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9.2. In the table, it is allowed to swap any two rows with each other and any two columns. Can we obtain the right table from the left one using several such operations?
| 1 | 2 | 3 | 1 | 4 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 4 | 5 | 6 | 2 | 5 | 8 |
| 7 | 8 | 9 | 3 | 6 | 9 | | Answer: No.
Solution. Note that when two rows or two columns are swapped, the numbers 1 and 2 remain in the same row. In the second table, this is not the case, so it cannot be obtained. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,524 |
9.3. Can the three parabolas shown in the figure be the graphs of the functions $y=\mathrm{a} x^{2}+b x+c, y=c x^{2}+a x+b$, $y=b x^{2}+c x+a$ for some $a$, $b, c$?
 | Answer: No.
Solution. Suppose the opposite. From the figure, it is clear that all quadratic trinomials have two roots, hence $b^{2}>4 c a, a^{2}>4 b c$ and $c^{2}>4 a b$, and $a >0, \mathrm{~b}>0, \mathrm{c}>0$ (the parabolas open upwards). Multiplying the obtained inequalities, we arrive at a contradiction: $a^{2} b^... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,525 |
9.4. There is a set of 2021 numbers. Moreover, it is known that if each number in the set is replaced by the sum of the others, the same set will be obtained. Prove that the set contains a zero. | Solution. Let the sum of the numbers in the set be $M$, then the number $a$ in the set is replaced by the number $b=M-a$. Summing these equations for all $a$:
$$
b_{1}+\ldots+b_{2021}=2021 M-\left(a_{1}+\ldots+a_{2021}\right)
$$
from which $M=0$, since $b_{1}+\ldots+b_{2021}=a_{1}+\ldots+a_{2021}=M$. Therefore, for a... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 18,526 |
9.5. On the hypotenuse $AB$ of the right triangle $ABC$, point $K$ is the midpoint, and on the leg $BC$, point $M$ is such that $BM: MC = 2: 1$. Let $P$ be the point of intersection of segments $AM$ and $CK$. Prove that the line $KM$ is tangent to the circumcircle of triangle $AKP$. | Solution. Let $T$ be the midpoint of segment $M B$, then by the converse of Thales' theorem, lines $A M$ and $K T$ are parallel, so $\angle K A M=\angle B K T$. Since point $K$ is the midpoint of the hypotenuse of the right triangle $A B C$, then $K C=K B$, and therefore $\angle K C B=\angle K B C$. Then triangles $T B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,527 |
9.1. Can twelve different integers be found, among which there are exactly six prime numbers, exactly nine odd numbers, exactly ten non-negative numbers, and exactly seven numbers greater than ten? | Solution. Yes, for example: $-8,-4,2,5,9,11,13,15,21,23,37,81$.
Comments. Any correct example is given - $\underline{7 \text{ points. }}$ | -8,-4,2,5,9,11,13,15,21,23,37,81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,528 |
9.2. Toshа travels from point A to point B via point C. From A to C, Toshа travels at an average speed of 75 km/h, and from C to B, Toshа travels at an average speed of 145 km/h. The entire journey from A to B took Toshа 4 hours and 48 minutes. The next day, Toshа travels back at an average speed of 100 km/h. The journ... | Answer: 290 km.
Solution. Let $\mathrm{X}$ km be the distance between B and C, and $y$ km be the distance between A and C. From the system of equations $x / 145 + y / 75 = 24 / 5$ and $(x + y) / 100 = 2 + y / 70$, we find: $x = 290$ and $y = 210$.
. Two chips are called "connected" if they are in the same column or row, and there are no other chips between them. What is the maximum number of chips that can be placed on the board so that each has no more than two "con... | Answer: 35.
Solution. Let $x_{1}, x_{2}, \ldots, x_{20}$ be the number of chips in rows $1,2, \ldots, 20$, and $y_{1}, y_{2}, \ldots, y_{15}$ be the number of chips in columns $1,2, \ldots, 15$. Then the total number of chips $S=x_{1}+x_{2}+\ldots+x_{20}=y_{1}+y_{2}+\ldots+y_{15}$. The number of "connectivities" $k \g... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,530 |
1. Variant 1. Petya thought of two numbers and wrote down their product. After that, he decreased the first of the thought numbers by 3, and increased the other by 3. It turned out that the product increased by 900. By how much would the product have decreased if Petya had done the opposite: increased the first number ... | Answer: 918.
Solution: Let these numbers be $a$ and $b$. Then, according to the condition, $(a-3)(b+3)-ab=600$. Expanding the brackets: $ab+3a-3b-9-ab=900$, so $a-b=303$. We need to find the difference $ab-(a+3)(b-3)=$ $ab-ab+3a-3b+9=3(a-b)+9=3 \cdot 303+9=918$. | 918 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,531 |
2. Variant 1. At the intersection of perpendicular roads, a highway from Moscow to Kazan and a road from Vladimir to Ryazan intersect. Dima and Tolya set out with constant speeds from Moscow to Kazan and from Vladimir to Ryazan, respectively. When Dima passed the intersection, Tolya had 3500 meters left to reach it. Wh... | Answer: 9100.
Solution. When Tolya has traveled 3500 meters, Dima will have traveled 4200 meters, so at the moment when Dima is 8400 meters past the intersection, Tolya will be 3500 meters past the intersection. By the Pythagorean theorem, the distance between the boys is 9100 meters. | 9100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,532 |
3. Variant 1. Find the largest root of the equation
$(x+1)(x+2)-(x+2)(x+3)+(x+3)(x+4)-(x+4)(x+5)+\cdots-(x+1000)(x+1001)=0$. | Answer: -501.
Solution. Let's break the terms into pairs of adjacent terms and factor out the common factor, we get $(x+2)(x+1-x-3)+\cdots+(x+1000)(x+999-x-1001)=-2 \cdot(x+2+x+4+\cdots+x+1000)=$ $-2 \cdot\left(500 x+\frac{2+1000}{2} \cdot 500\right)=-1000 \cdot(x+501)=0$. From this, $x=-501-$ the only root. | -501 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,533 |
7. Increase Grisha's catch by $40 \%$ and Vasya's catch by $20 \%$.
Grisha, the most resourceful of them, calculated that in the first case their total catch would increase by 1 kg; in the second case - decrease by 0.5 kg; in the third case - increase by 4 kg. What was the total catch of the friends (in kilograms) bef... | Answer: 15.
Solution: Let the harvest of Vasya, Misha, and Grisha be denoted by $x, y, z$ respectively. Then, $0.1 x + 0.2 y = 1; 0.1 y - 0.1 z = -0.5; 0.4 z + 0.2 x = 4$. Adding the equations, we get $0.3(x + y + z) = 4.5$, from which $x + y + z = 15$. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,534 |
# 5. Variant 1
It is known that $\cos \alpha+\cos \beta+\cos \gamma=\sqrt{\frac{1}{5}}, \sin \alpha+\sin \beta+\sin \gamma=\sqrt{\frac{4}{5}}$. Find $\cos (\alpha-\beta)+\cos (\beta-$ $\gamma)+\cos (\gamma-\alpha)$ | Answer: -1.
Solution. Consider the expression $(\cos \alpha+\cos \beta+\cos \gamma)^{2}+(\sin \alpha+\sin \beta+\sin \gamma)^{2}$ and expand the brackets: $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2(\cos \alpha \cdot \cos \beta+\cos \alpha \cdot \cos \gamma+\cos \beta \cdot \cos \gamma)+\sin ^{2} \alpha+\sin ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,535 |
# 6. Option 1
Initially, there were 20 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{25}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{16}... | Answer: 4.
Solution: Let there be $a$ white balls, $b$ blue balls, and $c$ red balls in the box. We can set up the following equations:
$$
\begin{gathered}
a+b+c=20 \\
\frac{a}{20}=\frac{a}{20+b}+\frac{1}{25} \\
\frac{b}{20}+\frac{1}{16}=\frac{b}{20-a}
\end{gathered}
$$
Transform the second equation: $\frac{a b}{20(... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,536 |
# 7. Variant 1
Unit cubes were used to assemble a large parallelepiped with sides greater than 4. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 6 neighbors is 836. Find the number of cubes that have no more th... | Answer: 144.
Solution. Let $a, b$ and $c$ be the lengths of the sides of the large parallelepiped. Then, the number of cubes with exactly 6 neighbors is: $(a-2)(b-2)(c-2)$. Since each of the factors $a-2, b-2$ and $c-2$ is greater than 2 and their product equals the product of four prime numbers $2,2,11$ and 19, we ha... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,537 |
# 8. Variant 1
At the base of the quadrilateral pyramid $S A B C D$ lies a square $A B C D, S A$ - the height of the pyramid. Let $M$ and $N$ be the midpoints of the edges $S C$ and $A D$. What is the maximum value that the area of triangle $B S A$ can have if $M N=3 ?$ | Answer: 9.
Solution: Let $O$ be the center of the square $ABCD$. Then $MO$ is the midline of the triangle $SAC$, so $SA = 2MO$. Similarly, $ON$ is the midline of the triangle $BDA$, so $AB = 2ON$. Therefore, $SA^2 + AB^2 = 4(MO^2 + ON^2) = MN^2 = 36$. Let $SA = x, AB = y$. From the formula $S = 0.5 \cdot SA \cdot AB =... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,538 |
10.2. Can the vertex of the parabola $y=4 x^{2}-4(a+1) x+a$ be located in the second coordinate quadrant for some value of $a$? | 10.2. Answer: it cannot. Solution. The discriminant of the quadratic trinomial $4 x^{2}-$ $4(a+1) x+a$, which is $16\left(a^{2}+a+1\right)$, is positive for any $a$, since $a^{2}+a+1=(a+0.5)^{2}+0.75$. This means that the given trinomial has two roots for any value of $a$. Since the branches of the parabola $y=4 x^{2}-... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,539 |
10.3. Each point of the plane is painted in one of three colors, and all three colors are used. Is it true that for any such coloring, one can choose a circle on which there are points of all three colors? | 10.3. Answer: correct. Solution. Suppose it is impossible to choose a circle on which there are points of all three colors. Let's select a point $A$ of the first color and a point $B$ of the second color and draw a line $l$ through them. If there is a point $C$ of the third color outside the line $l$, then on the circl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,540 |
10.4. A chord is at a distance $h$ from the center of a circle. In each of the two segments of the circle subtended by this chord, a square is inscribed such that a pair of its adjacent vertices lies on the chord, and the other pair of adjacent vertices lies on the corresponding arc of the circle. Find the difference i... | 10.4. Answer: $\frac{8}{5} h$. Solution. Let the side lengths of the larger and smaller squares be denoted by $2 x$ and $2 y$ respectively, and the radius of the circle by $R$. Then the distances from the center of the circle to the vertices of the inscribed squares, lying on the circle, give the expressions:
$$
(2 x-... | \frac{8}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,541 |
10.5. There are two piles of stones, one with 15 stones and the other with 20. Two players play the following game: they take turns, and on each turn, they can take any number of stones, but only from one pile. The player who cannot make a move loses. Who wins with correct play? | 10.5. Answer: First. Solution. We will describe the winning strategy for the first player. On the first move, he takes 5 stones from the pile that has 20 stones. Thus, after his move, each pile has 15 stones. With each subsequent move, the first player should take the same number of stones as the second player, but fro... | First | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,542 |
7.5 One natural number is 4 more than the other. Prove that their product does not end in 31. | Solution: Let $x$ be the smaller number. Then $x+4$ is the larger number. Suppose their product ends with the digits 31:
$$
x(x+4)=\ldots 31
$$
(dots represent the preceding digits, which are not important to us). Add 4 to both sides of the equation:
$$
x^{2}+4 x+4=\ldots 35
$$
But $x^{2}+4 x+4=(x+2)^{2}$. Thus, th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,543 |
9.7. First Solution. Let $\ell$ be the tangent to the circumcircle at point $B$, and $P$ and $Q$ be the points of intersection of $\ell$ with the rays $T M$ and $T N$, respectively. Denote by $K$ and $L$ the points where the rays $T M$ and $T N$ intersect the sides of the triangle (see Fig. 1).
Notice that quadrilater... | The second solution. We will prove the following well-known lemma.
Lemma. Let $P Q R$ be an equilateral triangle, and let point $W$ be chosen on the smaller arc $P R$ of its circumscribed circle. Then $P W+R W=Q W$.
Proof. Mark a point $V$ on the segment $Q W$ such that $V W = W R$ (see Fig. 3). We have $\angle V W R... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,545 |
9.1. In the village, seven people live. Some of them are liars (always lie), and the rest are knights (always tell the truth). Each of them said about each of the others whether they are a knight or a liar. Out of the 42 answers received, 24 were “He is a liar.” What is the smallest number of knights that can live in t... | Answer: 3.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of knight-liar, the phrase "He is a liar" will be said twice. Since this phr... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,546 |
9.2. Can a square grid $35 \times 35$ be cut into rectangular grid pieces such that the perimeter of each of them is 18, 22, or 26? | Answer: No.
Solution: Suppose such a cutting is possible. Since the perimeters of the rectangles are 18, 22, or 26, the sum of the length and width is 9, 11, or 13. This means that one of these values is even, and the other is odd. Therefore, the area of each rectangle in the cutting will be an even number. Thus, the ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,547 |
9.3. A positive number $a$ is the coefficient of $x^{2}$ in a quadratic trinomial $f(x)$, which has no roots. Prove that for any $x$ the inequality $f(x)+f(x-1)-f(x+1)>-4 a$ holds. | Solution. Let $f(x)=a x^{2}+b x+c$. Then the expression given in the problem has the form: $a x^{2}+b x+c+a(x-1)^{2}+b(x-1)+c-a(x+1)^{2}-b(x+1)-c=a x^{2}-4 a x+b x-2 b+c . \quad$ This expression can be transformed into $a(x-2)^{2}+b(x-2)+c-4 a=f(x-2)-4 a>-4 a$, since from the condition of the problem it follows that $f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,548 |
9.4. Quadrilateral $ABCD (AB > BC)$ is inscribed in circle $\Omega$. It is known that $AD = CD$. Prove that the bisector of angle $ADB$ cuts off an isosceles triangle from angle $BAC$. | Solution. Let the bisector of angle $A D B$ intersect the chords $A C$ and $A B$ at points $M$ and $N$, respectively, and intersect $\Omega$ at point $P$. To prove the statement of the problem, we need to show the equality of angles $A M N$ and $A N M$. From the equality of inscribed angles $A D P$ and $B D P$ (since $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,549 |
9.5. Three hundred non-zero integers are written in a circle such that each number is greater than the product of the next three numbers in the clockwise direction. What is the maximum number of positive numbers that can be among these 300 written numbers? | Answer: 200.
Solution: Note that three consecutive numbers cannot all be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number preceding them (counterclockwise) is also a natural number. Since it is greater than the product of these three natural numbers, it is greater ... | 200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,550 |
9.1. To the eight-digit number 20192020, append a digit on the left and a digit on the right so that the resulting 10-digit number is divisible by 72. List all possible solutions. | Answer: 2201920200 or 3201920208. Solution. Since $72=8 \cdot 9$, it is required to append digits so that the resulting number is divisible by both 8 and 9. Divisibility by 8 is determined by the last three digits: to twenty, we need to append a digit on the right so that a three-digit number divisible by 8 is obtained... | 2201920200or3201920208 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,551 |
9.2. The lengths $a, b, c$ of the sides of a triangle satisfy the inequality $c^{2} + ab < ca + cb$. Prove that the angle opposite side $c$ is acute. | Solution. Rewrite the inequality as $(c-a)(c-b)<0$. This means $a \neq b$, and the number $c$ lies in the interval between $a$ and $b$. If the angle opposite side $c$ were right or obtuse, the length of $c$ would be the greatest. Therefore, the angle opposite side $c$ is acute. | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,552 |
9.3. The graph of a reduced quadratic trinomial (parabola) with integer coefficients touches the Ox axis. Prove that on this parabola, there is a point with integer coordinates $(a, b)$ such that the graph $y=x^{2}+a x+b$ also touches the Ox axis. | Solution. Let $x^{2}+p x+q$ be the given quadratic polynomial. Then $p^{2}-4 q=0$ by the condition of the graph touching the Ox axis (this condition is equivalent to the discriminant of the quadratic polynomial being zero). Note that the point with coordinates $(-p, q)$ lies on the graph of the quadratic polynomial, si... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,553 |
9.4. Given a triangle with sides $a, b, c$. On its sides as diameters, semicircles are constructed outward, resulting in a figure $\Phi$ composed of the triangle and three semicircles. Find the diameter of $\Phi$ (the diameter of a set on a plane is the greatest distance between its points). | Answer: $\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$. Solution. Let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $B C, A C$, and $A B$ respectively. It is obvious that if $M N$ is the diameter of figure $\Phi$, then $M$ and $N$ are points on the boundary of $\Phi$ (otherwise, segment $M N$ could be extended to the b... | \frac{\mathrm{}+\mathrm{b}+\mathrm{}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,554 |
1.1. Misha and Grisha are writing numbers on the board. Misha writes threes, and Grisha writes fives. In total, they wrote 20 numbers. How many fives were written if the sum of all the numbers is 94? | Answer: 17
Solution. If all 20 numbers were "5", their sum would be 100. We will replace "5" with "3". With each replacement, the sum decreases by 2. Since $100-94=6, 6: 2=3$, we need 3 replacements. Therefore, the number of fives is $20-3=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,556 |
2.1. The seller has weights of $1, 2, 4, 8, 16, 32$ grams (one of each) and a balance scale. On the first pan, a candy weighing 25 grams and some three weights were placed, and on the second pan, the remaining three weights, with the scales coming into balance. Indicate the weights of all three weights on the second pa... | Answer: $4,8,32$ (All answers)
## Solution.
1st method. Let's find the total weight of all the weights and candies to understand how many grams we need to balance. $1+2+4+8+16+32+25=88.88: 2=44$ - on each pan. 44 can be obtained in only one way: $32+8+4$, which is the answer.
2nd method. Notice that in any set of we... | 4,8,32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,557 |
3.1. Five identical squares, standing in a row, were cut by two horizontal lines. The sum of the perimeters of the resulting 15 rectangles is 800 cm. Indicate in centimeters the length of the original squares.
| | | | | |
| :--- | :--- | :--- | :--- | :--- |
| | | | | | | # Answer: 20
Solution. Let's calculate how many times the side of the original square is repeated in the sum of all perimeters. The sides of the rectangle are counted once (a total of 12), and the crossbars are counted twice ($4 \cdot 2 + 10 \cdot 2 = 28$). In total, $40.800: 40=20$ cm - the side of the square. | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,558 |
4.1. In a row, 64 people are standing - each one is either a knight, who always tells the truth, or a liar, who always lies. One of the standing knights said that he stands next to a knight and a liar, and all the other 63 people repeated his phrase. Indicate how many of them were knights. | Answer: 42
Solution. The people at the ends must be liars, as they do not have a second neighbor, and by saying this phrase, those standing there lied. All others can be liars, in which case there are no knights at all, but this option contradicts the condition that at least one knight is present. If there is a knight... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,559 |
5.1. In a row, there are 27 matchboxes, each containing a certain number of matches. It is known that in any four consecutive boxes, the total is 25 matches, and in all of them, the total is 165. How many matches are in the eighth box? | Answer: 10
Solution. In the first 24 boxes, there are a total of $6 \cdot 25=150$ matches. In the last three boxes, there are 15 matches. Therefore, the 4th from the end (or 24th from the start) has 10 matches. Then, in boxes $24, 23, 22, 21$, there are 25 matches in total, meaning in boxes $21, 22, 23$, there are 15 ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,560 |
7.1. A certain number was written on the board, multiplied by 7, the last digit of the resulting product was erased, the resulting number was multiplied by 9, the last digit of the resulting number was erased again, and the result was 8. What numbers could have been written initially? | Answer: 13,14 (All answers)
Solution. Let $a$ be the initially written number. $7 a=10 x+y$, then after erasing the last digit, $x .9 x=10 p+q$ remains, where $q$ is the last digit, and after erasing, $p . p=8$ remains. Substitute into the number before erasing: $9 x=80+q$. Since $80+q$ is divisible by 9, then $q=1.9 ... | 13,14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,561 |
8.1. The numbers from 5 to 18 were written in the cells of the figure shown in the image without repetition. Then, all possible values of the sums in the $1 \times 3$ rectangles were calculated and added together. It turned out that the arrangement of the numbers gives the maximum value of this sum. What can the sum of... | Answer: 50, 51
Solution. Let's consider how many different rectangles each cell is in. The shaded cells (from top to bottom) are repeated 3, 4, 6 times. And the cell below will also be repeated 3 times. Thus, to arrange the numbers in the required manner, we need to place 18 and 17 in the cells that appear the most fr... | 50,51 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,562 |
2. On the eve of his birthday in 2015, Dima discovered an amazing fact: if you multiply the number (day of the month) of his birth by the month number of his birth and by the number of years he is turning, you get the year of his birth. Dima was born in the 21st century. Determine all possible dates of his birth.
Answ... | Brute force solution by iterating through years and incomplete iteration - no more than 3 points.
Translation of the text into English, preserving the original text's line breaks and format, as requested. | July22,2002orNovember14,2002 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,563 |
3. Two circles $\omega_{1}$ and $\omega_{2}$ of the same radius $R$ intersect and are positioned such that the center of one circle is outside the other. Let the center of circle $\omega_{1}$ be denoted by $O_{1}$, and the center of circle $\omega_{2}$ by $O_{2}$. Suppose $B$ is one of the two intersection points of th... | Solution 2. Let the second intersection point of the circles be denoted by the letter $A$. The circles are symmetric with respect to the line $A B$. Let the point of intersection of $A B$ and $\mathrm{O}_{1} \mathrm{O}_{2}$ be denoted by $D$. Due to symmetry, $A B$ and $\mathrm{O}_{1} \mathrm{O}_{2}$ are perpendicular,... | R\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,564 |
4. It is known that the sum of the numbers $a, b$ and $c$ is 1. Prove that $a b + b c + c a \leq 1 / 3$. | Solution. From the obvious inequality $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$, expanding the brackets, we get $a^{2}+b^{2}+c^{2} \geq a b+b c+c a, \quad$ from which $\quad a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \geq 3 a b+3 b c+3 c a \quad$ or $(a+b+c)^{2} \geq 3 a b+3 b c+3 c a$. Taking into account the condition of the p... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,565 |
5. A $7 \times 7$ board has a chessboard coloring. In one move, you can choose any $m \times n$ rectangle of cells and repaint all its cells to the opposite color (black cells become white, white cells become black). What is the minimum number of moves required to make the board monochromatic?
Answer: in 6 moves. | Solution. Consider segments of length equal to the side of a cell, separating pairs of cells adjacent to the side of the board. Along each side of the board, there are 6 such segments, totaling $6 \cdot 4=24$. Each of these segments separates cells that initially have different colors, so each segment must end up on th... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,566 |
1. The sum of two natural numbers is 2015. If the last digit of one of them is erased, the result is the second number. Find all such numbers. | Solution: Let b be the smaller number. Then the larger number a $=10 \mathrm{~b}+\mathrm{c}$, where c is the crossed-out digit, $0 \leq \mathrm{c} \leq 9$. According to the condition, $2015=\mathrm{a}+\mathrm{b}=11 \mathrm{~b}+\mathrm{c}$, so b is the quotient, and c is the remainder when 2015 is divided by 11, i.e., $... | 1832183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,567 |
3. Two parallel lines intersect the graph of the function $y=a x^{2}$. The first intersects it at points with abscissas $x_{1}$ and $x_{2}\left(x_{1}<x_{2}\right)$, the second - at points with abscissas $x_{3}$ and $x_{4}\left(x_{3}<x_{4}\right)$. Prove that ( $x_{3}-x_{1}$ ) $=$ $\left(\mathrm{x}_{2}-\mathrm{x}_{4}\ri... | Solution: Let the parallel lines have equations $\mathrm{y}=\mathrm{kx}+\mathrm{b}, \mathrm{y}=\mathrm{kx}+\mathrm{c}$. Since the first line intersects the graph of the function $y=a x^{2}$ at points with abscissas $x_{1}$ and $x_{2}$, then $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}=$ $\mathrm{kx}+$ b. By... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,569 |
5. There is a ruler 10 cm long without divisions. What is the smallest number of intermediate divisions that need to be made on the ruler so that segments of length 1 cm, 2 cm, 3 cm, ..., 10 cm can be laid off, applying the ruler in each case only once. | Answer: 4
Solution: First, let's prove that three divisions are not enough. There are a total of 10 segments with endpoints at five points. Therefore, if three divisions are made, each length from 1 to 10 should be obtained exactly once. If any division is made at a non-integer distance from the left end, there will b... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,571 |
10.1 It is known that $a+b+c=7, \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{7}{10}$. Find
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}
$$ | # Solution:
$$
\begin{aligned}
& (a+b+c) \cdot\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)= \\
& =\frac{c+(a+b)}{a+b}+\frac{a+(b+c)}{b+c}+\frac{b+(c+a)}{c+a}= \\
& =\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}+1
\end{aligned}
$$
Thus, the desired sum is $7 \cdot \frac{7}{10}-3=\frac{19}{10}$.
Answer: $\fr... | \frac{19}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,572 |
10.2 Does there exist a function $h(x)$ that is decreasing on the interval $[0, \infty)$ such that the function $f(x)=\left(x^{2}-x+1\right) h(x)$ is increasing on the interval $[0, \infty) ?$ | Answer: does not exist.
Solution: If such a function existed, then it would be $f(1)>f(0)$, i.e., $h(1)>h(0)$, which is impossible due to the decreasing nature of the function $h(x)$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,573 |
9.6. On an infinite strip of paper, all natural numbers whose digits sum to 2018 are written in ascending order. What number is written in the 225th position?
(Method Commission) | Answer: $3 \underbrace{999 \ldots 99} 8$.
223 nines
Solution. Since $2018=224 \cdot 9+2$, the smallest number with a digit sum of 2018 will be $2 \underbrace{999 \ldots 99}_{224 \text { nines }}$. This number has 225 digits. Consider 225-digit numbers where the leading digit is a three, and the rest are nines, except... | 3\underbrace{999\ldots99}_{223nines}8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,574 |
9.7. Initially, 40 blue, 30 red, and 20 green chips are arranged in a circle, with chips of each color grouped together. In one move, you can swap adjacent blue and red chips, or adjacent blue and green chips. Is it possible to achieve, after several such operations, that any two adjacent chips are of different colors?... | Answer. No.
Solution. Since red chips cannot swap places with green ones, their relative order will always remain the same as the initial one. In other words, if at any moment we remove the blue chips, there will be 30 red chips standing in a row and 20 green chips, also standing in a row. If the required result could... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,575 |
9.8. Seryozha chose two different prime numbers $p$ and $q$. He considers a natural number $n$ to be good if the number $p+q$ can be represented as the sum of exactly $q$ numbers, each of which has the form $n^{k}$ for an integer non-negative $k$. (For example, if Seryozha had chosen $p=7$ and $q=3$, he would have cons... | Solution. Let $n$ be a good number. Then $n>1$, and for some non-negative integers $k_{1}, k_{2}, \ldots, k_{q}$, the equality
$$
p+q=n^{k_{1}}+n^{k_{2}}+\ldots+n^{k_{q}} .
$$
holds. Consider the remainder of the right-hand side when divided by $n-1$. Since any power of $n$ gives a remainder of 1 when divided by $n-1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,576 |
9.9. In a circle $\omega$ with center at point $O$, non-intersecting chords $AB$ and $CD$ are drawn such that $\angle AOB = \angle COD = 120^{\circ}$. The tangent to $\omega$ at point $A$ intersects the ray $CD$ at point $X$, and the tangent to $\omega$ at point $B$ intersects the ray $DC$ at point $Y$. The line $\ell$... | The first solution. Let $Z$ be the point of intersection of the rays $X A$ and $Y B$ (see Fig. 1). The lines $Z A$ and $Z B$ are tangent to the circle $\omega$, so $\angle O A Z = \angle O B Z = 90^{\circ}$. Therefore, $\angle A Z B = 360^{\circ} - 90^{\circ} - 90^{\circ} - \angle A O B = 60^{\circ}$. Since the circle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,577 |
9.10. In the company, there are 100 children, some of whom are friends (friendship is always mutual). It is known that by selecting any child, the remaining 99 children can be divided into 33 groups of three such that in each group, all three are pairwise friends. Find the minimum possible number of pairs of friends.
... | Answer: 198.
Solution: Let's translate the problem into the language of graphs, associating each child with a vertex and each friendship with an edge. Then we know that in this graph with 100 vertices, after removing any vertex, the remaining vertices can be divided into 33 triples such that the vertices in each tripl... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,578 |
Problem 4.1. On nine cards, the numbers from 1 to 9 are written (each one only once). These cards were laid out in a row such that there are no three consecutive cards with numbers in ascending order, and there are no three consecutive cards with numbers in descending order. Then, three cards were flipped over, as show... | Answer: The number 5 is written on card $A$, the number -2 on card $B$, and the number -9 on card $C$.
Solution. The missing numbers are $-2, 5$, and 9.
If the number 5 is on card $B$, then we get the consecutive numbers 3, 4, 5. If the number 5 is on card $C$, then we get the consecutive numbers 8, 7, 5. Therefore, ... | 5,2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,580 |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,581 |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,582 |
Problem 4.8. The figure shows a road map between the houses of five children. The shortest distance by road from Asya to Galia is 12 km, from Galia to Borya - 10 km, from Asya to Borya 8 km, from Dasha to Galia - 15 km, from Vasya to Galia - 17 km. How many kilometers is the shortest distance by road from Dasha to Vasy... | Answer: 18.
Solution. Add the distances from Dasha to Gala and from Vasya to Gala: $15+17=32$. This will include the "main" road (from Dasha to Vasya) and twice the "branch" from it to Gala.
Add the distance from Asey to Bory to the obtained sum: $32+8=40$. Now all three branches (to Gala - twice) and the main road w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,583 |
Problem 5.3. The figure shows a plan of the road system of a certain city. In this city, there are 8 straight streets, and 11 intersections are named with Latin letters $A, B, C, \ldots, J, K$.
Three police officers need to be placed at some intersections so that at least one police officer is on each of the 8 streets... | Answer: $B, G, H$.
Solution. The option will work if police officers are placed at intersections $B, G, H$. It can be shown that this is the only possible option.
Since there are only three vertical streets, and each must have one police officer, there are definitely no police officers at intersections $C$ and $K$. T... | B,G,H | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,584 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,585 |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,586 |
Problem 6.3. In the cells of a $4 \times 4$ table, the numbers $1,2,3,4$ are arranged such that
- each number appears in each row and each column;
- in all four parts shown in the figure, the sums of the numbers are equal.
Determine in which cells the twos are located based on the two numbers in the figure.
![](http... | Answer: In row $A$, the two is in column 2, in row $B-1$, in row $C-4$, in row $D-3$.
Solution. First, let's find what the sum of the numbers in each of the parts into which the board is divided is. In each column, the numbers from 1 to 4 appear once. Their sum is $1+2+3+4=10$, and the sum of all numbers on the board ... | A2,B1,C4,D3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,587 |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,589 |
Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.
![](https://cdn.mathpix.com/c... | Answer: $52.5$.
Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.
Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.
![](https://cdn.mathpix.c... | 52.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,591 |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,592 |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,593 |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,594 |
Problem 9.5. Leonid has a white checkered rectangle. First, he painted every other column gray, starting with the leftmost one, and then every other row, starting with the topmost one. All cells adjacent to the border of the rectangle ended up being painted.
How many painted cells could there be in the rectangle if 74... | Answer: 301 or 373.
Solution. From the condition, it follows that the rectangle has an odd number of both rows and columns. Let's number the rows from top to bottom with the numbers $1,2, \ldots, 2 k+1$, and the columns from left to right with the numbers $1,2, \ldots, 2 l+1$ (for non-negative integers $k$ and $l$). W... | 301or373 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,595 |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,596 |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,598 |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,599 |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,600 |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
=0$. | Answer. $x_{1}=-1 ; x_{2}=2 ; x_{3}=\frac{\pi}{2}$. Solution. The expression under the root $-x^{2}+x+2$ gives two roots $x_{1}=-1 ; x_{2}=2$ and defines the domain of definition: $-1 \leq x \leq 2$. By equating the bracket $\sin 2 x-\pi \cos x=$ $\cos x(2 \sin x-\pi)$ to zero, we obtain the system of equations $\left[... | x_{1}=-1;x_{2}=2;x_{3}=\frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,603 |
11.3. How many points (different from the origin) are there on the parabola $y=x^{2}$, such that the tangent at these points intersects both coordinate axes at points with integer coordinates not exceeding 2020 in absolute value? | Answer: 44. Solution. The equation of the tangent to the parabola $y=x^{2}$ at the point $\left(x_{0}, y_{0}\right)$, where $y_{0}=x_{0}^{2}$, is $y-y_{0}=2 x_{0}\left(x-x_{0}\right)$. From this, we find the coordinates of the points of intersection of the tangent with the axes, namely $x_{1}=\frac{x_{0}}{2}, y_{1}=-x_... | 44 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 18,604 |
11.4. Given a tetrahedron \( SABC \) with mutually perpendicular edges \( SA, SB, SC \). Let \( O \) be the center of the sphere circumscribed around the tetrahedron. Prove that points \( S \) and \( O \) lie on opposite sides of the plane \( ABC \). | Solution. Let $S A=a, S B=b, S C=c$. Consider a Cartesian coordinate system in space with the origin at point $S$ and axes $x, y, z$ along $S A, S B$, and $S C$ respectively. Then the plane $A B C$ will have the equation $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ (this is the equation of a plane in intercepts; it can be d... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,605 |
11.5. Find the set of values of the function $y=\sqrt{x}-\sqrt{2-x}+2 \sin x$. | Answer: $[-\sqrt{2} ; \sqrt{2}+2 \sin 2]$. Solution. The domain of the function $y=f(x)$ is the interval [0,2]. We will prove that on the interval $(0 ; 2)$, the derivative $f'(x)$ is positive, and thus the function $f(x)$, continuous on $[0,2]$, is monotonically increasing. We have
$$
f'(x)=\frac{1}{2 \sqrt{x}}+\frac... | [-\sqrt{2};\sqrt{2}+2\sin2] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,606 |
1. An increasing arithmetic progression contains two natural numbers and the square of the smaller one. Prove that it also contains the square of the second number. | 1. Let the numbers $a$, $B$, and $a^2$ be part of an arithmetic progression. Then $B = a + nd$, $a^2 = a + md$. From this, $B - a = nd$, $B^2 = a^2 + B^2 - a^2 = a^2 + (B - a)(B + a) = a + md + nd(B + a) = a + kd$, i.e., $B^2$ is also part of the progression.
2. Let $m$ be the root equal to the sum of the coefficients... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,607 |
3. Let AB and CD be two perpendicular chords of a circle with center at point $\mathrm{R}$, intersecting inside the circle at point $\mathrm{F}$; let also $\mathrm{N}$ and $\mathrm{T}$ be the midpoints of segments AC and BD, respectively. Prove that quadrilateral FNRT is a parallelogram. | 3. Extend the segment TF until it intersects line AC at point K. Let $\angle \mathrm{CAB}=\alpha$. From the right triangle AFC we get $\angle \mathrm{ACF}=90^{\circ}-\alpha$. $\angle \mathrm{CDB}=\angle \mathrm{CAB}=\alpha$ (as inscribed angles subtending the same arc). On the other hand, FT is the median in the right ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,608 |
4. In a certain company, there are 100 shareholders, and any 66 of them own no less than $50 \%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? | 4. Let $M$ be the shareholder owning the largest percentage of shares - $\mathrm{x} \%$ of shares. Divide the other 99 shareholders into three groups A, B, C, each with 33 shareholders. Let them own a, b, c percentages of shares, respectively. Then $2(100-x)=2(a+b+c)=(a+b)+(b+c)+(c+a) \geq 50+50+50$, i.e., $x \leq 25$.... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,609 |
5. On the faces of each of the 27 cubes, all numbers from 1 to 6 are written arbitrarily. The boy then assembled a cube from these 27 cubes, such that on any two adjacent faces of the cubes, the numbers differ by exactly 1. After that, the boy calculated the sums of the numbers written on each of the faces. Could he ha... | 5. Note that the sum of the numbers written on two adjacent faces is odd. There are $3 \times 3 \times 6=54$ such pairs of faces in the cube, i.e., the sum of the numbers written on the faces located inside the cube is the sum of 54 odd numbers. Therefore, it is even. Let's denote it as $2 S$. Now, let's calculate the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,610 |
9.1. Determine whether there exist natural numbers $k$ such that the value of the expression $k^{3}-3 k^{2}+2 k+2$ is divisible by $2020 ?$ | Answer: such $k$ does not exist.
Solution. The expression $k^{3}-3 k^{2}+2 k+2$ can be represented as $(k-2)(k-1) k+2$. This is the product of three consecutive natural numbers, to which 2 is added.
The last digit of the product of three consecutive natural numbers $(k-2)(k-1) k$ can take one of the following values:... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,611 |
9.2. The age of a certain person in 1988 was equal to the sum of the digits of their birth year. How old was he
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
However, it seems there is a repetition. Here is the requested translation:
9.2. The age of a certain person in 1988 was equal to the sum of the digits of their birth y... | Answer: 22.
Solution. The number of years a person has lived is equal to the sum of the digits of a four-digit number, each of which is no more than 9. Therefore, he is no more than 36 years old, and he was born in the 20th century. Let $x$ be the number of tens, $y$ be the number of units in his birth year.
Then, ac... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,612 |
9.3. The road from Kimovsk to Moscow consists of three sections: Kimovsk - Novomoskovsk (35 km), Novomoskovsk - Tula (60 km), and Tula - Moscow (200 km). A bus, whose speed nowhere exceeded 60 km/h, traveled from
Kimovsk to Tula in 2 hours, and from Novomoskovsk to Moscow in 5 hours. How long could the bus have been ... | Answer: From $5 \frac{7}{12}$ to 6 hours.
Solution. The total travel time is obtained by subtracting from the sum of $2+5$ the time spent on the segment Novomoskovsk - Tula. According to the condition, the bus spent no less than an hour on this segment (60 km). On the other hand, the journey from Kimovsk to Novomoskov... | 5\frac{7}{12} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,613 |
9.4. Can a triangle be cut into four parts by two straight cuts passing through two of its vertices so that three of the triangles (among these parts) are equal in area? | Answer: No.
Solution. Suppose it is possible. Let in triangle $ABC$ points $A_{1}$ and $B_{1}$ lie on sides $BC$ and $AC$ respectively, segments $AA_{1}$ and $BB_{1}$ intersect at point $O$, and the areas of triangles $AOB_{1}$, $AOB$, and $BOA_{1}$ are equal.
Since triangles $AOB$ and $BOA_{1}$ have the same height ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,614 |
9.5. Varya and Miron are playing the following game. There are 10 piles of 10 stones each on the table. The players take turns, with Varya starting first. On their turn, a player divides any pile that has more than one stone into several equal piles. The player who cannot make a move (when there is exactly one stone in... | Answer: Miron.
Solution. We will call a pile with one stone "unitary," and a pile with a prime number of stones - "prime." A prime pile of $p$ stones can only be divided into $p$ unitary piles.
For Miron to win, it is sufficient for him to repeat Varya's moves until only one pile of 10 stones remains. At this point, ... | Miron | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,615 |
10.1 In a $3 \times 3$ table, non-repeating natural numbers not exceeding 20 were placed. Could it be that for any two adjacent cells by side, the ratio of the larger number to the smaller one is an integer? | Solution: Yes, it could, for example, like this:
| 10 | 2 | 6 |
| :---: | :---: | :---: |
| 5 | 1 | 3 |
| 20 | 4 | 12 |
Criteria: Full solution - 7 points. Attempts to explain how the example was obtained should not affect the final score if the correct example is provided. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,616 |
10.2 A group of friends went for a morning run around a lake. During the run, one by one they realized they had miscalculated their strength, and switched from running to walking. One of the friends calculated that he had run one-eighth of the total distance that the entire group had run, and walked one-tenth of the to... | Solution 1: Let the person who ran cover $x$ part of the road, then $0<x<1$, and he walked $(1-x)$ part of the way. If there were $n$ people in total, then according to the condition, the total distance covered by the group (expressed in terms of parts) is on one side $n$, and on the other side $8 x + 10(1-x) = 10 - 2 ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,617 |
10.3 Prove that for all positive $x$ the inequality
$$
(x+1) \sqrt{x+1} \geqslant \sqrt{2}(x+\sqrt{x})
$$
holds. | Solution 1: This inequality is the product of two inequalities: $\sqrt{x+1} \geqslant \frac{1}{\sqrt{2}}(1+\sqrt{x})$ and $x+1 \geqslant 2 \sqrt{x}$, both of which are equivalent to the inequality $(1-\sqrt{x})^{2} \geqslant 0$ (for this, the first inequality needs to be squared beforehand).
Solution 2: Both sides of ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,618 |
10.4 For each integer from 10 to 2021, we found the product of the digits, and then added all the obtained results. What is the sum that was obtained? | Solution: Consider the product $(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9)$. If we expand the brackets, we get the products of pairs of digits that form all two-digit numbers. Similarly, in the product
$$
(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9) \cdot(0+1+2+\cdots+9)
$$
after expanding the brackets, we get all combinations... | 184275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,619 |
10.5 In $\triangle A B C$, the angle bisectors $A A_{1}, B B_{1}$, and $C C_{1}$ intersect at point $O$. It turns out that the areas of $\triangle O C_{1} B, \triangle O B_{1} A$, and $\triangle O A_{1} C$ are equal. Is it true that $\triangle A B C$ is equilateral? | Solution 1: Let the lengths of the sides of the triangle be $AB = c, BC = a, CA = b$. For the three triangles in the condition, the areas and the heights from vertex $O$ (since $O$ is the center of the inscribed circle) are equal, which means the bases $AB_1 = BC_1 = CA_1$ are equal. Using the fact that the angle bisec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,620 |
8.1 Prove that if $\mathrm{ab}(\mathrm{a}+\mathrm{b})=1$, then $\frac{\mathrm{a}}{\mathrm{a}^{3}+\mathrm{a}+1}=\frac{\mathrm{b}}{\mathrm{b}^{3}+\mathrm{b}+1}$. | Solution. For example, one can act as follows:
$$
\begin{aligned}
& \frac{a}{a^{3}+a+1}=\frac{a}{a^{3}+a+ab(a+b)}=\frac{1}{a^{2}+1+ba+b^{2}} \\
& \frac{b}{b^{3}+b+1}=\frac{b}{b^{3}+b+ab(a+b)}=\frac{1}{b^{2}+1+a^{2}+ab}
\end{aligned}
$$
We obtained the same result. | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,621 |
8.2 Find all three-digit numbers $\mathrm{N}$ such that the sum of the digits of the number $\mathrm{N}$ is 11 times smaller than the number $\mathrm{N}$ itself. | Solution. Let a, b, c be the digits of the number N: $\mathrm{N}=100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}$, where $1 \leq \mathrm{a} \leq 9$, $0 \leq \mathrm{b}, \mathrm{c} \leq 9$. According to the condition, $100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}=11(\mathrm{a}+\mathrm{b}+\mathrm{c})$, i.e., $89 \mathrm{a}=\mathrm{... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,622 |
8.3 There are 12 natural numbers. It is known that the sum of any three of them is not less than 100. Prove that the sum of all the numbers is not less than 406. | Solution. Arrange these numbers in non-decreasing order: $\mathrm{a}_{1} \leq \mathrm{a}_{2} \leq \mathrm{a}_{3} \leq \mathrm{a}_{4} \leq \ldots \leq \mathrm{a}_{12}$. By the condition $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3} \geq 100$, therefore $3 \mathrm{a}_{3} \geq 100, \mathrm{a}_{3} \geq \frac{100}{3}>33$. S... | 406 | Inequalities | proof | Yes | Yes | olympiads | false | 18,623 |
8.4 The perpendicular bisector of side $\mathrm{BC}$ of triangle $\mathrm{ABC}$ intersects side $\mathrm{AC}$ at point $\mathrm{D}$, and the angle bisector of $\angle \mathrm{ABD}$ at point $\mathrm{P}$. Prove that points $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$, and $\mathrm{P}$ lie on the same circle. | Solution.
Triangle PCD is congruent to triangle PBD (by three sides: $\mathrm{DC}=\mathrm{DB}, \mathrm{PC}=\mathrm{PB}, \mathrm{PD}$ - common). Therefore, we have: $\angle \mathrm{PCA}=\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,624 |
8.58 schoolchildren solved 8 problems. It turned out that each problem was solved by 5 schoolchildren. Prove that there are such two schoolchildren that each problem was solved by at least one of them. | Solution. Let's construct a table (matrix) of results - an $8 \times 8$ table: if, for example, student number 2 solved problem number 5, then in the second row of this table in the column with number 5 we will write a one, if not, then a zero. Thus, in each column of this table there are 5 ones and 3 zeros. We need to... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,625 |
1. Solve the equation: $\sqrt{x^{2}-4} \cdot|x|+\sqrt{x^{2}-4} \cdot x=0$. | Solution. Since $\sqrt{x^{2}-4} \geq 0$, we consider two cases.
First case $\sqrt{x^{2}-4}=0$. Hence, $x= \pm 2-$ roots.
Second case. $\sqrt{x^{2}-4}>0$, hence $|x|>2$. Then the given equation can be rewritten as follows: $|x|+x=0$. From this, it follows that the solution to the last equation is the interval $x \leq ... | x\in(-\infty;-2]\cup{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,626 |
2. In triangle $ABC$, the bisector $BD$ is drawn. It is known that the center of the circumcircle around $\triangle ABC$ coincides with the center of the circle inscribed in $\triangle BCD$. Find the angles of $\triangle ABC$. | Solution. Let the common center of the circles be denoted by $I$, and the points of tangency of the incircle of $\triangle BCD$ with the sides $CD, BC$, and $BD$ be denoted by $E, F, G$ respectively (Fig. 1). The segments $IE$ and $IF$ are the perpendicular bisectors of the sides $AC$ and $BC$ respectively, so $AE = CE... | \angleA=\angleB=72,\angleC=36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,627 |
4. Let $P(n)$ and $S(n)$ denote the product and sum of the digits of the natural number $\mathbf{n}$, respectively. For example, $P(133)=9, S(133)=7$. Find all two-digit numbers $\boldsymbol{n}$ for which the equality $n=P(n)+S(n)$ holds. | Solution. Let the desired two-digit number be $n=\overline{a b}=10 a+b, a \neq 0$. The equation from the condition takes the following form:
$$
10 a+b=a b+a+b \Leftrightarrow 10 a=a(b+1) \Leftrightarrow 10=b+1 \Leftrightarrow b=9
$$
Therefore, the condition is satisfied by all two-digit numbers ending in 9, and only ... | 19,29,39,49,59,69,79,89,99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,629 |
11.8. First solution. Let $I$ be the center of the incircle of triangle $ABC$, and $J_{1}$ and $J_{2}$ be the centers of its excircles $\omega_{1}$ and $\omega_{2}$, touching sides $AB$ and $AC$, respectively. Line $AN$ is the external bisector of angle $BAC$, so points $J_{1}$ and $J_{2}$ lie on it. Let $K_{1}$ and $K... | The second solution. Note that right triangles $B M N$ and $C M N$ are symmetric with respect to $M N$. Let point $I_{2}^{\prime}$ be symmetric to $I_{2}$ with respect to $M N$. We have $\angle B M I_{1} + \angle B M I_{2}^{\prime} = \angle B M I_{1} + \angle C M I_{2} = \frac{1}{2}(\angle B M A + \angle C M A) = 90^{\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,631 |
1. The midpoints of adjacent sides of a rectangle with a perimeter of 32 were connected by segments. The same operation was performed on the resulting quadrilateral: the midpoints of adjacent sides were connected by segments (see figure). How many times in total does one need to perform such an operation so that the pe... | Answer: 11.
Solution: After two operations, a quadrilateral is obtained, the sides of which are the midlines of triangles with bases parallel to the sides of the original rectangle. Therefore, this quadrilateral is a rectangle, and each of its sides is half the length of the corresponding side of the original rectangl... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,632 |
2. Vasya arranges natural numbers from 1 to 10 in the product \(a^{b} b^{c} c^{d} d^{e} e^{f} f^{g} g^{k} k^{l} l^{m} m^{a}\) (identical letters denote identical digits, different letters denote different digits). What is the highest power of two that can divide this product? | Answer: $2^{69}$.
Solution. It is clear that in the highest possible power (10) there should be the digit $8=2^{3}$. In the power of 9, there should only be $4=2^{2}$, and the other even digits should be in the powers of 8, 7, and 6. Then the product will be divisible by a power of two equal to $3 \cdot 10+2 \cdot 9+8... | 2^{69} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,633 |
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