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# Task № 8.3
## Condition:
A board of size $2022 \times 2022$ is given. Liza and Varya take turns coloring $2 \times 2$ squares on it red and blue, with the agreement that each cell can be painted no more than once in blue and no more than once in red. Cells that are painted blue and then red (or vice versa) become p... | Solution by analogy with task №8. 1
# | 2022\cdot20202021\cdot2020 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 15,636 |
# Task № 8.4
## Condition:
A board of size $2022 \times 2022$ is given. Egor and Matvey take turns coloring $2 \times 2$ squares on it red and yellow, with the sides of the squares aligned with the grid lines. The boys agreed that each cell can be painted no more than once in blue and no more than once in red. Cells ... | Solution by analogy with task №8.1 | 2022\cdot20202021\cdot2020 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 15,637 |
5. Let's show that the number of brothers must equal the number of sisters. Let there be $m$ brothers and $n$ sisters. Since all the brothers are currently in a quarrel with a different number of sisters, there must be at least one pair that is still in a quarrel. Since each sister is currently in a quarrel with the sa... | # Answer: Three brothers and three sisters. | 3 | Combinatorics | proof | Yes | Yes | olympiads | false | 15,638 |
10.1. Nine skiers left the starting point one after another and covered the distance - each at their own constant speed. Could it have turned out that each skier participated in exactly four overtakes? (In each overtake, exactly two skiers participate - the one who overtakes and the one who is overtaken.) (S. Volchonko... | Answer: It could not.
Solution: Suppose it was possible. Since the speeds are constant, any two skiers could not meet more than once. Then the skier who started first could not overtake anyone; hence, he was overtaken by four others, and he finished fifth. On the other hand, the skier who started last could not be ove... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,639 |
10.2. Is it possible for some natural $k$ to divide all natural numbers from 1 to $k$ into two groups and write down the numbers in each group in a row in some order so that two identical numbers are obtained?
(N. Agakhanov) | Answer: No
Solution. Assume the opposite. Clearly, $k \geqslant 10$, since the set of digits from 1 to 9 contains no repetitions. Consider the highest power of ten $10^{n}$ that does not exceed $k$. The sequence of digits of the number $10^{n}$ will entirely fit into one of the formed numbers. But then the same sequen... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,640 |
1. Let $t=\log _{x^{2019}} 2018$, then the equation will take the form: $t+t^{2}+t^{3}+\ldots=2018$.
Simplify the left part of the last equation using the formula for the sum of an infinite decreasing geometric progression, we get: $\frac{t}{1-t}=2018$.
From which $t=\frac{2018}{2019}$.
We get, $\log _{x^{2019}} 201... | Answer: $\sqrt[2018]{2018}$. | \sqrt[2018]{2018} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,644 |
4. Let $A, B, C$ be the centers of the given circles, $A B=2, A C=2, B C=3$ (see Fig. 1).
Denote by $a, b$, and $c$ the radii of the circles centered at points $A, B$, and $C$ respectively, and find their values by solving the system of equations. We get:
$$
\left\{\begin{array}{l}
a + b = 2 \\
a + c = 2 \\
b + c = 3... | Answer: $10 \sqrt{7} \pi$ | 10\sqrt{7}\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,646 |
1. It is known that the numbers $a+b+c$ and $\frac{a b+b c+c a}{a+b+c}$ are integers. Prove that the number $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}$ is also an integer. | 1. $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}=\frac{(a+b+c)^{2}-2(a b+b c+c a)}{a+b+c}=(a+b+c)-2 \cdot \frac{a b+b c+c a}{a+b+c}$ - an integer. | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,648 |
2. Each of the two distinct roots of the quadratic trinomial $f(x)=x^{2}+p x+q$ and its value at $x=3$ are natural numbers, and the larger root of the trinomial and $f(3)$ are prime numbers. Find the roots of the trinomial $f(x)$. | 2. Answer. $x_{1}=4, x_{2}=5$ or $x_{1}=1, x_{2}=2$.
If $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, then the factorization
$f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)$ holds. Since the roots are distinct, we can assume that $x_{1}<x_{2}$. By the condition, $f(3)=\left(3-x_{1}\right)\left(3-... | x_{1}=4,x_{2}=5orx_{1}=1,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,649 |
3. Prove that if $a, b, c$ are positive real numbers such that $a < b + c$, then
$$
\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}
$$ | 3. First, let's prove that for positive numbers $x$ and $y$, the inequality $x<y$ implies the inequality $\frac{x}{1+x}<\frac{y}{1+y}$. Indeed, consider the difference
$$
\frac{x}{1+x}-\frac{y}{1+y}=\frac{x(1+y)-y(1+x)}{(1+x)(1+y)}=\frac{x-y}{(1+x)(1+y)}
$$
Given the condition $x-y<0$, we get
$$
\frac{x}{1+x}-\frac{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,650 |
4. Inside triangle $ABC$, where $\angle C=70^{\circ}, \angle B=80^{\circ}$, a point $M$ is taken such that triangle $CMB$ is equilateral. Find the angles $MAB$ and $MAC$. | 4. Answer. $\angle M A B=20^{\circ}, \angle M A C=10^{\circ}$.
Consider the circle with center at point $M$ and radius $R=M B=M C$ (see figure).
$\angle A=180^{\circ}-(\angle B+\angle C)=30... | \angleMAB=20,\angleMAC=10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,651 |
5. Given a sequence $x_{n}$ such that $x_{1}=1, x_{2}=2, x_{n+2}=\left|x_{n+1}\right|-x_{n}$. Find $x_{2015}$. | 5. Answer. ${ }^{x_{2015}}=0$.
We will show that the given sequence is periodic with a period of 9. Let's find
$x_{1}=1, x_{2}=2, x_{3}=1, x_{4}=-1, x_{5}=0, x_{6}=1, x_{7}=1, x_{8}=0, x_{9}=-1, x_{10}=1, x_{11}=2, \ldots$.
Since the sequence is completely determined by any two consecutive terms, we have obtained th... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,652 |
10.1. Positive numbers $a, b$, and $c$ are such that the point $A(1 ; 2)$ is located below the graph of the parabola $y=a x^{2}+b x+c$. Can it be uniquely determined how this point is positioned (above, below, or on) relative to the graph of the parabola $y=c x^{2}+b x+a$? Justify your answer. | Solution: Note that the graphs of both parabolas intersect at the point with abscissa 1, so any point with abscissa 1 is equally positioned relative to both parabolas, which means the answer to the question is positive.
Answer: Yes. Point $A$ is located lower.
Recommendations for checking:
| is in the work | points ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,653 |
10.2. In parallelogram $A B C D$, the angle $\angle A=\alpha$ and diagonal $B D=d$ are known. Let $M$ and $N$ be the feet of the perpendiculars dropped from vertex $B$ to lines $C D$ and $A D$, respectively. Find $M N$.
 Note that the equality $M N=d \sin \alpha$ holds. $\angle M B N=\angle B A D=\alpha$, since the sides of angle $M B N$ are perpendicular to the sides of angle $B A D$.
2) Consider the circle with diameter $B D=d$. Points $M$ and $N$ lie on this circle, as $\angle B M D=\angle B N D=90^{\circ}$.
3) By the ... | MN=\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,654 |
10.3. Several different real numbers are written on the board. It is known that the sum of any three of them is rational, while the sum of any two of them is irrational. What is the largest number of numbers that can be written on the board? Justify your answer. | Solution: The set of three numbers $\sqrt{2}, \sqrt{3}, -\sqrt{2}-\sqrt{3}$, as is easily verified, satisfies the condition of the problem. We will show that no more than three numbers can be written on the board.
Assume the contrary, and let $a_{i} (i=\overline{1,4})$ be some four numbers from this set. Then the numb... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,655 |
10.4. A point on the plane with coordinates $(x ; y)$ is allowed to be connected by a segment to a point with coordinates $(x+3 y ; y)$ or to a point with coordinates $(x ; y-2 x)$. Can points $A(19 ; 47)$ and $B(12 ; 17)$ be connected by a broken line under these conditions? Justify your answer. | Solution: It is obvious that if the coordinates of the initial point are integers, then the coordinates of all subsequent points will also be integers. The difference in the x-coordinates of the points that can be connected by a segment is either $3y$ or 0, and in both cases, it is divisible by 3. Therefore, the remain... | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,656 |
10.5. Let $a, b, c$ and $d$ be natural numbers and $\frac{a d-1}{a+1}+\frac{b d-1}{b+1}+\frac{c d-1}{c+1}=d$. Find all values that the number $d$ can take. | Solution: We will transform the equation in an equivalent manner:
\[
\begin{gathered}
\frac{a d-1}{a+1}+\frac{b d-1}{b+1}+\frac{c d-1}{c+1}=d \\
d-\frac{d+1}{a+1}+d-\frac{d+1}{b+1}+d-\frac{d+1}{c+1}=d \\
2 d=(d+1)\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right) \\
\frac{2 d}{d+1}=\frac{1}{a+1}+\frac{1}{b+1}+\fra... | \in{1,2,3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,657 |
10.6. At the factory, there are exactly 217 women, among whom 17 are brunettes, and the remaining 200 are blondes. Before New Year's, all of them dyed their hair, and each of these women wrote in "VK" the surnames of exactly 200 women from the factory, whom they believed to be definitely blondes. Each of the brunettes ... | Solution: According to the problem, the correct list of all 200 blondes will be on "Vkontakte" exactly for 17 female workers at the plant: brunettes will write exactly this list, and a blonde will never write it, as otherwise she would have to include herself in it. Therefore, if a certain list appears not 17 times but... | 13 | Combinatorics | proof | Yes | Yes | olympiads | false | 15,658 |
Problem 7.1. Denis thought of four different natural numbers. He claims that
- the product of the smallest and the largest numbers is 32;
- the product of the two remaining numbers is 14.
What is the sum of all four numbers? | Answer: 42.
Solution. There are two ways to represent the number 14 as the product of two numbers: $1 \cdot 14$ and $2 \cdot 7$; therefore, the second and third largest numbers are 1 and 14 or 2 and 7. The first case does not work for us, as 1 cannot be the second largest number. So, the second largest number is 2, an... | 42 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,659 |
Problem 7.2. Along the road stand the houses of Andrey, Boris, Vasya, and Gena (in that order). The distance between Andrey's and Gena's houses is 2450 meters. One day, the boys decided to have a 1 km race. They set the start halfway between Andrey's and Vasya's houses. As a result, the finish line was exactly halfway ... | Answer: 450.
Solution. Let's assume that the road goes from Andrey's house eastward to Gena's house; then we can say that Gena's house is east of Vasya's house, and Borya's house is east of Andrey's house. This means that the finish (the midpoint between Borya's and Gena's houses) is east of the start (the midpoint be... | 450 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,660 |
Problem 7.3. The numbers $1,2,4,5,8,10$ were placed in the cells of the figure shown in the image, such that the sums of the numbers in all columns (including the column with only one cell) were equal. What number can be in the topmost cell? List all possible options.
![](https://cdn.mathpix.com/cropped/2024_05_06_c7c... | Answer: 1, 4 or 5.
Solution. The total sum of the numbers is $1+2+4+5+8+10=30$. It follows that the sum of the numbers in each column is $\frac{30}{3}=10$; in particular, the number 10 must be in the column with only one cell.
Now, we need to notice that among the remaining five numbers, there is only one way to choo... | 1,4,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,661 |
Problem 7.4. On Monday, 5 people in the class received fives in math, on Tuesday, 8 people received fives, on Wednesday - 6 people, on Thursday - 4 people, on Friday - 9 people. No student received fives on two consecutive days. What is the minimum number of students that could have been in the class | Answer: 14.
Solution. Let's consider pairs of consecutive days.
- On Monday and Tuesday, 5s were received by $5+8=13$ people.
- On Tuesday and Wednesday, 5s were received by $8+6=14$ people.
- On Wednesday and Thursday, 5s were received by $6+4=10$ people.
- On Thursday and Friday, 5s were received by $4+9=13$ people... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,662 |
Problem 7.5. At a tribal council meeting, 60 people spoke in turn. Each of them said only one phrase. The first three speakers said the same thing: "I always tell the truth!" The next 57 speakers also said the same phrase: "Among the previous three speakers, exactly two told the truth." What is the maximum number of sp... | Answer: 45.
Solution. Note that among any four consecutive speakers, at least one lied (if the first three told the truth, then the fourth definitely lied). By dividing 60 people into 15 groups of four consecutive speakers, we get that at least 15 people lied, meaning no more than 45 people told the truth.
To constru... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,663 |
Problem 7.6. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively. It is known that $A B=B D, \angle A B D=46^{\circ}, \angle D E C=90^{\circ}$. Find $\angle B D E$, if it is known that $2 D E=A D$. | Answer: $67^{\circ}$.
Solution. Draw the height $B M$ in triangle $A B D$. Since this triangle is isosceles, $M$ bisects $A D$ (Fig. 7.6). This means that $M D = A D / 2 = D E$.
Then the right triangles $B D M$ and $B D E$ are congruent by the leg ( $D M = D E$ ) and the hypotenuse ( $B D$ - common). From this, it is... | 67 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,664 |
Problem 7.7. In the classroom, there are several single desks (no more than one person can sit at each desk; there are no other desks in the classroom). During the break, a quarter of the students went out into the corridor, and the number of people left in the classroom was equal to 4/7 of the total number of desks. H... | Answer: 21.
Solution. Note that the number of remaining students is $\frac{3}{4}$ of the initial number, so it must be divisible by 3. Let's denote it by $3x$. Let $y$ be the number of desks. Then
$$
3 x=\frac{4}{7} y
$$
from which $21 x=4 y$. Since 4 and 21 are coprime, $y$ must be divisible by 21. From the problem... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,665 |
Problem 7.8. Tanya and Vера are playing a game. Tanya has cards with numbers from 1 to 30. She arranges them in some order in a circle. For each pair of adjacent numbers, Vера calculates their difference, subtracting the smaller number from the larger one, and writes down the resulting 30 numbers in her notebook. After... | Answer: 14.
Solution. We will prove that Tanya cannot get 15 candies. Let's look at the card with the number 15. The number 15 differs from all the remaining numbers, except for the number 30, by no more than 14. Thus, at least one of the differences involving the number 15 will be no more than 14.
Let's provide an e... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,666 |
1. Do there exist such integers $a, b$ and $c$ that the discriminant of the quadratic equation $a x^{2}+b x+c=0$ is equal to 2019? | Solution: Suppose the discriminant of the given equation is equal to the number 2019. Then we can write: $\mathrm{b}^{2}-4 \mathrm{ac}=2019$, and $\mathrm{b}^{2}-1225=4 \mathrm{ac}+794$ or $(\mathrm{b}-35) \cdot(\mathrm{b}+35)=2(2 \mathrm{ac}+397)$. Note that $\mathrm{b}-35$ and $\mathrm{b}+35$ are numbers of the same ... | no | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,667 |
# 2. Find the sum:
$$
\frac{2}{1 \cdot 2 \cdot 3}+\frac{2}{2 \cdot 3 \cdot 4}+\frac{2}{3 \cdot 4 \cdot 5}+\ldots+\frac{2}{2008 \cdot 2009 \cdot 2010}
$$ | Solution:
Notice that $\frac{2}{n \cdot(n+1) \cdot(n+2)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+1}+\frac{1}{n+2}$.
From this, it follows that the required sum is:
$\frac{1}{1}-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{3}+\frac{1}{4}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+\ldots+\fra... | \frac{1009522}{2019045} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,668 |
3. Plot the graph of the function $y=(4 \sin 4 x-2 \cos 2 x+3)^{0.5}+(4 \cos 4 x+2 \cos 2 x+3)^{0.5}$. | Solution:
$\mathbf{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x+1}$
$\mathrm{y}=2 \sin ^{2} x+1+2 \cos ^{2} x+1, \m... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,669 |
5. Find the smallest natural number $A$, which is divisible by $p$, ends in $\boldsymbol{p}$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and is the cube of a natural number. | Solution. Let $2 \mathrm{p}+1=\mathrm{n}^{3}$. Then $(\mathrm{n}-1)\left(\mathrm{n}^{2}+n+1\right)=2 \mathrm{p}$. The number $2 \mathrm{p}$ can only have the following positive divisors: 1, 2, p, 2p. The number n is clearly odd, so n-1 is divisible by 2. The number $n^{2}+n+1$ is greater than 1, so $n-1=2, n^{2}+n+1=$ ... | 11713 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,670 |
2. Each Martian has three hands. Can 2021 Martians hold hands in such a way that each Martian uses all three hands? Each Martian can hold only one hand of another Martian with one of their hands. | Solution. In total, there are $2021 \cdot 3=6063$ hands - an odd number. If they were to hold hands, the hands would be paired (each pair consisting of two hands shaking each other), which is impossible. Answer: They cannot. | Theycannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,672 |
3. The numbers from 1 to 25 (each exactly once) were placed in the cells of a $5 \times 5$ square such that the sum of the numbers in each row and each column is the same. A boy erased all the numbers except for four (see the picture). Find the number marked with a circle $\bigcirc$.
=$ $65-51=14$. Answer: 14. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,673 |
4. In grandmother's garden, apples have ripened: Antonovka, Grushovka, and White Naliv. If there were three times as many Antonovka apples, the total number of apples would increase by $70 \%$. If there were three times as many Grushovka apples, it would increase by $50 \%$. By what percentage would the total number of... | Solution. First method. If the quantity of each type of apple were three times as much, the total number of apples would increase by $200 \%$. Of this increase, $70 \%$ is due to Antonovka, and $50 \%$ is due to Grushovka. Therefore, the increase due to White Naliv would be $200 \%-70 \%-50 \%=80 \%$.
Second method. S... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,674 |
5. Three friends gathered - Yaroslav, Kirill, and Andrey. One of them always tells the truth, another always lies, and the third one is a trickster, sometimes telling the truth and sometimes lying. When asked: "Who is Kirill?" - the boys answered as follows:
Yaroslav: "Kirill is a liar."
Kirill: "I am a trickster!"
... | Solution. Andrei cannot be honest, because then Kirill would also be honest, which is impossible. Therefore, Andrei is either a liar or a trickster. Let's assume Andrei is a liar. Then Kirill is a trickster (he cannot be a liar, as Andrei is the liar, and he cannot be honest, since his statement would then be false). B... | honest-Yaroslav,trickster-Andrei,liar-Kirill | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,675 |
3. Three mryak are more expensive than five bryak by 10 rubles. And six mryak are more expensive than eight bryak by 31 rubles. By how many rubles are seven mryak more expensive than nine bryak | 3. Answer: by 38 rubles. Solution. By adding 3 bryaks and myraks, we increase the price difference between myraks and bryaks by 21 rubles. This means that one myrak is 7 rubles more expensive than one bryak. Then seven myraks are more expensive than nine bryaks by $31+7=38$ rubles. Criteria: correct solution - 7 points... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,678 |
4. Two very small fleas are jumping on a large sheet of paper. The first jump of the fleas is along a straight line towards each other (their jumps may have different lengths). The first flea first jumps to the right, then up, then to the left, then down, then to the right again, and so on. Each jump is 1 cm longer tha... | 4. Answer: 2 meters. Solution: Let's observe the first four jumps of the first flea: 2 cm more to the left than to the right, and 2 cm more down than up. That is, after four jumps, the first flea moves 2 cm to the left and 2 cm down. Therefore, after 100 jumps, it will move 50 cm to the left and 50 cm down. Similarly, ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,679 |
10.1. The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13. Find the 2013-th term of the sequence. | Answer: 130
Solution. Let's calculate the first few terms of the sequence. We get: $a_{1}=934 ; a_{2}=16 \times 13=208$; $a_{3}=10 \times 13=130 ; a_{4}=4 \times 13=52 ; a_{5}=7 \times 13=91 ; a_{6}=10 \times 13=130=a_{3}$. Since each subsequent number is calculated using only the previous number, the terms of the seq... | 130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,681 |
10.2. The roots of the quadratic trinomial $f(x)=x^{2}+b x+c$ are $m_{1}$ and $m_{2}$, and the roots of the quadratic trinomial $g(x)=$ $=x^{2}+p x+q$ are $k_{1}$ and $k_{2}$. Prove that $f\left(k_{1}\right)+f\left(k_{2}\right)+g\left(m_{1}\right)+g\left(m_{2}\right) \geqslant 0$. | Solution. Let: $A=f\left(k_{1}\right)+f\left(k_{2}\right)+g\left(m_{1}\right)+g\left(m_{2}\right)$.
First method. Express $A$ through the roots of the given trinomials.
Since $f(x)=x^{2}+b x+c=\left(x-m_{1}\right)\left(x-m_{2}\right), g(x)=x^{2}+c x+d=\left(x-k_{1}\right)\left(x-k_{2}\right)$, then $f\left(k_{1}\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,682 |
10.3. Point $F$ is the midpoint of side $B C$ of square $A B C D$. A perpendicular $A E$ is drawn to segment $D F$. Find the angle $C E F$.
---
Translation provided as requested, maintaining the original format and line breaks. | Answer: $45^{\circ}$.
Solution. Let the line $A E$ intersect the side $C D$ of the square at point $M$ (see Fig. 10.3). Then triangles $A D M$ and $D C F$ are equal (by the leg and acute angle). Therefore, point $M$ is the midpoint of side $C D$. Then triangle $C F M$ is a right and isosceles triangle, so $\angle C M ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,683 |
10.4. Find the maximum value of the expression $a+b+c+d-ab-bc-cd-da$, if each of the numbers $a, b, c$ and $d$ belongs to the interval $[0 ; 1]$. | Answer: 2.
Solution. The first method. The value 2 is achieved, for example, if $a=c=1, b=d=0$. We will prove that with the given values of the variables $a+b+c+d-ab-bc-cd-da \leqslant 2$.
Notice that $a+b+c+d-ab-bc-cd-da=(a+c)+(b+d)-(a+c)(b+d)$. Let $a+c=x, b+d=y$, then we need to prove that $x+y-xy \leqslant 2$, if... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,684 |
10.5. On side $AB$ of triangle $ABC$, point $K$ is marked, and on side $AC$, point $M$ is marked. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find angle $BAC$. | Answer: $60^{\circ}$.
Solution. To both sides of the equality $S_{A K P M}=S_{B P C}$, add the area of triangle $B P K$ (see Fig. 10.5). We get that $S_{A B M}=S_{B C K}$. Therefore, $\frac{1}{2} B C \cdot B K \sin \angle B = \frac{1}{2} A B \cdot A M \sin \angle A$. Then $\frac{B K}{A M} = \frac{A B \sin \angle A}{B ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,685 |
10.6. In the cells of a $9 \times 9$ table, all natural numbers from 1 to 81 were placed. The products of the numbers in each row of the table were calculated, resulting in a set of nine numbers. Then, the products of the numbers in each column of the table were calculated, also resulting in a set of nine numbers. Coul... | Answer: No, they could not.
Solution. Each of the products of the numbers in the nine rows of the table can be represented as a product of prime factors. Let's list all prime numbers greater than 40 but less than 81: 41, 43, 47, 53, 59, 61, 67, 71, 73, 79. Note that each of these ten numbers can only appear in one of ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,686 |
114. Find the geometric locus of points $M$ on the coordinate plane such that two tangents drawn from $M$ to the parabola $y=x^{2}$ are perpendicular to each other. | 114. Find the geometric locus of points \(M\) on the coordinate plane such that two tangents drawn from \(M\) to the parabola \(y = x^2\) are perpendicular to each other.
Answer: the line \(y = -\frac{1}{4}\). Hint: Let \((a; b)\) be the coordinates of point \(M\), and \(M_{1}(x_{1}; x_{1}^{2}), M_{2}(x_{2}; x_{2}^{2}... | -\frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,688 |
1. Can the signs «+» or «-» be placed between every two adjacent digits of the number 20222023 so that the resulting expression equals zero? | Solution. Since among the digits of the given number there is only one (an odd number) odd digit, we will always get an odd sum when placing the signs «+» or «-» in any arrangement. Zero is an even number. Answer: it is not possible. | itisnotpossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,690 |
2. In the class, there are 39 students, all of whom were born in 2009. Is there a month in the year in which at least 4 students from this class celebrate their birthday? | Solution. Suppose that no such month can be found, then in each month, no more than three students have their birthdays. However, in this case, there would be no more than $3 \times 12=36$ students in the class. This contradiction proves that there will be a month in which at least four students celebrate their birthda... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,691 |
3. Cut the figure shown in the diagram into two parts that are identical in size and shape. You can cut along the grid lines and along the diagonals of the squares. Two parts are considered identical if, after cutting, they can be placed on top of each other so that they fit exactly. They can be rotated and flipped.
!... | Solution. One of the options is shown in the figure below.
 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,692 |
4. At the School of Sorcery and Wizardry, there are 13 students. Before the Divination exam, the teacher seated them at a round table and asked them to guess who would receive the Seer's diploma. Everyone modestly kept quiet about themselves and their two neighbors, but wrote about everyone else: "None of these ten wil... | Solution. Suppose no one received a diploma. Then the statement of each student is true. In this case, everyone should have received a diploma - a contradiction. Therefore, at least one of the students received a diploma of a seer. He told the truth, so no one except his neighbors received a diploma. If both neighbors ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,693 |
5. There was a whole number of cheese heads on the kitchen. At night, rats came and ate 10 heads, and everyone ate equally. Several rats got stomachaches from overeating. The remaining seven rats the next night finished off the remaining cheese, but each rat could eat only half as much cheese as the night before. How m... | Solution. Let there be $k$ rats in total $(k>7)$, then each rat ate $\frac{10}{k}$ pieces of cheese on the first night. On the second night, each rat ate half as much, that is, $\frac{5}{k}$ pieces. Then seven rats ate $\frac{35}{k}$ pieces. This is an integer. The only divisor of the number 35 that exceeds 7 is the nu... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,694 |
1. In an equilateral triangle, 3 points were marked on each side so that they divide the side into 4 equal segments. These points and the vertices of the triangle were painted. How many isosceles triangles exist with vertices at the painted points? | # Answer. 20
Solution. Let the vertices of the triangle be denoted as A, B, C. Suppose points M, N (M closer to A) are taken on side AB, points P, Q (P closer to A) on side AC, and points K, L (K closer to B) on side BC. Then the following cases are possible for isosceles triangles with the vertex at point A: ABC, AKL... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,695 |
2. Prove that in any right-angled triangle, the lengths of the legs \(a\) and \(b\) and the hypotenuse \(c\) satisfy the inequality \(a^{4} + b^{4} < c^{4}\). | Solution. Write the inequality as: $\left(\frac{a}{c}\right)^{4}+\left(\frac{b}{c}\right)^{4}<1$
For the sides of a right triangle, the following is true: $\left(\frac{a}{c}\right)^{2}<1,\left(\frac{b}{c}\right)^{2}<1,\left(\frac{a}{c}\right)^{2}+$ $\left(\frac{b}{c}\right)^{2}=1$. Multiply both sides of the first ine... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,696 |
3. From the set of natural numbers, the following sequence was formed: $1, 2+3, 4+5+6, 7+8+9+10$, and so on, where each subsequent sum has one more addend. What is the 2023rd term of the sequence? | Answer: $\frac{2023 *\left(2023^{2}+1\right)}{2}=4139594095$.
Solution. Let the $n$-th group start with the number $k$. Then these numbers are:
$k, k+1, \ldots k+(n-1)$ - a total of $n$ numbers. Their sum is the sum of an arithmetic progression with the first term equal to $k$, a common difference of 1, and can be ca... | \frac{2023\cdot(2023^2+1)}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,697 |
4. $\quad$ in triangle $ABC$ the sine of angle $A$ is equal to $3/5$. On side $AC$ a point $M$ is taken such that $CM=15$, on side $AB$ a point $N$ is taken such that $BN=7, AN=AM, T$ is the midpoint of $NC$, $P$ is the midpoint of $B$. Find the length of segment $PT$. | Answer: $\sqrt{26.5}, \sqrt{110.5}$
Solution. Let the length $\mathrm{AN}=\mathrm{AM}=\mathrm{x}$. Introduce a system of two unit vectors: let vector $\mathbf{~collinear~with~vector~} \mathrm{AB}$, and vector $\mathbf{c}$ be collinear with vector $\mathrm{AC}$. Then the following vector relations hold:
$$
\begin{gath... | \sqrt{26.5},\sqrt{110.5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,698 |
6. Find at least one solution to the equation
$$
567 x^{3}+171 x^{2}+15 x-777 \ldots 7555 \ldots 5333 \ldots 3=0
$$
where the constant term contains 2023 sevens, 2023 fives, and 2023 threes. | Answer. $\quad 111 \ldots 1_{w} 2023$.
Solution. Let the number whose decimal representation consists of 2023 ones be denoted by $x$ and note the following property of this number:
$$
111 \ldots 1_{\longleftarrow} 2023=x, \quad 9 x+1=10^{2023}
$$
Represent the number $777 \ldots 7555 \ldots 5333 \ldots 3$ as follows... | 111\ldots1_{w}2023 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,700 |
1. Is it possible to place 0 or 1 in each cell of a $5 \times 5$ square table so that the sum in each $2 \times 2$ square is divisible by 3, while both zeros and ones are present in the table? | Answer: Yes, it is possible.
Solution. For example, as follows:
| 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
Note. This arrangement is not unique.
Criteria. A correct arrangement without an explanation of how it is... | Yes,itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,701 |
2. At the Olympiad, for each solved problem, one could receive 3, 8, or 10 points. Vasya scored 45 points. What is the smallest number of problems he could have solved? (It is necessary to explain why he could not have solved fewer problems.) | Answer: 6 problems. Solution. Since 8 and 10 are even numbers, and 45 is an odd number, there must be a problem worth 3 points. It is impossible to score 42 points with 4 problems, even if each problem is worth 10 points. Therefore, a minimum of 5 problems are needed to sum up to 42 points, making the total number of p... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,702 |
11.1 From a three-digit number $A$, which does not contain zeros in its notation, a two-digit number $B$ was obtained by writing the sum of the first two digits instead of them (for example, the number 243 turns into 63). Find $A$ if it is known that $A=3 B$. | Answer: 135.
Solution: The last digit of the numbers $B$ and $A=3B$ is the same, so this digit is 5. Moreover, $A$ is divisible by 3, which means $B$ is also divisible by 3 (the sums of the digits are the same). Therefore, $B$ is one of the numbers $15, 45, 75$. By checking, we find that the number $B=45$ satisfies th... | 135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,703 |
11.2. For the number $x$, the inequalities $\sin x<\cos \frac{x}{2}<0$ are satisfied. Prove that $\cos x<\frac{1}{2}$. | Solution. Dividing the given inequality by the negative number $\cos \frac{x}{2}$, we get that $\sin \frac{x}{2}>\frac{1}{2}$. Therefore, $\cos x=1-2 \sin ^{2} \frac{x}{2}<\frac{1}{2}$.
Comment. Inequivalent transformation of the inequality - 0 points for the problem. | \cosx<\frac{1}{2} | Inequalities | proof | Yes | Yes | olympiads | false | 15,704 |
11.3. In the triangular pyramid $S A B C$, the heights $A A_{1}$ and $B B_{1}$ are drawn. It is known that the segment $A_{1} B_{1}$ is parallel to the edge $A B$. Prove that some two faces of the pyramid have the same area. | Solution. From the parallelism of the given segments, it follows that they lie in the same plane. Then, in this same plane lie the heights $A A_{1}$ and $B B_{1}$ of the pyramid. Let $P$ be the point of intersection of this plane with the edge $S C$. The line $A A_{1}$ is perpendicular to the plane $B S C$, so the line... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,705 |
11.4. Lines $l: y=k x+b, l_{1}: y=k_{1} x+b_{1}$ and $l_{2}: y=k_{2} x+b_{2}$ are tangent to the hyperbola $y=\frac{1}{x}$. It is known that $b=b_{1}+b_{2}$. Prove that $k \geq 2\left(k_{1}+k_{2}\right)$. | Solution. The condition of tangency means that the equation $\frac{1}{x}=k x+b$, that is, $k x^{2}+b x-1=0$ (and two similar equations) has a unique solution, which is equivalent to the discriminant being zero: $b^{2}+4 k=0$. Then $k=-\frac{1}{4} b^{2}$, similarly $k_{1}=-\frac{1}{4} b_{1}^{2}, k_{2}=-\frac{1}{4} b_{2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,706 |
11.5. Given a natural number $K>2$ and a set of $N$ cards, on which positive numbers are written. It turned out that from them one can choose several cards (possibly one) with the sum of numbers $K$, several cards with the sum of numbers $K^{2}, \ldots$, several cards with the sum of numbers $K^{K}$. Could it be that $... | Answer: It could not.
Solution: Suppose that $NK^{K} / K=K^{K-1}$. This means there is a card with a number greater than $K^{K-1}$. This card cannot participate in sums from $K$ to $K^{K-1}$. By considering cards with the sum $K^{K-1}$, we can show that there is a card with a number greater than $K^{K-2}$ but not grea... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,707 |
9.1. Given the expression $\frac{1}{x}+\frac{1}{y}$, where $x$ and $y$ are natural numbers. If the number $x$ is increased by 2, and the number $y$ is decreased by 2, the value of this expression does not change. Prove that $xy+1$ is a perfect square. | Solution. According to the condition $\frac{1}{x}+\frac{1}{y}=\frac{1}{x+2}+\frac{1}{y-2}$, hence $\frac{x+y}{x y}=\frac{x+y}{(x+2)(y-2)}$. Since $x+y$ is positive, then $x y=(x+2)(y-2)$. From this, $y=x+2$. Then $x y+1=x(x+2)+1=(x+1)^{2}$.
Comment. It is proven that $y=x+2-2$ points. | (x+1)^2 | Algebra | proof | Yes | Yes | olympiads | false | 15,708 |
9.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 4 blue points. And on any segment with endpoints at blue points, containing 3 blue points inside, there are at least 2 red point... | Answer: 4.
Solution: Note that on a segment with endpoints at red points, not containing other red points, there cannot be 5 blue points. Indeed, in this case, between the outermost blue points, there would be 3 blue points, which means there would be at least 2 more red points. Therefore, on such a segment, there are... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,709 |
9.3. The numbers $x$ and $y$ satisfy the inequalities $x^{3}>y^{2}$ and $y^{3}>x^{2}$. Prove that $y>1$. | Solution. The right-hand sides of the inequalities are non-negative, so both numbers $x$ and $y$ are positive. Multiplying the given inequalities and canceling the positive number $(x y)^{2}$, we get: $x y > 1$. This means that at least one of the (positive!) numbers $x$ and $y$ is greater than 1. If $y > 1$, the state... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,710 |
9.4. In the castle, there are 9 identical square rooms, forming a $3 \times 3$ square. Nine people, consisting of liars and knights (liars always lie, knights always tell the truth), each occupied one of these rooms. Each of these 9 people said: "At least one of the neighboring rooms to mine is occupied by a liar." Wha... | Answer: 6 knights.
Solution: Note that each knight must have at least one neighbor who is a liar. We will show that there must be at least 3 liars (thus showing that there are no more than 6 knights). Suppose there are no more than 2 liars, then there will be a "vertical row" of rooms where only knights live. But then... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,711 |
9.5. On the given circle $\omega$, a fixed point $A$ is chosen. We select two arbitrary points $B$ and $C$ on the circle, and find the point $D$ where the angle bisector of $\angle ABC$ intersects the circle $\omega$. Let $K$ be the point such that $D$ is the midpoint of segment $AK$. The line $KC$ intersects the circl... | The first solution. We will show that point $P$ is diametrically opposite to point $A$. This will mean that $P$ does not depend on the choice of points $B$ and $C$. Thus, we need to prove that $PD$ is perpendicular to $AK$. We will assume that point $C$ lies on the segment $KR$. The other case is considered similarly. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,712 |
1.1. The lines containing the bisectors of the exterior angles of a triangle with angles of 42 and 59 degrees intersected pairwise and formed a new triangle. Find the degree measure of its largest angle. | Answer: 69
Solution. Let the vertices of the given triangle be $A, B, C$, and its angles be $\alpha, \beta$, and $\gamma$, respectively. Additionally, let $A_{1}, B_{1}$, and $C_{1}$ be the points of intersection of the external angle bisectors of angles $B$ and $C$, $A$ and $C$, and $A$ and $B$, respectively. Then, $... | 69 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,713 |
2.1. For what largest $k$ can we assert that in any coloring of $k$ cells in black in a white $7 \times 7$ square, there will necessarily remain a completely white $3 \times 3$ square with sides along the grid lines? | Answer: 3
Solution. Let's highlight four $3 \times 3$ squares that are adjacent to the corners of the $7 \times 7$ square. These squares do not overlap, so if no more than three cells are sha... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,714 |
3.1. A pedestrian left point $A$ for point $B$. At the same time, a cyclist left point $B$ for point $A$. After one hour, the pedestrian was three times farther from the cyclist than from point $A$. Another 30 minutes later, they met, after which both continued their journey. How many hours did it take the pedestrian t... | Answer: 9
Solution. Let the distance from $A$ to $B$ be 1 km, the pedestrian moves at a speed of $x$ km/h, and the cyclist at $y$ km/h. Then in one hour, the pedestrian has walked $x$ km, the cyclist has traveled $y$ km, and the distance between them is $1-x-y$, which should be three times $x$. Therefore, $3 x=1-x-y, ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,715 |
4.1. Petya wrote a natural number $A$ on the board. If it is multiplied by 8, the result is the square of a natural number. How many three-digit numbers $B$ exist such that $A \cdot B$ is also a square of a natural number? | # Answer: 15
Solution. If $8 A$ is the square of a natural number, then any prime number greater than 2 must appear in $A$ with an even exponent, and the number 2 must appear with an odd exponent. Therefore, in $B$, any prime number greater than 2 must also appear with an even exponent, and the number 2 must appear wi... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,716 |
5.1. How many six-digit numbers exist that consist only of the digits 1 and 2, given that each of them appears? | Answer: 62
Solution. There are a total of $64=2^{6}$ six-digit numbers consisting of the digits 1 and 2, since there are 6 positions and for each position, there are two options to place a digit. However, two numbers - the one consisting only of ones and the one consisting only of twos - do not satisfy the condition, ... | 62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,717 |
6.1. Coins come in denominations of 50 kopecks, 1 ruble, 2 rubles, 5 rubles, 10 rubles. $V$ has several coins in his wallet. It is known that no matter which 20 coins are taken out of the wallet, there will be at least one 1-ruble coin, at least one 2-ruble coin, and at least one 5-ruble coin. What is the maximum numbe... | Answer: 28
Solution. Example: 9 coins of 1 ruble, 9 coins of 2 rubles, 9 coins of 5 rubles, and 1 coin of 10 rubles. Note that the wallet contains a total of $9+9+1=19$ coins that are not 1 ruble, so among any 20 coins, there will definitely be a 1-ruble coin. Similarly, this can be verified for all other denomination... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,718 |
7.1. In a football tournament, 20 teams participated. Each team played one match against each other. A team receives 3 points for a win, 1 point for a draw, and 0 points for a loss. After the tournament, Vasya calculated the total points scored by the teams. It turned out to be 500. How many matches ended in a draw? | Answer: 70
Solution. Each team played 19 matches, so the total number of matches played was $20 \cdot 19 / 2=190$ (we divide by two because each match is counted twice). If there were no draws, the total points would be $3 \cdot 190=570$. Each draw results in one point less in the total compared to a match that ends w... | 70 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,719 |
8.1. The real number a is such that $2a - 1/a = 3$. What is the value of $16a^4 + 1/a^4$? | Answer: 161
Solution. Squaring the equality from the condition, we get $4 a^{2}-4+1 / a^{2}=9$, that is, $4 a^{2}+1 / a^{2}=13$. Squaring again, we obtain $16 a^{4}+8+1 / a^{4}=169$. Therefore, $16 a^{4}+1 / a^{4}=161$. | 161 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,720 |
8-1. Kolya made a figure from four identical blocks as shown in the picture. What is the surface area of this figure? Express your answer in square centimeters.
 | Answer: 64.
Solution. The surface area of one block is 18 cm². Out of this area, 2 cm² is "lost" at the joints with other blocks, leaving a total area of $18-2=16$ cm². Since there are 4 blocks, the answer is $4 \cdot 16=64$ cm². | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,721 |
8-2. On a distant planet in the mangrove forests, there lives a population of frogs. The number of frogs born each year is one more than the difference between the number of frogs born in the previous two years (the larger number minus the smaller number).
For example, if 5 frogs were born last year and 2 in the year ... | Answer: 2033.
Solution: This is a fairly simple problem where the only thing you need to do is correctly understand the condition and apply it the necessary number of times. The ability to correctly understand the condition is an important skill in solving problems!
By applying the algorithm described in the conditio... | 2033 | Other | math-word-problem | Yes | Yes | olympiads | false | 15,722 |
8-3. Olya bought three gifts and packed them in three rectangular boxes: blue, red, and green. She tried to place these gifts in different ways: one on the table, and two on top of each other on the floor. Some distances are given in the diagram. Find the height of the table $h$. Express your answer in centimeters.
choose a... | AE:ED=2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,731 |
11.4. We will call a natural number interesting if it is the product of exactly two (distinct or equal) prime numbers. What is the greatest number of consecutive numbers, all of which are interesting | Solution. One of four consecutive numbers is divisible by 4. However, among the numbers divisible by 4, only the number 4 itself is interesting. But the numbers 3 and 5 are not interesting, so four consecutive interesting numbers do not exist. An example of three consecutive interesting numbers: $33,34,35$. Answer: thr... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,732 |
11.5. Functions $f(x)$ and $g(x)$ are defined for all $x$ in the interval $(2,4)$ and satisfy the conditions: $2<f(x)<4, 2<g(x)<4, f(g(x))=g(f(x))=x$, $f(x) g(x)=x^{2}$ for all $x \in(2,4)$. Prove that $f(3)=g(3)$. | Solution. Let $a \in(2,4)$. Set $x_{1}=a$ and construct the sequence $\left\{x_{n}\right\}$ according to the rule $x_{n+1}=f\left(x_{n}\right)$. Then $g\left(x_{n+1}\right)=g\left(f\left(x_{n}\right)\right)=x_{n}$, and therefore $x_{n+2} x_{n}=f\left(x_{n+1}\right) g\left(x_{n+1}\right)=x_{n+1}^{2}$. This means that th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,733 |
1. Positive numbers $\mathrm{x}$, y and $\mathrm{z}$ satisfy the inequality $\mathrm{xyz} \geq \mathrm{xy}+\mathrm{yz}+\mathrm{zx}$. Prove the inequality $\sqrt{x y z} \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$. | # 1. Solution.
By the inequality of means, we have
$x y+x z \geq 2 \sqrt{x y \cdot x} z$
$\mathrm{xy}+\mathrm{yz} \geq 2 \sqrt{\mathrm{xy} \cdot y z}$
$\mathrm{xy}+\mathrm{yz} \geq 2 \sqrt{\mathrm{xz} \cdot y z}$
Add these inequalities and divide the result by 2.
$\mathrm{xyz}>\mathrm{xy}+\mathrm{xz}+\mathrm{yz}>... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,734 |
2. The bisectors $\mathrm{AM}$ and $\mathrm{CN}$ of triangle $\mathrm{ABC}$ intersect at point L. The areas of triangle ALC and quadrilateral MLNB are equal. Prove that the lengths of the sides of triangle $\mathrm{ABC}$ form a geometric progression. | 2. Solution.
By the condition $S_{\text {ALC }}=S_{\text {mLNв }}$
Adding $S_{\text {сцм, }}$ to both sides of this equality, we get $S_{\text {дмс }}=S_{\text {слв }}$. On the other hand, a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,735 |
3. Given a rectangular grid of size 1 x 60. In how many ways can it be cut into grid rectangles of size 1 x 3 and 1 x 4? | 3. Answer: 45665.
Solution. Let $x$ be the number of rectangles of size 1 x $3, y$ be the number of rectangles of size $1 \times 4$. Then $3 x+4 y=60$, where $x, y$ are non-negative integers. The value $\mathrm{x}=\frac{60-4 \mathrm{y}}{3}=20-\frac{4 \mathrm{y}}{3}$ is an integer if $\mathrm{y}=3 \mathrm{t}(\mathrm{t}... | 45665 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,736 |
4. On the board, there are 2010 non-zero numbers $\mathrm{a}_{1}$, $\mathrm{a}_{2}, \ldots \mathrm{a}_{2010}$ and the products of all pairs of adjacent numbers: $\mathrm{a}_{1} \cdot \mathrm{a}_{2}, \mathrm{a}_{2} \cdot \mathrm{a}_{3}, \ldots \mathrm{a}_{2009} \cdot \mathrm{a}_{2010}$. What is the maximum number of neg... | 4. Answer: 3014.
Solution. Consider the triples of numbers ( $\left.\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{1} \cdot \mathrm{a}_{2}\right)$ ), ( $\left.\mathrm{a}_{3}, \mathrm{a}_{4}, \mathrm{a}_{3} \cdot \mathrm{a}_{4}\right), \ldots,\left(\mathrm{a}_{2009}, \mathrm{a}_{2010}, \mathrm{a}_{2009} \cdot \mathrm{a}_{... | 3014 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,737 |
5. Vasya came up with a new ship for a naval battle - the "combat donut." This ship consists of all the cells of a 3 x 3 square, except for its central cell. On an 8 x 8 field, one combat donut was placed. What is the minimum number of shots needed to guarantee hitting it? | 5. Answer: 8 shots.
Solution. Note that if the donut is placed on a 4x4 field, one shot is not enough to guarantee hitting it. Indeed, if the shot is fired into a cell adjacent to the side of the square, the donut can be placed next to the opposite side. If the shot is fired into one of the four central cells of the s... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,738 |
11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the required trio of people. How man... | Answer: 33.
Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,743 |
8.1. Is it possible to place natural numbers in the cells of a $7 \times 7$ table so that the sum of the numbers in any $2 \times 2$ square and any $3 \times 3$ square is odd? | Answer: No.
Solution. Suppose it is possible. Consider a square with a side of 6 cells. Since it can be divided into four $3 \times 3$ squares, the sum of the numbers in this square will be even. On the other hand, this same square can be divided into nine $2 \times 2$ squares, so this same sum will turn out to be odd... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,744 |
8.2. The part of the graph of a linear function located in the second coordinate quadrant, together with the coordinate axes, forms a triangle. By what factor will its area change if the slope of the function is doubled and the y-intercept is halved? | Answer: It will decrease by eight times.
Solution. Let the original linear function be defined by the equation $y=k x+b$. From the condition of the problem, it follows that $k>0$ and $b>0$ (see Fig. 8.2). The points of intersection of its graph with the axes are: $A(0 ; b)$ and $B\left(-\frac{b}{k} ; 0\right)$. Since ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,745 |
8.3. The height $CH$, dropped from the vertex of the right angle of triangle $ABC$, bisects the bisector $BL$ of this triangle. Find the angle $BAC$.
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The provided text has been translated from Russian to English, maintaining the original formatting and structure. | Answer: $30^{\circ}$.
Solution. Let $C H$ and $B L$ intersect at point $K$ (see Fig. 8.3). Then $C K$ is the median of the right triangle $B C L$, drawn to the hypotenuse, so $C K = 0.5 B L = B K$. Therefore, $\angle K C B = \angle K B C = \angle K B H$. Since the sum of these three angles is $90^{\circ}$ (from triang... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,746 |
8.4. On the Island of Liars and Knights, a circular arrangement is called correct if each person standing in the circle can say that among their two neighbors, there is a representative of their tribe. Once, 2019 natives formed a correct circular arrangement. A liar approached them and said: "Now we can also form a cor... | Answer: 1346.
Solution. We will prove that a correct arrangement in a circle is possible if and only if the number of knights is at least twice the number of liars.
Indeed, from the problem's condition, it follows that in such an arrangement, each liar has two knights as neighbors, and among the neighbors of any knig... | 1346 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,747 |
8.5. For a natural number $N$, all its divisors were listed, then the sum of digits for each of these divisors was calculated. It turned out that among these sums, all numbers from 1 to 9 were found. Find the smallest value of $N$. | Answer: 288.
Solution. Note that the number 288 has divisors $1,2,3,4,32,6,16,8,9$. Therefore, this number satisfies the condition of the problem. We will prove that there is no smaller number that satisfies the condition.
Indeed, since $N$ must have a divisor with the sum of digits 9, $N$ is divisible by 9. Now cons... | 288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,748 |
8.6. Inside an acute angle, there is a convex quadrilateral $A B C D$. It turns out that for each of the two lines containing the sides of the angle, the following condition is satisfied: the sum of the distances from vertices $A$ and $C$ to this line is equal to the sum of the distances from vertices $B$ and $D$ to th... | Solution. Let the line $m$ contain one of the sides of the given angle, and let $a$ and $c$ be the distances from vertices $A$ and $C$ to $m$, and $P$ be the midpoint of segment $AC$. Then the distance from $P$ to $m$ is $\frac{a+c}{2}$ (the midline of the trapezoid, see Fig. 8.6). Similarly, if $Q$ is the midpoint of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,749 |
8.1. Four girls were given candies. Masha said: "Katy and I have 12 more candies than Lena and Olya," and Katy said: "Lena and I have 7 fewer candies than Masha and Olya." Prove that one of the girls was wrong. | First solution. Suppose that none of the girls made a mistake. Then the total number of candies that Masha and Katya have is of the same parity as the total number of candies that Lena and Olya have (they differ by an even number 12). Therefore, the total number of candies that all four girls have is even. Similarly, w... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 15,750 |
8.2. During breaks, schoolchildren played table tennis. Any two schoolchildren played no more than one game with each other. By the end of the week, it turned out that Petya played half, Kolya - a third, and Vasya - a fifth of the total number of games played during the week. How many games could have been played durin... | Answer: 30.
Solution: From the condition, it follows that half, a third, and a fifth of the total number of games played are integers. Since the least common multiple of the denominators - the numbers 2, 3, 5 - is 30, the total number of games played is also a multiple of 30. Let this number be $30p$. Then Pete played... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,751 |
8.3. On the sides $AB$, $BC$, and $CA$ of triangle $ABC$, points $D$, $E$, and $F$ are chosen respectively such that $BE = BD$ and $AF = AD$. It is known that $ED$ is the bisector of angle $BEF$. Prove that $FD$ is the bisector of angle $AFE$. | Solution. From the equality of sides $B E$ and $B D$ of triangle $D B E$, it follows that $\angle B D E = \angle B E D$ (see Fig. 2). However, by the condition, $\angle F E D = \angle B E D$. Therefore, $\angle F E D = \angle B D E$. This means that lines $B A$ and $E F$ are parallel. Then $\angle E F D = \angle A D F$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,752 |
8.5. On an $8 \times 8$ chessboard, $k$ rooks and $k$ knights are placed such that no figure attacks any other. What is the largest $k$ for which this is possible? | Answer: 5.
Solution: From the condition, it follows that in one row (column) with a rook, no other figure can stand.
Suppose 6 rooks were placed on the board. Then they stand in 6 rows and 6 columns. Therefore, only 4 unpicked cells will remain (located at the intersection of two empty rows and two empty columns). It... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,754 |
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