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29. Through the given point, draw a line that cuts off a triangle of a given perimeter from the given angle.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
Note: The note about translation is not part of the translation but an ex... | 29. Let $S A F$ be the given angle, $M$ be the given point, and $2 p$ be the given perimeter. On the sides of the angle, lay off segments $A P = A Q = p$. Draw a circle $k$ that touches the sides of the angle at points $P$ and $Q$, and draw a line through point $M$ that is tangent to the circle $k$ at point $B$. This l... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,988 | |
30. Construct a circle passing through two given points $A$ and $B$ and tangent to a given line $l$. | 30. Let $X$ be the point of tangency of line $l$ with the circle, and $M$ be the point of intersection of lines $A B$ and $l$, then we have: $M A \cdot M B = M X^2$. From this, it is easy to find the point $X$ and, consequently, construct the desired circle. If $A B \parallel l$, the solution simplifies. If, however, p... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,989 |
31. Construct a triangle from three points that are symmetric to the orthocenter with respect to its sides. | 31. Let $H_{1}, H_{2}, H_{3}$ be the points symmetric to the orthocenter $H$ of the desired triangle $ABC$ with respect to its sides. We construct the triangle $H_{1} H_{2} H_{3}$ and draw its angle bisectors; denote the intersection point of the bisectors as $H$ (it coincides with the orthocenter of triangle $ABC$, an... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,990 |
32. Construct a triangle given its three altitudes $h_{a}$, $h_{b}$ and $h_{c}$. | 32. It is not difficult to prove that a triangle with sides $h_{b}, h_{a}$, and $\frac{h_{a} h_{b}}{h_{c}}$ will be similar to the desired one. From this, its construction follows. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,991 |
33. Prove that if two medians of a triangle are equal, then the triangle is isosceles. | 33. There are several proofs that if two medians of a triangle are equal, then the triangle is isosceles. For example, this can be proven using formulas for calculating medians, by expressing each median in terms of the three sides of the triangle and then equating these expressions; or this can be proven by relying on... | proof | Geometry | proof | Yes | Yes | olympiads | false | 34,992 |
34. Prove that the segment connecting the midpoints of the diagonals of a trapezoid is parallel to its bases and equal to their half-difference. | 34. The segment connecting the midpoints of the diagonals of a trapezoid lies on its midline, so it is parallel to the bases. To prove that it is equal to half the difference of the bases, let's denote the lower base of the trapezoid by $a$, the upper base by $b$, and the desired segment by $x$. It is easy to see that ... | \frac{-b}{2} | Geometry | proof | Yes | Yes | olympiads | false | 34,993 |
36. Cut a rectangle with sides 16 and 9 into two parts so that they can be assembled into a square. | 36. Let there be a rectangle $ABCD$ with sides $AB=9$ and $AD=16$ units of scale. The area of the rectangle $ABCD$ is 144 square units, so the side of the desired square should be 12 units long, i.e., it should be 3 units longer than the shorter side of the rectangle and 4 units shorter than the longer side. Divide $AB... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,995 |
37. Given a segment and a line parallel to it. Using only a ruler, divide the segment in half. | 37. Let a segment $A B$ and a line $l$ parallel to it be given. Take an arbitrary point $P$ outside the strip bounded by the lines $A B$ and $l$, and draw the lines $P A$ and $P B$, and denote by $D$ and $C$ the points of intersection of these lines with the line $l$. In the resulting trapezoid $A B C D$, draw the diag... | M | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,996 |
53. On the plane, there are two non-parallel lines $\boldsymbol{m}$ and $n$, the point of intersection of which is inaccessible (for example, it is outside the drawing). Through a given arbitrary point $K$, it is required to draw a line passing through the inaccessible point of intersection of lines $m$ and $n$. | 53. Let the sides of the angle be denoted by $a$ and $b$, and the inaccessible vertex of the angle by $O$. Take an arbitrary point $M$ on the line $b$ and draw a line $a^{\prime}$ through it, parallel to $a$. Then, from point $M$ as the center, draw a circle of arbitrary radius. This circle will intersect the lines $a^... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,012 |
57. Construct an isosceles triangle given the medians drawn to the lateral sides and the angle at the vertex. | 57. Let $A B C$ be the desired isosceles triangle, with $A B=B C=a$ being the lateral sides, $A D=\boldsymbol{h}_{a}$ the height, and $A E=m_{a}$ the median to the lateral side $B C$. We construct the triangle $A D E$ with the hypotenuse $A E=m_{a}$ and the leg $A D=h_{a}$. On the median, we mark the segment $A M=\frac... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,016 |
69. A quadrilateral $A B C D$ is circumscribed around a circle with center $O$. Prove that the sum of angles $A O B$ and $C O D$ is $180^{\circ}$. | 69. Dropping perpendiculars from the center $O$ to the sides of the quadrilateral $ABCD$, we obtain eight pairwise equal triangles. Renumbering the angles of these triangles formed at the vertices of the quadrilateral, we get $\angle 1 = \angle 2; \angle 3 = \angle 4; \angle 5 = \angle 6; \angle 7 = \angle 8;$ the sum ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,026 |
71. The diagonals of a cyclic quadrilateral are perpendicular. Prove that the midpoints of the sides of this quadrilateral and the feet of the perpendiculars dropped from the point of intersection of its diagonals to the sides lie on one circle. | 71. By connecting the midpoints of the sides of a given quadrilateral with straight line segments, we obtain a parallelogram. However, since its sides are parallel to the diagonals of the quadrilateral, this parallelogram is a rectangle, around which a circle can always be described. Now, let's drop perpendiculars from... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,028 |
72. Find, based on geometric considerations, the limit
$$
\lim _{n \rightarrow \infty} 2^{n} \sqrt{\underbrace{2-\sqrt{2+\sqrt{2+\sqrt{2+\ldots+V \sqrt{2}}}}}_{(n-1) \text { twos }}}
$$ | 72. The sides of regular inscribed polygons are expressed in terms of $R$ of the circumscribed circle as follows:
\[
\begin{aligned}
& a_{4}=R \sqrt{2}, \quad a_{8}=R \sqrt{2-\sqrt{2}}, \quad a_{16}=R \sqrt{2-\sqrt{2+\sqrt{2}}}, \ldots \\
& a_{2^{n+1}}=R \sqrt{2-\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}
\end{aligned}
\]
The... | \pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,029 |
1. Find the geometric locus of points: a) equidistant from the faces of a trihedral angle; b) equidistant from three planes passing through a given point and not having a common line. | 1. a) The line of intersection of the bisecting planes of the dihedral angles of a trihedral angle.
b) Four lines of intersection of the bisecting planes of the formed dihedral angles. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,032 |
3. A spatial quadrilateral is described around a sphere. Prove that the four points of tangency lie in the same plane. | 3. Let's pass a plane through any three points of tangency and use the theorem of equality of tangents drawn from an external point to a sphere, as well as the theorem that equal obliques drawn from one point form equal angles with the plane, and a larger oblique forms a smaller angle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,034 |
4. Prove that if all 6 faces of a parallelepiped are equal parallelograms, then they are all rhombuses. | 4. Let's consider the lateral faces of the parallelepiped. There are four of them, and according to the condition, they are equal to each other as parallelograms. Therefore, by the property of the parallelepiped, the lateral edges are equal, and the sides of the lower and upper bases are also equal. According to the co... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,035 |
5. A sphere with center at point $O$ is inscribed in a trihedral angle $S$. Prove that the plane passing through the three points of tangency is perpendicular to the line $S O$. | 5. The plane passing through the three points of tangency intersects the sphere along a circle, and the line $S O$, as it is easy to prove, passes through the center of this circle; therefore, it will be perpendicular to the plane in which this circle lies. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,036 |
8. What is the maximum number of parts into which the surface of a sphere (sphere) can be divided by 3, 4, and, generally, $n$ circles? | 8. One circle divides the sphere into 2 parts, two circles into 4 parts if they are not parallel and intersect, three circles can divide the sphere into a maximum of 8 parts, 4 circles can divide it into a maximum of 14 parts, and $n$ circles can divide the sphere into a maximum of $n^{2}-n+2$ parts. | n^{2}-n+2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,039 |
9. Given four balls of the same radius $r$, each touching the other three. Calculate the radius of the ball that touches all the given balls. | 9. There will be two such spheres. One of them will touch the four spheres internally, and the other externally. Let the radius of the sphere touching internally be denoted by $p$, and the radius of the sphere touching externally by $R$. Then it is not difficult to show that the centers of the last two spheres coincide... | R=r(\frac{\sqrt{6}}{2}+1)\rho=r(\frac{\sqrt{6}}{2}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,040 |
10. To inscribe a cylinder with the greatest total surface area in a cone with height $h$ and base radius $r$. | 10. Denoting the radius of the base of the desired cylinder by \( x \), and its height by \( y \), we first find that \(\frac{x}{r} = \frac{h-y}{h}\), hence \( y = \frac{h(r-x)}{r} \).
Let the total surface area be \( z \), then \( z = 2 \pi x^{2} + 2 \pi x y \), or
\( z = \frac{2 \pi r x^{2} + 2 \pi h r x - 2 \pi h ... | \frac{\pir^{2}}{2(-r)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,041 |
11. In an arbitrary triangular pyramid (tetrahedron), planes are drawn through each edge inside the pyramid, bisecting the corresponding dihedral angles. Prove that all six planes intersect at one point. | 11. The three bisecting planes of the dihedral angles of one of the trihedral angles of a tetrahedron intersect along a single line (see problem No. 1), which represents the geometric locus of points equidistant from the faces of this trihedral angle. Let's take any one of the three remaining bisecting planes; suppose ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,042 |
12. Determine the volume of the triangular pyramid $S A B C$, if in it $S A=S B=S C=1 ; \angle A S B=60^{\circ}, \angle A S C=90^{\circ}$ and $\angle B S C=120^{\circ}$. | 12. The volume of the triangular pyramid $S A B C$ is $v=\frac{\sqrt{6}}{6}$ (cubic units). | \frac{\sqrt{6}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,043 |
13. Two regular tetrahedra are inscribed in a cube such that four vertices of the cube serve as the vertices of one of them, and the four remaining vertices of the cube serve as the vertices of the other. What fraction of the volume of the cube is the volume of the common part of these tetrahedra? | 13. The volume of the common part of the specified tetrahedra is $v=\frac{1}{6} a^{3}$, i.e., one sixth of the volume of the cube. The common part of the tetrahedra forms a regular octahedron. | \frac{1}{6}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,044 |
15. Each edge of a triangular pyramid (tetrahedron) is equal to $a$. Find the maximum area that the orthogonal projection of this pyramid onto a horizontal plane can have. | 15. The orthogonal projection of a tetrahedron can be either a triangle or a quadrilateral. In the first case, a regular tetrahedron is projected into a triangle that coincides with the projection of one of its faces. The largest projection in this case will be when the plane of the face is parallel to the projection p... | \frac{^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,046 |
16. How should a rectangular parallelepiped be positioned in space so that the area of its projection onto a horizontal plane is the largest?
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 16. Let's conduct a plane through the ends of three edges of a parallelepiped, emanating from one vertex, and arrange it so that this plane is horizontal. Then the projection of the parallelepiped will be a hexagon, the area of which, as is easily seen, will always be twice the area of the triangle, which is the projec... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,047 | |
18. Prove that the three segments connecting the midpoints of opposite edges of a tetrahedron intersect at one point and are bisected there. | 18. Each tetrahedron can be considered as a combination of three spatial quadrilaterals. By connecting the midpoints of the sides of these quadrilaterals with segments, we obtain three parallelograms, and the considered three segments will be the diagonals of these parallelograms. Therefore, they intersect at one point... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,049 |
19. Prove that there is no polyhedron having 7 edges. | 19. There are several solutions to this problem. Let's consider one of the possible ones. A polyhedron with seven edges cannot have faces with 4 or more sides, since 4 sides and at least 4 more edges coming out of four vertices would already give 8 edges; but such a polyhedron cannot be composed of triangles either. In... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,050 |
20. Prove that the sum of the angles of a spatial quadrilateral (not planar) is less than $360^{\circ}$. | 20. Let $A B C D$ be a spatial quadrilateral, $B D$ - its diagonal. Based on the properties of a trihedral angle, for the trihedral angle $B A C D$ we have the following relation: $\angle A B C \leqslant \angle A B D+\angle D B C$, and for the trihedral angle $D A B C$ we have: $\angle A D C \leqslant \angle A D B+\ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,051 |
23. Given a material sphere. The internal space of it is inaccessible, but constructions on its surface are possible. How, using only a compass and a ruler, to construct on paper a segment equal to the diameter of the sphere? | 23. Let us draw an arbitrary circle on the surface of a sphere from any point $A$ as the center.
Take any three points $M, \Lambda^{\prime}, P$ on this circle and, measuring the distances between them with a compass, transfer these distances to a plane. We will obtain a triangle $M^{\prime} N^{\prime} P^{\prime}$ on t... | A^{\}B^{\}=AB | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,054 |
25. What is the maximum number of parts that $n$ planes can divide space into? | 25. The maximum number of parts obtained by dividing space with $n$ planes is achieved under the condition that no two of these planes are parallel, no three pass through the same line or are parallel to the same line, and no four pass through the same point. Then $n$ planes can divide space into $\frac{n^{3}+5 n}{6}+1... | \frac{n^{3}+5n}{6}+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,056 |
27. Determine the type of section of a cube by a plane passing through its center and perpendicular to its diagonal. | 27. Such a section of a cube is a regular hexagon. | regularhexagon | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,058 |
28. Prove that the sum of the solid angles (in steradians) of the dihedral angles of a tetrahedron minus the sum of the solid angles of its trihedral angles is equal to \(4 \pi\). | 28. 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 trihedral angles of a tetrahedron is equal to twice the sum of all its dihedral angles minus $4 \pi$. But since the solid angle of a dihedral angle is measured by twice the linear, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,059 |
29. Prove that all four lines connecting the vertices of a tetrahedron to the points of intersection of the medians of the opposite faces intersect at one point (the centroid of the tetrahedron). These lines are called the medians of the tetrahedron. | 29. It is preliminarily necessary to prove that any two medians of a tetrahedron intersect. Let the tetrahedron be denoted by \(A_{1} A_{2} A_{3} A_{4}\), and the medians by \(A_{1} M_{1}, A_{2} M_{2}\), \(A_{3} M_{3}, A_{4} M_{4}\). Consider the medians \(A_{1} M_{1}\) and \(A_{3} M_{3}\). Through the vertices \(A_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,060 |
30. Prove that if two altitudes of a tetrahedron intersect, then the other two altitudes also intersect. | 30. Let the tetrahedron be denoted by $ABCD$, and its altitudes by $AH_{1}$, $BH_{2}$, $CH_{3}$, and $DH_{4}$. Consider the altitudes $BH_{2}$ and $DH_{4}$. Suppose they intersect at point $O$. Draw a plane $\alpha$ through them and the edge $BD$. This plane will be perpendicular to both the face $ACD$ and the face $AB... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,061 |
31. How many planes of symmetry do the following have: a) a cube; b) a regular tetrahedron? | 31. a) A cube has 9 planes of symmetry; b) a regular tetrahedron has 6 planes of symmetry. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,062 |
32. How many axes of symmetry do the following have: a) a cube; b) a regular tetrahedron | 32. a) A cube has 13 axes of symmetry; b) a regular tetrahedron has 7 axes of symmetry. | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,063 |
33. Prove that a sphere can always be inscribed in a tetrahedron. | 33. The center of the sphere inscribed in a tetrahedron is the point of intersection of the bisecting planes of the dihedral angles of this tetrahedron. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,064 |
34. Prove that a sphere can always be circumscribed around a tetrahedron. | 34. The center of the sphere circumscribed around a tetrahedron is the point of intersection of planes, each of which is perpendicular to an edge of the tetrahedron and bisects this edge. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,065 |
36. Construct a convex polyhedron from regular tetrahedra and regular octahedra. | 36. Let's take a regular tetrahedron \(ABCD\). Let \(M_1, M_2, M_3, M_4, M_5, M_6\) be the midpoints of its edges. Through each set of three of these six midpoints, lying on the edges of one trihedral angle, we draw a plane parallel to the planes of the faces of the tetrahedron. Our tetrahedron is divided into four sma... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,067 |
37. Fill the space with regular tetrahedra and regular octahedra. | 37. The previous problem (No. 36) shows that space can be filled with regular tetrahedra and octahedra by taking an unlimited number of regular tetrahedra and octahedra.
$$
\text { 38. The volume of a tetrahedron } v=\frac{1}{3} \sqrt{\frac{a^{2}+b^{2}-c^{2}}{2} \cdot \frac{a^{2}-b^{2}+c^{2}}{2} \cdot \frac{b^{2}+c^{2... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,068 |
39. Through point $P$, draw a line so that it passes through the inaccessible intersection point of two given lines $l_{1}$ and $l_{2}$.
40. Construct a plane that bisects a given dihedral angle, the edge of which is inaccessible. | 39. Let $l_{1}$ and $l_{2}$ be given lines and $P$ be a given point. We construct an arbitrary triangle, one of whose vertices lies at point $P$, and the other two vertices $Q$ and $R$ lie on lines $l_{1}$ and $l_{2}$ (triangle $P Q R$). We construct triangle $P_{1} Q_{1} R_{1}$, two vertices of which $Q_{1}$ and $R_{1... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,069 |
1. On the plane, there are three points $A, B$, and $C$, not lying on the same line. Find a point $M$ on the line passing through points $B$ and $C$, such that the sum of the squares of the distances from $M$ to $A, B$, and $C$ is the smallest. | 1. Drop a perpendicular $A D$ from point $A$ to line $B C$ and introduce the notations $A D=a, D B=b, D C=c, M D=x$. The problem reduces to finding the minimum value of a quadratic trinomial. The answer depends on the position of point $D$ relative to points $B$ and $C$. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,070 |
3. Given two parallel lines and two points $A$ and $B$, which are outside these parallel lines and located on opposite sides of them. Find the broken line of the smallest length connecting points $A$ and $B$, if the vertices of this broken line lie on the given lines and the segment of the broken line between both line... | 3. If the direction $A B$ coincides with the specified direction, then the broken line becomes the segment $A B$ - this will be its shortest length. Let $A M N B$ be the desired broken line (Fig. 26). We construct the parallelogram $A M N C$. The broken line $A M N B$ has the shortest length when $C B$ is the shortest.... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,072 |
4. (Viviani's Problem). Given two parallel lines $AB$ and $CD$, the distance between which is $b$, and two points: $M$ on $AB$ and $N$ on $CD$. On $AB$ from point $M$, a segment $ME=a$ is marked. What point $L$ on the line $MN$ should be connected to point $E$ so that the sum of the areas of triangles $ELM$ and $NLF$ i... | 4. Let $\boldsymbol{h}$ be the height of triangle $M L E$. Then the height of triangle $F L N$ will be $b-h, F N=x, x=\frac{a(b-h)}{h}$.
The sought double area $2 S=a h+\frac{a(b-h)^{2}}{h}$,
i.e., $2 S=a\left(h+\frac{b^{2}}{h}-2 b+h\right)=a\left(2 h+\frac{b^{2}}{h}-2 b\right)$.
The minimum value of the sought area... | \frac{b\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,073 |
5. Given two parallel lines and a point $\boldsymbol{A}$ between them, which is the vertex of a right angle of a right triangle, with the other two vertices lying on each of the parallel lines. What position should the triangle be in to have the smallest area? | 5. From Fig. 27, it is visible that
$$
A C=\frac{b}{\cos \varphi} ; \quad A B=\frac{a}{\sin \varphi} ; \quad S_{A B C}=\frac{a b}{\sin 2 \varphi}
$$
The minimum value of the area $S=a b$ is obtained when $\varphi=\frac{\pi}{4}$. | \varphi=\frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,074 |
6. On one side of an acute angle, two points $A$ and $B$ are taken. Find a point $C$ on the other side of the angle such that the angle $A C B$ is the largest. Construct the point $C$ using a compass and a straightedge. | 6. Through points $A$ and $B$, we draw a circle that touches the second side of the angle. The point of tangency will be the desired one. Indeed, $O C^{2}=O B \cdot O A$. The problem has been reduced to constructing the geometric mean of segments $O B$ and $O A$. $M O N$. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,075 |
7. Given an angle $M O N$ and two points $A$ and $B$. Find points $C$ and $D$ on the lines $M O$ and $N O$ respectively, such that the broken line $A C D B$ has the smallest length. | 7. Let's consider two cases of the positions of points $A$ and $B$ relative to the angle $M O N$:
a) $A$ and $B$ are outside the angle $M O N$. By symmetry, we find a point $C$ on $M O$ such that the sum of the distances from $C$ to $A$ and $B$ is the smallest. Then, by symmetry, we find a point $D$ on $N O$ such that ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,076 |
10. Prove that in any triangle $\frac{h_{a}}{l_{a}}+\frac{h_{b}}{l_{b}}+\frac{h_{c}}{l_{c}} \geqslant 9$, where $h_{a}, h_{b}$ and $h_{c}$ are the altitudes of the triangle, and $l_{a}, l_{b}$ and $l_{c}$ are the distances from an arbitrary point inside the triangle to its sides. | 10. Based on the result of the previous task and the inequality $\left(a_{1}+a_{2}+\ldots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right) \geqslant n^{2} \quad$ we can write
$$
\left(\frac{l_{a}}{h_{a}}+\frac{l_{b}}{h_{b}}+\frac{l_{c}}{h_{c}}\right)\left(\frac{h_{a}}{l_{a}}+\frac{h_{b}}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,079 |
11. Find a point inside a triangle such that the product of its distances to the sides of the triangle is maximized. | 11. Let $x, y, z$ be the distances from a point to the sides of a triangle, then $a x + b y + c z = 2 S$, where $S$ is the area of the triangle. $x y z$ reaches its maximum value when $a x b y c z$ does, and therefore $a x = b y = c z = \frac{2}{3} S$.
$x = \frac{\boldsymbol{h}_{a}}{3} ; y = \frac{\boldsymbol{h}_{b}}{... | x=\frac{\boldsymbol{}_{}}{3};y=\frac{\boldsymbol{}_{b}}{3};z=\frac{\boldsymbol{}_{}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,080 |
12. Which of all the isosceles triangles inscribed in a given semicircle has the greatest base, if one of the equal sides lies on the diameter, and the other is a chord? | 12. Let the desired triangle be $ABC$ ($AC$ lies on the diameter) and in it $AB=AC=x, \quad BC=y$. By drawing $BD \perp AC$, we find that $y^{2}=2 x^{2}-2 x AD$, but $AD=\frac{x^{2}}{2 R}$, so $y^{2}=2 x^{2}-2 x \frac{x^{2}}{2 R}=\frac{1}{R} x^{2}(2 R-x)$.
The sum $x+(2 R-x)=2 R$ is a constant, therefore the product $... | \frac{4}{3}R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,081 |
13. Given a right triangle, one of whose acute angles is equal to $\alpha$. Find the ratio of the radii of the circumscribed and inscribed circles and determine for which $\alpha$ this ratio will be the smallest? | 13. $2 R$ is equal to the hypotenuse of the triangle, but
\[
2 R=r \operatorname{ctg} \frac{\alpha}{2}+r \operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)=\frac{2 r}{\sqrt{2} \cos \left(\alpha-\frac{\pi}{4}\right)-1}, \text{ hence }
\]
\[
\frac{R}{r}=\frac{1}{\sqrt{2} \cos \left(\alpha-\frac{\pi}{4}\right... | \sqrt{2}+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,082 |
14. Prove that among all triangles with a common angle $\alpha$ at the vertex and a given sum of the lengths of the lateral sides $a+b$, the isosceles triangle has the smallest base. | 14. Let $a+b=q$, then $c^{2}=a^{2}+(q-a)^{2}-2 a(q-a) \cos \alpha=$ $=q^{2} \frac{1-\cos \alpha}{2}+2(1+\cos \alpha)\left(a-\frac{q}{2}\right)^{2} ; q$ and $a$ are constant, $c$ has the smallest value when $a=\frac{q}{2}=\frac{a+b}{2}$, i.e., when $a=b$. | b | Geometry | proof | Yes | Yes | olympiads | false | 35,083 |
15. Among all triangles with the same base and the same angle at the vertex, find the triangle with the maximum perimeter. | 15. Construct a segment on the segment $b$ that accommodates the angle $\beta$.
$$
\begin{gathered}
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} ; 2 p=a+b+c \\
2 p=\frac{b \sin A}{\sin B}+b+\frac{b \sin (A+B)}{\sin B}=b+\frac{b}{\sin \frac{B}{2}} \sin \left(A+\frac{B}{2}\right)
\end{gathered}
$$
Since $b>0, \si... | \triangleABC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,084 |
16. In triangle $ABC$, a point $D$ is taken arbitrarily on the base $AB$ or on its extension, and circles are circumscribed around triangles $ACD$ and $BCD$. Prove that the ratio of the radii of these circles is a constant. Find the position of point $D$ for which these radii will have the smallest value. | 16. Let $R$ and $R_{1}$ be the radii of the circumcircles of triangles $A C D$ and $C D B$, respectively. Then
$$
2 R=\frac{b}{\sin \varphi} ; \quad 2 R_{1}=\frac{a}{\sin (\pi-\varphi)} ; \quad \frac{2 R}{2 R_{1}}=\frac{b \sin (\pi-\varphi)}{a \sin \varphi}=\frac{b}{a}
$$
$\frac{R}{R_{1}}=\frac{b}{a}$ is a constant v... | \frac{R}{R_{1}}=\frac{b}{} | Geometry | proof | Yes | Yes | olympiads | false | 35,085 |
17. Determine the angles of an isosceles triangle for which the ratio $\frac{r}{R}$ is the greatest (smallest). | 17. Since $S=p r$, then $r=\frac{S}{p}=\frac{a \sin \frac{\alpha}{2} \cos \alpha}{\cos \frac{\alpha}{2}}$ (where $\alpha$ is the angle at the base). $\quad R=\frac{a}{2 \sin \alpha}$, but $b=2 a \cos \alpha$, then $\frac{r}{R}=4 \sin ^{2} \frac{x}{2}\left(1-2 \sin ^{2} \frac{\alpha}{2}\right)$.
Since $0<\alpha<\pi$, $... | \alpha=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,086 |
18. Construct a triangle of the smallest perimeter, two vertices of which lie on the sides of a given angle, and the third one at a given point inside the angle. | 18. Let $A_{1}$ and $A_{2}$ be the points symmetric to point $A$ with respect to the sides of angle MON. The points of intersection of segment $A_{1} A_{2}$ with the sides of the angle give the vertices of the desired triangle. $A_{1} C=C A, D A=D A_{2}, A_{1} A_{2}$ is a straight line.
$2 p=A C+C D+D A=A_{1} C+C D+D ... | 2p=AC+CD+DA=A_{1}C+CD+DA_{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,087 |
19. In the plane of an equilateral triangle $A B C$, a point $P$ is given. It is known that $A P=2, B P=3$. Find the maximum length of the segment $P C$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 19. Let $P_{1}$ and $P_{2}$ be the points of intersection of the circles drawn from point $A$ with radius 2 and from point $B$ with radius 3. The problem is possible if the side of the triangle $a<5$. The greatest distance to $C$ will be from the point that lies outside the triangle; let this point be $P_{1}$, then
$$... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,088 | |
21. What is the maximum area that a triangle with sides \(a, b, c\) can have, given that the sides are within the following limits: \(0 \leqslant a \leqslant 1 \leqslant b \leqslant 2 \leqslant c \leqslant 3\) ? | 21. $S=\frac{1}{2} a b \sin \alpha$. Takes the maximum value when $a=1, b=2, \sin \alpha=1$,
$S=\frac{1}{2} \cdot 1 \cdot 2 \cdot 1=1$. Under these conditions, $c=\sqrt{1+4}=\sqrt{5}$, which satisfies the problem's condition ( $2 \leqslant c \leqslant 3$). | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,090 |
22. Given a regular pentagon. $M$ is an arbitrary point inside it (or on the boundary). Let's number the distances from point $M$ to the sides of the pentagon in ascending order: $r_{1} \leqslant r_{2} \leqslant r_{3} \leqslant$ $\leqslant r_{4} \leqslant r_{5}$. Find all positions of point $M$ for which the length $r_... | 22. Let $A B C D E$ be a regular pentagon. Draw radii from its center to all vertices and perpendiculars to the sides. Then the pentagon will be divided into 10 equal triangles. It is sufficient to consider the case when point $M$ lies inside or on the side of one of these triangles (for example, $O A K$, where $K$ is ... | M | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,091 |
23. From a rectangle with sides $a$ and $b$, a triangle with legs $a_{1}$ and $b_{1}$ has been cut off. How should the remaining part be trimmed to obtain a rectangle of the largest area with sides parallel to the sides of the original rectangle? | 23. Let a triangle with vertex $C$ be cut off from rectangle $ABCD$, resulting in pentagon $ABEFD$. The desired rectangle $AB_1C_1D_1$ should have vertex $C_1$ on line $EF$. The problem is to find the position of vertex $C_1$. Extend sides $AB$ and $AD$ of the rectangle to intersect with the extension of segment $EF$, ... | \frac{b}{2}+\frac{b_1}{2a_1}(-a_1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,092 |
25. From the circular sectors of a given perimeter, find the sector of the largest area. | 25. If $2 p$ is the perimeter, $R$ is the radius, and $l$ is the arc of the sector, then $2 p=2 R+l$, where $l=R \alpha, S=\frac{\pi R^{2}}{2 \pi} \alpha=R(p-R)$. Since the sum $R+(p-R)=p$ is constant, the sector has the maximum area when $R=\frac{p}{2}$, then $S=\frac{1}{4} p^{2}$. | \frac{1}{4}p^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,094 |
27. Find a circular segment such that its chord is equal to the length of the largest circle inscribed in it. | 27. Let $AB$ be a chord, $r$ be the radius of the largest inscribed circle, $\angle AOB = \alpha$, $N$ be the point where the circle touches the chord, and $M$ be the point of tangency with the larger circle. According to the problem, $AN = \pi r$, and on the other hand,
$$
AN^2 = MN(2R - MN), \quad \text{then} \quad ... | \alpha\approx12256' | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,096 |
30. For what dimensions does an open bathtub, having the shape of a half-cylinder with surface area $S$, have the greatest capacity | 30. Denoting the length of the bathtub by $l$, and the radius of the base by $r$, we can write
$$
S=\pi r^{2}+\pi r l, v=\frac{1}{2} \pi r^{2} l
$$
or $4 \pi v^{2}=\pi r^{2}\left(S-\pi r^{2}\right)^{2}$. Since the sum $\pi r^{2}+\left(S-\pi r^{2}\right)=S$ is a constant, $4 \pi v^{2}$, and therefore $v$, attains its ... | r=\sqrt{\frac{S}{2\pi}};=2r | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,099 |
31. Inscribe in the given sphere of radius $R$ a cylinder with the largest lateral surface area. | 31. $H=R \sqrt{2}$. | R\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,100 |
33. Determine $a$ so that the sum of the squares of the roots of the equation $x^{2}+(2-a) x-a-3=0$ is the smallest. | 33. $a=1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,102 |
34. Determine the coefficients of the quadratic trinomial $a x^{2}+$ $+b x+c$, knowing that its maximum value is 25 at $x=\frac{1}{2}$ and that the sum of the cubes of its roots is 19. | 34. $a=-4, b=4, c=24$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
34. $a=-4, b=4, c=24$. | =-4,b=4,=24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,103 |
35. Find the maximum value of the expression $\log _{2}^{4} x+12 \log _{2}^{2} x \log _{2} \frac{8}{x}$, assuming that $x$ varies from 1 to 64. | 35. Let's denote the given expression by $y$. Then it is easy to see that $y=\log _{2}^{2} x\left(6-\log _{2} x\right)^{2}$. But $1<x<64$, so $0<\log _{2} x<6$. The maximum value of $y$ coincides with the maximum value of the product. Since $\log _{2} x+(6-$ $\left.-\log _{2} x\right)=6$, the product will be the larges... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,104 |
36. Find the maximum value of the function $y=\frac{x}{a x^{2}+b}(a>0, b>0)$. | 36. $\frac{a x^{2}+b}{2} \geqslant \sqrt{a x^{2} b}=x \sqrt{a b}$ for all $x>0$, therefore
$$
y=\frac{a}{a x^{2}+b} \leqslant \frac{x}{2 x \sqrt{a b}}=\frac{1}{2 \sqrt{a b}}
$$
The maximum value of $y$ is achieved when the denominator $a x^{2}+b$ is at its minimum, i.e., when $a x^{2}+b=2 x \sqrt{a b}$. It is equal t... | \frac{1}{2\sqrt{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,105 |
37. Find the minimum value of the function $\varphi(x)=|x-a|+|x-b|+$ $+|x-c|+|x-d|$, where $a<b<c<d$ are fixed real numbers, and $x$ takes arbitrary real values.
Hint. It is convenient to reason by marking the numbers $a, b, c, d$ on the number line. | 37. We divide the number line by points $a, b, c, d$ into intervals and compute the value of $\varphi(x)$ on each interval. Let the number $x$ correspond to the point $M$.
a) $x \leqslant a, \varphi(x)=M A+M B+M C+M D=4 M A+3 A B+2 B C+C D$
The last three terms are constant, only the length $M A$ varies. $\varphi(x)$ ... | +-b- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,106 |
38. Among the complex numbers $Z$, satisfying the condition $|Z-25 i| \leqslant 15$, find the number with the smallest argument. Make a drawing. | 38. The modulus of the difference of two complex numbers $\left|z_{1}-z_{2}\right|$ is equal to the distance between the corresponding points in the complex plane. The condition $|z-25 i|<15$ is satisfied by points in the complex plane lying inside and on the boundary of a circle with center at the point $z_{0}=25 i$ a... | 12+16i | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,107 |
39. What is the greatest value that the modulus of a complex number $Z$ can take, satisfying the equation $\left|Z+\frac{1}{z}\right|=1$? | 39. Let $z=r(\cos \varphi+i \sin \varphi), z^{2}=r^{2}(\cos 2 \varphi+i \sin 2 \varphi)$,
$$
\begin{gathered}
\left|z^{2}+1\right|=\sqrt{\left(r^{2} \cos 2 \varphi+1\right)^{2}+\left(r^{2} \sin 2 \varphi\right)^{2}}=\sqrt{r^{4}+2 r^{2} \cos 2 \varphi+1} \\
\left|z+\frac{1}{z}\right|=\frac{\left|z^{2}+1\right|}{|z|}=\f... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,108 |
43. Points $A$ and $B$ are located on a straight highway running from west to east. Point $B$ is 9 km east of $A$. A car departs from point $A$ heading east at a speed of 40 km per hour. At the same time, a motorcycle departs from $B$ in the same direction with a constant acceleration of 32 km/hour ${ }^{2}$. Determine... | 43. At time $t$, the car is at a distance of $40 t$ km from point $A$, and the motorcycle is at a distance of $\left(\frac{32 t^{2}}{2}+9\right)$ km from the same point. The distance between them is $\left|16 t^{2}+9-40 t\right|$ (Fig. 29). The greatest distance of 16 km will be reached in 1 hour 15 minutes. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,111 |
44. Find the smallest positive period and the maximum value of the function
$$
y=3 \sin 2 x+4 \cos 2 x
$$ | 44. Let's transform the function as follows:
$$
y=3 \sin 2 x+4 \cos 2 x=5 \sin (2 x+\alpha)
$$
where $\alpha=\arcsin \frac{4}{5}$.
From this, it is clear that the maximum value of the function is 5. The smallest positive period is $\pi$. | 5,\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,112 |
46. Find the minimum and maximum values of the function $\varphi(x)=\sin ^{6} x+\cos ^{6} x$. | 46. $\varphi(x)=\frac{1+3 \cos ^{2} 2 x}{4}$. The maximum value of $\varphi(x)$ is 1, and the minimum value is $\frac{1}{4}$.
Fig. 29 | Themaximumvalueof\varphi(x)is1,theminimumvalueis\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,114 |
47. Find the smallest and largest values of the product $y=\cos ^{p} x \sin ^{q} x$ on the segment $0 \leqslant x \leqslant \frac{\pi}{2}$, where $p$ and $q$ are given positive rational numbers. | 47. On the segment $0 \leqslant x \leqslant \frac{\pi}{2}$, the values of $\cos x$ and $\sin x$ are positive, which means that $\cos ^{p} x$ and $\sin ^{q} x$ are positive, i.e., $\cos ^{p} x \sin ^{q} x \geqslant 0$. The minimum value $y=0$ is obtained at $x=0, x=\frac{\pi}{2}$. The maximum value of $y$ is the maximum... | \operatorname{arctg}\sqrt{\frac{q}{p}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,115 |
48. An electric lamp is suspended on a pulley above the center of a round table. At what height should this lamp be fixed to obtain the greatest illumination at the edges of the table? | 48. Illuminance is given by the formula $E=\frac{k \sin \alpha}{A C^{2}}$, where $k$ is the proportionality coefficient; $A C$ is the distance from the lamp to the edge of the table. But $A C=\frac{R}{\cos \alpha}$, where $\alpha$ is the angle of inclination of $A C$ to the plane of the table. Then
$$
E=\frac{k \sin \... | 0.7R | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,116 |
49. The picture is hung on the wall so that its lower edge is $a$ m, and the upper edge is $b$ m above the observer's eye. At what distance $x$ from the wall should the observer stand so that the picture is seen at the largest angle? | 49. Let $A C=b, \quad A B=a, \quad A D=x$. From Fig. 30, it is clear that $b=x \operatorname{tg}(\alpha+\beta)=$ $=x \cdot \frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}$, but $\operatorname{tg} \beta=\frac{a}{x}$, then $\operatorname{tg} \alpha=$ $=\frac{b-a... | \sqrt{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,117 |
2. On the table, a bottle, a mug, a cup, a glass, and a jug are placed in a row. They contain different drinks: tea, coffee, milk, lemonade, and mineral water. If the glass is moved so that it is between the containers with tea and milk, then the container with milk will be next to the container with lemonade, and the ... | 2. In the bottle, there is mineral water; in the mug, there is tea; in the cup, there is milk; in the glass, there is coffee; in the pitcher, there is lemonade. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,119 |
7. Out of three people standing next to each other, one always tells the truth (truth-teller), another always lies (liar), and the third, depending on the circumstances, tells the truth or lies (the "diplomat"). The person on the left was asked: "Who is standing next to you?" He answered: "The truth-teller." The person... | 7. The liar was in the center, to the right of him was the one speaking the truth, and to the left 一 "the diplomat". | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,124 |
12. In one of the institutes, 4 friends are studying. The youngest is studying on the I course, and the oldest - on the IV. Determine the name and surname of each of them, as well as the course they are studying, if it is known that Boris is a personal scholarship recipient, Vasily must go to Omsk for practice in the s... | 12. On the $1-\mathrm{st}$ year studies Krylov Petr, on the $2-\mathrm{nd}$ - Ivanov Nikolay, on the $3-\mathrm{rd}$ - Karpov Boris and on the $4-\mathrm{th}$ - Orlov Vasily. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,129 |
18. On the table, there are three identical boxes. In one of them, there are two black balls, in another - a black and a white ball, and in the third - two white balls. On the lids of the boxes, there are inscriptions: "2 black", "2 white", "White and black", but it is known that none of these inscriptions are correct.... | 18. You need to take the box labeled "black and white" and take out one ball. 1) If this ball turns out to be white, then the second ball in this box must also be white. The box labeled "2 black" should contain one white and one black ball, and the third box (labeled "2 white") should contain two black balls. 2) If the... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,135 |
19. After Don Diego's death, there were three chests of treasures and the following will: "The chests should be transferred to the ownership of Don Alvaro, Don Benito, and Don Vicente according to their choice, which is determined by the following conditions: 1) The one who saw me in a green cloak should not choose bef... | 19. The order of choosing the boxes should be as follows: first - Don Alvaro, then - Don Vincente, and last - Don Benito. Don Vincente or Don Benito could have seen Don Diego in a green cloak, the snuffbox was given to him by Don Benito or Don Vincente, and the first to carry a sword was Don Alvaro. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,136 |
21. Four athletes competed in a running race. After the competition, each of them was asked what place they took. The first athlete answered: "I was neither first nor last." The second: "I was not first." The third: "I was first." The fourth: "I was last." Only three of these answers were correct. Who lied? Who took th... | 21. The fourth one told a lie. The first place was taken by the third athlete. | The\third\athlete | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,138 |
23. Five schoolchildren participated in a cycling race. Before the race, four fans made the following assumptions about the distribution of places: 1) Sergey will take second place, and Kolya third; 2) Nadya will take third place, and Tolya fifth; 3) Tolya will take first place, and Nadya second; 4) Sergey will take se... | 23. 1st place was taken by Sergei, 2nd - Nadya, 3rd - Kolya, 4th - Vanya, 5th - Tolya. | 1stplacewastakenSergei,2nd-Nadya,3rd-Kolya,4th-Vanya,5th-Tolya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,140 |
24. When taking exams with an electrical examiner, 5 questions appear on the screen. For each question, Petya must answer "yes" or "no" by pressing buttons. Observing the examiner's operation, he determined that there are always more questions that require an affirmative answer and that there are never three consecutiv... | 24. According to the condition of the problem, the possible sequences of answers are: 1) no, yes, yes, no, yes; 2) no, yes, no, yes, yes; 3) yes, yes, no, yes, no; 120
4) yes, no, yes, yes, no. Of these, only one contains a different answer to the second question (the 4th). This is what Pete used. (He needed to answer ... | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,141 |
26. Three pioneers - Burov, Gridnev, and Klimenko, named Kolya, Petya, and Grisha, arrived at the pioneer camp. It is known that: 1) Burov is not Kolya; 2) the brother of Burov's mother has the surname Serov; 3) Petya started school at the age of seven, always studied well, and this year will begin studying algebra and... | 26. Burov - Grisha, he is 13 years old. Gridnev - Kolya, he is 13 years old. Klymenko Petya, he is 12 years old. | Burov-Grisha,13;Gridnev-Kolya,13;KlymenkoPetya,12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,143 |
27. The names of three friends are Kostya, Vasya, and Kolya. Their surnames are Semyonov, Burov, and Nikolaev. Whose surname is whose - is unknown. But it is known that Semyonov's grandfather is a biological brother of their neighbor Petrov, Kostya is a year older than Kolya, and Kolya is a year older than Nikolaev. Th... | 27. From the condition of the problem, we immediately conclude that Nikolaev's name is Vasya. Next, we note that Kolya's mother is Korobova, and therefore the mother of the third boy, Kostya, must have a maiden name different from Korobova. However, Semyonov's mother's maiden name is not Korobova, but Petrova, since he... | Vasya:16,Kolya:17,Kostya:18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,144 |
28. Vasya, Petya, and Grisha are inseparable friends. Their last names are Lepeshkin, Vatrushkin, and Bublikov. One day, Vasya and Grisha went to Petya's house. "Let's solve chess problems," Vasya suggested. Just as the friends set up the pieces, there was a knock at the door and the postman came in. "A letter for Vatr... | 28. Vasya's last name is Bublikov, Petya's is Vatrushkin, and Grisha's is Lepeshkin. | Vasya'lastnameisBublikov,Petya'isVatrushkin,Grisha'isLepeshkin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,145 |
29. Three people are sitting next to each other. Each of them answers any question asked either always correctly or always incorrectly. I approached them and asked the middle one if he always answers questions correctly. But I didn't hear the answer and so I asked the ones sitting on the left and right what the middle ... | 29. The one standing in the middle, when asked: "Does he answer questions correctly?" - will answer that he does, regardless of whether he always tells the truth or always lies. This means that the one standing on the left always gives correct answers, while the one standing on the right gives incorrect answers. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,146 |
30. Several families lived in one house. In total, there are more children in these families than adults: there are more adults than boys; more boys than girls; and more girls than families. There are no childless families, and no families have the same number of children. Each girl has at least one brother and at most... | 30. Three families lived in the house. In one of these families, there is a single child, a boy. In another family - two girls and a boy. In the third - two girls and three boys. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,147 |
31. Sergey is twice as old as Sasha will be when Tolya is as old as Sergey is now. Who is the oldest among them? Who is the youngest? Who is the middle one in terms of age? | 31. The oldest is Sergei, the middle one by age is Tolya, and the youngest is Sasha. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,148 |
32. During the lunch break, the members of the communist labor brigade started talking about how many newspapers each of them reads. It turned out that each member subscribes to and reads exactly two newspapers, each newspaper is read by five people, and any combination of two newspapers is read by one person. How many... | 32. 6 newspaper names, 15 members. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,149 |
39. Objects $A, B, C$ and $D$ need to be compared by weight using a two-pan balance. When $A$ and $B$ were placed on one pan and $C$ and $D$ on the other, equilibrium was established. However, when $C$ and $A$ were moved from one pan to the other, the pan with $D$ and $A$ sank. When $B$ was placed on one pan and $A$ an... | 39. In ascending order by weight, the items are arranged as follows: $C, A, B, D$.
40, 41. We should use the Pigeonhole Principle. Let's solve, for example, problem 41. Let $n_{1}, n_{2}, n_{3}, \ldots, n_{100}$ be arbitrary integers. Consider the following 100 numbers: $n_{1}, n_{1}+n_{2}, n_{1}+n_{2}+n_{3}, \ldots, ... | C,A,B,D | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,156 |
46. 200 students lined up in a rectangle with 10 people in each transverse row and 20 people in each longitudinal row. In each longitudinal row, the tallest student was chosen, and from them, the shortest one was selected. Similarly, in each transverse row, the shortest student was chosen, and from them, the tallest on... | 46. Let's introduce the following notations: $A$ - the shortest of all the tallest students, $B$ - the tallest of all the shortest students. There are 3 and only 3 cases. 1) $A$ and $B$ stood in the same transverse row. In this case, $B$ is the shortest in this row. Therefore, $B$ is shorter than $A$. 2) $A$ and $B$ st... | A>B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,159 |
47. On a plane, there are 25 points. Among any 3 of them, there are 2 whose distance is less than 1. Prove that there exists a circle of radius 2 that contains at least 13 of these points. | 47. Let's take two of the given points, the distance between which is greater than 1, and, taking them as centers, describe circles with radii equal to 1. Each of the remaining 23 points must be at a distance of less than 1 from at least one of the two initially chosen points, that is, it will lie in one of the two con... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,160 |
48. On the plane, there are $2 n$ points. Among any three of them, there are two such that the distance between them is no more than 1. Prove that the plane can be covered by two circles of radius 1, which will cover all these points. | 48. From the given points, you need to choose two points such that the distance between them is the greatest (more than 1), and, taking each of them as the center, describe a circle with a radius of 1. Then proceed to reason as in solving problem № 47. | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,161 |
49. On a plane, there are $n$ points, no three of which lie on the same straight line. Prove that it is possible to construct a non-self-intersecting closed broken line with vertices at these points. | 49. One of the solutions might be, for example, as follows: choose one of the given points, draw a ray from it, and rotate the ray. The rotating ray will sequentially pass through the given points. It remains to connect the given points with segments in the order in which they are passed by the ray, with the last point... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,162 |
51. There are three villages: $A, B$, and $C$. The inhabitants of $A$ always tell the truth, the inhabitants of $B$ always lie, and the inhabitants of $C$, depending on their mood, tell the truth or lie. Since the villages are located close to each other, the inhabitants visit each other. A tourist ended up in one of t... | 51. Four questions: 1) Am I in one of the settlements $A$ and $B$? 2) Am I in settlement C? 3) Do you live in settlement C? 4) Am I in settlement $A^{*}$? | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,164 |
52. My friend has thought of an integer between 10 and 19. To guess which number he has in mind, I can ask him questions that he will answer with "yes" or "no": What is the smallest number of questions and which specific questions can I ask to determine which number he thought of? | 52. The least number of questions is three. The first question: "Is the number you are thinking of among the first four numbers (11-14)?" If the answer is "Yes," then the second question can be: "Is the number you are thinking of among the numbers 11 and 12?" If the answer is "No," then the third question can be: "Is t... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,165 |
53. To find out my friend's birthday, I should ask him questions to which he will only answer "yes" or "no." What is the minimum number of questions I need to ask? What questions should I ask? | 53. Nine questions: four questions to find out the month, and five questions to find out the day. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,166 |
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