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# Problem 11.4
Let $M$ and $N$ be the midpoints of the hypotenuse $AB$ and the leg $BC$ of the right triangle $ABC$, respectively. The excircle of triangle $ACM$ touches the side $AM$ at point $Q$ and the line $AC$ at point $P$. Prove that points $P$, $Q$, and $N$ lie on the same line.
## Number of points 7
# | # Solution
Let $D$ be the center of the excircle of triangle $A C M$, then $P$ and $Q$ are the projections of point $D$ onto lines $A C$ and $A B$ respectively (see figure).
Since $M N$ is t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,402 |
# Task 11.5
The king traversed the chessboard, visiting each square exactly once and returning to the starting square on the last move. (The king moves according to the usual rules: in one move, he can move to any adjacent square horizontally, vertically, or diagonally.) When his path was drawn by sequentially connect... | # Solution
The king made 64 moves. Since the length of each move is either one (a straight move) or $\sqrt{2}$ (a diagonal move), the total length of the path is at least 64. A path of length 64 is shown in the figure.
We will show that the length of the king's path cannot be greater than $28 + 36 \sqrt{2}$.
Conside... | 64,28+36\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,403 |
1.1. January first of a certain non-leap year fell on a Saturday. And how many Fridays are there in this year? | Answer: 52
Solution. In a non-leap year, there are 365 days. The first two days fell on Saturday and Sunday, followed by 51 full weeks ( $51 \cdot 7=357$ days) and 6 more days. In total, there are 52 Fridays. | 52 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,404 |
2.1. Lisa wrote a quadratic equation. Artem erased its free term, so the equation now looks like $\operatorname{\operatorname {mak}} 2 x^{2}+20 x+\ldots=0$. Lisa doesn't remember what number Artem erased, but she remembers that the equation has exactly one real root. What is this root? | # Answer: -5
Solution. A quadratic equation has one root if and only if its discriminant is 0. And if the discriminant is 0, then the root is calculated using the formula $x_{1}=-b /(2 a)=-20 / 4=-5$. | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,405 |
3.1. Lisa drew graphs of all functions of the form $y=a x+b$, where $a$ and $b$ take all natural values from 1 to 100. How many of these graphs pass through the point $(3,333)$? | Answer: 23
Solution. We are looking for such $a$ and $b$ that $333=3a+b$. That is, $b=3(111-a)$. Therefore, $0<111-a \leqslant 33$, $78 \leqslant a$. There are exactly 23 such $a$ and for each of them, $b$ can be uniquely determined. | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,406 |
4.1. The sea includes a bay with more saline water. The salinity of the water in the sea is 120 per mille, in the bay 240 per mille, in the part of the sea not including the bay - 110 per mille. How many times is the volume of water in the sea larger than the volume of water in the bay? The volume of water is considere... | Answer: 13
Solution. Let the volume of salt in the bay be $s_{1}$, and the volume of water $v_{1}$; in the part of the sea not including the bay, the volume of salt is $s_{2}$, and the total volume is $v_{2}$. We have the equations
$$
\frac{s_{1}}{v_{1}}=\frac{240}{1000} ; \quad \frac{s_{2}}{v_{2}}=\frac{110}{1000} ;... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,407 |
5.1. A paper rectangle $4 \times 8$ was folded along the diagonal as shown in the figure. What is the area of the triangle that is covered twice?
 | Answer: 10.
Solution. By the Pythagorean theorem, $A C=\sqrt{A B^{2}+B C^{2}}=4 \sqrt{5}$. Triangle $A E C$ is isosceles, and if we drop the height $E H$ in it, then triangle $A E H$ will be similar to triangle $A C D$. Therefore, $A H / E H=A D / D C=2 . A H=2 \sqrt{5}$, since the height in an isosceles triangle is a... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,408 |
6.1. In the club, there are 9 people. Every day, some three of them went to the cafe together, while the others did not go to the cafe. After 360 days, it turned out that any two people from the club had been to the cafe together the same number of times. How many times? | Answer: 30
Solution. The total number of pairs of people in the circle is $9 \cdot 8 / 2=36$. Over 360 days, the cafe was visited by $360 \cdot 3$ pairs (since three new pairs are added each day). Since all pairs visited the cafe an equal number of times, this number is $360 \cdot 3 / 36=30$. | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,409 |
7.1. Real $x, y, z$ are such that $x y + x z + y z + x + y + z = -3, x^{2} + y^{2} + z^{2} = 5$. What is $x + y + z$? | Answer: -1
Solution. Add twice the first equation to the second, we get $(x+y+z)^{2}+2(x+y+z)=-1$. Therefore, $(x+y+z+1)^{2}=0, x+y+z=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,410 |
8.1. Of all numbers with the sum of digits equal to 25, find the one whose product of digits is maximal. If there are several such numbers, write the smallest of them in the answer. | Answer: 33333334
Solution. Obviously, there is no 0 in the number. If the number contains the digit 1, then it can be removed and one of the remaining digits can be increased by 1, which does not change the sum, but increases the product. If the number contains a digit $x \geqslant 5$, then it can be replaced by the d... | 33333334 | Other | math-word-problem | Yes | Yes | olympiads | false | 18,411 |
11.1 The number 890 has the property that by changing any of its digits to 1 (increasing or decreasing), you can obtain a number that is a multiple of 11. Find the smallest three-digit number with the same property. | Answer: 120.
Solution. Since the last digit of the number must be changed to obtain a number divisible by 11, the required number should differ from it by 1. The smallest three-digit number divisible by 11 is 110. However, the numbers adjacent to it, 109 and 111, do not have the required property. Indeed, if the secon... | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,412 |
11.3. Solve the system of equations $\left\{\begin{array}{l}\sin x \cos y=\sin z \\ \cos x \sin y=\cos z\end{array}\right.$, if the numbers $x, y$ and $z$ lie in the interval $\left[0 ; \frac{\pi}{2}\right]$ | Answer: $\left(\frac{\pi}{2} ; 0 ; \frac{\pi}{2}\right),\left(0 ; \frac{\pi}{2} ; 0\right)$
Solution. First method. Square both equations. We get: $\left\{\begin{array}{l}\sin ^{2} x \cos ^{2} y=\sin ^{2} z \\ \cos ^{2} x \sin ^{2} y=\cos ^{2} z\end{array}\right.$
Add the equations:
$\sin ^{2} x \cos ^{2} y+\cos ^{2... | (\frac{\pi}{2};0;\frac{\pi}{2}),(0;\frac{\pi}{2};0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,414 |
11.4. Quadrilateral $ABCD$ is inscribed in a circle, $BC=CD, AC=c, \angle BAD=2\alpha$. Find the area of this quadrilateral. | Answer: $c^{2} \sin \alpha \cos \alpha$.
Solution. Since the chords $B C$ and $C D$ are equal, the arcs $B C$ and $C D$, not containing point $A$, are also equal. Therefore, $\angle B A C=\angle D A C=\alpha$.
First method. Let $\angle A B C=\beta, \angle A D C=\gamma$. Translate triangle $A B C$ so that the equal se... | ^{2}\sin\alpha\cos\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,415 |
11.5. A $7 \times 7$ checkered board was assembled using three types of figures (see the image), not necessarily all. How many figures, composed of four cells, could have been used?
 | Answer: only one.
Solution. We will prove that only one figure consisting of four cells can be used. We will color the cells of the board as shown in Fig. 11.5a: Each of the given figures can cover no more than one shaded cell, therefore, the number of figures must be no less than 16.
Since 16 three-cell figures cove... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,416 |
11.6. Given a tetrahedron \(ABCD\), all of whose faces are similar right triangles with acute angles at vertices \(A\) and \(B\). The edge \(AB\) is equal to 1. Find the length of the shortest edge of the tetrahedron. | Answer: $\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{3}{2}}$.
Solution. Let $\angle C B D=\alpha, \angle C D B=\beta, \alpha+\beta=90^{\circ}$ (see Fig. 11.6a). Right triangles $D B C$ and $A B C$ share the side $B C$. If $\angle A B C=\alpha$, then triangles $A B C$ and $D B C$ are congruent, which implies $C D=A C$, wh... | (\frac{\sqrt{5}-1}{2})^{\frac{3}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,417 |
1. All graduates of the mathematics school took the Unified State Exam (USE) in mathematics and physical education. Each student's result in mathematics turned out to be equal to the sum of the results of all other students in physical education. How many graduates are there in the school if the total number of points ... | # Answer: 51.
## Solution:
Let the total number of graduates be $n$. Denote the total score in mathematics by $M$. This sum is equal to the sum of the physical education scores of all students, counted $(n-1)$ times.
Thus, we have the equation: $M=50 \frac{M}{n-1}$.
From this, the answer is
Instructions for checki... | 51 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,418 |
4. The letters А, Б, К, М, П, У, Ш were encoded with sequences of zeros and ones (each with its own). Then, in the word ПАПАМАМАБАБУШКА, the letters were replaced with their codes. Could the length of the resulting sequence be shorter than 40 characters, if the sequence can be uniquely decoded? | Answer: She could.
## Solution:
Here is an example of a code table:
| A | B | K | M | P | U | Sh |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 110 | 1111 | 100 | 101 | 11100 | 11101 |
The word will look like this: 10101010100010001100110111001110111110 - a total of 38 characters. Decoding is una... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,420 |
11.2. A travel agency ran a promotion: "Buy a trip to Egypt, bring four friends who also buy a trip, and get the cost of your trip back." During the promotion, 13 customers came on their own, and the rest were brought by friends. Some of them brought exactly four new customers, while the other 100 did not bring anyone.... | Answer: 29.
Let each of the $x$ potential "lucky ones" bring 4 friends. Then the number of "brought" customers is $4 x$, and 13 came on their own, so the total number of tourists was $13+4 x$.
On the other hand, $x$ people brought new customers, while 100 people did not, so the total number of tourists was $x+100$.
... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,425 |
11.3. The function $f(x)$ is such that for all values of $x$ the equality $f(x+1)=f(x)+$ $+2x+3$ holds. It is known that $f(0)=1$. Find $f(2012)$. | Answer: $4052169=2013^{2}$.
Write the condition of the problem as: $f(x+1)-f(x)=2 x+3$. Substitute for $x$ the numbers $0,1,2$, ..., 2011. We get: $f(1)-f(0)=2 \cdot 0+3, f(2)-f(1)=2 \cdot 1+3, \ldots, f(2012)-f(2011)=2 \cdot 2011+3$.
Add the obtained equations term by term: $f(2012)-f(0)=2 \cdot(1+2+\ldots+2011)+3 \... | 4052169=2013^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,426 |
11.4. Point $X$ is located on the diameter $A B$ of a circle with radius $R$. Points $K$ and $N$ lie on the circle in the same half-plane relative to $A B$, and $\angle K X A=\angle N X B=60^{\circ}$. Find the length of the segment $K N$. | Answer: $K N=R$.
First method (axial symmetry). Consider the point $K_{1}$, symmetric to the point $K$ with respect to the diameter $A B$ (see Fig. 11.4a). It lies on the same circle and $\angle K X A=60^{\circ}$. Then the sum of the three angles with the vertex at point $X$ (marked on the drawing) is $180^{\circ}$. T... | KN=R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,427 |
11.5. In the decimal representation of a certain number, the digits are arranged from left to right in descending order. Can this number be a multiple of $111?$ | Answer: No, it cannot.
First method. Suppose there exists a number $\overline{a_{1} a_{2} \ldots a_{n}}$ such that $a_{1}>a_{2}>\ldots>a_{n}$ and it is divisible by 111. Then note that:
1) if in such a number the last digit is 0, then it can be crossed out and the new number $\overline{a_{1} a_{2} \ldots a_{n-1}}$ wi... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,428 |
3. Papa always picked up Misha from kindergarten at the same time and in the same car. But today, after firmly establishing that there would be pea porridge for dinner, Misha resolutely left the kindergarten and set off down the road to meet Papa. Having walked 5 km, Misha met Papa, got into the car, and they arrived h... | 3. Speed: 60 km/h (or 1 km/min). Today, Misha and his dad saved 10 minutes that were previously spent driving 10 km (from the meeting point to the kindergarten and back). This means that in 1 minute, Misha's dad drives 1 km, i.e., his speed is 60 km/h.
Criteria: Correct reasoning and correct answer - 7 points; only an... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,429 |
4. A string contains several even natural numbers. Each number, except the leftmost one, is obtained by subtracting the largest digit of its left neighbor from it. Can the string contain more than 12 numbers? | 4. Answer. Yes, it can. Solution: For example, the following sequence of 13 numbers works: 888, 880, 872, 864, 856, 848, 840, 832, 824, 816, 808, 800, 792. Note that other examples are not excluded, and it is necessary to carefully check the condition that all numbers are even.
Criteria: any correct example of a seque... | Yes,itcan | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,430 |
5. On New Year's Day, January 1st, the Elderly Sage was reflecting on his life. He noticed that over the past 5 years, all days of the week had been equally represented, and 10 years ago, the New Year's celebration fell on a Friday. On which day of the week did the sage's somber reflections take place? It is known that... | 5. Offset. For Thursday. Solution In a common year, there are 365 days, which is 52 weeks and one extra day. In a leap year, there are 52 weeks and two extra days. From the condition, it follows that over the last five years of the Sage's life, due to the additional days, exactly one week has accumulated. Therefore, du... | Thursday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,431 |
2. How many natural numbers greater than one exist, the product of which with their smallest prime divisor does not exceed 100? | Answer: 33.
Sketch of the solution. The smallest prime divisor can be 2. These are the numbers: $2, 4, \ldots, 50$ (25 numbers). The smallest prime divisor can be 3. These are the numbers: 3, 9, 15, 21, 27, 33 (6 numbers). The smallest prime divisor can be 5. This number: 5 (1 number). The smallest prime divisor can b... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,432 |
4. The cells of an $n \times n$ square are painted in $n$ colors, with $n$ cells of each color. For which $n$ can the square be painted in such a way that in each row and each column there are cells of exactly two colors? | Answer: 2, 3, 4. Sketch of the solution.
Examples for 2, 3, and 4 are shown in
the figure on the right. Colors
are numbered by numbers.
Estimation. Let $n \geq 5$ and it was possible to c... | 2,3,4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,433 |
6. Every week, listeners choose the ten most popular songs via SMS. It is known that 1) the same set of songs in the same order is never chosen twice in a row; 2) a song that has once dropped in the ranking will not rise again in the future. What is the maximum number of weeks that the same 10 songs can remain in the r... | Answer: 46 weeks.
Sketch of the solution. Suppose that for two consecutive weeks, the same set of 10 songs was on the list. Since the order of the songs is different, at least one song in the list has moved up in the ranking, and at least one has moved down. Since a song, once it has moved down, does not move up in th... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,434 |
1. Two classes with the same number of students took a test. After checking the tests, the strict director, Alexander Vladimirovich, said that he had given 11 more twos than all other grades. Did Alexander Vladimirovich make a mistake? | Solution. Let there be $x$ students in each class, then there are $-2 x$ students in two classes (an even number). Let Alexander Vladimirovich give $y$ other grades (not twos), then he gave $y+11$ twos. Thus, he gave a total of $2 y+11$ grades (an odd number). Since the number of students must match the number of grade... | mademistake | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,435 |
3. In a certain state, there were initially 10 banks. Since the restructuring of society, everyone wanted to become bankers. However, by law, a new bank could only be opened by dividing an existing bank into 8 new banks. After 2 years, the finance minister reported to the president that there were already 2023 banks op... | Solution. Note that with the given algorithm for opening new banks, the remainder of their total number when divided by 7 does not change. Indeed, let at the current moment there be $N$ banks open, then with the next opening, their number will be $N-1+8=N+7$, that is, the remainder when divided by 7 has not changed. No... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,437 |
4. A student's income comes from three sources: scholarship, temporary part-time work, and parental support. If the government doubles the scholarship, the income will increase by $5 \%$. If the time spent on part-time work is doubled, the income will increase by $15 \%$. By what percentage will the student's income in... | Solution. Let $\mathrm{S}$ be the student's monthly income, $a, b$ and $c$ be the amounts of the scholarship, part-time job, and parental support, respectively (expressed, for example, in rubles). Clearly, $S=a+b+c$. Then, according to the conditions, $2 a+b+c=1.05 S$ and $a+2 b+c=1.15 S$. From the first equation, $a=0... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,438 |
1. In the basket, there are fruits (no less than five). If you randomly pick three fruits, there will definitely be an apple among them. If you randomly pick four fruits, there will definitely be a pear among them. What fruits can be picked and in what quantities if you randomly pick five fruits? | Solution. From the condition of the problem, it follows that the "non-apples" in the box are no more than two fruits (otherwise, you could pull out 3 fruits, none of which would be apples). Similarly, "non-pears" are no more than three fruits (otherwise, you could pull out 4 fruits, none of which would be pears). Thus,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,439 |
2. At one vertex of a cube, the number 2021 is written, and zeros are written at the other vertices. You can add one to the numbers at the ends of any edge. Is it possible to achieve that all numbers standing at the vertices of the cube are divisible by $2?$ | Solution. Initially, the sum of the numbers at the vertices of the cube is 2021, which is odd. Each time we add one to two vertices, the sum remains odd. Since the sum of eight integers is odd (in this problem, the oddness of the sum is an invariant), the numbers at all vertices cannot become even. Answer: it is imposs... | itisimpossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,440 |
3. In square $A B C D$, the center is marked as point $O$, and points $M$ and $N$ are marked on sides $A B$ and $B C$ such that $\angle M O N=90^{\circ}$. Prove that $M B+B N=A B$. | Solution. Drop perpendiculars from $O R$ and $O T$ to sides $A B$ and $B C$. Triangles $R O M$ and $T O N$ are equal by the side and acute angle of the right triangle. Therefore, $R M = T N$ and $M B = R B - R M = \frac{1}{2} A B - R M, B N = B T + T N = \frac{1}{2} A B + R M$. Adding these last expressions, we get $M ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,441 |
4. Given a natural number $N$. Vera performs the following operations with it: first, she adds 3 until the resulting number is divisible by 5 (if $N$ is initially divisible by 5, nothing needs to be added). The resulting number is then divided by 5. She then performs the same operations with the new number, and so on. ... | Solution. These are numbers of the form $3 k, k \in \mathbb{N}$. Indeed, from numbers that are multiples of 3, the number 1 cannot be obtained, because if a number $N$ is a multiple of 3, then $N+3$ is also a multiple of 3, and if $N=5 k$ is a multiple of 3, then $k$ is a multiple of 3, since 3 and 5 are coprime. There... | 3k,k\in\mathbb{N} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,442 |
5. Two magicians show the following trick to the audience. The audience has 48 cards numbered from 1 to 48. They choose 25 cards and hand them to the first magician. The first magician returns two of them to the audience. The audience adds one of the remaining 23 cards to these two and, after shuffling, hands these thr... | Solution. The first magician must return to the viewer two cards, the sum of the numbers on which is 49 (such a pair will always be found, see Lemma). The second magician, knowing this, will infallibly distinguish the pair of cards from the first magician from the card of the viewer, since out of three cards there can ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,443 |
1. Is the number $\underbrace{444 . . .4}_{2 n \text { digits }}-\underbrace{88 \ldots 8}_{\text {n digits }}$ a perfect square for any natural $n$? | Solution. The number $K=\underbrace{444 ...4}_{2 n \text{ digits }}-\underbrace{88 \ldots 8}_{n \text{ digits }}$ is a perfect square. Indeed.
$\underbrace{444 \ldots 4}_{2 n \text{ digits }}=4\left(10^{2 n}+10^{2 n-1}+\ldots+10+1\right)=4 \frac{10^{2 n}-1}{10-1}=4 \frac{10^{2 n}-1}{9}$
$\underbrace{88 \ldots ...8}_{... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,444 |
2. Solve the equation: $\cos ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)+\sin ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)=\sin ^{-2}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$. | # Solution.
Let's make the substitution $t=\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$. Clearly, $0<t \leq \frac{\pi}{2}$.
$\cos ^{4} t+\sin ^{4} t=\sin ^{-2} t$.
$\cos ^{4} t+\sin ^{4} t=\left(\cos ^{2} t+\sin ^{2} t\right)^{2}-2 \sin ^{2} t \cdot \cos ^{2} t=1-2 \sin ^{2} t \cdot \cos ^{2} t \leq 1 . \frac{1}{\sin ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,445 |
3. It is known that on a chessboard, 8 rooks can be placed so that they do not attack each other. Schoolboy Pete does not like the chessboard coloring, and he colored the board in 32 colors, so that there are exactly two cells of each color. Will he now be able to place 8 rooks so that they do not attack each other and... | # Answer: Yes
Solution.
Note that the total number of ways to place 8 rooks on the board so that they do not attack each other is $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 8$! (8 ways on the first vertical, 7 on the second, and so on.)
Now let's count the number of ways to place 8 rooks so that so... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,446 |
4. Find all triples of natural numbers $x, y$ and $z$ such that $(x+1)(y+1)(z+1)=3 x y z$ | Solution. Since the left side of the equation is divisible by 3, without loss of generality, let $\mathrm{x}=3 \mathrm{k}-1$, where $k \in N$, and we obtain the following estimate:
$3 k(y+1)(z+1)=3(3 k-1) y z \geq 3(3 k-k) y z$ or $(y+1)(z+1) \geq 2 y z$. Further, we have
$y z+y+z+1 \geq 2 y z \Leftrightarrow y+z+1 \g... | (x,y,z)\in{(2,2,3),(2,3,2),(3,2,2),(5,1,4),(5,4,1),(4,1,5),(4,5,1),(1,4,5),(1,5,4),(8,1,3),(8,3,1),(3,1,8),(3,8,1),(1,3,8),(1,8} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,447 |
5. Given a triangular prism with an equilateral base $m$. The lateral edges are equal to the side of the base. One of the vertices is equidistant from the vertices of the base. Find the largest radius of the sphere that can fit inside the prism.
# | # Solution.
Let vertex $A_{1}$ be equidistant from the vertices of the triangle $A, B, C$. Then $A_{1} A B C$ is a tetrahedron with height $A_{1} O=\sqrt{2 / 3} m$, and the apothem $B M=M C=\sqrt{3} / 2 m$. We need to find the radius of the sphere that touches three planes: $A B B_{1}, A C C_{1}, B C B_{1}$. If we dra... | \frac{\sqrt{6}-\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,448 |
1. Do there exist integers $a_{1}, a_{2}, \ldots, a_{2016}$ such that $a_{1}+a_{2}^{2}+\ldots+$ $a_{2016}^{2016}=20152016$ and $a_{1}^{3}+a_{2}^{4}+\ldots+a_{2016}^{2018}=20162017$ | # Answer: do not exist
Solution: Consider the difference $a_{1}^{3}+a_{2}^{4}+\ldots+a_{2016}^{2018}-\left(a_{1}+a_{2}^{2}+\ldots+a_{2016}^{2016}\right)=a_{1}\left(a_{1}-1\right)\left(a_{1}+1\right)+a_{2}^{2}\left(a_{2}-1\right)\left(a_{2}+1\right)+\ldots+a_{2016}^{2016}\left(a_{2016}-1\right)\left(a_{2016}+1\right)$.... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,449 |
2. In the children's creativity center, there are 32 clubs. It is known that 6 children attend each club, and in any two clubs, the total number of children is 13. How many children in total attend the children's creativity center.
# | # Answer: There are no solutions.
Solution (Attention to detail problem): Since six children attend each circle, the total number of children attending two circles does not exceed 12, while the problem states there are 13. Therefore, this is impossible.
## Recommendations for checking.
Only the answer - 0 points.
3... | \frac{\pi}{4},2\pik,k\inZ | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,450 |
4. The sequence of numbers $a_{1}, a_{2}, \ldots, a_{2016}$ is a geometric progression, and the sequence $b_{1}, b_{2}, \ldots, b_{2016}$ is an arithmetic progression. It is known that among all the quadratic polynomials $P_{i}(x)=x^{2}+a_{i} x+b_{i}, i=1, \ldots, 2016$ only one quadratic polynomial $\mathrm{P}_{k}(x)$... | Solution: If $a_{1}=3, a_{2}=1, \ldots, a_{2016}=\frac{1}{3^{2016}}, b_{1}=1, b_{2}=2, \ldots, b_{2016}=2016$, then $k=1$. If $a_{1}=\frac{1}{3^{2016}}, a_{2}=\frac{1}{3^{2015}}, \ldots, a_{2015}=1, a_{2016}=3$, $b_{1}=2016, b_{2}=2015, \ldots, b_{2015}=2, b_{2016}=1$, then $k=2016$.
Now let $1<k<2016$. Then the quadr... | 1;2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,451 |
5. Find the height of the pyramid if, by cutting it only along the lateral edges and unfolding the lateral faces onto the plane of the base, outside of it, you get a square with a side of 18. If it is impossible to get such a square, explain why.
# | # Solution:
The figure shows such a net. If we flip the pyramid and place it on the face $A B C$, it is easy to calculate the volume of the pyramid $V=\frac{1}{3} \cdot \frac{1}{2} \cdot 9 \c... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,452 |
10.1. In each cell of a table consisting of 10 columns and $n$ rows, a digit is written. It is known that for any row $A$ and any two columns, there exists a row that differs from $A$ exactly in these two columns. Prove that $n \geqslant 512$.
(P. Karasev) | 10.1. Let $R_{0}$ be the first row of the table. Consider any set of an even number of columns and number them from left to right: $C_{1}, \ldots, C_{2 m}$. Then there is a row $R_{1}$ that differs from $R_{0}$ exactly in columns $C_{1}$ and $C_{2}$; further, there is a row $R_{2}$ that differs from $R_{1}$ exactly in ... | 512 | Combinatorics | proof | Yes | Yes | olympiads | false | 18,453 |
10.2. On the board, nine quadratic trinomials are written: $x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \ldots, x^{2}+a_{9} x+b_{9}$. It is known that the sequences $a_{1}, a_{2}, \ldots, a_{9}$ and $b_{1}, b_{2}, \ldots, b_{9}$ are arithmetic progressions. It turned out that the sum of all nine trinomials has at least o... | 10.2. Answer. 4.
Let $P_{i}(x)=x^{2}+a_{i} x+b_{i}, P(x)=P_{1}(x)+\ldots+P_{9}(x)$. Notice that $P_{i}(x)+P_{10-i}(x)=2 x^{2}+\left(a_{i}+a_{10-i}\right) x+\left(b_{i}+b_{10-i}\right)=$ $=2 P_{5}(x)$. Therefore, $P(x)=9 P_{5}(x)$, and the condition is equivalent to $P_{5}(x)$ having at least one root.
Let $x_{0}$ be ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,454 |
10.3. Let's call a company $k$-indivisible if in any partition of it into $k$ groups, there will be two acquaintances in one of the groups. Given a 3-indivisible company in which there are no four people who are all mutually acquainted. Prove that it can be divided into two companies, one of which is 2-indivisible, and... | 10.3. Suppose the opposite. Consider a graph $G$ where people are vertices, and two people are connected by an edge if they are acquainted. Then the graph is $k$-colorable if its vertices can be properly colored with $k$ colors (i.e., colored such that adjacent vertices have different colors). We will use the following... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,455 |
10.4. The perimeter of triangle $ABC$ is 4. Points $X$ and $Y$ are marked on rays $AB$ and $AC$ such that $AX = AY = 1$. Segments $BC$ and $XY$ intersect at point $M$. Prove that the perimeter of one of the triangles $ABM$ and $ACM$ is 2.
(V. Shmarov) | 10.4. Since segments $B C$ and $X Y$ intersect, without loss of generality, we can assume that $A B > A X$ and $A C < A Y$.
Let the exscribed circle $\omega$ of the original triangle touch side $B C$ at point $R$, and the extensions of sides $A B$ and $A C$ at points $P$ and $Q$ respectively. Let $J$ be the center of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,456 |
5. Plot the points \(A(4 ; 1), B(3 ; 5), C(-1 ; 4), D(0 ; 0)\) on the coordinate plane. Prove that these points are the vertices of a square. | Solution. Let's plot the given points on the coordinate plane (see figure). Let $A_{1}$ and $C_{1}$ be the projections of points $A$ and $C$ onto the $O x$ axis.
Using the Pythagorean theore... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,459 |
10.5. A circle of 101 sages stands. Each of them either believes that the Earth revolves around Jupiter, or believes that Jupiter revolves around the Earth. Once a minute, all sages simultaneously announce their opinions. Immediately after this, each sage whose both neighbors think differently from him changes his opin... | Solution. Let's associate each sage with a certain opinion with the sign "+", and with the opposite opinion - the sign "-". Then the arrangement of sages corresponds to the arrangement of 101 signs in a circle.
Suppose at some moment two identical signs stand next to each other. Then in the next minute they will not c... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 18,460 |
10.6. Do there exist natural numbers $a, b, c$, greater than $10^{10}$, such that their product is divisible by any of them increased by 2012?
(V. Senderov) | Answer. They exist.
First solution. Choose some $t>10^{10}$ and set $a=b=2012 t, c=2012\left(t^{2}-1\right)$. Then $c \vdots 2012(t+1)=a+2012=$ $=b+2012$, and $a b=2012^{2} t^{2}: 2012 t^{2}=c+2012$. It follows that $a, b, c$ form the desired triplet.
Second solution. It is sufficient to find numbers $a^{\prime}, b^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,461 |
10.7. On the coordinate plane, $n$ parabolas are drawn, which are graphs of quadratic trinomials; no two of them touch. They divide the plane into several regions, one of which is located above all the parabolas. Prove that the boundary of this region has no more than $2(n-1)$ vertices (i.e., points of intersection of ... | Solution. We will prove the statement of the problem by induction on $n$. For $n=1$ it is obvious. Let now $f_{1}(x), f_{2}(x), \ldots, f_{n}(x)$ be the given quadratic trinomials $(n \geqslant 2)$, and let $f_{n}(x)$ be the trinomial with the smallest leading coefficient (if there are several, then any of them). Denot... | 2(n-1) | Geometry | proof | Yes | Yes | olympiads | false | 18,462 |
10.8. Point $E$ is the midpoint of the segment connecting the orthocenter of an acute scalene triangle $ABC$ with its vertex $A$. The incircle of this triangle touches the sides $AB$ and $AC$ at points $C'$ and $B'$, respectively. Prove that the point $F$, symmetric to point $E$ with respect to the line $B'C'$, lies on... | The first solution. Let's assume that $A B > A C$ (see Fig. 3). Let $B B_{1}$ and $C C_{1}$ be the altitudes of triangle $\triangle A B C$, $H$ be the point of their intersection, $O$ and $I$ be the centers of the inscribed and circumscribed circles of triangle $A B C$, and $r$ be the radius of its inscribed circle; le... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,463 |
1. Write the number 2021 using each of the digits from 0 to 9 exactly once, along with parentheses and arithmetic operations. (Parentheses and arithmetic operations can be used in any quantity. "Sticking" digits together to form a single number is allowed) | For example, $43 \cdot (8 \cdot 5 + 7) + 0 \cdot 1 \cdot 2 \cdot 6 \cdot 9 = 2021$. | 2021 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,464 |
2. Anya has a chocolate bar of size 5 x 6 squares. It contains 6 squares, forming a $2 \times 3$ rectangle, in which there are nuts (the rectangle can be positioned either vertically or horizontally). Anya does not know exactly where the nuts are. She wants to eat the smallest number of squares, but in such a way that ... | 2. Answer: 5.
Solution. An example of which slices to eat.
Evaluation. Let's number the slices as shown in the figure.
| 1 | 1 | 2 | 2 | 3 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- |
... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,465 |
3. Let $a, b, c$ be integers such that $(a+b+c)^{2}=-(a b+a c+b c)$ 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. | 3. Solution. $(a+b)(a+c)=a^{2}+a b+a c+b c=a^{2}-(a+b+c)^{2}=-(b+c)(2 a+b+c)$.
Other cases are considered similarly. | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,466 |
4. The perpendicular bisectors of sides $A B$ and $C D$ of quadrilateral $A B C D$ intersect on side $A D$. Prove that if angle $A$ equals angle $D$, then the diagonals of quadrilateral $A B C D$ are equal. | 4. Solution. Let $E$ be the point of intersection of the perpendicular bisectors of $AB$ and $CD$. Triangle $CED$ is isosceles with base $DC$, since $E$ lies on the perpendicular bisector of $CD$. Similarly, triangle $BEA$ is isosceles. Therefore, $\angle ECD = \angle EDC = \angle EAB = \angle ABE = a$. From triangle $... | AC=BD | Geometry | proof | Yes | Yes | olympiads | false | 18,467 |
5. There are four locks and 6 keys. It is known that each key opens exactly 2 locks. Moreover, there are no two keys that open exactly the same two locks. Determine which key opens which two locks in no more than 11 checks. In one check, you can insert a key into a lock and check if it opens it or not. | 5. Solution. Note that there are a total of 6 different pairs of locks. This means that each pair of locks corresponds to its own key.
Let's number the locks $1,2,3,4$. Then the pairs of locks will be: $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)$. Note that each lock can be opened by exactly three keys.
First 5 attempts: tr... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,468 |
1. A mathematician and a physicist started running on a running track towards the finish line at the same time. After finishing, the mathematician said: "If I had run twice as fast, I would have beaten the physicist by 12 seconds." And after finishing, the physicist said: "If I had run twice as fast, I would have beate... | # Answer: 16.
Solution: Let the time it took the mathematician to run the entire distance be $2 x$ seconds, and the time it took the physicist be $2 y$ seconds. Then, under the condition that the mathematician ran twice as fast, we get $2 y-x=12$. And from the condition that the physicist ran twice as fast, we get $2 ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,469 |
2. On the island, there live 7 natives who know mathematics and physics, 6 natives who know physics and chemistry, 3 natives who know chemistry and mathematics, and 4 natives who know physics and biology. In how many ways can a team of three people be formed who together know at least three subjects out of the four? Th... | # Answer: 1080.
## Solution.
Let's choose three people. If this trio includes representatives of at least two different groups of aborigines, then together they know no fewer than three subjects.
If this trio includes representatives of only one group, then together they know only two subjects.
Therefore, we will f... | 1080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,470 |
3. Prove that for any positive numbers $x, y$ and $z$ the following inequality holds
# | # Solution.
$$
\frac{x}{x+2 y+3 z}+\frac{y}{y+2 z+3 x}+\frac{z}{z+2 x+3 y} \leq \frac{4}{3} .
$$
We use the lemma.
Lemma. If $a>0, b>c>0$, then $\frac{c}{b}<\frac{c+a}{b+a}$.
Then we get
$$
\begin{aligned}
& \frac{x}{x+2 y+3 z} \leq \frac{3 x+y}{3 x+3 y+3 z} \\
& \frac{y}{y+2 z+3 x} \leq \frac{3 y+z}{3 y+3 z+3 x} ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,471 |
4. The perpendicular bisector of side $A B$ of the inscribed quadrilateral $A B C D$ intersects the diagonal $A C$ at point $P$. The perpendicular bisector of side $C D$ intersects the diagonal $B D$ at point $Q$. Prove that $P Q \| A D$. | Solution. Since quadrilateral $ABCD$ is inscribed, and triangles $ABP$ and $DCQ$ are isosceles, then $\angle ABP = \angle BAP = \angle CDQ = \angle DCQ$ (see Fig. 1). Angles $\angle BPC$ and $\angle CQB$ are external to triangles $ABP$ and $DCQ$ respectively, so $\angle BPC = \angle ABP + \angle BAP = \angle DCQ + \ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,472 |
5. Islandland consists of ten islands, some of which are connected by two-way air routes. If you choose any 9 islands, you can fly around them one by one and return to the starting island at the end. Find the minimum number of air routes that can exist in this country. | Answer: 15 airlines.
Solution. Evaluation. Let islands $A$ and $B$ be connected by an airline. If we choose 9 islands for the tour, excluding $A$, then a circular path passes through $B$, meaning $B$ is connected by airlines to at least two other islands. Thus, at least three airlines depart from $B$, and a similar st... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,473 |
Problem 5.3. In five of the nine circles in the picture, the numbers 1, 2, 3, 4, 5 are written. Replace the digits $6, 7, 8, 9$ in the remaining circles $A, B, C, D$ so that the sums of the four numbers along each of the three sides of the triangle are the same.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,477 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots marked on one face, 5 dots on another, ..., and 1 dot on the remaining face. These cubes were glued together to form the figure shown in the image.
How many dots are on the four left faces?
![](https://cdn.mathpix.com/cropped/2024_05_06_43b... | Answer: On face $A$ there are 3 points, on face $B-5$, on face $C-6$, on face $D-5$. Solution. Let's consider the arrangement of the faces on one die. We will denote the faces by numbers corresponding to the number of dots on them.
From the picture, it is clear that face 1 borders with faces $2,3,4$ and 5. Therefore, ... | 3,5,6,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,478 |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,479 |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,480 |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | In\\A\there\is\the\\6,\in\B-2,\in\C-4,\in\D-5,\in\E-3,\in\F-8,\in\G-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,482 |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
=9 \cdot(18+S)$. Solving this linear equation, we get $S=13.5$. | 13.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,484 |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,486 |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,487 |
Problem 8.6. Vasya thought of three natural numbers with a sum of 1003. Calculating their product, Vasya noticed that it ends with $N$ zeros. What is the maximum value that
$N$ can take? | Answer: 7.
Solution. The product of the three thought-of numbers could end with 7 zeros, for example, if these were the numbers $625, 250, 128$. Indeed, $625+250+128=1003$ and
$$
625 \cdot 250 \cdot 128=5^{4} \cdot\left(2 \cdot 5^{3}\right) \cdot 2^{7}=2^{8} \cdot 5^{7}=2 \cdot 10^{7}=20000000
$$
Suppose there exist... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,488 |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,491 |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we get $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,494 |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,495 |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
 | Solution. Let $n, n+1, n+2, n+3$ be the given numbers. The sum of the three smallest of them is $3n+3=3(n+1)$, and the sum of the three largest numbers is $3(n+2)$. But at least one of the numbers $n+1$ and $n+2$ is even, that is, equal to the product of the numbers 2 and $k$, where $k>3$. Therefore, the given sum can ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,497 |
10.7. Given real numbers $a$ and $b$, where $b > a > 1$. Let
$$
x_{n}=2^{n}(\sqrt[2^{n}]{b}-\sqrt[2^{n}]{a})
$$
Prove that the sequence $x_{1}, x_{2}, \ldots$ is decreasing.
(A. Khryabrov) | Solution. We will prove that $x_{n}>x_{n+1}$. Let $A=\sqrt[n^{n+1}]{a}$ and $B=\sqrt[2^{n+1}]{b}$. It is easy to see that $B>A>1$, hence $\frac{A+B}{2}>1$. Then we have
$$
\begin{gathered}
x_{n+1}=2^{n+1}(B-A)>0 \\
x_{n}=2^{n}\left(B^{2}-A^{2}\right)=2^{n+1}(B-A) \frac{A+B}{2}=x_{n+1} \cdot \frac{A+B}{2}>x_{n+1}
\end{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,498 |
10.8. In an acute-angled triangle $ABC$, the altitude $BH$ is drawn. Points $M$ and $N$ are the midpoints of segments $AH$ and $CH$ respectively. In the circumcircle $\Omega$ of triangle $BMN$, a diameter $BB'$ is drawn. Prove that $AB'=CB'$.
( $A$. Kuznetsov) | Solution. Let $O$ be the center of the circle $\Omega$; in other words, $O$ is the midpoint of the diameter $B B'$. Denote by $H'$ and $P$ the projections of points $B'$ and $O$ respectively onto the line $AC$ (see Fig. 5). Since $O$ lies on the perpendicular bisector of $MN$, we get that $P$ is the midpoint of $MN$. S... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,499 |
10.9. On the board, a convex $n$-gon ($n \geqslant 4$) is drawn. Each of its vertices must be painted either black or white. A diagonal is called bicolored if its ends are painted in different colors. A coloring of the vertices is called good if the $n$-gon can be divided into triangles by bicolored diagonals that do n... | Answer: $n^{2}-n$.
Solution. Immediately note that coloring all vertices in one color is not a good coloring; such colorings are not considered in the further solution.
Let's call a side of the polygon bicolored if its ends are colored in different colors (i.e., extend the definition of bicolored to sides). We will c... | n^2-n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,500 |
10.10. The sequence of natural numbers $a_{1}, a_{2}, \ldots$ is defined by the conditions $a_{1}=1, a_{n+1}=a_{n}+a_{[\sqrt{n}]}$ for all natural $n \geqslant 1$. Prove that for each natural $k$, there is a term in this sequence that is divisible by $k$. (As usual, $[x]$ denotes the greatest integer not exceeding $x$.... | Solution. Among the numbers $a_{k}, a_{k+1}, a_{k+2}, \ldots$ we choose the number that gives the smallest remainder when divided by $k$. Let this number be $a_{m}$, and let it give a remainder $r$ when divided by $k$. If $r=0$, then $a_{m}$ is the term we need. Suppose now that $0<r<k$.
Since $m \geqslant k$, we have... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,501 |
1. (7 points) Provide an example of two common fractions, the difference of which is three times greater than their product. Provide calculations that justify this property. | Answer. For example, $\frac{1}{2}$ and $\frac{1}{5}$.
Solution.
$\frac{1}{2}-\frac{1}{5}=\frac{5-2}{10}=\frac{3}{10}, \quad \frac{1}{2} \times \frac{1}{5}=\frac{1}{10}$.
Any fractions of the form $\frac{1}{n}$ and $\frac{1}{n+3}$ work, and there are other solutions as well.
## Grading Criteria.
- Any complete corr... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,502 |
3. (7 points) The number 49 is written on the board. In one move, it is allowed to either double the number or erase its last digit. Is it possible to get the number 50 in several moves? | Answer. Possible.
Solution. The number 50 can be obtained by doubling 25, and 25 can be obtained by erasing the last digit of the number 256, which is a power of two. Thus, the necessary chain of transformations can look like this:
$$
49 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 32 \rightarrow 64 \righta... | 49arrow4arrow8arrow16arrow32arrow64arrow128arrow256arrow25arrow50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,504 |
4. (7 points) One of the three friends: Andrey, Boris, or Vladimir - is the strongest, another is the smartest, and the third is the kindest. One day, they said the following:
Andrey: Vladimir is stronger than me.
Boris: I am smarter than Vladimir.
Vladimir: Boris is smarter than me.
It is known that the strongest ... | Answer: Yes.
Solution. Let's denote: A - Andrey, B - Boris, V - Vladimir. Statements B and V repeat each other, and since there is only one false statement among the three, B and V told the truth, and A lied. Therefore, A is the smartest (by condition), A is stronger than V (since A lied), and B is smarter than V (sin... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,505 |
5. (7 points) Mom is walking around a lake with a stroller and completes a full lap around the lake in 12 minutes. Vanya rides a scooter on the same path in the same direction and meets (overtakes) mom every 12 minutes. At what intervals will Vanya meet mom if he rides at the same speed but in the opposite direction? | Answer: Every 4 minutes.
Solution. Since Mom completes a full lap around the lake in 12 minutes and meets Vanya every 12 minutes, in 12 minutes Vanya rides around the lake exactly 2 times, while Mom completes one lap. Therefore, Vanya's speed is twice that of Mom's. From this, it follows that when Vanya was riding in ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,506 |
# 8.1. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 3 meters away from me."
- The rightmost in the row said: "My fellow ... | # Answer:
The second islander -2 m
The third islander -1 m; 3 m; 4 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then, from his statement, it follows that the fourth one must also be a knight. The fourth one said that his fellow islander stands 2 meters away, and ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,512 |
# 8.2. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 2 meters away from me."
- The rightmost in the row said: "My fellow ... | # Answer:
The second islander is 1 m
The third islander is 1 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then from his statement, it follows that the third one is also a knight; by the principle of exclusion, the second and fourth must be liars. The fourth said ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,513 |
Problem 8.2.1. Given trapezoid $A B C D(A D \| B C)$. It turns out that $\angle A B D=\angle B C D$. Find the length of segment $B D$, if $B C=36$ and $A D=64$.
 | Answer: 48.
Solution. Since $A D \| B C$, we have $\angle C B D=\angle B D A$. Then triangles $A B D$ and $D C B$ are similar by the first criterion. Therefore, $\frac{64}{B D}=\frac{A D}{B D}=\frac{B D}{B C}=\frac{B D}{36}$, from which we find $B D=\sqrt{64 \cdot 36}=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,514 |
Problem 8.4.1. A rectangle was cut into six smaller rectangles, the areas of five of them are marked on the diagram. Find the area of the remaining rectangle.
 | Answer: 101.
Solution. Let's introduce the notation as shown in the figure.
Notice that
$$
2=\frac{40}{20}=\frac{S_{H I L K}}{S_{D E I H}}=\frac{H K \cdot H I}{H D \cdot H I}=\frac{H K}{H ... | 101 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,515 |
Problem 8.7.1. Given trapezoid $A B C D (B C \| A D)$. Point $H$ on side $A B$ is such that $\angle D H A=$ $90^{\circ}$. It is known that $C H=C D=13$ and $A D=19$. Find the length of segment $B C$.
.
Figure for the solution to problem 8.7.1
In the isosceles triangle $HCD$, we have $\angle C... | 9.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,516 |
Task 7.1. How to cut off 50 cm from a thread, the length of which is $2 / 3$ meters, without having any measuring tools? ( ( points)
# | # Solution.
If a quarter of the shoelace is cut off (for which it needs to be folded in half twice), exactly 50 cm will remain. Indeed, $3 / 4 \cdot 2 / 3=1 / 2$.
Comment. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,517 |
Problem 7.2. For natural numbers $m$ and $n$, it is known that if their sum is multiplied by the absolute value of their difference, the result is 2021. What values can the numbers $m$ and $n$ take? (7 points)
# | # Solution.
Let $\mathrm{m} \geq \mathrm{n}$, then the modulus of their difference is (m-n). Since $\mathrm{m}$ and $\mathrm{n}$ are natural numbers, their sum and the modulus of their difference are also natural numbers. The number 2021 is divisible only by $1,43,47$
and 2021. Therefore, there are only four possible... | =1011,n=1010=45,n=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,518 |
Task 7.3. Tolya made four tokens for the game with denominations of $1, 2, 3, 5$ bokins, which should weigh $1, 2, 3, 5$ grams respectively. But one of these tokens he made poorly - with the wrong weight. How can you determine the "incorrect" token in two weighings on a balance scale without weights? (7 points)
# | # Solution.
First, place the tokens worth 1 bokins and 2 bokins on one pan, and the token worth 3 bokins on the other; then place the tokens worth 2 bokins and 3 bokins on one pan, and the token worth 5 bokins on the other.
If the scales show equality in one of the weighings, then the defective token is the one that ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,519 |
Task 7.4. 17 peony bushes are planted in a circle.
a) Prove that there will definitely be two adjacent bushes, the total number of buds on which is even.
b) Is it always possible to find two adjacent bushes, the total number of buds on which is divisible by 3? (7 points)
# | # Solution.
a) First method. Suppose that it is impossible to find such a pair of neighboring peonies. This means that the bushes with an even and odd number of buds alternate. But an odd number (17) of bushes cannot alternate.
Second method. Consider all 17 "neighboring" sums. Their total sum is twice the number of ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,520 |
Task 7.5. Two people are writing, a) a 60-digit; b) a 14-digit number, using only the digits $1,2,3,4,5$. The first digit is written by the first person, the second by the second, the third by the first, and so on. Can the second person ensure that the resulting number is divisible by 9, if the first person tries to pr... | # Solution.
Answer. a) Can; b) cannot.
a) The second player's strategy: write a digit so that the sum of it and the previous digit equals 6. Since (60/2)*6 is divisible by 9, the resulting 60-digit number will be divisible by 9.
b) The first player's strategy: first write 1, and then write a digit so that the sum of... | )Can;b)cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,521 |
9.1. Find all possible pairs of prime numbers \( m \) and \( n \), such that the equation \( x^{2} - m x - n = 0 \) has a prime root. | Answer: $m=2, n=3$.
Solution. Suppose that $x=p$ is a prime root, then $p^{2}-m p-n=0$, from which it follows that $n=p(p-m)$. Since $n$ and $p$ are prime numbers, then $n=p$ and $p-m=1$, that is, $p=m+1$. The number $m$ can only be 2, because in any other case $m+1$ would be even, and then $p$ would not be a prime nu... | =2,n=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,523 |
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