problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|---|
19. Prove that the lateral surface area of a regular pyramid is equal to the area of the base divided by the cosine of the dihedral angle between the plane of the lateral face and the plane of the base. | 19. In the school course, it is proved that the area of the orthogonal projection of a planar polygon is equal to the area of this polygon multiplied by the cosine of the dihedral angle between the plane of the figure and the plane of projection.
When projecting the lateral faces of a regular $n$-sided pyramid onto th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,468 |
20. What are the necessary and sufficient conditions for a sphere to be circumscribed around a pyramid? | 20. If a sphere is circumscribed around a pyramid, the base of the pyramid is an inscribed polygon, and the center of the sphere is equidistant from all vertices of the polygon and from the apex of the pyramid.
Conversely: if the base of the pyramid is an inscribed polygon, then by drawing a perpendicular from the cen... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,469 |
21. What are the necessary and sufficient conditions for a sphere to be inscribed in a pyramid? | 21. All faces of the described pyramid are equidistant from the center of the inscribed sphere. At this point, the bisector planes of all dihedral angles of the pyramid intersect.
Conversely: the existence of a common point of all bisector planes of the pyramid is a sufficient condition for the existence of a sphere i... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,470 |
22. Prove that the volume of a polyhedron into which a sphere is inscribed is one third of the product of the radius of the sphere and the surface area of the polyhedron. | 22. Let's divide such a polyhedron into pyramids, the bases of which are the faces of the polyhedron, and the common vertex is the center of the inscribed sphere. If the areas of the faces are $S_{1}, S_{2}, \ldots, S_{n}$, then the volume of the polyhedron is $V=\frac{1}{3} S_{1} R+\frac{1}{3} S_{2} R+\cdots+\frac{1}{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,471 |
23. A pyramid is intersected by a plane parallel to the base, so that its lateral surface is divided into two parts of equal area. In what ratio does this plane divide the lateral edges of the pyramid? | 23. When intersecting a pyramid with a plane parallel to the base, we obtain a pyramid similar to the given one with the center at the vertex of the pyramid. The areas of similar polygons are proportional to the squares of the corresponding sides. In this case, the latter ratio is equal to the ratio of the squares of t... | \frac{}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,472 |
24*. Consider an arbitrary convex polyhedron. We will color its faces blue and red such that no two adjacent faces, i.e., faces sharing a common edge, are colored blue (and each face is colored with one color). Prove that if more than half of all the faces are colored blue in such a coloring, then it is impossible to i... | 24. Let a polyhedron have $n$ faces, and a sphere is inscribed in it. Then there exists a unique point of tangency of the sphere with each face. By connecting each of the $m$ vertices of this face to the point of tangency, we obtain $m$ planar angles with the vertex at the point of tangency, which we will call "central... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,473 |
25. Two spheres are "inscribed" in a regular truncated triangular pyramid: one touches all its faces, the other touches all its edges.
Calculate the dihedral angle between the planes of the base of the pyramid and its lateral face. | 25. Let $a$ denote the length of the side of the upper base, and $b$ - the length of the side of the lower base of the pyramid $(a<b)$. The sphere that touches the edges of the pyramid intersects the lateral face (an isosceles trapezoid) along a circle inscribed in this face. By the property of the sides of a circumscr... | 3515^{\}52^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,474 |
26. The center of the cube is reflected off all its faces, and the resulting 6 points together with the vertices of the cube determine 14 vertices of a polyhedron, all 12 faces of which are congruent rhombuses. (This polyhedron is called a rhombic dodecahedron—natural monocrystals of garnet have this shape.) Calculate ... | 26. The twelve faces of the rhombic dodecahedron (Fig. 100) are rhombuses, one diagonal of which is equal to $a$, and the other is $-a \sqrt{2}$ (why?). The area of each rhombus is $a^{2} \frac{\sqrt{2}}{2}$,
$. | 30. Let's number the faces of the polyhedron with indices \(i=1,2, \ldots, f\) and denote by \(m_{i}\) the number of sides of the \(i\)-th face. Then the sum of all dihedral angles can be expressed by the formula:
\[
S=\sum_{i=1}^{f} \pi\left(m_{i}-2\right)=\pi \sum_{i=1}^{f} m_{i}-2 \pi f
\]
But from problem 27 we h... | 2\pi(-f) | Geometry | proof | Yes | Yes | olympiads | false | 44,479 |
31*. Does there exist a convex polyhedron having 7 edges? | 31. From the formula $2 a=3 f_{3}+4 f_{4}+5 f_{5}+\cdots$, setting $a=7$, we get:
$$
14=3 f_{3}+4 f_{4}+\cdots
$$
The number of faces of a polyhedron cannot be less than four:
$$
f_{3}+f_{4}+\cdots \geqslant 4
$$
This means that the polyhedron can only have triangular and quadrilateral faces, because if it had at l... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,480 |
32*. For an arbitrary point $S$ of a convex polyhedron, consider the corresponding polyhedral angles with a common vertex $S$, "resting" on the faces of the polyhedron. These polyhedral angles in their union give the entire space. Taking this into account and using the formulas derived in previous problems, prove Euler... | 32. The sum of the values of polyhedral angles with a common vertex $S$, resting on the faces of a polyhedron and filling all space, is $4 \pi$ (see § 6, problem 34). This same sum can be obtained by subtracting from the sum of the dihedral angles of all polyhedral angles the sum of the internal (planar) angles of all ... | p+2 | Geometry | proof | Yes | Yes | olympiads | false | 44,481 |
33*. Prove that there does not exist a convex polyhedron that has neither any triangular faces nor any trihedral angles. | 33. Suppose there exists a convex polyhedron in which there is no triangular face and no trihedral angle. This means that in the formulas from problems 27 and 28, we can set $f_{3}=p_{3}=0$, and we get:
$$
\begin{gathered}
2 a=4 f_{4}+5 f_{5}+\cdots \\
2 a=7 p_{4}+5 p_{5}+\cdots
\end{gathered}
$$
Since $f_{4}+f_{5}+\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,482 |
34*. a) Prove that there does not exist a convex polyhedron all of whose faces have more than 5 sides.
b) Prove that there does not exist a convex polyhedron all of whose polyhedral angles have more than 5 faces. | 34. a) Suppose in the formula of problem 27 we are given: $f_{3}=f_{4}=f_{5}=0$. Then we get: $2 a=6 f_{6}+7 f_{7}+\cdots, 2 a \geqslant 6 f$. On the other hand, from the formula of problem 28, applying the conclusions of the previous problem, we obtain:
$$
p_{3} \neq 0, 2 a \geqslant 3 p, 4 a \geqslant 6 p
$$
Adding... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,483 |
35*. Prove that the sum of the magnitudes of the plane angles of all the faces of a convex polyhedron is twice the magnitude of the sum of the interior angles of a plane polygon, the number of vertices of which is equal to the number of vertices of the polyhedron. | 35. In problem 30, the formula for the sum of all dihedral angles of a polyhedron was obtained: $S=2 \pi(a-f)$. Since by Euler's theorem $p-2=a-f$, we get: $S=2 \pi(p-2)$. | 2\pi(p-2) | Geometry | proof | Yes | Yes | olympiads | false | 44,484 |
36*. A polyhedron is called regular if all its faces are regular and congruent polygons to each other and all its polyhedral angles are also regular and congruent to each other ${ }^{2}$. Investigate the possibility of constructing simple regular polyhedra with $n$-sided faces and $m$-sided angles and determine the pos... | 36. Let the faces of a regular polyhedron be regular $n$-gons, and the polyhedral angles at the vertices be regular $m$-gonal angles; obviously, $m \geqslant 3, n \geqslant 3$. Divide the space into $f$ congruent regular $n$-gonal angles with a common vertex $S$ such that around each edge $[S A)$ there are $m$ angles. ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,485 |
37. A line $l$ is called an axis of rotational symmetry of a spatial figure $\Phi$ if, under a certain rotation of space around the axis $l$, the figure $\Phi$ is mapped onto itself. If the smallest angle of such a rotation is $\frac{2 \pi}{n}$, then $l$ is called an axis of symmetry of order $n$. (For example, a regul... | 37. In a tetrahedron, there are four axes of third-order symmetry passing through the vertices and the centers of the opposite faces, and three axes of second-order symmetry passing through the midpoints of opposite edges. The other regular polyhedra have the following axes of symmetry.
Hexahedron (cube):
3 axes of 4... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,486 |
38*. Prove that for each regular polyhedron, three concentric spheres can be constructed: one passing through all its vertices (circumscribed sphere); another touching all its edges; a third touching all its faces (inscribed sphere). The common center of these spheres is called the center of the regular polyhedron. | 38. Since the correct polyhedron is composed of $f$ congruent regular pyramids with a common vertex $S$, and in a regular pyramid, the vertex is equidistant from all vertices of the base and from the midpoints of all edges of the base, and the heights of all $f$ pyramids are equal to each other, it follows that point $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,487 |
39*. Prove that the number of rotations of space under which a regular polyhedron is mapped onto itself is twice the number of its edges. (In this case, the identity mapping of space should also be considered a rotation - by an angle of $0^{\circ}$.) | 39. A rotation of space about an axis, under which a polyhedron is mapped onto itself, will be called simply a rotation of the polyhedron. If the polyhedron is regular, then a fixed edge $[A B]$ can be mapped to any edge $\left[A^{\prime} B^{\prime}\right]$ out of the total number $a$ by two rotations of the polyhedron... | 2a | Geometry | proof | Yes | Yes | olympiads | false | 44,488 |
40. Prove that the sum of vectors $\overrightarrow{O A}_{1}+\overrightarrow{O A}_{2}+\cdots+\overrightarrow{O A}_{n}$, where $O$ is the center of a regular polyhedron, and $A_{1}, A_{2}, \ldots, A_{n}$ are all its vertices, is equal to zero. | 40. Suppose the sum of the given vectors is a vector $\vec{m} \neq \overrightarrow{0}$. Consider the rotation of the polyhedron about one of its axes of symmetry, not parallel to $\vec{m}$. On the one hand, the considered sum should not change; on the other hand, the vector $\vec{m}$ will rotate and transform into a ne... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,489 |
41. Prove that the sum of the distances from an internal point of a regular polyhedron to the planes of its faces does not depend on the choice of the point. | 41. Let point $P$ be at distances $h_{1}, h_{2}, \ldots, h_{f}$ from the plane of $f$ faces of a regular polyhedron. By dividing the polyhedron into $f$ pyramids with a common vertex $P$ and bases as the faces of the polyhedron, we obtain that the volume of the polyhedron is $V=\frac{Q}{3}\left(h_{1}+h_{2}+\cdots+h_{f}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,490 |
43. Prove that the centers of the faces of each regular polyhedron serve as the vertices of another regular polyhedron, and that the centers of the faces of this second polyhedron are the vertices of a third polyhedron, homothetic to the original. Two types of regular polyhedra, connected by this construction, are call... | 43. A ray emanating from the center of a regular polyhedron to any of its vertices is an axis of symmetry of order $m$ of the regular $m$-sided angle. Upon rotation about this axis by an angle $\frac{2 \pi}{m}$, the adjacent faces and the centers of the corresponding regular $n$-gons coincide with each other. Therefore... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,492 |
44. Prove that from the eight vertices of a cube, four vertices can be chosen in two ways such that they form the vertices of a regular tetrahedron. Calculate the ratio of the volume of such a tetrahedron to the volume of the cube. Also, calculate the volume of the figure that is the intersection of these two tetrahedr... | 44. In the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the face diagonals $[A C],\left[C D^{\prime}\right],\left[D^{\prime} A\right]$, $\left[A B^{\prime}\right],\left[B^{\prime} C\right],\left[B^{\prime} D^{\prime}\right]$ serve as the edges of a regular tetrahedron, while the other 6 face diagonals se... | \frac{1}{3}^{3},\frac{1}{6}^{3},\frac{1}{2}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,493 |
45*. Which four of the eight faces of an octahedron need to be extended so that in the intersection of the four half-spaces defined by the resulting planes and containing the octahedron, a regular tetrahedron is obtained? What is the ratio of the surface areas and volumes of the tetrahedron and the octahedron? | 45. In the previous problem, an octahedron inscribed in a tetrahedron was considered. This tetrahedron can be obtained by extending four non-adjacent and non-opposite faces of the octahedron. As already proven in the previous problem, the volume of the resulting tetrahedron is twice the volume of the octahedron. Each f... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,494 |
46*. Inscribed a cube in a dodecahedron so that all eight vertices of the cube are vertices of the dodecahedron. How many solutions does the problem have? | 46. Consider a dodecahedron with center $S$ (Fig. 101). Due to the congruence of the faces, their diagonals are also congruent: $[A B] \cong|B C| \cong \cong[C D] \cong[D A]$. In the regular triangular pyramid $P A B Q$, the edge $[P Q]$ is perpendicular to the base edge $[A B]$ (see problem 8 from § 5). But $(P Q)\|(A... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,495 |
48*. To intersect an octahedron with a plane so that the intersection is a regular hexagon. Calculate the area of this hexagon if the edge of the octahedron has length $a$. | 48. The axis of symmetry of a third-order octahedron passes through the centers of two opposite faces. Let $A B C$ be one of these faces, and $A^{\prime}, B^{\prime}, C^{\prime}$ be the vertices of the second face, symmetric to the corresponding vertices of the first face relative to the center $S$ of the octahedron. S... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,497 | |
49*. Find two points on each of the six faces of a cube so that these points become the vertices of an icosahedron inscribed in the cube. | 49. Place one segment of equal length \( m \) on the middle lines of the faces of a cube, with the ends at equal distances from the edges. Arrange the segments and choose their length such that, by connecting the ends of the segment on one face with the end of the segment on another face, we form an equilateral triangl... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,498 | |
51*. Consider a sphere that touches all the edges of a regular polyhedron, and through the midpoint of each edge, draw a line perpendicular to this edge and the radius of the sphere connecting its center to the midpoint of the edge. Prove that each of these lines contains an edge of the polyhedron dual to the given one... | 51. Let us draw an axis of symmetry of order $n$ through the center of a regular polyhedron and the center of one of its faces. The lines passing through the midpoints of the edges of this face and perpendicular to the edge lie in the plane of symmetry of the polyhedron, which passes through the drawn axis and the midp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,500 |
52*. Prove that any two regular polyhedra of the same type are similar to each other: each of them is congruent to a polyhedron homothetic to the other. | 52. From the method of constructing a regular polyhedron described in problem 36, it follows that each polyhedron from its center $S$ can be divided into $f$ regular polyhedral angles that fill all space. These angles in like-named polyhedra are congruent. Therefore, by superimposing the centers of two regular like-nam... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,501 |
1. What are the necessary and sufficient conditions for a right prism to be inscribed in a cylinder? | 1. A right prism inscribed in a cylinder has equal heights with it, and the base of the prism is an inscribed polygon.
Conversely: if the base of a right prism is an inscribed polygon, then by describing a circle around it, we obtain the base of the cylinder, the generators of the lateral surface of which are the edge... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,502 |
4. The figure is obtained by rotating a rectangle about an axis parallel to one of its sides and lying in the plane of the rectangle, but not intersecting it. Prove that the volume of this figure is equal to the product of the area of the rectangle and the length of the circle described by the center of the rectangle d... | 4. Let a rectangle have sides of lengths $a$ and $b$. Let the axis of rotation $l$ be parallel to the side of length $a$ and be at a distance $r$ from the center of the rectangle. Then the volume of the solid of revolution will be equal to the difference in volumes of two cylinders: one with a base radius of $r+\frac{b... | \cdot2\pir | Geometry | proof | Yes | Yes | olympiads | false | 44,505 |
5. A cylinder is called equilateral if its height is equal to the diameter of its base. An equilateral cylinder is inscribed in a regular tetrahedron with edge $a$ such that one of the cylinder's bases lies in one of the tetrahedron's faces, and the circle of the second base touches the three other faces. Calculate the... | 5. Place a right circular cylinder inside a tetrahedron such that its base coincides with the base of the tetrahedron, and its axis coincides with the height of the tetrahedron. Now, intersect the entire figure with a plane passing through a lateral edge and the height of the tetrahedron. Then, in
}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,506 |
6. Inscribed in a cube with edge a is a right circular cylinder such that the axis of the cylinder contains the diagonal of the cube, and the circles of the cylinder's bases touch the faces of the cube. Calculate the volume of this cylinder. | 6. Place a right circular cylinder so that its axis coincides with the diagonal of a cube, and the center of the cube is equidistant from the bases of the cylinder. Draw a diagonal section through this diagonal of the cube, and we will obtain the figure shown in Figure 105. Take the center of the cube as the center of ... | \frac{\sqrt{3}}{2(1+\sqrt{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,507 |
7. The diameter of the base of the cylinder was doubled and at the same time its height was halved. How did the area of the lateral surface and the volume of the cylinder change? | 7. The length of the circumference of the base of the cylinder has doubled, and the height has been halved, so the lateral surface area remains unchanged. When the radius of the base is doubled, its area increases fourfold, and when the height is halved at the same time, the volume of the cylinder doubles.
. A section plane, parallel to the axes and passing at a distance $x$ from point $O$, intersects the cylinders to form strips of width $2 \sqrt{R^{2}-x^{2}}$, and in the intersection with the considered body, it forms a square. The area of ... | \frac{16}{3}R^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,512 |
12. The vertex of a surface of revolution that is conical coincides with the origin $O$ of a rectangular coordinate system $O x y z$, the axis of rotation coincides with the positive direction of the $O z$ axis, and the ray, by the rotation of which the surface is obtained, forms an angle $\varphi$ with the $O z$ axis.... | 12. Let the point $M(x ; y ; z)$ be at a distance $r$ from the $O z$ axis. Then $x^{2}+y^{2}=r^{2}$, since the point $M^{\prime}$ - the projection of $M$ on $O x y$ is at a distance $r$ from the point $O$. But $r=z \operatorname{tg} \varphi$. Therefore, the equation of the conical surface will be the relation $x^{2}+y^... | x^{2}+y^{2}-z^{2}\operatorname{tg}^{2}\varphi=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,513 |
13. Construct a surface of rotation having as its generators three given lines passing through one point and not lying in the same plane. How many solutions does the problem have? Do the resulting conical surfaces have common generators? | 13. Let $S$ be the given point where the lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ intersect. From point $S$, six rays originate: $S A, S A^{\prime}, S B, S B^{\prime}$, $S C$ and $S C^{\prime}$. For each triplet of these rays, for example $S A, S B$, $S C$, a line can be constructed, the points of which are equ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,514 |
14*. What are the necessary and sufficient conditions for a given convex dihedral angle to be inscribed in a surface of a cone of revolution so that the edges of the angle are generators of the conical surface? | 14. Let a convex quadrilateral angle be inscribed in a surface of a cone of revolution. Since the axis of rotation $l$ lies in the plane of symmetry of each pair of generators, the dihedral angles formed by the plane of the face and two planes passing through the axis and adjacent edges are congruent to each other. Let... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,515 |
15*. What are the necessary and sufficient conditions for a cone of revolution to be inscribed in a given convex dihedral angle so that the angle and the conical surface intersect exactly along four generatrices of the surface - lying in the faces of the angle? | 15. Let a surface of revolution be inscribed in a tetrahedral angle $S A B C D$. By intersecting the figure with a plane perpendicular to the axis of rotation $l$, we obtain the figure shown in Figure 111. Points $A, B, C, D$ belong to the edges; $O$ is the projection of the vertex $S$ onto this plane; $m$, $n, p, q$ a... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,516 |
17. In what case does the area of the triangle formed by the intersection of a cone with a plane passing through the vertex of the cone have the greatest magnitude? | 17. If in the axial section of a cone the angle between the generators does not exceed $90^{\circ}$, then this angle will be the largest among the angles between the generators. Consequently, the area of this (axial) section will also be the largest, since this area is equal to half the product of the square of the len... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,518 |
18. In a cone of height $h$ and base radius $R$, inscribe a cylinder with the maximum lateral surface area and find this area. | 18. If a cylinder is inscribed in a cone such that one of its bases lies on the base of the cone, and the other touches the lateral surface of the cone, then in the axial section of the cone, we will obtain an isosceles triangle with an inscribed rectangle. Let the distance from the vertex of the cone to the upper base... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 44,519 | |
19. Given a sphere of radius $R$. At a distance equal to $2R$ from the center of the sphere, a point $S$ is taken, and from it, all lines tangent to the sphere (i.e., having exactly one common point with it) are drawn. What does the union of these tangents represent? Calculate the area of the surface formed by the segm... | 19. All tangents have a common point and are at the same distance from the center of the sphere. Therefore, the union of these tangents forms a surface of a cone of revolution. The length of the generatrix $l$ from point $S$ to the points of tangency is calculated using the Pythagorean theorem: $l^{2}=4 R^{2}-R^{2}, l=... | \frac{3\piR^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,520 |
20. Prove that for a frustum of a cone to have a sphere inscribed in it, touching the bases and each generatrix of the cone, it is necessary and sufficient that the height of the cone be the mean proportional between the diameters of the upper and lower bases of the cone: $2 R: h = h: 2 r$ (Here $r$ is the radius of th... | 20. If a sphere is inscribed in a truncated cone, then by making a section of the cone with a plane passing through its axis, we obtain an isosceles trapezoid with an inscribed circle in the section. The diameter of the circle is the height of the cone, and the bases of the trapezoid are equal to the diameters of the b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,521 |
22*. All four sides of a spatial quadrilateral (see § 6, problem 15) touch a sphere. Prove that the four points of tangency lie in the same plane. | 22. Let $ABCD$ be a given spatial quadrilateral: from each of its vertices, two sides touch a sphere. The points of tangency of the sphere with the lines $AB, BC, CD, DA$ are denoted by $P, Q, R, T$ respectively. We draw a plane through two such tangents from one vertex, which intersects the sphere in a circle; the sam... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,523 |
23. Consider a hexahedron (six-sided polyhedron), all faces of which are quadrilaterals that can be inscribed in a circle. Prove that there exists a unique sphere passing through all 8 vertices of this polyhedron. | 23. Let $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a hexahedron, where $A B C D$ is a face, $\left[A A^{\prime}\right],\left[B B^{\prime}\right],\left[C C^{\prime}\right],\left[D D^{\prime}\right]$ are edges. A circle passes through $A B C D$. This circle and the point $A^{\prime}$ outside the plane $A B ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,524 |
24*. What does the set of points in space represent, not lying in one plane, not coinciding with the entire space, and such that through any three of its points there passes a circle contained in this set? | 24. Let's take three arbitrary points $A, B, C$ of the given set. Any point of the circle passing through $A, B$, and $C$ also belongs to this set. Let $D$ be another point of the same set, $D \notin (ABC)$. The circle passing through
, $B$ and $D$, also lies within this set. Since $D \notin (ABC)$, the planes of both... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,525 |
25. Four pairwise non-intersecting spheres of equal radii are given, with centers not lying in the same plane. a) How many spheres exist that touch all four spheres simultaneously? b) How to construct these spheres? | 25. a) The contact of spheres can be of two kinds: 1) external contact, when the spheres lie one outside the other and the distance between their centers is equal to the sum of the radii; 2) internal contact, when one sphere is inside the other and the distance between the centers is equal to the difference of the radi... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,526 |
26. What does the set of midpoints of all chords represent, which are obtained by the intersection of a given sphere with all lines passing through a given point? Consider the cases when the common point of the lines is outside the sphere, belongs to the sphere, and lies inside the sphere. | 26. Let's conduct a plane through the center of the sphere $O$ and the common point $S$ of the considered lines. In the section, we will obtain a family of lines passing through $S$ and intersecting the circle with center $O$. If $P$ is the midpoint of one of the resulting chords, then $(O P) \perp(P S)$ and therefore ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,527 |
27. What does the set of centers of circles represent, which are obtained by the intersection of a sphere with all possible planes passing through a given line $l$? | 27. The problem is solved in the same way as the previous one - by considering the image in the plane passing through the center of the given sphere and perpendicular to the line $l$. The centers of the circle sections form a circle located inside the sphere if the line intersects the given sphere. These centers form a... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,528 |
28. Inscribed a regular tetrahedron in a sphere of radius $R$ (describe the construction). Find the volume of this tetrahedron. | 28. The center $O$ of a regular tetrahedron divides its height in the ratio $1:3$. The center of the circumscribed sphere coincides with the center of the tetrahedron. Therefore, to construct a tetrahedron inscribed in a sphere, we choose any diameter of the sphere and at a distance of $\frac{1}{3} R$ from the center, ... | \frac{8R^3\sqrt{3}}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,529 |
29. Inscribe a cube in a sphere of radius $R$ and find its volume. | 29. The diameter of the sphere circumscribed around a cube is equal to the diagonal of the cube: \(2 R = a \sqrt{3}\). Therefore, the edge of the inscribed cube is not difficult to construct (for example, as the radius of the circle circumscribed around an equilateral triangle with sides equal to \(2 R\)). Knowing the ... | \frac{8R^{3}}{3\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,530 |
30. Describe a regular tetrahedron around a sphere of radius $R$ and find its volume. | 30. To describe a regular tetrahedron around a sphere, we inscribe a regular tetrahedron in it and draw planes through its vertices parallel to the opposite faces. These planes are perpendicular to the radii leading to the vertices and therefore touch the sphere. The intersection of these four planes results in a tetra... | 8R^{3}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,531 |
31. Describe a regular octahedron around a sphere of radius $R$ and find its volume. | 31. The three diagonals of an octahedron are mutually perpendicular. Let \( a \) be half the length of a diagonal, \( r \) be the radius of the inscribed sphere, and \( O \) be the common center of the octahedron and the inscribed sphere. If we take the diagonals as coordinate axes and point \( O \) as the origin, then... | 4r^{3}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,532 |
33. The intersection of a sphere with a dihedral angle, the edge of which passes through the center of the sphere, is called a spherical digon. The two semicircles bounding it are called its sides, and the magnitude of the dihedral angle, equal to the angle between the tangents drawn to the semicircles at their point o... | 33. a) Two biangles with equal angles on a sphere can be mapped onto each other by a composition of rotations about axes passing through the center of the sphere: first, by a rotation around an axis perpendicular to the diameters of the biangles, these diameters are aligned, and then by a rotation about the resulting c... | S_{\alpha}=2R^{2}\alpha | Geometry | proof | Yes | Yes | olympiads | false | 44,534 |
34. A spherical triangle is the intersection of a sphere and a trihedral angle with its vertex at the center of the sphere (or a part of the sphere bounded by three arcs of great circles) (Fig. 6). Calculate the area of the spherical triangle \(ABC\), considering it as the intersection of three dihedral angles: one wit... | 34. Figure 6 shows a hemisphere, the area of the surface of which is $2 \pi R^{2}$. This hemisphere is divided into three spherical digons with areas $S_{\alpha}, S_{\beta}, S_{\gamma}$. Here, the digons $S_{\alpha}$ and $S_{\beta}$ are entirely contained within the hemisphere, while the digon $S_{\gamma}$ is "partiall... | (\alpha+\beta+\gamma-\pi)R^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,535 |
36. Each edge of a cube is divided into three congruent parts. Prove that the twenty-four points of division lie on one sphere. Calculate the surface area of this sphere if the edge length of the cube is $a$. | 36. Consider the diagonal section of a cube. By connecting its center $O$ with one of the division points on the edge of the cube and with the midpoint of the edge, we obtain a right triangle with legs of length $\frac{a}{6}$ and $\frac{a \sqrt{2}}{2}$. The distance from the center $O$ to each of the division points is... | \frac{19}{9}\pi^{2} | Geometry | proof | Yes | Yes | olympiads | false | 44,536 |
37. The faces of a parallelepiped with edges of length $a$ are rhombuses with an acute angle of $60^{\circ}$. Calculate the volume of the sphere inscribed in the parallelepiped. | 37. Given that the magnitude of the acute angles of all faces is $60^{\circ}$, the diagonals of the faces are congruent to the lateral edges of the parallelepiped. Let $A B C D$ be one of the faces, $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ - the parallel face, and $A$ - the vertex of three acute angles. If two reg... | \frac{2}{9}\pi\frac{^{3}}{\sqrt{6}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,537 |
38. Two congruent spheres are inscribed in opposite trihedral angles of a cube and touch each other. Calculate the volumes of the spheres if the length of the edge of the cube is $a$. | 38. To construct tangent spheres inscribed in opposite trihedral angles of a cube, consider the diagonal section of the cube $A A^{\prime} C^{\prime} C$ (Fig. 116). The spheres can be constructed using the method of homothety. From the similarity of triangles $A A^{\prime} C^{\prime}$ and $O P C^{\prime}$, we obtain:
... | \frac{\sqrt{3}}{2(\sqrt{3}+1)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,538 |
39*. Output Simpson's formula: if the area $S(x)$ of a cross-section of a spatial figure by a plane $P_{x}$, perpendicular to a certain coordinate line $O x$ through a point with coordinate $x$, is expressed as a polynomial of variable $x$ not higher than the second degree, then the volume of the figure can be calculat... | 39. If \( S(x) = a^2 x + b x + c \), then the volume \( V \) of the body is determined by the integral: \( V = \int_{0}^{h} \left( a x^2 + b x + c \right) d x = \frac{a h^3}{3} + \frac{b h^2}{2} + c h \), where \( h \) is the height of the body.
To derive the Simpson's formula from this, substitute the values \( x = 0... | proof | Calculus | proof | Yes | Yes | olympiads | false | 44,539 |
Example 2. From the vertex $B$ of an equilateral triangle $A B C$, a perpendicular $B K$ is erected to the plane $A B C$, such that $|B K|=|A B|$. We will find the tangent of the acute angle between the lines $A K$ and $B C$.
Translated as requested, preserving the original text's line breaks and format. | Solution. Let's construct the image of the given figure and find its parametric number (Fig. 4).
Considering an arbitrary triangle $ABC$ as the image of an equilateral triangle,
Fig. 4, w... | \sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,541 |
Example 8. In a parallelogram with sides equal to $a$ and $b$, and the angle between these sides equal to $\alpha$, the bisectors of the four angles are drawn. Find the area of the quadrilateral bounded by the bisectors.
Given (Fig. 10): $ABCD$ is a parallelogram, $l_{A}, l_{B}, l_{C}, l_{D}$ are the bisectors of angl... | Solution. First, let's determine the type of quadrilateral $P M E K$. Since $(A D) \|(B C)$, $\widehat{M A D}=\frac{\alpha}{2}$ and $\widehat{B C K}=\frac{\alpha}{2}$, then $(A M) \|(C K)$. Similarly, $(B K) \|(D M)$, i.e., $P M E K$ is a parallelogram.
Furthermore, since $\widehat{B A M}=\frac{\alpha}{2}$ and $\wideh... | \frac{(-b)^{2}\sin\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,544 |
E x a m p l e 9. The angle between the sides of a triangle, the lengths of which are $a$ and $b$, is $\varphi$. Let's find the length of the segment of the bisector of this angle, lying inside the triangle.
G i v e n (Fig. 11): $\triangle A B C,|A C|=b,|B C|=a, \widehat{A C B}=\varphi$, $\widehat{A C D}=\widehat{B C D... | Solution. Let $|CD| = x$. Since $S_{\triangle ABC} = S_{\triangle ACD} + S_{\triangle BCD} = \frac{1}{2} b x \sin \frac{\phi}{2} + \frac{1}{2} a x \sin \frac{\varphi}{2} = \frac{1}{2} x(a+b) \cdot \sin \frac{\varphi}{2}$ and $S_{\triangle ABC} = \frac{1}{2} a b \sin \varphi$, we obtain the following equation:
.
Let $|A B|=x$, then $|B C|=x+2$. Introduce an auxiliary unknown: $\widehat{A C B}=y$. Then $\widehat{B A C}=2 y$.
By the Law of Sines, we have:
$$
\frac{x}{\sin y}=\frac{x+2}{\sin 2 y}=\frac{5}{\sin \left(180^{\circ}-3 y\right)}
$$
or
$$
\left\{\begin{array}{l}
\frac{x}{\sin y}=\frac{x+2}{\sin ... | |AB|=4\mathrm{~},|BC|=6\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,546 |
Example 17. In a right-angled triangle, from the vertex of the right angle, a height and a median are drawn. The ratio of their lengths is $40: 41$. Find the ratio of the lengths of the legs.
Given (Fig. 19): $\triangle A B C, \widehat{A C B}=90^{\circ},[C K]-$ height, $[C M]-$ median, $|C K|:|C M|=40: 41$.
To find: ... | Solution. Introduce an auxiliary parameter, setting $|C K|=h$. Then $|C M|=\frac{41 h}{40}$. We will find the desired ratio by direct calculation.
From the right triangle $C M K$ we have: $|M K|=\frac{9 h}{40}$, and since $[C M]$ is the median of the right triangle $A B C$, then $|A M|=|B M|=|C M|=\frac{41 h}{40}$.
!... | |AC|:|BC|=4:5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,549 |
Example 18. A line is drawn through the centroid of an equilateral triangle in the plane of this triangle. We will prove that the sum of the squares of the distances from the vertices of the triangle to this line does not depend on the choice of the line.
Given (Fig. 20): $\triangle A B C,|A B|=|B C|=|A C|$. Point $O$... | Proof. Consider the case when the line $l$ does not contain the median of $\triangle ABC$. Construct $[AO], [BO]$, and $[CO]$. Since $\triangle ABC$ is equilateral, these segments are congruent. Let $|AO| = a$. Then $|BO| = |CO| = a$. As an auxiliary parameter, we could naturally take the length of the side of $\triang... | ((ABC),\widehat{,}P)=\arcsin\sqrt{\sin^2\alpha+\sin^2\beta} | Geometry | proof | Yes | Yes | olympiads | false | 44,550 |
Example 20. The radius of the circumscribed circle around a right-angled triangle relates to the radius of the circle inscribed in this triangle as $5: 2$. Find the acute angles of the triangle.
Given (Fig. 22): $\triangle A B C, \widehat{A C B}=90^{\circ}, \omega_{1}$ - the circumscribed circle, $\omega_{2}$ - the in... | Proof. We will prove that $\widehat{A C B}=90^{\circ}$. It is clear that the assumption that $\widehat{C A B}=90^{\circ}$ is inconsistent, because then the segment $C K$ would coincide with the side $A C$ and therefore must coincide with the side $B C$ to form the required congruent angles. However, the median cannot c... | \arcsin\frac{3}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,551 |
Example 22. The centers of the inscribed and circumscribed circles are symmetric with respect to one of the sides of the triangle. Find the angles of the triangle.
Given (Fig. 25): $\triangle A B C, \omega_{1}\left(O_{1},\left|O_{1} C\right|\right)$ - circumscribed circle, $\omega_{2}\left(O_{2},\left|O_{2} M\right|\r... | Solution. First, we will prove that triangle $ABC$ is isosceles. Indeed, since point $O_1$ is the center of circle $\omega_1$, we have $|O_1A| = |O_1C|$. Since $(O_1O_2) \perp (AC)$ and $|O_1M| = |O_2M|$, we have $\widehat{O_2CM} = \widehat{O_1CM}$. Similarly, $\widehat{O_2AM} = \widehat{O_1AM}$. But point $O_2$ is the... | \widehat{BCA}=\widehat{BAC}=36,\widehat{ABC}=108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,552 |
Example 23. Prove that the square of the length of the bisector drawn from the vertex of an arbitrary triangle is equal to the product of the lengths of the lateral sides minus the product of the lengths of the segments of the base.
Given (Fig. 26): $\triangle A B C, [B D]$ - bisector.
To prove: $|B D|^{2}=|A B| \cdo... | Proof. 1st method (algebraic). For brevity, let $|A B|=a,|B C|=b,|A D|=m,|C D|=n$, $|B D|=l, \widehat{A D B}=\alpha$. Clearly, then $\widehat{C D B}=180^{\circ}-\alpha$. Let's calculate $a^{2}$ and $b^{2}$. By the cosine theorem for triangles $A B D$ and $C B D$ we have:
$$
\begin{aligned}
& a^{2}=l^{2}+m^{2}-2 l m \c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,553 |
Example 25. In $\triangle ABC$, $[BD]$ is a median, point $P \in [BD]$ and $|BP|:|PD|=3:1$; $(AP) \cap [BC]=K$. Find the ratio of the area of $\triangle ABK$ to the area of $\triangle ACK$. | S olution. Let $(A P) \cap(B C)=K$ (Fig. 29). We need to find $S_{\triangle A B K}: S_{\triangle A C K}$. Drop perpendiculars $D M_{1}, B M_{2}, C M_{3}$ to the line $A K$. From the similarity of triangles $B P M_{2}$ and $D P M_{1}$, we write: $\left|B M_{2}\right|:\left|D M_{1}\right|=3: 1$, and from the similarity o... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,554 |
Example 26. The ratio of the radius of the circle inscribed in an isosceles triangle to the radius of the circle circumscribed around this triangle is $k$. Find the cosine of the angle at the base of the triangle. | Solution (Fig. 30). Let $\widehat{B C A}=x$ and introduce an auxiliary parameter: $|M K|=r$. Then
Fig. 30
$$
|O B|=r k, \cos x=|C K|:|B C| .
$$
From triangle $M C K$ we have: $|C K|=$ $=... | \cosx=\frac{1\\sqrt{1-2k}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,555 |
Example 27. In a regular triangular pyramid, a plane is drawn through the side of the base, perpendicular to the opposite lateral edge. Let's find the area of the resulting section, if the base is equal to $a$, and the height of the pyramid is equal to $h$. | Solution. Let $SABC$ be the given pyramid and $[SO]$ its height (Fig. 31). Draw the apothem $SD$ and connect point $D$ with point $C$ and with point $M$ - the vertex of triangle
Fig. 31
$... | \frac{3^{2}}{4\sqrt{^{2}+3^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,556 |
Example 28. In a right triangle $A B C$, from the vertex $C$ of the right angle, the bisector $C K$ and the median $C M$ are drawn. Find the legs of triangle $A B C$ if $|K M|=a,|C M|=m$. | Solution (Fig. 32). This is a problem of type $F$. We will solve it algebraically by setting up a system of equations with respect to $|A C|=x$ and $|B C|=y$. For definiteness, let's assume that $|A C| \leqslant|B C|$.
Since $\triangle A B C$ is a right triangle and $[C M]$ is the median to the hypotenuse $A B$, then ... | \frac{(-)\sqrt{2}}{\sqrt{^{2}+^{2}}}\text{}\frac{(+)\sqrt{2}}{\sqrt{^{2}+^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,557 |
Example 29. In a trapezoid, the bases of which are respectively equal to $a$ and $b(a>b)$, a line parallel to the bases is drawn through the point of intersection of the diagonals. We will find the length of the segment of this line, cut off by the lateral sides of the trapezoid.
Given (Fig. 33): $A B C D$ - trapezoid... | S o l u t i o n. We will find $|P Q|$ by direct calculation. From $\triangle B C M \sim \triangle A M D$ we have:
$$
\frac{|B M|}{|M D|}=\frac{b}{a}
$$
from which, by the property of the derived proportion, we can write:
$$
\frac{|B M|}{|B M|+|M D|}=\frac{b}{a+b}, \text { or } \frac{|B M|}{|B D|}=\frac{b}{a+b}
$$
F... | |PQ|=\frac{2}{+b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,558 |
Example 33. A line $l$ is drawn through the centroid of triangle $A B C$, intersecting sides $A B$ and $B C$. We will prove that the sum of the distances from points $A$ and $C$ to the line $l$ is equal to the distance from point $B$ to this line.
Given (Fig. 37): $\triangle A B C, O$ - centroid of $\triangle A B C, l... | Proof. Let's consider a purely geometric proof.
Draw the median $BD$ of triangle $ABC$. Since point $O$ is the centroid of $\triangle ABC$, then $O \in [BD]$. Drop a perpendicular $DD_1$ from point $D$ to line $l$. Since $(AA_1) \perp l$ and $(CC_1) \perp l$, then $(AA_1) \parallel (CC_1)$, i.e., quadrilateral $AA_1CC... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,560 |
Example 34. In an isosceles triangle $A B C$, where the angle $C$ at the vertex is $100^{\circ}$, two rays are constructed: one starting from point $A$ at an angle of $30^{\circ}$ to ray $A B$, and the other starting from point $B$ at an angle of $20^{\circ}$ to ray $B A$. The constructed rays intersect at point $M$, w... | Solution (Fig. 38). Let's provide an algebraic solution.
Construct $[C M],\left[M A_{1}\right] \perp(B C),\left[M B_{1}\right] \perp(A C),\left[M C_{1}\right] \perp(A B)$. Let $\widehat{A C M}=x$, then $\widehat{B C M}=100^{\circ}-x$. Introduce an auxiliary parameter, setting $|C M|=k$, and calculate $\left|M C_{1}\ri... | \widehat{ACM}=20,\widehat{BCM}=80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,561 |
Example 35. On the sides $A B$ and $B C$ of triangle $A B C$, squares $A B D E$ and $B C K M$ are constructed (outside the triangle). We will prove that the segment $D M$ is twice the length of the median $\quad B P \quad$ of triangle $A B C$. | P r o o f (Fig. 39). This is a problem of type $M$, let's solve it geometrically.
Let $\widehat{A B C} = \alpha$, then $\widehat{D B M} = 180^{\circ} - \alpha$.
56
On the line $B P$, lay off the segment $P T$ such that $|P T| = |B P|$, and connect point $T$ with points $C$ and $A$.
Since in the quadrilateral $A B C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,562 |
Example 36. In triangle $A B C \widehat{C}=90^{\circ},[A D)$ and $[B K)-$ are the bisectors of the interior angles. Find $\widehat{B A C}$ and $\widehat{A B C}$, if $2|A D| \cdot|B K|=|A B|^{2}$ | The solution is (Fig. 40). Introduce an auxiliary parameter, setting $|A B|=c$, and let $\widehat{D A C}=x$. Then $\widehat{B A C}=2 x, \widehat{A B C}=$ $=90^{\circ}-2 x$ and $\widehat{C B} K=45^{\circ}-x$.
From the right triangle $A B C$, we find that $|A C|=$ $=c \cos 2 x$. From the right triangle $A D C$, we have:... | \widehat{BAC}=45+\arccos\frac{1+2\sqrt{2}}{4},\widehat{ABC}=45-\arccos\frac{1+2\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,563 |
Example 37. We will find the length of the diagonal of a square, two vertices of which lie on a circle of radius $R$, and the other two - on a tangent to this circle.
Given (Fig. 41): $\omega(0,|O A|)$ - circle, $|O A|=R$, $A B C D$ - square, $A \in \omega, D \in \omega,(B C)$ - tangent to $\omega,[A C]-$ diagonal of ... | S o l u t i o n. It is clear that this is a problem of type $F$. We will use the algebraic method to solve it. Let the line $B C$ touch the circle $\omega$ at point $M$, and the line $C D$ intersect it at point $F$. Draw $[OM]$, $[O F]$, and $[O K]$, where $[O K] \perp[C D]$. Note that
$$
|C F| \cdot|C D|=|C M|^{2}
$$... | \frac{3R\sqrt{57}}{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,564 |
Example 39. An isosceles triangle is inscribed in a circle, with the base equal to $a$, and the angle at the base equal to $\beta$. We will find the radius of the circle that is tangent to the lateral sides of the triangle and the first circle.
Given (Fig. 43): $\omega_{1}\left(O_{1},\left|O_{1} A\right|\right), \tria... | Solution. Construct point $D$ - the midpoint of the base $BC$ of triangle $ABC$. Since $|AB|=|AC|$, $(AD)$ is the perpendicular bisector of segment $BC$. Since $\omega_{1}$ is circumscribed around $\triangle ABC$, its center - point $O_{1} \in (AD)$.
It is clear that the bisector of angle $BAC$ lies on line $AD$. The ... | \frac{}{4\cos^2\frac{\beta}{2}\sin\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,565 |
Example 41. Points $A, M$, and $B$ are taken on the circle $\omega$, with $-A M \cong -B M$. On the arc $A M B$, a point $K$ is taken. Points $M$ and $K$ are connected to points $A$ and $B$. We will prove that
$$
|A K| \cdot |B K| = |A M|^2 - |K M|^2.
$$
. We will solve this problem of type $M$ algebraically. Introduce an auxiliary parameter, setting the radius of the circle $\omega$ to $R$. To simplify further calculations, let $\widehat{A B K}=\alpha, \widehat{K B M}=\beta$.
Since $\triangle A B K$ is inscribed in the circle $\omega$, we have $\frac{|... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,566 |
Example 42. The sides of a parallelogram are equal to $a$ and $b(a>b)$. Find the area of the parallelogram, knowing that the acute angle between its diagonals is $\alpha$.
Given (Fig. 46): $ABCD$ is a parallelogram, $|AD|=a$, $|AB|=b, \widehat{COD}=\alpha$.
Find: $S_{ABCD}$. | S o l u t i o n. This is a problem of type $F$. To find $S_{A B C D}$, we will use the direct counting method. We write:
$$
S_{A B C D}=\frac{1}{2}|A C| \cdot|B D| \cdot \sin \alpha
$$
For brevity, let $|A C|=d_{1},|B D|=d_{2}$. We will find $d_{1} d_{2}$ by applying the cosine theorem to $\triangle A O D$ and $\tria... | S_{ABCD}=\frac{1}{2}(^2-b^2)\tan\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,567 |
Example 43. Let's find the area of a right-angled triangle if its hypotenuse is equal to $c$, and the sum of the sines of the acute angles is $q$.
Given (Fig. 47): $\triangle A B C, \widehat{C}=90^{\circ},|A B|=c, \sin \widehat{A}+\sin \widehat{B}=q$.
Find: $S_{\triangle A B C}$. | Solution. $\quad S_{\triangle A B C}=\frac{1}{2}|A C| \cdot|B C|$. But $\quad|A C|=c \sin \widehat{B}$, and $|B C|=c \sin \widehat{A}$. Then
$$
|A C|+|B C|=c q
$$
Moreover,
$$
|A C|^{2}+|B C|^{2}=c^{2}
$$
Raising both sides of equation (43.1) to the second power and subtracting equation (43.2) term by term, we find... | S_{\triangleABC}=\frac{1}{4}^{2}(q^{2}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,568 |
Example 44. In triangle $A B C$, the ratio $|B C|:|A C|=3$ and $\widehat{A C B}=\alpha$. Points $D$ and $K$ are taken on side $A B$ such that $\widehat{A C D}=\widehat{D C K}=\widehat{K C B}$. Find: $|C D|:|C K|$. | S o l u t i o n (Fig. 48). The problem under consideration is of type $M$. We will solve it by introducing an auxiliary parameter. Let, for example, $|A C|=b$, then $|B C|=3 b$.
Using the area $S_{\triangle A B C}$ as a reference element (see example 9, p. 22), we will express it in two ways. On the one hand,
$$
\beg... | \frac{2\cos\frac{\alpha}{3}+3}{1+6\cos\frac{\alpha}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,569 |
Example 45. An isosceles trapezoid is circumscribed around a circle. Find the ratio of the area of the trapezoid to the area of the circle, given that the distance between the points of tangency of the circle with the lateral sides of the trapezoid is to the radius of the circle as $\sqrt{3}: 1$.
Given (Fig. 49): $A B... | S o l u t i o n. As in Example 44, this is a problem of type $M$. We will solve it similarly by introducing an auxiliary parameter. Let $|O K|=R$. We will calculate the area of trapezoid $A B C D$, making some additional constructions beforehand. From point $B$, drop $[B L] \perp [A D]$, and from point $K-[K N] \perp [... | \frac{8\sqrt{3}}{3\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,570 |
Example 46. Through the given point $A$, let's draw a plane parallel to the given plane $P$. | Solution. Analysis. Let $Q$ be the desired plane (Fig. 50), i.e., $Q \| P$ and $A \in Q$. Construct different lines $l$ and $m$ in the plane $Q$ passing through point $A$, and take an arbitrary point $A^{\prime}$ in the plane $P$. Construct the planes $\left(A^{\prime}, l\right)$ and $\left(A^{\prime}, m\right)$. Since... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,571 |
Example 47. Let's find the set of points in space that are equidistant from two given intersecting planes $P_{1}$ and $P_{2}$. | Solution. 1) Let point $M$ be equidistant from $P_{1}$ and $P_{2}$ (Fig. 51). Construct a plane $Q$ passing through point $M$ and perpendicular to the line $a = P_{1} \cap P_{2}$. For this, drop a perpendicular $M L$ from point $M$ to the line $a$, then through point $L$ draw $l_{1} \perp a$ in the plane $P_{1}$. The p... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,572 |
Example 48. Let us construct the set of points belonging to a given plane $P$ and equidistant from given points $A$ and $B$, which do not lie in the plane $P$. | S o l u t i o n. Analysis. Let point $M$ (Fig. 52) belong to the desired set of points. Since $|A M|=|B M|$, point $M$ lies in the plane $Q$ passing through point $C$ - the midpoint of segment $A B$ and perpendicular to segment $A B$. Since, in addition, $M \in P$, the desired set of points is $P \cap Q$.
Let $P \cap ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,573 |
Example 54. The plane angles at one of the vertices of a triangular pyramid are right angles. Prove that the height of the pyramid, drawn from this vertex, passes through the point of intersection of the heights of the opposite face.
Construction of the image. Let the quadrilateral $S A B C$ with its diagonals be the ... | Proof. Since $[S A] \perp[S B]$ and $[S A] \perp[S C]$, then $[S A] \perp(S B C)$. But $(B C) \subset(S B C)$, then $[S A] \perp[B C]$. Draw a line $A D$ through point $O$ - the foot of the altitude $S O$. Since $[S O]-\mathrm{L}(A B C)$, then $[A O]$ is the projection of segment $S A$ on plane $A B C$. But $[S O] \per... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,575 |
Example 55. In a regular tetrahedron $S A B C$, the segment $D O$ connects the midpoint $D$ of edge $S A$ with the centroid $O$ of face $A B C$, and point $E$ is the midpoint of edge $S B$. We will find the angle between the lines $D O$ and $C E$.
Construction of the image. Let the quadrilateral $S A B C$ with its dia... | The problem is solved. Let's first perform the following additional constructions:
1) $K:|A K|=|K B|$
2) $[K E]$;
3) $M:|K M|=\frac{1}{3}|K E|$
4) $(O M); 5)[D M]$.
Fig. 61
Since in $\tr... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,576 | |
Example 57. At the base of the pyramid lies an equilateral triangle. One of the lateral faces is perpendicular to the base plane, while the other two are inclined to it at angles equal to $\alpha$. We need to find the angles formed by the lateral edges with the base plane.
Construction of the image. Let the quadrilate... | S o l u t i o n. To introduce the given and sought angles into the drawing, we will perform the necessary additional constructions. We will construct $[S D]$ - the height of $\triangle S A B$.
Then, since $(S A B) \perp(A B C)$, we have $(S D) \perp(A B C)$, and therefore $[A D]$ is the projection of the edge $S A$ on... | ((SA)\widehat{(ABC)})=\operatorname{arctg}(\frac{1}{2}\operatorname{tg}\alpha),((SB)\widehat{,}(ABC))=((SC),(ABC))=\operatorname{arctg}(\frac{\sqrt{3}}{2}\operatorname{tg}\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,577 |
Example 59. On the edges of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, points $P, Q$ and $R$ are given such that $|A P|=\frac{1}{3}\left|A A_{1}\right|,\left|B_{1} Q\right|=\frac{1}{2}\left|B_{1} C_{1}\right|$ and $|C R|=$ $=\frac{1}{3}|C D|$. Let's construct the section of the cube by the plane $P Q R$. | Solution. First, let's determine whether this problem is solvable. Suppose the figure $A B C D A_{1} B_{1} C_{1} D_{1}$ is an image of a cube (Fig. 65). This image is complete. It is also clear that, given points $P, Q$, and $R$ - projections of points $P_{0}, Q_{0}$, and $R_{0}$, we can find the secondary projections ... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,578 | |
Example 62. At the base of the pyramid $SABC$ lies a right-angled triangle $ABC$. The edge $SA$ is perpendicular to the base plane, $|SA|=|AB|=|BC|=a$. Through the midpoint of the edge $AB$, a plane is drawn perpendicular to the edge $SC$, and we need to find the area of the resulting section.
Construction of the imag... | The problem is solved. To calculate the desired area, we first determine the type of quadrilateral $M K L N$.
From the right triangles $A B P$ and $S A B$, we have respectively:
$$
M K=\frac{1}{2}|B P| \text { and }|M N|=\frac{1}{2}|A D| .
$$
But $|B P|=|A D|=\frac{a \sqrt{2}}{2}$. Thus, $|M K|=|M N|=$ $=\frac{a \sq... | \frac{^{2}\sqrt{3}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,580 |
Example 63. The base of a right prism, whose height is 1 cm, is a rhombus with sides equal to 2 cm and an acute angle of $30^{\circ}$. A cutting plane is drawn through the side of the base, the angle between which and the base plane is $60^{\circ}$. We need to find the area of the section.
Construction of the image. L... | Solution. It is clear that the quadrilateral $A Q P B$ is a parallelogram. Since $[M K] \perp [A B]$, then $[N K]$ is the height of this parallelogram, where $N=(M K) \cap (P Q)$. Thus, $S_{A Q P B} = |A B| \cdot |N K|$. Let's find $|N K|$.
In the right-angled $\triangle M D K$ we have: $|M K| = 2 \, \text{cm}, |M D| ... | \frac{4\sqrt{3}}{3}\,^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,581 |
Example 65. One edge of a triangular pyramid is equal to $a$, and each of the others is equal to $b$. Let's find the volume of the pyramid. | Solution. Without dwelling on the construction of the image (Fig. 74) and other stages of the solution, let us note that if we solve this example using the formula \( V = \frac{1}{3} S_{\text{base}} \cdot H \), then finding \( H \) would require quite complex calculations. We will take a different path. We will constru... | \frac{}{12}\sqrt{3b^2-^2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,582 |
Example 66. In the base of the right prism $A B C A_{1} B_{1} C_{1}$ lies a triangle, where $|A B|=|A C|$ and $\widehat{A B C}=\alpha$. It is also known that point $D$ is the midpoint of edge $A A_{1}, \widehat{D C A}=$ $=\beta$ and $|C D|=b$. Find the lateral surface area of the prism.
Construction of the image. Let ... | Solution. It is clear that this problem belongs to type $F$. Let's solve it by direct calculation. Since $A B C A_{1} B_{1} C_{1}$ is a right prism, then
$$
S_{\text {side }}=P \cdot H, \text { where } P=2|A C|+|B C|, \text { and } H=\left|A A_{1}\right| .
$$
Since $\triangle A C D$ is a right triangle, then $|A C|=b... | 4b^{2}\sin2\beta\cos^{2}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,583 |
Example 67. The base of the pyramid is an equilateral triangle with a side length of $a$. One of the lateral faces, which is perpendicular to the plane of the base, is also an equilateral triangle. Find the area of the lateral surface of the pyramid.
Construction of the image. Let the quadrilateral $S A B C$ with its ... | Solution. $S_{\text {side }}=S_{\triangle S A B}+$ $+S_{\triangle S A C}+S_{\triangle S B C}$. But $\triangle S A B \cong$ $\cong \triangle S A C$ (by three sides), i.e.
$$
S_{\text {side }}=2 S_{\triangle S A B}+S_{\triangle S B C}
$$
Since $S_{\triangle S A B}=\frac{1}{2}|A B| \times$ $\times|S K|$, we will first f... | \frac{^{2}}{4}(\sqrt{15}+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,584 |
Example 76. The side of the base of a regular triangular pyramid is equal to $a$; the height dropped from the vertex of the base to the opposite lateral face is equal to $b$. Let's find the volume of the pyramid.
Construction of the image. Let the quadrilateral $S A B C$ with its diagonals be the image of the given py... | Solution. Since $V=\frac{1}{3} S_{\triangle A B C} \cdot|S O|$, where $S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}$, it is sufficient to calculate the length of the height $S O$.
From the similarity of right triangles $S O D$ and $A D K$, we get:
$$
\frac{|S O|}{|A K|}=\frac{|O D|}{|D K|}
$$
from which $|S O|=\frac... | \frac{^{3}b}{12\sqrt{3^{2}-4b^{2}}},\text{where}\frac{\sqrt{2}}{2}\leqslantb<\frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,588 |
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