problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
284. In space, there is a triangle $A B C$. Find the geometric locus of points $M$ in space such that the line connecting the center of the sphere circumscribed around $A B C M$ with the point $G$ - the centroid of the tetrahedron $A B C M$, is perpendicular to the plane $A M G$. | 284. If $K$ and $L$ are the midpoints of $B C$ and $A M$, and $O$ is the center of the sphere circumscribed around $A B C M$, then, since $G$ is the midpoint of $L K$ and $O G$ is perpendicular to $L K$, $|O L|=|O K|$. From this, it follows that $|A M|=|B C|$, i.e., $M$ lies on the surface of the sphere with center at ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,978 |
285. A circle of constant radius moves, touching the edges of a trihedral angle, all plane angles of which are right angles. Find the geometric locus of the centers of these circles. | 285. Introduce a rectangular coordinate system, taking the origin $O$ at the vertex of the trihedral angle, and directing the axes along the edges of this angle. Let the plane of the circle form angles $\alpha, \beta$, and $\gamma$ with the coordinate planes $X O Y, Y O Z$, and $Z O X$. Then the point $O_{1}$ - the cen... | |OO_{1}|=R\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,979 |
286. A spider is sitting at one of the vertices of a cube with an edge length of 1 cm. The spider crawls along the surface of the cube at a speed of 1 cm per second. Find the geometric locus of points on the surface of the cube that it can reach in 2 seconds. | 286. Let a spider be located at vertex $A$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Consider the triangle $D C C_{1}$. It is not difficult to prove that the shortest path connecting $A$ with any point inside the triangle $D C C_{1}$ intersects the edge $D C$. In this case, if the faces $A B C D$ and $D C C_{1} D_... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,980 |
287. Given a trihedral angle, all plane angles of which are right angles, $O$ - the vertex of this angle. Consider all possible broken lines of length $a$, starting at point $O$ and such that any plane parallel to one of the faces of the angle intersects this broken line in no more than one point. Find the geometric lo... | 287. Let the edges of a trihedral angle be the coordinate axes. Let $(x, y, z)$ be the coordinates of the vector $\bar{O} \vec{A},\left(x_{i}, y_{i}, z_{i}\right)$ be the coordinates of the $i$-th side of the broken line. Each side of the broken line is considered as a vector. Then
$$
x=\Sigma x_{i}, y=\Sigma y_{i}, z... | |x|+|y|+|z|=|OA|\leqslant | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,981 |
288. Given a sphere with center $O$. Let $A B C D$ be a pyramid circumscribed around it, for which the inequalities $|O A| \geqslant|O B| \geqslant|O C| \geqslant|O D|$ hold. Find the geometric loci of points $A, B, C$, and $D$. | 288. First, note that if $r$ is the radius of the sphere inscribed in $ABCD$, then, firstly, all edges of the tetrahedron $ABCD$ are greater than $2r$, and, secondly, the radius of the circle inscribed in any face of the tetrahedron is greater than $r$. The first statement is obvious. To prove the second, we draw a pla... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,982 |
290. On the surface of the Earth, all possible points are taken, the geographical latitude of which is equal to their longitude. Find the geometric locus of the projections of these points onto the equatorial plane. | 290. Let $O$ be the center of the Earth, $A$ be the point on the equator corresponding to the zero meridian, $M$ be a point on the surface of the Earth with longitude and latitude equal to $\varphi$, and $N$ be the projection of $M$ onto the equatorial plane. By introducing a rectangular coordinate system in the equato... | (x-\frac{R}{2})^2+y^2=\frac{R^2}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,984 |
291. Given a right circular cone and a point $A$ outside it, located at a distance from the plane of its base equal to the height of the cone. Let $M$ be a point on the surface of the cone such that a light ray coming from $A$ to $M$, reflecting off the surface of the cone, will be parallel to the plane of the base. Fi... | 291. Let us introduce the following notations: $\quad S$ - the vertex of the cone, $N$ - the projection of point $M$ onto the plane passing through points $S$ and $A$ parallel to the base of the cone, $P$ - a point on the line $S N$ such that $\widehat{S M P}=90^{\circ}$ (Fig. 56), $MP$ is the normal to the surface of ... | \sin2\alpha\cos\varphi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,985 |
292. Through a fixed point $P$ inside a sphere, three mutually perpendicular rays are drawn arbitrarily, intersecting the surface of the sphere at points $A, B$, and $C$. Prove that the point of intersection of the medians of triangle $ABC$ and the projection of point $P$ onto the plane $ABC$ describe the same spherica... | 292. To solve the problem, we need the following statements from planimetry. If through a point $P$, located at a distance $d$ from the center of a circle of radius $R$, two mutually perpendicular chords $A D$ and $B E$ are drawn, then
a) $|A D|^{2}+|B E|^{2}=8 R^{2}-4 d^{2}$,
b) the perpendicular dropped from $P$ to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,986 |
293. Given a trihedral angle with vertex $O$ and a point $N$. An arbitrary sphere passes through $O$ and $N$ and intersects the edges of the trihedral angle at points $A, B$, and $C$. Find the geometric locus of the centroids of triangles $A B C$.
Prove the same for a tetrahedron. | 293. Let $\boldsymbol{a}, \boldsymbol{b}$ and $\boldsymbol{c}$ be unit vectors directed along the edges of a trihedral angle. Let, further, $\overrightarrow{O N}=\boldsymbol{e}, P$ be the center of the sphere, $\overrightarrow{O P}=\boldsymbol{u}, \overrightarrow{O A}=x \boldsymbol{a}, \overrightarrow{O B}=y \boldsymbo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,987 |
294. Given an arbitrary tetrahedron and a point $N$. Prove that six planes, each of which passes through one edge of the tetrahedron and is parallel to the line connecting $N$ with the midpoint of the opposite edge, intersect at one point. | 294. Prove that each of these planes passes through the point symmetric to point $N$ with respect to the centroid of the tetrahedron. | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,988 |
296. Prove that if the Monge point (see problem 295) lies in the plane of any face of the tetrahedron, then the foot of the perpendicular dropped onto this face is located on the circumcircle of the face. | 296. In solving problem 295, we proved that the Monge point is symmetric to the center of the circumscribed sphere of the tetrahedron relative to the centroid of the tetrahedron. Therefore, if the Monge point lies in the plane of any face of the tetrahedron, then the center of the circumscribed sphere is removed from t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,990 |
297. Prove that the sum of the squares of the distances from an arbitrary point in space to the vertices of a tetrahedron is equal to the sum of the squares of the distances between the midpoints of opposite edges and four times the square of the distance from this point to the centroid of the tetrahedron. | 297. Use the equality
$$
|M A|^{2}+|M B|^{2}=\frac{4|M D|^{2}+|A B|^{2}}{2}
$$
where $D$ is the midpoint of $A B$, and the fact that in any tetrahedron, the sum of the squares of its opposite edges is equal to twice the sum of the squares of the distances between the midpoints of two pairs of its other edges (see pro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,991 |
298. Prove that for any tetrahedron, there will be no fewer than five and no more than eight spheres, each of which touches the planes of all its faces. | 298. Let the areas of the faces of a tetrahedron be denoted by $S_{1}, S_{2}, S_{3}, S_{4}$; $V$ - the volume of the tetrahedron. If $r$ is the radius of the sphere that touches all the planes forming the tetrahedron, then with the appropriate choice of signs $\varepsilon_{i}= \pm 1, i=1,2,3,4$, the equality $\left.\va... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,992 |
299. $A B C D$ is a spatial quadrilateral (points $A, B, C$ and $D$ do not lie in the same plane). Prove that there exist at least eight spheres that touch the lines $A B, B C, C D$ and $D A$. Also prove that if the sum of any two sides of the quadrilateral equals the sum of the other two sides, then there are infinite... | 299. For any two adjacent sides of a quadrilateral, there exist two planes equidistant from them (the bisector planes of the angle of the quadrilateral and the adjacent angle). In this case, if three such planes corresponding to three vertices of the quadrilateral intersect at some point, then through this point passes... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,993 |
300. Prove that the product of the lengths of two opposite edges of a tetrahedron, divided by the product of the sines of the dihedral angles of the tetrahedron corresponding to these edges, is constant for a given tetrahedron (the sine theorem). | 300. Using the formula for the volume of a tetrahedron from problem 11, we will prove that each of the considered ratios is equal to $\frac{4 S_{1} S_{2} S_{3} S_{4}}{9 V^{2}}$, where $S_{\mathrm{i}}, S_{2}, S_{3}, S_{4}$ are the areas of the faces of the tetrahedron, and $V$ is its volume. | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,994 |
301. Let $S_{i}, R_{i}, l_{i} (i=1,2,3,4)$ be the areas of the faces, the radii of the circles circumscribed around these faces, and the distances from the centers of these circles to the opposite vertices of the tetrahedron, respectively. Prove that the volume of the tetrahedron is given by the formula
$$
V=\frac{1}{... | 301. If $h_{i}(i=1,2,3,4)$ is the height of the corresponding face of the tetrahedron, then
$$
\begin{array}{r}
\frac{1}{3} \sqrt{\frac{1}{2} \sum_{i=1}^{4} S_{i}^{2}\left(l_{i}^{3}-R_{i}^{2}\right)}=\frac{1}{3} \sqrt{\frac{1}{2} \sum_{i=1}^{4} S_{i}^{2} h_{i}^{2} \frac{l_{i}^{2}-P_{i}^{2}}{h_{i}^{2}}}= \\
=V \sqrt{\f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,995 |
302. Given an arbitrary tetrahedron. Prove that there exists a triangle, the lengths of whose sides are numerically equal to the products of the lengths of the opposite edges of the tetrahedron. Let \( S \) be the area of this triangle, \( V \) the volume of the tetrahedron, and \( R \) the radius of the circumscribed ... | 302. Let the lengths of the edges of the tetrahedron \(ABCD\) be denoted as shown in Fig. 58, a. We draw a plane through vertex \(A\) that is tangent to the sphere circumscribed around the tetrahedron \(ABCD\). The tetrahedron \(AB C_1 D_1\) in the figure is formed by this tangent plane, the planes \(ABC\), \(ABD\), an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,996 |
303. Let $a$ and $b$ be the lengths of two skew edges of a tetrahedron, $\alpha$ and $\beta$ be the measures of the corresponding dihedral angles: Prove that the expression
$$
a^{2}+b^{2}+2 a b \operatorname{ctg} \alpha \operatorname{ctg} \beta
$$
is independent of the choice of edges. (Bretschneider's theorem.)
Equ... | 303. Let $S_{1}$ and $S_{2}$ be the areas of the faces sharing a common edge $a$, and $S_{3}$ and $S_{4}$ be the areas of the two remaining faces. Let $a, m$, and $n$ be the lengths of the edges forming the face $S_{1}$, and $\alpha, \gamma$, and $\delta$ be the dihedral angles adjacent to them, and $V$ be the volume o... | ^{2}+b^{2}+2\operatorname{ctg}\alpha\operatorname{ctg}\beta=\frac{1}{9V^{2}}(2Q-T) | Geometry | proof | Yes | Yes | olympiads | false | 29,997 |
305. Prove that the sum of the cosines of the dihedral angles of a tetrahedron is positive and does not exceed 2, and that the equality of this sum to 2 holds for equifacial tetrahedra and only for them. | 305. Let $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ and $\boldsymbol{d}$ be vectors perpendicular to the faces of a tetrahedron, directed outward and equal in length to the areas of the corresponding faces, and let $e_{a}, e_{b}, e_{c}, e_{d}$ be unit vectors having the same directions as $\boldsymbol{a}, \boldsy... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,999 |
306. The sum of the planar angles of a trihedral angle is $180^{\circ}$. Find the sum of the cosines of the dihedral angles of this trihedral angle. | 306. Consider a tetrahedron, all faces of which are equal triangles, the angles of which are respectively equal to the plane angles of a given trihedral angle. (Prove that such a tetrahedron exists.) All trihedral angles of this tetrahedron are equal to the given trihedral angle. The sum of the cosines of the dihedral ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,000 |
307. Prove that for a regular tetrahedron
a) the radius of the inscribed sphere is half the radius of the sphere that touches one face of the tetrahedron and the extensions of the other three (such a sphere is called an exsphere),
b) the centers of the four exspheres are the vertices of a tetrahedron equal to the giv... | 307. By completing the tetrahedron to a parallelepiped, by drawing through each edge a plane parallel to the opposite edge, we obtain, for a regular tetrahedron, as is known, a rectangular parallelepiped.
The center of the inscribed sphere coincides with the center of the parallelepiped, and the centers of the exscrib... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,001 |
308. Let $h$ be the height of a regular tetrahedron, $h_{1}$ and $h_{2}$ be the segments into which one of the heights of a face is divided by the point of intersection of the heights of this face. Prove that $h^{2}=4 h_{1} h_{2}$. Also prove that the base of the height of the tetrahedron and the point of intersection ... | 308. Let $ABCD=$ be a tetrahedron, $DH$ - its height, $DA_{1}$, $DB_{1}$, and $DC_{1}$ - the heights of the faces dropped from vertex $D$ to sides $BC$, $CA$, and $AB$. If we cut the surface of the tetrahedron along edges $DA$, $DB$, and $DC$ and unfold it (see Fig. 60), it is clear that $H$ is the point
. | 309. Solving problem 308, we proved that the center of the sphere circumscribed around a tetrahedron projects onto each face at the midpoint of the segment whose ends are the base of the height dropped onto this face and the point of intersection of the heights of this face. Since the distance from the center of the ci... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,003 |
310. On a plane, there is a circle and a point $M$ at a distance from its center, less than $1 / 3$ of its radius. Let $A B C$ be an arbitrary triangle inscribed in the given circle, with its centroid at point $M$. Prove that there exist two fixed points in space - $D$ and $D_{i}^{\prime}$ symmetric with respect to the... | 310. First of all, note that all triangles $ABC$ are acute-angled. Indeed, if $H$ is the point of intersection of the altitudes of $\triangle ABC$, and $O$ is the center of the circumscribed circle, then $|OH| = 3|OM|$, where $M$ is between $O$ and $H$, i.e., $H$ is inside the circumscribed circle of $\triangle ABC$, w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,004 |
311. On the plane, there is a square $A B C D$. On the sides $B C$ and $C D$, points $P$ and $Q$ are taken such that $|C P|+|C Q|=$ $=|A B|$. Let $M$ be a point in space such that in the tetrahedron $A P Q M$, all faces are equal triangles. Determine the geometric locus of the projections of points $M$ onto the plane p... | 311. Consider a cube $A E F G A_{1} E_{1} F_{1} G_{1}$ with an edge equal to the side of the square $A B C D$. Take points $P$ and $Q$ on the edges $A_{1} E_{1}$ and $A_{1} G_{1}$ such that $\left|A_{1} P\right|=|B P|=|C Q|$, $\left|A_{1} Q\right|=|Q D|=|P C|$ (Fig. $\left.62, a\right)$. Consider the rectangle $A_{1} P... | twosegmentsemanatingfromthemidpointofACatanangle\varphitoACsuchthat\cos\varphi=\frac{1}{\sqrt{3}},havinglengthof\frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,005 |
312. For the heights of a tetrahedron to intersect at one point (such a tetrahedron is called orthocentric), it is necessary and sufficient that:
a) the opposite edges of the tetrahedron are perpendicular
b) one height of the tetrahedron passes through the point of intersection of the heights of the base;
c) the sum... | 312. a) Let $ABCD$ be the given tetrahedron. If its altitudes intersect at point $H$, then $DH$ is perpendicular to the plane $ABC$ and, therefore, $DH$ is perpendicular to $BC$. Similarly, $AH$ is perpendicular to $BC$. Thus, the plane $DAH$ is perpendicular to $BC$, meaning that the edges $DA$ and $BC$ are perpendicu... | proof | Geometry | MCQ | Yes | Yes | olympiads | false | 30,006 |
313. Prove that in an orthocentric tetrahedron, the centroid is located at the midpoint of the segment connecting the center of the circumscribed sphere with the point of intersection of the altitudes. | 313. Let $A B C D$ be the given tetrahedron. Extend it in the usual way to form a parallelepiped. Since $A B C D$ is orthocentric, all edges of the parallelepiped will be equal to each other. Let $A_{i} B_{i}$ be the diagonal of the face of the parallelepiped parallel to $A B, O$ be the center of the sphere circumscrib... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,007 |
314. Prove that in an orthocentric tetrahedron, the following relation holds:
$$
|O H|^{2}=4 R^{2}-3 l^{2}
$$
where $O$ is the center of the circumscribed sphere, $H$ is the point of intersection of the altitudes, $R$ is the radius of the circumscribed sphere, and $l$ is the distance between the midpoints of the skew... | 314. Let's introduce the same notations as in the previous problem. Let $K$ and $L$ be the midpoints of $A B$ and $A_{1} B_{1}$. Then $K O L H$ is a parallelogram. Therefore,
$$
\begin{aligned}
& |O H|^{2}=2|O K|^{2}+2|O L|^{2}-|K L|^{2}= \\
& \begin{aligned}
=2\left(R^{2}-\frac{|A B|^{2}}{4}\right)+ & 2\left(R^{2}-\f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,008 |
315. Prove that in an orthocentric tetrahedron, all dihedral angles adjacent to one vertex are either simultaneously acute or obtuse. | 315. If $A B C D$ is an orthodiagonal tetrahedron, then (see problem
$$
|A B|^{2}+|C D|^{2}=|A D|^{2}+|B C|^{2}
$$
from which
$$
|A B|^{2}+|A C|^{2}-|B C|^{2}=|A D|^{2}+|A C|^{2}-|C D|^{2}
$$
i.e., the angles $\overparen{B A C}$ and $\widehat{D A C}$ are either both acute or both obtuse. | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,009 |
316. Prove that for an orthocentric tetrahedron, the nine-point circles of each face belong to one sphere (the sphere of 24 points). | 316. The section of an orthocentric tetrahedron by any plane parallel to opposite edges and passing at equal distances from these edges is a rectangle, the diagonals of which are equal to the distance between the midpoints of the opposite edges of the tetrahedron (all these distances are equal to each other, see proble... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,010 |
317. Prove that for an orthocentric tetrahedron, the centroids and the points of intersection of the altitudes of the faces, as well as the points that divide the segments of each altitude of the tetrahedron from the vertex to the point of intersection of the altitudes in the ratio $2: 1$, lie on one sphere (the sphere... | 317. Let $O, M$ and $H$ be the center of the circumscribed sphere, the centroid, and the orthocenter (the point of intersection of the altitudes) of an orthocentric tetrahedron; $M$ is the midpoint of the segment $O H$ (see problem 313). The centroids of the faces of the tetrahedron serve as the vertices of a tetrahedr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,011 |
318. Let $H$ be the point of intersection of the altitudes of an orthocentric tetrahedron, $M$ be the centroid of one of its faces, and $N$ be one of the points of intersection of the line $H M$ with the circumscribed sphere of the tetrahedron (with $M$ between $H$ and $N$). Prove that $|M N|=2|H M|$. | 318. The centers of gravity of the faces of an orthocentric tetrahedron lie on the surface of a sphere that is homothetic to the sphere circumscribed around the tetrahedron, with the center of homothety at point $M$ and the coefficient $1 / 3$ (see the solution to problem 317). This leads to the statement of the proble... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,012 |
319. Let $G$ be the centroid of an orthocentric tetrahedron, $F$ the foot of one of its altitudes, and $K$ one of the points of intersection of the line $F G$ with the circumscribed sphere of the tetrahedron (with $G$ between $K$ and $F$). Prove that $|K G| = 3|F G|$.
An arbitrary polyhedron.
Sphere. | 319. The bases of the altitudes of an orthocentric tetrahedron lie on the surface of a sphere that is homothetic to the sphere circumscribed around the tetrahedron, with the center of homothety at point $G$ and the homothety coefficient $-\frac{1}{3}$ (see the solution to problem 317). From this, the statement of the p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,013 |
320. Prove that it is impossible to place three arcs of great circles, each $300^{\circ}$ long, on a sphere such that no two of them have any points in common. | 320. Suppose the opposite. Let the planes in which the arcs are located intersect pairwise on the surface of the sphere at points \(A\) and \(A_{1}, B\) and \(B_{1}, C\) and \(C_{1}\) (Fig. 65). Since each arc is greater than \(180^{\circ}\), it must contain at least one of any two opposite points of the circle on whic... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,014 |
321. Prove that the shortest line connecting two points on the surface of a sphere is the smaller arc of the great circle passing through these points. (Lines going along the surface of the sphere are considered.) | 321. Let $A$ and $B$ be two points on the surface of a sphere, and $C$ be a point on the shorter arc of the great circle passing through $A$ and $B$.
We will prove that the shortest path from $A$ to $B$ must pass through $C$. Consider two circles $\alpha$ and $\beta$ on the surface of the sphere, passing through $C$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,015 |
322. Given a polyhedron, all of whose edges are equal to each other and touch a certain sphere. Does there always exist a sphere circumscribed around this polyhedron? | 322. The described sphere may not exist. An example can be a polyhedron constructed as follows. Take a cube and on its faces, as bases, construct regular quadrilateral pyramids outward with dihedral angles at the base equal to $45^{\circ}$. As a result, we obtain a 12-faced polyhedron (the edges of the cube are not edg... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,016 |
323. Find the area of the triangle formed by the intersection of a sphere of radius $R$ with a trihedral angle, the dihedral angles of which are equal to $\alpha, \beta$ and $\gamma$, and the vertex coincides with the center of the sphere. | 323. First of all, note that the area of the spherical "digon" formed by the intersection of the surface of the sphere with the faces of a dihedral angle of magnitude $\alpha$, whose edge passes through the center of the sphere, is equal to $2 \alpha R^{2}$. This follows from the fact that this area is proportional to ... | S_{\Delta}=R^{2}(\alpha+\beta+\gamma-\pi) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,017 |
324. Let $M$ be the number of faces, $K$ be the number of edges, and $N$ be the number of vertices of a convex polyhedron. Prove that
$$
M-K+N=2
$$
(This relation was first obtained by Euler, and it is valid not only for convex polyhedra but also for a broader class of so-called simply connected polyhedra.) | 324. Consider a sphere with its center inside a polyhedron and project the edges of the polyhedron from the center of the sphere onto the surface of the sphere.
The surface of the sphere will be divided into polygons. If \( n_{k} \) is the number of sides of the \( k \)-th polygon, \( A_{k} \) is the sum of its angles... | N-K+M=2 | Geometry | proof | Yes | Yes | olympiads | false | 30,018 |
325. On the surface of a sphere, a circle is given. Prove that among all spherical $n$-gons containing the given circle inside them, the one with the smallest area is the regular spherical $n$-gon. | 325. Let $\alpha$ be the central angle corresponding to the spherical radius of the circle (the angle between the radii of the sphere drawn from the center of the sphere to the center of the circle and a point on the circle).
Consider a spherical triangle corresponding to a trihedral angle with its vertex at the cente... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,019 |
327. Prove that in any convex polyhedron, there will be either a triangular face or a vertex where three edges meet. | 327. If each face has more than three sides and more than three edges come out of each vertex, then (notations from problem 324)
$$
K \geqslant 2 M, K \geqslant 2 N
$$
and $N-K+M \leqslant 0$, which is impossible. | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,021 |
328. Prove that a convex polyhedron cannot have seven edges. Also prove that for any $n \geqslant 6, n \neq 7$ there exists a polyhedron having $n$ edges. | 328. If all faces are triangles, then the number of edges is a multiple of 3. If there is at least one face with more than three sides, then the number of edges is not less than eight. $2 n$ edges ($n \geqslant 3$) has an $n$-sided pyramid. $(2 n+3)$ edges ($n \geqslant 3$) has a polyhedron that results from cutting of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,022 |
330. Inside a sphere of radius 1, there is a convex polyhedron, all dihedral angles of which are less than $2 \pi / 3$. Prove that the sum of the lengths of the edges of this polyhedron is less than 24. | 330. Consider the so-called $d$-neighborhood of our polyhedron, i.e., the set of points, each of which is at a distance of no more than $d$ from at least one point of the polyhedron. The surface of the resulting body consists of flat parts equal to the corresponding faces of the polyhedron, cylindrical parts correspond... | \Sigmal_{i}<24 | Geometry | proof | Yes | Yes | olympiads | false | 30,024 |
331. The center of a sphere with radius $R$ is located outside a dihedral angle of magnitude $\alpha$ at a distance $a(a<R)$ from its edge and is situated in the plane of one of its faces. Find the area of the part of the sphere's surface that is inside the angle. | 331. In Fig. 66, O is the center of the sphere, A and B are the points of intersection of the edge of the dihedral angle with the surface of the sphere, D and C are the midpoints of the arcs $\overline{A D B}$ and $\overline{A C B}$, the plane $A D B$ passes through $O, E$ is the vertex of the segment cut off by the pl... | 2R^{2}\arccos\frac{R\cos\alpha}{\sqrt{R^{2}-^{2}\sin^{2}\alpha}}-2R\sin\alpha\arccos\frac{\cos\alpha}{\sqrt{R^{2}-^{2}\sin^{2}\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,025 |
332. A sphere of radius $R$ touches the edges of a trihedral angle, all plane angles of which are equal to $60^{\circ}$. The surface of the sphere inside the angle consists of two curvilinear quadrilaterals. Find the areas of these quadrilaterals. | 332. Consider a regular octahedron with edge $2 R$. A sphere that touches all its edges has a radius $R$. The surface of the sphere is divided by the surface of the octahedron into eight spherical segments and six curvilinear quadrilaterals, equal to the smaller of the two sought.
$$
\text { Answer: } \frac{2 \pi R^{2... | \frac{2\piR^{2}}{3}(4\sqrt{\frac{2}{3}}-3),\piR^{2}(\frac{16}{3}\sqrt{\frac{2}{3}}-2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,026 |
333. Given a cube with edge $a$. Determine the areas of the parts of the sphere circumscribed around this cube, into which it is divided by the planes of the cube's faces. | 333. Twelve "digons" with an area of $\frac{\pi a^{2}(2-\sqrt{3)}}{4}$ and six curvilinear quadrilaterals with an area of $\frac{\pi a^{2}(\sqrt{3}-1)}{2}$. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,027 |
334. Given a convex polyhedron. Some of its faces are painted black, and no two painted faces share a common edge, and their number is more than half the number of all the polyhedron's faces. Prove that a sphere cannot be inscribed in this polyhedron. | 334. Suppose that a sphere can be inscribed in this polyhedron. Connect the point of tangency of the sphere with any face to all the vertices of this face. Each face will be divided into triangles. Triangles located in adjacent faces, having a common edge, are equal. Therefore, each "black" triangle corresponds to an e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,028 |
335. What is the greatest number of balls of radius 7 that can simultaneously touch, without intersecting, a ball of radius 3?
Transition to space. | 335. We will prove that there cannot be more than six such spheres. Suppose there are seven. We connect the centers of all seven spheres with the center of the given sphere and denote by $O_{1}, O_{2}, \ldots, O_{7}$ the points of intersection of these segments with the surface of the given sphere. For each point $O_{i... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,029 |
336. On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $N$ are taken such that $|C M| + |C N| = |A B|$. The lines $A M$ and $A N$ divide the diagonal $B D$ into three segments. Prove that a triangle can always be formed from these segments, and one of the angles of this triangle is $60^{\circ}$. | 336. Consider the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Take points $K$ and $L$ on the edges $A_{1} B$ and $A_{1} D$ such that $\left|A_{1} K\right|=|C M|,\left|A_{1} L\right|=$ $=|C N|$. Let $P$ and $Q$ be the points of intersection of the lines $A K$ and $B A_{1}$, $A L$ and $D A_{1}$.
The sides of the triangle $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,030 |
338. Prove that the diagonals connecting opposite vertices of a hexagon circumscribed around a circle intersect at one point. (Brianchon's Theorem.) | 338. Let $A B C D E F$ be a planar hexagon circumscribed around a circle. Consider an arbitrary spatial hexagon $A_{1} B_{1} C_{1} D_{1} E_{1} F_{1}$ (Fig. 67), different from $A B C D E F$, whose projection onto our plane is the hexagon $A B C D E F$, and whose corresponding sides pass through the points of tangency o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,032 |
340. Three planes in space intersect along a single line. Three trihedral angles are arranged such that their vertices lie on this line, and their edges are in the given planes (it is assumed that the corresponding edges, i.e., edges located in the same plane, do not intersect at a single point). Prove that the three p... | 340. This problem represents one of the possible four-dimensional analogs of Desargues' theorem (see problem 339). To solve it, it is convenient to "step out" into four-dimensional space.
First, let's talk about some properties of this space.
The simplest figures in four-dimensional space will be: a point, a line, a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 30,034 |
1. Three bandits want to divide the loot. Each is sure that he would divide the loot into equal parts, but the others do not trust him. If there were only two bandits, it would be easy to get out of the situation: one would divide the loot into two parts, and the other would take the part that seems larger to him. Indi... | 1. Let the first bandit divide the loot into three parts, which he considers to be equal, and let the second and third bandits point to the part that seems larger to them. If they point to different parts, then each takes the part that they consider to be larger, and the first takes the remaining part. If they point to... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 30,035 |
4. The logician found himself on an island inhabited by two tribes. Representatives of one tribe always tell the truth, while representatives of the other always lie. The traveler approached a fork in the road and had to ask a local who happened to be nearby which of the two roads led to the village. He did not know wh... | 4. Here is one variant of the solution. The logician asks, pointing to one of the roads: "If I ask any representative of your tribe, - Does this road lead to the village? - will he answer affirmatively?" If this road does indeed lead to the village, then both the liar and the truth-teller will answer: "Yes." If, howeve... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 30,037 |
6. There are 10 bags of coins. In nine bags, the coins are genuine (weighing 10g each), and in one bag, all the coins are counterfeit (weighing 11g each). Determine which bag contains the counterfeit coins with one weighing. | 6. Let's number the bags from 1 to 10. Take one coin from the first bag, two from the second, ..., and ten from the tenth, and determine their total weight. Let this weight be $P$. If all the coins were genuine, they would weigh $10+20+\ldots+100=550$ g. The excess $P-550$ obviously coincides with the number of the bag... | P-550 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 30,039 |
7. In a bus without a conductor, there were 20 people. Although they only had coins worth 10, 15, and 20 kopecks, each of them paid for the fare and received the change they were due. How could this have happened? Prove that they had no fewer than 25 coins. (One ticket costs 5 kopecks.) | 7. Let's first prove that the passengers had no less than 25 coins. Indeed, each passenger should receive change, i.e., no fewer than twenty coins should remain in the hands of the passengers. One ruble was dropped into the cash box, so there must be no fewer than five coins (since the largest coin available is 20 kope... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 30,041 |
8. Prove that in any group of six schoolchildren, there are always either three schoolchildren who are all acquainted with each other, or three schoolchildren, each of whom is not acquainted with the other two. | 8. Let $A$ be one of our six schoolchildren. If $A$ is acquainted with no more than two schoolchildren from our group, then there are three schoolchildren in the group who are not acquainted with $A$. If all these schoolchildren are acquainted with each other, then they already form a trio of mutually acquainted indivi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,042 |
9. In a chess tournament, $n$ chess players, including grandmasters and masters, participated. After the tournament, it turned out that each participant scored exactly half of their points in games against masters. Prove that $\sqrt{n}$ is an integer. | 9. The number of games played by $k$ participants among themselves is $k(k-1) / 2$. Indeed, each of the $k$ participants must play with each of the $k-1$ others, but in the product $k(k-1)$, each game is counted twice, as it involves two players.
Let the number of masters be $m$, the number of grandmasters be $g$, so ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,043 |
10. The bus network in the city of Lisse is arranged in such a way that: a) each route has three stops; b) any two routes either have no common stops at all or have only one common stop. What is the maximum number of routes that can exist in this city, given that there are only nine different stops? | 10. Let's consider some stop $A$. Define how many routes can pass through it. Besides $A$, there are eight other stops in the city. On each route passing through $A_{\text {r }}$, there are two more stops. Since no two of these routes can have common stops other than $A$, a total of no more than $8: 2=4$ routes can pas... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,044 |
12. Prove that in a game of tic-tac-toe, the second player, no matter how well he plays, cannot expect more than a draw if his opponent plays correctly. | 12. Let's prove that the beginner (X) has a way to win or achieve a draw. Suppose this is not the case. Then, no matter how the beginner plays, "O" will always win by applying the best strategy. However, the beginner, after placing the first X anywhere, can then use the best strategy of O, mentally swapping X and O. Th... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,046 |
13. Let the game be played on a sheet of graph paper (of the usual school format) until four signs in a row. Prove that with the correct strategy, crosses win in no more than six moves. | 13. A beginner should place the first cross not too close to the border (no closer than seven cells). It is easy to understand that the best method of defense for the second player is to place the first zero next to the cross. There are two fundamentally different possibilities (Fig. 15, a and b); others reduce to the ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,047 |
14**. Let the game be played until three signs in a row are achieved. What is the minimum number of cells the board should contain so that the first player can win, regardless of how his opponent plays? [Draw a board (of arbitrary shape) with the minimum number of cells and prove that on any board with fewer cells, the... | 14. On a board of seven cells, shown in Fig. $16, a$, the first player can win regardless of how the opponent plays: first, he places a cross at the intersection of rows, and then on one of the middle cells in the row that does not have a zero.
No matter what move the opponent makes, the first player can win on the th... | 7 | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,048 |
15. In the country of Lemniscata, a competition has been announced for a modular apartment project designed for occupancy by one, two, three, or four families. It is required that the number of rooms in the apartment be the smallest possible and that the living space can be divided equally among the families moving in.... | 15. Let's denote the area of the apartment by 1.- It is not difficult to verify that six rooms, the areas of which are equal to
$$
\frac{1}{4}, \frac{1}{4}, \frac{1}{6}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12},
$$
satisfy the conditions of the problem. Five rooms are insufficient. Indeed, when resettling three famili... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 30,049 |
16. Two players play on a sheet of grid paper according to the following rules. The first player draws a segment coinciding with the side of a cell. The second player continues this line by drawing from the end of the first segment his own segment, also coinciding with the side of some cell. Then the turn passes to the... | 16. Suppose the game ended with a result: on the $n$-th move, one of the players lost. Then before the ( $n-1$ )-th move, the situation was as follows (Fig. $17, a$ ): the player who made the ( $n-1$ )-th move drew the segment $O A$, after which all segments emanating from point $O$ were occupied. Point $O$ can only be... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,050 |
17. Prove that in the previous game, the beginner can play in such a way that the game ends in a draw. | 17. The first player must make all their moves in one direction (say, upwards). In this case, the broken line will not intersect itself and will reach the edge of the sheet. The verification of this statement is left to the reader. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 30,051 |
18. Prince Gvidon had three sons. Among his descendants, 93 each had two sons and no daughters, while all the others died childless. How many descendants did Prince Gvidon have in total? | 18. Each of the descendants who have children adds two more descendants to the three descendants of the first generation. Since 93 descendants had children, this added $2 \cdot 93 = 186$ descendants, making a total of $3 + 186 = 189$. This problem ad-
, in each sector such that: a) any two numbers will be different, b) any two numbers in adjacent sectors will differ in exactly one digit. | 22. We will prove by induction that for any $n$ it is possible to write numbers composed of $n$ ones and twos in $2^{n}$ sectors of a circle, so that conditions a) and b) formulated in the problem statement are satisfied.
For $n=1$, the statement is trivial. Suppose it is true for $n=k$. To obtain the required arrange... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,056 |
24. The width of a rectangle is called the length of its shortest side *). In how many different ways can a rectangle of width 3 be cut out from a square sheet of paper consisting of 100 cells? (Cuts must be made only along the cell boundaries.) | 24. Let's call the length of a rectangle the length of its longer side (the length of a square is equal to its width). Let's calculate the number of different rectangles of length $l$, where $l=4,5,6,7,8,9,10$. Such rectangles can be either horizontal or vertical, depending on whether their longer side is horizontal or... | 8^3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,058 |
25. A chocolate bar consists of $5 \times 8$ square pieces. The bar is broken along straight lines, separating the pieces, until 40 individual pieces are obtained. How many times will the bar need to be broken? (Find all solutions.) | 25. Each time we break one piece, we get two smaller pieces, i.e., the total number of pieces increases by one. Initially, there was one piece. Therefore, we will have to break the chocolate bar 39 times.
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。 | 27. Let's put three matches on the table and arrange the coins in a row so that the coins of the first schoolboy lie before the first match, between the first and second - the coins of the second, between the second and third - the coins of the third, and finally, after the third match - the coins of the fourth schoolb... | 3276 | Number Theory | proof | Yes | Yes | olympiads | false | 30,061 |
28**. In the Martian language, the alphabet consists of the letters $A$ and O. Any two words of the same length differ in at least three positions. Prove that the number of words of length $n$ is no more than $2^{n} /(n+1)$. | 28. Suppose there are only $k$ Martian words of length $n$. Let's write these words in a string. Under each word, we will write sets of $n$ letters obtained by changing one letter in the word. Obviously, the sets from two different columns differ in at least one letter. We have a total of $k$ columns, each with $n+1$ s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,062 |
29**. A cube was painted white on the outside and then cut into 64 small cubes. Then the small cubes were randomly assembled into a large cube. (The cubes could not only be rearranged but also rotated.) What is the probability that it will be white on the outside? (All ways of assembling the large cube are considered e... | 29. Suppose first that only one small cube, say, the one standing in the corner (Fig. 29), has been moved. In how many ways can it be put back in place? Any of the eight vertices of the cube can be placed at point $A$. Each of these can be rotated in three different ways. This gives a total of 24 options. The same numb... | p<10^{-83} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,063 |
30. There are 111 lamps, and each lamp has its own switch. It is allowed to simultaneously switch 13 of them. At the initial moment, some lamps are on, and some are off.
a) Is it possible to turn off all the lamps?
b) How many switches will be required for this if all the lamps were initially on? | 30. Let's show that we can always turn off all the lamps. We can assume that initially more than 13 lamps were on (otherwise, we would have turned on 13 lamps from the extinguished ones). Moreover, for the same reason, we can assume that the number of extinguished lamps is more than 6. We will select 13 lamps such that... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,064 |
31. The city of Sigmagrad has the shape of a square with a side of 5 km (Fig. 5). Streets divide it into blocks, which are squares with a side of 200 m. What is the largest area that can be covered by walking 10 km along the streets of this city and returning to the starting point?
 be the "topmost" street we visited, \( DC \) the "bottommost", \( AD \) the "leftmost", and \( BC \) the "rightmost". To g... | 6.24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,065 |
32. On a $10 \times 10$ board for playing "Battleship," a four-cell "ship" $\square \square$ ( $\square$ is located. What is the minimum number of "shots" needed to hit the ship? (Indicate the method of delivering this number of shots and prove that with fewer shots, the ship can always be placed in such a way that it ... | 32. Let's color the fields of the board as shown in Fig. 31 (instead of colors, we will write letters: k - red, s - blue, 3 - green, y - yellow). A direct count shows that the board has 24 yellow, 26 blue, 25 green, and 25 red fields. - It is clear that each four-cell ship occupies exactly one cell of each color. There... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,066 |
35**. Prove that there are exactly eight ways for a knight to tour a $3 \times 4$ chessboard, visiting each square exactly once, and that there is no way for the knight to return to the starting square on its last move. (Two tours of the board that are mirror images of each other, obtained by making the same moves in r... | 35. Fig. 35 shows a $3 \times 4$ chessboard, the squares of which are labeled with letters $A, D, B$, etc. In one move,
Fig. 35.
![](https://cdn.mathpix.com/cropped/2024_05_21_7d0959e1f193b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,068 |
38. In class 4A, there are 30 students. During a dictation, one student made 12 mistakes, and the others made fewer. Prove that there are at least three students in the class who made the same number of mistakes. | 38. Let's divide all the students in the class into 13 groups: the first group will include students who wrote the dictation without errors, the second group will include those who made one error, the third group will include those who made two errors, and so on, and finally, the thirteenth group will include students ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,071 |
39. 30 teams participate in a football championship. Each pair of teams must play one match against each other. Prove that at any moment during the competitions, there are two teams that have played the same number of matches up to that point. | 39. Proof by contradiction. Suppose that by some point in the competitions, the teams have played a different number of matches. Assign each team a number equal to the number of matches played by that team plus one. Clearly, the number can be any integer from 1 to 30. Since there are also 30 teams and no two teams will... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,072 |
40. Prove that on October 23, 1965, in the cinema "Mir" during the first session, there were at least two viewers who had the same number of acquaintances among those sitting in the hall (it is known that 5 people were sitting in the first row). | 40. According to the condition, there were $k \geqslant 5$ people in the hall. Suppose initially that each viewer has acquaintances among those sitting in the hall. The maximum number of acquaintances (for one person) does not exceed $k-1$. If $k \geqslant 2$, then among the $k$ numbers between 1 and $k-1$, there must ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,073 |
41. Given 20 distinct positive integers, all less than 70. Prove that among their differences, there will be 4 identical ones. | 41. Let's write down all our twenty numbers in ascending order, starting from the smallest number $a$ and ending with the largest $A$. Subtract each number from the previous one. If there are no four identical differences among them, then among the nineteen resulting numbers, there are no more than three ones, twos, th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,074 |
42. Schoolboy solves problems throughout the year; every day - at least one problem. Each week, to avoid overworking, he solves no more than 12 problems. Prove that there will be several consecutive days during which he will solve exactly 20 problems. | 42. Suppose that on the first day, the student solved $a_{1}$ problems, in the first two days $a_{2}$ problems, and in the first 77 days (11 weeks) $a_{77}$ problems.
Consider the numbers
$$
\begin{array}{ll}
a_{1}, & a_{2}, \ldots, \quad a_{77} \\
a_{1}+20, & a_{2}+20, \ldots a_{77}+20
\end{array}
$$
There are 154 ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,075 |
43. In the city of Lisse, there are 10,000 telephones, the numbers of which are defined by four-digit numbers. In the central district, more than half of all telephones are installed. Prove that at least one of the central telephone numbers is equal to the sum of two other central telephone numbers (or double the numbe... | 43. In the central district, no less than 5001 telephone sets have been installed. Let $A$ be the smallest of their numbers. Subtract $A$ from the remaining central numbers. We will write down two groups of numbers: the first group consists of the numbers of central telephones, the second - from the resulting differenc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,076 |
44. There are $2 k+1$ cards, numbered with consecutive natural numbers from 1 to $2 k+1$. What is the maximum number of cards that can be selected so that no one of the selected numbers is equal to the sum of two other selected numbers? | 44. The cards that satisfy the condition of the problem have numbers $k+1, k+2, \ldots, 2 k+1$. Their number is $k+1$. We will prove that it is impossible to choose a greater number of cards. Suppose the opposite, i.e., that it is possible to choose $k+r$ cards that satisfy the condition of the problem, where $r>1$. Le... | k+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,077 |
45. A sequence of $k$ numbers is written in a row. Prove that there will always be several consecutive numbers whose sum is divisible by $k$. | 45. Consider the first number, its sum with the second, the sum of the first three numbers, and so on. If one of the obtained numbers is divisible by $k$ without a remainder, then everything is proven. If not, then dividing these numbers by $k$, we get $k$ remainders. All remainders are between 1 and $k-1$, so there ar... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,078 |
52. Prove that if the numbers $m$ and $m^{2}+2$ are prime, then the number $m^{3}+2$ is also prime. | 52. Any prime number $m$, different from 3, can be represented in the form $3n+1$ or in the form $3n-1$, where $n$ is some integer. In the first case, we can write
$$
m^{2}+2=9n^{2}+6n+3
$$
in the second case $m^{2}+2=9n^{2}-6n+3$.
Since $m \geqslant 2$, in any case the number $m^{2}+2$ is greater than 3 and divisib... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,084 |
58. The numbers $1,2,3, \ldots, 1000000000$ are written down. Then each number is replaced by the sum of its digits, and so on, until only single-digit numbers remain in the sequence. Which digit appears more frequently in this sequence: ones or fives? | 58. From the divisibility rule for 9, which we have used several times, it follows that the numbers in the last row give the same remainders when divided by 9 as the numbers in the first row above them. From this, it easily follows that in the first 999999999 positions of the last row, the digits $1,2,3,4,5,6$, $7,8,9$... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 30,089 |
59. Prove that if a number is divisible by 99, then the sum of its digits is not less than 18. | 59. If a number is divisible by 11, then the sum of the digits in the even positions differs from the sum of the digits in the odd positions by a number that is a multiple of 11. Additionally, the sum of all the digits of the number is divisible by 9. However, the number 9 cannot obviously be represented as the sum of ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,090 |
60. Prove that a number, the decimal representation of which consists of $3^{n}$ ones, is divisible by $3^{n}$. | 60. We will prove this by mathematical induction. The number 111 is divisible by 3. Assume that our statement is true for $n=k$, and we will prove it for $n=k+1$. Represent the number $\underbrace{111 \ldots 1}_{3^{k+1}}$ as
$$
\underbrace{111 \ldots 1}_{3^{k}} \times 1 \underbrace{00 \ldots 0}_{3^{k}-1} 1 \underbrace... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,091 |
61. Prove that if a number $l$ is not divisible by 2 or 5, then there exists a number divisible by $l$ whose decimal representation consists entirely of ones. | 61. Consider the numbers $1, 11, \ldots, \underbrace{11 \ldots 1}_{\text{l}}$. If one of them is divisible by $l$, then everything is proven. Otherwise, these numbers give remainders $1 = r_{1}, r_{2}, \ldots, r_{l}$ when divided by $l$, each of which is either 1, 2, ..., or $l - 1$. Since there are $l$ remainders and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,092 |
62. Prove that if the natural numbers from 1 to 1967 inclusive are written down in any order, the resulting number is not a perfect cube. | 62. From remark a), given on p. 52, it follows that any number $N$, obtained by writing down the natural numbers from 1 to 1967 in some order, has the same remainders when divided by 3 and 9 as the number
$$
1+2+\ldots+1967=\frac{1967 \cdot 1968}{2}
$$
But the latter number is divisible by 3 and not divisible by 9. T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,093 |
63*. Prove that for no polynomial $f(x)$ with integer coefficients can the equalities
$$
f(7)=11, \quad f(11)=13
$$
hold.
64*. Let $P(x)$ be a polynomial of degree 4 with integer coefficients. Suppose that for any integer value of $x$, $P(x)$ is an integer divisible by 5. Prove that all coefficients of the polynomia... | 63. Let $f(7)=11, f(11)=13$. Then $f(11)-f(7)$ must be divisible by $11-7=4$. Indeed, if $f(x)=$ $=a x^{n}+b x^{n-1}+\ldots+g x+h, \quad$ then $f(11)-f(7)=a\left(11^{n}-\right.$ $\left.-7^{n}\right)+b\left(11^{n-1}-7^{n-1}\right)+\ldots+g(11-7)$, and each of the expressions in parentheses is divisible by $11-7=4 *$ ). ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,094 |
65*. There is a polynomial $P(x)$ of the 4th degree, which takes integer values at integer values of $x$. Prove that all coefficients of the polynomial $24 P(x)$ are integers.
The content of the following three problems constitutes a proof of the theorem, which generalizes the result of the previous problem to the cas... | 65. The reasoning of the previous problem goes almost without change in this case. Indeed, let
$$
P(x)=a x^{4}+b x^{3}+c x^{2}+d x+e
$$
Substituting $x=0, \pm 1, \pm 2$, we get that the numbers
$$
e, a+b+c+d+e, a-b+c-d+e
$$
$$
16 a+8 b+4 c+2 d+e, \quad 16 a-8 b+4 c-2 d+e
$$
are integers. Adding and subtracting the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 30,095 |
66a*. Prove that the polynomial
$$
Q_{n}(x)=\frac{x(x-1) \ldots(x-n+1)}{n!}
$$
takes integer values for all integer $x$. | a. For an integer $x \geqslant n$, $Q_{n}(x)=C_{x}^{n}$ is an integer (see Corollary 3 from the combinatorial lemma, p. 36). For $x=0,1,2, \ldots, n-1$, $Q_{n}(x)=0$. Finally, if $x$ is a negative integer, then $Q_{n}(x)=(-1)^{n} C_{n-x-1}^{n}$ is again an integer.
Another solution to the problem may use the method of... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,096 |
67*. Prove that the number $1967 k+3$ for any integer $k$ is not a perfect cube.
1 68. A sequence is given: 1, 1, 2, 3, 7, 22 ... Each term is equal to the product of the previous two terms plus 1. Prove that no term of the sequence is divisible by 4.
- 69**. Let $\sin x=3 / 5$. Prove that $5^{25} \sin 25 x$ is an in... | 67. To solve the problem, it is sufficient to check that numbers of the form $l^{3}-3$ are not divisible by 1967 for any integer $l$. Note that $1967=7 \cdot 281$. We will prove that numbers of the form $l^{3}-3$ are not even divisible by 7. Any number $l$ can be represented as $l=7 m+r$, where $r=0, \pm 1, \pm 2, \pm ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,097 |
70. In Shvambania, bus tickets have numbers from 000001 to 999999. Shvambrians consider tickets lucky if the sum of the first three digits of the number equals the sum of the last three digits. Prove that the sum of the numbers of all lucky tickets is divisible by 13. | 70. Note that if the number $\overline{a b c d e f}$ is lucky, i.e., $a+b+c=d+e+f$, then the number $999999-\overline{a b c d e f}$ is also lucky, because the sum of its first three digits $27-(a+b+c)$ equals the sum of the last three $27-(d+e+f)$. All lucky numbers, except for the number 999 999, can be divided into p... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,098 |
71. Kolya, Lena, and Misha pooled their money and bought a football. The amount of money each of them contributed does not exceed half of the total amount contributed by the other two. How much money did Misha contribute if the ball cost 6 rubles? | 71. According to the condition, the doubled amount of money invested by each boy does not exceed the sum invested by the other two. If one of the boys had given more than two rubles, then the other two would have given less than four, i.e., less than the doubled amount of money of the first. Therefore, each gave no mor... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 30,099 |
72. Seven mushroom pickers collected a total of 100 mushrooms, and each of the seven collected a different number of mushrooms. Prove that there are three mushroom pickers who together collected no fewer than 50 mushrooms. | 72. Arrange the mushroom pickers by the number of mushrooms found, so that the first one collected the most mushrooms, and the seventh one collected the least. If the fourth one collected no less than 15 mushrooms, then the first three collected no less than
$$
16+17+18=51 \text { mushrooms. }
$$
If the fourth one co... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,100 |
73. On the board, there were five integers. By adding them in pairs, the following ten numbers were obtained:
$$
0,2,4,4,6,8,9,11,13,15
$$
What five numbers were written on the board? Can the following ten numbers be obtained in this way:
$$
12,13,14,15,16,16,17,17,18,20 ?
$$ | 73. Adding all 10 sums, we get 72. Since each of the five original numbers appears in four sums, the sum of the desired numbers is $72: 4=18$. The sum of the two smallest, obviously, is 0, and the sum of the two largest is 15. Therefore, the third largest number is $18-0-15=3$. In the sequence of sums, the second place... | -1,1,3,5,10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,101 |
74. 30 students from five courses came up with 40 problems for the olympiad, with students from the same course coming up with the same number of problems, and students from different courses coming up with a different number of problems. How many students came up with exactly one problem? | 74. Let's choose 5 students, one from each year. Each of them came up with a different number of problems. Therefore, the total number of problems proposed by them is no less than $1+2+3+4+5=15$. The remaining 25 students came up with no more than $40-15=25$ problems. It is clear that each of them came up with one prob... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,102 |
75. Prove that in any six-digit number, the digits can be rearranged so that the sum of the first three digits of the new number differs from the sum of the last three digits by less than 10. | 75. Consider an arbitrary six-digit number. Let the digits of this number (in descending order) be
$$
\begin{gathered}
a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} \\
a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant a_{4} \geqslant a_{5} \geqslant a_{6}
\end{gathered}
$$
We will prove that the number $\overline{a_{1} a_{3... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 30,103 |
76**. Several numbers are arranged around a circle, the sum of which is positive. Prove that one can choose such a number that it itself is positive, the sum of it and the next one in the clockwise direction is positive, and so on. | 76. Let, for example, the numbers $a, b, c$ stand in a clockwise direction around a circle. Write these numbers twice in a row: $a b c a b c$, and then write down six consecutive sums
\[
\begin{aligned}
& s_{1}=a \\
& s_{2}=a+b \\
& s_{3}=a+b+c \\
& s_{4}=a+b+c+a \\
& s_{5}=a+b+c+a+b \\
& s_{6}=a+b+c+a+b+c
\end{aligne... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 30,104 |
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