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11.2. Given 2011 non-zero integers. It is known that the sum of any one of them with the product of the remaining 2010 numbers is negative. Prove that if all the given numbers are arbitrarily divided into two groups and the numbers in the groups are multiplied, then the sum of the two resulting products will also be ne... | Solution. Suppose that among the given numbers there is an even number of negative ones. Then among them there is a positive number $a$, and the product of all numbers except $a$ is positive. This contradicts the condition.
Therefore, among the given numbers there is an odd number of negative ones. Let $x_{1}, x_{2}, ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,518 |
11.3. On the circle circumscribed around rectangle $A B C D$, a point $K$ is chosen. It turns out that the line $C K$ intersects the segment $A D$ at a point $M$ such that $A M: M D=2$. Let $O$ be the center of the rectangle. Prove that the point of intersection of the medians of triangle $O K D$ lies on the circle cir... | Solution. Mark a point $T$ on the extension of segment $A D$ such that $A T = D M$. Then the right triangles $C D M$ and $B A T$ are equal, which means $B T \parallel C M$. Note that $D T = D A + A T = 3 D M + D M = 4 D M$. By Thales' theorem, the line $C M$ intersects segment $B D$ at a point $N$ such that $D B = 4 D ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,519 |
11.4. 2011 Warehouses are connected by roads in such a way that from any warehouse you can drive to any other, possibly by driving along several roads. On the warehouses, there are $x_{1}, \ldots, x_{2011}$ kg of cement respectively. In one trip, you can transport an arbitrary amount of cement from any warehouse to ano... | Answer. In 2010 trips.
Solution. First, we show that the plan cannot always be completed in 2009 trips. Suppose (with any road scheme) that initially all the cement is located at one warehouse $S$, and it needs to be evenly distributed to all warehouses. Then, cement must be delivered to each warehouse, except $S$, in... | 2010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,520 |
10.1. Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits of $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? Find all p... | Answer. The digit 6.
Solution. By the condition, $M=3 N$, so the number $A=M-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6.
We will show that $N$ cannot end in 1. If $N$ e... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,521 |
10.2. In an acute-angled triangle $A B C$, the altitudes $A A_{1}$ and $C C_{1}$ are drawn. The circumcircle $\Omega$ of triangle $A B C$ intersects the line $A_{1} C_{1}$ at points $A^{\prime}$ and $C^{\prime}$. The tangents to $\Omega$ at points $A^{\prime}$ and $C^{\prime}$ intersect at point $B^{\prime}$. Prove tha... | Solution. Since $\angle A A_{1} C = \angle A C_{1} C = 90^{\circ}$, the points $A, C_{1}, A_{1}$, and $C$ lie on a circle with diameter $A C$, hence $\angle B A_{1} C_{1} = 180^{\circ} - \angle C A_{1} C_{1} = \angle B A C$ (see Fig. 2). Then $\angle B A_{1} C_{1} = \angle B A_{1} C^{\prime} = \frac{1}{2}(\overline{B C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,522 |
10.3. Given three quadratic trinomials $P(x), Q(x)$, and $R(x)$ with positive leading coefficients, each having two distinct roots. It turns out that substituting the roots of the trinomial $R(x)$ into the polynomial $P(x)+Q(x)$ yields equal values. Similarly, substituting the roots of the trinomial $P(x)$ into the pol... | The first solution. Let $a_{1}$ and $a_{2}, b_{1}$ and $b_{2}, c_{1}$ and $c_{2}$ be the pairs of roots of the quadratic polynomials $P(x), Q(x)$, and $R(x)$, respectively. Consider the quadratic polynomial $S(x)=P(x)+Q(x)+R(x)$. Its values at points $c_{1}$ and $c_{2}$ coincide with the values at these points of the p... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,523 |
10.4. Can the set of all natural numbers be partitioned into non-intersecting finite subsets $A_{1}, A_{2}, A_{3}, \ldots$ such that for any natural number $k$, the sum of all numbers in the subset $A_{k}$ equals $k+2013$?
(R. Zhenodarov) | Answer: No.
First solution. Suppose the desired partition exists. Let's call a set $A_{k}$ large if it contains more than one element. We will prove that for any $n$ there are $n$ large sets, by induction on $n$. For $n=1$, consider the set $A_{k_{1}}$ containing the number one; the sum of the numbers in it is $k_{1}+... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,524 |
4. $\quad$ Given $\mathrm{a}^{2}<4 \mathrm{~b}, \mathrm{c}^{2}<4 \mathrm{~d}$. We will show that then $\left(\frac{\mathrm{a}+\mathrm{c}}{2}\right)^{2}<4\left(\frac{b+d}{2}\right)$. We have $\left(\frac{a+c}{2}\right)^{2}=\frac{1}{4}\left(a^{2}+2 a c+c^{2}\right) \leq \frac{1}{4}\left(a^{2}+\left(a^{2}+c^{2}\right)+c^{... | Answer: This is a correct statement | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,526 |
10.1. A first-grader made two triangles from six sticks. Then he disassembled the triangles and divided the six sticks into two groups of three sticks: in the first group, there were the three longest sticks, and in the second group, the three shortest sticks. Is it necessarily possible to form a triangle from the thre... | Answer. 1) Yes, necessarily. 2) No, not necessarily.
Solution. 1) Let $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant a_{4} \geqslant a_{5} \geqslant a_{6}$ be the given lengths of the sticks. Since $a_{1}$ was part of a triangle with some two other sticks, $a_{1}$ is less than their sum, and therefore less than the ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,527 |
10.2. Non-zero numbers $x$ and $y$ satisfy the inequalities $x^{4}-y^{4}>x$ and $y^{4}-x^{4}>y$. Can the product $xy$ be a negative number?
(N. Agakhanov) | Answer: It cannot.
First solution. We will prove that $xy > 0$. Assume the opposite: $xy \leq 0$. Then, without loss of generality, we can assume $x \leq 0$ and $y \geq 0$. Consider the inequalities $x^4 - x > y^4$ and $y^4 - y > x^4$. We multiply these (this is permissible because their right-hand sides are positive)... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 12,528 |
10.4. The incircle touches the sides $AB$, $BC$, and $CA$ of the non-isosceles triangle $ABC$ at points $C_1$, $A_1$, and $B_1$ respectively. Let $m$ be the midline of triangle $A_1B_1C_1$ parallel to side $B_1C_1$. The angle bisector of $\angle B_1A_1C_1$ intersects $m$ at point $K$. Prove that the circumcircle of tri... | Solution. Let $m$ intersect $B C$ at point $X$, and $P$ and $Q$ be the midpoints of $A_{1} B_{1}$ and $A_{1} C_{1}$, respectively. For definiteness, let $X$ and $C$ lie on the same side of $A_{1}$. To solve the problem, it is sufficient to prove that $X B \cdot X C = X K^{2}$.
 .
$$
For the equation to have an integer root, the discriminant must be a perfect square. However, the number $2^{2021}\left(2^{2015}-1\right)$ is not a perfect square, since the power of 2 i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,532 |
Problem 2. Pentagon $A B C D E$ is inscribed in circle $\omega$. Diagonal $A C$ is the diameter of circle $\omega$. Find $\angle B E C$, if $\angle A D B=20^{\circ}$. | Answer: $70^{\circ}$.
Solution. Fig. 4. Since $\angle A D B=20^{\circ}$, the arc $A B$ is $40^{\circ}$. Since $A C$ is a diameter, the arc $A B C$ is $180^{\circ}$, so the arc $B C$ is $180^{\circ}-40^{\circ}=140^{\circ}$. The angle $B E C$ subtends the arc $B C$, which means it is equal to $140^{\circ} / 2=70^{\circ}... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,533 |
Problem 4. The number zero is written on the board. Peter is allowed to perform the following operations:
- apply a trigonometric (sin, $\cos$, $\operatorname{tg}$, or ctg) or inverse trigonometric (arcsin, arccos, $\operatorname{arctg}$, or arcctg) function to one of the numbers written on the board and write the res... | Solution. Petr can, for example, perform the following calculations:
- $\cos 0=1$
- $\arctan 1=\frac{\pi}{4}$
- $\arccos 0=\frac{\pi}{2} ;$
- $\frac{\pi}{4}: \frac{\pi}{2}=\frac{1}{2}$
- $\arccos \frac{1}{2}=\frac{\pi}{3}$
- $\operatorname{tg} \frac{\pi}{3}=\sqrt{3}$.
Remark. Many sequences of operations can lead to ... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,535 |
Problem 5. On the edge $A A^{\prime}$ of the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge length 2, a point $K$ is marked. In space, a point $T$ is marked such that $T B=\sqrt{11}$ and $T C=\sqrt{15}$. Find the length of the height of the tetrahedron $T B C K$, dropped from vertex $C$. | Answer: 2.
Fig. 5: to problem 5
Solution. Notice that
$$
T B^{2}+B C^{2}=11+4=15=T C^{2}
$$
From this, by the converse of the Pythagorean theorem, it follows that angle $T B C$ is a right... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,536 |
Problem 6. Inside the magician's hat, there live 100 rabbits: white, blue, and green. It is known that if 81 rabbits are randomly pulled out of the hat, there will definitely be three of different colors among them. What is the minimum number of rabbits that need to be taken out of the hat to ensure that there are defi... | # Answer: 61.
Solution. We will prove that if 61 rabbits are randomly pulled out of the hat, then among them there will be two of different colors. Suppose the opposite: let there be $a \geqslant 61$ rabbits of some color (for example, white). Let the second color by number of rabbits be blue. Then there are at least ... | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,537 |
1. Find any four different natural numbers that have the following property: if the product of any two of them is added to the product of the other two numbers, the result is a prime number. | 1. For example, the numbers $1,2,3$ and 5 will work. Indeed, the values of all three expressions $1 \cdot 2+3 \cdot 5=17,1 \cdot 3+2 \cdot 5=13$ and $1 \cdot 5+2 \cdot 3=11$ are prime numbers. | 1,2,3,5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,538 |
2. In a convex quadrilateral $A B C D$, point $K$ is the midpoint of $A B$, point $L$ is the midpoint of $B C$, point $M$ is the midpoint of $C D$, and point $N$ is the midpoint of $D A$. For some point $S$ lying inside the quadrilateral $A B C D$, it turns out that $K S=L S$ and $N S=M S$. Prove that $\angle K S N=\an... | 2. Note that the segment $K N$ is the midline of triangle $B A D$. Therefore, $K N = \frac{B D}{2}$. Similarly, from triangle $B C D$ we get $L M = \frac{B D}{2}$. Thus, triangles $K S N$ and $M S L$ are equal by three sides (see the figure). From this, the required equality of the corresponding angles follows.
. ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,543 |
4. A square piece of paper $A B C D$ with side 1 is folded along a line so that vertex $A$ coincides with the midpoint of side $C D$. What is the area of the resulting hexagon? | Answer: $61 / 96$.
Solution. The diagram is as follows:
Let $E$ be the midpoint of side $CD$, $Q$ the midpoint of $AE$, $FQ \perp AE$, $R$ the intersection point of line $FQ$ with side $BC$... | \frac{61}{96} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,544 |
5. In an $n \times n$ square, there are 1014 dominoes (each covering two adjacent cells). No two dominoes share any points (even corner points). For what smallest $n$ is this possible? | Answer. For $n=77$.
Solution. Attach four cells to the right and below each domino so that they form a $2 \times 3$ rectangle (if the domino is vertical) or a $3 \times 2$ rectangle (if it is horizontal). If the rectangles of two dominoes have at least one common cell, then the dominoes have a common point. Therefore,... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,545 |
1. (7 points) Percival's castle had a square shape. One day, Percival decided to expand his domain and added a square extension to the castle. As a result, the perimeter of the castle increased by $10 \%$. By what percentage did the area of the castle increase? | Answer: $4 \%$.
Solution. Let the width of the castle be $a$, and the width of the extension be $b$. Then the original perimeter is $4 a$, and the final perimeter is $4 a+2 b$. Therefore:
$$
1.1 \cdot 4 a=4 a+2 b \Leftrightarrow b=0.2 a
$$
From this, the area of the castle becomes $a^{2}+(0.2 a)^{2}=1.04 a^{2}$, whi... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,546 |
2. (7 points) It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)$ take?
## Answer: 0. | Solution. Note that the equality $a^{2}+b=b^{2}+c$ can be written as: $a^{2}-b^{2}=c-b$. Similarly, we have $b^{2}-c^{2}=a-c, c^{2}-a^{2}=b-a$. Substituting these equalities into the desired expressions, we get that
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,547 |
3. (7 points) On the board, the numbers from 1 to 2017 are written in some arbitrary order. Two numbers can be swapped if one of them is divisible by the other. Prove that after several such operations, the numbers can be arranged in ascending order. | Solution. We will show how to place the number $k \neq 1$ in the $k$-th position. Suppose the number $n$ is in the $k$-th position. First, we swap $n$ with 1, then we swap $k$ with 1. Thus, $k$ will indeed be in its correct position.
By sequentially placing the numbers $2017, 2016, \ldots$ in their correct positions, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,548 |
4. (7 points) Compare the magnitudes of angles $B A C$ and $C E D$ (see the figure). Justify your answer.
 | Answer: these angles are equal.
Solution. Let $K$ be the foot of the perpendicular dropped from $B$ to $AC$.
Consider triangles $ABK$ and $EDC$. Both are right triangles, and their legs are... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,549 |
5. (7 points) Lёsha did not hesitate to calculate the sum
$$
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017}
$$
and wrote it on the board. How many times is the digit 1 written in the final result? | Answer: 2013.
Solution. Transform the expression:
$$
\begin{aligned}
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017} & =(10-1)+(100-1)+\ldots+\left(10^{2017}-1\right)= \\
& =\underbrace{1 \ldots 10}_{2017}-2017=\underbrace{1 \ldots 1}_{2013} 09093 .
\end{aligned}
$$
Criteria. Any correct solution: 7 points.
It is sh... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,550 |
6. (7 points) Several sages lined up in a column. All of them wore either black or white caps. It turned out that among any 10 consecutive sages, there were an equal number of sages with white and black caps, while among any 12 consecutive sages - not an equal number. What is the maximum number of sages that could be
... | Solution. We will prove that there cannot be more than 15 sages. Suppose the opposite, that there are at least 16 sages. Sequentially number all the sages. Consider nine consecutive sages. If we add one of the two neighboring sages to them, then among them there will be an equal number of sages with white and black hat... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,551 |
3. By the meaning of the problem, $x \geq 0, y \geq 0$.
Square both sides, we get: $x+y+2 \sqrt{x y}=2017$.
Consider the expression $(2017+x-y)^{2}$, we get
$$
(2017+x-y)^{2}=(x+y+2 \sqrt{x y}+x-y)^{2}=4 x(\sqrt{x}+\sqrt{y})^{2}=4 \cdot 2017 \cdot x .
$$
Since the expression $(2017+x-y)^{2}$ is a perfect square, th... | Answer: ( 0 2017), ( 2017,0$). | (0,2017),(2017,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,552 |
5. Note that $2017=13 \cdot 155+2$, i.e., all the stones on the table can be divided into 155 equal piles of 13 stones each, plus 2 additional stones.
The first player takes 2 stones on the first move, and then proceeds as follows:
1) if the second player takes $k$ stones from the table, the first player takes $13-k$... | Answer: the first player wins. | thefirstplayerwins | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,553 |
1. Variant 1.
Petya has four cards with digits $1,2,3,4$. Each digit appears exactly once. How many natural numbers greater than 2222 can Petya make from these cards? | Answer: 16.
Solution: Let's find out how many different numbers can be formed from these cards: the first digit can be chosen in 4 ways, the second can be appended in 3 ways, the third in 2 ways, and the last one is uniquely determined. That is, a total of 24 different numbers can be obtained (it is also possible to v... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,554 |
2. Variant 1.
Nine integers from 1 to 5 are written on the board. It is known that seven of them are not less than 2, six are greater than 2, three are not less than 4, and one is not less than 5. Find the sum of all the numbers. | Answer: 26.
Solution. A number not less than 5 is 5. There is exactly one number 5. Three numbers are not less than 4, so exactly two numbers are equal to 4. Six numbers are greater than 2, meaning all of them are not less than 3. Therefore, exactly three numbers are equal to 3. Seven numbers are not less than 2, so o... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,555 |
# 3. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya said that from April 1 to April 20, the café was open for 17 days, and from April 10 to April 30, it was open for 18 days. It is known that he made a mistake once. What was the date of the last Tuesday in April? | Answer: 29
Solution: Since there are exactly 21 days from April 10 to April 30, each day of the week occurred exactly 3 times during this period. Therefore, this statement cannot be false. This means the first statement is false, and there were only 2 Mondays from April 1 to April 20 (there could not have been four, a... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,556 |
4. Variant 1.
A rectangle was cut into three rectangles, two of which have dimensions 9 m $\times$ 12 m and 10 m $\times$ 15 m. What is the maximum area the original rectangle could have had? Express your answer in square meters. | Answer: 330
Solution. Since the sizes of the two rectangles are fixed, in order for the original rectangle to have the maximum area, the third rectangle must have the largest area. Since the two given rectangles do not have the same sides, the largest area will be obtained by attaching the smaller side of one rectangl... | 330 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,557 |
5. Variant 1.
In the addition example, where the numbers were written on cards, two cards were swapped, resulting in the incorrect expression: $37541+43839=80280$. Find the error and write down the correct sum. | Answer: 80380
Solution. Let's start checking the example from right to left. There are no errors in the units and tens place, but an error appears in the hundreds place. This means that one of the digits in this place $-2, 8$ or $5-$ is transposed.
Let's consider the following cases: | 80380 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,558 |
# 6. Option 1.
Nезнайка named four numbers, and Ponchik wrote down all their pairwise sums on six cards. Then he lost one card, and the numbers left on the remaining cards were $270, 360, 390, 500, 620$. What number did Ponchik write on the lost card? | Answer: 530.
Solution. Let the original numbers be $a \leq b \leq c \leq d$. Suppose the card with the maximum sum is lost. Then this sum is $c+d$. Therefore, $a+b=270$ and $a+b+c+d>270+620=890$. On the other hand, the sum of all numbers on the cards is $3a+3b+2c+2d=270+360+390+500+620=2140$. We get that $2140>1780+2c... | 530 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,559 |
7. Variant 1.
101 natural numbers are written in a circle. It is known that among any 3 consecutive numbers, there is at least one even number. What is the smallest number of even numbers that can be among the written numbers? | Answer: 34.
Solution: Consider any 3 consecutive numbers. Among them, there is an even number. Fix this number and its neighbor, and divide the remaining 99 into 33 sets of 3 consecutive numbers. In each such set, there is at least one even number. Thus, the total number of even numbers is no less than $1+33=34$. Such... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,560 |
# 8. Variant 1.
Identical coins are laid out on a table in the shape of a hexagon. If they are laid out so that the side of the hexagon consists of 2 coins, then 7 coins are enough, and if the side consists of 3 coins, then a total of 19 coins are required. How many coins are needed to build a hexagon with a side cons... | Answer: 271.
Solution.
Method 1.
We will divide the coins into layers (contours) from the center. The first layer contains 1 coin, the second layer contains 6, and so on. Notice that each new layer contains 6 more coins than the previous one (if we remove the coins at the vertices, we get exactly as many coins as th... | 271 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,561 |
1. There were 10 chatterboxes sitting in a circle. At first, one of them told one joke, the next one clockwise told two jokes, the next one told three, and so on around the circle, until one of them told 100 jokes at once. At this point, the chatterboxes got tired, and the next one clockwise told 99 jokes, the next one... | Answer: 1000 jokes.
Solution. Let's number the chatterboxes from 1 to 10 clockwise, starting with the one who told the first joke. Then for any pair of chatterboxes with numbers $k$ and $k+1 (1 \leq k \leq 9)$, the $(k+1)$-th chatterbox initially tells one more joke per round than the $k$-th, for 10 rounds. After the ... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,562 |
2. A divisor of a natural number is called proper if it is not equal to the number itself and 1. Find all such natural numbers for which the largest proper divisor differs by 3 (in either direction) from the cube of the smallest proper divisor. | Answer: 10 and 22.
Solution. Let $x$ be the smallest proper divisor of the number $N$. Then its largest proper divisor is $x^{3}+3$ or $x^{3}-3$. Both pairs of numbers, $x$ and $x^{3}+3$, as well as $x$ and $x^{3}-3$, are of different parity, so one of them is even. Both numbers are divisors of $N$, hence $N$ is even.... | 1022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,563 |
3. In triangle $ABC$, the bisector $BD$ was drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \| AC$. Find the angle $DEF$.
| Answer: $45^{\circ}$.
Solution. Let segments $B D$ and $E F$ intersect at point $G$. From the condition, we have $\angle E D G = \angle E D A = \angle D E G$, hence $G E = G D$. Similarly, $G F = G D$. Therefore, $G E = G F$, which means $B G$ is the bisector and median, and thus also the altitude in triangle $B E F$.... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,564 |
4. Prove that the expression $x^{5}-4 x^{4} y-5 y^{2} x^{3}+20 y^{3} x^{2}+4 y^{4} x-16 y^{5}$ is not equal to 77 for any integer values of $x$ and $y$. | Solution. Let's factor the given expression:
$x^{5}-4 x^{4} y-5 y^{2} x^{3}+20 y^{3} x^{2}+4 y^{4} x-16 y^{5}=x^{4}(x-4 y)-5 x^{2} y^{2}(x-4 y)+4 y^{4}(x-4 y)$ $=(x-4 y)\left(x^{4}-5 x^{2} y^{2}+4 y^{4}\right)=(x-4 y)(x+2 y)(x-2 y)(x-y)(x+y)$.
We need to check that the factors are pairwise distinct. They can coincide... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,565 |
5. In a football tournament where each team played against each other once, teams A, B, C, D, and E participated. For a win, a team received 3 points, for a draw 1 point, and for a loss 0 points. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\m... | Answer: 7 points.
Solution: In a match where one of the teams won, the teams together score 3 points, in a match that ended in a draw - 2 points. Since 7 is not divisible by 3, the team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament.... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,566 |
1. How many different triangles can be formed from: a) 40 matches; b) 43 matches? Solution. We need to find the number of triples of natural numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ such that $\mathrm{x} \leq \mathrm{y} \leq \mathrm{z}$, $\mathrm{x}+\mathrm{y}+\mathrm{z}=40$ and $\mathrm{x}+\mathrm{y}>\mathrm{z}$. ... | Answer: a) $33 ;$ b) 44. | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,567 |
# 2. Solve the equation
$$
\sqrt{3 x-2-x^{2}}+\sqrt{x^{2}-4 x+3}=\sqrt{2}(1-\sqrt{x})
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{c}
3 x-2-x^{2} \geq 0 \\
x^{2}-4 x+3 \geq 0
\end{array}\right.
$$
we obtain that the domain of the function on the left side of the equation is $\{1\}$. The domain of the function on the right side of the equation is the numerical ray $[0 ;+\infty)$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,568 |
3. In quadrilateral $A B C D A D=B D=C D$. Angle $B D C$ is twice the angle $B D A$, and angles $B D A$ and $C A D$ are in the ratio $2: 3$. Find the angles of the quadrilateral. | Solution. Let angle $B D A$ be $x$ (see the figure), then by the condition angle $B D C$ is $2 x$, angle $C A D$ is $1.5 x$. Since $A D=B D=C D$, point $D$ is the center of the circle passing through points $A, B$, and $C$.
Triangle $\triangle A D B$ is isosceles $(A D=B D)$, so angle $D A B$ is $90^{\circ}-0.5 x$. Th... | 75,135,60,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,569 |
5. Prove that the polynomial $x^{95}+x^{94}+x^{93}+\ldots+x^{2}+x+1$ is divisible by the polynomial $x^{31}+x^{30}+x^{29}+\ldots+x^{2}+x+1$. | Solution. Transform:
$$
\begin{aligned}
& x^{95}+x^{94}+x^{93}+\ldots+x^{2}+x+1= \\
& =\left(x^{95}+x^{94}+x^{93}+\ldots+x^{64}\right)+\left(x^{63}+x^{62}+\ldots+x^{32}\right)+\left(x^{31}+x^{30}+\ldots+1\right)= \\
& =x^{64}\left(x^{31}+x^{30}+\ldots+1\right)+x^{32}\left(x^{31}+x^{30}+\ldots+1\right)+\left(x^{31}+x^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,571 |
8.6. First solution. Divide all the coins into two parts of 20 coins each and weigh them. Since the number of counterfeit coins is odd, one of the piles will weigh more. This means that there is no more than one counterfeit coin in it. Divide it into two piles of 10 coins and weigh them. If the scales are in balance, t... | The second solution. Divide all the coins into five equal piles, each containing 8 coins, and number them. Place the 1st and 2nd piles on one side of the scales, and the 3rd and 4th piles on the other.
Consider the first case - the scales balance. Then either there is one fake coin on each side, or all the coins being... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,572 |
8.7. First solution. Let the midpoint of segment $B H$ be denoted by $K$, and the point symmetric to it with respect to $B C$ by $L$ (see Fig. 3). Since $K L$ and $H A_{1}$ are perpendicular to $B C$, $K L$ is parallel to $H A_{1}$. Moreover, since $K$ is the midpoint of $B H$, $K L$ contains the midline of triangle $B... | The second solution. Let the midpoint of segment $B H$ be $K$, and the point symmetric to it with respect to $B C$ be $L$. Since $A_{1} K$ is the median of the right triangle $B A_{1} H$, we have $K A_{1}=K B$, hence $\angle K B A_{1}=\angle K A_{1} B$. Due to the symmetry of points $K$ and $L$ with respect to the line... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,573 |
1. In 2022, knights and liars lined up, with the extreme left and right being liars. Everyone except the extremes made the statement: “There are 42 times more liars to the right of me than to the left.” Provide an example of a row where there is exactly one knight. | Solution. Note that $2021=43 * 47$, so if we place the knight as the forty-eighth from the left, there will be 47 liars to his left, and to his right there will be 2021-47=42*47 liars.
Grading criteria: Correct example - 7 points, if "left" and "right" are switched, do not deduct points. In other cases - 0 points. | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,577 |
2. Prove that $4 \cos (\mathrm{x}) \cos \left(\mathrm{x}+\frac{\pi}{3}\right) \cos \left(\mathrm{x}-\frac{\pi}{3}\right)$ equals $\cos (3 \mathrm{x})$ for all $\mathrm{x}$? | Solution.
$\cos (3 x)=\cos (x) \cos (2 x)-\sin (x) \sin (2 x)=\cos ^{3}(x)-\sin ^{2}(x) \cos (x)-$
$2 \sin ^{2}(x) \cos (x)=\cos (x)\left(\cos ^{2}(x)-3 \sin ^{2}(x)\right)$
On the other hand
$4 \cos \left(x+\frac{\pi}{3}\right) \cos \left(x-\frac{\pi}{3}\right)=4\left(\frac{1}{2} \cos (x)-\frac{\sqrt{3}}{2} \sin (... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,578 |
3. Postman Pechkin is riding a bicycle along a highway. He noticed that every 4.5 kilometers, a suburban bus overtakes him, and every 9 minutes, a suburban bus passes him in the opposite direction. The interval of bus movement in both directions is 12 minutes. At what speed is Pechkin riding? | Answer: 15 km/h
Solution: Let x (km/h) and y (km/h) be the speeds of the cyclist and the bus, respectively. Since the interval between bus movements is 12 minutes (1/5 hour), the distance between two consecutive buses is y/5 km. Therefore, at the moment the cyclist meets a bus, the distance between the cyclist and the... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,579 |
4. Prove that if $\mathrm{n}>4$ and ( $\mathrm{n}-1)$ ! does not divide $\mathrm{n}$, then $\mathrm{n}$ is a prime number. | Solution. Let n not be a prime number, then n has a prime divisor, denote it by p. Obviously, if $p \neq 2$, then $p > 2$, and thus $2p < p^2 = n$. Therefore, in the product (n-1)!, the factors p and 2p are present, and hence (n-1)! is divisible by $2p * p = 2n$, which contradicts the condition.
Grading criteria: Corr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,580 |
5. $A B C D$ is an isosceles trapezoid, and angle $A B D$ is $90^{\circ}$. On the larger base $A D$, as a diameter, a circle $\Omega$ is constructed. An isosceles triangle KLM (KL=LM) is inscribed in the circle $\Omega$, such that $K L \| A B$ and $L M \| C D$. Prove that the areas of triangle KLM and trapezoid ABCD ar... | Solution. Since angle $\mathrm{ABD}$ is equal to $90^{\circ}$, then $\mathrm{B}$ belongs to the circle $\Omega$, and since the trapezoid is isosceles, then $\mathrm{C}$ also belongs to $\Omega$, that is, $\mathrm{ABCD}$ is a cyclic quadrilateral. Let $\alpha$ be the angle at the base of the isosceles triangle KLM, R - ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,581 |
6. On the table, there are 2021 piles of nuts, with one nut in each pile at the beginning. Petya and Vasya are playing the following game (Petya moves first). Each move consists of combining three piles that have the same number of nuts. The player who cannot make a move loses. Who will win with correct play, and how s... | Answer: Petya.
Solution. Note that at the end of the game, the number of nuts in each pile will be equal to a power of three, and there will be no more than two piles of each type. Such a decomposition is unique, and the number of piles with $3^{\text {s }}$ nuts will necessarily be equal to the corresponding digit in... | Petya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,582 |
9.5. 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 opini... | 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 | 12,583 |
9.6. Points $A_{1}, B_{1}, C_{1}$ are chosen on the sides $B C, C A$, and $A B$ of triangle $A B C$ respectively. It turns out that $A B_{1} - A C_{1} = C A_{1} - C B_{1} = B C_{1} - B A_{1}$. Let $I_{A}, I_{B}$, and $I_{C}$ be the centers of the circles inscribed in triangles $A B_{1} C_{1}, A_{1} B C_{1}$, and $A_{1}... | Solution. Let $I$ be the center of the circle inscribed in triangle $ABC$, and let $A_{0}, B_{0}, C_{0}$ be the points of tangency of the circle with sides $BC, CA, AB$, respectively. We will assume that point $A_{1}$ lies on the segment $A_{0}B$ (see Fig. 1). Note that $C A_{0} + A C_{0} = C B_{0} + A B_{0} = C A$. Fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,584 |
9.7. Initially, 10 consecutive natural numbers are written on the board. In one operation, it is allowed to choose any two numbers on the board (denote them as $a$ and $b$) and replace them with the numbers $a^{2}-2011 b^{2}$ and $a b$. After several such operations, none of the original numbers remain on the board. Co... | Answer: Could not.
Solution: Suppose that after several operations, ten consecutive natural numbers were obtained again, and each of the original numbers participated in at least one operation.
Lemma. For any natural number $k$, the number of numbers on the board divisible by $k$ does not decrease when an operation i... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,585 |
9.8. In a certain city, the network of bus routes is arranged in such a way that any two routes have exactly one common stop, and each route has at least 4 stops. Prove that all stops can be distributed between two companies
so that on each route there will be stops of both companies. | Solution. Consider any two routes $\ell_{1}$ and $\ell_{2}$; let $A$ be their common stop. If stop $A$ is on all routes, then we can assign it to one company and all other stops to the other; it is clear that in this case, each route will have stops of both companies.
Now suppose there exists a route $\ell_{3}$ that d... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,586 |
1. Can we use the signs «+», «-», «$\cdot$» and several expressions $a^{4}$ and $a^{6}-1$ to obtain $a^{6} ?$ | Solution. Yes, $-\left(a^{6}-1\right) \cdot\left(a^{6}-1\right)-\left(a^{6}-1\right)+a^{4} \cdot a^{4} \cdot a^{4}=a^{6}$. | ^{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,587 |
4. Given a square $A B C D$. Point $N$ lies on side $A D$, such that $A N$ : $N D=2: 3$, point $F$ lies on side $C D$ and $D F: F C=1: 4$, point $K$ lies on side $A B$, such that $A K: K B=1: 4$. Find the angle $K N F$.
 there is an odd number of checkers? | Solution. Place the checkers on all the fields. Consider the diagonals of the chessboard. Since the colored and uncolored diagonals do not share any common cells, it is sufficient to consider only the diagonals of one color. Let these be the colored diagonals. We have an odd number of diagonals with an even number of c... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,590 |
7.1 The number 100 was divided by some number less than 50, with a remainder of 6. By what number could the division have occurred? | Let $x$ be the number by which they divided, $m$ be the quotient: $100 = m \cdot x + 6$ (since 6 is the remainder, then $x > 6$, and $m \geq 2$, since $x < 50$) $m \cdot x = 94 = 2 \cdot 47$. From this, it is clear that the numbers $m$ and $x$ are exactly the numbers 2 and 47, i.e., $m = 2$, $x = 47$. (since the number... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,591 |
7.2 Several numbers are written on the board. It is allowed to erase two numbers $X$ and $Y$ and write $X-2$ and $Y+1$ instead. Prove that after several such operations, a negative number will inevitably appear on the board. | Solution We will keep track of the sum $\mathrm{S}$ of all numbers on the board. Each operation $(\mathrm{x}, \mathrm{y}) \rightarrow(\mathrm{x}-2, \mathrm{y}+1)$ reduces $\mathrm{S}$ by one. Therefore, after a certain number of operations, the sum $S$ will become negative. This means that there are negative numbers am... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,592 |
7.4 Can the cells of a $5 \times 8$ table be filled with the numbers 1 and 3 so that the sum of the numbers in each row and each column is divisible by 7? | Solution: Suppose it was possible to arrange the numbers 1 and 3 in the table such that their sum in each of the 5 rows and in each of the 8 columns is divisible by 7. The sum in any row is not less than $8 \cdot 1=8$ and not more than $8 \cdot 3=24$. The numbers in the range from 8 to 24 that are divisible by 7 are 14... | impossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,593 |
# Task 9.1
All three-digit numbers are written in a row: $100101102 \ldots 998$ 999. How many times in this row does a zero follow a two?
## Number of points 7 Answer: 19
# | # Solution
Since a three-digit number cannot start with zero, the two followed by a zero cannot be in the units place of any three-digit number in the sequence. Let's assume the two is in the tens place of a three-digit number. Then the zero following it is in the units place of the same number, i.e., the number ends ... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,594 |
# Task 9.2
The sum of the squares of p prime numbers, each of which is greater than 5, is divisible by 6. Prove that n is also divisible by 6.
## Number of points 7
# | # Solution
If the sum of several numbers is divisible by six, then the sum of their remainders when divided by six will also be divisible by 6. A prime number greater than five, when divided by 6, can only have remainders of 1 or 5 (otherwise, the number would be divisible by 2 or 3). Therefore, the square of any prim... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,595 |
# Task 9.4
A quadrilateral is inscribed in a circle of radius 17, with diagonals that are perpendicular to each other and are 8 and 9 units away from the center of the circle. Find the sides of the quadrilateral.
Number of points 7 | Answer:
$$
4(2 \sqrt{13} \pm 1), \quad 4(8 \pm 2 \sqrt{13})
$$
## Solution
Let $O$ be the center of the circumscribed circle of quadrilateral $ABCD$, $M$ be the intersection point of its d... | 4(2\sqrt{13}\1),\quad4(8\2\sqrt{13}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,597 |
# Task 9.5
In a $7 \times 7$ square grid, 16 tiles of size $1 \times 3$ and one tile of size $1 \times 1$ are placed. Prove that the $1 \times 1$ tile either lies in the center or is adjacent to the borders of the square.
## Number of points 7 | Solution
Let's color the $7 \times 7$ square as shown in the left figure. Each tile $1 \times 3$ covers exactly one shaded cell. Since there are 16 tiles $1 \times 3$ and 17 cells are shaded, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,598 |
7.1. Draw four rays $O A, O B, O C$ and $O D$ with a common origin such that the drawing contains angles of $100^{\circ}, 110^{\circ}, 120^{\circ}, 130^{\circ}$ and $140^{\circ}$. Write down which angles have the specified measures. | Answer: $\angle A O D=100^{\circ}, \angle A O C=110^{\circ}, \angle A O B=120^{\circ}, \angle B O C=130^{\circ}$, $\angle B O D=140^{\circ} \quad$ (see Fig. 7.1).
For example, one can proceed as follows: first, draw rays $O A, O B$ and $O D$ such that $\angle A O B=120^{\circ}, \angle B O D=140^{\circ}, \angle A O D=1... | \ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,599 |
7.2. Harry, Ron, and Hermione wanted to buy identical waterproof cloaks. However, they lacked the money: Ron was short by a third of the cloak's price, Hermione by a quarter, and Harry by one fifth of the cloak's price. When the price of the cloak dropped by 9.4 sickles during a sale, the friends pooled their savings a... | Answer: 36 sickles.
Solution. Let the initial cost of the cloak be $x$ sickles, then Ron had $\frac{2}{3} x$ sickles, Hermione had $\frac{3}{4} x$ sickles, and Harry had $\frac{4}{5} x$ sickles. During the sale, the cloak cost $(x-9.4)$ sickles, and three cloaks cost $3(x-9.4)$ sickles. Since the friends bought three ... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,600 |
7.3. Each of the thirteen dwarfs is either a knight, who always tells the truth, or a liar, who always lies. One day, all the dwarfs in turn made the statement: “Among the statements made previously, there are exactly two more false ones than true ones.” How many knights could there have been among the dwarfs? | Answer: 6
Solution. The first two statements are obviously false, as there were fewer than two statements made before each of them. The third statement is true, as there were 2 false statements and zero true statements made before it. The fourth statement is false, as it adds one true statement to the two false ones, ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,601 |
7.4. On graph paper, a large square was drawn. It was cut into several identical medium squares. One of the medium squares was cut into several identical small squares. The sides of all squares run along the grid lines. Find the side lengths of the large, medium, and small squares if the sum of their areas is 154. | Answer: 12, 3, and 1 respectively.
Solution. From the problem statement, it follows that the side length of each square is a natural number, and the side length of each square is a divisor of the side length of the previous one. Let the side length of the small square be $a$, the medium square be $k a$, and the large ... | 12,3,1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,602 |
7.5. At each vertex of a cube lives a number, not necessarily positive. All eight numbers are distinct. If a number is equal to the sum of the three numbers living in the adjacent vertices, then it is happy. What is the maximum number of happy numbers that can live at the vertices of the cube? | Answer: 8.
Solution. See, for example, Fig. 7.5a. It is easy to verify that each vertex of the cube contains a lucky number.
There are other examples as well. Let's understand how they are structured (this was not required of the olympiad participants). Let's denote the numbers at the vertices of the cube (see Fig. 7... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,603 |
7.2 In a three-digit number, the first digit was crossed out, resulting in a two-digit number. If the original number is divided by this two-digit number, the quotient is 9 and the remainder is 8. Find the original number. (Provide all possible solutions.) | Answer: 4 possible numbers: 224; 449; 674; 899. Hint Let $x$ be the first digit of the original number, and $y$ be the two-digit number after erasing $x$. Then we have the equation $100 x+y=9 y+8 \Leftrightarrow 2 y+2=25 x$. Therefore, $x$ is an even number: $x=2 z$ for some $z$ ( $z \leq 4$ ). Thus, $y+1=25 z$, and th... | 224;449;674;899 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,604 |
6.1. In the numerical example АБВ $+9=$ ГДЕ, the letters А, Б, В, Г, Д, and Е represent six different digits. What digit is represented by the letter Д? | Answer: 0.
Solution: In the addition, the second digit of the first addend АБВ has changed (Д instead of Б). This could only happen if 1 was carried over from the units place to the tens place during the addition. However, the first digit also changed (Г instead of А). This means that 1 was also carried over from the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,605 |
6.2. Lines parallel to the sides of a square form a smaller square, the center of which coincides with the center of the original square. It is known that the area of the cross formed by the smaller square (see the figure on the right) is 17 times the area of the smaller square. How many times is the area of the origin... | Answer. 81 times.
Solution. Let the square have dimensions of 1 cm $\times 1$ cm, and the larger square have dimensions of $n$ cm $\times n$ cm. Then the area of the cross is $(2 n-1)$ cm $^{2}$ (the vertical column has dimensions $n \times 1$, the horizontal row has dimensions $1 \times n$, and the area of the square... | 81 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,606 |
6.3. The boys brought candies to the class and distributed them to the girls. Petya said that he brought exactly half of the total number of candies. Kolya said that he brought exactly one third of the total number of candies, and he gave his candies only to Masha and Tanya, with Masha receiving 3 more candies than Tan... | Solution. Suppose that both boys did not make a mistake. Since Petya did not make a mistake, the total number of candies brought is even (twice the number of candies brought by Petya). A third of an even number is also an even number. Therefore, the number of candies brought by Kolya is even. However, according to his ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,607 |
6.4. Vasya wrote 999 products on the board: $1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \ldots, 999 \cdot 1000$. Is it true that the sum of some two of these products equals 20000? | Answer. Correct.
Solution. For example, $99 \cdot 100 + 100 \cdot 101 = 100(99 + 101) = 100 \cdot 200 = 20000$.
Remark. Only three more examples are possible: $54 \cdot 55 + 130 \cdot 131$, $40 \cdot 41 + 135 \cdot 136$, $89 \cdot 90 + 109 \cdot 110$.
Comment. A correct answer without examples of pairs of products -... | 99\cdot100+100\cdot101=20000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,608 |
6.5. In the room, there are 10 people - liars and knights (liars always lie, and knights always tell the truth). The first said: "In this room, there is at least 1 liar." The second said: "In this room, there are at least 2 liars." The third said: "In this room, there are at least 3 liars." And so on,
up to the tenth,... | Answer: 5.
Solution: Let there be $k$ liars in the room. Then the first $k$ people told the truth (and thus were knights), while the remaining $(10-k)$ lied (and were liars). Therefore, $k=10-k$, from which $k=5$.
Comment: The answer is obtained by considering an example -3 points. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,609 |
4. On the island, there live knights who always tell the truth and liars who always lie. In the island's football team, there are 11 people. Player number 1 said: "In our team, there are as many knights as there are liars." Player number 2 said: "In our team, the number of knights and the number of liars differ by one,... | 4. Answer: There are either no knights at all, or there is only one and he plays under number 10. The two answers cannot both be true, as they contradict each other. This means there can be no more than one true answer. If there is no true answer, then the team consists entirely of liars, in which case indeed none of t... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,611 |
5. In triangle $ABC$, the median $BM$ is drawn. Given that $\angle ABM=40^{\circ}$, $\angle \mathrm{CBM}=70^{\circ}$. Prove that $\mathrm{AB}=2 \mathrm{BM}$. | 5. Continue the median BM and lay off the segment MK=BM. Then triangle AMK is equal to triangle BMC by two sides and the angle between them, $\angle \mathrm{CBM}=\angle \mathrm{MKA}=70^{\circ}$. In triangle $\mathrm{ABK} \angle \mathrm{BAK}=180^{\circ}-40^{\circ}-70^{\circ}=70^{\circ}$, so $A B=B K=2 B M$.
Criteria: c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,612 |
6. Vasya has 19950 chips, painted in 200 different colors. He arranged them in a circle. Prove that there is such a pair of colors that the chips of these colors do not stand next to each other. | 6. A color will be found in which fewer than 100 chips are painted - otherwise there would be no fewer than 20000 chips. Chips of this color are adjacent to no more than 198 other chips. Therefore, there is at least one color among the 199 other colors with which the chips of this color do not share adjacency.
Criteri... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,613 |
7.1 Rays OA and OB form a right angle. Seventh-grader Petya drew rays OC and OD inside this angle, forming an angle of $10^{\circ}$, and then calculated all the acute angles between any pairs of the drawn rays (not only adjacent ones). It turned out that the sum of the largest and the smallest of the found angles is $8... | Solution: See fig.
If the smallest angle turned out to be $\alpha$ or $\beta$, for example $\alpha$, then the largest angle would be the angle $\beta+10^{\circ}$, and their sum would be $90^{\circ}$, which contradicts the condition of the problem. Therefore, the smallest angle is the angle $10^{\circ}$, and the larges... | 65,15,10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,615 |
7.2 Out of 24 matches, a figure in the form of a $3 \times 3$ square is laid out (see figure), the side length of each small square is equal to the length of a match. What is the smallest number of matches that can be removed so that there are no whole $1 \times 1$ squares left, formed from matches.
![](https://cdn.ma... | Answer: 5 matches.
Solution: Estimation: It is impossible to manage with four matches, as by removing a match, we "ruin" no more than two squares (each match is a side of one or two adjacent squares), but initially, we have 9 small squares. Example for 5 matches:
 and canceling $a^{2} b^{2... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,620 |
9.3. From the natural numbers $1,2, \ldots, 101$, a group of numbers is chosen such that the greatest common divisor of any two numbers in the group is greater than two. What is the maximum number of numbers that can be in such a group? | Answer: 33. Solution: Estimation. Let's divide the numbers $1,2, \ldots, 101$ into 34 sets: $A_{0}=\{1,2\}, A_{1}=\{3,4,5\}$, $A_{2}=\{6,7,8\}, \ldots, A_{33}=\{99,100,101\}$ (i.e., $A_{k}$ for $k \geq 1$ consists of three numbers $3 k, 3 k+1,3 k+2$). In the desired group of numbers, there cannot be a number from $A_{0... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,622 |
9.5. On a plane, there are 99 segments, and all their intersection points are marked. Could it be that a) on any segment, there are exactly three marked points? b) each segment intersects exactly with three other segments? | Answer: a) could; b) could not. Solution. a) An example of such an arrangement can be, for instance, as follows: make 19 copies of the five segments of the right part of the figure from problem 7.5 and add four segments of the right part of the figure from problem 8.5. b) Suppose, for the sake of contradiction, that su... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,624 |
9.1. Winnie-the-Pooh and Piglet divided a cake between themselves. Piglet cried that he got too little. Then Pooh gave him a third of his share. This tripled the amount of cake Piglet had. What part of the cake did Pooh have initially, and what part did Piglet have? | Answer: Winnie-the-Pooh and Piglet originally had $6 / 7$ and $1 / 7$ of the cake respectively.
Let $a$ and $b$ be the shares of the cake for Winnie-the-Pooh and Piglet, respectively. Then $a+b=1$. From the condition $a / 3+b=3 b$, it follows that $a=6 b, 7 b=1, b=1 / 7$ and $a=6 / 7$. | =6/7,b=1/7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,625 |
9.3. Find any five consecutive natural numbers less than 100, the product of which is divisible by 2014. | Answer: 53, 54, 55, 56, 57.
This example can be obtained by noting that $2014=2 \cdot 19 \cdot 53$.
Remark. This example of such numbers is unique. Only the answer without verification or explanation - 5 points; the answer with verification or explanation - 7 points. | 53,54,55,56,57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,626 |
9.4. Given a rectangle $A B C D$. On the ray $D C$, segment $D K$ is laid off, equal to $D B$. $M$ is the midpoint of segment $B K$. Prove that $A M$ is the bisector of angle $B A C$. | First solution. Triangle BDK is isosceles by construction, so its median $DM$ is also its altitude and bisector. Consider the circle constructed on $BD$ as the diameter. Points $M$, $C$, and $A$ lie on this circle because angles $BMD$, $BCD$, and $BAD$ are right angles. Angles $BAM$ and $BDM$ are equal, as inscribed an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,627 |
9.5. In the equation $* x^{2}+* x+*=0$ two players take turns replacing any asterisk with an arbitrary number (zero cannot be placed before $x^{2}$). The first player wins if the resulting equation has no roots, and the second player wins otherwise. Can either of them win, regardless of the opponent's play? | Answer: The second player wins regardless of the first player's moves.
If the first player does not place a number $c$, different from zero, in the position of the free term, then the second player will place the number 0 in this position, and regardless of the further game, the final equation will have a root $x=0$. ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,628 |
6.1. Given the incorrect equation $1 \times 2 \times 3+4 \times 5=50$. Insert parentheses to make it correct. | Solution. $(1 \times 2 \times 3+4) \times 5=50$.
Comment. Any correct example - 7 points. | (1\times2\times3+4)\times5=50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,629 |
6.2. There were 76 people in two rooms. When 30 people left one room and 40 people left the second room, the number of people remaining in the rooms became equal. How many people were initially in the rooms? (Don't forget to explain your answer.) | Answer: 33 and 43 people. Solution: After 70 people left, 6 people remained in the rooms (3 in each room), which means there were initially $30+3=33$ and $40+3=43$ people in the rooms.
Comment. Correct answer only - 3 points; answer with explanations or calculations leading to the answer - 7 points. | 3343 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,630 |
6.3. Can three fractions be chosen from $1 / 2017, 2 / 2016, 3 / 2015, \ldots, 2017 / 1$ such that their product equals 1? | Answer: It can be. Solution. $1=(1 / 2017) \times(1009 / 1009) \times(2017 / 1)$.
Comment. At least one example is provided - 7 points. | (1/2017)\times(1009/1009)\times(2017/1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,631 |
6.4. At a round table, 120 people are sitting - knights, who always tell the truth, and liars, who always lie. Each of them made exactly one of the following two statements: 1) "My right neighbor is a knight"; 2) "The person who is one to the right of my right neighbor is a knight." How many liars could there be at the... | Answer: $0,60,120$. Solution. The cases where all sitting at the table are liars, or all sitting at the table are knights, are obviously valid. Suppose there are both knights and liars sitting at the table. Then there is a liar (L) whose right neighbor is a knight (R). L could not have made statement 1). Therefore, he ... | 0,60,120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,632 |
11.1. The graphs of the functions $y=a x^{2}, y=b x$ and $y=c$ intersect at a point located above the x-axis. Determine how many roots the equation $a x^{2}+b x+c=0$ can have. | Answer: No roots.
Solution. From the condition of the problem, it follows that the graphs intersect at the point $(m ; c)$, where $c$ > 0. Then the equalities $b m=c$ and $a m^{2}=c$ hold, so $m \neq 0$. Therefore, the discriminant of the given equation $D=b^{2}-4 a c=\frac{c^{2}}{m^{2}}-\frac{4 c^{2}}{m^{2}}=-\frac{3... | Noroots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,633 |
11.2. Does there exist a triangle in which the sum of the cosines of the interior angles is 1? | Answer: does not exist.
Solution. First method. Suppose such a triangle $ABC$ exists, that is, $\cos A + \cos B + \cos C = 1$. Since $\cos C = \cos (180^\circ - A - B) = -\cos (A + B)$, we have $\cos A + \cos B = 1 + \cos (A + B)$, which leads to $2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} = 2 \cos^2 \frac{A + B}{2}$... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,634 |
11.3. In a regular hexagonal pyramid SABCDEF (ABCDEF is the base), the lateral edge is equal to $a$, and the plane angle at vertex $S$ is $10^{\circ}$. An ant crawls along the surface of the pyramid from vertex $A$, aiming to visit all lateral edges (possibly at the vertices) and return to point $A$. What is the length... | Answer: $a$.
Solution. "Cut" the pyramid SABCDEF along the edge SA and make a net (see Fig. 11.3).
Then any path on the lateral surface of the pyramid, satisfying the condition, will be a broken line on the net, connecting points $A$ and $A_{1}$. The shortest path from $A$ to $A_{1}$ is the length of the segment $A A... | a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,635 |
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