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20.19*. Prove that if the center of the inscribed circle of a quadrilateral coincides with the point of intersection of the diagonals, then the quadrilateral is a rhombus. | 20.19. Let $O$ be the point of intersection of the diagonals of quadrilateral $ABCD$. For definiteness, we can assume that $AO \geqslant CO$ and $DO \geqslant BO$. Let points $B_{1}$ and $C_{1}$ be symmetric to points $B$ and $C$
 of the polygon. Draw lines $a$ and $b$ through points $A$ and $B$, perpendicular to line $AB$. If $X$ is a vertex of the polygon, then $AX \leqslant AB$ and $XB \leqslant AB$, therefore
 be a vertex of this convex hull. Then \( A \) is a vertex of two different squares - a cardboard one and a plastic ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,922 |
20.26*. On a plane, there are $n \geqslant 4$ points, such that no three of them lie on the same line. Prove that if for any three of them there exists a fourth (also from the given points) such that they form the vertices of a parallelogram, then $n=4$.
## §6. Various Problems | 20.26. Let's consider the convex hull of the given points. There are two possible cases.
1. The convex hull is a parallelogram $A B C D$. If point $M$ lies inside the parallelogram $A B C D$, then the vertices of all three parallelograms with vertices $A, B$, and $M$ lie outside $A B C D$ (Fig. 20.8). Therefore, in th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,923 |
20.27. On a plane, there is a finite set of polygons (not necessarily convex), each two of which have a common point. Prove that there exists a line that has common points with all these polygons. | 20.27. Let's take an arbitrary line $l$ in the plane and project all polygons onto it. In this case, we will obtain several segments, any two of which have a common point. Consider the left ends of these segments and choose the rightmost one (to make it clear what "right" and "left" mean, a direction needs to be define... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,924 |
20.29. On a plane, there are four points that do not lie on the same straight line. Prove that at least one of the triangles with vertices at these points is not acute-angled. | 20.29. There are two possible arrangements of four points.
1. The points are the vertices of a convex quadrilateral $A B C D$. Choose the largest angle among its vertices. Let this be angle $A B C$. Then $\angle A B C \geqslant 90^{\circ}$, i.e., triangle $A B C$ is not acute-angled.
$, $(0, m)$, $(n, 0)$, $(n, m)$, where $n$ and $m$ are positive integers (different for each rectangle). Prove that from these rectangles, two can be chosen such that one is contained ... | 20.30. For a rectangle with vertices at points $(0,0),(0, m),(n, 0)$ and $(n, m)$, the horizontal side is equal to $n$, and the vertical side is equal to $m$. We select from this set of rectangles the one with the smallest horizontal side. Let its vertical side be $m_{1}$. Consider any $m_{1}$ of the remaining rectangl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,927 |
20.31*. Given a convex polygon $A_{1} \ldots A_{n}$. Prove that the circumcircle of some triangle $A_{i} A_{i+1} A_{i+2}$ contains the entire polygon. | 20.31. Consider all circles passing through two adjacent vertices $A_{i}$ and $A_{i+1}$ and such a vertex $A_{j}$ that $\angle A_{i} A_{j} A_{i+1}<90^{\circ}$. At least one such circle exists. Indeed, one of the angles $A_{i} A_{i+2} A_{i+1}$ and $A_{i+1} A_{i} A_{i+2}$ is less than $90^{\circ}$; in the first case, let... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,928 |
21.1. The nodes of an infinite grid paper are colored in two colors. Prove that there exist two horizontal and two vertical lines, at the intersections of which lie points of the same color. | 21.1. Let's take three vertical lines and nine horizontal ones. We will only consider the points of intersection of these lines. Since there are only $2^{3}=8$ ways to color three points in two colors, there will be two horizontal lines on which there are identically colored triplets of points. Among the three points c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,929 |
21.2. Inside an equilateral triangle with a side of 1, five points are placed. Prove that the distance between some two of them
is less than 0.5. | 21.2. The medians of an equilateral triangle with a side length of 1 divide it into four smaller equilateral triangles with a side length of 0.5. Therefore, at least two of the given points lie within one of these smaller triangles, and these points cannot be at the vertices of the triangle. The distance between these ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,930 |
21.3. In a rectangle of size $3 \times 4$, 6 points are placed. Prove that among them, there are two points whose distance does not exceed $\sqrt{5}$. | 21.3. Let's cut a rectangle into five figures as shown in Fig. 21.2. At least two points will fall into one of them, and the distance between any two points in each of these figures does not exceed $\sqrt{5}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,931 |
21.4. On an $8 \times 8$ chessboard, the centers of all the squares are marked. Can thirteen lines be used to divide the board into parts such that no more than one marked point lies inside each part? | 21.4. $\mathrm{K}$ along the edge of an $8 \times 8$ chessboard, there are 28 fields. We draw 28 segments connecting the centers of adjacent edge fields. Each line can intersect no more than two such segments, so 13 lines can intersect no more than 26 segments, i.e., there will be at least 2 segments that do not inters... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,932 |
21.5. On a plane, there are 25 points, and among any three of them, there are two at a distance less than 1. Prove that there exists a circle of radius 1 containing at least 13 of these points. | 21.5. Let $A$ be one of the given points. If all other points lie within the circle $S_{1}$ of radius 1 centered at $A$, there is nothing more to prove. Now let $B$ be a given point lying outside the circle $S_{1}$, i.e., $A B>1$. Consider the circle $S_{2}$ of radius 1 centered at $B$. Among the points $A, B$, and $C$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,933 |
21.6*. In a square with side 1, there are 51 points. Prove that some three of them can be covered by a circle of radius $1 / 7$. | 21.6. Let's divide the given square into 25 identical smaller squares with a side length of 0.2. In one of these smaller squares, there are no fewer than three points. The radius of the circumscribed circle of a square with a side length of 0.2 is $1 / 5 \sqrt{2}<1 / 7$, so it can be covered by a circle with a radius o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,934 |
21.7*. Each of the two disks is divided into 1985 equal sectors, and 200 sectors on each are painted in an arbitrary color (one color). The disks are placed on top of each other, and one is rotated by angles that are multiples of $360^{\circ} / 1985$. Prove that there are at least 80 positions in which no more than 20 ... | 21.7. Let's take 1985 disks colored the same as the second of our disks and place them on the first disk so that they occupy all possible positions. Then above each colored sector of the first disk, there are 200 colored sectors, i.e., there are a total of $200^{2}$ pairs of coinciding colored sectors. Let there be $n$... | 81 | Combinatorics | proof | Yes | Yes | olympiads | false | 27,935 |
21.8*. Each of the nine lines divides the square into two quadrilaterals, the areas of which are in the ratio $2: 3$. Prove that at least three of these nine lines pass through one point. | 21.8. These lines cannot intersect the adjacent sides of the square $ABCD$, otherwise they would form not two quadrilaterals, but a triangle and a pentagon. Let the line intersect sides $BC$ and $AD$ at points $M$ and $N$. The trapezoids $ABMN$ and $CDNM$ have equal heights, so their areas are in the ratio of their mid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,936 |
21.9*. In the park, there are 10000 trees planted in a square grid pattern (100 rows of 100 trees each). What is the maximum number of trees that can be cut down so that the following condition is met: if you stand on any stump, you will not see any other stump? (The trees can be considered thin enough.) | 21.9. Let's divide the trees into 2500 quartets, as shown in Fig. 21.3. In each such quartet, no more than one tree can be cut down. On the other hand, all the trees growing in the upper left corners of the squares formed by our quartets of trees can be cut down. Therefore, the maximum number of trees that can be cut d... | 2500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 27,937 |
21.10*. What is the smallest number of points that need to be marked inside a convex $n$-gon so that within any triangle with vertices at the vertices of the $n$-gon, there is at least one marked point? | 21.10. Since the diagonals emanating from one vertex divide an $n$-sided polygon into $n-2$ triangles, $n-2$ points are necessary.
Fig. 21.3
/2 = n(2n-3)$. It is easy to verify that the number of diagonals parallel to a given side does not exceed $n-2$. Therefore, the total number of diagonals parallel to the sides does not exceed $2n(n-2)$. Since $2n(n-2) < n(2n-3)$, there exists a diagonal that is n... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,940 |
21.13*. Nodes of an infinite grid paper are colored in three colors. Prove that there exists an isosceles right triangle with vertices of the same color.
## §2. Angles and Lengths | 21.13. Suppose there is no isosceles right triangle with legs parallel to the sides of the cells and vertices of the same color. For convenience, we can assume that the cells, not the nodes, are colored. We divide the sheet into squares with side length 4; then on the diagonal of each such square, there will be two cel... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,941 |
21.14. On a plane, there are $n$ pairwise non-parallel lines. Prove that the angle between some two of them is not greater than $180^{\circ} / n$. | 21.14. Let's take an arbitrary point on the plane and draw lines through it parallel to the given ones. They will divide the plane into \(2n\) angles, the sum of which is \(360^{\circ}\). Therefore, one of these angles does not exceed \(180^{\circ} / n\). | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,942 |
21.15. In a circle of radius 1, several chords are drawn. Prove that if each diameter intersects no more than \( k \) chords, then the sum of the lengths of the chords is less than \( \pi k \). | 21.15. Suppose that the sum of the lengths of the chords is not less than $\pi k$, and we will prove that then there exists a diameter that intersects at least $k+1$ chords. Since the length of the arc subtended by a chord is greater than the length of the chord itself, the sum of the lengths of the arcs subtended by t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,943 |
21.16. A point $O$ is marked on a plane. Is it possible to place on the plane: a) five circles; b) four circles, not covering the point $O$, such that any ray starting from point $O$ intersects at least two circles? ("Intersects" - has a common point.) | 21.16. a) It is possible. Let $O$ be the center of the regular pentagon $ABCDE$. Then the circles inscribed in the angles $AOC, BOD, COE, DOA$, and $EOB$ have the required property.
b) It is not possible. Consider for each of the four circles the angle formed by the tangents to it passing through the point $O$. Since ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,944 |
21.17*. Inside a circle of radius $n$, there are $4 n$ segments of length 1. Prove that one can draw a line parallel or perpendicular to a given line $l$ and intersecting at least two of the given segments. | 21.17. Let $l_{1}$ be an arbitrary line perpendicular to $l$. Denote the lengths of the projections of the $i$-th segment onto the lines $l$ and $l_{1}$ by $a_{i}$ and $b_{i}$, respectively. Since the length of each segment is 1, we have $a_{i}+b_{i} \geqslant 1$. Therefore, $\left(a_{1}+\ldots+a_{4 n}\right)+$ $+\left... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,945 |
21.18*. Inside a square with side 1, there are several circles, the sum of the lengths of which is 10. Prove that there is a straight line that intersects at least four of these circles. | 21.18. Project all the data of the circle onto the side $A B$ of the square $A B C D$. The projection of a circle of length $l$ is a segment of length $l / \pi$. Therefore, the sum of the lengths of the projections of all the given circles is $10 / \pi$. Since $10 / \pi > 3 = 3 A B$, there is a point on the segment $A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,946 |
21.19*. On a segment of length 1, several segments are painted, and the distance between any two painted points is not equal to 0.1. Prove that the sum of the lengths of the painted segments does not exceed 0.5. | 21.19. Let's divide a segment into ten segments
Fig. 21.7 with a length of 0.1, stack them on top of each other, and project them onto a segment of the same length (Fig. 21.7). Since the di... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,947 |
21.20*. Two circles are given, the length of each being 100 cm. On one of them, 100 points are marked, and on the other, several arcs are marked, the total length of which is less than 1 cm. Prove that these circles can be superimposed so that no marked point falls on a marked arc. | 21.20. Let's combine these circles and place a painter at a fixed point on one of them. We will rotate this circle and instruct the painter to paint the point on the circle that he passes by every time any marked point lies on the marked arc. We need to prove that after a full rotation, part of the circle will remain u... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,948 |
21.21*. Given two identical circles. On each of them, $k$ arcs are marked, the angular measure of each of which is less than $\frac{1}{k^{2}-k+1} \cdot 180^{\circ}$, and the circles can be superimposed so that the marked arcs of one circle coincide with the marked arcs of the other. Prove that these circles can be supe... | 21.21. Let us combine these circles and place a painter at a fixed point on one of them. We will rotate this circle and instruct the painter to paint the point of the circle that he passes by every time any marked arcs intersect. We need to prove that after a full rotation, part of the circle will remain unpainted. The... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,949 |
21.22. In a square with a side of 15, 20 non-overlapping smaller squares with a side of 1 are placed. Prove that a circle of radius 1 can be placed in the larger square so that it does not intersect any of the smaller squares. | 21.22. Consider a figure consisting of all points that are no more than 1 unit away from a square with side length 1 (Fig. 21.8). It is clear that a circle of radius 1, whose center is located outside this figure, does not intersect the square. The area of such a figure is $\pi+5$. The center of the required circle mus... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,950 |
21.23*. Given an infinite grid paper and a figure whose area is less than the area of a cell. Prove that this figure can be placed on the paper without covering any vertex of a cell. | 21.23. Attach the figure to a grid paper in any way, cut the paper along the cells, and stack them, moving them parallel and without flipping. Project this stack onto a cell. The projections of parts of the figure cannot cover the entire cell, as their area is smaller. Now, recall how the figure was positioned on the g... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,951 |
21.24*. We will call a cross the figure formed by the diagonals of a square with side 1 (Fig. 21.1). Prove that in a circle of radius 100, only a finite number of non-overlapping crosses can be placed. | 21.24. For each cross, consider a circle of radius $1 / 2 \sqrt{2}$ centered at the center of the cross. We will prove that if two such circles intersect, then the crosses themselves intersect. The distance between the centers of intersecting equal circles does not exceed their doubled radius, so the distance between t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,952 |
21.25*. The pairwise distances between points $A_{1}, \ldots, A_{n}$ are greater than 2. Prove that any figure with an area less than $\pi$ can be shifted by a vector of length no more than 1 so that it does not contain the points $A_{1}, \ldots, A_{n}$.
 $S=M_{1}-M_{2}+M_{3}-\ldot... | 21.27. a) Let $C_{n}^{k}$ be the number of ways to choose $k$ elements from $n$. It can be verified that $(x+y)^{n}=\sum_{k=0}^{n} C_{n}^{k} x^{k} y^{n-k}$ (Binomial Theorem).
Denote by $W_{m}$ the area of the part of the plane covered by exactly $m$ figures. This part consists of pieces, each of which is covered by s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,955 |
21.28*. a) In a square with an area of 6, three polygons with an area of 3 are placed. Prove that among them, there are two polygons whose common area is not less than 1.
b) In a square with an area of 5, nine polygons with an area of 1 are placed. Prove that among them, there are two polygons whose common area is not... | 21.28. a) According to problem 21.27, a) $6=9-\left(S_{12}+S_{23}+S_{13}\right)+S_{123}$, i.e., $S_{12}+S_{23}+S_{13}=3+S_{123} \geqslant 3$. Therefore, one of the numbers $S_{12}, S_{23}, S_{13}$ is not less than 1.
b) According to problem 21.27, b) $5 \geqslant 9-M_{2}$, i.e., $M_{2} \geqslant 4$. Since from nine po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,956 |
21.29*. A kaftan with an area of 1 has five patches, and the area of each of them is not less than 0.5. Prove that there will be two patches, the area of the common part of which is not less than 0.2. | 21.29. Let the area of the caftan be $M$, the area of the intersection of patches numbered $i_{1}, \ldots, i_{k}$ be $S_{i_{1} \ldots i_{k}}$, and $M_{k}=\sum S_{i_{1} \ldots i_{k}}$. According to problem 21.27, a) $M-M_{1}+M_{2}-M_{3}+M_{4}-M_{5} \geqslant 0$, since $M \geqslant S$. Similar inequalities can be written... | 0.2 | Combinatorics | proof | Yes | Yes | olympiads | false | 27,957 |
21.30*. On a segment of length 1, there are pairwise non-intersecting segments, the sum of whose lengths is $p$. Denote this system of segments by $A$. Let $B$ be the complementary system of segments (segments of systems $A$ and $B$ do not have common interior points and completely cover the given segment). Prove that ... | 21.30. Let $-1 \leqslant c \leqslant 1$. We will shift the given segment by $c$ along itself, and then shift it by $c$ in an orthogonal direction. The shaded area in Fig. 21.10 corresponds to the intersection of segments $A_{i}$ and $B_{j}$. Its area is equal to the product of the lengths of these segments. If we consi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,958 |
22.1. On a plane, there are $n$ points, and any four of them are vertices of a convex quadrilateral. Prove that these points are vertices of a convex $n$-gon. | 22.1. Consider the convex hull of the given points. It is a convex polygon. We need to prove that all the given points are its vertices. Suppose that one of the given points (point $A$) is not a vertex, i.e., it lies inside or on a side of this polygon. By cutting the convex hull with diagonals emanating from one verte... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,959 |
22.2. On a plane, there are five points, and no three of them lie on the same line. Prove that four of these points are located at the vertices of a convex quadrilateral. | 22.2. Consider the convex hull of the given points. If it is a quadrilateral or pentagon, everything is clear. Now suppose the convex hull is a triangle $ABC$, and points $D$ and $E$ lie inside it. Point $E$ lies inside one of the triangles $ABD$, $BCD$, or $CAD$; let's assume for definiteness that it lies inside trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,960 |
22.3. On the plane, there are several regular $n$-gons. Prove that the convex hull of their vertices has at least $n$ angles. | 22.3. Let the convex hull of the vertices of $n$-gons be an $m$-gon and $\varphi_{1}, \ldots, \varphi_{m}$ be its angles. Since each angle of the convex hull is adjacent to an angle of a regular $n$-gon, $\varphi_{i} \geqslant(1-(2 / n)) \pi$ (on the right is the measure of the angle of a regular $n$-gon). Therefore, $... | \geqslantn | Geometry | proof | Yes | Yes | olympiads | false | 27,961 |
22.4. Among all such numbers $n$ that any convex 100-gon can be represented as the intersection (i.e., common part) of $n$ triangles,
find the smallest. | 22.4. First, note that 50 triangles are sufficient. Indeed, let $\Delta_{k}$ be the triangle whose sides lie on the rays $A_{k} A_{k-1}$ and $A_{k} A_{k+1}$ and which contains the convex polygon $A_{1} \ldots A_{100}$. Then this polygon is the intersection of the triangles $\Delta_{2}, \Delta_{4}, \ldots, \Delta_{100}$... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,962 |
22.5. We will call a convex heptagon special if three of its diagonals intersect at one point. Prove that by slightly moving one of the vertices of a special heptagon, one can obtain a non-special heptagon. | 22.5. Let $P$ be the intersection point of the diagonals $A_{1} A_{4}$ and $A_{2} A_{5}$ of the convex heptagon $A_{1} \ldots A_{7}$. One of the diagonals $A_{3} A_{7}$ and $A_{3} A_{6}$, for definiteness the diagonal $A_{3} A_{6}$, does not pass through the point $P$. The number of intersection points of the diagonals... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,963 |
22.6*. On a plane, there are two convex polygons $F$ and $G$. Denote by $H$ the set of midpoints of segments, one end of each of which belongs to $F$, and the other to $G$. Prove that $H$ is a convex polygon.
a) How many sides can $H$ have if $F$ and $G$ have $n_{1}$ and $n_{2}$ sides respectively?
b) What can the pe... | 22.6. First, let's prove that $H$ is a convex figure. Let points $A$ and $B$ belong to $H$, i.e., $A$ and $B$ are the midpoints of segments $C_{1} D_{1}$ and $C_{2} D_{2}$, where $C_{1}$ and $C_{2}$ belong to $F$, and $D_{1}$ and $D_{2}$ belong to $G$. We need to prove that the entire segment $A B$ belongs to $H$. It i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,964 |
22.7*. Prove that there exists a number $N$ such that among any $N$ points, no three of which lie on the same line, one can select 100 points that are the vertices of a convex polygon. | 22.7. We will prove a more general statement. Let $p, q$ and $r$ be natural numbers, with $p, q \geqslant r$. Then there exists a number $N=N(p, q, r)$, possessing the following property: if all $r$-element subsets of an $N$-element set $S$ are arbitrarily partitioned into two non-intersecting families $\alpha$ and $\b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,965 |
22.8*. A convex $n$-gon is divided into triangles by non-intersecting diagonals. Consider a transformation of such a partition, where triangles $A B C$ and $A C D$ are replaced by triangles $A B D$ and $B C D$. Let $P(n)$ be the smallest number of transformations required to convert any partition into any other. Prove ... | 22.8. a) Let $A$ and $B$ be adjacent vertices of an $n$-gon. Consider the partition of the $n$-gon by diagonals emanating from vertex $A$, and the partition by diagonals emanating from vertex $B$. These partitions have no common diagonals, and each transformation changes only one diagonal.
b) By induction on $n$, it i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,966 |
22.9*. Prove that in any convex polygon, except for a parallelogram, three sides can be chosen such that their extensions form a triangle that encompasses the given polygon.
保留源文本的换行和格式，这里指的是保持原文的换行和格式不变。由于原文只有一行，因此翻译结果也只有一行。如果原文有多行，翻译时也会保持多行的格式。 | 22.9. If a polygon is neither a triangle nor a pa-
Fig. 22.5 rallelogram, then it will have two non-parallel non-adjacent sides. Extending them until they intersect, we obtain a new polygon... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,967 |
22.10*. Given a convex $n$-gon, no two sides of which are parallel. Prove that the number of different triangles mentioned in problem 22.9 is at least $n-2$.
Translate the above text into English, please retain the line breaks and format of the source text, and output the translation result directly. | 22.10. We will prove the statement by induction on $n$. For $n=3$, the statement is obvious. According to problem 22.9, there exist lines $a, b$, and $c$, which are extensions of the sides of the given $n$-gon and form a triangle $T$ that contains the given $n$-gon. Let line $l$ be the extension of another side of the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 27,968 |
22.11*. Point $O$ lies inside a convex $n$-gon $A_{1} \ldots A_{n}$. Prove that among the angles $A_{i} O A_{j}$, not less than $n-1$ are not acute. | 22.11. We will prove this by induction on $n$. For $n=3$, the proof is obvious. Now consider an $n$-gon $A_{1} \ldots A_{n}$, where $n \geqslant 4$. The point $O$ lies inside some triangle $A_{p} A_{q} A_{r}$. Let $A_{k}$ be a vertex of this $n$-gon, different from the points $A_{p}, A_{q}$, and $A_{r}$. By removing th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,969 |
22.12*. A convex $n$-gon $A_{1} \ldots A_{n}$ is inscribed in a circle, and among its vertices, there are no diametrically opposite points. Prove that if there is at least one acute triangle $A_{p} A_{q} A_{r}$, then there are at least $n-2$ such acute triangles.
## §2. Helly's Theorem | 22.12. We will prove the statement by induction on $n$. For $n=3$, the statement is obvious. Let $n \geqslant 4$. Fix one acute triangle $A_{p} A_{q} A_{r}$ and remove a vertex $A_{k}$, different from the vertices of this triangle. We can apply the induction hypothesis to the resulting $(n-1)$-gon. Moreover, if, for ex... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,970 |
22.13*. a) On a plane, there are four convex figures, and any three of them have a common point. Prove that then all of them have a common point.
b) On a plane, there are $n$ convex figures, and any three of them have a common point. Prove that all $n$ figures have a common point (Helly's theorem). | 22.13. a) Let's denote the given figures as $M_{1}, M_{2}, M_{3}$, and $M_{4}$. Let $A_{i}$ be the point of intersection of all figures except $M_{i}$. There are two possible arrangements of the points $A$.
1. One of the points, for example, $A_{4}$, lies inside the triangle formed by the other points. Since the point... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,971 |
22.14*. On a plane, there are $n$ points, and any three of them can be covered by a circle of radius 1. Prove that then all $n$ points can be covered by a circle of radius 1. | 22.14. A circle of radius 1 with center $O$ covers some points if and only if circles of radius 1 with centers at these points contain the point $O$. Therefore, our problem can be restated as follows: "On a plane, there are $n$ points, and any three circles of radius 1 with centers at these points have a common point. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,972 |
22.15*. Prove that inside any convex heptagon there is a point not belonging to any of the quadrilaterals formed by its four consecutive vertices. | 22.15. Consider the pentagons that remain when pairs of adjacent vertices of a heptagon are discarded. It is sufficient to check that any three of them have a common point. For three pentagons, no more than six different vertices are discarded, i.e., one vertex remains. If vertex $A$ is not discarded, then the shaded t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,973 |
22.16*. Given several parallel segments, and for any three of them, there exists a line intersecting them. Prove that there exists a line intersecting all segments.
## §3. Non-Convex Polygons | 22.16. Let's introduce a coordinate system with the $O y$ axis parallel to the given segments. For each segment, consider the set of all points $(a, b)$ such that the line $y=a x+b$ intersects it. It is sufficient to check that these sets are convex and apply Helly's theorem to them. For a segment with endpoints $\left... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,974 |
22.19. Prove that if a polygon is such that its entire contour is visible from some point $O$, then from any point in the plane at least one of its sides is completely visible. | 22.19. Let the entire contour of the polygon $A_{1} \ldots A_{n}$ be visible from point $O$. Then the angle $A_{i} O A_{i+1}$ does not contain any other sides of the polygon except $A_{i} A_{i+1}$, and therefore point $O$ lies inside the polygon (Fig. 22.9). Any point $X$ on the plane belongs to one of the angles $A_{i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,977 |
22.20. Prove that the sum of the exterior angles of any polygon, adjacent to internal angles less than $180^{\circ}$, is not less than $360^{\circ}$. | 22.20. Since in a convex $n$-gon all interior angles are less than $180^{\circ}$ and their sum is $(n-2) \cdot 180^{\circ}$, the sum of the exterior angles is $360^{\circ}$, i.e., in the case of a convex polygon, equality is achieved.
Let $M$ now be the convex hull of the polygon $N$. Each angle of $M$ contains a less... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,978 |
22.21*. a) Prove that any $n$-gon ( $n \geqslant 4$ ) has at least one diagonal that lies entirely inside it.
b) Determine the smallest number of such diagonals that an $n$-gon can have. | 22.21. a) If the polygon is convex, the statement is obvious. Suppose now that the interior angle of the polygon at vertex $A$ is greater than $180^{\circ}$. The visible part of the side is seen from point $A$ at an angle less than $180^{\circ}$, so from point $A$ at least parts of two sides are visible. Therefore, the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,979 |
22.23*. Prove that any $n$-gon can be cut into triangles by non-intersecting diagonals. | 22.23. We will prove this statement by induction on $n$. For $n=3$, it is obvious. Suppose the statement is proven for all $k$-gons, where $k<n$, and we will prove it for any $n$-gon. Any $n$-gon can be cut by a diagonal into two polygons (see problem 22.21, a)), and the number of vertices of each of them is strictly l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,981 |
22.24*. Prove that the sum of the interior angles of any $n$-sided polygon is $(n-2) 180^{\circ}$. | 22.24. We will prove this statement by induction. For $n=3$ it is obvious. Suppose
Fig. 22.13
that it is proven for all $k$-gons, where $k<n$, and we will prove it for any $n$-gon. Any $n$-... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,982 |
22.25*. Prove that the number of triangles into which non-intersecting diagonals divide an $n$-gon is $n-2$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 22.25. The sum of the angles of the obtained triangles is equal to the sum of the angles of the polygon, i.e., it is equal to $(n-2) \cdot 180^{\circ}$ (see problem 22.24$). Therefore, the number of triangles is $n-2$. | Number Theory | proof | Yes | Yes | olympiads | false | 27,983 | |
22.27*. Prove that for any tridecagon (13-sided polygon), there exists a line containing exactly one of its sides; however, for any $n>13$, there exists an $n$-sided polygon for which this is not true. | 22.27. Suppose there exists a tridecagon (13-sided polygon) such that on any line containing a side, there is at least one more side. Draw lines through all the sides of this tridecagon. Since it has thirteen sides, at least one of these lines must contain an odd number of sides, i.e., at least three sides lie on one l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,985 |
22.28*. What is the maximum number of acute angles in a non-convex $n$-gon
保留源文本的换行和格式，翻译结果如下：
22.28*. What is the maximum number of acute angles in a non-convex $n$-gon | 22.28. Let $k$ be the number of acute angles in an $n$-sided polygon. Then the sum of its angles is less than $k \cdot 90^{\circ} + (n-k) \cdot 360^{\circ}$. On the other hand, the sum of the angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$ (see problem 22.24), so $k \cdot 90^{\circ} + (n-k) \cdot 360^{\circ... | k\leqslant[2n/3]+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,986 |
23.1. Can a line intersect (at interior points) all sides of a non-convex: a) $(2 n+1)$-gon; b) $2 n$-gon | 23.1. a) Let a line intersect all sides of a polygon. Consider all vertices lying on one side of it. Each of these vertices can be associated with a pair of sides emanating from it. In this way, we obtain a partition of all the sides of the polygon into pairs. Therefore, if a line intersects all sides of an $m$-gon, th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,989 |
23.2. On a plane, there is a closed broken line with a finite number of segments. A line $l$ intersects it at 1985 points. Prove that there exists a line
intersecting this broken line at more than 1985 points. | 23.2. A line $l$ defines two half-planes; we will call one of them the upper half-plane and the other the lower half-plane. Let $n_{1}$ (respectively $n_{2}$) be the number of vertices of the broken line lying on the line $l$, for which both segments emanating from them lie in the upper (respectively lower) half-plane,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,990 |
23.3. On a plane, there are three pucks $A, B$, and $C$. A hockey player hits one of the pucks so that it passes between the other two and stops at some point. Can all the pucks return to their original positions after 25 hits? | 23.3. No, they cannot. After each strike, the orientation (i.e., the direction of traversal) of triangle $A B C$ changes. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 27,991 |
23.4. Is it possible to color 25 cells on a grid paper so that each of them has an odd number of colored neighbors? (Neighboring cells are those that share a common side.) | 23.4. Let several cells be colored on a grid paper, and let $n_{k}$ be the number of colored cells that have exactly $k$ colored neighbors. Let $N$ be the number of common sides of colored cells. Since each of them belongs to exactly two colored cells, $N=\left(n_{1}+2 n_{2}+3 n_{3}+4 n_{4}\right) / 2=\left(n_{1}+n_{3}... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 27,992 |
23.5*. A circle is divided by points into $3 k$ arcs: $k$ arcs of length 1, 2, and 3. Prove that there will be two diametrically opposite division points. | 23.5. Suppose the circle is divided into arcs in the specified manner, with no diametrically opposite division points. Then, opposite the ends of any arc of length 1, there are no division points, so opposite it lies an arc of length 3. We will remove one of the arcs of length 1 and the opposite arc of length 3. As a r... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,993 |
23.6*. On a plane, there is a non-self-intersecting closed broken line, no three vertices of which lie on the same straight line. We will call a pair of non-adjacent segments of the broken line special if the extension of one of them intersects the other. Prove that the number of special pairs is even. | 23.6. Let's take the adjacent segments $A B$ and $B C$ and call the corner the angle symmetric to angle $A B C$ with respect to point $B$ (in Fig. 23.7, the corner is shaded). Similar corners can be considered for all vertices of the broken line. It is clear that the number of special pairs is equal to the number of po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,994 |
23.7*. The vertices of a triangle are marked with the numbers 0, 1, and 2. This triangle is divided into several smaller triangles in such a way that no vertex of one triangle lies on the side of another. The vertices of the original triangle retain their original labels, and additional vertices receive the numbers $0,... | 23.7. Consider the segments into which side 01 is divided. Let $a$ be the number of segments of the form 00,
Fig. 23.7 $b$ be the number of segments of the form 01. For each segment, consi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 27,995 |
23.8*. The vertices of a regular $2 n$-gon $A_{1} \ldots A_{2 n}$ are divided into $n$ pairs. Prove that if $n=4 m+2$ or $n=4 m+3$, then two pairs of vertices are the endpoints of equal segments.
## §2. Divisibility | 23.8. Suppose that all pairs of vertices define segments of different lengths. To the segment \(A_{p} A_{q}\), we assign the minimum of the numbers \(|p-q|\) and \(2 n-|p-q|\). As a result, for the given \(n\) pairs of vertices, we obtain the numbers \(1, 2, \ldots, n\); let there be \(k\) even numbers and \(n-k\) odd ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,996 |
23.10*. A square sheet of grid paper is divided into smaller squares by segments running along the sides of the cells. Prove that the sum of the lengths of these segments is divisible by 4. (The side length of a cell is 1.)
## §3. Invariants | 23.10. Let $Q$ be a square sheet of paper, and $L(Q)$ be the sum of the lengths of the sides of the cells that lie inside it. Then $L(Q)$ is divisible by 4, since all the considered sides are divided into quartets of sides obtained from each other by rotations of $\pm 90^{\circ}$ and $180^{\circ}$ about the center of t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,998 |
23.12. Given a chessboard. It is allowed to repaint at once all the cells located inside a $2 \times 2$ square in another color. Can there be exactly one black cell left on the board? | 23.12. When repainting a $2 \times 2$ square containing $k$ black and $4-k$ white cells, it will result in $4-k$ black and $k$ white cells. Therefore, the number of black cells will change by ( $4-k$ ) $-k=4-2k$, i.e., by an even number. Since the parity of the number of black cells is preserved, from the initial 32 bl... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 28,000 |
23.13*. Given a convex $2 m$-gon $A_{1} \ldots A_{2 m}$. Inside it, a point $P$ is taken, which does not lie on any of the diagonals. Prove that point $P$ belongs to an even number of triangles with vertices at points $A_{1}, \ldots, A_{2 m}$. | 23.13. Diagonals divide a polygon into several parts. We will call those parts neighboring if they share a common side. It is clear that from any internal point of the polygon, one can reach any other point by moving only from one neighboring part to another. The part of the plane lying outside the polygon can also be ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,001 |
23.14*. In the center of each cell of a chessboard, there is a chip. The chips were rearranged so that the pairwise distances between them did not decrease. Prove that in fact, the pairwise distances did not change. | 23.14. If at least one of the distances between the chips increased, then the sum of all pairwise distances between the chips would also increase, but the sum of all pairwise distances between the chips does not change with any permutation. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,002 |
23.15*. A polygon is cut into several polygons, and neither on the sides of the original polygon nor on the sides of the resulting polygons do the vertices of the resulting polygons lie. Let $p$ be the number of resulting polygons, $q$ be the number of segments that are their sides, and $r$ be the number of points that... | 23.15. Let $n$ be the number of vertices of the original polygon, and $n_{1}, \ldots, n_{p}$ the number of vertices of the obtained polygons. On the one hand, the sum of the angles of all the obtained polygons is $\sum_{i=1}^{p}\left(n_{i}-2\right) \pi=\sum_{i=1}^{p} n_{i} \pi-2 p \pi$. On the other hand, it is equal t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,003 |
23.16*. A square field is divided into 100 equal square plots, 9 of which are overgrown with weeds. It is known that weeds spread to those and only those plots that have at least two adjacent (i.e., sharing a side) plots already overgrown with weeds. Prove that the field will never be completely overgrown with weeds. | 23.16. It is easy to check that the length of the boundary of the entire area overgrown with weeds (or several areas) does not increase. At the initial moment, it does not exceed \(9 \cdot 4=36\), therefore, at the final moment, it cannot be equal to 40.
, which can be translated into each other by parallel translation. | 23.17. We fix a certain ray $A B$ on a plane. For any polygon $M$, we will associate a number $F(M)$ (depending on $A B$) as follows. Consider all sides of $M$ that are perpendicular to $A B$, and to each of these sides, we will assign a number $\pm l$, where $l$ is the length of this side and the sign “plus” is taken ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,005 |
23.18*. Prove that a convex polygon cannot be cut into a finite number of non-convex quadrilaterals. | 23.18. Suppose that a convex polygon $M$ is cut into non-convex quadrilaterals $M_{1}, \ldots, M_{n}$. To each polygon $N$, we assign a number $f(N)$, which is the difference between the sum of its internal angles less than $180^{\circ}$ and the sum of the angles that complement to $360^{\circ}$ its angles greater than... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,006 |
23.19*. Given points $A_{1}, \ldots, A_{n}$. Consider a circle of radius $R$ containing some of them. We then construct a circle of radius $R$ centered at the center of mass of the points lying inside the first circle, and so on. Prove that this process will stop, i.e., the circles will start to coincide.
## §4. Auxil... | 23.19. Let $S_{n}$ be the circle constructed at the $n$-th step, and $O_{n}$ be its center. Consider the quantity $F_{n}=\sum\left(R^{2}-O_{n} A_{i}^{2}\right)$, where the summation is taken only over the points that lie inside the circle $S_{n}$. We will denote the points lying inside the circles $S_{n}$ and $S_{n+1}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,007 |
23.20. A beetle sits in each cell of a $5 \times 5$ board. At some moment, all the beetles crawl to adjacent (horizontally or vertically) cells. Will there necessarily be an empty cell? | 23.20. Since the total number of cells on a $5 \times 5$ chessboard is odd, the number of black and white cells cannot be equal. Let's assume, for definiteness, that there are more black cells. Then the number of beetles sitting on white cells is less than the number of black cells. Therefore, at least one of the black... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,008 |
23.21. Can a chessboard of size $8 \times 8$, from which two opposite corner squares have been cut out, be tiled with domino tiles of size $1 \times 2$? | 23.21. Squares of the same color, let's say black, have been cut out. Therefore, 32 white and 30 black squares remain. Since a domino tile always covers one white and one black square, it is impossible to tile the $8 \times 8$ chessboard with dominoes if two opposite corner squares have been cut out. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,009 |
23.24*. Three grasshoppers are located at three vertices of a square, playing leapfrog. If grasshopper $A$ jumps over grasshopper $B$, it ends up at the same distance from $B$ but, naturally, on the other side and on the same line. Can one of the grasshoppers end up at the fourth vertex of the square after several jump... | 23.24. Consider the lattice shown in Fig. 23.11, and color it in two colors as shown in this figure (white nodes on this figure are not filled; the original square is shaded, with grasshoppers sitting at its white vertices). We will prove that the grasshoppers can only land on white nodes, i.e., under symmetry
. Since the ends of any unit segment are of different colors, a broken line with monochromatic ends contains an odd number of nodes, and one with bichromatic ends contains an even number. Suppose that from all the nodes on the boundary (... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,013 |
23.26. A regular triangle is divided into $n^{2}$ identical regular triangles (Fig. 23.4). Some of them are numbered with the numbers $1,2, \ldots, m$, such that triangles with consecutive numbers have adjacent sides. Prove that $m \leqslant n^{2}-n+1$. | 23.26. Let's color the triangles as shown in Fig. 23.13. Then the number of black triangles will be \(1+2+\ldots+n=n(n+1) / 2\), and the number of white triangles will be \(1+2+\ldots+(n-1)=n(n-1) / 2\). It is clear that two triangles with consecutive numbers are of different colors. Therefore, among the numbered trian... | \leqslantn^{2}-n+1 | Combinatorics | proof | Yes | Yes | olympiads | false | 28,014 |
23.27. The bottom of a rectangular box is paved with tiles of size $2 \times 2$ and $1 \times 4$. The tiles were spilled out of the box and one $2 \times 2$ tile was lost. Instead of it, a $1 \times 4$ tile was taken out. Prove that it is now impossible to pave the bottom of the box with the tiles. | 23.27. Let's paint the bottom of the box in two colors as shown in Fig. 23.14. Then each $2 \times 2$ tile covers exactly one black cell, while a $1 \times 4$ tile covers 2 or 0. Therefore, the parity of the number of black cells on the bottom of the box matches the parity of the number of $2 \times 2$ tiles. Since rep... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,015 |
23.28. From a sheet of graph paper measuring $29 \times 29$ cells, 99 squares of size $2 \times 2$ cells have been cut out. Prove that it is possible to cut out one more such square. | 23.28. On the given square sheet of paper, we will shade the $2 \times 2$ squares as shown in Fig. 23.15. As a result, 100 shaded squares will be obtained. Each cut-out square touches exactly one shaded square, so at least one shaded square remains intact and can be cut out.
Fig. 23.4
... | 23.29. If a polygon is divided into parts by several diagonals, then these parts can be colored in two colors so that parts sharing a common side are of different colors. This can be done as follows. We will sequentially draw the diagonals. Each diagonal divides the polygon into two parts. In one of them, we preserve t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,017 |
23.30. Can a checkerboard of size $10 \times 10$ be tiled with tiles of size $1 \times 4$? | 23.30. We will color the board in four colors as shown in Fig. 23.17. It is easy to count that there are 26 cells of the second color, and 24 of the fourth. Each tile $1 \times 4$ covers one cell of each color. Therefore, it is impossible to tile the $10 \times 10$ board with $1 \times 4$ tiles, since otherwise there w... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,018 |
23.31. On a grid paper, there are $n$ arbitrary cells. Prove that from them, one can select at least $n / 4$ cells that do not share any common points. | 23.31. We will color the grid paper in four colors as shown in Fig. 23.18. Among the given $n$ cells, there will be no less than $n / 4$ cells of the same color, and the cells of the same color do not share any points.
. The vertices of the $n$-gon can be colored in three colors such that all vertices of each of the resulting triangles will be of different colors (see problem 23.40). The vertices of some color will not exceed $[n / 3]... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,022 |
23.35. A plane is colored in two colors. Prove that there are two points of the same color, the distance between which is 1. | 23.35. Consider an equilateral triangle with side length 1. All three of its vertices cannot be of different colors, so two vertices have the same color; the distance between them is 1. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,023 |
23.36*. A plane is painted in three colors. Prove that there are two points of the same color, the distance between which is 1. | 23.36. Suppose that any two points lying at a distance of 1 are colored in different colors. Consider an equilateral triangle $A B C$ with side 1; all its vertices are of different colors. Let point $A_{1}$ be symmetric to $A$ with respect to the line $B C$. Since $A_{1} B=A_{1} C=1$, the color of point $A_{1}$ is diff... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,024 |
23.37*. A plane is painted in seven colors. Is it necessarily true that there will be two points of the same color, the distance between which is $1$? | 23.37. Let's provide an example of coloring the plane in seven colors such that the distance between any two points of the same color is not equal to 1. We will divide the plane into equal hexagons with side length $a$ and color them as shown in Fig. 23.20 (points belonging to two or three hexagons can be colored in an... | 1/\sqrt{7}<1/2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 28,025 |
23.38*. The points on the sides of an equilateral triangle are colored in two colors. Prove that there exists a right triangle with vertices of the same color.
$$
* * *
$$ | 23.38. Suppose there is no right-angled triangle with vertices of the same color. Divide each side of an equilateral triangle into three equal parts with two points. These points form a regular hexagon. If two opposite vertices of the hexagon are of the same color, then all other vertices will be of the second color, w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 28,026 |
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