problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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345. A pair (of distinct) natural numbers $m$ and $n$ is called "good" if these numbers consist of the same prime factors, taken, however, in (in general) different powers (example: $90=2 \cdot 3^{2} \cdot 5$ and $150=2 \cdot 3 \cdot 5^{2}$), and "very good" if both the pair $m, n$ and the pair $m+1, n+1$ are "good" (e... | 345. It is clear that the pair of natural numbers $n$ and $n^{2}$ is "good" for any $n>1$. On the other hand, the pair of numbers $n-1$ and $n^{2}-1=(n-1)(n+1)$ is probably "good" if the number $n+1$ is a power of two, i.e., $n+1=2^{k}$, where $k \geqslant 1$ is an integer: indeed, in this case the numbers
$$
n-1 \tex... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,601 |
346. Let $a, a+d, a+2d, a+3d, \ldots$ be an arbitrary (infinite) arithmetic progression, where the initial term $a$ and the common difference $d$ are natural numbers. Prove that the progression contains infinitely many terms whose prime factorization consists of the same prime factors (but taken, of course, in differen... | 346. It is clear that the initial term $a$ and the difference $d$ of the progression can be considered coprime numbers: for if both $a$ and $d$ were divisible by some number $k>1$, we could simply divide all terms of the progression by $k$. If, however, $a$ and $d$ are coprime, then by Euler's theorem (problem 341), th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,602 |
347. Wilson's Theorem). Prove that if the integer $p$ is prime, then the number $(p-1)!+1$ is divisible by $p$; if, however, $p$ is composite, then $(p-1)!+1$ is not divisible by $p$. | 347. Let $a$ be any of the numbers in the sequence $2,3, \ldots, p-2$. Consider the numbers
$$
a, 2a, \ldots, (p-1)a
$$
No two of these numbers can give the same remainder when divided by $p$; therefore, these numbers give remainders $1,2, \ldots, p-1$ when divided by $p$, and each of these remainders occurs only onc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,603 |
348. Prove that
a) for each prime number $p$, there exist such integers $x$ and $y$ that $x^{2}+y^{2}+1$ is divisible by $p$;
b)* if a prime number $p$ gives a remainder of 1 when divided by 4 (and for odd prime numbers - only in this case), there exists such an integer $x$ that $x^{2}+1$ is divisible by $p$.
Proble... | 348. a) If $p=2$, then $p=1^{2}+0^{2}+1$. Let now the prime number $p$ be odd; we will show that it is possible to find two integers $x$ and $y$, both less than $\frac{p}{2}$, satisfying the condition of the problem.
Consider $\frac{p+1}{2}$ numbers $0,1,2, \ldots, \frac{p-1}{2}$. The squares of any two of these numbe... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,604 |
349. Prove that there are infinitely many prime numbers. | 349. The existence of an infinite number of prime numbers follows from the result of problem 234 (this result even shows that prime numbers occur in the sequence of all integers quite "frequently," for example, "more frequently" than squares; see the note to this problem). From the result of problem 90, it can also be ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,605 |
1. Prove that the sum of the angles of a spatial quadrilateral does not exceed $360^{\circ}$. | 1. Let $A B C D$ be an arbitrary spatial quadrilateral, $B D$ - its diagonal (Fig. 32). We will compare the sum of the angles of the quadrilateral $A B C D$ with the sum of the angles of triangles $A B D$ and $B C D$, which is $4 d$. Consider the trihedral angles $D A B C$ and $B A C D$ with vertices at $D$ and $B$ res... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,607 |
2. In space, two intersecting planes $\tau$ and $\sigma$ are given. On the line of their intersection, a point $A$ is taken. Prove that among all lines lying in plane $\tau$ and passing through point $A$, the one that forms the greatest angle with plane $\sigma$ is the one that is perpendicular to the line of intersect... | 2. The angle between a line and a plane and the angle between the line and the perpendicular to this plane together make up $90^{\circ}$. Therefore, among the lines lying in the plane
Fig. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,608 |
3. Prove that the area of the orthogonal projection of a planar polygon is equal to the area of the polygon multiplied by the cosine of the angle between the plane of the polygon and the plane of the projection. | 3. We will prove the required statement for a triangle. Let triangle $A B C$ be orthogonally projected onto plane $\alpha$ as triangle $A^{\prime} B^{\prime} C^{\prime}$, and let the plane of triangle $A B C$ intersect plane $\alpha$ along line $m$.
4 D. O. Shklyarsky et al.
(figure 34). Through the vertices of triang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,609 |
4. In space, there are two intersecting perpendicular lines. Find the geometric locus of the midpoints of segments of a given length $d$, the ends of which lie on these lines. | 4. Let's construct a plane $\sigma$, parallel to the given lines $l$ and $m$ and equidistant from them (Fig. 35). Consider the projections $l_{1}$ and $m_{1}$ of the given lines $l$ and $m$ onto the plane $\sigma$ and the projection $C D$ of one of the considered segments $A B$ of a given length $d$. We have: $A C = B ... | \frac{1}{2}\sqrt{^{2}-^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,610 |
5. Two segments slide along two intersecting lines (one along the first, and the other along the second line). Prove that the volume of the tetrahedron ${ }^{1}$) with vertices at the ends of these segments does not depend on their positions.
1) In this book, the word tetrahedron (from Greek, meaning four-faced) is use... | 5. Let segments $AB$ and $CD$ slide respectively along lines $l_{1}$ and $l_{2}$. Through $l_{2}$, draw a plane $\pi$ parallel to $l_{1}$ (Fig. 36).
Fig. 36. In the plane $\pi$, draw lines ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,611 |
6. A sphere is inscribed in an $n$-sided pyramid. Prove that if all the lateral faces of the pyramid are combined with the plane of the base by rotating them around the corresponding edges of the base (Fig. 1), then all points of tangency of these faces with the sphere will merge into one point $H$, and the vertices of... | 6. Let $O$ be the center of the sphere inscribed in the pyramid $S A B C D E$, $H$ and $P$ be the points of tangency of the sphere with the base of the pyramid and with one of its lateral faces $S A B$ (Fig. 37). In this case, $O H$ is perpendicular to the plane of the base, and therefore to the line $A B$; $O P$ is pe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,612 |
7. Given a regular pyramid. From point $N$ of its base (Fig. 2), a perpendicular to the plane of the base is erected. Prove that the sum of the segments from point $N$ to the points of intersection of this perpendicular with the planes of all lateral faces of the pyramid does not depend on the position of point $N$ in ... | 7. Let $N_{1}$ (Fig. 38) be the point of intersection of the perpendicular $N N_{1}$ to the base of the pyramid with its face $S A B$. Drop
Fig. 38. a perpendicular $N M$ from $N$ to the e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,613 |
8. a) Prove that if a certain number of lines have the property that any two of them intersect, then either all of them pass through one common point, or all lie in one plane.
b) Prove that if a certain number of circles have the property that any two of them intersect at two points, then either all the circles pass t... | 8. a) If all lines pass through one common point, the statement of the problem is satisfied. If this is not the case, then there will be three lines \( l_{1}, l_{2} \) and \( l_{3} \) that do not have a common point. Draw a plane \( \pi \) through the lines \( l_{1} \) and \( l_{2} \). Then the line \( l_{3} \) has two... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,614 |
10. Find the geometric locus of the centers of circles formed by the intersection of a given sphere $U$ with planes passing through:
a) a given line $a$;
b) a given point $H$. | 10. a) Let $O$ be the center of the sphere, $O_{\alpha}$ the center of the circle obtained by cutting the sphere $U$ with a plane $\alpha$ passing through a given line $a$ (Fig. 43). Draw $O H \perp a$ and connect $O$ with $O_{\alpha}$. By a known property of the sphere, $O O_{\alpha}$ is perpendicular to the plane $\a... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,616 |
11*. Prove that if all sections of a certain body by planes passing through a given fixed point $P$ are circles, then this body is a sphere.
Note. The statement of problem 11 is a strengthening of the following known theorem (see, for example, B. N. Delone and O. K. Zhitomirskii, "Problem Book in Geometry", problem 38... | 11. Let us use the blinding lemma. Suppose in space there are two intersecting circles $K$ and $K_{1}$ at two points, and let $P$ be a point on their common chord $A B$ (or on its extension). Then, if point $M$ does not lie in the plane of either circle, there exists a plane $\alpha$ passing through the line $M P$ and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,617 |
12. Among all orthogonal projections of a regular tetrahedron onto different planes, find the one that has the largest area. | 12. The projection of any polyhedron is a polygon, the vertices of which are the projections of the vertices of the polyhedron. Since a tetrahedron has four vertices, depending on the choice of the projection plane, its projection can be either a triangle or a quadrilateral. (It cannot be a segment or a point, as a reg... | \frac{^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,618 |
13. On the surface of a cube, find the points from which a given diagonal of the cube is visible at the smallest angle ${ }^{1}$ ). | 13. Let's describe a sphere around a cube (Fig. 48). Its diameter will be the given diagonal. From all points of the sphere, this diagonal
Fig. 48. is seen at an angle of $90^{\circ}$. Inde... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,619 |
14. Determine the type of section of a cube by a plane passing through its center and perpendicular to its diagonal. | 14. A cube is a convex hexahedron with a center of symmetry. Therefore, any section of it by a plane passing through its center is a convex, centrally symmetric polygon with no more than six sides. On the other hand, when the cube is rotated around its diagonal by \(120^\circ\), \(240^\circ\), and \(360^\circ\), it coi... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,620 |
16*. Prove that a through-hole can be cut in a cube through which a cube of the same size can be passed. | 16. Let's take the diagonal of the cube as the axis of the hole. If we project the cube onto a plane perpendicular to the diagonal, then due to the symmetry of the cube's edges, those edges of the cube that do not share endpoints with this diagonal will project into the sides of a regular hexagon (this hexagon will be ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,622 |
17. Are there polyhedra other than the cube, all faces of which are equal squares?
1) The ends of the main diagonal are not considered in this case. | 17. There exists an infinite number of non-convex polyhedra that satisfy the condition of the problem. An example is the polyhedron depicted in Fig. 55. However, no convex polyhedra other than the cube exist with equal square faces. Indeed, the number \( n \) of faces of a convex polyhedral angle with planar angles equ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,623 |
18. Place three cylindrical cylinders of height $a$ and diameter $\frac{a}{2}$ inside a cube with edge length $a$, so that they cannot move inside the cube. | 18. It is easy to prove that if the bases of a cylinder lie on the faces of a cube, then the direction of the cylinder's axis will remain unchanged during all its movements inside the cube.
Consider the possible movements of two cylinders placed in a cube as shown in Fig. 56.
During any movements, their axes will rem... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,624 |
20*. What figures can the central projection of a triangle be? (It is assumed that the center of projection does not lie in either the projection plane or the plane of the triangle.)
保留了源文本的换行和格式。 | 20. Let the plane of projections be denoted by the letter $\sigma$, the vertices of the projected triangle by the letters $A, B, C$, and the center of projections by the letter $S$. Consider the "complete" trihedral angle $S A B C$, consisting of two vertical trihedral angles, the edges of which are the lines $A_{1} A,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,626 |
21*. $n$ wire triangles are arranged in space such that:
$1^{\circ}$. Any two of them share one common vertex.
$2^{\circ}$. At each vertex, the same number $k$ of triangles meet.
Find all values of $k$ and $n$ for which the specified arrangement is possible. | 21. Let $k$ be the number of triangles converging at one vertex, and we will prove that $k \leqslant 3$. Indeed, suppose that $k>3$. We will number our triangles with Roman numerals, and their vertices with Arabic numerals. Consider one of the triangles, for example, triangle $I$; its vertices will be denoted by the nu... | k=1,n=1;k=2,n=4;k=3,n=7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29,627 |
22. A spatial quadrilateral is circumscribed around a sphere. Prove that the four points of tangency lie in the same plane. | 22. Let the quadrilateral $ABCD$ be circumscribed around a sphere, and let $K, L, M$, and $N$ be the points of tangency of the lines $AB, BC, CD$, and $DA$ respectively (Fig. 66a).
By the property of tangents to a sphere, we have
$$
AK = AN, BK = BL, CL = CM, DM = DN
$$
Draw a plane $\sigma$ through the points $K, L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,628 |
24. Can a plane intersect a trihedral angle in such a way that the section is a triangle equal to a given triangle? | 24. First solution. Let the section of a right trihedral angle $S$ by an arbitrary plane result in a triangle $A B C$, the sides of which are $B C=a, A C=b, A B=c$.
Fig. 71.
(Fig. 71). By ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,630 |
26. A sphere is drawn through some three vertices of the given tetrahedron $A B C D$, intersecting the edges emanating from the fourth vertex at points $M, N$, and $P$ (Fig. 4). Prove that the shape of triangle $M N P$ depends only on the tetrahedron $A B C D$, but not on the radius of the sphere or on which three vert... | 26. Consider two triangles $M N P$ and $M^{\prime} N^{\prime} P^{\prime}$ (Fig. 76), obtained as specified in the problem. Clearly, there will be two vertices belonging to the first and second triangles, respectively, which lie on one edge of the tetrahedron $A B C D$ ( $N$ and $N^{\prime}$ in Fig. 76). We will prove t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,632 |
27. Given two triangular pyramids $A B C D$ and $A^{\prime} B C D$ with a common base $B C D$, and the point $A^{\prime}$ is inside the pyramid $A B C D$.
a) Prove that the sum of the plane angles at vertex $A^{\prime}$ of the pyramid $A^{\prime} B C D$ is greater than the sum of the plane angles at vertex $A$ of the ... | 27. a) Let's first prove the required statement for the case when point $A^{\prime}$ is located on an edge of the tetrahedron $A B C D$.
Fig. 77.
Suppose, for example, that point $A^{\prim... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,633 |
28*. Prove that if all dihedral angles of a tetrahedron are acute, then all planar angles are also acute. | 28. First, let's prove an auxiliary statement: if all dihedral angles of a convex trihedral angle are acute, then all plane angles are also acute. (This statement can be proven without assuming the convexity of the trihedral angle in advance, but then the proof becomes more complicated.)
Let \( SABC \) be the given co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,634 |
29*. Prove that if all four faces of a tetrahedron have the same area, then they are necessarily equal to each other. | 29. Let \(ABCD\) (Fig. 80) be a tetrahedron satisfying the conditions of the problem. Draw a plane \(\pi\) through \(AB\) parallel to \(CD\). Let \(C'\) and \(D'\) be the projections of points \(C\) and \(D\) onto the plane \(\pi\), and let \(E\) be the point of intersection of segments \(AB\) and \(C'D'\).
Drop perpe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,635 |
30. a) Prove that four perpendiculars erected to the planes of the faces of a tetrahedron from the centers of the circles circumscribed around the faces intersect at one point (the center of the circumscribed sphere).
b) Derive from this that through two circles, not lying in the same plane and intersecting at two poi... | 30. a) Each of the considered perpendiculars is, obviously, the geometric locus of points equidistant from the vertices of the corresponding face. Let's draw a plane \(\pi\) through the midpoint of the segment \(A_{1} A_{4}\), perpendicular to this segment (Fig. 81). Clearly, this plane is not parallel to the perpendic... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,636 |
31. The bisecting plane of a dihedral angle is a plane that divides this angle in half. Prove that
a) The three bisecting planes of the dihedral angles of a trihedral angle intersect along a single line. This line is called the bisector of the trihedral angle.
b) The four bisectors of the trihedral angles of a tetrah... | 31. a) Let's prove that the bisector plane is the geometric locus of points lying inside a dihedral angle and equidistant from its faces.
Fig. 83.
Indeed, let's take an arbitrary point \(... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,637 |
32. a) A median of a tetrahedron is a line connecting a vertex of the tetrahedron to the point of intersection of the medians of the opposite face. Prove that the four medians of a tetrahedron intersect at one point (the center of gravity of the tetrahedron).
b) Prove that the three lines connecting the midpoints of op... | 32. a) Let's show that any two medians, for example, $A_{1} M_{1}$ and $A_{3} M_{3}$ of an arbitrary tetrahedron $A_{1} A_{2} A_{3} A_{4}$, intersect (Fig. $86, a$). Through the vertices $A_{1}$ and $A_{3}$ of the tetrahedron and through the point $D_{24}$ - the midpoint of the edge $A_{2} A_{4}$ - we draw the plane $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,638 |
33. Prove that if two altitudes of a tetrahedron intersect, then the other two altitudes also intersect. | 33. Let in the tetrahedron $A B C D$ the heights $D H_{4}$ and $B H_{2}$ intersect (Fig. 87). Then a plane $\alpha$ can be drawn through them, which will also pass through the edge $B D$. This plane is perpendicular to both planes $A B C$ and $A D C$, as it passes through the perpendiculars to them, $D H_{4}$ and $B H_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,639 |
34. A tetrahedron, the opposite edges of which are pairwise perpendicular, is called orthogonal.
Prove that
a) If two pairs of opposite edges of a tetrahedron are mutually perpendicular, then the third pair of edges is also mutually perpendicular, i.e., such a tetrahedron is orthogonal.
b) The four altitudes of an o... | 34. a) Let in the tetrahedron $ABCD$ (Fig. 88) $AB \perp CD$ and $AC \perp BD$. Drop a perpendicular $DH_4$ from vertex $D$ to the plane of triangle $ABC$. Then $BH_4$ is the projection of edge $BD$ onto the plane $ABC$, and $CH_4$ is the projection of edge $CD$ onto the same plane. Based on the theorem of three perpen... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,640 |
35. Prove that
a) Planes passing through each edge of a trihedral angle and perpendicular to the opposite face intersect along a single line (the altitude line of the trihedral angle).
b) If two altitude lines of a tetrahedron intersect, then the other two also intersect.
c) If three altitude lines of a tetrahedron ... | 35. a) Let in the trihedral angle $SABC$ through each edge a plane is drawn perpendicular to the opposite face. Let planes $\beta$ and $\gamma$, passing through edges $SB$ and $SC$ perpendicular to faces $SAC$ and $SAB$ respectively, intersect along line $SH$ (Fig. 89). Further, let $SC'$ be the intersection of plane $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,641 |
36. Prove that the sums of the squares of the lengths of pairs of opposite edges of an orthogonal tetrahedron are equal; conversely, if the sums of the squares of the lengths of pairs of opposite edges of a tetrahedron are equal, then the tetrahedron is orthogonal. | 36. Let $ABCD$ be a tetrahedron and let plane $\alpha$, passing through edge $DB$ and perpendicular to edge $CA$, intersect it at point $H$, and let plane $\beta$, passing through edge $AC$ and perpendicular to edge $BD$, intersect edge $BD$ at point $E$ (see Fig. 87). From triangles $ADH$ and $BCH$ we have
$$
\begin{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,642 |
37. Prove that
a) Planes passing through the bisectors of the dihedral angles of the corresponding faces of a trihedral angle, perpendicular to the faces, intersect along a single line (the bisector line of the trihedral angle).
b) If two bisector lines of a tetrahedron intersect, then the other two also intersect.
... | 37. a) Let $S A B C$ (Fig. 90) be a given trihedral angle. The plane $\alpha$, passing through the bisector $S a$ of the angle $B S C$ and perpendicular to the plane $S B C$, is the geometric locus of points equally distant from the rays $S B$ and $S C$ or their extensions ${ }^{1}$. Similarly, the plane $\beta$, passi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,643 |
38*. Prove that if the sum of two opposite edges of a tetrahedron is equal to the sum of its two other opposite edges, then the sums of the corresponding dihedral angles are also equal. | 38. Let in the tetrahedron $A_{1} A_{2} A_{3} A_{4}$ (Fig. 93)
$$
A_{1} A_{3}+A_{2} A_{4}=A_{1} A_{4}+A_{3} A_{2}
$$
Then the bisector lines emerging from vertices $A_{3}$ and $A_{4}$ intersect at some point $H$. Denote by $P, Q, R$ and $S$ the projections of point $H$ onto the faces $A_{2} A_{3} A_{4}$, $A_{1} A_{3}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,644 |
39. Prove that if the centers of the inscribed and circumscribed spheres of a tetrahedron coincide, then all faces are equal triangles. | 39. Let the center of the circumscribed sphere of the tetrahedron coincide with the center of the inscribed sphere. Then the planes of all faces are equidistant from this center. Therefore, all circles obtained by intersecting the circumscribed sphere with the planes of the faces are equal. In equal circles, equal chor... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,645 |
40. Prove that the center of the sphere inscribed in a tetrahedron lies inside the tetrahedron formed by the points of tangency.
## 2. THEORY OF POLYHEDRA
Below is a series of problems on the theory of polyhedra. The theory of polyhedra (mainly the theory of convex polyhedra) constitutes a large branch of mathematics... | 40. Let us first prove the following statement. If a sphere is inscribed in a trihedral angle $SABC$, and $A'$, $B'$, $C'$ are the points of tangency of this sphere with the corresponding faces of the trihedral angle, then the plane $A'B'C'$ is perpendicular to the line $SO$ and intersects it within the segment $SO$ (F... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,646 |
41. a) Prove that in any polyhedron, the number of faces with an odd number of sides is even (for example, a 19-faced polyhedron where all faces are triangles cannot exist).
b) Prove that in any polyhedron, the number of vertices from which an odd number of edges emanate is even. | 41. a) Let's calculate the number of sides of each face of our polyhedron, sum all the obtained numbers, and denote the resulting sum by $N$. Obviously, the number $N$ is even or odd depending on whether the number of faces with an odd number of sides is even or odd. On the other hand, the number $N$ is always even. In... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,647 |
42. a) A tetrahedron has six edges. Prove that a polyhedron with fewer edges does not exist.
b) Prove that a polyhedron with seven edges does not exist, but polyhedra with any greater number of edges do exist. | 42. a) Let $W$ be some polyhedron and $\mu$ be one of its faces. Since the smallest number of sides of a polygon is three, $\mu$ has at least three sides. Furthermore, since our polyhedron does not reduce to the single face $\mu$, there exists a vertex $A$ that lies outside the plane of this face. In each vertex of the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,648 |
43. Prove that there does not exist a 7-faced polyhedron, all of whose faces are quadrilaterals.[^1] | 43. First of all, note that if two faces of a polyhedron have two common edges or a common edge and a common vertex not belonging to this edge, then they lie in the same plane. We will not consider such polyhedra and exclude them from consideration ${ }^{1}$ ).
Now suppose there exists a heptahedron $W$, each face of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,649 |
44. Prove that in any convex polyhedron, there will be either a triangular face or a trihedral angle.
Note. A refinement of this result is given by the theorem of problem 51. | 44. Suppose there exists a polyhedron $W$ without any triangular faces or trihedral angles. Let $\mathscr{B}$ be the number of vertices of this polyhedron. Choose a point $O_{k}$ inside each face $\mu_{k}$ and connect it to all the vertices of the face $\mu_{k}$ (Fig. 98). Then all the faces will be divided into triang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,650 |
50. Prove that in any simply connected polyhedron, the following relations hold:
a) $\frac{3}{2} \leqslant \frac{\mathscr{T}}{\mathscr{G}}<3$;
b) $\frac{3}{2} \leqslant \frac{\mathscr{P}}{\mathscr{B}}<3$. | 50. a) Each face of a polyhedron contains at least three edges. Each edge belongs to two faces. Hence, it follows that
$$
\mathscr{P} \geqslant \frac{3}{2} \mathscr{F}
$$
Similarly, since each edge connects two vertices and each vertex belongs to at least three faces, we have \(\mathscr{P} \geqslant \frac{3}{2} \math... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,654 |
51. Prove that in any simply connected polyhedron, the sum of the number of triangular faces and the number of trihedral angles is not less than 8. | 51. We will denote by $\sigma_{3}$ the number of triangular faces of a polyhedron, by $\mathscr{I}_{4}$ - the number of quadrilateral faces, $\overbrace{5}$ - the number of pentagonal faces, and so on. Then the total number of faces will be $\mathscr{\sigma}=\mathscr{\sigma}_{3}+\mathscr{F}_{4}+\ldots$ Each triangular ... | \mathscr{B}_{3}+\mathscr{\sigma}_{3}\geqslant8 | Geometry | proof | Yes | Yes | olympiads | false | 29,655 |
52. Prove that a simply-connected polyhedron must have either triangular, quadrilateral, or pentagonal faces. In this case,
a) if the polyhedron has neither quadrilateral nor pentagonal faces, then it must have at least four triangular faces (for example, a regular tetrahedron has neither quadrilateral nor pentagonal ... | 52. Let the given polyhedron have $\sigma_{3}$ triangular, $\sigma_{4}$ quadrilateral, $\sigma_{5}$ pentagonal, and so on, faces.
In this case, the total number of $\mathscr{\sigma}$ faces is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,656 |
53. Prove that for any convex polyhedron with a sufficiently large number of faces, at least one face has no fewer than six neighbors. Find the minimum number $N$ such that for any convex polyhedron with a number of faces not less than $N$, this statement is true. | 53. Let a certain convex polyhedron $W$ have the number of neighbors of each face less than six, i.e., not exceeding five.
It is easy to see that in each vertex of such a polyhedron, no more than five edges (and thus faces) can meet. To show this, consider any face of the polyhedral angle. This face has at least three... | 12 | Geometry | proof | Yes | Yes | olympiads | false | 29,657 |
58**. a) Cauchy's Theorem. If two convex polyhedra are composed of the same corresponding equal faces, then either these polyhedra are equal, or they are mirror images (i.e., each is equal to a polyhedron that is symmetric to the other with respect to some plane).
b) A.D. Alexandrov's Theorem. If two convex polyhedra ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,662 | |
64. Find the value of the solid angle of a circular cone with an angle a at the vertex. | 64. Consider a circular cone with vertex $S$, axis $S O$ and angle $z$ at the vertex. Take any point $O$ on the axis $S O$ inside the cone and drop a perpendicular from $O$ to any of the generators (Fig. 125). All other perpendiculars to the generators can be obtained by rotating the constructed perpendicular around th... | 4\pi\sin^2\frac{}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,663 |
65. Theorem on the sum of the solid angles of a tetrahedron. Prove that the sum of the dihedral angles (in steradians) of a tetrahedron minus the sum of the solid angles of its trihedral angles equals \(4 \pi\). | 65. The solid angle of a trihedral angle is equal to the sum of its dihedral angles minus $\pi$.
Therefore, the sum of all solid angles of the trihedral angles of a tetrahedron is equal to twice the sum of all its dihedral angles minus $4 \pi$. However, since the solid angle of a dihedral angle is measured by twice th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,664 |
67. a) Prove that in a regular polyhedron, all faces, all edges, all plane angles, all dihedral angles, and all polyhedral angles are equal to each other.
b) Prove that a regular polyhedron can be superimposed on itself in such a way that any face \(\mu_{1}\) and any edge \(m_{1}\) belonging to this face coincide resp... | 67. a) According to the definition of a regular polyhedron, the edges belonging to any one face are equal to each other. Since each edge belongs to two faces, the edges of any face must be equal to the edges of all adjacent faces, and thus to the edges of those adjacent to these adjacent faces, and so on. Therefore, al... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,666 |
68. Prove that for any regular polyhedron:
a) there exists a sphere passing through all its vertices (circumscribed sphere, Fig. 18, a);
b) there exists a sphere tangent to all its faces (inscribed sphere, Fig. 18, b);
 First solution. With the help of Fig. 127, it can be easily proven that the perpendiculars to two adjacent faces
Fig. 127. $\alpha$ and $\beta$ of the regular polyhedron $T$, erected... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,667 |
69*. Prove that two regular polyhedra of the same combinatorial type, having faces and polyhedral angles of the same kind, are similar ${ }^{1}$ ). | 69. Let us establish some facts concerning regular polygons. As is known, a circle can be circumscribed around any regular polygon (the proof given in Kiselev's textbook for convex polygons remains valid for star polygons as well). In this case, as can be easily seen, the vertices of the polygon divide the circle into ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,668 |
71. Prove that the polyhedron, reciprocal to a regular one, is also regular.
Let the regular polyhedron $T_{1}$ be reciprocal to the polyhedron $T$. Then the regular polyhedron $r_{2}$, similar to the polyhedron $T_{1}$, is called the polyhedron dual to the polyhedron $T$. Dual polyhedra are a special case of polar po... | 71. Let $T^{\prime}$ be a polyhedron, dual to the regular polyhedron $T$. Let also $\alpha^{\prime}$ and $\beta^{\prime}$ be any faces of the polyhedron $T^{\prime}$, $l^{\prime}$ and $m^{\prime}$ be any sides of the faces $\alpha^{\prime}$ and $\beta^{\prime}$, and $A, B, l$ and $m$ be the vertices and edges of the po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,670 |
72. a) Prove that in any regular polyhedron, the centers of the faces of one polyhedral angle lie in the same plane.
b) Let $T$ be a regular polyhedron. Prove that the polyhedron, whose vertices are the centers of the faces of polyhedron $T$, edges are segments connecting the centers of adjacent faces of $T$, and face... | 72. a) The proof is conducted in the same way as the proof of problem 70 b): if \( O \) is the center of the polyhedron, then the centers \( O_{1}, O_{2}, \ldots, O_{n} \) of the faces adjacent to vertex \( A \) lie in a plane perpendicular to \( O A \) and at a distance from \( O \) equal to \( r \cdot \cos \alpha \),... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,671 |
73. Let $T_{1}$ and $T_{2}$ be two dual regular polyhedra. Prove that if the spheres inscribed in the polyhedra $T_{1}$ and $T_{2}$ are equal, then the circumscribed spheres are also equal. | 73. Let $R_{1}, r_{1}, p_{1}$ be the radii of the circumscribed, inscribed, and semi-inscribed spheres of the polyhedron $T_{1}$. Let, further, $T^{\prime}$ be the reciprocal polyhedron of $T_{1}$, and $R^{\prime}, r^{\prime}$ be the radii of the circumscribed and inscribed spheres of the polyhedron $T^{\prime}$. Then
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,672 |
76. a) Prove that there is only one type of regular tetrahedron, namely, the regular convex tetrahedron.
b) Prove that there is only one type of regular hexahedron, namely, the regular convex hexahedron, or cube.
c) Prove that there is only one type of regular octahedron, namely, the regular convex octahedron. | 76. a) Any two faces of a convex tetrahedron are adjacent. The edges belonging to one face form a single polygon - the triangle of that face. Therefore, it follows that a convex tetrahedron cannot be the kernel of any other polyhedron.
b) A cube also cannot be the kernel of any other polyhedron. Indeed, the only face ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,675 |
78. Let the number of faces, edges, and vertices of one of two mutually regular polyhedra be $\mathscr{G}, \mathscr{P}$, and $\mathscr{B}$, respectively, and let $n$ edges meet at each vertex of this polyhedron. Find the number of faces, edges, and vertices of the corresponding quasi-regular polyhedron. | 78. Let each face of the polyhedron $T$ have $k$ sides. Since at each vertex of the polyhedron $T$ there are $n$ edges converging, each face of the polyhedron $T^{\prime}$, reciprocal to $T$, has $n$ sides, and the number of these faces is $\mathscr{B}$. By the remarks made on page 35, the almost regular polyhedron obt... | \mathscr{B}+\math | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,677 |
81. From which regular polygons of the same type can parquet be made
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instruction and should not be part of the translation. Here is the correct translation:
81. From which regular polygons of the same type can parquet be made | 81. Let the pavement be composed of regular $n$-gons. Such a pavement can have nodes of two kinds: a) in the node lie
a)
 with triangles equal to an arbitrarily given triangle;
b) with quadrilaterals equal to an arbitrarily given quadrilateral;
c) with hexagons equal to an arbitrarily given centrally symmetric hexagon. | 82. a) From two equal triangles, one can form a parallelogram, and it is easy to cover the plane with parallelograms (Fig. 149).
b) Let a quadrilateral \(ABCD\) be given. Rotate it by \(180^\circ\) around the midpoint \(O\) of side \(CD\). We obtain a centrally symmetric hexagon \(ABCA'B'D\) (Fig. 150a), consisting of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,681 |
84. Prove that any triangle can be cut into four parts and that these parts can be used to form two triangles similar to the original one. | 84. Let's cut the given triangle $ABC$ along the line $DE$, parallel to the base $AC$ and such that $AD: DB=1: 4$ (see Fig. 153, a). Triangles $DBE$ and $ABC$ are obviously similar.
, all the angles of the polygon will be divided into parts, which will
. Now it is enough to transform the resulting rectangle into a square. Since the area of the desired square is obviously equal to the area of the rectangle, one side of the rectangle is smaller than the side of the square, and the o... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,688 |
92. Cut a rectangular parallelepiped with side ratios $8: 8: 27$ into four parts, from which a cube can be assembled. | 92. The given parallelepiped is cut into two equal stepped parts, as shown in Fig. 165, $a$ (the height of the step is 9, and the width is 4). From these parts, we form a new
Fig. 165.
par... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,691 |
95. Prove that two polygons, each equal in composition to a third, are equal in composition to each other. | 95. Let two polygons $P$ and $Q$ be equicomposable with a third polygon $R$ separately. In polygon $R$, we draw all the cuts that divide it into parts from which $Q$ is composed (Fig. 168), and all the cuts that divide it into parts from which $P$ is composed (Fig. 169). All these cuts together will divide polygon $R$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,694 |
96. a) Prove that any two equal-area rectangles are equicompositional.
b) Prove that any two equal-area triangles are equicompositional. | 96. a) Let rectangles $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$ be equal in area, and let, for example, $A B > A_{1} B_{1}, \quad B C > 2 A_{1} B_{1}$. Then, we can obviously construct a chain of equal-area rectangles $A_{1} B_{1} C_{1} D_{1}, A_{2} B_{2} C_{2} D_{2}, A_{3} B_{3} C_{3} D_{3}, \ldots, A_{k} B_{k} C_{k} D_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,695 |
97*. Bolyai-Gerwien Theorem. Prove that any two equal-area polygons are equidecomposable.
When proving the equality of areas of two polygons in an elementary geometry course, two methods are used: either both polygons are divided into smaller, respectively equal polygons (for example, when proving that a triangle is e... | 97. First, let us show that any polygon \( M \) is equicomposable with some rectangle. For this, we will cut the polygon \( M \) along all possible lines on which the sides of the polygon lie; in this case, the polygon \( M \) will break down into a series of convex polygons (if the polygon \( M \) is convex, this step... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,696 |
101. Prove that if a convex polygon $P$ can be composed of a finite number of non-overlapping rectangles $P_{1}, P_{2}, \ldots, P_{n}$, then $P$ is a rectangle. | 101. It is not hard to see that if some convex polygon $P$ is covered by rectangles without gaps or double coverings, then the sides of all these rectangles are mutually parallel. Indeed, let us choose any side $l$ of the polygon $P$. Obviously, all rectangles adjacent to $l$ have a side directed along $l$; all rectang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,700 |
103*. a) Show that there exist nine pairwise distinct squares from which a certain rectangle can be formed.
b) From 10 pairwise distinct squares, form a rectangle with sides 47 and 65.
c) Show that for any number $n$, greater than eight, it is always possible to find $n$ pairwise distinct squares from which a rectang... | 103. a) The configuration of 8 squares from drawing $184, a$ turned out to be impossible, as the inequality $a_{7}a_{5}$, and if we added square $Q_{\text {t }}$ as shown in drawing 188, no contradiction would arise.
Now let's show that for a certain specific size
 (Fig. 215) be cut into parallelograms. Consider the side \( A_{1} B_{1} \) of one of the parallelograms in the partition, adjacent to the side \( A B \) of the polygon. The parallelogram to which this side belongs has a side \( A_{1}^{\prime} B_{1}^{\prime} \) parallel to \( A_{1} B_{1} \)... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,708 |
113. a) Prove that any convex centrally symmetric polygon can be cut into parallelograms.
b) What is the smallest possible number of parallelograms into which a centrally symmetric convex $2k$-gon can be divided?
Note. Obviously, the number of sides of any centrally symmetric polygon is even, i.e., has the form $2k$ ... | 113. a) Let \( A_{1} A_{2} \ldots A_{2 k} \) be a centrally symmetric convex polygon. (The number of sides of a centrally symmetric polygon is necessarily even, as all its sides are divided into pairs of symmetric, equal, and parallel sides, — see Fig. 216.)
}{2} | Geometry | proof | Yes | Yes | olympiads | false | 29,709 |
114. Prove that if for each side of a convex polygon there can be found an equal and parallel side, then the polygon has a center of symmetry.
Is this statement true for non-convex polygons? | 114. First of all, let's prove that if in a convex polygon \(A_{1} A_{2} \ldots A_{n}\) for each side \(l_{i}\) there is a parallel and equal side \(l_{k}\), and if \(l_{i-1}\) and \(l_{i+1}\) are the sides adjacent to the side \(l_{i}=A_{i-1} A_{i}\), then the sides \(\bar{l}_{i-1}\) and \(\bar{l}_{i+1}\) parallel to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,710 |
115. Prove that if a convex polygon can be divided into smaller polygons, each of which has a center of symmetry, then the polygon itself has a center of symmetry.
Does this statement remain true for non-convex polygons?
Note. From the theorems of problems 113 a) and 115, it follows in particular that a convex polygo... | 115. Consider a series of polygons adjacent along some side $AB$ of the polygon $M$ under consideration (Fig. 220, a). It is clear that for each segment of this side, which is divided by the vertices of the centrally symmetric polygons of the partition, there will be an equal and parallel segment $a^{\prime}$ - the opp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,711 |
116**. Theorem of A. D. Alexandrov. If in a convex polyhedron all faces have centers of symmetry, then the polyhedron itself has a center of symmetry.
Does this theorem remain valid for non-convex polyhedra? | 116. Let us take an arbitrary edge $r_{1}$ of some face $L_{1}$ of the considered polyhedron; since the face $L_{1}$ is centrally symmetric, it has an edge $r_{2}$, equal and parallel to the edge $r_{1}$. Let $L_{2}$ be the face that is adjacent to the edge $r_{2}$ on the other side; then in the face $L_{2}$ there will... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,712 |
117*. Prove that every convex polyhedron with centrally symmetric faces can be divided into parallelepipeds
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 117. Consider some closed "zone" of edges (see the solution to the previous problem). All vertices of the polyhedron are divided into three groups, entering this zone and belonging to one or the other "cap." From all the vertices of one cap, we draw segments inside the polyhedron, equal and parallel to the edges of the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,713 |
118. Prove that if a convex polyhedron can be divided into smaller polyhedra with centrally symmetric faces, then all faces of the larger polyhedron also have centers of symmetry.
Does this statement remain true for non-convex polyhedra?
Note. From the theorems of problems 117 and 118, it follows in particular that a... | 118. Suppose that the polyhedron $M$ is divided into a series of polyhedra with centrally symmetric faces. In this case, each face of the polyhedron $M$ will be divided into a series of centrally symmetric polygons - the faces of the polyhedra of the partition.
1) This operation can be imagined as a parallel translatio... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,714 |
7. Find the distance between the intersecting diagonals of two adjacent faces of a cube with edge $a$. In what ratio does the common perpendicular to these diagonals divide each of them? | 7. a \sqrt{\frac{1}{3}}, \frac{1}{2}. | \sqrt{\frac{1}{3}},\frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,718 |
8. Prove that the area of the projection of a polygon located in the plane $\alpha$ onto the plane $\beta$ is equal to $S \cos \varphi$, where $S$ is the area of the polygon, and $\varphi$ is the angle between the planes $\alpha$ and $\beta$. | 8. The statement of the problem is obvious for a triangle, one side of which lies on the line of intersection of planes $\alpha$ and $\beta$. Then, it can be proven to be valid for an arbitrary triangle, and subsequently for an arbitrary polygon. | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,719 |
12. Prove that for the volume of an arbitrary tetrahedron \( V \), the formula \( V = \frac{1}{6} a b d \sin \varphi \) holds, where \( a \) and \( b \) are two opposite edges of the tetrahedron, \( d \) is the distance between them, and \( \varphi \) is the angle between them. | 12. Consider a parallelepiped formed by planes passing through the edges of a tetrahedron and parallel to the opposite edges. (This method of extending a tetrahedron to a parallelepiped will be repeatedly used in the future.) The volume of the tetrahedron is $1 / 3$ of the volume of the parallelepiped (the planes of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,722 |
13. Prove that the plane, bisecting a dihedral angle at some edge of a tetrahedron, divides the opposite edge into parts proportional to the areas of the faces enclosing this angle. | 13. It is easy to see that each of these ratios (of the areas of the faces and the segments of the edge) is equal to the ratio of the volumes of the two tetrahedra into which the bisector plane divides the given tetrahedron. | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,723 |
14. Prove that for the volume $V$ of a polyhedron circumscribed about a sphere of radius $R$, the equality $V=\frac{1}{3} S_{n} R$ holds, where $S_{n}$ is the total surface area of the polyhedron. | 14. By connecting the center of the sphere with the vertices of the polyhedron, we can divide it into pyramids, the bases of which are the faces of the polyhedron, and the heights are equal to the radius of the sphere. | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,724 |
15. Given a convex polyhedron, all vertices of which are located in two parallel planes. Prove that its volume can be calculated using the formula
$$
V=\frac{h}{6}\left(S_{1}+S_{2}+4 S\right)
$$
where $S_{1}$ is the area of the face located in one plane, $S_{2}$ is the area of the face located in the other plane, $S$... | 15. It is easy to verify the validity of the given formula for a tetrahedron. For this, we need to consider two cases: 1) three vertices of the tetrahedron are located in one plane and one vertex - in another; 2) two vertices of the tetrahedron are located in one plane, and two - in another. In the second case, for the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,725 |
17. Prove that the area of the part of the sphere's surface enclosed between two parallel planes, not intersecting the sphere, can be found using the formula
\[
S=2 \pi R h
\]
where \( R \) is the radius of the sphere, and \( h \) is the distance between the planes. | 17. First, let's prove the following auxiliary statement. Let segment \(AB\) rotate around line \(l\) (line \(l\) does not intersect segment \(AB\)). The perpendicular erected to \(AB\) at point \(C\), the midpoint of \(AB\), intersects line \(l\) at point \(O\), and \(MN\) is the projection of \(AB\) onto line \(l\). ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,727 |
18. Prove that the volume of the body obtained by rotating a circular segment around a diameter not intersecting it can be calculated using the formula
$$
V=\frac{1}{6} \pi a^{2} h
$$
where $a$ is the length of the chord of this segment, and $h$ is the projection of this chord onto the diameter. | 18. Let $AB$ be a chord of a given segment, $O$ the center of the circle. Denote by $x$ the distance from $O$ to $AB$, and by $R$ the radius of the circle. Then the volume of the solid obtained by rotating the sector $AOB$ around the diameter is equal to the product of the surface area obtained by rotating the arc $\ov... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,728 |
19. Prove that the segments connecting the vertices of a tetrahedron with the points of intersection of the medians of the opposite faces intersect at one point (the centroid of the tetrahedron) and are divided in this point in the ratio $3: 1$ (counting from the vertices).
Prove also that in this same point intersect... | 19. By placing equal weights at the vertices of a pyramid, to find the center of gravity of the system, one can first find the center of gravity of three weights, and then, by placing a tripled weight at the found point, find the center of gravity of the entire system. Alternatively, one can first find the center of gr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,729 |
21. Prove that the sum of the squares of the lengths of the edges of a tetrahedron is four times greater than the sum of the squares of the distances between the midpoints of its skew edges. | 21. Let's pass through each edge of the tetrahedron a plane parallel to the opposite edge (see the solution to problem 12). These planes form a parallelepiped, the edges of which are equal to the distances between the midpoints of the skew edges of the tetrahedron, and the edges of the tetrahedron are the diagonals of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,730 |
22. Given a cube $\left.A B C D A_{1} B_{1} C_{1} D_{1} *\right)$ with edge $a$, $K$ is the midpoint of edge $D D_{1}$. Find the angle and the distance between the lines $C K$ and $A_{1} D$. | 22. If $M$ is the midpoint of $B B_{1}$, then $A_{1} M \| C K$. Therefore, the required angle is equal to the angle $M A_{1} D$. On the other hand, the plane $A_{1} D M$ is parallel to $C K$, so the distance between $C K$ and $A_{1} D$ is equal to the distance from point $K$ to the plane $A_{1} D M$. Let the required d... | \arccos\frac{1}{\sqrt{10}},\frac{}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,731 |
23. Find the angle and the distance between two intersecting medians of two lateral faces of a regular tetrahedron with edge $a$. | 23. The problem can be solved in the same way as problem 22. Let's propose another method for determining the distance between skew medians. Let \(ABCD\) be the given tetrahedron, \(K\) the midpoint of \(AB\), and \(M\) the midpoint of \(AC\). Project the tetrahedron onto a plane passing through \(AB\) and perpendicula... | \arccos\frac{1}{6},\sqrt{\frac{2}{35}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,732 |
25. The base of the pyramid is an equilateral triangle with side $a$, and the lateral edges have length $b$. Find the radius of the sphere that is tangent to all the edges of the pyramid or their extensions. | 25. $\frac{(2 b \pm a) a}{2 \sqrt{3 b^{2}-a^{2}}}, | \frac{(2)}{2\sqrt{3b^{2}-^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,734 |
26. A sphere passes through the vertices of one face of a cube and touches the sides of the opposite face of the cube. Find the ratio of the volumes of the sphere and the cube. | 26. \frac{41 \pi \sqrt{41}}{384}, | \frac{41\pi\sqrt{41}}{384} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,735 |
27. The edge of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ is $a$. Find the radius of the sphere passing through the midpoints of the edges $A A_{1}, B B_{1}$ and through the vertices $A$ and $C_{1}$.[^0] | 27. a \sqrt{\frac{7}{8}}$. | \sqrt{\frac{7}{8}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,736 |
29. A regular triangular prism with a base side of $a$ is inscribed in a sphere of radius $R$. Find the area of the section of the prism by a plane passing through the center of the sphere and a side of the base of the prism. | 29. $\frac{2 a}{3} \sqrt{4 R^{2}-a^{2}}.
.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
29. $\frac{2 a}{3} \sqrt{4 R^{2}-a^{2}}. | \frac{2}{3}\sqrt{4R^{2}-^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,738 |
30. Two balls of one radius and two of another are arranged so that each ball touches three others and a given plane. Find the ratio of the radius of the larger ball to the smaller one. | 30. 2+\sqrt{3}. | 2+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,739 |
31. Given a regular tetrahedron \(ABCD\) with edge length \(a\). Find the radius of the sphere passing through vertices \(C\) and \(D\) and the midpoints of edges \(AB\) and \(AC\). | 31. $\frac{a \sqrt{22}}{8}$, | \frac{\sqrt{22}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,740 |
32. One face of the cube lies in the plane of the base of a regular triangular pyramid, two vertices of the cube lie on one of the lateral faces of the pyramid, and one vertex on each of the other two. Find the edge of the cube, if the side of the base of the pyramid is $a$, and the height of the pyramid is $h$. | 32. $\frac{3 a h}{3 a+h(3+2 \sqrt{3})}$. | \frac{3}{3+(3+2\sqrt{3})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,741 |
33. The dihedral angle at the base of a regular $n$-sided pyramid is $\alpha$. Find the dihedral angle between adjacent lateral faces. | 33. $2 \arccos \left(\sin \alpha \sin \frac{\pi}{n}\right)$. | 2\arccos(\sin\alpha\sin\frac{\pi}{n}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,742 |
34. In a triangular prism $A B C A_{1} B_{1} C_{1}{ }^{*}$, two planes are drawn: one passes through the vertices $A, B$ and $C_{1}$, and the other through the vertices $A_{1}, B_{1}$ and $C$. These planes divide the prism into four parts. The volume of the smallest of these parts is $V$. Find the volume of the prism. | 34. $12 V$.
The above text has been translated into English, maintaining the original text's line breaks and format. | 12V | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,743 |
35. Through a point located at a distance $a$ from the center of a sphere of radius $R(R>a)$, three mutually perpendicular chords are drawn. Find the sum of the squares of the segments of the chords, into which the given point divides them. | 35. $6 R^{2}-2 a^{2}$. | 6R^{2}-2^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,744 |
37. The side of the base of a regular quadrilateral pyramid is equal to the apothem of the lateral face. A section is made through the side of the base, dividing the surface of the pyramid in half. Find the angle between the plane of the section and the plane of the base of the pyramid. | 37. $\operatorname{arctg}(2-\sqrt{\overline{3}})$; | \operatorname{arctg}(2-\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,746 |
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