problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
44*. Let $T_{1}$ and $T_{2}$ be two triangles with sides $a_{1}, b_{1}, c_{1}$, and $a_{2}, b_{2}, c_{2}$, respectively, and let $a=\sqrt{\frac{a_{1}^{2}+a_{2}^{2}}{2}}$, $b=\sqrt{\frac{b_{1}^{2}+b_{2}^{2}}{2}}, \quad c=\sqrt{\frac{c_{1}^{2}+c_{2}^{2}}{2}} . \quad$ Prove that there exists a triangle $T$ with sides $a, ... | 44. In solving this problem, the algebraic inequality (err ${ }^{1}$ )
$$
(A X+B Y)^{2} \leqslant\left(A^{2}+B^{2}\right)\left(X^{2}+Y^{2}\right)
$$
(where $A, B, X, Y$ are positive numbers), the validity of which follows from the fact that
$$
\begin{aligned}
\left(A^{2}+B^{2}\right)\left(X^{2}+Y^{2}\right)- & (A X+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,141 |
46. The projections of the polygon $M$ on the $O x$ axis, on the bisector of the 1st and 3rd coordinate angles, on the $O y$ axis, and on the bisector of the 2nd and 4th coordinate angles are respectively $4, 3 \sqrt{2}, 5, 4 \sqrt{2}$. The area of the polygon is $S$. Prove that:
a) $S \leqslant 17.5$
b) if the polyg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,143 | |
48. Two equal rectangles are placed on a plane such that their contours intersect at 8 points. Prove that the area of the common part of these rectangles is greater than half the area of each of them. | 48. Let the rectangle $A_{1} B_{1} C_{1} D_{1}$ with sides $A_{1} B_{1}=a$ and $A_{1} D_{1}=b$ and area $S$ cut off triangles $A K L, B M N, C P Q$ and $D R T$ from an equal rectangle $A B C D$ (Fig. 101, a). Extend these four triangles to rectangles $A K L A^{\prime}, B M N B^{\prime}, C P Q C^{\prime}$ and $D R T D$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,145 |
49. A raft is floating through a straight channel that is 1 unit wide and turns $90^{\circ}$ from its initial direction (see Fig. 12). Prove that if the area of the raft is $2 \sqrt{2}$ (or larger), it will definitely not be able to turn the corner and continue moving through the channel.
The estimate obtained in prob... | 49. Enclose the raft $F$, which we assume has an area $\geqslant 2 \sqrt{2}$ and has not yet reached the bend of the channel, in the smallest possible rectangle $P \equiv \Lambda_{1} B_{1} B_{2} A_{2}$,
 by area is smaller than at least one of its faces. | 50. Let us preliminarily prove that if in space there are two non-parallel lines \(a\) and \(b\), and from three consecutive points \(A_1, A_2, A_3\) on line \(a\) perpendiculars \(A_1B_1, A_2B_2, A_3B_3\) are dropped to line \(b\), then \(A_2B_2\) is less than one of the two segments \(A_1B_1\) and \(A_3B_3\). Indeed,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,147 |
52. Given a triangle $T$.
a) Place a centrally symmetric polygon $m$ of the largest possible area inside $T$.
What is the area of $m$ if the area of $T$ is 1?
b) Enclose $T$ in a convex centrally symmetric polygon $M$ of the smallest possible area.
What is the area of $M$ if the area of $T$ is 1? | 52. a) If $O$ is the center of the sought centrally symmetric polygon $m$, then by symmetrically reflecting the given $\triangle ABC \equiv T$ together with $m$ (which will transform into itself), we can see that $m$ is also inscribed in $\triangle A_{1} B_{1} C_{1} \equiv T_{1}$, which is symmetric to $\triangle ABC$ ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,149 |
54. a) In a unit square $K$, there is some figure $\Phi$ (which may consist of several separate pieces). Prove that if $\Phi$ does not contain a pair of points whose distance is 0.001, then the area of $\Phi$ does not exceed 0.34.
b) Can you improve the estimate of the area of $\Phi$ given in the condition of part a)? | 54. a) Let's consider the sides of the square $K$ to be horizontal and vertical, and parallel shift the figure $\Phi$ by a distance of 0.001 in the direction of one of the sides of the square, say, from left to right. Since $\Phi$ does not contain points that are 0.001 units apart, the figure $\Phi'$ obtained by this s... | 0.287 | Geometry | proof | Yes | Yes | olympiads | false | 25,151 |
56. a) A segment of length 1 is completely covered by a certain number of smaller segments. Prove that from these segments, one can select several non-overlapping segments such that the sum of the lengths of the selected segments is $\geqslant \frac{1}{2}$.
b) A square with side 1 is covered by a certain number of sma... | 56. a) First, we will exclude from consideration all those segments of the covering that are entirely covered by one or several of these segments. After this, we will enumerate all the remaining segments in a certain order as follows. We will assume that our original segment \( O \) of length 1 is positioned horizontal... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,153 |
58. In a rectangle of area 5 sq. units, nine polygons of area 1 are placed; prove that among them there are two polygons, the area of the common part of which is not less than $\frac{1}{9}$. | 58. Let the areas of all pairwise intersections of the considered polygons $M_{1}, M_{2}, \ldots, M_{9}$ be less than $\frac{1}{9}$; then the area of the part of polygon $M_{2}$ not covered by polygon $M_{1}$ is greater than $1-\frac{1}{9}=\frac{8}{9}$; the area of the part of polygon $M_{3}$ not covered by either poly... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,155 |
61. Prove that
a) every convex polygon of area 1 can be enclosed in a parallelogram of area 2;
b) a triangle of area 1 cannot be enclosed in a parallelogram of area $<2$.
$62^{*}$. Prove that
a) every convex polygon of area 1 can be enclosed in a triangle of area 2;
b) a parallelogram of area 1 cannot be enclosed ... | 61. a) Let $AB$ be a side of a convex polygon $M$ of area $1$, and $C$ be the point of the polygon $M$ farthest from $AB$ (or one of such points if $M$ has a side parallel to $AB$). The line $AC$ divides the polygon $M$ into two parts $M_{1}$ and $M_{2}$ (Fig. 119; one of these parts may not exist). If $D_{1}$ and $D_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,158 |
63. Let $M$ be a convex polygon of area $S$ and $l$ be an arbitrary line. Prove that
a) in $M$ one can inscribe a triangle, one side of which is parallel to $l$ and the area of which is $\geqslant \frac{3}{8} S$;
1) See, for example, the book [22], entirely devoted to solving the following rather specific problem: wh... | 63. a) Let's draw two lines parallel to $l$ and located on opposite sides of $M$, and then shift them so that they pass through some vertices $A$ and $B$ of the polygon; in this way, we obtain a "strip" around $M$ bounded by two parallel lines $l_{1}$ and $l_{2}$ passing through $A$ and $B$ (Fig. 123). Let the distance... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,159 |
65. In the same circle $K$, a convex polygon $M=A_{1} A_{2} \ldots A_{n}$ and a convex polygon $M^{\prime}=A_{1}^{\prime} A_{2}^{\prime} \ldots A_{n}^{\prime}$ are inscribed, the vertices of which are the midpoints of the arcs subtended by the sides of the $n$-gon $M$ (Fig. 23, b). Prove that if $P$ and $S$, respective... | 65. Since $\triangle O A_{1} A_{2}$, formed by the center $O$ of the circle $K$ with radius 1 and two points $A_{1}$ and $A_{2}$ on it, such that $\angle A_{1} O A_{2}=2 \alpha$ (see Fig. 23,6 on p. 58), obviously has the base $A_{1} A_{2}=$ $=2 \sin \alpha_{1}$ and area $S_{O A_{1} A_{2}}=\frac{1}{2} \sin 2 \alpha_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,161 |
67. a) Two points $A$ and $B$ in a plane, the distance between which is 1, are connected by a convex broken line $A P_{1} \ldots P_{n} B$ (i.e., such a broken line that the polygon $A P_{1} P_{2} \ldots P_{n} B$ is convex). Prove that if the sum of the exterior angles of the broken line at points $P_{1}, P_{2}, \ldots,... | 67. a) First of all, note that if the considered broken line consists of only two segments \( A P \) and \( P B \) (Fig. 129, a), then the theorem is valid. Indeed, extend the line \( A P \) beyond point \( P \) and on the extension, lay off the segment \( P Q = P B \). Since \(\triangle B P Q\) is isosceles, \(\angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,163 |
70**. a) Prove that among all convex quadrilaterals with given angles and a given perimeter, the one with the largest area is the quadrilateral in which a circle can be inscribed.
b) Prove that among all convex $n$-gons with given angles and a given perimeter, the one with the largest area is the $n$-gon in which a ci... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,166 | |
71. Prove that among all $n$-gons inscribed in a given circle $S$, the regular $n$-gon has
a) the largest area;
b) the largest perimeter. | 71. a) If an n-sided polygon \( M \) inscribed in a circle \( S \) is not regular, then it must have a side that is smaller than the side \( a \) of the regular n-sided polygon \( M_{0} \) inscribed in \( S \). Furthermore, we can also assume that \( M \) has a side larger than \( a \): indeed, this is not the case onl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,167 |
73. Let $R$ and $r$ be the radii of the circumscribed and inscribed circles of a convex $n$-gon (see Fig. 24, a and b). Prove that
$$
\frac{R}{r} \geqslant \sec \frac{180^{\circ}}{n}
$$
In which case is the ratio $\frac{R}{r}$ exactly equal to $\sec \frac{180^{\circ}}{n}$?
In the case $n=3$, from the theorem of prob... | 73. From the formulas (see p. 65)
$$
2 n R \sin \frac{180^{\circ}}{n} \geqslant P \quad \text { and } \quad P \geqslant 2 n r \operatorname{tg} \frac{180^{\circ}}{n}
$$
where $P$ is the perimeter of the $n$-sided polygon $M$, it directly follows that
$$
2 n R \sin \frac{180^{\circ}}{n} \geqslant 2 n r \operatorname{... | \frac{R}{r}\geqslant\\frac{180}{n} | Geometry | proof | Yes | Yes | olympiads | false | 25,169 |
76. Prove that a circle has a greater area than any convex polygon with the same perimeter.
The result of problem 76 makes the following (famous!) "isoperimetric theorem for convex figures" highly plausible: among all plane convex figures of a given perimeter ${}^{1}$), the circle has the largest area—indeed, by the t... | 76. We have already noted several times that for a regular
$$
\frac{P}{S^{2}}=4 n \operatorname{tg} \frac{180^{\circ}}{n}
$$
(see p. 67), and for an $n$-sided polygon that is not regular,
$$
\frac{P^{2}}{S}>4 n \operatorname{tg} \frac{180^{\circ}}{n}
$$
On the other hand, for a circle with area $S=\pi r^{2}$ and pe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,171 |
80. Let $P$ be the perimeter of a convex polyhedron $M$, i.e., the sum of the lengths of all its edges, and $D$ be the diameter of $M$. Prove that $P > 3D$. | 80. Let $A_{1}$ and $A_{2}$ be two farthest vertices of $M$; $A_{1} A_{2}=D$ is the diameter of $M$ (see problem 78). Draw planes perpendicular to $A_{1} A_{2}$ through all the vertices of the polyhedron; it is clear that the "outermost" of these will be the planes $\pi_{1}$ and $\pi_{2}$, passing through $A_{1}$ and $... | P>3D | Geometry | proof | Yes | Yes | olympiads | false | 25,174 |
81. A branch of diameter $d$ (it can have any shape; we neglect the thickness of the branch) is floating through a channel of width 1, which at a certain point turns $90^{\circ}$ relative to its initial direction (Fig. 33). What is the maximum $d$ for which the branch can have any shape and still pass the turn without ... | 81. First solution. Draw a circle through the "inner angle" M of the channel, touching its "outer" banks at points \(A_{0}\) and \(B_{0}\) (Fig. 142a); it is clear that the radius \(R = O A_{0} = O M = O B_{0}\) of this circle is determined from the relationship (note that the side of the square Oaмb is \(R-1\))
\[
2(... | 2\sqrt{2}+2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,175 |
84. All links of a closed n-link broken line (possibly self-intersecting, but "non-degenerate", meaning all vertices of the broken line are distinct and all links belong to different lines) are equal to 1.
a) What is the smallest possible value of the diameter of the broken line if \( n=2,3,4,5,6 \) ?
b) For which \(... | 84. a) It is clear that the diameter of the considered broken line cannot be less than 1. It is not difficult to see that for \( n=3 \) and \( n=5 \) it can equal 1 - examples can be a regular triangle with side 1 (Fig. 145, \(\alpha\)) and a regular pentagram (Fig. 145, \(\beta\); note that for \( n=3 \) the diameter ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,178 |
88. Let $A_{1}, A_{2}, \ldots, A_{n}$ be $n$ consecutive points on a straight line; we will assume that $A_{1} A_{n}=1$ and define some number $a$ (where $0 \leqslant a \leqslant 1$). What is the maximum possible number of those from $C_{n}^{2}=\frac{n(n-1)}{2}$ segments $A_{i} A_{j}$ (where $i, j=1,2, \ldots$ or $n$),... | 88. We will use the notations specified on p. 78, not forgetting that we are talking about points on one line. To grasp the general pattern, we will start with small values of $n$.
$1^{\circ}$. It is clear that for $n=2$ we have: $N(a, 2)=1$ for all $a$.
$2^{\circ}$. Of course, $N(1, n)=1$ for any number of points $n... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,182 |
90. On a plane, there are 4 points $A_{1}, A_{2}, A_{3}, A_{4}$, the distance between any two of which is not less than 1. What is the maximum possible number of line segments $A_{i} A_{j}$ of length 1 connecting these points pairwise? | 90. The number $v_{2}(4)<6$, otherwise the number of diameters of a system of four points on a plane could equal the "complete" number of six segments connecting our points, and by the result of problem 87 a) this is not the case. However, the number of segments equal to 1, connecting the points of our system of four p... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,184 |
97. a) It is clear that any plane broken line of length 1 can be enclosed in a circle of radius 1: for this, it is sufficient for the center $O$ of the circle to coincide with one of the ends of the broken line. What is the radius of the smallest circle in which any broken line of length $l$ can be enclosed?
b) It is ... | 97. a) Let $S$ be the midpoint of the broken line $AB$, i.e., such a point that the broken lines $AS$ and $SB$ have the same length $1/2$. In this case, each point $M$ of the broken line is such that the length of the broken line $SM$ does not exceed $1/2$, and even more so, the distance $SM \leqslant 1/2$, from which ... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,191 |
103. Let $K_p$, $K_{v}$, and $T p$ be a circle, a square, and an equilateral triangle of diameter 1. Prove that:
a) $\delta(1, K_p)=\delta(2, K_p)=1 ; \delta(3, K_p)=\frac{\sqrt{3}}{2}(\approx 0.86) ;$ $\delta(4, K_p)=\frac{\sqrt{2}}{2}(\approx 0.71) ; \delta(5, K_p)=\frac{1}{4} \sqrt{10-2 \sqrt{5}}(\approx 0.59) ;$ $... | 103. a) This problem is about dividing a circle $Kp$ of unit diameter into $n$ "possibly smaller" parts $F_{1}, \ldots, F_{n}$, where the phrase "possibly smaller" has the following precise meaning: it is required that the diameter of the largest (by diameter) of the parts be as small as possible (i.e., we are looking ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,195 |
104*. Prove that any plane figure of diameter 1 can be divided
a) into 3 parts of diameter $\leqslant \frac{\sqrt{3}}{2}(\approx 0.86)$;
b) into 4 parts of diameter $\leqslant \frac{\sqrt{2}}{2}(\approx 0.71)$;
c) into 7 parts of diameter $\leqslant \frac{1}{2}(=0.5)$.
Thus, we have
$$
\delta_{2}(2)=1, \quad \delt... | 104. a) Any plane figure $F$ of diameter 1 can be enclosed in a regular hexagon $Ш \equiv A B C D E G$ with side $\frac{\sqrt{3}}{3}$ and inscribed circle diameter 1 (see Fig. $190, a$; see problem 93 b) above). It is easy to see that each of the pentagons OMBCN, ONDEP, and OPGAM depicted in Fig. 190 has a diameter of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,196 |
108. Let a circle $K$ of radius 1 and smaller circles $K_{1}, K_{2}$, $K_{3}, \ldots$ be given. Prove that
a) the smallest number of circles $K_{1}, K_{2}, \ldots$, needed to cover circle $K$, is always $\geqslant 3$;
b) if the radii of circles $K_{1}, K_{2}, \ldots$ are less than $\frac{\sqrt{3}}{2}$, then this numb... | 108. a) Since the circle $K$ cannot be divided into two parts with a diameter less than 1 (see problem 103 a) above), it cannot be covered by two circles $K_{1}$ and $K_{2}$ with diameters $<1$, otherwise the diameter of the part of $K$ covered by $K_{1}$ and the diameter of the "remainder" (entirely covered by the cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,200 |
116. Let $m$ be a convex polygon, and $M$ (the larger) a convex polygon obtained from $m$ by shifting all sides of $m$ outward by a distance of 1 (Fig. 56). Prove that
$$
n(1, M) \geqslant v(2, m)
$$
$ non-intersecting circles $k_{1}$, $k_{2}, \ldots, k_{n}$ of radius 1, located inside the polygon $M$; it is not difficult to see that the centers of all these circles will belong to the polygon $m$ (why?). Now replace these $n$ circles with concentric circles $K_{1}, K_{2}, \ldots, K_{n}$ of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,206 |
1. Given a figure consisting of 16 segments (Fig. 1). Prove that it is impossible to draw a broken line that intersects each segment exactly once. (The broken line can be open and self-intersecting, but its vertices should not lie on the segments, and its sides should not pass through the vertices of the figure.)
 Given a quadruple of positive numbers ( $a, b$, $c, d$ ). From it, a new quadruple ( $a b, b c, c d, d a$ ) is obtained according to the following rule: each number is multiplied by the next one, the fourth by the first. From the new quadruple, a third one is obtained by the same rule, and so on. Prove that in th... | 5. a) Suppose the quadruple ( $a, b, c, d$ ) has reappeared.
First, let's prove that in this case $a b c d=1$.
Let $a b c d=p$. Then the product of the numbers in the second quadruple is $p^{2}$, in the third $-p^{4}$, in the fourth $p^{8}, \ldots$ Clearly, if $p \neq 1$, the sequence of products will not have two id... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,214 |
6. a) Points $A$ and $B$ move uniformly and with equal angular velocities along circles centered at $O_{1}$ and $O_{2}$, respectively (clockwise). Prove that vertex $C$ of the equilateral triangle $A B C$ also moves uniformly along some circle.
b) The distances from a fixed point $P$ on the plane to two vertices $A, B... | 6. a) Let $\overrightarrow{\mathrm{O}_{1} \mathrm{O}_{3}}$ be the vector obtained by rotating $\overrightarrow{O_{1} O_{2}}$ by $60^{\circ}$ (in the same direction as the rotation that transforms $\overrightarrow{A B}$ into $\overrightarrow{A C}$). Points $A^{\prime}$ and $B^{\prime}$ are the images of $A$ and $B$ unde... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,215 |
7. In the cells of an $m \times n$ table, some numbers are written. It is allowed to simultaneously change the sign of all numbers in a certain column or a certain row. Prove that by repeatedly performing this operation, the given table can be transformed into one where the sums of the numbers in any column and any row... | 7. Among all the tables that can be obtained from the given variables by changing signs in rows and columns, we take the one for which the sum $\Sigma$ is maximal. Such a table T exists, since the number of ways to place signs before the numbers in the $m \times n$ table is finite - $2^{m n}$ (the number of ways to cha... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,216 |
8. $n$ points are connected by non-intersecting segments in such a way that from each point, one can travel to any other point via these segments, and there are no two points that are connected by two different paths. Prove that the total number of segments is $n-1$. | 8. Let's call one of the $n$ points the "root". For each of the remaining $n-1$ vertices, we will associate the last segment (the only one by condition) of the path leading to this point from the "root". This correspondence between
 b / d=p k . \quad$ Thus, $a k+b l$ is divisible by $p$. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,218 | |
10. Kolya and Petya are dividing $2 n+1$ nuts, $n \geqslant 2$, with each wanting to get as many as possible. There are three ways of dividing (each goes through three stages).
1st stage: Petya divides all the nuts into two parts, each containing no fewer than two nuts.
2nd stage: Kolya divides each part again into t... | 10. Answer: the most profitable way for Kolya is the first one; with the second and third, he will get one nut less with correct play. (In general, as we will see, the dispute in this division is over one nut.)
No matter what piles (of $a$ and $b$ nuts, $a<b$) form after Petya's first move, Kolya can split the larger ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,219 |
11. Prove that for any three infinite sequences of natural numbers
$$
\begin{aligned}
& a_{1}, a_{2}, \ldots, a_{n}, \ldots \\
& b_{1}, \quad b_{2}, \ldots, b_{n}, \ldots \\
& c_{1}, \quad c_{2}, \ldots, c_{n}, \ldots
\end{aligned}
$$
there exist indices $p$ and $q$ such that $a_{p} \geqslant a_{q}, b_{p} \geqslant b... | 11. Since the numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ are natural, there exists a sequence of indices $i_{1}, i_{2}, \ldots, i_{n}, \ldots$, such that $a_{i_{1}} \leqslant a_{i_{2}} \leqslant \ldots \leqslant a_{i_{n}} \leqslant \cdots \quad\left(a_{l_{1}}\right.$ is the smallest number in the sequence $a_{n}, a_... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,220 |
12. In a rectangle with sides 20 and 25, 120 squares with a side length of 1 are thrown. Prove that it is possible to place a circle with a diameter of 1 in the rectangle, such that it does not intersect with any of the squares.
Second All-Russian Olympiad,
1962 (Moscow)
| Class | | | | | |
| :---: | :---: | :-... | 12. The center of a circle with a diameter of 1, which is entirely contained within a rectangle, must be located at a distance greater than
Fig. 26
than $1 / 2$ from any side of the recta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,221 |
13. On the extensions of sides $A B, B C, C D$ and $D A$ of a convex quadrilateral $A B C D$, points $A^{\prime}$, $B^{\prime}, C^{\prime}, D^{\prime}$ are taken such that $\overrightarrow{B B^{\prime}}=\overrightarrow{A B}, \overrightarrow{C C^{\prime}}=\overrightarrow{B C}, \overrightarrow{D D^{\prime}}=\overrightarr... | 13. The median divides the area of the triangle in half. Therefore (Fig. 27) $S_{A B C}=S_{C B B^{\prime}}=S_{C B^{\prime} C^{\prime}}$. From this it follows that $S_{B B^{\prime} C^{\prime}}=2 S_{A B C}$.
Similarly, the equalities $S_{C C^{\prime} D^{\prime}}=2 S_{B C D^{\prime}}$; $S_{D D^{\prime} A}=2 S_{C D A^{\pr... | 5S | Geometry | proof | Yes | Yes | olympiads | false | 25,222 |
14. On a plane, a circle $s$ and a line $l$ passing through the center $O$ of the circle $s$ are given. Through the point $O$, an arbitrary circle $s^{\prime}$ with its center on the line $l$ is drawn. Find the set of points $M$ where the common tangent of the circles $s$ and $s^{\prime}$ touches the circle $s^{\prime}... | 14. Answer: the sought set of points $M$ of tangency is a pair of lines tangent to the circle $s$ at points $P, Q$ of its intersection with the line $l$ (excluding the points $P$ and $Q$ themselves).
Let the common tangent of $s$ and $s^{\prime}$ touch the circle $s$ at point $N$. Then $\angle N O M=\angle O M O_{1}=\... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,223 |
15. Given positive integers $a_{0}, a_{1}$, $a_{2}, \ldots, a_{99}, a_{100}$. It is known that $a_{1}>a_{0}, a_{2}=3 a_{1}-$ $-2 a_{0}, \quad a_{3}=3 a_{2}-2 a_{1}, \ldots, a_{100}=3 a_{99}-2 a_{98}$. Prove that $a_{100}>2^{99}$. | 15. By the condition $a_{1}-a_{0} \geqslant 1$. Further, $a_{2}-a_{1}=2\left(a_{1}-a_{0}\right)$, $a_{3}-a_{2}=2\left(a_{2}-a_{1}\right), \ldots, a_{100}-a_{99}=2\left(a_{99}-a_{98}\right) . \quad$ Multiplying these 99 equalities, both sides of which are positive, and canceling the common factors $a_{2}-a_{1}, a_{3}-a_... | a_{100}\geqslant2^{100} | Algebra | proof | Yes | Yes | olympiads | false | 25,224 |
16. Prove that there do not exist integers $a$, $b, c, d$ such that the expression $a x^{3}+b x^{2}+c x+d$ equals 1 when $x=19$ and equals 2 when $x=62$. | 16. Let $P(x)=a x^{3}+b x^{2}+c x+d$. The difference
$$
P(62)-P(19)=a\left(62^{3}-19^{3}\right)+b\left(62^{2}-19^{2}\right)+c(62-19)
$$
is divisible by 43 and cannot equal 1.
$\nabla$ In general, for any polynomial $P(x)$ with integer coefficients, the difference $P\left(x_{1}\right)-P\left(x_{2}\right)$, where $x_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,225 |
18. Construct a triangle given two sides so that the medians drawn to these sides intersect at a right angle. | 18. Let the lengths $a, b$ of sides $B C$ and $A C$ of a triangle be known, and we want its medians $A D$ and $B E$ to intersect at a right angle. Take a point $F$ on segment $A C$ of length $b$ such that $A F: F C=3$; then $F D \| B E$ and $\angle A D F$ should be $90^{\circ}$, i.e., point $D$ should lie on the circle... | ^{2}=(^{2}+b^{2})/5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,227 |
19. $a, b, c, d$ - positive numbers, the product of which is equal to 1. Prove that
$$
a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d \geqslant 10
$$ | 19. From the inequality $x^{2}+y^{2} \geqslant 2 x y$ it follows that
$$
\begin{gathered}
\left(a^{2}+b^{2}\right)+\left(c^{2}+d^{2}\right) \geqslant 2(a b+c d) \geqslant 4 \sqrt{a b c d} \geqslant 4 \\
a b+c d \geqslant 2, \quad b c+a d \geqslant 2, \quad a c+b d \geqslant 2
\end{gathered}
$$
$\nabla$ Using the gene... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,228 |
20. Given a regular pentagon; $M$ is an arbitrary point inside it (or on its boundary). Number the distances from point $M$ to the sides of the pentagon (or their extensions) in ascending order: $r_{1} \leqslant r_{2} \leqslant r_{3} \leqslant r_{4} \leqslant r_{5}$. Find all positions of point $M$ for which the value ... | 20. Answer: $r_{3}$ is the largest for the vertices of the pentagon, and the smallest - for the midpoints of the sides. By dividing the regular pentagon $A B C D E$ with its axes of symmetry into 10 triangles, it is sufficient to investigate points $M$ inside and on the boundary of one of them, for example, $A O K$ (Fi... | r_{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,229 |
21. Let's take any 1962-digit number divisible by 9. The sum of its digits will be denoted by $a$, the sum of the digits of $a$ by $b$, and the sum of the digits of $b$ by $c$. What is $c$?[^0] | 21. Answer. $c=9$. The sum of the digits of any number gives the same remainder when divided by 9 as the number itself (PZ). On the other
Fig. 30 hand, $a \leqslant 1962.9<19999$, so $b \l... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,230 |
22. From the midpoint $M$ of the base $AC$ of an isosceles triangle $ABC$, a perpendicular $MH$ is dropped to the side $BC$. Point $P$ is the midpoint of the segment $MH$. Prove that $AH \perp BP$. | 22. Drop the altitude $A D$ in $\triangle A B C$ (Fig. 30). Then in $\triangle A D C$, the segment $M H$ will be the midline, i.e., $D H=C H$. The right triangles $B H M$ and $A D C$ are similar, since $\angle D A C=$ $=90^{\circ}-\angle C=\angle H B M$, and after rotating one of them by $90^{\circ}$, the corresponding... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,231 |
23. What is the maximum area that a triangle with sides \(a, b, c\) can have, given the following constraints:
\[
0 \leqslant a \leqslant 1 \leqslant b \leqslant 2 \leqslant c \leqslant 3 \text { ? }
\] | 23. Answer: 1. Among triangles with two sides $a, b$ that satisfy the conditions $0<a \leqslant 1,1 \leqslant b \leqslant 2$, the one with the largest area is the right triangle with legs $a=1, b=2$ (indeed, $s \leqslant a b / 2 \leqslant 1$, since the height dropped to side $b$ is no more than $a$). The third side of ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,232 |
24. $x, y, z$ - arbitrary pairwise distinct integers. Prove that $(x-y)^{5}+(y-z)^{5}+$ $+(z-x)^{5}$ is divisible by $5(y-z)(z-x)(x-y)$. | 24. The quotient is equal to $x^{2}+y^{2}+z^{2}-2 x y-2 y z-2 x z$. To verify the required identity, it is convenient to set $x-y=u_{0}$, $y-z=v$, then $z-x=-(u+v)$, and prove the identity
$$
(u+v)^{5}=u^{5}+v^{5}+5 u v(u+v)\left(u^{2}+u v+v^{2}\right)
$$
using the binomial formula (P6):
$$
(u+v)^{5}=u^{5}+5 u^{4} v... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,233 |
25. For the numbers $a_{0}, a_{1}, \ldots, a_{n}$, it is known that $a_{0}=$ $=a_{n}=0$ and that $a_{k-1}-2 a_{k}+a_{k+1} \geqslant 0$ for all $k$ $(k=1,2, \ldots, n-1)$. Prove that all $a_{k}$ are non-positive. | 25. Solving this problem is aided by Fig. 31: the broken line with vertices at points ( $k, a_{k}$ ) is "convex" because $a_{k+1}-a_{k} \geqslant a_{k}-a_{k-1}$ (i.e., the slope of each subsequent segment is greater than the previous one), so the entire line, except for the ends, lies below the base $0 k$
Suppose that... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,234 |
26. Given positive numbers $a_{1}, a_{2}, \ldots, a_{m}$, $b_{1}, b_{2}, \ldots, b_{n}$, such that $a_{1}+a_{2}+\ldots+a_{m}=b_{1}+$ $+b_{2}+\ldots+b_{n}$. Prove that in an empty table with $m$ rows and $n$ columns, no more than $m+n-1$ positive numbers can be placed so that the sum of the numbers in the $i$-th row equ... | 26. Consider the table $m \times n$
Fig. 31
and let's reason by induction, assuming that the statement is already proven for tables with a smaller sum $m+n$. For a $1 \times 1$ table, the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,235 |
27. Out of five given circles, any four pass through one point. Prove that there exists a point through which all five circles pass. | 27. Let the 1st, 2nd, 4th, and 5th circles pass through point $A$; the 1st, 3rd, 4th, and 5th circles through point $B$; and the 2nd, 3rd, 4th, and 5th circles through point $C$.
We see that all three points $A, B$, and $C$ cannot be distinct, as they lie on the 4th and 5th circles, and two circles have no more than t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,236 |
29. a) Each diagonal of a convex quadrilateral $ABCD$ divides its area in half. Prove that $ABCD$ is a parallelogram.
b) Given a convex hexagon $ABCDEF$. It is known that each of the diagonals $AD, BE$, and $CF$ divides its area in half. Prove that these diagonals intersect at one point. | 29. a) Let $O$ be the point of intersection of the diagonals of quadrilateral $ABCD$. Since triangles $ABD$ and $BCD$ have equal areas and a common base $BD$, their heights dropped from vertices $A$ and $C$ are equal, i.e., points $A$ and $C$ are equidistant from $BD$, hence
^{2}-$ $-\left(a^{2}+b^{2}\right)=2 a b$ is also divisible by $d$. Therefore, $2 a^{2}=2 a(a+b)$ $-2 a b$ and $2 b^{2}=2 b(a+b)-2 a b$ are divisible by $d$.
But if $a$ and $b$ are coprime, then $a^{2}$ and $b^{2}$ are also coprime, so $2 a^{2}$ and $2 b^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,239 |
31. Two points $A$ and $B$ are fixed on a circle, while point $M$ runs through the entire circle. From the midpoint $K$ of segment $M B$, a perpendicular $K P$ is dropped onto the line $M A$.
a) Prove that all lines $K P$ pass through one point,
b) Find the set of points $P$. | 31. a) Let $C$ be the point diametrically opposite to point $A$. The angle $A M C=$ is right. Since $M K=K B$ and $P K$ is a median, the line $P K$ intersects the segment $B C$ at its midpoint: $B H=H C$. Therefore, all lines $P K$ pass through the point $H$, the midpoint of segment $B C$.
b) From a) it follows that t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,240 |
32. Given an equilateral triangle with side 1. What is the smallest $d$ such that a segment of length $d$ can, by sliding its ends along the sides of the triangle, sweep over the entire triangle? | 32. Answer: $d=2 / 3$.
If a segment sweeps the entire triangle, then in some position it passes through its center. We will prove that among all segments with endpoints on the sides and passing through the center $O$ of an equilateral triangle, the shortest one is the segment $A B$ parallel to its side.
Let $A^{\prim... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,241 |
33. A chessboard of size $6 \times 6$ is covered with 18 dominoes of size $2 \times 1$ (each domino covers two cells). Prove that for any such covering, the board can be cut into two parts along a horizontal or vertical line without cutting any domino. | 33. Suppose that a set of dominoes can be laid on a board in such a way that each of the horizontal and vertical lines dividing the board into cells intersects at least one domino.
There are 10 such lines in total. Each of these lines divides the board into 2 parts, each consisting of an even number of cells. In any o... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,242 |
34. Given $n$ distinct positive numbers $a_{1}$, $a_{2}, \ldots, a_{n}$. From them, all possible sums with any number of addends (from 1 to $n$) are formed. Prove that among these sums, there will be at least $n(n+1) / 2$ pairwise distinct sums. | 34. We can assume that the numbers are arranged in ascending order: $a_{1}<a_{2}<\ldots<a_{n}$. Consider the numbers:
$$
\begin{array}{llll}
a_{1}, & a_{2}, & \ldots, a_{n-2}, & a_{n-1}, \\
a_{1}+a_{n}, & a_{2}+a_{n}, \ldots, a_{n-2}+a_{n}, \\
a_{1}+a_{n-1}+a_{n}, & \ldots, & \ldots, a_{n-2}+a_{n-1}+a_{n}, \\
\ldots &... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,243 |
35. In triangle $ABC$, the angle bisectors from vertices $A$ and $B$ are drawn. Then, from vertex $C$, lines parallel to these bisectors are drawn. Points $D$ and $E$ are the intersections of these lines with the bisectors, and these points are connected. It turns out that lines $DE$ and $AB$ are parallel. Prove that t... | 35. From the equality of angles (Fig. 34), it follows that point $A$ lies on the circle passing through points $C, D$, and $E$, and also that point $B$ lies on this circle, with $D E \| A B$.
^{2}=$ $=m^{2}+2 m k d+k^{2} d^{2}=a+d\left(2 k m+k^{2}\right)$ is also a term of the progression for any natural $k$. We have thus indicated an infinite number of terms of the progr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,245 |
37. Given a regular 45-gon. Is it possible to place the digits $0,1, \ldots, 9$ at its vertices such that for any pair of different digits, there is a side whose endpoints are labeled with these digits? | 37. Answer: no. The digit a forms 9 pairs (with each of the nine other digits). To find a side of the 45-gon numbered with the corresponding digits for all these pairs, it is necessary to place a in at least five of its vertices. Since there are only ten digits, it is necessary to have 50 places for their arrangement.
... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,246 |
38. Find such real numbers $a, b, p, q$ so that the equality
$$
(2 x-1)^{20}-(a x+b)^{20}=\left(x^{2}+p x+q\right)^{10}
$$
is satisfied for any $x$. | 38. Answer: $\quad a_{1,2}= \pm \sqrt[20]{2^{20}-1} ; \quad b_{1,2}=\mp \sqrt[20]{2^{20}-1} / 2$, $p=-1, q=1 / 4$ (two sets of coefficients in total).
Substituting $x=1 / 2$, we get
$$
\left(\frac{a}{2}+b\right)^{20}+\left(\frac{1}{4}+\frac{p}{2}+q\right)^{10}=0
$$
from which $a=-2 b$.
Now the identity takes the fo... | a_{1,2}=\\sqrt[20]{2^{20}-1};\quadb_{1,2}=\\sqrt[20]{2^{20}-1}/2,\quadp=-1,\quadq=1/4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,247 |
39. At the ends of the diameter of a circle, units are placed. Each of the resulting semicircles is divided in half, and in its middle, the sum of the numbers at the ends is written (first step). Then each of the four resulting arcs is divided in half, and in its middle, the number equal to the sum of the numbers at th... | 39. Answer: $2 \cdot 3^{\text{n }}$. After the first step, the sum of all numbers will be $6=2 \cdot 3$.
Let $S_{n}$ be the sum of all numbers after the $n$-th step. It is not hard to prove that after the $(n+1)$-th step, the sum will become $2 S_{n}+S_{n}=3 S_{n}$.
Thus, the sum of all numbers triples each time, so ... | 2\cdot3^{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,248 |
40. Given an isosceles triangle. Find the set of points inside the triangle such that the distance from these points to the base is equal to the geometric mean of the distances to the lateral sides.
Fourth All-Russian Olympiad, 1964 (Moscow)
| Class | | | | | |
| :---: | :--- | :--- | :--- | :--- | :--- |
| 8 | ... | 40. Answer: the arc of a circle touching the lateral sides of the triangle at vertices $A$ and $C$ of the base.
Let $A B C$ be the given triangle, $A B=B C$ and $M$ be some point of the desired set, $E, F$ and $H$ be its projections onto the sides $A B, B C$ and $A C$ respectively (Fig. 35). Quadrilaterals $A E M H$ a... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,249 |
41. In a triangle, two altitudes are not less than the sides to which they are dropped. Find the angles of the triangle. | 41. Answer: angles of the triangle $-90^{\circ}, 45^{\circ}, 45^{\circ}$.
Let $h_{a}$ and $h_{b}$ be the heights dropped to sides $a$ and $b$. By the condition, $a \leqslant h_{a}, b \leqslant h_{b}$, but in any triangle $h_{a} \leqslant b_{2} h_{b} \leqslant$ $\leqslant a$, so $a \leqslant h_{a} \leqslant b \leqslant... | 90,45,45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,250 |
42. Prove that $m(m+1)$ is not a power of an integer for any natural $m$. | 42. Suppose that the number $m(m+1)$ is the $k$-th power of some natural number. Since the numbers $m$ and $m+1$ are coprime, each of them must be the $k$-th power of a natural number. But this is impossible: if $\boldsymbol{m}=\boldsymbol{a}^{\text {k }}$, then already $(a+1)^{k}>(a+1) a^{k-1}=a^{k}+a^{k-1}>m+1(k>1)$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,251 |
43. For each of the numbers from 1 to 1000000000, the sum of its digits is calculated, and for each of the resulting billion numbers, the sum of its digits is calculated again, and so on, until a billion single-digit numbers are obtained. Which number will there be more of: 1 or 2? | 43. Answer: the number of ones will be one more than the number of twos.
Any number gives the same remainder when divided by 9 as the sum of its digits (DS). Therefore, in our problem, ones come from numbers that give a remainder of 1 when divided by 9, i.e., from the numbers 1, 10, 19, 28, ..., 999999991, 1000000000,... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,252 |
44. Given an arbitrary set of $n$ integers $a_{1}, a_{2}, \ldots, a_{2 k+1}$. From it, a new set is obtained: $\frac{a_{1}+a_{2}}{2}, \quad \frac{a_{2}+a_{3}}{2}, \ldots, \frac{a_{n-1}+a_{n}}{2}, \quad \frac{a_{n}+a_{1}}{2} ;$ from this set - the next one by the same rule and so on. Prove that if all the resulting numb... | 44. No matter what set of $n$ numbers $x_{1}, x_{2}, \ldots, x_{n}$ (not all equal to each other) we take, after a certain number of steps, the maximum number in the set will decrease, and the minimum number will increase.
From this, it is clear that the maximum number cannot remain an integer all the time, unless we ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,253 |
45. a) In a convex hexagon $A B C D E F$ all angles are equal. Prove that
$$
A B-D E=E F-B C=C D-F A
$$
b) Conversely, prove that from six segments, the lengths of which are $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ and satisfy the relations $a_{1}-a_{4}=a_{5}-a_{2}=a_{3}-a_{6}, \quad$ a convex hexagon with equal an... | 45. a) The measure of each angle of the hexagon is $120^{\circ}$, and therefore, its opposite sides are parallel. We can assume that $A B \geqslant D E$ (Fig. 36).
Now, let's construct parallelograms $A B C K, C D E L, A F E M$. If points $K, L, M$ do not coincide, then all angles of triangle $K L M$ are $60^{\circ}$,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,254 |
46. Solve the equation in integers
$$
\sqrt{\sqrt{x+\sqrt{x+\sqrt{x+\ldots+\sqrt{x}}}}}=y
$$ | 46. Answer: the only solution is $x=y=0$.
Let $\boldsymbol{x}$ and $\boldsymbol{y}$ be integers satisfying the condition. After a series of squaring, we are convinced that $\sqrt{x+\sqrt{x}}=\boldsymbol{m}$ and $\sqrt{x}=k$ are integers, and
$$
m^{2}=k(k+1)
$$
If $k>0$, then it must be that $k^{2}<m^{2}<(k+1)^{2}$, ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,255 |
47. Perpendiculars are dropped from the vertices of a convex quadrilateral $ABCD$ to its diagonals. Prove that the quadrilateral formed by their feet is similar to the original one. | 47. Let $A_{1}, B_{1}, C_{1}, D_{1}$ be the bases of the perpendiculars dropped from the vertices $A, B, C$, and $D$ to the diagonals $B D$ and $A C$, $O$ be the point of intersection of the diagonals, and $\alpha$ be the measure of the acute angle between them (Fig. 37). Then $O A_{1} = O A \cdot \cos \alpha ; O B_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,256 |
48. Find all odd natural $n$ for which $(n-1)!$ does not divide $n^2$.
untranslated part:
( $n-1$ )! does not divide $n^2$. | 48. Answer: $n$ is a prime number and $n=9$.
If $n$ can be represented in the form $n=ab$, where $a \geqslant 3, b \geqslant 3$, and $a$ is, then the product $1.2 \cdot \ldots \cdot(n-1)$ includes the factors $a, 2a$ and $b, 2b$, so $(n-1)$ is divisible by $a^2 b^2=n^2$.
If $n \Rightarrow p^2$, where $p$ is a prime n... | nisn=9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,257 |
51. $a, b$ and $n$ are fixed natural numbers. It is known that for any natural $k (k \neq b)$,
the number $a-k^{n}$ is divisible without remainder by $b-k$. Prove that $a=b^{n}$. | 51. Since $k^{n}-b^{n}=(k-b)\left(k^{n-1}+k^{n-2}b+\ldots+b^{n-1}\right)$, then $k^{n} - b^{2}$ is divisible by $k-b$.
Thus, $\left(k^{2}-b^{n}\right)-\left(k^{n}-a\right)=a-b^{n}$ is divisible by $\boldsymbol{k}-\boldsymbol{b}$ for any $k \neq b$, but this can only be true when $a=b^{n}$. | b^{n} | Number Theory | proof | Yes | Yes | olympiads | false | 25,260 |
52. In the expression $x_{1}: x_{2}: \ldots: x_{n}$ for indicating the order of operations, parentheses are placed and the result is written as a fraction:
$$
\frac{x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}}{x_{i_{1}} x_{l_{2}} \ldots x_{i_{n-k}}}
$$
(every letter $x_{1}, x_{2}, \ldots, x_{n}$ stands either in the numerat... | 52. Answer: $2^{n-2}$ fractions.
First of all, it is clear that in the resulting fraction, $x_{1}$ will be in the numerator. It is almost as obvious that $x_{2}$ will end up in the denominator in any arrangement of parentheses (the division sign in front of $x_{2}$ relates either to $x_{2}$ itself or to some expressio... | 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,261 |
53. What is the smallest number of non-overlapping tetrahedra into which a cube can be divided? | 53. It is easy to see that a cube can be divided into 5 tetrahedra. In Fig. 40, these are the tetrahedra $A A^{\prime} B^{\prime} D^{\prime}, A B^{\prime} B C, A C D D^{\prime}, B^{\prime} C^{\prime} D^{\prime} C$, and $A C D^{\prime} B^{\prime}$.
Let us now prove that it is impossible to divide the cube into fewer te... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,262 |
54. Find the largest perfect square such that after erasing its last two digits, a perfect square is obtained again. (It is assumed that one of the erased digits is not zero.) | 54. Answer: $41^{2}=1681$.
Let the number $n^{2}$ satisfy the condition of the problem, then $n^{2}=$ $=100 a^{2}+b$, where $0<10 a$ and therefore, $n \geqslant 10 a+1$. This means that $b=n^{2}-100 a^{2} \geqslant 20 a+1$, from which $20 a+141$, then $n^{2}-40^{2} \geqslant 42^{2}-40^{2}>100$.
$\nabla$ This topic is... | 41^{2}=1681 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,263 |
55. $ABCD$ is a circumscribed trapezoid; $E$ is the point of intersection of the diagonals $AC$ and $BD$; $r_{1}, r_{2}, r_{3}, r_{4}$ are the radii of the circles inscribed in triangles $ABE, BCE$, $CDE, DAE$ respectively. Prove that
$$
\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}
$$
Fifth All-Rus... | 55. Let $S_{1}, S_{2}, S_{3}, S_{4}$ and $2 p_{1}, 2 p_{2}, 2 p_{3}, 2 p_{4}$ be the areas and perimeters of triangles $A B E, B C E, C D E$ and $D A E$ respectively. We need to prove the equality
$$
\frac{p_{1}}{S_{1}}+\frac{p_{3}}{S_{3}}=\frac{p_{2}}{S_{2}}+\frac{p_{4}}{S_{4}}
$$
Since $A B C D$ is a trapezoid, $S_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,264 |
56. a) Each of the numbers $x_{1}, x_{2}, \ldots, x_{n}$ can independently take the value 1, 0, or -1. What is the smallest value that the sum of all possible pairwise products of these $n$ numbers can have?
b) What is the smallest value that the sum of all possible pairwise products of $n$ numbers $x_{1}, x_{2}, \ldo... | 56. Answer: the minimum value $s_{\mathrm{min}}$ is $-[n / 2]$, i.e., $-n / 2$ if $n$ is even, and $-(n-1) / 2$ if $n$ is odd.
a) The sum $s$ of all possible pairwise products of numbers $x_{1}, x_{2}, \ldots, x_{n}$ can be written as:
$$
s=\frac{1}{2}\left(\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2}-x_{1}^{2}-x_{2}^{2... | -[n/2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,265 |
57. There is a $3 \times 3$ board and 9 cards of one cell size, on which some numbers are written. Two players take turns placing these cards on the board. After all the cards are placed, the first player (the one who starts) calculates the sum of the six numbers in the top and bottom rows, and the second player calcul... | 57. Let \(a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{9}\) be numbers written on cards. If \(a_{1} + a_{9} > a_{2} + a_{8}\), then the first player places the number \(a_{9}\) in cell 2 (Fig. 41), and in their second move - the number \(a_{2}\) (or \(a_{1}\)) in one of the cells 1 or 4. If \(a_{1} + a_{9} < a_{... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,266 |
58. A circle is circumscribed around triangle $A B C$. Chords connecting the midpoint of arc $A C$ with the midpoints of arcs $A B$ and $B C$ intersect sides $A B$ and $B C$ at points $D$ and $E$. Prove that segment $D E$ is parallel to side $A C$ and passes through the center of the inscribed circle of triangle $A B C... | 58. Let $L, M$ and $N$ be the midpoints of the arcs $AB, BC$ and $CA$, $O$ be the center of the inscribed circle of triangle $ABC$, and $D$ and $K$ be the points of intersection of the segment $LN$ with the sides $AB$ and $AC$ (Fig. 42).
We will prove that the quadrilateral $ADOK$ is a rhombus (at the same time, we wi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,267 |
61. In the people's militia, there are 100 people, and every evening three people go on duty. Prove that it is impossible to organize the duty schedule so that any two people are on duty together exactly once. | 61. Let's take one of the militiamen. If the required distribution of duties were possible, the remaining 99 militiamen would have to break into pairs, each of which would stand duty with the chosen militiaman, but this is impossible since 99 is an odd number.
$\nabla$ The problem of for which $n$ it is possible to se... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,269 |
62. What is the greatest value that the length of the segment cut off by the lateral sides of the triangle on the tangent to the inscribed circle, drawn parallel to the base, can take if the perimeter of the triangle is $2 p$? | 62. Let $x$ be the length of the desired segment, $b$ be the length of the base $AC$ of triangle $ABC$ (Fig. 44). The perimeter of triangle $BDE$
Fig. 44
is $2p - 2b$ (convince yourself o... | \frac{p}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,270 |
64*. Can 1965 points be placed in a square with a side of 1 so that any rectangle of area $1 / 200$ with sides parallel to the sides of the square contains at least one of these points? | 64. Answer: it is possible.
In the square $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$ and on the line $y=1 / 2$, uniformly place $c_{0}=200$ points: ( $\left.k / 201 ; 1 / 2\right), k=1,2, \ldots$ $\ldots, 200$. Then on each of the lines $y=1 / 4$ and $y=3 / 4$, place $c_{1}=100$ points ( $\left.k / 101 ; 1... | 1704 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,272 |
65*. Rounding a number means replacing it with one of the two nearest integers.
Given $n$ numbers. Prove that they can be rounded in such a way that the sum of any $m$ rounded numbers ( $1 \leqslant m \leqslant n$ ) differs from the sum of these same unrounded numbers by no more than $(n+1) / 4$.
| 65. Let $x_{1}, x_{2}, \ldots x_{n}$ be given numbers, numbered in the order of increasing their fractional parts $\alpha_{1}=x_{1}-\left[x_{i}\right]$.
We will round the first $k$ of these numbers down, and the rest up; $k$ will be chosen below. It is easy to see that the largest error in rounding will accumulate in ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,273 |
66. A tourist who arrived in Moscow by train wandered around the city all day. After having dinner at a cafe on one of the squares, he decided to return to the station and do so only by streets he had walked an odd number of times before. Prove that he can always do this. | 66. A tourist can walk from the cafe along any route with one condition: upon reaching a square, they choose the next street from those they have walked an odd number of times before. It is not hard to see that such a street exists for any intersection, except for the station (indeed, the tourist approached the interse... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,274 |
67. a) A certain commission met 40 times. Each time, 10 people were present at the meetings, and no two members of the commission were together at more than one meeting.
Prove that the number of commission members is greater than 60.
b) Prove that from 25 people, it is impossible to form more than 30 commissions of 5... | 67. a) First solution. Each pair of commission members could meet at no more than one meeting. At each meeting, there were $C_{10}^{2}=45$ pairs. Since there were 40 meetings, at least $1800=45 \cdot 40$ pairs could be formed from the commission members. However, from 60 (no less) people, no more than $C_{60}^{2}=60 \c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,275 |
68*. Given two coprime numbers \( p > 0 \) and \( q > 0 \). An integer \( n \) is called good if it can be represented as \( n = px + qy \), where \( x \) and \( y \) are non-negative integers, and bad otherwise. a) Prove that there exists an integer \( c \) such that out of two integers \( n \) and \( c - n \), one is... | 68. Answer: b) $(p-1)(q-1) / 2$. If $p$ and $q$ are coprime natural numbers, then every integer $z$ can be represented as $z=p x+q y$ (P2). Every such representation is derived from some fixed $z p a+q b$ by the general formula $z=p(a-q t)+q(b+p t)$, where $t$ is an integer, and there exists a unique representation for... | (p-1)(q-1)/2 | Number Theory | proof | Yes | Yes | olympiads | false | 25,276 |
69. A reconnaissance aircraft is flying in a circle centered at point $A$. The radius of the circle is $10 \mathrm{km}$, and the speed of the aircraft is 1000 km/h. At some moment, a missile is launched from point $A$ with the same speed as the aircraft, and it is controlled such that it always remains on the line conn... | 69. An s w e r: in $\frac{\pi}{200}$ h. It is necessary to notice that the trajectory of the imaginary "rocket" in the conditions of the problem is a circle of half the radius (Fig. 46): the degree measure of the arc $A R$ is twice that of $\angle Q A P$ (the angle between the tangent and the chord), i.e., twice the de... | \frac{\pi}{200} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,277 |
70. Prove that the sum of the lengths of the edges of a polyhedron is greater than $3 d$, where $d$ is the distance between the most distant vertices of the polyhedron. | 70. Let $A$ and $B$ be two vertices of a polyhedron, separated by a distance $d$. Through points $A$ and $B$, we draw planes perpendicular to the line $A B$.
It is clear that the entire polyhedron is contained between these two planes. Through each vertex of the polyhedron, we draw a plane perpendicular to $A B$. Cons... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,278 |
72. On each of the planets in a certain system, there is an astronomer observing the nearest planet. The distances between the planets are all different. Prove that if the number of planets is odd, then some planet is not observed by anyone. | 72. Let's take the two closest planets to each other. Clearly, the astronomers on these planets are looking at each other.
There are still $n-2$ planets and $n-2$ astronomers left. If at least one of them is looking at one of the already chosen planets, then there won't be enough astronomers for one of the $n-2$ plane... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,280 |
73. a) Points $B$ and $C$ lie inside the segment $A D$. Prove that if $A B$ equals $C D$, then for any point $P$ on the plane, the inequality $P A + P D \geqslant P B + P C$ holds.
b) Points $A, B, C$, and $D$ are marked on the plane. It is known that for any point $P$, the inequality $P A + P D \geqslant P B + P C$ h... | 73. a) If point $P$ lies on line $A D$, the statement is obvious.
Let $P$ not lie on $A D$ and $O$ be the midpoint of segment $A D$. Let $P^{\prime}$ be the point symmetric to point $P$ with respect to $O$. Quadrilaterals $B P C P^{\prime}$ and $P^{\prime} A P D$ are parallelograms and $A P + P D = A P^{\prime} + A P ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,281 |
74. Do there exist such natural numbers $x$ and $y$, for which $x^{2}+y$ and $y^{2}+x$ are squares of integers? | 74. Answer: there are no such numbers $x$ and $y$.
Indeed, let $x \geqslant y$. Then $x^{2}<x^{2}+y \leqslant x^{2}+x<$ $<(x+1)^{2}$, i.e., $x^{2}+y$ is not a square of an integer. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,282 |
75. a) Eighth-graders are lined up in a row. In front of each of them stands a seventh-grader who is shorter than them. Prove that if the row of eighth-graders is arranged by height and the row of seventh-graders is also arranged by height, then each eighth-grader will still be taller than the seventh-grader standing i... | 75. a) It is sufficient to prove that the $k$-th tallest eighth-grader is taller than the $k$-th tallest seventh-grader. Let $A$ be the $k$-th tallest seventh-grader. Then there are at least $k$ seventh-graders (including $A$) who are not shorter than $A$. All $k$ eighth-graders standing behind them are taller than $A$... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,283 |
76. On a grid paper, a rectangle $A B C D$ is drawn, with its sides lying on the grid lines, and $A D$ is $k$ times greater than $A B$ ($k$ is an integer). Consider all possible paths that follow the grid lines and lead from $A$ to $C$ in the shortest way. Prove that among these paths, those in which the first segment ... | 76. Let there be a rectangle $ABCD$ on a grid paper of size $m \times n$ cells (Fig. 48). Denote the number of paths leading from vertices $B_{1}, D_{1}, B_{2}, D_{2}$, and $A_{2}$ to $C$ along the grid lines by $b_{1}, d_{1}, b_{2}, d_{2}$, and $a_{2}$ respectively. The statement of the problem is that $d_{1} / b_{1} ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,284 |
78. Prove that in a convex quadrilateral of area $S$ and perimeter $P$, a circle of radius $S / P$ can be inscribed.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 78. Let's solve this problem for an arbitrary convex $n$-gon with area $S$ and perimeter $P$.
Let the lengths of the sides of this polygon be $a_{1}, a_{2}, \ldots a_{n}$ $\left(a_{1}+a_{2}+\ldots+a_{n}=P\right)$. Consider a rectangle with area $S$ and base $P$, the height of which is obviously $S / P$.
Now, let's cu... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 25,286 |
79. In a certain city, for any three intersections $A, B$, and $C$, there is a path leading from $A$ to $B$ that does not pass through $C$. Prove that from any intersection to any other, there are at least two non-intersecting paths (an intersection is a place where at least two streets meet; the city has at least two ... | 79. Let's call the length of a path the number of segments it consists of. Let $n-$ be the length of the shortest path from $A$ to $B$. We will prove the statement of the problem by induction on $n$.
For $n=1$, besides the shortest path $AB$, there exists a path that goes from $A$ to a crossroad $C \neq A$, which is 1... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,287 |
84. a) In an acute-angled triangle \(ABC\), the greatest height \(AH\) is equal to the median \(BM\). Prove that the angle \(ABC\) is not greater than \(60^\circ\).
b) In an acute-angled triangle \(ABC\), the height \(AH\) is equal to the median \(BM\) and is also equal to the angle bisector \(CD\). Prove that triangl... | 84. a) The measure of angle $B$ in a right triangle is equal to (or less than) $30^{\circ}$ if the length of the leg opposite angle $B$ is equal to (or less than) half the length of the hypotenuse.
 In a certain natural number, the digits were arbitrarily rearranged. Prove that the sum of the obtained number and the original number is not equal to $\underbrace{999 \ldots 9}_{1967 \text { digits }}$.
b) The digits of a certain number were rearranged and the obtained number was added to the original number. ... | 85. a) If the sum of two numbers $a$ and $b$ consists of only nines, then there cannot be any carry-over of units to the next place when adding these numbers. Therefore, $S(a+b)=S(a)+S(b)$, where $S(x)$ is the sum of the digits of the number $x$. If $S(a)=S(b)$, then $S(a+b)$ is even and cannot equal $9 \cdot 1967$.
b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,291 |
87. a) Is it possible to arrange the numbers $0,1,2, \ldots, 9$ on a circle so that any two adjacent numbers differ by 3, 4, or 5?
b) Is it possible to arrange the numbers $1,2,3, \ldots, 13$ on a circle so that any two adjacent numbers differ by 3, 4, or 5? | 87. Answer: a) no. Note that any two of the numbers $0,1,2,8,9$ cannot stand next to each other. Therefore, they must stand every other place. However, the number 7 cannot end up in any of the 5 remaining places, since next to it, only 2 can stand from the listed numbers.
b) Answer: no. The reasoning is similar to a).... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,292 |
88. Prove that there exists a number divisible by $5^{1000}$ and not containing any zeros in its decimal representation. | 88. The number $5^{1000}$ ends with 5. Let the $k$-th digit from the end in the decimal representation of this number be 0, and all subsequent digits be different from 0. Add $5^{1000} \cdot 10^{k-1}$ to this number. As a result, we get a number divisible by $5^{1000}$, in which the last $k$ digits are different from 0... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,293 |
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