is trying to make the amount of milk equal in all glasses. For this, he takes any two glasses and pours milk from one to the other until the amount of milk in them is equal. Is it possible to pour milk into the glasses in such a way that the boy cannot achieve his goal, no mat... | 432. Answer: it is possible. You need to pour 100 g of milk into all the glasses except one, and into the remaining one - 200 g
256 | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
456. Every evening, Uncle Chernomor appoints 9 or 10 out of 33 bogatyrs for duty, at his discretion. What is the smallest number of days after which it can happen that each of the bogatyrs has been on duty the same number of times? | 456. Answer: 7 days. Let $k$ be the number of days when 9 bogatyrs were on duty, and $l$ be the number of days when 10 bogatyrs were on duty, with each of them being on duty $m$ times. Then $9 k + 10 l = 33 m$. For $m=1$ there are no solutions, but for $m=2$ we have $k=4$ and $l=3$. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.3. Replace the ellipsis with such a natural number $p$ so that the answer to the following question is unique: how many lines are drawn on the plane, if it is known that they intersect at ... different points? | 1.3. Answer. $p=2$.
We will prove that if there are two intersection points, there can only be three lines. Suppose there are only two intersection points $A$ and $B$, and let $l_{1}$ and $l_{2}$ be the lines intersecting at point $A$, and $l_{3}$ be a line not passing through $A$ (such a line exists, otherwise all li... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. Six pennies lie on the table, forming a closed chain (i.e., the first penny touches the second, the second touches the third, and so on, the sixth touches the first). A seventh penny, also lying on the table, rolls without slipping along the outer side of the chain, touching each of the six pennies in the chain in... | 6.3. Answer. 4 turns.
From figure $a$, it can be seen that during the time the moving coin, depicted with a dashed line, rolls along the arc $\alpha$ of the stationary coin with center $O$, it rotates by an angle of $2 \alpha$: in this figure, $M^{\prime} A^{\prime}$ is the new position of the radius $M A$, the radii ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. On the board in the laboratory, two numbers are written. Every day, senior researcher Petya erases both numbers from the board and writes down their arithmetic mean and harmonic mean instead. On the morning of the first day, the numbers 1 and 2 were written on the board. Find the product of the numbers written on th... | 1. The product of the numbers on the board does not change. Indeed,
$$
\frac{a+b}{2} \cdot \frac{2}{1 / a+1 / b}=\frac{a+b}{2} \cdot \frac{2 a b}{a+b}=a b
$$
Therefore, on the 1999th day, the product will be the same as it was on the first day. See also fact 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3 *. Find all pairs of natural numbers $x, y$ such that $x^{3}+y$ and $y^{3}+x$ are divisible by $x^{2}+y^{2}$. | 3. Let's first prove that $x$ and $y$ are coprime. Assume the opposite. Then $x$ and $y$ are divisible by some prime number $p$. Let $p$ enter the prime factorizations of $x$ and $y$ with powers $a \geqslant 1$ and $b \geqslant 1$ respectively (see fact 10). Without loss of generality, we can assume that $a \geqslant b... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4*. Find all such positive integers $k$ for which the number
$$
\underbrace{1 \ldots 1 \overbrace{2 \ldots 2}^{k}}_{2000}-\underbrace{2 \ldots 2}_{1001}
$$
is a perfect square. | 4. Let $n=1000$. Consider two cases.
$1^{\circ} . k>n$. Then
$$
\underbrace{1 \ldots 12 \ldots 2}_{2 n}-\underbrace{2 \ldots 2}_{n+1}=\underbrace{1 \ldots 1}_{2 n-k} \overbrace{2 \ldots 2}^{k-(n+1)} \underbrace{0 \ldots 0}_{n+1} .
$$
This number ends with $n+1=1001$ zeros. But if a number is a square of a natural nu... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a convex quadrilateral $A B C D$, points $E$ and $F$ are the midpoints of sides $B C$ and $C D$ respectively. Segments $A E, A F$, and $E F$ divide the quadrilateral into 4 triangles, the areas of which are consecutive natural numbers. What is the maximum possible value of the area of triangle $A B D$? | 3. Let the areas of the triangles be $n, n+1, n+2$, $n+3$. Then the area of the quadrilateral $A B C D$ is $4 n+6$. It is easy to see that the area of triangle $B C D$ is four times the area of triangle $E C F$, so this area is at least $4 n$. Therefore,
$$
S_{A B D}=S_{A B C D}-S_{B C D} \leqslant(4 n+6)-4 n=6 .
$$
... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. For a convex polyhedron, the internal dihedral angle at each edge is acute. How many faces can the polyhedron have? | 5. For each of the faces, consider the vector of the external normal, i.e., a vector perpendicular to this face and directed outward from the polyhedron.
$1^{\circ}$. Let's prove that the angle between any two external normals is obtuse or straight. Suppose this is not the case, and there exist two faces $\Gamma_{1}$ ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. A square sheet of checkered paper $8 \times 8$ was folded several times along the grid lines so that a $1 \times 1$ square was obtained. It was then cut along a segment connecting the midpoints of two opposite sides of the square. Into how many pieces could the square have split as a result? | 2. Let the cut be vertical (the case of a horizontal cut is analogous). Draw vertical segments in all $1 \times 1$ squares, connecting the midpoints of opposite sides. Notice that when folding along the grid lines, these
. | 35. Since the length of the segment $X Y$ is equal to $a$, it is necessary that the sum $A X + B Y$ be minimal. Suppose that the segment $X Y$ is laid out. The sliding symmetry with the axis $l$ "of the parallel translation magnitude $a$ translates point $B$ to $B'$, and point $Y$ to $X$ (Fig. 144); therefore, $B Y = B... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. How many real solutions does the system of two equations with three unknowns have:
$$
\left\{\begin{aligned}
x+y & =2 \\
x y-z^{2} & =1 ?
\end{aligned}\right.
$$ | Sol 1. From the second equation, it follows that $x y \geqslant 1$. The numbers $x$ and $y$ cannot both be negative, since their sum is 2. Therefore, the numbers $x$ and $y$ are positive and $x+y \geqslant 2 \sqrt{x y} \geqslant 2$, and the equality $x+y=2$ is possible only when $x=y=1$. In this case, $z=0$.
Part 2. S... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-5. The city's bus network is organized as follows: 1) from any stop to any other stop, you can get without transferring;
2) for any pair of routes, there is, and only one, stop where you can transfer from one of these routes to the other;
3) on each route, there are exactly three stops.
How many bus routes are ther... | Solve 5. Answer: 7. We will prove that if the given conditions are satisfied, then the number of stops $n$ and the number of routes $N$ are related by the formula $N=n(n-1)+1$. Let $a$ be one of the routes, and $B$ be a stop that route $a$ does not pass through. Each route passing through $B$ intersects route $a$. Ther... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. In a chess tournament, students from grades IX and X participated. There were 10 times more students from grade X than from grade IX, and they scored 4.5 times more points in total than all the students from grade IX. How many points did the students from grade IX score? Find all solutions. | Solution 1. Let $x$ be the number of ninth graders in the tournament. Then there were a total of $11x$ participants, and they scored $\frac{11x(11x-1)}{2}$ points. According to the problem, the ratio of the number of points scored by the ninth graders to the number of points scored by the tenth graders is $1:4.5$. Ther... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-4. In the city, there are 57 bus routes. It is known that:
1) from any stop to any other stop, one can travel without transferring;
2) for any pair of routes, there is one, and only one, stop where one can transfer from one of these routes to the other;
3) each route has no fewer than three stops.
How many stops do... | Solution 4. Answer: 8. We will prove that if the given conditions are satisfied, then the number of stops $n$ and the number of routes $N$ are related by the formula $N=n(n-1)+1$. First, we will show that if a route has $n$ stops, then any other route also has $n$ stops, and furthermore, each stop is served by $n$ rout... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-5. There is a piece of chain consisting of 60 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, \(\ldots, 60\) g can be formed (an unbuckled link also weighs 1 g) | Solution 5. Answer: 3 links. Let's determine the greatest \(n\) for which it is sufficient to break \(k\) links of an \(n\)-link chain so that all weights from 1 to \(n\) can be formed from the resulting parts. If \(k\) links are broken, then any number of links from 1 to \(k\) can be formed from them. But \(k+1\) link... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3-4. There is a piece of chain consisting of 150 links, each weighing 1 g. What is the smallest number of links that need to be unbuckled so that from the resulting parts, all weights of 1 g, 2 g, 3 g, \(\ldots, 150\) g can be formed (an unbuckled link also weighs 1 g)? | Solution 4. Answer: 4 links. According to the solution of problem 5 for grades \(7-8\) for a chain consisting of \(n\) links, where \(64 \leqslant n \leqslant 159\), it is sufficient to unfasten 4 links. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3-rd 5. When dividing the polynomial \(x^{1951}-1\) by \(x^{4}+x^{3}+2 x^{2}+x+1\), a quotient and a remainder are obtained. Find the coefficient of \(x^{14}\) in the quotient. | Solve 5. Answer: -1. The equalities \(x^{4}+x^{3}+2 x^{2}+x+1=\left(x^{2}+1\right)\left(x^{2}+x+1\right)\) and \(x^{12}-1=(x-1)\left(x^{2}+x+\right.\)
1) \(\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)\) show that
\[
\begin{aligned}
x^{4}+x^{3}+2 x^{2}+x+1 & =\frac{x^{12}-1}{(x-1)\left(x^{3}+1\ri... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-5. Five people play several games of dominoes (two on two) so that each player has each other as a partner once and as an opponent twice. Find the number of games played and all possible ways of distributing the players. | Solution 5. Answer: 5 games; the distribution of players is unique (up to their numbering). First, let's see who the 1st player played with. We can assume that the 1st and 2nd played against the 3rd and 4th, and the 1st and 5th played against the 2nd and 3rd (this can be achieved by swapping the 3rd and 4th). The 3rd p... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-2. Points \(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\) divide a circle of radius 1 into six equal parts. From \(A_{1}\), a ray \(l_{1}\) is drawn in the direction of \(A_{2}\), from \(A_{2}\) - a ray \(l_{2}\) in the direction of \(A_{3}, \ldots\), from \(A_{6}\) - a ray \(l_{6}\) in the direction of \(A_{1}\). From ... | Solve 2. Answer: \(B_{1} A_{1}=2\).
Let \(B_{2}, B_{3}, \ldots, B_{7}\) be the feet of the perpendiculars dropped from \(l_{6}, l_{5}, \ldots, l_{1} ; x_{1}=A_{2} B_{1}, x_{2}=\) \(A_{1} B_{2}, x_{3}=A_{6} B_{3}, \ldots, x_{7}=A_{2} B_{7}\). Then \(x_{k+1}=\frac{1}{2}\left(1+x_{k}\right)\). By the condition \(x_{1}=x_... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3-3. We took three numbers \(x, y, z\). We calculated the absolute values of the pairwise differences \(x_{1}=|x-y|\), \(y_{1}=|y-z|, z_{1}=|z-x|\). In the same way, from the numbers \(x_{1}, y_{1}, z_{1}\) we constructed the numbers \(x_{2}, y_{2}, z_{2}\) and so on. It turned out that for some \(n\), \(x_{n}=x\), \(y... | Solve 3. Answer: \(y=z=0\).
The numbers \(x_{n}, y_{n}, z_{n}\) are non-negative, so the numbers \(x, y, z\) are also non-negative. If all the numbers \(x\), \(y, z\) were positive, then the largest of the numbers \(x_{1}, y_{1}, z_{1}\) would be strictly less than the largest of the numbers \(x, y, z\), and then the ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-4. In a rectangular table composed of positive numbers, the product of the sum of the numbers in any column and the sum of the numbers in any row equals the number at their intersection. Prove that the sum of all numbers in the table is one. | Solution 4. First solution. Let \(x_{1}, \ldots, x_{n}\) be the sums of the numbers in the rows, and \(y_{1}, \ldots, y_{m}\) be the sums of the numbers in the columns. At the intersection of the \(i\)-th row and the \(j\)-th column, the number \(x_{i} y_{j}\) is placed. Therefore, the sum of the numbers in the \(i\)-t... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3-6. All values of the quadratic trinomial \(a x^{2}+b x+c\) on the interval \([0,1]\) do not exceed 1 in absolute value. What is the maximum value that the quantity \(|a|+|b|+|c|\) can have in this case? | Solve 6.
\section*{9th Grade} | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.90. What is the maximum number of acute angles a convex polygon can have
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6.90. Let a convex $n$-gon have $k$ acute angles. Then the sum of its angles is less than $k \cdot 90^{\circ} + (n-k) \cdot 180^{\circ}$. On the other hand, the sum of the angles of an $n$-gon is $(n-2) \cdot 180^{\circ}$. Therefore, $(n-2) \cdot 180^{\circ} < k \cdot 90^{\circ} + (n-k) \cdot 180^{\circ}$, i.e., $k < 4... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8.68. Using a compass and a ruler, divide an angle of $19^{\circ}$ into 19 equal parts. | 8.68. If there is an angle of magnitude $\alpha$, then angles of magnitude $2 \alpha, 3 \alpha$, etc., can be constructed. Since $19 \cdot 19^{\circ}=361$, an angle of $361^{\circ}$, which coincides with an angle of $1^{\circ}$, can be constructed. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.18*. How many sides can a convex polygon have if all its diagonals have the same length? | 9.18. We will prove that the number of sides of such a polygon does not exceed 5. Suppose that all diagonals of the polygon $A_{1} \ldots A_{n}$ have the same length and $n \geqslant 6$. Then the segments $A_{1} A_{4}, A_{1} A_{5}, A_{2} A_{4}$, and $A_{2} A_{5}$ have the same length, as they are diagonals of this poly... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.22. In a triangle, the lengths of two sides are 3.14 and 0.67. Find the length of the third side, given that it is an integer. | 9.22. Let the length of the third side be $n$. By the triangle inequality, $3.14-0.67<n<3.14+$ $+0.67$. Since $n$ is an integer, then $n=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.32. The perimeter of a convex quadrilateral is 4. Prove that its area does not exceed 1. | 9.32. According to problem $9.31 S_{A B C D} \leqslant(A B+C D)(B C+A D) / 4$. Since $a b \leqslant(a+$ $+b)^{2} / 4$, then $S_{A B C D} \leqslant(A B+C D+A D+B C)^{2} / 16=1$. | 1 | Geometry | proof | Yes | Yes | olympiads | false |
25.25*. A regular octagon with side 1 is cut into parallelograms. Prove that among them there are at least two rectangles, and the sum of the areas of all rectangles is 2.
## §5. Plane, cut by lines
Let $n$ pairwise non-parallel lines be drawn on the plane, with no three intersecting at the same point. Problems 25.26... | 25.25. Let us consider two mutually perpendicular pairs of opposite sides in a regular octagon and, as in problem 25.1, chains of parallelograms connecting opposite sides. At the intersections of these chains, there are rectangles. By considering two other pairs of opposite sides, we will obtain at least one more recta... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
43. It is obvious that any figure with a diameter of 1 can be enclosed within a square with a side of 2: for this, it is sufficient for the center of the square to coincide with any point of the figure. What is the side of the smallest square that can enclose any figure with a diameter of 1?
Note. One can also pose a ... | 43. It is obvious that a circle with a diameter of 1 cannot be enclosed in any square whose side is less than 1. On the other hand, it is almost equally clear that any figure with a diameter of 1 can be enclosed in a square with a side of 1. Indeed, any square that encloses some figure $\Phi$ can always be reduced so t... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
49. How many circles of radius 1 can be applied ${ }^{1}$ ) to a given unit circle $\mathcal{S}$ so that no two of these circles intersect? So that no one of these circles contains the center of another circle inside itself? | 49. Since a unit circle tangent to $S$ is seen from the center $O$ of circle $S$ at an angle of $60^{\circ}$ (Fig. $132, a$), no more than $6\left(=\frac{360}{60^{\circ}}\right)$ non-overlapping unit circles can be applied to $S$. Six circles can obviously be applied (Fig. 132, b).
Further, if $O_{1}$ and $O_{2}$ are ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
51. What is the greatest number of squares with side 1 that can be placed ${ }^{1}$ ) next to a given unit square $K$ so that no two of them intersect?
$^{1}$) The superscript "1" is kept as is, since it might refer to a footnote or additional information in the original text. | 51. First solution. Let $O$ be the center of the main square $K$, and $O_{1}$ and $O_{2}$ be the centers of the non-overlapping squares $K_{1}$ and $K_{2}$ attached to it (Fig. $135, a$). Since the smallest distance from the center of a unit square to its boundary is $\frac{1}{2}$, the segments $O O_{1}$, $O O_{2}$, an... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
54. What is the smallest number of circles with which a circle of twice the radius can be completely covered? | 54. Since the diameter of the smaller circles is equal to the radius $R$ of the larger circle, each such circle intersects the circumference of the larger circle at points no more than $R$ apart, and thus covers an arc of this circumference no greater than $60^{\circ}$. It follows that at least six smaller circles are ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
102. Given a triangle $ABC$. A point $M$, located inside the triangle, moves parallel to side $BC$ until it intersects side $CA$, then moves parallel to $AB$ until it intersects side $BC$, then - parallel to $AC$ until it intersects $AB$, and so on. Prove that after a certain number of such steps, the point will return... | 102. If point $M$ lies on any of the medians of triangle $ABC$, for example, on $M_{3} M_{1}$, then it is obvious that after four steps it will return to its initial position (Fig. $227, a$). Now suppose that $M$ does not lie on any of the three medians of triangle $ABC$.
Let $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6},... | 7 | Geometry | proof | Yes | Yes | olympiads | false |
31. a) The third term of an arithmetic progression is equal to 0. Find the sum of the first 5 terms.
b) The third term of a geometric progression is equal to 4. Find the product of the first 5 terms. | 31. a) We know that for an arithmetic progression $u_{1}=u_{3}-2 d ; u_{2}=u_{3}-d$; $u_{4}=u_{3}+d ; \quad u_{5}=u_{3}+2 d$. Therefore, the sum of the first five terms is
$$
\begin{aligned}
& \delta_{5}=u_{1}+u_{2}+u_{3}+u_{4}+u_{5}=\left(u_{3}-2 d\right)+ \\
& +\left(u_{3}-d\right)+u_{3}+\left(u_{3}+d\right)+ \\
& +... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
56. Mom has two apples and three pears. Every day for five consecutive days, she gives out one fruit. In how many ways can this be done? | 56. Here is one way to distribute apples and pears: 000 'on the 1st and 3rd day - apples, on the 2nd, 4th, and 5th days - pears),
## 000
(on the first three days - pears, on the last two days - apples). Thus, we need to count all the tables with two light and three dark circles; but we have already done this when sol... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
81. Determine the sum of the coefficients of the polynomial that results from expanding and combining like terms in the expression $\left(1+x-3 x^{2}\right)^{1965}$. | 81. If we expand the expression $\left(1+x-3 x^{2}\right)^{1965}$ and combine like terms, we get a polynomial $a_{0}+a_{1} \cdot x+$ $+a_{2} \cdot x^{2}+a_{3} \cdot x^{3}+\ldots$. Note that the sum of its coefficients is equal to the value of the polynomial at $x=1$
$$
\begin{aligned}
& a_{0}+a_{1} \cdot 1+a_{2} \cdot... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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