problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Problem 5. The numbers $a, b$, and $c$ satisfy the equation $\sqrt{a}=\sqrt{b}+\sqrt{c}$. Find $a$, if $b=52-30 \sqrt{3}$ and $c=a-2$. | Answer: $a=27$.
Solution. We have
$$
\sqrt{b}=\sqrt{52-30 \sqrt{3}}=\sqrt{27-2 \cdot 5 \cdot 3 \sqrt{3}+25}=3 \sqrt{3}-5
$$
Therefore, $\sqrt{a}-\sqrt{a-2}=3 \sqrt{3}-5, \frac{2}{\sqrt{a}+\sqrt{a-2}}=\frac{2}{3 \sqrt{3}+5}, \sqrt{a}+\sqrt{a-2}=\sqrt{27}+\sqrt{25}$. Since the function $f(a)=\sqrt{a}+\sqrt{a-2}$ is in... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,557 |
Problem 6. Lёsha's cottage plot has the shape of a nonagon, which has three pairs of equal and parallel sides (see figure). Lёsha knows that the area of the triangle with vertices at the midpoints of the remaining sides of the nonagon is 12 acres. Help him find the area of the entire cottage plot.
 sides. From the condition, it follows that the quadrilateral $D F G E$ is a parallelogram, since its opposite sides $D E$ and $F G$ are equal and parallel. In... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,558 |
Problem 7. Which number is greater: $\frac{1}{99}$ or
$$
\frac{1}{9903}+\frac{1}{9903+200}+\frac{1}{9903+200+202}+\ldots+\frac{1}{9903+200+202+\ldots+2018} ?
$$ | Answer: The first.
Solution. Let $A$ be the second of the given numbers. If we decrease all denominators of the number $A$ by 3, the resulting number $B$ will be greater than $A$:
$$
A<B=\frac{1}{9900}+\frac{1}{9900+200}+\frac{1}{9900+200+202}+\ldots+\frac{1}{9900+200+202+\ldots+2018}
$$
Consider the denominators of... | \frac{1}{99} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,559 |
Problem 8. In a right triangle $ABC$ with a right angle at $C$, points $P$ and $Q$ are the midpoints of the angle bisectors drawn from vertices $A$ and $B$. The inscribed circle of the triangle touches the hypotenuse at point $H$. Find the angle $PHQ$. | Answer: $90^{\circ}$.
Solution. First, let's prove the following lemma.
Lemma. Let $ABC (\angle C=90^{\circ})$ be a right triangle, $I$ the intersection point of the angle bisectors $AM$ and $BK$, $S$ the midpoint of $KM$, and the inscribed circle of the triangle touches the hypotenuse at point $H$. Then the points $... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,560 |
5.1. $[5-6.4$ (а. - 10 points, б. - 20 points)] On a grid paper, a figure is drawn (see the picture). It is required to cut it into several parts and assemble a square from them (parts can be rotated, but not flipped). Is it possible to do this under the condition that a) there are no more than four parts; b) there are... | Answer: a) yes; b) yes.
Solution. Possible cutting options for parts a) and b) are shown in Fig. 5, a) and b).
Fig. 5, a)
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,561 |
Task 1. On the faces of a six-faced dice, the numbers 1, 2, 3, 4, 5, and 6 are placed. The dice is thrown and falls on the table. After this, the numbers on all faces except one are visible. The numbers on the five visible faces are multiplied. Find the probability that this product is divisible by 16.
Answer: $0.5$. | Solution. If an odd number is on the invisible face, then in the product of the remaining five numbers, 2, 4, and 6 are present, and it is divisible by 16. If an even number is on the invisible face, the product of the remaining five digits is not divisible by 16 - its prime factorization will not have four twos. | 0.5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,564 |
Task 2. Find the number of natural numbers not exceeding 2022 and not belonging to either the arithmetic progression $1,3,5, \ldots$ or the arithmetic progression $1,4,7, \ldots$ | Answer: 674.
Solution. These two progressions define numbers of the form $1+2n$ and $1+3n$. This indicates that the desired numbers are of the form $6n$ and $6n-4, n \in \mathbb{N}$. Since 2022 is divisible by 6, the numbers of the form $6n$ will be $\frac{2022}{6}=337$, and there will be as many numbers of the form $... | 674 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,565 |
Task 3. Find the three last digits of the number $10^{2022}-9^{2022}$. | Answer: 119.
Solution. Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10-1$, then $A(\bmod 1000) \equiv-C_{2022}^{2} \cdot 100+C_{2022}^{1} \cdot 10-1(\bmod 1000) \equiv$ $-\frac{202... | 119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,566 |
Problem 4. The Dursleys are hiding Harry Potter on an island that is 9 km from the shore. The shore is straight. On the shore, 15 kilometers from the point on the shore closest to the island, Hagrid is on a magical motorcycle and wants to reach Harry as quickly as possible. The motorcycle travels along the shore at a s... | # Answer: 3.
Solution. Let $A$ be the point from which Hagrid starts, $B$ be the island, and $C$ be the point where the motorcycle crashes into the sea. The travel time is $t=\frac{A C}{50}+\frac{B C}{40}=p A C+q B C=q\left(\frac{p}{q} A C+B C\right)$, where $p=\frac{1}{50}, q=\frac{1}{40}$. Thus, we aim to minimize $... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,567 |
Problem 6. Point $A$ on the plane is located at the same distance from all points of intersection of two parabolas given in the Cartesian coordinate system on the plane by the equations $y=-3 x^{2}+2$ and $x=-4 y^{2}+2$. Find this distance. | Answer: $\frac{\sqrt{697}}{24}$.
Solution. By plotting the parabolas with the given equations, it is easy to notice that they have 4 common points. Therefore, a point equidistant from them can be at most one. Rewrite the equations of the parabolas as $x^{2}+\frac{1}{3} y-\frac{2}{3}=0$ and $y^{2}+\frac{1}{4} x-\frac{1... | \frac{\sqrt{697}}{24} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,569 |
Problem 7. There is a certain number of identical plastic bags that can be placed inside each other. If all the other bags end up inside one of the bags, we will call this situation a "bag of bags." Calculate the number of ways to form a "bag of bags" from 10 bags.
Explanation. Denote the bag with parentheses.
If we ... | Answer: 719.
Solution. If $\Pi_{n}$ denotes the number of ways for $n$ packages, then:
$$
\begin{gathered}
\Pi_{1}=1, \Pi_{2}=1, \Pi_{3}=2, \Pi_{4}=4, \Pi_{5}=9, \Pi_{6}=20, \Pi_{7}=48, \Pi_{8}=115, \Pi_{9}=286 \\
\Pi_{10}=719
\end{gathered}
$$
The problem is solved by enumerating the cases. For example, if we take ... | 719 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,570 |
10.1. Propose a word problem that reduces to solving the inequality
$$
\frac{11}{x+1.5}+\frac{8}{x} \geqslant \frac{12}{x+2}+2
$$
Write the problem statement, its solution, and the answer.
Example of the required problem. Points A and B are connected by two roads: one is 19 km long, and the other is 12 km long. At 1... | Answer: no more than 4 km/h. | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,571 |
1.1. (2 points) How many roots does the equation $x^{2}-x \sqrt{5}+\sqrt{2}=0$ have? | Answer: 0.
Solution. Since $D=5-4 \sqrt{2}<0$, the equation has no roots. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,572 |
2.1. (2 points) Find the area of the triangle bounded by the line $y=9-3x$ and the coordinate axes. | Answer: 13.5.
Solution. The line intersects the coordinate axes at points $(0,9)$ and $(3,0)$, so $S=\frac{3 \cdot 9}{2}=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,573 |
3.1. (12 points) Find $x$, if $\frac{1}{2+\frac{1}{37}}=\frac{16}{37}$. | Answer: -0.25.
Solution. We find sequentially:
$$
2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}}=\frac{37}{16}, \quad 3+\frac{1}{4+\frac{1}{5+\frac{1}{x}}}=\frac{16}{5}, \quad 4+\frac{1}{5+\frac{1}{x}}=5, \quad 5+\frac{1}{x}=1, \quad x=-\frac{1}{4}
$$ | -0.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,574 |
4.1. (12 points) In trapezoid $A B C D$ with bases $A D=17$ and $B C=9$, points $E$ and $F$ are marked on the bases respectively such that $M E N F$ is a rectangle, where $M$ and $N$ are the midpoints of the diagonals of the trapezoid. Find the length of segment $E F$. | Answer: 4.
Solution. The segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of the bases, so $M N=\frac{A D-B C}{2}=4$. Since $M E N F$ is a rectangle, its diagonals are equal. Therefore, $E F=M N=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,576 |
5.1. (12 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offer... | Answer: 199.
Solution. If the guard asks for 199 coins, then the outsider, agreeing, will give him this amount, but will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. ... | 199 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,577 |
6.1. (12 points) How many minutes after 17:00 will the angle between the hour and minute hands be exactly the same again
保留源文本的换行和格式，直接输出翻译结果。
6.1. (12 points) How many minutes after 17:00 will the angle between the hour and minute hands be exactly the same again | Answer: $54 \frac{6}{11} \approx 54.55$.
Solution. At 17:00, the angle between the hour and minute hands is $\frac{5}{12} \cdot 360^{\circ}=150^{\circ}$. The next moment when the angle will be the same will occur within the next hour, after the minute hand has overtaken the hour hand. Let this happen after $x$ minutes... | 54\frac{6}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,578 |
6.2. How many minutes before $16:00$ was the angle between the hour and minute hands exactly the same in the previous instance | Answer: $21 \frac{9}{11} \approx 21.82$. | 21\frac{9}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,579 |
8.1. (12 points) Find the smallest natural number such that after multiplying it by 9, the result is a number written with the same digits but in some different order. | Answer: 1089.
Solution. Note that the number must start with one, otherwise multiplying by 9 would increase the number of digits. After multiplying 1 by 9, we get 9, so the original number must contain the digit 9. The number 19 does not work, so two-digit numbers do not work. Let's consider three-digit numbers. The s... | 1089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,582 |
10.1. (12 points) The surface of a round table is divided into $n$ identical sectors, in which numbers from 1 to $n (n \geqslant 4)$ are written sequentially clockwise. Around the table sit $n$ players with numbers $1,2, \ldots, n$, going clockwise. The table can rotate around its axis in both directions, while the pla... | Answer: 69.
Solution. Let player No. 3 get one coin exactly $k$ times, then we get the equation $(m-k)-(n-1) k=50$. Exactly in $7 k$ cases, one coin was received by someone from players $2,3,4$, then we get the equation $3(m-7 k)-7 k(n-3)=74$. We expand the brackets, combine like terms, and get a system of two equatio... | 69 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,585 |
1. Let's call a natural number $n$ squareable if the numbers from 1 to $n$ can be arranged in such an order that each member of the sequence, when added to its position, results in a perfect square. For example, the number 5 is squareable, as the numbers can be arranged as: 32154, in which $3+1=2+2=1+3=4$ and $5+4=4+5=... | Answer: 9 and 15.
Solution. The number 7 cannot be squareable, since both numbers 1 and 6 must be in the third position, which is impossible.
The number 9 is squareable, as the numbers from 1 to 9 can be arranged in the following order: $8,2,6$, $5,4,3,9,1,7$ thus the required condition is satisfied.
The number 11 i... | 915 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,586 |
2. Find the area of a right triangle if the height drawn from the right angle divides it into two triangles, the radii of the inscribed circles of which are 3 and 4. | Answer: 150.
Solution. The ratio of the radii of the inscribed circles is equal to the similarity coefficient of the right triangles into which the height divides the original triangle. This coefficient is equal to the ratio of the legs of the original triangle. Let these legs be denoted as $3x$ and $4x$. By the Pytha... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,587 |
3. Find the minimum value of the expression $\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}}$ under the conditions $|x|<1,|y|<2$ and $x y=1$.
Answer: 4. | Solution. Using the inequality between the arithmetic mean and the geometric mean, under the given conditions on $x$ and $y$, we obtain
$$
\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}} \geqslant 2 \sqrt{\frac{1}{1-x^{2}} \cdot \frac{4}{4-y^{2}}}=\frac{4}{\sqrt{5-4 x^{2}-y^{2}}}=\frac{4}{\sqrt{1-(2 x-y)^{2}}} \geqslant 4
$$
and... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,588 |
5. Find all natural numbers $k$ for which the number $k^{2}-101 k$ is a perfect square, i.e., the square of an integer. | Answer: 101 or 2601.
Solution. Let $k^{2}-101 k=(k-m)^{2}, m \geqslant 1$, then $k^{2}-101 k=k^{2}-2 m k+m^{2}$, from which $m(2 k-m)=101 k$. Since the number 101 is prime, one of the numbers $m$ or $2 k-m$ is divisible by 101.
If $m=101 s$, then we get $s(2 k-101 s)=k$, from which
$$
k=\frac{101 s^{2}}{2 s-1}=101\l... | 101or2601 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,589 |
6. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place
# | # Answer: 41000.
Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number a... | 41000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,590 |
8. A moth made a hole in the carpet in the shape of a rectangle with sides 10 cm and 4 cm. Find the smallest size of a square patch that can cover this hole (the patch covers the hole if all points of the rectangle lie inside the square or on its boundary). | Answer: $7 \sqrt{2} \times 7 \sqrt{2}$ cm.
Solution. We need to find the smallest square that can contain a rectangle of $10 \times 4$ cm. If fewer than 4 vertices of the rectangle lie on the sides of the square, the size of the square can be reduced by rotating it relative to the rectangle. Let the vertices of the re... | 7\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,591 |
1. Two pieces of cheese have the shape of a rectangular parallelepiped each. The length of the first piece is $50\%$ greater than the length of the second piece, while the width and height of the first piece are $20\%$ and $30\%$ less than the width and height of the second piece, respectively. Which piece of cheese ha... | Answer: The second one is $19 \frac{1}{21} \%$ more (the first one is $16 \%$ less).
Solution. Let $a, b$ and $c$ be the length, width, and height of the second piece of cheese, respectively. Then its volume is $V_{2}=a b c$. According to the problem, the volume of the first piece of cheese is $V_{1}=\frac{3}{2} a \cd... | 19\frac{1}{21} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,592 |
2. Solve the inequality $\sqrt{x^{2}-1} \leqslant \sqrt{5 x^{2}-1-4 x-x^{3}}$. | Answer: $(-\infty ;-1] \cup\{2\}$.
Solution. $\sqrt{x^{2}-1} \leqslant \sqrt{5 x^{2}-1-4 x-x^{3}} \Leftrightarrow 0 \leqslant x^{2}-1 \leqslant 5 x^{2}-1-4 x-x^{3} \Leftrightarrow$
$$
\Leftrightarrow\left\{\begin{array}{l}
x \in(-\infty ;-1] \cup[1 ;+\infty), \\
x(x-2)^{2} \leqslant 0
\end{array} \Leftrightarrow x \i... | (-\infty;-1]\cup{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 6,593 |
3. Find all two-digit numbers of the form $\overline{X Y}$, if the number with a six-digit decimal representation $\overline{64 X 72 Y}$ is divisible by 72. | Answer: $80,98$.
Solution. Since $\overline{64 X 72 Y}$ is divisible by 8, then $Y \in\{0,8\}$, and since $\overline{64 X 72 Y}$ is divisible by 9, then $6+4+X+7+2+Y=19+X+Y$ is a multiple of 9. If $Y=0$, then $X=8$, and if $Y=8$, then either $X=0$ (cannot be the first digit of the number), or $X=9$. Therefore, the des... | 80,98 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,594 |
4. Having walked $2 / 5$ of the length of a narrow bridge, a pedestrian noticed that a car was approaching the bridge from behind. Then he walked back and met the car at the beginning of the bridge. If the pedestrian had continued walking forward, the car would have caught up with him at the end of the bridge. Find the... | Answer: 5.
Solution. In the time $t$ that the pedestrian walked towards the car until they met at the beginning of the bridge, he covered $2 / 5$ of the bridge's length. Therefore, if the pedestrian continued walking forward, in time $t$ he would have covered another $2 / 5$ of the bridge's length, and he would have $... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,595 |
5. Find all solutions of the equation $|\sin 2 x-\cos x|=|| \sin 2 x|-| \cos x||$ on the interval $(-2 \pi ; 2 \pi]$. Answer: $(-2 \pi ;-\pi] \cup[0 ; \pi] \cup\left\{-\frac{\pi}{2} ; \frac{3 \pi}{2} ; 2 \pi\right\}$. | Solution.
$$
\begin{aligned}
& |\sin 2 x-\cos x|=|| \sin 2 x|-| \cos x|| \Leftrightarrow(\sin 2 x-\cos x)^{2}=(|\sin 2 x|-|\cos x|)^{2} \Leftrightarrow \\
& \Leftrightarrow \sin 2 x \cos x=|\sin 2 x| \cdot|\cos x| \Leftrightarrow \sin 2 x \cos x \geqslant 0 \Leftrightarrow \sin x \cos ^{2} x \geqslant 0 \Leftrightarro... | (-2\pi;-\pi]\cup[0;\pi]\cup{-\frac{\pi}{2};\frac{3\pi}{2};2\pi} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,596 |
7. How many solutions does the equation
\[
\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}=\frac{2}{x^{2}} ?
\]
 | Answer: 1.
Solution. Let's introduce the functions $f(x)=\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}$ and $g(x)=\frac{2}{x^{2}}$.
For $x<0$, both functions $f$ and $g$ are positive, but $f(x)$ increases from $+\infty$ to 2 (not inclusive), since $f^{\prime}(x)=\frac{-2}{(x-1)^{3}}-\frac{2}{(x-2)^{3}}<0 \quad \text { for ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,598 |
8. Given three points, the distances between which are 4, 6, and 7. How many pairwise distinct triangles exist for which each of these points is either a vertex or the midpoint of a side? | Answer: 11.
Solution. Let's list all the constructions of triangles that satisfy the condition of the problem, with
| | Description of the triangle | | Side lengths |
| :---: | :---: | :---: | :---: |
| №1 | All points are vertices | | $4,6,7$ |
| | One point is a vertex, two are midpoints of sides | | |
| №2 |... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,599 |
9. To what power must the root $x_{0}$ of the equation $x^{11} + x^{7} + x^{3} = 1$ be raised to obtain the number $x_{0}^{4} + x_{0}^{3} - 1 ?$ | Answer: 15.
Solution. If $x_{0}=1$, then $x_{0}^{4}+x_{0}^{3}-1=1$, so in this case the degree can be any. But the number $x_{0}=1$ does not satisfy the equation $x^{11}+x^{7}+x^{3}=1$, therefore $x_{0} \neq 1$.
Since $1=x_{0}^{11}+x_{0}^{7}+x_{0}^{3}$, we get
$$
x_{0}^{4}+x_{0}^{3}-1=x_{0}^{4}+x_{0}^{3}-x_{0}^{11}-... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,600 |
10. A sphere touches all the edges of the pyramid $S A B C$, and the lateral edges $S A, S B$, and $S C$ at points $A^{\prime}, B^{\prime}$, and $C^{\prime}$. Find the volume of the pyramid $S A^{\prime} B^{\prime} C^{\prime}$, if $A B=B C=S B=5$ and $A C=4$. | Answer: $\frac{2 \sqrt{59}}{15}$.
Solution. Since the sphere touches all edges, the pyramid has the property that
$$
B C+A S=A B+S C=A C+S B=9
$$
from which $A S=S C=4$, so $\triangle A S C$ is equilateral.
 | Answer: 24. Solution: Let $m$ be the number of boys and $d$ be the number of girls who came to the disco. Denote the boys by blue dots, the girls by red dots, and connect the boys and girls who did not dance with each other by segments. Let there be a total of $k$ segments. Since 4 segments come out of each red dot and... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,602 |
2. Write down four numbers (not necessarily integers), none of which are the same, such that if a number \( x \) is among the written numbers, then at least one of the numbers \( x-1 \) or \( 6x-1 \) is also among the written numbers. (I. Rubanov, S. Berlov) | Answer. For example, $1 / 5, 6 / 5, 11 / 5, 16 / 5$. Solution. Let the numbers written be $x, x+1, x+2, x+3$. For the last three numbers, the condition of the problem is obviously satisfied. To satisfy it for the first one as well, we choose $x$ so that the equation $x=6 x-1$ holds. Solving the obtained equation, we fi... | \frac{1}{5},\frac{6}{5},\frac{11}{5},\frac{16}{5} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,603 |
3. Inside side $BC$ of the convex quadrilateral $ABCD$, there is a point $E$ such that the line $AE$ divides the quadrilateral into two equal area parts. Which vertex of the quadrilateral is farthest from the line $AE$? (I. Rubanov, D. Shiryayev) | Answer. Vertex $B$. Solution. Drop perpendiculars $B B_{1}, C C_{1}, D D_{1}$ to the line $A E$. Then
$$
D D_{1} \cdot A E=2 S(A D E)<2 S(A D C E)=2 S(A B E)=B B_{1} \cdot A E
$$
from which $D D_{1}<B B_{1}$. Similarly, from the inequality $2 S(A C E)<2 S(A D C E)$, we get that $C C_{1}<B B_{1}$. Since the distance f... | B | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,604 |
4. Pete and Vasya are playing such a game. Initially, there is one matchstick in each of 2022 boxes. In one move, you can transfer all the matchsticks from any non-empty box to any other non-empty box. They take turns, with Pete starting. The winner is the one who, after their move, has at least half of all the matchst... | Answer: Vasya. Solution: Note that it does not matter whether matches are moved from box A to box B or from B to A: in both cases, one of the boxes becomes empty, and the other ends up with all the matches that were in both boxes. On his first move, Petya adds one match to one of the boxes. We will mark this box and co... | 1011 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,605 |
5. The divisors of a natural number $n$ (including $n$ and 1), which has more than three divisors, were listed in ascending order: $1=d_{1}<d_{2} \ldots<d_{k}=n$. The differences $u_{1}=d_{2}-d_{1}, u_{2}=d_{3}-d_{2}, \ldots, u_{k-1}=d_{k}-d_{k-1}$ turned out to be such that $u_{2}-u_{1}=u_{3}-u_{2}=\ldots=u_{k-1}-u_{k... | Answer: 10. Solution: Let $n$ be odd. Then $u_{k-1}=d_{k}-d_{k-1} \geq n-n / 3=2 n / 3$. In this case, $u_{k-2}=d_{k-1}-d_{k-2}n / 3$, but $u_{k-1}-u_{k-2}=u_{k-2}-u_{k-3}<u_{k-2}<n / 3$ - a contradiction. In the case of even $n$, we get $u_{k-1}=d_{k}-d_{k-1}=n / 2, u_{k-2}=d_{k-1}-d_{k-2}=n / 2-d_{k-2}$, so $u_{k-1}-... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,606 |
1. The numbers from 1 to 2150 are written on a board. Every minute, each number undergoes the following operation: if the number is divisible by 100, it is divided by 100; if it is not divisible by 100, 1 is subtracted from it. Find the largest number on the board after 87 minutes. | Answer: 2012. Solution. All numbers, the last two digits of which are 86 or less, will transform into numbers ending in 00 within 87 minutes, and in the next step, they will decrease by a factor of 100. In the end, all such numbers will be no more than $2100 / 100=21$ after 87 minutes. Those numbers that end in 87 or m... | 2012 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,607 |
2. The navigator in businessman Boris Mikhailovich's "Lexus" informs him how much longer it will take to reach the destination if he continues at the average speed from the start of the journey to the present moment. Boris Mikhailovich set off from home to the cottage. Halfway through the journey, the navigator reporte... | Answer: In 5 hours. First solution. Let the distance from home to the cottage be 4 units of length. Then the average speed of the "Lexus" on the first half of the journey was 2 units/hour. Let $v$ units/hour be the speed of the tractor. Then the "Lexus" spent $1 / v$ hours on the third quarter of the journey, and its a... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,608 |
3. Point $E$ is the midpoint of the base $A D$ of trapezoid $A B C D$. Segments $B D$ and $C E$ intersect at point $F$. It is known that $A F \perp B D$. Prove that $B C = F C$. | Solution. $F E$ is the median drawn to the hypotenuse of the right triangle $A F D$. Therefore, $F E=A D / 2=E D$ and $\angle E D F=\angle D F E$. But angles $E D F$ and $C B F$ are equal as alternate interior angles when the parallel lines $A D$ and $B C$ are cut by a transversal, and angles $D F E$ and $B F C$ are eq... | BC=FC | Geometry | proof | Yes | Yes | olympiads | false | 6,609 |
4. A square $20 \times 20$ is divided into unit squares. Some sides of the unit squares have been erased, with the erased segments having no common endpoints, and there are no erased segments on the top and right sides of the square. Prove that it is possible to travel from the bottom left corner of the square to the t... | Solution. From each cell corner, except the upper right corner of the $20 \times 20$ square, a move can be made either to the right or upwards - otherwise, two erased segments would have a common end. Therefore, starting from the lower left corner of the $20 \times 20$ square and making 40 such moves, we will end up at... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,610 |
5. Vasya calculated the sums of digits of 200 consecutive natural numbers and wrote these sums in a row in some order. Petya wrote the sums of digits of another 200 consecutive natural numbers under them (also in some order). Then Tanya multiplied each of Vasya's numbers by the number written under it and obtained 200 ... | Solution. A number is divisible by 3 or 9 if and only if the sum of its digits is divisible by 3 or 9, respectively. Among 200 consecutive numbers, 66 or 67 are divisible by 3. Therefore, among the sums of their digits, the same is true. Let's say that among Vasya's numbers divisible by 3, exactly $k$ of Petya's number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,611 |
1. Prove that for any natural number $n>1$ there exist such natural numbers $a, b, c, d$ that $a+b=c+d=ab-cd=4n$. (S. Berlov) | Solution. Let $a=2 n+x, b=2 n-x, c=2 n+y, d=2 n-y$. Then the equation can be rewritten as $y^{2}-x^{2}=4 n$. Now suppose that $y+x=2 n, y-x=2$. We get $x=n-1, y=n+1$, from which $a=3 n-1, b=n+1, c=3 n+1, d=n-1$. | =3n-1,b=n+1,=3n+1,=n-1 | Number Theory | proof | Yes | Yes | olympiads | false | 6,612 |
2. At a round table, 40 people are sitting. Could it happen that any two of them, between whom an even number of people sit, have a common acquaintance at the table, while any two, between whom an odd number of people sit, do not have a common acquaintance? (A. Shapovalov) | Answer. It cannot. First solution. Note that if there are $a$ people sitting between A and B, and $b$ people sitting between B and C, then there are $a+b+1$ or $|a-b|-1$ people sitting between A and C. Therefore, if the numbers $a$ and $b$ have the same parity, then there is an odd number of people sitting between A an... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,613 |
4. Inside a convex quadrilateral $A B C D$, in which $A B=C D$, a point $P$ is chosen such that the sum of the angles $P B A$ and $P C D$ is 180 degrees. Prove that $P B+P C<A D$. (S. Berlov) | Solution. Construct point $K$ on the extension of ray $PC$ beyond point $C$ such that $CK=BP$. Then triangles $ABP$ and $DCK$ will be equal by two sides and the included angle. Therefore, $DK=AP$ and $\angle BAP=\angle CDK$. Construct parallelogram $PKDL$. Since $\angle ABC + \angle BCD > 180^{\circ}$, then $\angle BAD... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,615 |
5. In handball, 2 points are awarded for a win, 1 point for a draw, and 0 points for a loss. 14 handball teams conducted a tournament where each team played against each other once. It turned out that no two teams scored the same number of points. Could it have happened that each of the teams that took the top three pl... | Answer: It could not. Solution: Let it could. We will show that the best of the last three teams scored no less than 9 points. Indeed, the last three teams in games among themselves played for 6 points, plus another 18 points taken from the top three. Therefore, together they scored no less than 24 points, and if each ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,616 |
6. In a convex quadrilateral $ABCD$, where $AB=CD$, points $K$ and $M$ are chosen on sides $AB$ and $CD$ respectively. It turns out that $AM = KC$, $BM = KD$. Prove that the angle between lines $AB$ and $KM$ is equal to the angle between lines $KM$ and $CD$. (S. Berlov) | First solution. Triangles $A B M$ and $C D K$ are equal by three sides, so their heights $K K_{1}$ and $M M_{1}$ are also equal. If they are equal to $K M$, then they coincide with $K M$, and the statement of the problem is obvious. If these heights are less than $K M$, then the right triangles $K K_{1} M$ and $M M_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,617 |
7. On a board, a sequence of n consecutive natural numbers in ascending order is written in a row. Under each of these numbers, a divisor is written, which is less than the number and greater than 1. It turns out that these divisors also form a sequence of consecutive natural numbers in ascending order. Prove that each... | Solution. Notice that all differences between the numbers and their divisors written below them are equal to the same number $c$. Let $p$ be a prime number less than $n$, and $p^{s}$ be the highest power of it not exceeding $n$. Then among the numbers in the second row, there will be one divisible by $p^{s}$. But then ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,618 |
8. 99 wise men sat at a round table. They know that fifty of them are wearing hats of one of two colors, and the other forty-nine are wearing hats of the other color (but it is not known in advance which of the two colors 50 hats are, and which 49 are). Each wise man can see the colors of all the hats except his own. A... | Solution. Let there be 50 white and 49 black hats among the hats. It is clear that the 49 sages who see 50 white and 48 black hats know that they are wearing black hats. Now let each of those who see 49 white and black hats name the color that predominates among the 49 people following them clockwise. If A is one of th... | 74 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,619 |
1. Is it possible to number the vertices, edges, and faces of a cube with distinct integers from -12 to 13 such that the number of each vertex is equal to the sum of the numbers of the edges converging at it, and the number of each face is equal to the sum of the numbers of the edges bounding it? (I. Rubanov) | Answer: No. Solution: Add up all the assigned numbers, replacing the numbers of the vertices and faces with the sums of the numbers of the edges adjacent to them. Then the number of each edge would appear in the resulting sum five times: once by itself, and as part of the numbers of its two ends and as part of the numb... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,620 |
2. King Hiero has 13 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, 2, ..., 13 kg. He also has a device into which he can place one or several of the 13 ingots, and it will signal if their total weight is exactly 46 kg. Archimedes, knowing the weights of ... | Solution. Let Archimedes first place the four heaviest ingots in the device. Their total weight is $-10+11+12+13=46$ kg, and the device will activate. There are no other sets of four ingots with a total weight of 46 kg, so he has shown Hiero which four ingots are the heaviest. Then he places nine ingots weighing $1, \l... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,621 |
3. On side $BC$ of triangle $ABC$, a point $D$ is marked. On side $AB$, a point $P$ is chosen. Segments $PC$ and $AD$ intersect at point $Q$. Point $R$ is the midpoint of segment $AP$. Prove that there exists a fixed point $X$ through which the line $RQ$ passes for any choice of point $P$. (A. Kuznetsov) | Solution. Draw a line through point $C$ parallel to line $A B$, and let $E$ be the point of its intersection with line $A D$. The desired point $X$ is the midpoint of segment $C E$. Indeed, points $R, Q$, and $X$ lie on the same line for any choice of point $P$ as the midpoint of the sides and the point of intersection... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,622 |
4. Natural numbers $a$, $b$, and $c$, greater than 2022, are such that $a+b$ is divisible by $c-2022$, $a+c$ is divisible by $b-2022$, and $b+c$ is divisible by $a-2022$. What is the greatest value that the number $a+b+c$ can take? (S. Berlov) | Answer: 2022.85. Solution. Lemma. For any natural numbers $d_{1}, d_{2}, d_{3}$, if $1 / d_{1}+1 / d_{2}+1 / d_{3}<2$, then $d$ does not exceed $1 / 3+1 / 3+1 / 4=11 / 12$. If $d_{1}=2$ and $d_{2}>3$, then $d$ does not exceed $1 / 2+1 / 4+1 / 5=19 / 20$. Finally, if $d_{1}=2$ and $d_{2}=3$, then $d$ does not exceed $1 ... | 2022\cdot85 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,623 |
1. Let's call two positive integers almost adjacent if each of them is divisible (without a remainder) by their difference. During a math lesson, Vova was asked to write down in his notebook all numbers that are almost adjacent to $2^{10}$. How many numbers will he have to write down? | Answer: 21. Solution. The number $2^{10}$ is divisible only by powers of two: from $2^{0}$ to $2^{10}$. Therefore, the numbers that are almost adjacent to it can only be $2^{10}-2^{9}, 2^{10}-2^{8}, \ldots, 2^{10}-2^{0}, 2^{10}+2^{0}, \ldots, 2^{10}+2^{10}$ (the number $0=2^{10}-2^{10}$ does not count, as it is not pos... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,624 |
2. Points E and $\mathrm{F}$ are the midpoints of sides BC and CD, respectively, of rectangle $A B C D$. Prove that $A E<2 E F$. | First solution. $2 E F = B D = A C > A E$. The last inequality follows from the fact that angle $A E C$ is obtuse. Second solution. Let $B C = 2 x, C D = 2 y$. Then $A E^{2} = x^{2} + 4 y^{2} < 4 x^{2} + 4 y^{2} = (2 E F)^{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,625 |
3. Let $a, b, c$ be integers such that $(a+b+c)^{2}=-(ab+ac+bc)$ and the numbers $a+b, b+c, a+c$ are not equal to 0. Prove that the product of any two of the numbers $a+b, a+c, b+c$ is divisible by the third. | Solution. $(a+b)(a+c)=a^{2}+a b+a c+b c=a^{2}-(a+b+c)^{2}=-(b+c)(2 a+b+c)$. The other two cases are obtained by permuting the letters. | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,626 |
4. A row of 101 cards is laid out. On each of the 50 cards lying in this row at even positions, a symbol > or < is drawn. Prove that, no matter how these symbols are drawn, the remaining cards can be filled with the numbers 1, 2, ... 51 (using each exactly once) so that all the resulting inequalities are true. | First solution. First, write a number above all the cards lying in odd positions. Above the first card, write the number 0. Then we will move to the right and each time after the sign “>” write a number that is 1 less than the previous one, and after the sign “b$ the inequality is true (since by construction $a$ is gre... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,627 |
5. Alina outlined 22 different (but possibly overlapping) three-cell rectangles on a chessboard (8x8), and Polina - 22 non-overlapping two-cell rectangles (but possibly overlapping with Alina's rectangles). Prove that a cross made of 5 cells can be placed on the board, completely covering at least two of the outlined f... | Solution. Note that a cross covers a 1x3 rectangle if their central cells coincide, and covers a 1x2 rectangle if the central cell of the cross lies within the rectangle. Mark the central cells of the long rectangles and all cells of the short ones; in total, 66 cells are marked, so some cell is marked twice. If we mak... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,628 |
1. Pete and Vasya ran the same distance. Vasya ran twice as fast as Pete, but started a minute later, and Pete finished first. Prove that Pete ran the distance in less than two minutes. (I. Rubanov) | Solution. Vasya runs twice as fast as Petya, and therefore covers as much distance in one minute as Petya does in two minutes. This means that if Petya does not finish the race within two minutes after the start, Vasya will catch up with him. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,629 |
3. Two numbers are such that their sum, the sum of their squares, and the sum of their cubes are all equal to the same number t. Prove that the sum of their fourth powers is also equal to t. (I. Rubanov) | First solution. Let $a+b=a^{2}+b^{2}=a^{3}+b^{3}=m$. Then $m=a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)=m(m-a b)$, from which $m=0$ or $a b=m-1$. If $m=a^{2}+b^{2}=0$, then $a=b=0$ and $a^{4}+b^{4}=0=m$. If $a b=m-1$, then
$$
m^{2}=(a+b)\left(a^{3}+b^{3}\right)=a^{4}+b^{4}+a b\left(a^{2}+b^{2}\right)=a^{4}+b^{4}+(m... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,631 |
4. In triangle $A B C$, the bisector $B E$ and the perpendicular bisector $t$ of side $A B$ were drawn. It turned out that $B E=E C$, and line $t$ intersects side $B C$. Prove that angle $C$ is less than 36 degrees. (I. Rubanov) | Solution. Let $\angle A B C=2 \beta$, and the line $m$ intersects side $B C$ at point $D$. By the property of the perpendicular bisector, $D A=D B$, hence $\angle D A B=\angle A B D=2 \beta$. Therefore, by the exterior angle theorem of a triangle, $\angle A D C=\angle D A B+\angle A B D=4 \beta$. Moreover, since $B E=E... | \beta<36 | Geometry | proof | Yes | Yes | olympiads | false | 6,632 |
5. From the odd natural numbers from 1 to 47, 12 fractions less than 1 were formed, using each number exactly once. The resulting fractions were divided into groups of equal values. What is the smallest number of groups that could have been obtained? (I. Rubanov) | Answer: 7. Solution: Evaluation. A fraction containing at least one of the prime numbers 17, 19, 23, 29, 31, 37, 41, 43, 47 cannot equal any of our other fractions, because already $17 \cdot 3 > 47$. Since there are no more than nine such "lonely" fractions, among the constructed fractions there are some that are not e... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,633 |
2. (A. Shapovalov) For any natural number n greater than 2009, from the fractions $\frac{1}{n}, \frac{2}{n-1}, \frac{3}{n-2}, \ldots, \frac{n-1}{2}, \frac{n}{1}$, can two pairs of fractions with the same sums always be selected? | Answer: Yes. Solution. Each of the given fractions has the form $\frac{n+1-a}{a}=\frac{n+1}{a}-1$, where $1 \leq a \leq n$. Therefore, we need to find such distinct natural numbers $a, b$, $c$ and $d$, not greater than 2009, for which $\left(\frac{n+1}{a}-1\right)+\left(\frac{n+1}{b}-1\right)=\left(\frac{n+1}{c}-1\righ... | \frac{1}{12}+\frac{1}{3}=\frac{1}{6}+\frac{1}{4} | Number Theory | proof | Yes | Yes | olympiads | false | 6,634 |
3. (S. Berlov) In triangle $ABC$, sides $AB$ and $BC$ are equal. Point $D$ inside the triangle is such that angle $ADC$ is twice the angle $ABC$. Prove that twice the distance from point $B$ to the line bisecting the angles adjacent to angle $ADC$ is equal to $AD + DC$. | Solution. Let $l$ be the bisector of the angles adjacent to angle $A D C$, point $K$ be the projection of $B$ onto $l$, and points $B^{\prime}$ and $C^{\prime}$ be symmetric to points $A, B$, and $C$ respectively with respect to $l$. Then
 In the country of Leonardia, all roads are one-way. Each road connects two cities and does not pass through other cities. The Department of Statistics calculated for each city the total number of residents in the cities from which roads lead to it, and the total number of residents in the cities to which... | First solution. Construct a graph where the vertices correspond to the residents of the country, and two vertices are connected by a directed edge if and only if their cities are connected by a road (the direction of the edge will be the same as the direction of the road between the cities). For each vertex $v$ of this... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,636 |
1. The natural numbers from 1 to 1000 are written on a board, each exactly once. Vasya can erase any two numbers and write down one instead: their greatest common divisor or their least common multiple. After 999 such operations, one number remains on the board, which is a natural power of ten. What is the largest valu... | Answer. The fourth. Solution. Example. First, we get $10^{4}=\operatorname{LCM}(16,625)$. Then, sequentially taking $\operatorname{GCD}(1, n)=1$ for all remaining $n$ from 2 to 1000, we leave only $10^{4}$ and 1 on the board, and finally, take $\operatorname{LCM}\left(10^{4}, 1\right)=10^{4}$. Evaluation. Note that if ... | 10^4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,637 |
3. Among the natural numbers $a_{1}, \ldots, a_{k}$, there are no identical ones, and the difference between the largest and the smallest of them is less than 1000. For what largest $k$ can it happen that all quadratic equations $a_{i} x^{2}+2 a_{i+1} x+a_{i+2}=0$, where $1 \leq i \leq k-2$, have no roots? (I. Bogdanov... | Answer. For $k=88$. Solution. Let the sequence $a_{1}, \ldots, a_{k}$ satisfy the condition of the problem. The absence of roots in the equations specified in the condition is equivalent to the inequality $\left(a_{i+1}\right)^{2}0$, from which $e-d>0$.
Let $a_{m}$ be the smallest number in the sequence. Obviously, on... | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,639 |
4. In 2n barrels, 2n different reagents are filled (one reagent in each). They are divided into p pairs of conflicting reagents, but it is unknown which barrel conflicts with which. The engineer needs to find this partition. He has n empty test tubes. In one action, he can add to any test tube (empty or non-empty) a re... | Solution. Let's number the test tubes from 1 to $n$ and the barrels from 1 to $2n$. We will call operation $k (k \leq n)$ the sequential pouring of reagent from the $k$-th barrel into test tubes numbered $k$, $k-1, \ldots$, 1 (in that order). Operation $n+1$ will be the sequential pouring of reagent from the $(n+1)$-th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,640 |
5. The integers $a, b$, and $c$ and the natural number $n$ are such that $a+b+c=1$ and $a^{2}+b^{2}+c^{2}=2 n+1$. Prove that $a^{3}+b^{2}-a^{2}-b^{3}$ is divisible by $n$. (N. Agakhanov) | Solution. From the condition, it follows that $a^{2}+b^{2}+(1-a-b)^{2}=2 n+1$, from which $a^{2}+b^{2}+a b-(a+b)=n$. Therefore, $a^{3}+b^{2}-a^{2}-b^{3}=\left(a^{3}-b^{3}\right)+\left(b^{2}-a^{2}\right)=(a-b)\left(a^{2}+b^{2}+a b-(a+b)\right)=(a-b) \cdot n$, which is what we needed to prove. | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,641 |
6. Among ten people, exactly one is a liar and 9 are knights. Knights always tell the truth, and liars always lie. Each of them was given a card with a natural number from 1 to 10, and all the numbers on the cards are different. You can ask anyone the question: "Is it true that the number on your card is M?" (M can onl... | Answer. Correct. Solution. Ask one person 9 questions - about numbers from 1 to 9. If he is a liar, we will get no less than eight positive answers and thus find him. Otherwise, he is a knight, and we will learn what is written on his card: if the number is from 1 to 9, we will get an affirmative answer to the correspo... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,642 |
7. From a $70 \times 70$ checkerboard, 2018 cells were cut out. Prove that the board has split into no more than 2018 pieces. Two pieces that do not share any points other than the vertices of the cells are considered not connected to each other. (I. Rubanov) | Solution. It is not difficult to construct a cycle that passes through each cell of a $70 \times 70$ board exactly once, such that adjacent cells in the cycle share a common side: for example, one can walk the entire first vertical column from the bottom cell to the top, then move along the vertical columns in a "zigza... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,643 |
8. Vertex $F$ of parallelogram $A C E F$ lies on side $B C$ of parallelogram $A B C D$. It is known that $A C=A D$ and $A E=2 C D$. Prove that $\angle C D E=\angle B E F$. (A. Kuznetsov) | First solution. Let $M$ be the midpoint of segment $CF$. Since quadrilateral $ACEF$ is a parallelogram, point $M$ is the midpoint of segment $AE$. Denote $\angle MAC = \angle MEF = \alpha$ and $\angle ABC = \angle ADC = \angle ACD = \beta$. Since $AM = AE / 2 = CD$, $AMCD$ is an isosceles trapezoid, from which we obtai... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,644 |
5. Nезнayka wrote down 11 natural numbers in a circle. For each pair of adjacent numbers, he calculated their difference (subtracting the smaller from the larger). As a result, among the differences found, there were four ones, four twos, and three threes. Prove that Nезнayka made a mistake somewhere. ( | Solution. Let's write each of our differences with a plus sign if in the corresponding pair of numbers the larger one stands before the smaller one clockwise, and with a minus sign otherwise. We have 11 differences between a number and the one following it clockwise; thus, the sum of all these numbers is zero, which is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,645 |
6. In a company of six people, any five can sit at a round table such that every two neighbors are acquainted. Prove that the entire company can also be seated at a round table such that every two neighbors are acquainted. (S. Volchonkov) | Solution. Note that each person in the company has at least three acquaintances. Indeed, if someone $X$ were acquainted with fewer than three people, then by excluding one of his acquaintances from the company, we would end up with a group of five people in which $X$ has no more than one acquaintance, which means it wo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,646 |
7. For what largest $n$ can the numbers $1,2, \ldots, 14$ be colored red and blue so that for any number $k=1,2, \ldots, n$ there exist a pair of blue numbers whose difference is $k$, and a pair of red numbers whose difference is also $k$? (D. Khramtsov) | Answer: $n=11$. Solution. Obviously, $\mathrm{n} \leq 12$, since there is only one pair of numbers with a difference of 13. Suppose the required is possible for $n=12$. The number 12 can be represented as the difference of numbers from 1 to 14 in exactly two ways: 13-1 and 14-2. Let the number 1 be red for definiteness... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,647 |
8. The angle bisectors of angles $A$ and $C$ of trapezoid $A B C D$ intersect at point $P$, and the angle bisectors of angles $B$ and $D$ intersect at point $Q$, different from $P$. Prove that if segment $P Q$ is parallel to base $A D$, then the trapezoid is isosceles. (L. Emelyanov) | Solution. Let $\rho(X, Y Z)$ denote the distance from point $X$ to line $Y Z$. Since point $P$ lies on the bisector of angle $C$, we have $\rho(P, B C)=\rho(P, C D)$. Similarly, $\rho(Q, A B)=\rho(Q, B C)$. But since $Q P \| B C$, we have $\rho(Q, B C)=\rho(P, B C)$, hence $\rho(Q, A B)=\rho(P, C D)$ (see figure). Simi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,648 |
1. On the board, three quadrilaterals are drawn. Petya said: "At least two trapezoids are drawn on the board." Vasya said: "At least two rectangles are drawn on the board." Kolya said: "At least two rhombuses are drawn on the board." It is known that one of the boys lied, while the other two told the truth. Prove that ... | Solution. A trapezoid cannot be a parallelogram. Therefore, if Petya is right, then there is no more than one parallelogram drawn on the board, and Vasya and Kolya are both wrong. But according to the condition, only one person lied. So, it must be Petya, and Vasya and Kolya told the truth. This means that at least one... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,649 |
2. It is known about three positive numbers that if you choose one of them and add to it the sum of the squares of the other two, you will get the same sum, regardless of which number you choose. Prove that at least two of the original numbers are the same. (L. Emelyanov) | First solution. Let our numbers be $a, b, c$. Then $a+b^{2}+c^{2}=a^{2}+b+c^{2}=a^{2}+b^{2}+c$. From the first two equalities, we have $a^{2}-a=b^{2}-b$, which is equivalent to $(a-b)(a+b-1)=0$. Therefore, $a=b$ or $b=1-a$. Similarly, $a=c$ or $c=1-a$. Consequently, if $a \neq b$ and $a \neq c$, then $b=1-a=c$, meaning... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,650 |
3. Given a convex quadrilateral $A B C D$ such that $A D=A B+C D$. It turns out that the bisector of angle $A$ passes through the midpoint of side $B C$. Prove that the bisector of angle $D$ also passes through the midpoint of $B C$.
 and the angle between them. Therefore, $EF = BE = EC$. Now we get that triangles $DEF$ and $DEC$ are equ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,651 |
4. A closed broken line without self-intersections is drawn through the centers of some cells of an 8x8 chessboard. Each segment of the broken line connects the centers of adjacent cells, either horizontally, vertically, or diagonally. Prove that the total area of the black pieces within the region bounded by the broke... | First solution. Draw dashed vertical and horizontal lines through the centers of the cells of the board. On the resulting dashed grid, each segment of our broken line connects nodes that are adjacent vertically, horizontally, or diagonally. Therefore, the dashed lines divide the area bounded by the broken line into uni... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,652 |
5. Businessman Boris Mikhailovich decided to have a race with tractor driver Vasya on the highway. Since his "Lexus" is ten times faster than Vasya's tractor, he gave Vasya a head start and set off an hour after Vasya. After Vasya's tractor had driven exactly half of the planned route, a spring fell off, so he covered ... | First solution. Since Boris met with Vasya's spring, by the halfway point, he had not yet caught up with Vasya. Clearly, under normal conditions, he would have caught up with Vasya in $10 / 9$ hours; thus, Vasya took $10/9 - a$ hours to travel the first half of the distance for some $a \geq 0$, and then the entire jour... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,653 |
6. The number 1 is written on the board. If the number a is written on the board, it can be replaced by any number of the form a+d, where $d$ is coprime with a and $10 \leq d \leq 20$. Is it possible to obtain the number $18!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 18$ on the board after several such operations? (I. Ruban... | First solution. Note that the number 18!-19 ends with 1. We will add 10 to the number on the board. Each time, we will get a number ending in 1, and thus coprime with the number 10, so the operation is possible. Eventually, the number 18!-19 will appear on the board. We will add 19 to it and get 18!.
Second solution. ... | 18! | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,654 |
7. For four different integers, all their pairwise sums and pairwise products were calculated and written on the board. What is the smallest number of different numbers that could have appeared on the board? (I. Rubanov) | Answer: 6. Solution: If we take the numbers $-1,0,1,2$, it is easy to verify that each of the numbers written on the board will be equal to -2, -1, 0, 1, 2, or 3 - a total of 6 different values. We will show that fewer than six different numbers could not appear on the board. Let the taken numbers be $ac+d$. If $ac+d$.... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,655 |
8. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $A B$ respectively. On the median $B M$, a point $P$ is chosen, which does not lie on $C N$. It turns out that $P C=2 P N$. Prove that $A P=B C$. (C. Berlov) | First solution. Mark a point $K$ such that $A K B P$ is a parallelogram. Then its diagonals are bisected by the point of intersection, that is, they intersect at point $N$ and $P K=2 P N=P C$. Let the lines $M B$ and $C K$ intersect at point $T$. Since $M T \| A K$ and $M$ is the midpoint of $A C$, then $M T$ is the mi... | AP=BC | Geometry | proof | Yes | Yes | olympiads | false | 6,656 |
1. Fedia has several weights, the masses of which in kilograms are integers less than 10. Can it happen that with these weights, one can make up the weights of 100, 102, 103, and 104 kg, but not the weights of 101 and 105 kg? | Answer: It can. Example (there are others!). 11 weights of 9 kg each and one weight each of 1, 3, and 4 kg. It is easy to check that all the required weights can be assembled. Further, since $1+3+4+90<100$, to assemble a weight of 101 or 105 kg, all 11 nine-kilogram weights must be used. However, the missing weights of... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,657 |
2. In each of the 111 families, there are three people: a father, a mother, and a child. All 333 people lined up in a row. It turned out that the parents of each child are standing on different sides of the child (but not necessarily next to the child). Prove that among the central 111 people in this row, there is at l... | Solution. Let's call a parent left (right) if their child stands to the left (right) of them in the row. Suppose all the central 111 people are parents. Then among them, either the left or the right parents will be more than 55, that is, at least 56. Suppose there are at least 56 left parents. Then all 56 of their chil... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,658 |
3. Point $D$ lies on the hypotenuse $AB$ of a right triangle $ABC$, but does not coincide with its midpoint. Prove that among the segments $AD, BD$, and $CD$, there are no equal ones. | Solution. $A D$ is not equal to $B D$ by the condition. Suppose $A D = C D$. Then the angles $D A C$ and $A C D$ are equal. Let each of them be $x$. But then each of the angles $D A B$ and $A B D$ is $90^{\circ} - x$, which implies $A D = B D = C D$ - a contradiction. Similarly, $C D$ cannot equal $B D$.
Grading Guide... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,659 |
4. A five-digit number x starts with 4 and ends with 7, and a five-digit number y starts with 9 and ends with 3. It is known that the numbers x and y have a common five-digit divisor. Prove that $2y - x$ is divisible by 11. | Solution. Let $x=a z, y=b z$, where $z$ is a five-digit common divisor of the numbers $x$ and $y$. Obviously, $a$ and $b$ are odd, $b>a, a \leq 4$ and $b \leq 9$. From this it follows that $a$ is 1 or 3. If $a=1$, then $z=x, y=b x, b \geq 3$, but already $3 x>120000-$ a six-digit number. Therefore, $a=3$. The only digi... | 2y-x=11z | Number Theory | proof | Yes | Yes | olympiads | false | 6,660 |
5. There are 2009 piles, each containing 2 stones. It is allowed to take the largest pile from those in which the number of stones is even (if there are several, then any of them), and move exactly half of the stones from it to any other pile. What is the maximum number of stones that can be obtained in one pile using ... | Answer: 2010. Solution. The operations described in the condition do not reduce the numbers of stones in the heaps, so at any moment, there is at least one stone in each of them. Therefore, it is impossible to accumulate more than $2009 \cdot 2$ $2008=2010$ stones in one heap.
We will show how to obtain a heap with 20... | 2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,661 |
1. Let's call good rectangles a square with side 2 and a rectangle with sides 1 and 11. Prove that any rectangle with integer sides, greater than 100, can be cut into good rectangles. (S. Volchendov) | Solution. A rectangle $2 n \times 2 m$ can be cut into squares $2 \times 2$. A rectangle $(2 n+1) \times 2 m$ can first be cut into rectangles $11 \times 2 m$ and $(2 n-10) \times 2 m$, then the first one can be cut into rectangles $1 \times 11$, and the second one - into squares $2 \times 2$. Finally, a rectangle $(2 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,662 |
2. In triangle $ABC$, side $AB$ is greater than side $BC$. On the extension of side $BC$ beyond point $C$, point $N$ is marked such that $2BN = AB + BC$. Let $BS$ be the bisector of triangle $ABC$, $M$ be the midpoint of side $AC$, and $L$ be a point on segment $BS$ such that $ML \parallel AB$. Prove that $2LN = AC$. (... | Solution. Extend $BN$ beyond $N$ and $BL$ beyond $L$ to segments $NN' = BN$ and $LL' = BL$ respectively. Since $M$ is the midpoint of $AC$ and $ML \parallel AB$, the line $ML$ contains the midline $MK$ of triangle $ABC$. Since $L$ is the midpoint of $BL$, this line also contains the midline $LK$ of triangle $BCL$; henc... | 2LN=AC | Geometry | proof | Yes | Yes | olympiads | false | 6,663 |
3. Twenty fifteen positive numbers are written in a circle. The sum of any two adjacent numbers is greater than the sum of the reciprocals of the next two numbers in the clockwise direction. Prove that the product of all these numbers is greater than 1. (A. Gолованов, S. Berlov) | Solution. Let the numbers $x_{1}, x_{2}, \ldots, x_{2015}$ stand in a circle. Notice that $a+b>\frac{1}{c}+\frac{1}{d} \cdot(a+b) c d>c+d(*)$. Writing all 2015 inequalities given in the problem in the form (*), and multiplying them, we get the inequality $X \cdot\left(x_{1} x_{2} \ldots x_{2015}\right)^{2}>X$, where $X... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 6,664 |
4. On each side of the square, 100 points are chosen, and from each chosen point, a segment perpendicular to the corresponding side of the square is drawn inside the square. It turns out that no two of the drawn segments lie on the same line. Mark all the intersection points of these segments. For what largest $k<200$ ... | Answer. For $k=150$. Solution. Estimation. Suppose there is an example with $k>150$. We will associate it with a $200 \times 200$ table, the rows of which correspond to horizontal segments (ordered from bottom to top), and the columns to vertical segments (ordered from left to right). In the cell of the table, there is... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,665 |
1. Prove that the prime factorization of the product of ten consecutive three-digit numbers includes no more than 23 different prime numbers. (I. Rubanov) | Solution. Note that in the prime factorization of a three-digit number, there are no more than two factors greater than 10 - otherwise, the product would be greater than 1000. Moreover, among 10 consecutive natural numbers, there is a number divisible by 10, in the factorization of which there is at most one factor gre... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,666 |
2. On the side AB of triangle ABC with an angle of $100^{\circ}$ at vertex C, points $P$ and $Q$ are taken such that $A P=B C$ and $B Q=A C$. Let $M, N, K-$ be the midpoints of segments $A B, C P, C Q$ respectively. Find the angle NМК. (M. Kungozhin + jury) | Answer: $40^{\circ}$. Solution: Extend the triangle to form a parallelogram $A C B D$. Then $M$ is the midpoint of segment $C D$.
Since $A P = B C = A D$ and $B Q = A C = B D$, triangles $A P... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,667 |
3. On the hundredth year of his reign, the Immortal Treasurer decided to start issuing new coins. In this year, he put into circulation an unlimited supply of coins with a value of $2^{100}-1$, the following year - with a value of $2^{101}-1$, and so on. As soon as the value of the next new coin can be exactly matched ... | Answer. On the two hundredth. Solution. Let on the $k$-th year of reign $2^{k}-1$ can be made up of previously issued coins: $2^{k}-1=a_{1}+\ldots+a_{n}=N-n$, where $N$ is the sum of powers of two, each of which is divisible by $2^{100}$. Since $2^{k}$ is also divisible by $2^{100}$, the number $n-1$ must be divisible ... | 200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,668 |
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