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6.9. In parallelogram $A B C D$, diagonal $A C$ is greater than diagonal $B D$. A circle is drawn through points $B, C, D$, intersecting diagonal $A C$ at point $M$. Prove that $B D$ is a common tangent to the circumcircles of triangles $A M B$ and $A M D$.
bases of heights | 6.9. Consider the circle (see figure) passing through points $B, C$, and $D$. The inscribed angles $M B D$ and $M C D$, subtending the arc $M D$ of this circle, are equal, and since $\angle M C D = \angle M A B$, it follows that $\angle M A B = \angle M B D$ (see figure). From this, deduce that the line $B D$ is tangen... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,691 |
7.1. Divide a regular $n$-gon $(n \geqslant 5)$ into the smallest number of:
a) acute triangles,
b) obtuse triangles.
Each side of any triangle in the partition must simultaneously be a side of another triangle in the partition or a side of the polygon.
It is important that no vertex of one triangle divides the sid... | 7.1. a) The segments connecting the center of the given $n$-gon with its vertices divide it into $n$ acute triangles. It is impossible to divide the given polygon into a smaller number of acute triangles, since two different sides of the $n$-gon cannot belong to one acute triangle (for $n \geqslant 5$, all angles of a ... | )n,\quadb)n | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,693 |
7.3. a) Given an arbitrary quadrilateral.
Prove that the plane can be covered without gaps or overlaps using quadrilaterals equal to the given one.
b) Does there exist a pentagon with which the plane can be tiled using its copies?
c) The same question for a pentagon, no two sides of which are parallel. | 7.3. a) The method of laying ash as shown in figure a. It is crucial here that the sum of the angles of a quadrilateral is $360^{\circ}$.
b) Although a regular pentagon is not suitable for this purpose, there are many different types of pentagons that are suitable for "tessellation" (see, for example, the tessellation... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,695 |
7.4. Prove that the plane cannot be covered with equal convex heptagons. | 7.4. This task is much more complicated than the previous ones. The idea of the solution is as follows. Suppose that such a covering is possible. We will calculate the average angle of a heptagon in two ways. On the one hand, the average angle in a heptagon is $\frac{5 \cdot 180^{\circ}}{7}$. On the other hand, at each... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,696 |
7.5. a) Prove that the plane can be covered by convex $n$-gons ($n \geqslant 7$), if it is not required that they be equal to each other.
b) Prove that the plane can be covered by equal $n$-gons ($n \geqslant 7$), if it is not required that they be convex.
COVERINGS | 7.5. a) The construction process can be understood from figure a. The covering polygons become increasingly narrow. This is precisely why the result of problem 7.5a) does not contradict the reasoning of problem 7.4,
![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-57.jpg?height=370&width=449&top_le... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,697 |
7.6. A circle is covered by several arcs. These arcs may overlap each other, but none of them covers the entire circle. Prove that it is possible to select several of these arcs such that they also cover the entire circle and their total length does not exceed $720^{\circ}$. | 7.6. The ends of all arcs divide the circle into several parts (we take the "smallest" parts, so that none of them contain the ends of the arcs inside and, by condition, are covered by one or several arcs). If any of these parts is covered by three or more arcs, then we select from these arcs two that extend the farthe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,698 |
7.7. The corridor is completely covered by several carpet runners, the width of which is equal to the width of the corridor. Prove that some of the runners can be removed so that the remaining ones do not overlap and cover at least half of the corridor. (It is assumed that any runner can be removed without changing the... | 7.7. Reasoning as in the previous problem, first prove that some of the paths can be removed so that the remaining ones cover the corridor, and no more than two layers.
Now, let's number the paths from left to right in the order of their left ends. All the odd-numbered paths do not intersect with each other, and the s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,699 |
7.8. The floor of a rectangular room $6 \times 3 \mu^{2}$ is completely covered by square carpets of different sizes (the edges of the carpets are parallel to the walls). Prove that some of the carpets can be removed so that the remaining carpets do not overlap and cover more than $2 \mu^{2}$.
## SNAIL AND OBSERVERS | 7.8. This problem resembles the previous one, but the previous reasoning does not apply. Specifically, if several carpets cover one area, it is generally not possible to select a smaller number of carpets that cover all the others (figure $a$;
. The observation time will be represented by a segment \(AB\) of length 6. We will also mark segments of length 1 on the OT axis, representing the observation intervals of each observer.
 The bottom of a rectangular box is paved with tiles of size $2 \times 2$ and $1 \times 4$. The tiles were spilled out of the box and one $2 \times 2$ tile was lost. Instead, a $1 \times 4$ tile was retrieved. Prove that it is now impossible to pave the bottom of the box with the tiles.
b) The bottom of a recta... | 8.3. a) We will shade cells in every other odd horizontal row (see figure $a$). Notice that
a)
 / 2$, if $n$ is odd.
Examples where exactly this number of cells are shaded are easy to construct; you need to shade all rows with odd numbers: 1st, 3rd, etc. (check this).
We also need to prove that it is impossible to shade more cells. For even $n=... | n^{2}/2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,705 |
8.5. In a city, there are 10 streets parallel to each other and 10 others intersect them at right angles. What is the minimum number of turns a closed route can have, passing through all intersections? | 8.5. Answer. 20.
It is easy to provide an example with 20 turns. We will prove that fewer than 20 turns are not possible. Consider 10 streets of a certain direction. If the route passes through each of them, then there are already at least two turns of the route on each of them, and the proof is complete. If there is ... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,706 |
8.6. On a grid paper with a cell side of 1, a circle of radius 10 is drawn. Prove that inside this circle there are at least 250 grid nodes. (Nodes are called points where the grid lines intersect.) | 8.6. Construct a circle of radius 9, concentric with the given circle. Consider all the $1 \times 1$ squares, the sides of which are parallel to the grid lines, and the centers of which lie at the grid nodes and inside the circle of radius 10. Prove that these squares completely cover the constructed circle of radius 9... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,707 |
8.8. On graph paper, a closed broken line is drawn, all segments of which have the same length, and all vertices lie at the grid nodes. Prove that the number of segments of such a broken line is necessarily even.
BEE AND HONEYCOMBS | 8.8. Introduce a coordinate system, taking any two mutually perpendicular lines on the grid paper as coordinate axes, and the width of a cell as 1. Let $x_{i}$ and $y_{i}$ be the projections of the segments of the broken line onto the coordinate axes (taking into account the sign). Then $x_{1}+x_{2}+\ldots$ $\ldots+x_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,709 |
9.1. In a square with side 1, 51 points were thrown. Prove that some three of them can be covered by a circle of radius $1 / 7$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 9.1. Let's divide the square by lines parallel to its sides into 25 smaller squares with a side length of $1 / 5$. If no more than two points fell into each small square, then there would be no more than 50 points in total across the 25 small squares. Therefore, there must be a small square that contains three points. ... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,711 | |
9.2. On a plane, there are $n$ points such that any triangle with vertices at these points has an area no greater than 1. Prove that all these points can be placed in a triangle of area 4. | 9.2. Among all triangles with vertices at the given points, we select the triangle of the largest area $s \leqslant 1$. Through each of its vertices, we draw a line parallel to the opposite side. The area of the triangle formed by these lines, as it is easy to see, is $4 s \leqslant 4$.
Prove that this triangle covers... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,712 |
9.3. a) Prove that in a circle of radius 1, it is impossible to select more than five points such that all pairwise distances between them are greater than 1.
b) Prove that in a circle of radius 10, it is impossible to place 450 points such that the distance between any two points is greater than 1.
c) Prove that it i... | 9.3. a) Suppose this is possible, and let $A_{1}, A_{2}, \ldots, A_{6}$ be six such points inside a circle of radius 1, the pairwise distances between which are greater than 1. Draw radii from the center of the circle through each of these points. (Clearly, no two points lie on the same radius.) At least 2 of these 6 r... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,713 |
9.4. Given a convex polygon $M$ with area $S$ and perimeter $P$. Consider the set $\Phi(x)$ of points that are at a distance of no more than $x$ from $M$ (it is said that a point $A$ is at a distance of no more than $r$ from the polygon $M$ if there exists a point $B$ of the polygon $M$ such that $|A B| \leqslant r$). ... | 9.4. Prove that the geometric locus $\Phi(x)$ consists of the following pieces, as shown in the figure:
a) polygon $M$, b) rectangles of height $x$, adjacent to each side of $M$ from the outside, c) sectors of radius $x$, adjacent to each vertex.
Using the theorem on the sum of the angles of a polygon, prove that the... | (x)=+Px+\pix^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,714 |
9.6. A paratrooper is somewhere in a forest of area $S$. The shape of the forest is unknown to him, but he knows that there are no clearings in the forest. Prove that he can get out of the forest by traveling a distance of no more than $2 \sqrt{\pi S}$ (it is assumed that the paratrooper can move along a path of a pre-... | 9.6. The paratrooper will certainly exit the forest if he walks along a circle of radius \( r = \sqrt{S / \pi} \). Indeed, if the entire circumference of this circle were inside the forest, the area of the forest would be greater than \( \pi r^2 = S \).
Formally, the problem is solved, but we will show how one could n... | 2\sqrt{\piS} | Geometry | proof | Yes | Yes | olympiads | false | 25,716 |
9.7. Prove that if in the conditions of the previous problem the forest is convex, then there exists a path guaranteeing an exit from the forest with a length of no more than $\sqrt{2 \pi S}$. | 9.7. The paratrooper must move along a semicircle of radius $\sqrt{2 s / \pi}$. The length of this semicircle,
guaranteeing an exit, is exactly $\sqrt{2 \pi S}$. The proof that this path works is obvious (cf. problem 9.6). | \sqrt{2\piS} | Geometry | proof | Yes | Yes | olympiads | false | 25,717 |
9.9. In the forest, all trees are taller than 10 meters and shorter than 50 meters, and the distance between any two trees is no more than the difference in their heights. Prove that this forest can be surrounded by a fence of length $80 \mathrm{~m}$.
convex quadrilateral | 9.9. Notice that the path starting from the tallest tree then goes straight to the next tallest tree, then to the third tallest, and so on, to the shortest tree, has a length of less than $40 \mathrm{~m}$. By fencing this path, which connects all the trees, with a fence (80 m long), we will thereby enclose the entire f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,719 |
10.2. The diagonal ' $A C$ divides the area of the convex quadrilateral $A B C D$ into two equal parts. Prove that if $A B>A D$, then $B C<D C$.
is the angle obtuse | 10.2. Notice that the midpoint $K$ of segment $B D$ lies on the line $A C$. Compare the angles $\angle A K B=\angle C K D$ and $\angle A K D=\angle C K B$ (see figure). | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,720 |
10.3. On the extension of side $A C$ of triangle $A B C$, segment $C D=C B$ is laid off from $C$. Prove that if $A C>B C$, then angle $A B D$ is obtuse. | 10.3. Let's draw a circle with radius \( CD = BC \) centered at point \( C \). It will intersect segment \( AC \) at some point \( E \) (since \( AC > BC \)). The inscribed angle \( EBD \) is a right angle (it subtends the diameter), so angle \( ABD \) is obtuse.
Another solution can be obtained by noting that the bis... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,721 |
10.4. a) The heights of a triangle are 3, 4, and 5. Is this triangle a right triangle, an acute triangle, or an obtuse triangle?
b) The same question, if the medians of the triangle are 3, 4, and 5.
## VERY Acute angle | 10.4. a) Let \(a, b, c\) be the sides of the triangle, \(h_{a}=3, h_{b}=4, h_{c}=5\) be the heights dropped to these sides, and \(s\) be the area of the triangle. Then \(a h_{a} = b h_{b} = c h_{c} = 2 s\), from which \(a = \frac{2 s}{h_{a}}, b = \frac{2 s}{h_{b}}, c = \frac{2 s}{h_{c}}\), i.e., the given triangle is s... | ) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,722 |
10.5. Prove that in any convex 11-gon, there exist two diagonals such that the smaller angle between the lines on which they lie does not exceed $5^{\circ}$ (the angle between parallel lines is considered to be $0^{\circ}$).
## PERIMETER ESTIMATION | 10.5. First, let's calculate the number of diagonals in a convex $n$-gon. In each of the $n$ vertices, $(n-3)$ diagonals converge, but in the product $n(n-3)$, each diagonal is counted twice, as it connects 2 vertices. Therefore, a convex $n$-gon has $n(n-3) / 2$ diagonals. In particular, an 11-gon has $\frac{11 \cdot ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,723 |
10.6. The midpoints of adjacent sides of a convex polygon are connected by segments. Prove that the perimeter of the polygon formed by these segments is not less than half the perimeter of the original polygon.
## SQUARE IN A TRIANGLE | 10.6. For a triangle, the statement of the problem is obvious (the inequality turns into an equality), so we will assume that the number \( n \) of the polygon's sides is at least four.
Draw all \( n \) diagonals in this \( n \)-sided polygon, connecting its vertices every other one. Each of these diagonals is twice a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,724 |
10.7. Prove that the area of a square lying inside a triangle does not exceed half the area of this triangle.
## piece OF hExAgOn
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 10.7. We will prove a more general statement, namely, that the area of any parallelogram lying inside a triangle does not exceed half the area of this triangle. Extend the opposite sides $M N$ and $P Q$ of the parallelogram $M N P Q$ until they intersect the sides of the triangle $A B C$ (figure a). At least two of the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,725 |
10.8. Prove that in any convex hexagon, there exists a diagonal that cuts off a triangle with an area no greater than one-sixth of the area of the hexagon.
PIECE OF CAKE | 10.8. Let \( A B C D E F \) be an arbitrary convex hexagon (see figure). Draw the diagonals \( A D, B E \), and \( C F \). Let \( M, N, P \) be the points of intersection of the diagonals (they may coincide). The hexagon is divided into 3 quadrilaterals \( A B M F \), \( B C D N \), \( D E F P \), and the triangle \( M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,726 |
11.1. On a plane, an angle with vertex $A$ is drawn, and a point $M$ is given inside it. Indicate points $B$ and $C$ on the sides of the angle such that segments $A B$ and $A C$ are equal and the sum $M B + M C$ is minimized. | 11.1. Let $M^{\prime}$ be the point obtained from point $M$ by rotating it around point $A$ by a given angle,
Fig. 11.1. Then the point $C$ should be the intersection of the segment $M M^{\... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,728 |
11.3. How to cut out a rhombus of the largest area from a rectangle?
## POINT IN A TRIANGLE | 11.3. The largest area is that of a rhombus, one of whose diagonals coincides with the diagonal of a rectangle, and the ends of the other lie on the sides of the rectangle.
Prove that any other rhombus has a smaller area. Let \(ABCD\) be the given rectangle (see figure), \(AD\) and \(BC\) be its larger sides, and let
... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,730 |
11.4. Given a triangle $A B C$. Where in its plane should a point $M$ be chosen so that the sum of the radii of the circumcircles of triangles $A B M$ and $B C M$ is minimized? | 11.4. Answer. Point $M$ is the foot of the altitude $B M$ of triangle $A B C$. Notice that the diameters of the circles mentioned in the condition cannot be less than the chords $A B$ and $B C$ respectively. | Point\M\is\the\foot\of\the\altitude\BM\of\triangle\ABC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,731 |
11.6. Inside a circle with center $O$, there is a point $A$ different from the center $O$. Find a point $M$ on the circle for which the angle $A M O$ is maximized. | 11.6. Answer. The angle OMA is maximal when the angle $O A M$ is a right angle.
Let $M$ be an arbitrary point on the circle and $O K$ be the perpendicular dropped from point $O$ to the line $M A$ (see figure). In triangle KMO, the hypotenuse $O M$ is constant (equal to the radius), therefore, the angle $A M O$ is grea... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,733 |
11.7. Given two concentric circles with center $O$. Through a point $A$, lying inside the smaller of them, draw a ray so that its segment, enclosed between the circles, is
a) the smallest,
b) the largest. | 11.7. Answer. a) rays lying on the line $O A$, b) rays perpendicular to $O A$.
Prove that the segment $M P$, enclosed between the circles (see figure), is the larger, the larger the angle $A M O$. After this, problem b) reduces to problem 11.6.
## ADDITIONAL PROBLEMS
1. a) Into how many parts can four different line... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,734 | |
1. Let $S(x)$ denote the sum of the digits of a natural number $x$. Solve the equations:
a) $x+S(x)+S(S(x))=1993$
b)* $x+S(x)+S(S(x))+S(S(S(x)))=1993$.
2*. It is known that the number $n$ is the sum of the squares of three natural numbers. Show that the number $n^{2}$ is also the sum of the squares of three natural ... | 1. a) According to the divisibility rule for 3 (see fact 6), the numbers $x$ and $S(x)$ give the same remainder when divided by 3. The same remainder will also be given by the number $S(S(x))$. Therefore, the sum
$$
x + S(x) + S(S(x))
$$
is divisible by 3 (since it is the sum of three numbers with the same remainder ... | 1963 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,735 |
1. For two given distinct points $A$ and $B$ on a plane, find the geometric locus of points $C$ such that triangle $ABC$ is acute-angled, and its angle $A$ is the middle in size.
Comment. By the middle in size angle, we mean an angle that is not greater than one of the angles and not less than the other. For example, ... | 1. Draw a line through point $A$ perpendicular to segment $A B$. It is clear that $\angle B A C < 90^{\circ}$ if and only if points $B$ and $C$ lie on the same side of this line. Now it is clear that the set of points such that $\angle A < 90^{\circ}$ and $\angle B < 90^{\circ}$ is a strip, the boundaries of which pass... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,736 |
1. When the numbers $A$ and $B$ are expressed as infinite decimal fractions, the lengths of the minimal periods of these fractions are 6 and 12, respectively. What can the length of the minimal period of the number $A+B$ be? | 1. As is known, the length of the minimal period of a fraction is a divisor of the length of any other period of the fraction, see fact 4 (the length of the minimal period of a finite decimal fraction will be considered equal to one).
Let's prove the following statement: if $k$ is the length of one of the periods (not... | 12or4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,737 |
1. It is known that $\operatorname{tg} \alpha+\operatorname{tg} \beta=p, \operatorname{ctg} \alpha+\operatorname{ctg} \beta=q$. Find $\operatorname{tg}(\alpha+\beta)$. | 1. If $\operatorname{tg}(\alpha+\beta)$ is defined, then
$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=\frac{p}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
The product of the tangents is related to ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,738 | |
3. On a line, there are two chips, a red one on the left and a blue one on the right. It is allowed to perform any of two operations: inserting two chips of the same color consecutively at any point on the line and removing any two adjacent chips of the same color. Is it possible to leave exactly two chips on the line ... | 3. The first method. We will call two chips, where the left one is red and the right one is blue, an inversion. Consider the number of all inversions (for example, in Fig. 20, the number of inversions is 3). It turns out that the parity of this indicator
 700 meters, b)* 800 meters from each of the banks?
Note. It is known that the river connects two... | 3. а) The example is shown in Fig. 30. Here $A C=1000$ m, $A B>1400$ m, $C D=1$ m. The ship's route must intersect the segment $A B$, but the distance from any point on $A B$ to one of the shores is more than $700 \mathrm{m}$.
б) This problem turned out to be unexpectedly difficult. Let's first formulate two
. Then the sequence $x_{n}=\{a n+b\}$ corresponds to a sequence of points on the circle obtained ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,745 |
5. Does there exist a finite word made of letters from the Russian alphabet, in which there are no two adjacent identical substrings, but such appear when appending (either to the right or to the left) any letter of the Russian alphabet.
C om m entar y. By a word, we mean any sequence of letters from the Russian alpha... | 5. Consider the sequence of words:
$$
\text { A, ABA, ABAVABA, ABAVABAGABAVABA, ... }
$$
The next word is obtained from the previous one as follows: the previous word is written, then the first of the letters that are not in it, and then the same word again.
We will prove by complete induction (see fact 24) the foll... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,747 |
5. Each pair of numbers $x$ and $y$ is assigned a number $x * y$. Find $1993 * 1935$, given that for any three numbers $x, y$ and $z$ the identities $x * x=0$ and $x *(y * z)=(x * y)+z$ are satisfied. | 5. Let's take $y=z$ in the second identity. Then we get
$$
(x * y)+y=x *(y * y)=x * 0
$$
Thus, $x * y=x * 0 - y$. It remains to compute $x * 0$. For this, take $x=y=z$ in the second identity:
$$
x * 0=x *(x * x)=x * x + x=0 + x=x.
$$
Thus, $x * y=x * 0 - y=x - y$. Therefore, $1993 * 1935=1993 - 1935=58$.
Comment. ... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,748 |
5. In a botanical key, plants are described by one hundred characteristics. Each characteristic can either be present or absent. A key is considered good if any two plants differ in more than half of the characteristics. Prove that in a good key, no more than 50 plants can be described. | 5. Let $m$ be the number of plants in a good determinant. We estimate the total number of differences between all pairs of plants across all characteristics. The number of pairs of plants is $\frac{m(m-1)}{2}$, and each pair differs in at least 51 characteristics, so the total number of differences $S \geqslant 51 \fra... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,749 |
5. a) It is known that the domain of the function $f(x)$ is the interval $[-1 ; 1]$, and $f(f(x)) = -x$ for all $x$, and its graph is the union of a finite number of points and intervals. Draw the graph of the function $f(x)$.
b) Can this be done if the domain of the function is the interval $(-1 ; 1)$? The entire num... | 5. a) For example, we can take:
$$
f(x)= \begin{cases}-\frac{1}{2}-x & \text { for } x \in\left[-1 ;-\frac{1}{2}\right) ; \\ x-\frac{1}{2} & \text { for } x \in\left[-\frac{1}{2} ; 0\right) ; \\ 0 & \text { for } x=0 ; \\ x+\frac{1}{2} & \text { for } x \in\left(0 ; \frac{1}{2}\right] ; \\ \frac{1}{2}-x & \text { for ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,750 |
6. A circle with center $D$ passes through points $A, B$ and the center $O$ of the excircle of triangle $A B C$, which touches its side $B C$ and the extensions of sides $A B$ and $A C$. Prove that points $A, B, C$ and $D$ lie on the same circle.
## 9 t h g r a d e | 6. Let $\alpha, \beta, \gamma$ be the angles at the vertices of $\triangle ABC$ (Fig. 23), then
$$
\angle BAO = \frac{\alpha}{2}, \quad \angle CBO = 90^{\circ} - \frac{\beta}{2}
$$
(since point $O$ lies on the bisector of angle $A$ and on the bisector of the external angle at vertex $B$, see fact 16),
$$
\angle ABO ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,751 |
6. Given a convex quadrilateral $A B M C$, in which $A B=B C, \angle B A M=30^{\circ}, \angle A C M=150^{\circ}$. Prove that $A M$ is the bisector of angle $B M C$.
## 10th grade | 6. First method. Let $B^{\prime}$ be the point symmetric to point $B$ with respect to the line $A M$ (Fig. 27). Then $A B=A B^{\prime}, \quad \angle B A B^{\prime}=60^{\circ}$, and triangle
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,752 |
6. On the side $A B$ of triangle $A B C$, a square is constructed externally with center $O$. Points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively, and the lengths of these sides are $a$ and $b$ respectively. Find the maximum of the sum $O M+O N$ as the angle $A C B$ changes.
## 11 k l a ss | 6. Let $D$ and $E$ be the remaining vertices of the square $ABDE$ and $\gamma = \angle ACB$ (Fig. 32).
Then (by the midline theorem of a triangle) from $\triangle ACD$ we get: $CD = 2OM$. Similarly, $CE = 2ON$. Therefore, it is sufficient to find the maximum of $CD + CE = 2(OM + ON)$.
First method. On the side $BC$ o... | \frac{1+\sqrt{2}}{2}(b) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,753 |
1. A cooperative receives apple and grape juice in identical barrels and produces an apple-grape drink in identical cans. One barrel of apple juice is enough for exactly 6 cans of the drink, and one barrel of grape juice is enough for exactly 10 cans. When the recipe for the drink was changed, one barrel of apple juice... | 1. The first method. For one can of the drink, $\frac{1}{6}$ of a barrel of apple juice and $\frac{1}{10}$ of a barrel of grape juice are used, so the volume of the can is
$$
\frac{1}{6}+\frac{1}{10}=\frac{4}{15}
$$
of the barrel's volume.
After changing the recipe, for one can of the drink, $\frac{1}{5}$ of a barre... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,758 |
1. Does there exist a non-convex pentagon in which no two of the five diagonals have any points in common (except at the vertices)? | 1. Let five points $A, B, C, D$ and $E$, two of which $D$ and $E$ lie inside the triangle $A B C$, be connected by segments. These 10 segments can be divided into two non-self-intersecting broken lines, each consisting of 5 segments (Fig. 45). Any one of these broken lines can be taken as a pentagon. Then the second br... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,759 |
1. A student did not notice the multiplication sign between two seven-digit numbers and wrote a single fourteen-digit number, which turned out to be three times their product. Find these numbers. | 1. A similar problem was in the 8th grade (problem 2). More detailed solutions are provided there.
First method. Let $x$ and $y$ be the required numbers, then we can form the equation: $3 \cdot x \cdot y = 10^7 x + y$ (see fact 11), $3 y = 10^7 + \frac{y}{x}$. Since $\frac{y}{x}$ is a number between 0 and 10, we have ... | x=1666667,y=3333334 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,760 |
1. Invent a polyhedron that does not have three faces with the same number of sides. | 1. We will look for a polyhedron with the smallest number of faces. A polyhedron cannot be made from two triangles and two squares.
A hexahedron with two triangular, two quadrilateral, and two pentagonal faces can be constructed: place the two pentagons with a common edge in the shape of an open shell, and fill the ga... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,761 |
2. A student did not notice the multiplication sign between two three-digit numbers and wrote a single six-digit number, which turned out to be seven times greater than their product. Find these numbers. | 2. First method. Let $x, y$ be the desired three-digit numbers. If we append three zeros to the number $x$, we get the number $1000 x$, and if we append $y$, we get $1000 x + y$ (see fact 11).
Thus, the student wrote the number $1000 x + y$. According to the problem, this number is seven times greater than $x \cdot y$... | 143 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,762 |
2. Kolya has a segment of length $k$, and Lev has a segment of length $l$. First, Kolya divides his segment into three parts, and then Lev divides his segment into three parts. If two triangles can be formed from the resulting six segments, Lev wins; otherwise, Kolya wins. Depending on the ratio $k / l$, who can ensure... | 2. First case. If $k > l$, then Kolya wins: he can cut off a part from $k$ that will be greater than the sum of all the others.
For example, he can cut $k$ into parts (Fig. 46)
$$
l+\frac{2}{3}(k-l), \frac{1}{6}(k-l), \frac{1}{6}(k-l)
$$
 in the case b) and only in the case, if $x_{1}$ is rational. | 2. a) If $0 \leqslant x_{n} \leqslant 1$, then $0 \leqslant x_{n+1} \leqslant 1$. Indeed,
$$
\begin{aligned}
0 \leqslant x_{n} \leqslant 1 \Rightarrow -1 \leqslant 1-2 x_{n} & \leqslant 1 \Rightarrow \\
& \Rightarrow 0 \leqslant\left|1-2 x_{n}\right| \leqslant 1 \Rightarrow 0 \leqslant x_{n+1} \leqslant 1
\end{aligned... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,764 |
3. In triangle $ABC$, the angle bisectors of angles $A$ and $C$ are drawn. Points $P$ and $Q$ are the feet of the perpendiculars dropped from vertex $B$ to these bisectors. Prove that segment $PQ$ is parallel to side $AC$. | 3. Extend segments $B Q$ and $B P$ until they intersect line $A C$ at points $A_{1}$ and $C_{1}$, respectively. In triangle $A B C_{1}$, segment $A P$ is both a bisector and an altitude (Fig. 41). Therefore, this triangle is isosceles $\left(A B=A C_{1}\right)$. Then segment $A P$ is also a median: $B P=C_{1} P$. Simil... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,766 |
3. Prove that the equation $x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}$ has an infinite number of solutions in integers $x, y, z$. | 3. For example: $x=k\left(2 k^{2}+1\right), y=2 k^{2}+1, z=-k\left(2 k^{2}+1\right)$.
How to come up with these formulas? By eliminating two cubes using the substitution $z=-x$, we obtain the equation $2 x^{2}+y^{2}=y^{3}, 2 x^{2}=(y-1) y^{2}$. If $\frac{y-1}{2}=k^{2}$ is a perfect square, then the solution can be fou... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,767 |
3. A cherry, a sphere of radius $r$, is dropped into a round glass, the axial cross-section of which is the graph of the function $y=x^{4}$. For what largest $r$ will the sphere touch the lowest point of the bottom? (In other words, what is the maximum radius $r$ of a circle lying in the region $y \geqslant x^{4}$ and ... | 3. Let's first solve another problem: construct a circle with its center on the $y$-axis, which touches the $x$-axis, and determine the smallest radius $r$ for which it has a common point with the curve $y=x^{4}$, different from the origin (Fig. 57). In other words, for what smallest $r$ does the system of equations
$... | \frac{3\sqrt[3]{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,769 |
4. Four grasshoppers sit at the vertices of a square. Every minute, one of them jumps to a point symmetric to it relative to another grasshopper. Prove that the grasshoppers cannot end up at the vertices of a larger square at some point. | 4. $1^{\circ}$. Let's imagine that the square, with grasshoppers sitting at its vertices, is a square on a grid paper (the size of the square is $1 \times 1$). Notice that the grasshoppers always jump to the vertices of the cells: if grasshoppers are placed at the vertices of the cells on the grid paper (these vertices... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,770 |
4. Two circles intersect at points $A$ and $B$. Tangents are drawn to both circles at point $A$, intersecting the circles at points $M$ and $N$. Lines $B M$ and $B N$ intersect the circles again at points $P$ and $Q$ ( $P$ - on line $B M$, $Q$ - on line $B N$ ). Prove that segments $M P$ and $N Q$ are equal. | 4. The first method. Let the circles be arranged as in Fig. 48. The case of a different arrangement of the circles is considered similarly (see comment).
It is sufficient to prove that
Fig... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,771 |
4. $D$ is a point on side $B C$ of triangle $A B C$. Incircles are inscribed in triangles $A B D, A C D$, and a common external tangent (different from $B C$) is drawn to them, intersecting $A D$ at point $K$. Prove that the length of segment $A K$ does not depend on the position of point $D$ on $B C$. | 4. Let's try to guess what the length of segment $A K$ is. Consider the limiting case when point $D$ approaches $C$, then the second circle shrinks to a point, and segment $A K$ turns into a tangent segment to the first circle. It is easy to see that in this case $A K = \frac{A B + A C - B C}{2}$. We will prove that th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,772 |
4. From a convex polyhedron with 9 vertices, one of which is $A$, by parallel translations that map $A$ to each of the other vertices, 8 equal polyhedra are formed. Prove that at least two of these 8 polyhedra intersect (at interior points). | 4. Let's perform a homothety of the original polyhedron $M$ with the center at point $A$ and a coefficient of 2. The volume of the stretched polyhedron $M^{\prime}$ will be 8 times the volume of the polyhedron $M$. We will prove that all 8 "transferred" polyhedra are contained in $M^{\prime}$.
Let vertex $A$ be transf... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,773 |
5. The court astrologer calls a moment of time good if the hour, minute, and second hands of a clock are on the same side of some diameter of the clock face (the hands rotate on a common axis and do not make jumps). Is there more good or bad time in a day? | 5. Let's make several remarks, each of which is obvious.
$1^{\circ}$. Any two arrows define a "bad" sector between their extensions, entering which, the third arrow creates a bad moment in time. This sector does not exceed $180^{\circ}$ - see Fig. 43.
$2^{\circ}$. After an integer number of hours, the positions of th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 25,774 |
5*. Find the largest natural number, not ending in zero, which, when one (not the first) digit is erased, decreases by an integer factor.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 5. Let $x$ be the deleted digit, $a$ be the part of the number to the left of $x$, and $c$ be the part of the number to the right of $x$. Then the number has the form $\overline{a x c}$, see fact 11. Suppose the digit $x$ is in the $(n+1)$-th place (counting from the right). Then
$$
\overline{a x c}=a \cdot 10^{n+1}+x... | 180625 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,775 |
5*. The extensions of sides $A B$ and $C D$ of a convex quadrilateral $A B C D$ intersect at point $P$, and the extensions of sides $B C$ and $A D$ intersect at point $Q$. Prove that if each of the three pairs of angle bisectors: of the external angles of the quadrilateral at vertices $A$ and $C$, of the external angle... | 5*. Drawing a diagram for the problem is very difficult (the points of intersection are far from each other), so we will look for a solution from general considerations. Let's recall that points on the bisector of an angle are equidistant from the sides of this angle.
For each line containing a side of the quadrilater... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,777 |
6. In a square of grid paper $10 \times 10$, one ship $1 \times 4$, two $1 \times 3$, three $1 \times 2$, and four $1 \times 1$ need to be placed. Ships should not share any points (even vertices) with each other, but they can touch the borders of the square. Prove that
a) if they are placed in the order specified abo... | 6. In task "b", it is easy to provide an example of an "unextendable" arrangement of nine ships (Fig. $50, a$).
In task "a", there is a "hidden rock": it seems sufficient to prove that there will be a place for the last single-cell ship. However, in reality, it is necessary to prove that in the process of placement, t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 25,779 |
6*. Does there exist a polynomial $P(x)$ that has a negative coefficient, while all coefficients of any of its powers $(P(x))^{n}, n>1,$ are positive?
## 11th grade | 6. An important observation is that it is sufficient to find a polynomial such that the coefficients of its square and cube are positive. Any other power can be represented as a product of squares and cubes.
We will call a polynomial
$$
a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}
$$
positive if all its coeffici... | (x)=x^{4}+x^{3}-\varepsilonx^{2}+x+1 | Algebra | proof | Yes | Yes | olympiads | false | 25,780 |
6. Prove that for any $k>1$ there exists a power of 2 such that among the $k$ last digits of this power, at least half are nines. (For example, $2^{12}=\ldots 96, 2^{53}=\ldots 992$.) | 6. What can cause a lot of nines in a row in powers of two? - If a power of two is "slightly" less than a number divisible by a high power of ten. For example, $2^{12}+4$ is divisible by $100$, and $2^{53}+8$ is divisible by $1000$.
Let's first try to find numbers of the form $2^{n}+1$ that are divisible by a high pow... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,781 |
1. M. V. Lomonosov spent one coin on bread and kvass. When prices increased by $20 \%$, he could buy half a loaf of bread and kvass with that same coin. Would that same coin be enough to buy at least kvass if prices rise by another $20 \%$? | 1. Let the price of bread be $x$, and the price of kvass be $y$ (before the price increase). Then, according to the condition: money $=x+y$. After the price increase, bread costs $1.2 x$, and kvass costs $1.2 y$. Therefore,
$$
\text { money }=0.5 \text { bread }+ \text { kvass }=0.6 x+1.2 y .
$$
From these equations,... | 1.5y>1.44y | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,782 |
1. Prove that if any number of threes are inserted between the zeros in the number 12008, the resulting number is divisible by 19. | 1. We will prove the statement by induction (see fact 24). Base of induction: 12008 is divisible by 19, indeed, $12008=19 \cdot 632$.
Induction step. We will show that if a number of the given form is divisible by 19, then the next one is also divisible by 19. For this, it is sufficient to prove that the difference of... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,783 |
1. Given the value of $\sin \alpha$. What is the maximum number of values that a) $\sin \frac{\alpha}{2}$, b)* $\sin \frac{\alpha}{3}$ can take? | 1. a) First, let's show that $\sin \frac{\alpha}{2}$ cannot take more than 4 values. Indeed, if $\sin \alpha = \sin \beta$, then $\beta = \alpha + 2\pi k$ or $\beta = \pi - \alpha + 2\pi k$ (where $k$ is an integer). Accordingly, $\frac{\beta}{2} = \frac{\alpha}{2} + \pi k$ or $\frac{\beta}{2} = \frac{\pi - \alpha}{2} ... | )4,b)3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,784 |
1. Prove that
$$
|x|+|y|+|z| \leqslant|x+y-z|+|x-y+z|+|-x+y+z|
$$
where $x, y, z$ are real numbers. | 1. Since the modulus of the sum does not exceed the sum of the moduli (see comment), we have:
$$
|x+y-z|+|x-y+z| \geqslant|(x+y-z)+(x-y+z)|=2|x| \text {. }
$$
Similarly, we obtain the inequalities
$$
\begin{aligned}
& |x-y+z|+|-x+y+z| \geqslant 2|z| \\
& |-x+y+z|+|x+y-z| \geqslant 2|y|
\end{aligned}
$$
Adding all t... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,785 |
2. Prove that all numbers 10017, 100117, 1001117, ... are divisible by 53. | 2. We will prove the statement by induction (see fact 24).
Base of induction: 10017 is divisible by 53. Indeed, $10017=53 \cdot 189$.
Step of induction. We will show that if a number of the given form is divisible by 53, then the next one is also divisible by 53. For this, we will compute the difference between two c... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,786 |
2. Given an equilateral triangle $A B C$. For an arbitrary point $P$ inside the triangle, consider the points $A^{\prime}$ and $C^{\prime}$ as the intersections of lines $A P$ with $B C$ and $C P$ with $B A$ respectively. Find the geometric locus of points $P$ for which segments $A A^{\prime}$ and $C C^{\prime}$ are eq... | 2. Let's choose an arbitrary point $A^{\prime}$ on $B C$ and draw the segment $A A^{\prime}$ (Fig. $61, a$). We will prove that among the segments with the starting point at $C$ and the endpoint on side $A B$, there are only two equal to segment $A A^{\{\prime}$ - these are such segments $C C_{1}$ and $C C_{2}$, that
... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,787 |
2. Can the edges of an $n$-sided prism be colored in 3 colors so that each face has all 3 colors and at each vertex, edges of different colors meet, if a) $n=1995$; b) $n=1996$. | 2. а) Let's paint the edges of the lower base in a circle: the 1st in the first color, the 2nd in the second color, the 3rd in the third color, the 4th again in the first color, and so on (1995 is divisible by 3, so the chain will close). Each edge of the upper base will be painted in the color of the corresponding edg... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,789 |
3. Given a convex quadrilateral $A B C D$ and a point $O$ inside it. It is known that $\angle A O B=\angle C O D=120^{\circ}, A O=O B$ and $C O=O D$. Let $K, L$ and $M$ be the midpoints of segments $A B, B C$ and $C D$ respectively. Prove that a) $K L=L M$; b) triangle $K L M$ is equilateral. | 3. а) When rotated $120^{\circ}$ clockwise around point $O$, segment $A C$ transitions to $B D$, which means their lengths are equal (Fig. 59). Segment $K L$
Fig. 59 is the midline of tria... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,790 |
3. A rectangle of size $1 \times k$ for any natural number $k$ will be called a strip. For which natural numbers $n$ can a rectangle of size $1995 \times n$ be cut into pairwise distinct strips? | 3. Idea of the solution: take the maximum strip (equal to the maximum side of the rectangle). The remaining strips will be combined in pairs, giving the sum of the maximum strip. If we have filled the rectangle, the problem is solved; otherwise, reasoning with areas shows that the rectangle cannot be cut into different... | 3989 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 25,791 |
3. The diagonals of trapezoid $ABCD$ intersect at point $K$. Circles are constructed on the lateral sides of the trapezoid as diameters. Point $K$ lies outside these circles. Prove that the lengths of the tangents drawn from point $K$ to these circles are equal. | 3. Let the lateral sides of the trapezoid be sides $AB$ and $CD$. Denote by $M$ and $N$ the second points of intersection of the lines $AC$ and $BD$ with the circles having diameters $AB$ and $CD$, respectively (Fig. 64). If the line $AC$ is tangent to the circle with diameter $AB$, we set $M=A$; similarly, if the line... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,792 |
3. In triangle $ABC$, $AA_{1}$ is a median, $AA_{2}$ is a bisector, $K$ is a point on $AA_{1}$ such that $KA_{2} \| AC$. Prove that $AA_{2} \perp KC$.
In triangle $ABC$, $AA_{1}$ is a median, $AA_{2}$ is a bisector, $K$ is a point on $AA_{1}$ such that $KA_{2} \| AC$. Prove that $AA_{2} \perp KC$. | 3. First with the allowance. Extend the median $A A_{1}$ by its length and complete the triangle to a parallelogram $A B D C$ (Fig. 66). Let's recall that the bisector divides the base of the triangle into segments proportional to the sides:
$$
\frac{A_{2} B}{A_{2} C}=\frac{A B}{A C}
$$
From the generalized Thales' t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,793 |
4. Is it enough to make a closed rectangular box from all sides, enclosing no less than 1995 unit cubes, a) 962; b) 960; c) 958 square units of material? | 4. It is sufficient to take a box of size $11 \times 13 \times 14$. Its volume is 2002, which is sufficient; and the total area of its walls is $2 \cdot(11 \cdot 13+11 \cdot 14+13 \cdot 14)=958$.
Comments. $1^{\circ}$. How can one guess such a solution? It is known that among all parallelepipeds with a given volume, t... | 958 | Geometry | MCQ | Yes | Yes | olympiads | false | 25,794 |
4. Natural numbers $a, b, c, d$ are such that $a b=c d$. Can the number $a+b+c+d$ be prime? | 4. First solution. From the condition, it follows that
$$
a+b+c+d=a+b+c+\frac{a b}{c}=\frac{(a+c)(b+c)}{c}
$$
- is an integer. Therefore, the fraction is reducible. Since both factors in the numerator are greater than the denominator, after reduction, each of them will remain a number greater than one. Thus, the numb... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 25,795 |
5. Several settlements are connected to the city by roads, but there are no roads between them. A car departs from the city with cargos for all the settlements at once. The cost of each trip is equal to the product of the weight of all cargos in the truck and the distance. Prove that if the weight of each cargo is nume... | 5. First with the allowance. First, let's prove that the cost of transportation does not change from the permutation of the order of two consecutive trips. In other words, let the car visit some populated point $M$, and then the populated point $N$. We will show that if it had first gone to $N$, and then to $M$, the co... | proof | Algebra | proof | Yes | Yes | olympiads | false | 25,798 |
5. Initially, four identical right-angled triangles are given. With each move, one of the existing triangles is cut along the altitude (drawn from the right angle) into two other triangles. Prove that after any number of moves, there will be two identical triangles among the triangles. | 5. The first method. First, let's outline the idea of the solution. Note that the order in which the triangles are cut does not matter (in the sense that the final result does not depend on it).
Since we initially have four identical triangles, three of them will have to be cut (Fig. 63). Let's make these three cuts f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,799 |
6. A line cuts off triangle $A K N$ from a regular hexagon $A B C D E F$ such that $A K+A N=A B$. Find the sum of the angles under which segment $K N$ is seen from the vertices of the hexagon ( $\angle K A N+\angle K B N+\angle K C N+\angle K D N+\angle K E N+$ $+\angle K F N$).
## 9 t h g r a d e | 6. Let's assume that $N$ lies on $A B$, and $K$ lies on $A F$ (Fig. 60). Note that $F K = A N$. We choose point $P$ on $B C$, point $R$ on $C D$, point $S$ on $D E$, and point $T$ on $E F$ such that the equalities $F K = A N = B P = C R = D S = E T$ hold. Then $\angle K B N = \angle T A K$, $\angle K C N = \angle S A T... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,801 |
6. Geologists took 80 cans of preserves on an expedition, the weights of which are all known and different (there is a list). After some time, the labels on the cans became unreadable, and only the custodian knows what is where. He can prove this to everyone (i.e., justify which can contains what),
without opening the ... | 6. a) It will be more convenient for us to work with a number of cans divisible by 3. Therefore, we will introduce one more "fictitious" can with zero weight.
Let's assume that the caretaker first puts the 27 lightest cans on the left pan and the 27 heaviest on the right. This will convince all the geologists that the... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 25,802 |
6*. Prove that there are infinitely many composite $n$ such that $3^{n-1}-2^{n-1}$ is divisible by $n$.
保留源文本的换行和格式，直接输出翻译结果。 | 6. Let's take
$$
n=3^{2^{t}}-2^{2^{t}}
$$
and prove that it satisfies the condition of the problem for any natural $t$.
By fact 8, it is sufficient to prove that $n-1$ is divisible by $2^{t}$, i.e., that $3^{2^{t}}-1$ is divisible by $2^{t}$ (since $2^{2^{t}}$ is divisible by $2^{t}$).
We will prove by induction (s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,804 |
7. Does there exist a polyhedron and a point outside it such that from this point none of its vertices are visible? | 7. It is possible to create a spatial "cross" from 6 "pencils" - long, thin parallelepipeds, adjacent to the faces of a unit cube (the cube is only needed to explain the structure). The pencils lie symmetrically relative to the center of the cube, one on each face, with the center of the pencil's face coinciding with t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,805 |
1. It is known that $a+\frac{b^{2}}{a}=b+\frac{a^{2}}{b}$. Is it true that $a=b$? | 1. Bring the left side to a common denominator: $a+$ $+\frac{b^{2}}{a}=\frac{a^{2}+b^{2}}{a}$. Similarly, we handle the right side. We obtain the equality:
$$
\frac{a^{2}+b^{2}}{a}=\frac{a^{2}+b^{2}}{b}
$$
From the condition, it is clear that $a \neq 0, b \neq 0$. Therefore, $a^{2}+b^{2}>0$ (see comment). Thus, both ... | b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 25,806 |
1. Prove that in any convex polygon, there are no more than 35 angles less than $170^{\circ}$. | 1. First method. Assume the opposite. Then, in some convex $n$-gon, there are at least 36 angles less than $170^{\circ}$ (the remaining $n-36$ angles do not exceed $180^{\circ}$). Therefore, the sum of all angles of such a polygon is less than $36 \cdot 170^{\circ} + (n-36) \cdot 180^{\circ}$. However, as is known, the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,807 |
1. Positive numbers $a, b$ and $c$ are such that $a^{2}+b^{2}-$ $-a b=c^{2}$. Prove that $(a-c)(b-c) \leqslant 0$. | 1. The first method. We can assume that $a \geqslant b$ (the case $a \leqslant b$ is analogous). Then $b^{2} \leqslant a b, a^{2} \geqslant a b$, therefore
$$
a^{2} \geqslant a^{2}+b^{2}-a b \geqslant b^{2}
$$
from which $a \geqslant c \geqslant b$. Thus, the first factor in the expression $(a-c)(b-c)$ is non-negativ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,808 |
2. Ten iron weights are arranged in a circle. Between each pair of adjacent weights, there is a bronze ball. The mass of each ball is equal to the difference in mass between its adjacent weights. Prove that the balls can be divided into two groups to balance the scales. | 2. Let the masses of the weights be denoted by $m_{i}$, and the masses of the balls by $x_{i}$. We have
$$
\left(m_{1}-m_{2}\right)+\left(m_{2}-m_{3}\right)+\ldots+\left(m_{9}-m_{10}\right)+\left(m_{10}-m_{1}\right)=0
$$
Indeed, each $m_{i}$ appears in this sum twice: once with a positive sign and once with a negativ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 25,810 |
2. Prove that if for numbers $a, b$ and $c$ the inequalities $|a-b| \geqslant |c|, |b-c| \geqslant |a|, |c-a| \geqslant |b|$ hold, then one of these numbers is equal to the sum of the other two. | 2. First method. Suppose, first, that one of the numbers is zero. Let, for example, $a=0$ (other cases are similar). Then we get the inequalities: $|b| \geqslant|c|$ and $|c| \geqslant|b|$, from which $|b|=|c|$, i.e., $b=c$ or $b=-c$. In the first case, $b=a+c$, in the second case, $a=b+c$. Everything is proven.
Now l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,811 |
2. In a $10 \times 10$ grid, the centers of all unit squares are marked (a total of 100 points). What is the minimum number of lines, not parallel to the sides of the square, needed to cross out all the marked points? | 2. Let's draw all lines parallel to one of the diagonals of the square and containing more than one of the marked points - there are 17 such lines. The un-
Fig. 77 erased will be the two c... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,812 |
3. A circle is circumscribed around triangle $A B C$, and tangents are drawn through points $A$ and $B$, intersecting at point $M$. Point $N$ lies on side $B C$, and line $M N$ is parallel to side $A C$. Prove that $A N=N C$. | 3. Angles $B N M$ and $N C A$ are equal because lines $M N$ and $A C$ are parallel (Fig. 74). Angle $N C A$ subtends arc $A B$, so it is equal to half of this arc. According to the theorem about the angle between a tangent and a chord (see fact 15), angle $B A M$ is also equal to half of arc $A B$. Therefore, $\angle N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,815 |
3. Points $P_{1}, P_{2}, \ldots, P_{n-1}$ divide the side $B C$ of an equilateral triangle $A B C$ into $n$ equal parts: $B P_{1}=P_{1} P_{2}=\ldots=P_{n-1} C$. Point $M$ is chosen on the side $A C$ such that $A M=B P_{1}$. Prove that $\angle A P_{1} M + \angle A P_{2} M + \ldots + \angle A P_{n-1} M = 30^{\circ}$, if ... | 3. The external angle of a triangle is equal to the sum of the two internal angles not adjacent to it. Therefore, considering triangles $A P_{k} M$, we get $\angle P_{k} M C=\angle P_{k} A C+\angle A P_{k} M$, i.e., $\angle A P_{k} M=\angle P_{k} M C-\angle P_{k} A C$. Adding all such equalities, we see that the desire... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,816 |
3. In space, there are eight parallel planes such that the distances between any two adjacent ones are equal. A point is chosen on each of the planes. Can the selected points be the vertices of a cube. | 3. Let's first solve the inverse problem: take an arbitrary cube and draw planes through its vertices, which are parallel to each other and equidistant.
This can be done as follows. Introduce a coordinate system such that the cube under consideration is a unit cube, i.e., its vertices have coordinates 0 or 1. Then the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,817 |
4. Given an equilateral triangle $A B C$. Side $B C$ is divided into three equal parts by points $K$ and $L$, and point $M$ divides side $A C$ in the ratio $1: 2$, starting from vertex $A$. Prove that the sum of angles $A K M$ and $A L M$ is $30^{\circ}$. | 4. Without loss of generality, we can assume that point $K$ is closer to point $B$ than point $L$ (Fig. 72). Then triangle $M K C$ is equilateral (since $M C = K C, \angle M C K = 60^{\circ}$). Therefore, $A B \| M K$ (since $\left.\angle M K C = \angle A B C = 60^{\circ}\right)$. This means that angles $A K M$ and $B ... | 30 | Geometry | proof | Yes | Yes | olympiads | false | 25,818 |
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