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5. In rectangle $A B C D$, a point $K$ is marked on the diagonal $A C$ such that $C K=B C$. A point $M$ is marked on the side $B C$ such that $K M=C M$. Prove that $A K+B M=C M$. | Solution: On the extension of side $B C$ beyond point $B$, mark a point $E$ such that $B E = A K$. Then, $C E = C B + B E = C K + K A = C A$ (see Fig. 8.5).
In triangles $E K C$ and $A B C: C E = C A, C K = C B$, and angle $E C A$ is common, so these triangles are equal. Therefore, $\angle E K C = \angle A B C = 90^{\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,597 |
6. What is the maximum number of natural numbers not exceeding 2016 that can be marked so that the product of any two marked numbers is a perfect square? | Answer: 44.
Solution. Let's find the number of natural numbers whose squares are no greater than 2016. There are 44 such numbers, since $44^{2}=1936 < 2016$. Since the product of two perfect squares is a perfect square, the numbers $1=1^{2}, 4=2^{2}, \ldots, 1936=44^{2}$ can be marked.
We will prove that it is imposs... | 44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,598 |
1.1. If you add the last digit of a number to the number itself, you get 5574, and if you add the second-to-last digit, you get 557. What is this number? | Answer: 5567
Solution. Note that the guessed number differs from 5574 by one digit, so the guessed number is not less than $5574-9=5565$ and not more than 5574. Then the second to last digit of the number is 6 or 7, so our number is 5566 or 5567. The first one does not fit, but the second one does. | 5567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,599 |
2.1. In the table, there are numbers. It turned out that six sums: three sums by rows and three sums by columns are the same. The minimum of the numbers $a, b$, and $c$ is 101. Find the sum $a+b+c$.
| $\mathrm{a}$ | 1 | 4 |
| :---: | :---: | :---: |
| 2 | $\mathrm{~b}$ | 5 |
| 3 | 6 | $\mathrm{c}$ | | Answer: 309
Solution. Note that $a+5=b+7=c+9$, therefore, the number $c$ is the smallest, and the others are 2 and 4 more, that is, 103 and 105. The sum of all three numbers $101+103+105=309$ | 309 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,600 |
3.1. Alina and Masha wanted to create an interesting version of the school tour olympiad. Masha proposed several problems and rejected every second problem of Alina's (exactly half), Alina also proposed several problems and did not reject only every third problem of Masha's (exactly a third). In the end, there were 10 ... | Answer: 15
Solution. In total, 17 tasks were rejected. Note that among the tasks proposed by Alina, an equal number were included and rejected, and among the tasks proposed by Masha, the number of rejected tasks is twice the number of tasks that remained in the variant. Since the number of rejected tasks is 7 more tha... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,601 |
4.1. On a grid sheet, a $1 \times 5$ rectangle was painted. Each minute, all those uncolored cells that have at least one side-adjacent cell already colored are colored. For example, after one minute, 17 cells will be colored. How many cells will be colored after 5 minutes? | Answer: 105
Solution 1. Note that after 5 minutes, we will get a stepped figure, in which the rows will have $5,7,9,11,13,15,13,11,9,7$ and 5 cells. In total, this is 105.
Solution 2. Note that if we shorten the strip to $1 \times 1$, 44 cells will disappear, which is $4 \cdot 11=44$. If there is only one cell at the... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,602 |
5.1. The girl always tells the truth on Thursdays and Fridays, and always lies on Mondays. On the other days of the week, she may tell the truth or lie. One day, she was asked for seven days in a row, in which year she was born. For the first six days, the answers received were 2010, 2011, 2012, 2013, 2002, 2011. What ... | Answer: 2011 or 2010
Solution. Note that if among the specified 6 days there were a Thursday and a Friday, then some two adjacent answers would be the same. Therefore, either the first day is Friday, or the last day is Thursday. From this, we get the answer options. | 2011or2010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,603 |
6.1. A cube lies on a plane. Each face is marked with 1 to 6 points such that the sum of points on opposite faces is always 7. The cube is rolled over the plane as shown in the picture. How many points will be on the top face of the cube when it lands on the last cell?
![](https://cdn.mathpix.com/cropped/2024_05_06_ff... | Answer: 5
Solution. To find out the answer, you need to imagine rolling the die in the opposite direction and each time keep track of where the face that will end up on top will be. It turns out that this face will be at the bottom in the initial position of the die. Since the top face has 2 dots, the bottom face now ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,604 |
7.1. Six consecutive numbers were written on the board. When one of them was erased and the remaining ones were added up, the result was 10085. Which number could have been erased? List all possible options. | Answer: 2020 or 2014
Solution. Note that the sum of five consecutive numbers is equal to the middle number multiplied by 5. If we have five out of six consecutive numbers, and they are not consecutive, then we can increase 1, 2, 3, or 4 of them by 1 so that the numbers become consecutive. Since the sum of the given nu... | 2020or2014 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,606 |
8.1. A four-digit number is called "beautiful" if it is impossible to append a digit to the right so that the resulting five-digit number is divisible by 11. How many beautiful numbers are there that are greater than 3100 and less than 3600? | Answer: 46
Solution. Note that the number 3101 is beautiful because 31009 and 31020 are divisible by 11, so 3101* cannot be divisible by 11. Suppose the number is beautiful, then we have found 10 consecutive numbers, none of which are divisible by 11, but among 11 consecutive numbers, at least one is divisible by 11, ... | 46 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,607 |
1. Find the minimum value of the expression $\frac{25 x^{2} \sin ^{2} x+16}{x \sin x}$ for $0<x<\pi$. | Answer: 40.
Solution. Let $y = x \sin x$. Then the expression can be written as
$$
\frac{25 y^{2} + 16}{y} = 25 y + \frac{16}{y}.
$$
Note that $y > 0$ (since $x > 0$ and $\sin x > 0$), so we can apply the inequality between the means:
$$
25 y + \frac{16}{y} \geq 2 \sqrt{25 y \cdot \frac{16}{y}} = 40.
$$
Equality h... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,610 |
2. In a right-angled triangle, the lengths of the leg $a$ and the hypotenuse $c$ are expressed as two-digit integers, and both numbers use the same set of digits. The length of the leg $b$ is a positive rational number. What can the length of the hypotenuse be? (Provide all answers and prove that there are no others.) | Answer: 65.
Solution. Since $a \neq c$, in two-digit numbers, two different digits must be used, and they will stand in different places. Let $a=10 e+d, c=10 d+e$. Then $b=\sqrt{(10 d+e)^{2}-(10 e+d)^{2}}=3 \sqrt{11(d-e)(d+e)}$.
By the condition, $b$ is rational, so it cannot contain square roots. The number 11 is pr... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,611 |
3. Find the smallest natural number $N$, which is divisible by $r$, ends in $r$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and $2p+1$ is the cube of a natural number. | Answer: 11713.
Solution. Let $2 p+1=n^{3}$. Then $(n-1)\left(n^{2}+n+1\right)=2 p$. The number $2 p$ can only have the following positive divisors: $1,2, p, 2 p$. The number $n$ is obviously odd, so $n-1$ is divisible by 2. The number $n^{2}+n+1$ is greater than 1, so $n-1=2, n^{2}+n+1=p$. From this, $n=3, p=13$.
The... | 11713 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,612 |
4. A paper triangle with sides $34, 30, 8 \sqrt{13}$ was folded along the midlines and formed into a triangular pyramid. Through the opposite edges of the pyramid, parallel planes were drawn (a total of 6 planes). Prove that the parallelepiped formed by the intersection of these planes is a rectangular parallelepiped, ... | Answer: 1224.
Solution: All four faces of the pyramid are equal, so the opposite edges of the pyramid are pairwise equal. On opposite faces of the parallelepiped, different diagonals are equal, as they are the opposite edges of the pyramid (see figure). Thus, in the parallelogram, which is a face of the parallelepiped... | 1224 | Geometry | proof | Yes | Yes | olympiads | false | 17,613 |
5. Twenty people, including $A, B, C$, sit down at a round table at random. What is the probability that at least two of $A, B, C$ sit next to each other? | Answer: $\frac{17}{57}$.
Solution: Let's number the seats around the table clockwise from 1 to 20. We will consider any (unordered) set of three seat numbers where $A, B, C$ are sitting as an outcome, and a favorable outcome as an unordered set of three seat numbers where $A, B, C$ are sitting if the numbers are not c... | \frac{17}{57} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,614 |
Problem 2. The teacher wrote two numbers on the board. Petya divided the first number by the second. Vasya added both numbers and divided the resulting sum by the doubled first number. It turned out that Petya and Vasya got the same result, which is not equal to 1. What is this result? | Answer: -1/2. Solution: Let the first number be $a$, the second number be $b$. According to the condition, $a / b = (a + b) / 2a$. Multiplying both sides of this equation by $2ab$, we get $2a^2 = ab + b^2 \Rightarrow a^2 - b^2 + a^2 - ab = 0 \Rightarrow (a - b)(2a + b) = 0$. Note that $a \neq b$, since according to the... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,615 |
Problem 5. Can 100 lines be chosen in space so that they intersect at exactly 2022 points? | Answer. It is possible. Solution. Note that $43 \cdot 47=2021$. Therefore, if we draw 43 parallel lines in one direction and 47 parallel lines in another direction in some plane $\alpha$, we will already have 2021 intersection points. We will draw one more line so that it intersects the plane $\alpha$ at a point lying ... | itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,616 |
Problem 6. On the board, there are five "equations" of the form $x^{2}+\ldots x+\ldots=0$. Two players take turns filling in the dots with natural numbers from 1 to 10, with each number being used only once. The game ends when all numbers are filled in. The player who makes the first move wants to have as many equation... | Answer. 3. Solution. To obtain three equations, each having two distinct roots, the first player needs to use the three initial moves to write the largest of the numbers not yet written in the place of the coefficient of $x$ in the "equation" where no number has yet been written. This is possible because if no more tha... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,617 |
1. Can a square be cut into ten squares? Provide an example or prove that there is no way to cut it.
Can a square be cut into ten squares? Provide an example or prove that there is no way to cut it. | Solution. It is enough to provide a drawing. Two methods of solution are shown.
 | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,618 |
3. In the store, apples were sold. On the second day, they sold a quarter of the amount of apples sold on the first day, and an additional eight kilograms. On the third day, they sold a quarter of the amount of apples sold on the second day, and an additional eight kilograms. How many kilograms of apples were sold on t... | Solution. Let's reason from the end. On the third day, 18 kg of apples were sold. If we subtract 8 kilograms, the remaining 10 kg will be a quarter of the amount sold on the second day. Therefore, 40 kilograms of apples were sold on the second day. Of these, 32 kilograms are a quarter of the amount sold on the first da... | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,620 |
4. On the table, there are three piles of candies, the first has 10 pieces, the second has 20 pieces, and the third has 30 pieces. Masha and Rita are playing a game - they take turns dividing any pile of candies on the table into two parts. The game ends when one of the girls cannot make a move and thus loses. Masha go... | Solution. Any pile can be divided, including one consisting of two candies. Therefore, the game will end when all candies are divided into piles of one candy each. Previous moves do not affect the next move - as long as there are piles left, one candy can be separated from the others. To divide 60 candies, initially in... | Mashawillwin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,621 |
5. Ten football teams conducted two tournaments over two years. Each year, the results of the tournament were recorded in a table, listing the teams from first to tenth place. A team is considered stable if there is at least one other team that ranks lower than it in both tables. What is the minimum and maximum number ... | Answer: The minimum number is 0 stable teams, the maximum number is 9 stable teams. | 09 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,622 |
6. Write all prime numbers that cannot be represented as the sum of two composite numbers. | Solution. The number 2 cannot be represented as the sum of two composite numbers (the number 1 is neither prime nor composite). From now on, we only consider odd prime numbers.
The difference between any prime number greater than 13 and the number 13 is an even number. Let this difference for a given prime number $p$ ... | 2,3,5,7,11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,623 |
# Task 10.1
Prove that for any prime number $p>2$ the numerator of the fraction
$\frac{m}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{p-3}+\frac{1}{p-2}+\frac{1}{p-1}$ is divisible by $p$.
Number of points 7
Hint
$\frac{1}{k}+\frac{1}{p-k}=\frac{p}{k(p-k)}$ | # Solution
Let's break down the fractions into pairs with denominators $k$ and $p-k$.
The sum of such fractions has a numerator of $p$, and its denominator is not divisible by $p$, since $p$ is a prime number.
Thus, after adding all the fractions, we get a fraction with a numerator that is a multiple of $p$, and a d... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,624 |
# Task 10.2
In the decimal representation of the number ${ }^{1 / 7}$, the 2021st digit after the decimal point was crossed out (and the other digits were not changed). Did the number increase or decrease?
## Number of points 7
# | # Answer:
increased
## Solution
By dividing the numerator by the denominator, we get that $1 / 7 = 0.(142857)$.
This means that the period of the resulting fraction contains 6 digits.
Since the number 2021, when divided by 6, gives a remainder of 5, the 2021st digit after the decimal point in the decimal represent... | increased | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,625 |
# Task 10.3
Two identical gears with 13 teeth each are placed on top of each other so that the teeth align (so the projection on the plane looks like one gear). After this, four pairs of aligned teeth are cut out. Is it always possible to rotate these gears relative to each other so that the projection on the plane lo... | # Answer:
not always.
## Solution
Let's construct a counterexample:
number the teeth in order and remove the teeth numbered $1,2,5,7$.
It is easy to verify that the differences of these numbers give all 13 numbers $0,1,2 \ldots, 12(\bmod 13)$.
Therefore, with each rotation, exactly one pair of "removed teeth" coi... | notalways | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,626 |
# Task 10.4
In the cells of a $4 \times 4$ square table, numbers +1 and -1 are arranged as shown in the figure. It is allowed to simultaneously change the signs of the numbers in all cells located in one row, one column, or on a line parallel to one of the diagonals (in particular, the sign of the number in any corner... | # Solution
Mark 8 cells in the table (see the figure on the right).
Each row and each column contain 2 marked cells, and each diagonal contains 0 or 2. Therefore, the parity of the number of negative numbers in the marked cells does not change.
Thus, according to the condition, the number of negative numbers in thes... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 17,627 |
# Task 10.5
The altitudes AD and $\mathrm{BE}$ of an acute-angled triangle $\mathrm{ABC}$ intersect at point $\mathrm{H}$.
The circumcircle of triangle $\mathrm{ABH}$ intersects sides $\mathrm{AC}$ and $\mathrm{BC}$ at points $\mathrm{F}$ and $\mathrm{G}$, respectively. Find FG, if $\mathrm{DE}=5$ cm.
## Number of p... | # Answer:
$\mathrm{FG}=10 \mathrm{~cm}$
## Solution
We have only one numerical given.
So, the length of FG will either be equal to ED or a multiple of it. It doesn't seem to be equal. It should be larger (judging by the diagram), by some factor. What do we know about such relationships? There is the midline of a tr... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,628 |
1. Cut a $7 \times 9$ rectangle into 7 squares, not necessarily equal to each other. Cuts can only be made along the grid lines.
 | Solution. One of the possible cuts is shown in the figure.
 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,629 |
3. Compare the numbers $x=2 \cdot 20212021 \cdot 1011 \cdot 202320232023$ and $y=43 \cdot 47 \cdot 20232023 \cdot 202220222022$. | Solution. Let's factorize the numbers $x=2022 \cdot 2021 \cdot 10001 \cdot 2023 \cdot 100010001$, $y=2021 \cdot 2023 \cdot 10001 \cdot 2022 \cdot 100010001$. From the factorizations, it is clear that the numbers are equal.
Answer. $x=y$. | y | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,631 |
5. On the swamp, there are 64 bumps, they are arranged in an $8 \times 8$ square. On each bump sits a frog or a toad. Frogs never lie, while toads only tell lies. Each of them, both frogs and toads, croaked: “At least one of the neighboring bumps has a toad.” What is the maximum number of toads that could sit on these ... | Solution. Note that frogs cannot sit on adjacent lily pads, otherwise they would be telling the truth. If we divide the lily pads into pairs, there will be no more than one frog in each pair, otherwise they would be telling the truth. Therefore, there can be no more than 32 frogs, as the lily pads can be divided into 3... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,632 |
6. Vanya thought of a seven-digit number, and then subtracted from it the sum of all the digits except one. He got 9875352. What number would Vanya have gotten if he had subtracted the sum of all the digits except the second from the left? | Solution. Since the number is a seven-digit number, the sum of its digits is no more than 63. Adding 63 to 9875352, we get an upper estimate for the original number: 9875415. We see that in the intended number, after subtracting the sum of all digits, the first four digits do not change (the hundreds place initially ha... | 9875357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,633 |
1. In three boxes, there is one ball each: white, red, and blue. The first box has the inscription "white," the second - "red," and the third - "white or blue." None of the inscriptions are correct. Where is what located? | 1. Answer: blue, white, red.
The third box does not contain a white or blue ball. Therefore, it is red. The first one is not white, nor is it red. Therefore, it is blue. It follows that the second box contains the white ball. | blue,white,red | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,634 |
2. Solve the numerical riddle
$$
\text { BLOW + BLOW = FIGHT }
$$
where different letters represent different digits. | 2. Answer: $8126+8126=16252$.
It is clear that D=1 as the first digit of the five-digit sum. Consider the hundreds place. For it, $\mathrm{A}=D+D=2$, since the digit $\mathrm{A}$ is at the end of the sum, and it is even. Therefore, $\mathrm{P}+\mathrm{P}$ ends in 2, so $\mathrm{P}=6$, as the digit 1 is already occupie... | 8126+8126=16252 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,635 |
3. Find the largest natural number without repeating digits, in which the product of any two consecutive digits is divisible by 6. | 3. Answer: 894326705.
Among two adjacent digits, there is certainly one divisible by 3, and there are only four such digits. Therefore, a 10-digit number is impossible. Let's try to construct a 9-digit number. The digits $0,3,6,9$, which are multiples of three, must be in even positions. Moreover, next to 3 and 9, the... | 894326705 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,636 |
4. How many four-digit numbers are there for which the sum of the digits is a multiple of ten? | 4. Answer: 900.
To any three-digit number, you can append exactly one decimal digit on the right so that the sum of the digits becomes a multiple of 10. For example, to 780, we append 5; to 202, we append 6; to 334, we append 0, and so on. Since three-digit numbers range from 100 to 999 inclusive, there are exactly $9... | 900 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,637 |
5. Cut a $3 \times 9$ rectangle into 8 squares.
7 points are awarded for a complete solution to each problem
The maximum total score is 35 | 5. First, cut the rectangle into three squares of size $3 \times 3$. Leave two of them, and from the third, cut out a square of size $2 \times 2$. Cut the remaining part into 5 squares of size $1 \times 1$. In total, you will have 8 squares. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,638 |
Problem 8.1. The cells of the pyramid are filled according to the following rule: above every two adjacent numbers, their arithmetic mean is written. Some numbers were erased, and the structure shown in the figure was obtained. What number was in the bottom right cell? (The arithmetic mean of two numbers is their sum d... | Answer: 6.
Solution. Let's restore the numbers in the table by going through it from top to bottom. For example, if the numbers 21 and $x$ are in the second row, then from $18=\frac{1}{2}(21+x)$ we get $x=15$. Similarly, in the third row, we get that next to the number 14 is 16, and next to it is -26; in the last row,... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,639 |
Problem 8.3. On the side $B C$ of rectangle $A B C D$, a point $K$ is marked. Point $H$ on segment $A K$ is such that $\angle A H D=90^{\circ}$. It turns out that $A K=B C$. How many degrees does angle $A D H$ measure if $\angle C K D=71^{\circ}$?
Fig. 5: to the solution of problem 8.3
Solution. Since $B C \| A D$, we get $\angle A D K=\angle C K D=71^{\circ}$ (Fig. 5).
Since $A K=B C=A D$, triangle $A K D$ is isosceles a... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,641 |
Problem 8.4. Thirty-six children are standing in a circle, each dressed in a red or blue sweater. It is known that next to each boy stands a girl, and next to each girl stands a person in a blue sweater. Find the maximum possible number of girls in red sweaters.
# | # Answer: 24.
Solution. Note that there will not be 3 girls in red sweaters standing in a row (otherwise, the condition would not be satisfied for the middle one). By dividing 36 children into 12 triplets, we get that in each of them there are no more than 2 girls in red sweaters, and the total number of girls in red ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,642 |
Problem 8.5. A car left the city for the village at the same time a cyclist left the village for the city. When the car and the cyclist met, the car immediately turned around and headed back to the city. As a result, the cyclist arrived in the city 35 minutes after the car. How many minutes did the cyclist spend on the... | Answer: 55.
Fig. 7: to the solution of problem 8.5
Solution. In Fig. 7, let's mark village $A$, city $B$, point $P$ where the car and the cyclist meet, and point $Q$, where the cyclist was ... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,643 |
Problem 8.7. In an acute-angled triangle $A B C$, the altitude $B H$ is drawn. It turns out that $C H=A B+A H$. How many degrees does the angle $B A C$ measure if $\angle A B C=84^{\circ}$ ?
 | Answer: 64.
Fig. 8: to the solution of problem 8.7
Solution. Mark a point \( K \) on the segment \( CH \) such that \( AH = HK \). Then, from the condition, it follows that \( AB = CK \) (Fig... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,644 |
Problem 8.8. On an island, there live knights who always tell the truth, and liars who always lie.
One day, 10 inhabitants of the island gathered, each wearing a T-shirt with a number from 1 to 10 (different inhabitants had different numbers). Each of them said one of the following phrases:
- "Among those gathered, t... | Answer: $1,2,3,4,5,6$.
Solution. Consider the people who said the first phrase. Among them, there is no more than one knight (otherwise, the knight with the smallest number among them would have lied). Thus, the total number of knights is no more than 6.
Also, among all present, there is at least 1 knight (otherwise,... | 1,2,3,4,5,6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,645 |
11.3. Find the roots of the equation: $\left(x^{3}-2\right)\left(2^{\sin x}-1\right)+\left(2^{x^{3}}-4\right) \sin x=0$. | Answer. $\sqrt[3]{2}, \pi n(n \in \mathbf{Z})$.
Solution. From the fact that the function $\mathrm{y}=2^{\mathrm{t}}$ is increasing, it follows:
1) if $\sin x>0$, then $2^{\sin x}-1>0$; if $\sin x0$, then $2^{x^{3}}-4>0$; if $x^{3}-20$, then $\left(2^{x^{3}}-4\right) \cdot \sin x>0$; if $\left(x^{3}-2\right)\left(2^{... | \sqrt[3]{2},\pin(n\in{Z}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,646 |
11.4. Quadrilateral $M N P Q$ is inscribed in a circle and there exists a point $S$ inside it, the distances from which to the lines passing through the sides of the given quadrilateral are proportional to the corresponding sides. Prove that $S$ is the intersection point of the diagonals of $M N P Q$. | # Solution.
First method. The set of points whose distances to the lines $M N$ and $P Q$ are proportional to the corresponding sides is a line passing through the intersection point of $M N$ and $P Q$. Since the quadrilateral $M N P Q$ is cyclic, triangles $L M N$ and $L P Q$, where $L$ is the intersection point of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,647 |
11.5. In 50 baskets, there are cucumbers, eggplants, and tomatoes. Prove that it is possible to choose 26 baskets such that they contain no less than half of all cucumbers (by count), no less than half of all eggplants, and no less than half of all tomatoes (also by count).
# | # Solution.
Lemma. Any $2 n$ pairs of positive numbers ( $a_{i}, b_{i}$ ) can be divided into two groups of $n$ pairs each, such that the sum of $a_{i}$ in the first group differs from the sum of $a_{i}$ in the second group by no more than the maximum $a_{i}$, and the sum of $b_{i}$ in the first group differs from the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,648 |
1. (7 points) On the coordinate plane, the graph of the function $y=a x^{2}+b x+c$ is shown (see figure). On the same coordinate plane, schematically draw the graph of the function $y=c x^{2}+2 b x+a$. Explain your answer.
$; $(1 ;-2)$, and $(-1 ; 2)$. Considering that $y(0)=c, y(1)=a+b+c, y(-1)=a-b+c$, we get the system of equations
$$
\left\{\begin{array}{l}
c=1 \\
a+b+c=-2 \\
a-b+c=2
\end{ar... | (x-2)^{2}-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,649 |
2. (7 points) Prove that if the product of two positive numbers is 1, then the sum of these numbers is not less than 2.
# | # Solution.
Method 1. Consider two numbers $x$ and $\frac{1}{x}$, satisfying the condition, and $x>0$, which means $\frac{1}{x}>0$. Compare their sum with 2, assuming that the sum is not less than 2:
$$
\begin{gathered}
x+\frac{1}{x} \geq 2 \\
\frac{x^{2}-2 x+1}{x} \geq 0 \\
\frac{(x-1)^{2}}{x} \geq 0
\end{gathered}
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,650 |
3. (7 points) In triangle $ABC$, any point $K$ is chosen. Line $AK$ intersects side $BC$ at point $A_{1}$, line $BK$ intersects $AC$ at point $B_{1}$, and line $CK$ intersects $AB$ at point $C_{1}$. Prove that $\frac{AK}{KA_{1}}=\frac{AB_{1}}{B_{1}C}+\frac{AC_{1}}{BC_{1}}$. | # Solution.
 | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,651 |
6. (7 points) Prove that the points of intersection of the lines $x+2 y=19$ and $y+2 x=98$ with the hyperbola $y=\frac{1}{x}$ lie on the same circle.
# | # Solution.
Let $A, B, C$ and $D$ be the points of intersection of the lines with the hyperbola $y=\frac{1}{x}$.
We will rewrite each of the line equations as follows:
$$
x+2 y-19=0, \quad y+2 x-98=0
$$
Multiplying the left and right sides of the resulting equations, we get:
$$
(x+2 y-19)(y+2 x-98)=0
$$
We will t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,652 |
5. (7 points) The king decided to test his hundred sages and announced that the next day he would line them up with their eyes blindfolded and put a black or white hat on each of them. After their eyes are uncovered, each, starting from the last in line, will name the supposed color of their hat. If he fails to guess c... | Solution. Let's describe the strategy that the sages should adhere to. The last in line looks ahead, counts the number of black hats, and says "black" if this number is even. In doing so, he cannot save himself for sure. However, the 99th, 98th, ..., 1st in line receive very important information. Thus, the 99th counts... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,653 |
1. The parabola $y=x^{2}-20 x+c$, where $c \neq 0$, intersects the $O x$ axis at points $A$ and $B$, and the $O y$ axis at point $C$. It is known that points $A$ and $C$ are symmetric with respect to the line $y=-x$. Find the area of triangle $A B C$. | Answer: 231.
Solution. Since $y(0)=c$, we have $C(0, c)$ and $A(-c, 0)$. Therefore, one of the roots of the equation $x^{2}-20 x+c=0$ is $-c$, i.e., $c^{2}+20 c+c=0$, from which $c=-21, x_{1}=21$. By Vieta's theorem, $x_{1}+x_{2}=20$. Thus, $x_{2}=-1$. The length of the base of the triangle is $x_{1}-x_{2}=22$, and th... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,654 |
2. Find all pairs of natural numbers $n$ and $m$ for which
$$
(n+1)(2n+1)=18m^2
$$ | Answer: there are none.
Solution. Since $18 m^{2}=(n+1)(2 n+1)>2 n^{2}, 9 m^{2}>n^{2}$ and $n<3 m-1$. It follows that $3 m-1<n<3 m$. This is impossible.
Remark. Other solutions are possible. For example, if the equation is considered as a quadratic in $n$, then the discriminant $144 m^{2}+1$ is not a square for natur... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,655 |
8.1. Three natural numbers are written on the board. Tanya calculated the product of the numbers written on the board and got a number ending in 1010. And Vanya calculated their sum and got the number 9999. Did Vanya make a mistake? | Answer: Mistaken.
Solution: If the sum of three integers is 9999, then either they are all odd (in which case their product ends in an odd digit), or two of them are even and one is odd (in which case their product is divisible by 4, while a number ending in 10 is not divisible by 4).
# Criteria:
7 points - complet... | Mistaken | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,656 |
8.2. Around a table sit 7 natives: knights, liars, and bores (there is at least one representative of each tribe at the table). A knight always tells the truth, a liar always lies. A bore lies if at least one knight sits next to them, and in other cases says whatever they want.
How many bores could there be at the tab... | Answer: 2, 3, or 4.
## Solution:
One bore could not sit at the table alone, because there must be a knight at the table who tells the truth, which means there are at least two bores.
Since representatives of all tribes are sitting at the table, there are no more than five bores.
Let's try to seat 5 bores. The remai... | 2,3,or4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,657 |
8.3. It is known that both КРУГ and КУБ are cubes of some numbers. Find all possible values of the numbers КРУГ and КУБ. List them all and explain why there are no others. (Identical digits are replaced by the same letters, different digits - by different letters.) | Answer: 1728 and 125.
Solution: Notice that in the words КУБ (CUBE) and КРУГ (CIRCLE), two digits repeat - К and У.
Let's list all possible three-digit numbers that are cubes of some numbers:
$5^{3}=125, 6^{3}=216, 7^{3}=343, 8^{3}=512, 9^{3}=729$ (4^3=64 - two-digit; 10^3=1000 - four-digit). Notice that 343 does no... | 1728125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,658 |
8.4. In quadrilateral $A B C D$, where $\angle A = \angle D$, the perpendicular bisectors of sides $A B$ and $C D$ were drawn. It turned out that they intersect on side $A D$. Prove that the diagonals of quadrilateral $A B C D$ are equal. | # Solution:
Let E be the point of intersection of the perpendicular bisectors. Then $\mathrm{AE}=\mathrm{EB}, \mathrm{CE}=\mathrm{ED}$ and $\angle \mathrm{EBA}=\angle \mathrm{EAB}=\angle \mathrm{ECD}=\angle \mathrm{EDC}$. Therefore, $\angle \mathrm{AEB}=$ $\angle \mathrm{CED}$, which means $\angle \mathrm{AEC}=\angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,659 |
8.5. A group of children, after the municipal stage of the mathematics competition, formed a circle, discussing the solved problems. It turned out that there are exactly 20 future prizewinners and exactly 25 future winners of the municipal stage such that each of them has at least one neighbor who is a participant who ... | Solution. If among the children standing in a circle, several participants stand in a row, none of whom are winners or prizewinners (we will simply call them participants), we will expel all of them except one. Obviously, this will not affect the conditions of the problem. Now, each participant has only prizewinners or... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,660 |
9.1. Find $x^{2}+y^{2}+z^{2}$, if $x+y+z=2, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$. | # 9.1. Answer: 4.
## 9th grade
By getting rid of the denominators in the second equation, we get $x y + y z + z x = 0$. Squaring the first equation, we get $x^{2} + y^{2} + z^{2} + 2(x y + y z + z x) = 4$. From this, we obtain the answer. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,661 |
9.2. Prove that all parabolas, which are graphs of functions of the form $y=x^{2}+p x+q$, where $p+q=2019$, intersect at one point. | 9.2. $y(1)=1+p+q=2020$. This means, the graph of each function passes through the point $(1 ; 2020)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,662 |
9.4. Given 9 different natural numbers, not greater than 19. Prove that among their pairwise differences, there will be three that are the same in absolute value. | 9.4. The number of pairwise differences is $9 \cdot 8 / 2=36$ (each of the nine numbers forms a difference with eight others, and each difference is counted twice). The maximum absolute value of the difference is 18, the minimum is -1, and there are 18 different values. However, the difference of 18 can be obtained in ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,664 |
9.5. In rectangle $A B C D$, vertex $C$ is connected to the midpoint $K$ of side $A D$. It turns out that $C K \perp B D$. Let $H$ be the point of intersection of $B D$ and $C K$. Prove that triangle $A H B$ is isosceles. | 9.5. Angles $A B D$ and $B C K$ are equal as acute angles with mutually perpendicular sides. Points $A$ and $H$ lie on the circle with diameter $B K$, so $\square A H B$ $=\square A K B$. From the equality of triangles $C K D$ and $B K A$ it follows that $\angle$ $A K B=\square C K D$, and $\angle C K D=\square B C K$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,665 |
9.1. Petya had a large wooden cube with dimensions $30 \times 30 \times 30$ cm $^{3}$. Petya decided to paint the entire cube, which took 100 grams of paint. Later, Petya needed smaller cubes, so he cut the large cube with 6 cuts, parallel to the faces of the cube (2 cuts parallel to each pair of faces), into 27 cubes ... | Answer: 200 grams.
First solution. Each cut increases the unpainted surface area by an amount equal to twice the area of the cut. It is clear that the area of the cut is the same as the area of a face. Therefore, it is necessary to paint an additional surface area equal to \(6 \cdot 2 = 12\) faces, which is twice the ... | 200 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,666 |
9.2. For positive real numbers $p$ and $q$, it is known that each of the equations $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0$ has two distinct real roots. Prove that the equation $x^{2} + (p+q) x + 2(p+q) = 0$ also has two distinct real roots. | Solution. From the condition, it follows that $p^{2}>4 q$ and $q^{2}>4 p$. From this, we get $p^{4}>16 q^{2}>64 p, p^{3}>64, p>4$. Similarly, it can be proven that $q>4$. The discriminant $D$ of the equation $x^{2}+(p+q) x+2(p+q)=0$ is $(p+q)^{2}-8(p+q)=(p+q)(p+q-8)$. Since $p+q>0$ and $p+q-8=(p-4)+(q-4)>0$, $D$ is the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,667 |
9.3. The incircle $\omega$ of triangle $ABC$ touches sides $BC$ and $CA$ at points $A_1$ and $B_1$, respectively. On the ray $AB$ beyond point $B$, a point $A_2$ is taken such that $BA_1 = BA_2$. On the ray $BA$ beyond point $A$, a point $B_2$ is taken such that $AB_1 = AB_2$. Prove that the intersection point of lines... | Solution. Let $\omega$ touch side $AB$ at point $C_{1}$, and the line $A_{1} A_{2}$ intersects $\omega$ again at point $N$. Since $B C_{1} = B A_{1}$ (segments of tangents drawn from the same point), then $B C_{1} = B A_{1} = B A_{2}$. If $\angle A_{1} B C_{1} = 2x$ and $\angle A_{1} B A_{2} = 2y$, then $2x + 2y = 180^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,668 |
9.4. For four consecutive natural numbers, it is known that the largest of them is a divisor of the product of the other three. Find all values that the largest of these numbers can take. | Answer: 6.
First solution. Let our numbers be $n-3, n-2, n-1, n$, where $n-3 \geqslant 1, n \geqslant 4$. Since the numbers $n-1$ and $n$ are coprime, $(n-3)(n-2)$ is divisible by $n$. Note that $(n-3)(n-2) = n^2 - 5n + 6 = n(n-5) + 6$, and since both numbers $(n-3)(n-2)$ and $n(n-5)$ are divisible by $n$, 6 is also d... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,669 |
9.5. Consider the set of all natural numbers whose representation does not contain digits greater than 1. Is it possible to color each number in this set either blue or red so that the sum of any two different numbers of the same color contains at least two ones in its representation? | Answer. Yes.
Solution. It is clear that the representation of our numbers contains no digits other than 1 and 0. We will paint all numbers with an odd number of 1s in blue, and all numbers with an even number of 1s in red. Consider two different numbers of the same color. If one is shorter than the other, we will pad ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,670 |
10.1. If you divide 2014 by 105, the quotient is 19 and the remainder is also 19. What other natural numbers can 2014 be divided by so that the quotient and the remainder are the same? (List all possible options and prove that there are no others.) | Answer: also by 52, by 1006, and by 2013.
Solution. If 2014 is divided by a natural number $N$ and the quotient and the remainder are both natural numbers $k$, then $2014=N k+k=k(N+1)$, where $k<N$. Therefore, $k$ is a divisor of the number 2014. From the example given in the problem, it follows that $2014=105 \cdot 1... | 52,1006,2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,671 |
10.2. Prove that if the expression $\left(x^{2}-x+1\right)^{2014}$ is expanded and like terms are combined, then some coefficient of the resulting polynomial will be negative. | Solution. First method. We will prove that the coefficient of $x$ in the obtained polynomial will be negative. Similar terms with the variable part $x$ are formed when multiplying 2014 identical brackets in the following way: in one of the brackets, the term $(-x)$ is taken, and in the remaining brackets, the term 1. T... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,672 |
10.3. In space (but not in the same plane), six different points are located: $A, B, C, D$, $E$ and $F$. It is known that the segments $A B$ and $D E$, $B C$ and $E F$, $C D$ and $F A$ are pairwise parallel. Prove that these segments are also pairwise equal. | Solution. First method. In the plane $ABC$, there is a pair of intersecting lines $AB$ and $BC$, which are respectively parallel to the lines $DE$ and $EF$ in the plane $DEF$. Therefore, the planes $ABC$ and $DEF$ are parallel (by the criterion of parallel planes), and $CD$ and $AF$ are segments of parallel lines enclo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,673 |
10.4. Every day, from Monday to Friday, the old man went to the blue sea and cast his net into the water. Each day, the net caught no more fish than the previous day. In total, over the five days, the old man caught exactly 100 fish. What is the smallest total number of fish he could have caught on the three days - Mon... | Answer: 50.
Solution. If the old man caught 25 fish each of the first four days and caught nothing on Friday, the conditions of the problem are met, and exactly 50 fish were caught over the specified three days.
Let's prove that in the specified days, fewer than 50 fish could not have been caught. Indeed, suppose few... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,674 |
10.5. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$. | Answer: $45^{\circ}$.
Solution. Let $H$ be the orthocenter (the point of intersection of the altitudes) of triangle $ABC$. Then the altitude $AT$ of triangle $ABC$ contains the median of triangle $AMN$, that is, it intersects segment $MN$ at its midpoint - point $E$ (see Fig. 10.5a, b). We can reason in different ways... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,675 |
10.6. In one of the vertices of a hexagon, there is a gold coin, and there is nothing in the others. Koschei the Deathless is pining over the gold and every morning takes an arbitrary number of coins from one vertex, after which he immediately places six times more coins in the adjacent vertex. If by the end of some da... | Solution. Let's number the vertices of the hexagon starting from the one where the coin lies, with consecutive natural numbers from 1 to 6 (moving, for example, counterclockwise). Denote by \( n_{1}, n_{2}, \ldots, n_{6} \) the number of coins lying in vertices \( 1, 2, \ldots, 6 \) respectively. Let \( N_{1} = n_{1} +... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,676 |
1. In the drawer, there are 23 socks: 8 white and 15 black. Every minute, Marina approaches the drawer and pulls out a sock. If at any moment Marina pulls out more black socks than white ones, she exclaims: "Finally!" - and ends the process.
What is the maximum number of socks Marina can pull out before she exclaims: ... | Answer: 17.
Solution: If Marina takes out 17 socks, then among them there will be no more than 8 white ones, which means there will be at least 9 black ones. Therefore, at this moment, she will definitely exclaim, "Finally!"
On the other hand, if she takes out only 16 socks, she might be unlucky: if she sequentially ... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,677 |
2. Ingots of highly enriched uranium must be stored in such a way that the distance to the nearest ingot is as large as possible. If there are several suitable places, any one of them can be chosen. Unfortunately, no one knows how many ingots will need to be stored. Ingots arrive one by one. Once an ingot is placed in ... | Answer: $12,34,56$ or 78.
Solution. In this problem, it is enough to read the condition and perform the actions described in the condition.
The third ingot will be placed in the middle between the first and second: i.e., in cell $(1+89) / 2=45$. Where can the fourth one be placed? Currently, the ingots are in the 1st... | 12,34,56,78 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,678 |
3. A natural number $n$ is such that $n+2$ is divisible by 3, and $n+3$ is divisible by 4. Which of the following statements are guaranteed to be true, i.e., hold for all $n$ that meet the condition?
a) $n+4$ is divisible by 5;
b) $n+5$ is divisible by 6;
c) $n+6$ is divisible by 7;
d) $n+7$ is divisible by 8. | Answer. Only b).
Solution. First, note that b) works: $n+5=(n+2)+3-$ is divisible by $3, n+5=(n+3)+2-$ is divisible by 2. Since the number is divisible by 2 and by 3, it is also divisible by 6.
Now let's prove that none of the other options work. For this, in each case, we need to provide an example of $n$ that meets... | b) | Number Theory | MCQ | Yes | Yes | olympiads | false | 17,679 |
4. On the circumscribed circle of square $A B C D$, a point $M$ is marked such that $\angle M A B=18^{\circ}$. Segment $M D$ intersects diagonal $A C$ at point $Q$.
What values can the measure of angle $\angle A Q D$ take? Express your answer in degrees. List all possible answer options. | Answer: 72 or 108.
Solution. The first thing to remember in this problem is that for point $M$, there are two possible positions: in the same half-plane relative to line $AB$ as the square (point $M_{1}$ in the diagram), and in the other half-plane (point $M_{2}$ in the diagram).
-\left(... | \frac{\sin3}{\sin4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 17,688 |
2. Is it true that in three-dimensional space, two angles with respectively perpendicular sides are either equal or sum up to $180^{\circ}$? | Solution. Consider, for example, the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ (see figure). The angles $A B C$ and $B C^{\{\prime\}} C$ satisfy the condition, since $A B \perp B C^{\{\prime\}}$ (by the theorem of three perpendiculars) and $B C \perp C^{\{\prime\}} C$. In this case, the angle $A B C$ i... | incorrect | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,689 |
3. In the cells of a $10 \times 10$ square table, non-zero digits are placed. In each row and each column, a ten-digit number is formed arbitrarily from the digits present there. Can it happen that out of the twenty resulting numbers, exactly one is not divisible by 3? | Solution. Suppose there is exactly one number that is not divisible by 3, and it is formed by the digits of one of the columns. Then the sum of the digits of this number is also not divisible by 3. In all other columns, the numbers are divisible by 3, and the sums of their digits are divisible by 3. Therefore, the sum ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,690 |
4. Can three different numbers of the form $2^{n}+1$, where $n$ is a natural number, be consecutive terms of a geometric progression | Solution. Suppose there exist such distinct natural numbers $k, m$, and $n$ that $2^{k}+1, 2^{m}+1$, and $2^{n}+1$ are consecutive terms of some geometric progression. Then $\left(2^{m}+1\right)^{2}=\left(2^{k}+1\right)\left(2^{n}+1\right)$, which means $2^{2 m}+2^{m+1}=2^{k+n}+2^{k}+2^{n}$. However, this equality is i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,691 |
5. The value of the parameter $a$ is chosen so that the number of roots of the first equation $4^{x}-4^{-x}=2 \cos a x, 4^{x}+4^{-x}=2 \cos a x+4$ is 2021. How many roots does the second equation have for the same $a$? | Solution. Transform the second equation:
$\left.4^{x}+4^{-x}=2 \cos a x+4 \Leftrightarrow 4^{x}-2+4^{-x}=2(1+\cos a x) \Leftrightarrow\left(2^{x}-2^{-x}\right)^{2}=4 \cos ^{2 a x} / 2\right) \Leftrightarrow$
$$
\left[\begin{array} { l }
{ 4 ^ { x / 2 } - 4 ^ { - x / 2 } = 2 \operatorname { cos } \frac { a x } { 2 } ... | 4042 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,692 |
11.1. Four numbers whose sum is zero are placed on a circle. For each pair of adjacent numbers, their product was found and the four resulting numbers were summed. Can the resulting sum be positive? | Answer: No, it cannot.
Solution 1. Let $a, b, c, d$ be the recorded numbers. $a+b+c+d=0$. Then $a b+b c+c d+d a=a b+b c+c(-a-b-c)+a(-a-b-c)=a b+b c-a c-b c-c^{2}-a^{2}-a b-a c=-(a+c)^{2} \leq 0$.
Solution 2. Let $a, b, c, d$ be the recorded numbers. $a+b+c+d=0$. Then $a b+b c+c d+d a=a(b+d)+c(d+b)=(a+c)(d+b)=-(a+c)^{... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,698 |
11.2. For the angles of triangle $ABC$, it is known that $\sin \angle A + \cos \angle B = \sqrt{2}$ and $\cos \angle A + \sin \angle B = \sqrt{2}$. Find the measure of angle $C$. | Answer: $90^{\circ}$.
Solution. Square both equalities and add them. Then use the fundamental trigonometric identity, and after transformations, we get $2 \sin \angle A \cos \angle \mathrm{B}+2 \sin \angle B \cos \angle \mathrm{A}=2 \sin (\angle A+\angle \mathrm{B})=2 \quad, \quad$ that is, $\quad \sin (\angle A+\angl... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,699 |
11.3. Find all natural numbers $n$ such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number less than 2219. | Answer: $n=1, n=2^{8}$ and $n=3^{8}$.
Solution. Transform the expression $\sqrt{n \sqrt{n \sqrt{n}}}=\left(n\left(n(n)^{\frac{1}{2}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(n\left(n^{\frac{3}{2}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=$ $=\left(n(n)^{\frac{3}{4}}\right)^{\frac{1}{2}}=\left(n^{\frac{7}{4}}\r... | n=1,n=2^{8},n=3^{8} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,700 |
11.4. Fill the cells of an $8 \times 8$ table with integers such that the sum of all numbers in the table is positive, while the sum of the numbers in any $3 \times 3$ square is negative. | Answer: For example,
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | -9 | 1 | 1 | -9 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | -9 | 1 | 1 | -9 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,701 |
11.5. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, two diagonals of the lateral faces $A_{1} C_{1}$ and $C_{1} D$ are drawn. Find the angle between them, given that the diagonal $A_{1} C_{1}$ is equal in length to one of the edges of the parallelepiped, and the diagonal $C_{1} D$ forms an angle of... | Answer: $\arccos \frac{\sqrt{3}}{6}$.
Solution. It is clear that the diagonal $A_{1} C_{1}$ can only be equal to the edges $A A_{1}=B B_{1}=$ $C C_{1}=D D_{1}$, because otherwise the leg of the right triangle $A_{1} D_{1} C_{1}$ would be equal to its hypotenuse. Let $D_{1} C_{1}=a$. Then, in the triangle $D D_{1} C_{1... | \arccos\frac{\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,702 |
10.1 A weirdo didn't mind and recorded 2022 numbers in a circle, and it turned out that each number coincides with the absolute difference of its two neighbors. Determine which numbers were recorded if their sum is 2022. | Solution 1: Note that all the written numbers are non-negative. Let the largest of them be $N$. The difference between two numbers from 0 to $N$ can only be $N$ when these numbers are 0 and $N$, so on one side of the number $N$ stands 0, and on the other side - also $N$. If on one side of zero stands the number $N$, th... | \frac{3}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,703 |
10.2 About numbers a and b it is known that the system of equations
$$
\left\{\begin{array}{l}
y=x^{2}+a x+b \\
x=y^{2}+a y+b
\end{array}\right.
$$
has no solutions. Prove that $b>0$. | Solution 1: The graphs of the equations $x=y^{2}+a y+b$ and $y=x^{2}+a x+b$ are symmetric with respect to the line $y=x$. If the graph of $y=x^{2}+a x+b$ intersects the axis of symmetry, then the second graph also passes through the point of intersection, and the system has a solution. Since for large $x$ the graph lie... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,704 |
10.3 On the board, an acute-angled $\triangle A B C$ was drawn, and the feet of the altitudes $A_{1}, B_{1}$ and $C_{1}$ were marked. Then the entire drawing was erased, except for the points $A_{1}, B_{1}$ and $C_{1}$. Is it possible to reconstruct the original $\triangle A B C$ using a compass and a ruler? | Solution: Angles $\angle A A_{1} B$ and $\angle B B_{1} A$ are equal, so the quadrilateral $A B_{1} A_{1} B$ is cyclic, $\angle B_{1} A_{1} C=\angle A$. Similarly, $\angle C_{1} A_{1} B=\angle A$. Therefore, the sides of triangle $A B C$ are the external angle bisectors of triangle $A_{1} B_{1} C_{1}$. This constructio... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,705 |
10.4 The Oddball marked the centers of 17 cells in an $N \times N$ grid such that the distance between any two marked points is greater than 2. What is the smallest value that $N$ can take? | Solution: We will show that in an $8 \times 8$ square (and then in any smaller size), it is impossible to mark the cells in this way. Indeed, let's divide the square into $2 \times 2$ squares. In each of them, the pairwise distances between the centers of the cells do not exceed $\sqrt{2}$, so no more than one cell is ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,706 |
8.1. From 1812 identical squares with a side of 1 mm, a rectangular frame for a group photograph was made (see figure, the boundaries of the photograph coincide with the inner boundaries of the frame). Then the photograph was cut along the lines of the millimeter grid into two rectangular parts. Now two frames were nee... | Answer: $101 \times 803$ (mm).
Solution. Without loss of generality, we can assume that the cut was made vertically. After the cut, two vertical sides of the frame, including the corner squares, were added, which used $2018-1812=206$ squares. Therefore, each side used $206: 2 = 103$ squares, and the same number of squ... | 101\times803 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,707 |
8.2. Four different natural numbers are recorded. It turns out that the sum of their reciprocals is equal to 1. Can the number $2$ be absent among the recorded numbers? | Answer: cannot.
Solution. If at least one of the written numbers is 1, then the reciprocal of it is also 1, then the sum of the reciprocals of the numbers is greater than 1. Therefore, among the written numbers there is no 1. Suppose that among them there is no number 2. Consider the four smallest natural numbers from... | cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,708 |
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