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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 |
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 |
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 |
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 |
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 |
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 |
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 $D Z$.
. In which year will this property reoccur for the first time?
(I. V. Raskina) | Answer: in 2022.
Solution. In 2010, 2011, ..., 2019, and in 2021, the year number contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be decreased by rearranging the digits.
## Comment.
3 points. For th... | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. What angle do the clock hands form at 12:20? | Answer: $110^{\circ}$.
Solution. At 12:00, the clock hands coincide. After this, in 20 minutes, the minute hand travels $1 / 3$ of the circumference, i.e., it describes an angle of $120^{\circ}$. The hour hand moves 12 times slower than the minute hand (since it describes one circle in 12 hours). Therefore, in 20 minu... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.5. Grandfather is 31 times older than his grandson. In how many years will he be 7 times older than his grandson, given that the grandfather is more than 50 but less than 90 years old? | Answer: In 8 years.
Solution: Grandfather's age is divisible by 31. But the only such number greater than 50 and less than 90 is 62. Therefore, the grandfather is 62 years old, and the grandson is 2 years old. In x years, the grandfather will be x+62 years old, and the grandson will be x+2 years old. If at that time h... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(x^{2}-|x|-2\right)^{2}+\sqrt{|2 x+1|-3}=0$. | # Solution
Since each term of the original equation is non-negative, the equation can have solutions if and only if both terms are equal to zero. We obtain the equivalent system of equations
$$
\left\{\begin{array}{l}
x^{2}-|x|-2=0 \\
|2 x+1|-3=0
\end{array}\right.
$$
The solution to this system is $x=-2$.
## Crite... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.5. A square plot of 14 by 14 cells needs to be paved with rectangular tiles of size $1 \times 4$. The tiles can only be laid along the grid (not diagonally), and the tiles cannot be broken. What is the maximum number of tiles required? Will there be any uncovered area left? | # Solution:
Evaluation. The total number of cells on the plot is $14 \times 14=196$. Dividing by the number of cells in one tile, 196:4 = 49. Therefore, the number of tiles that can be cut from the $14 \times 14$ plot is no more than 49.
We will color the cells of the plot in 4 colors, as shown in the diagram. Clearl... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A $3 \times 3$ table was filled with prime numbers. It turned out that the sum of the numbers in any two adjacent cells is a prime number. What is the maximum number of different numbers that can be in the table? Provide an example and prove that there cannot be more different numbers.
Fill the $3 \times 3$ table w... | 3. Answer: 6.
Example:
| 5 | 2 | 29 | |
| :---: | :---: | :---: | :---: |
| 2 | 3 | 2 | |
| 11 | 2 | 17 | |
| | -2 | | 0 |
We will prove that it cannot be more. Divide the cells of th... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Six numbers are written in a row on the board. It is known that each number, starting from the third, is equal to the product of the two preceding numbers, and the fifth number is equal to 108. Find the product of all six numbers in this row. | 4. Answer: 136048896.
Solution. Let the first number be $x$, and the second number be $y$. Then the third number is $xy$.
The fourth number is $xy^2$. The fifth number is $x^2 y^3 = 108$. The sixth number is $x^3 y^3$.
The product of all six numbers is $x^8 y^{12} = (x^2 y^3)^4 = 108^4 = 136048896$.
## Grading Crit... | 136048896 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In Grandfather Frost's bag, there are chocolate and jelly candies, a total of 2023 pieces. Chocolate candies make up 75% of the jelly candies. How many chocolate candies are in Grandfather Frost's bag? | Answer: 867. Solution. Chocolate candies make up 3/4 of the quantity of jelly candies. Then the total number of candies is 7/4 of jelly candies. Therefore, the number of jelly candies is 2023×4:7=1156, and the number of chocolate candies is 2023-1156=867.
Grading criteria. A correct and justified solution - 7 points. ... | 867 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. What is the greatest length of an arithmetic progression of natural numbers
$a_{1}, a_{2}, \ldots, a_{n}$ with a difference of 2, in which for all $k=1,2 \ldots, n$ all numbers $a_{k}^{2}+1$ are prime | Solution.
All members of the progression must be even numbers. Consider the remainders when divided by 5. Since the common difference of the arithmetic progression is 2, the remainders in the progression will appear in the order
$\ldots 2,4,1,3,0,2,4,1,3,0, \ldots$ This means that among the terms of the progression, ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A $10 \times 10$ square was cut into 17 rectangles, each with both side lengths greater than 1. What is the smallest number of squares that could be among these rectangles? Provide an example of such a cutting. | Answer: one square.
Solution. Suppose that among the rectangles, there is not a single square. Let $a$ and $b$ be the sides of an arbitrary rectangle, with $a > b$. Since the integer $b$ is greater than 1, then $b \geqslant 2$, and thus $a \geqslant 3$. Therefore, the area of each such rectangle $a \times b$ is at lea... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The diagonals of quadrilateral $A B C D$ intersect at point $O$. It is known that $A B=B C=$ $=C D, A O=8$ and $\angle B O C=120^{\circ}$. What is $D O ?$ | Answer: $D O=8$.
Solution. Mark a point $E$ on the line $A C$ such that triangle $B O E$ is equilateral (Fig. 2). We will prove the equality of triangles $B A E$ and $B C O$. Indeed, since $A B=B C$, triangle $A B C$ is isosceles, and thus $\angle B A C=\angle B C A$. Additionally, note another pair of equal angles $\... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A seagull is being fed from a moving boat. A piece of bread is thrown down, the seagull takes 3 seconds to pick up the piece from the sea surface, and then it takes 12 seconds to catch up with the boat. Upon entering the bay, the boat reduces its speed by half. How much time will it now take the seagull to catch up ... | Solution. Let the speed of the seagull relative to the boat be $x$ (m/s). Then in 12 seconds, it flies $12 x$ (m). The boat covers this distance in 3 seconds, so its speed is $4 x$ (m/s), and the speed of the seagull is $5 x$ (m/s). In the bay, the boat's speed becomes $2 x$ (m/s). In 3 seconds, it travels $6 x$ (m). T... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. On a rectangular table of size $x$ cm $\times 80$ cm, identical sheets of paper of size 5 cm $\times 8$ cm are placed. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right ... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9-1. Segment $P Q$ is divided into several smaller segments. On each of them, a square is constructed (see figure).
What is the length of the path along the arrows if the length of segment ... | Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$. | 219 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.8. A numismatist has 100 coins that look identical. He knows that among them, 30 are genuine and 70 are counterfeit. Moreover, he knows that the weights of all genuine coins are the same, while the weights of all counterfeit coins are different, and any counterfeit coin is heavier than a genuine one; however, the ex... | Answer: 70.
Solution: 1. We will show that the numismatist can find the genuine coin in 70 weighings. Stack all 100 coins in a pile. With each weighing, he will select two coins from the pile and compare them. If their weights are equal, then both coins are genuine, and the required coin is found. If not, then the hea... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. Twelve different natural numbers are written in a circle, one of which is 1. Any two adjacent numbers differ by either 10 or 7. What is the greatest value that the largest written number can take? | Answer: 58.
Solution. For convenience, let's number all the numbers in a circle clockwise, starting from the number 1: $a_{1}=1, a_{2}, \ldots, a_{12}$. Notice that for all $1 \leqslant i \leqslant 6$ we have
$$
\begin{gathered}
a_{i+1}-a_{1}=\left(a_{i+1}-a_{i}\right)+\left(a_{i}-a_{i-1}\right)+\ldots+\left(a_{2}-a_... | 58 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. Let $\alpha$ and $\beta$ be the real roots of the equation $x^{2}-x-2021=0$, with $\alpha>\beta$. Denote
$$
A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7
$$
Find the greatest integer not exceeding $A$. | Answer: -6055.
Solution. By Vieta's theorem, we have $\alpha+\beta=1$ and $\alpha \beta=-2021$. Also, $\beta^{2}-\beta-2021=0$, since $\beta$ is a root of the equation. Therefore,
$A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7=(\alpha+\beta)^{2}-3\left(\beta^{2}-\beta-2021\right)-6063+7=1^{2}-3 \cdot 0-6063+7=-60... | -6055 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. Let $k_{1}$ be the smallest natural number that is a root of the equation
$$
\sin k^{\circ}=\sin 334 k^{\circ}
$$
(a) (2 points) Find $k_{1}$.
(b) (2 points) Find the smallest root of this equation that is a natural number greater than $k_{1}$. | # Answer:
(a) (2 points) 36.
(b) (2 points) 40.
Solution.
$$
0=\sin 334 k^{\circ}-\sin k^{\circ}=2 \sin \frac{333 k^{\circ}}{2} \cos \frac{335 k^{\circ}}{2}
$$
- Let $\sin \frac{333 k^{\circ}}{2}=0$. This is equivalent to $\frac{333 k}{2}=180 \mathrm{~m}$ for some integer $m$, that is, $333 k=360 m$ and $37 k=40 m... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. In a sports school, 55 people are training, each of whom is either a tennis player or a chess player. It is known that there are no four chess players who would have the same number of friends among the tennis players. What is the maximum number of chess players that can train in this school? | Answer: 42.
Solution. Let there be $a$ tennis players and $55-a$ chess players in the school. Each of the chess players has a number of tennis player friends that is no less than 0 and no more than $a$, meaning it can take on $a+1$ values. If there were more than $3(a+1)$ chess players, by the pigeonhole principle, th... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Points $A, B, C, D, E, F$ are arranged clockwise on a circle, as shown in the figure. Chords $A D$ and $C E$ intersect at point $X$ at a right angle, and chords $A D$ and $B F$ intersect at point $Y$.
It is known that $C X=12, X E=27, X Y=15, B Y=9, Y F=11$.
 (2 points) 36.
(b) (2 points) 19.5.
Solution. Let $A Y=a, X D=b$ (Fig. 8). We use the fact that the products of the segments of intersecting chords through a given point inside a circle are equal. This means that $324=12 \cdot 27=a(15+b)$ and $99=9 \cdot 11=b(a+15)$. Subtracting the second equation fro... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Given the set of numbers $\{-1,-2,-3, \ldots,-26\}$. On the board, all possible subsets of this set were written, each containing at least 2 numbers. For each written subset, the product of all numbers belonging to that subset was calculated. What is the sum of all these products? | Answer: 350.
Solution. Consider the expression $(1-1)(1-2)(1-3) \ldots(1-26)$, the value of which is 0. Expand this expression without combining like terms, resulting in the sum $1^{26}-1^{25}(1+2+3+\ldots+26)+S$. Notice that the sum $S$ coincides with the sum from the problem's condition. Therefore, it equals $0-1+(1... | 350 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.7. All vertices of a regular tetrahedron $ABCD$ are on the same side of the plane $\alpha$. It turns out that the projections of the vertices of the tetrahedron onto the plane $\alpha$ form the vertices of a certain square. Find the value of $AB^2$ if it is known that the distances from points $A$ and $B$ to... | Answer: 32.
Fig. 9: to the solution of problem 11.7
Solution. Let $A_{1}, B_{1}, C_{1}, D_{1}$ be the projections of points $A, B, C, D$ onto the plane $\alpha$. Let $M$ be the midpoint of s... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. In each cell of a strip $1 \times N$ there is either a plus or a minus. Vanya can perform the following operation: choose any three cells (not necessarily consecutive), one of which is exactly in the middle between the other two cells, and change the three signs in these cells to their opposites. A number... | Answer: 1396.
Solution. We will prove that all the numbers under consideration are positive, except for 4 and 5. Then the answer to the problem will be $1398-2=1396$.
For each $N$, we will number the cells of the strip $1 \times N$ from left to right with numbers from 1 to $N$.
- Let $N=3$. By applying the operation... | 1396 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
This results in 8 rectan... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. In the parliament of the island state of Promenade-and-Tornado, only the indigenous inhabitants of the island can be elected, who are divided into knights and liars: knights always tell the truth, liars always lie. A secret ballot on 8.09.19 re-elected 2019 deputies. At the first meeting, all deputies were present... | Solution. A rectangle $42 \times 48$ can be tiled with $3 \times 3$ squares (224 are required). In each such square, there must be at least one knight (otherwise - if the square contains only liars, the liar in the central cell would have told the truth, which is impossible). Thus, the minimum number of knights in the ... | 225 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. It is known that $\frac{1}{\cos (2022 x)}+\operatorname{tg}(2022 x)=\frac{1}{2022}$.
Find $\frac{1}{\cos (2022 x)}-\operatorname{tg}(2022 x)$. | Answer. 2022.
## Solution.
Consider the product $\left(\frac{1}{\cos \alpha}+\operatorname{tg} \alpha\right)\left(\frac{1}{\cos \alpha}-\operatorname{tg} \alpha\right)=\frac{1}{\cos ^{2} \alpha}-\operatorname{tg}^{2} \alpha=\frac{1-\sin ^{2} \alpha}{\cos ^{2} \alpha}=1$. Therefore, the desired expression is 2022.
Co... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. One hundred and one non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of negative numbers that can be among these 101 written numbers? | Answer: 67.
Solution: Consider any 3 consecutive numbers. All of them cannot be negative. Among them, there is a positive one. Fix this positive number and its neighbor (this number can be negative), and divide the remaining 99 numbers into 33 groups of 3 consecutive numbers. In each such group, there will be no more ... | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. The lateral faces of the pentagonal pyramid $S A B C D E$ are acute-angled triangles. We will call a lateral edge of the pyramid good if it is equal to the height of the opposite lateral face, drawn from the vertex of the pyramid (for example, the edge $S A$ is good if it is equal to the height of the triangle $S... | # Answer. 2.
Solution. We will show that two non-adjacent lateral edges of the pyramid cannot both be good. Suppose this is not the case, and, for example, edges $S A$ and $S C$ are good, meaning $S A = S P$ is the height of face $S C D$, and $S C = S Q$ is the height of face $S A E$. However, by the property of the h... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Martians love to dance dances where they have to hold hands. In the dance "Pyramid," no more than 7 Martians can participate, each of whom has no more than three hands. What is the maximum number of hands that the dancers can have if any hand of one Martian holds exactly one hand of another Martian? | Solution. In each handshake, two hands are involved, so the total number of hands shaken will be even. The maximum variant (7 Martians with three hands) gives 21 hands. Therefore, the greatest number of hands will be achieved if 6 Martians have three hands each and one has two hands. The dancers have 20 hands.
 or on the right (seats 3 and 4). Each couple can sit in two ways (boy on the left or right), and the couples themselves can... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4.1.2. A sports team consists of two boys and two girls. In the relay, the stages run by girls must alternate with the stages run by boys. In how many ways can the stages be distributed
# | # Answer: 8
Solution. Boys can run stages 1 and 3, or stages 2 and 4. If they run stages 1 and 3, they can be distributed in two ways, and girls can be distributed in two ways, making a total of 4 ways. In the second case, there are another 4 ways, making a total of 8 ways. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4.1.3 Two mothers with their children want to sit on a bench with 4 seats. In how many ways can they sit so that each mother sits next to her child? Each mother is walking with one child.
# | # Answer: 8
Solution. Let's number the seats from 1 to 4. If someone from a family sits in seat number 1, then the second person must sit in seat 2. Thus, each family either sits on the left (seats 1 and 2) or on the right (seats 3 and 4). Each family can sit in two ways (child on the left or right), and the families ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.1 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first one said that, in his opinion, there were 20 apples, the second thought there were 22, and the third said 25. When the friends opened the bag, it turned out that one of them was off by 1, another by 3, a... | # Answer: 19
Solution. Note that the first and second friends give even predictions, which means they make an error of the same parity, while the third makes an error of a different parity. Therefore, the third friend was off by 6 apples, so there were either 19 or 31 apples in the bag. Checking both options, we get t... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.2 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first said that, in his opinion, there were 19 apples, the second thought there were 22, and the third said 23. When the friends opened the bag, it turned out that one of them was off by 1, another by 2, and t... | Answer: 20
Solution. Note that the first and third friends give odd predictions, which means they make errors of one parity, while the second makes errors of the other parity. Therefore, the second friend was off by 2 apples, so there are either 20 or 24 apples in the bag. Checking both options, we get the answer. | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.3 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first one said that, in his opinion, there were 16 apples, the second thought there were 19, and the third said 25. When the friends opened the bag, it turned out that one of them was off by 2, another by 4, a... | # Answer: 21
Solution. Notice that the second and third friends give odd predictions, which means they make an error of one parity, while the first makes an error of the other parity. Therefore, the first one was off by 5 apples, and the number of apples in the bag is either 11 or 21. Checking both options, we get the... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.4 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first one said that, in his opinion, there were 18 apples, the second thought there were 21, and the third said 22. When the friends opened the bag, it turned out that one of them was off by 2, another by 3, a... | Answer: 24 apples
Solution. Note that the first and third give even predictions, which means they make an error of one parity, while the third makes an error of the other parity. Therefore, the second one was off by 3 apples, so there are either 18 or 24 apples in the bag. Checking both options, we get the answer. | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4.1 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared twice as many balloons as Piglet, and Owl prepared three times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a hurry, stumbled, and some of the balloons burst. Eeyore received a ... | # Answer: 1
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 6 times more balloons than Piglet prepared. The next number after 31 that is a multiple of 6 is $36, 36: 6=6$, which means Piglet prepared no less than 6 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4.2 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared three times as many balloons as Piglet, and Owl prepared four times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a great hurry, stumbled, and some of the balloons burst. Eeyore ... | # Answer: 4
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 8 times more balloons than Piglet prepared. The next number after 60 that is a multiple of 8 is $64, 64: 8=8$, which means Piglet prepared no less than 8 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.3 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared twice as many balloons as Piglet, and Owl prepared four times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a great hurry, stumbled, and some of the balloons burst. Eeyore receiv... | # Answer: 2
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 7 times more balloons than Piglet prepared. The next number after 44 that is a multiple of 7 is 49, $49: 7=7$, which means Piglet prepared no less than 6 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1 Six children stood in a circle and whispered their favorite number to the neighbors on their right and left. Each child added the two numbers they heard and said them out loud. The picture shows what three of the children said. What number did the girl think of, next to whom there is a question mark?
 for a game. Each had to say one of the following phrases: "There is a liar below me" or "There is a k... | Answer: 17
Solution. The formation cannot consist of only liars (in such a case, the 3-7th person would be telling the truth). Consider the tallest knight, let's call him P. He could only say the first phrase, so the 1st and 2nd places must be occupied by liars, and P stands from the 3rd to the 7th place. Moreover, th... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2 On the Island of Truth and Lies, there are knights who always tell the truth, and liars who always lie. One day, 15 residents of the island lined up by height (from tallest to shortest, the tallest being the first) for a game. Each had to say one of the following phrases: "There is a liar below me" or "There is a k... | # Answer: 11
Solution. The formation cannot consist of only liars (in such a case, 4-8 would be telling the truth). Consider the tallest knight, let's call him R. He could only say the first phrase, so the positions 1, 2, and 3 must be occupied by liars, and R stands from 4 to 8. In addition, there is a liar who is sh... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3 On the Island of Truth and Lies, there are knights who always tell the truth, and liars who always lie. One day, 17 residents of the island lined up by height (from tallest to shortest, the tallest being the first) for a game. Each had to say one of the phrases: "There is a liar below me" or "There is a knight abov... | Answer: 14
Solution. The formation cannot consist of only liars (in such a case, 3-6 would be telling the truth). Consider the tallest knight, let's call him P. He could only say the first phrase, so the 1st and 2nd positions must be occupied by liars, and P stands from the 3rd to the 6th position. Moreover, there is ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. Each student in the 7th grade attends a math club or an informatics club, and two students attend both the math club and the informatics club. Two-thirds of all students attend the math club, and $40 \%$ attend the informatics club. How many students are there in this class? | Answer: 30.
Solution: Note that $40\%$ is $\frac{2}{5}$ of all students. Adding $\frac{2}{3}$ and $\frac{2}{5}$ gives $\frac{16}{15}$. The two students who are involved in both the math and informatics clubs were counted twice. Therefore, these two students represent $\frac{1}{15}$ of the entire class, meaning there a... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. How many seven-digit natural numbers exist, for which the product of the first three digits is 30, the product of the three digits in the center is 7, and the product of the last three digits is 15? | Answer: 4.
Solution: Let the number be $\overline{\operatorname{abcdefg}}$. According to the condition, $c d e=7$, which means one of these digits is 7, and the other two are 1. Since 30 and 15 are not divisible by $7, d=7, c=e=1$. The number $30=5 \cdot 6 \cdot 1$, so we get two three-digit numbers $\overline{a b c}$... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.5. At a round table, 10 people are sitting, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each of them was given a token. Then each of them passed their token to one of their two neighbors. After that, 5 people said: "I have one token," while the other 5 said: "I have no tok... | # Answer. 7.
Solution. Evaluation. After passing the chips, each person sitting at the table can have 0, 1, or 2 chips. The total number of chips will be 10. Note that if a person lies, they will state a number of chips that differs from the actual number by 1 or 2. Since the total number of chips according to the ans... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Having walked $4 / 9$ of the length of the bridge, the traveler noticed that a car was catching up to him, but it had not yet entered the bridge. Then he turned back and met the car at the beginning of the bridge. If he had continued his movement, the car would have caught up with him at the end of the bridge. Find ... | # Solution
From the condition, it follows that the time it takes for the car to approach the bridge is equal to the time it takes for the traveler to walk $4 / 9$ of the bridge. Therefore, if the traveler continues moving, by the time the car reaches the bridge, he will have walked ${ }^{8} / 9$ of the bridge. This me... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Vovochka is playing a computer game. If he scores less than 1000 points, the computer will add $20 \%$ of his score. If he scores from 1000 to 2000 points, the computer will add $20 \%$ of the first thousand points and $30 \%$ of the remaining points. If Petya scores more than 2000 points, the computer will add $20 ... | # Solution
It is clear that Petya scored more than 1000 points (otherwise his result would not exceed 1200) and less than 2000 (otherwise the result would not be less than 2500). Let's discard the 1200 points (the first thousand plus the prize for it). The remaining 1170 points constitute $130 \%$ of the points Petya ... | 470 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given various real numbers $p$ and $q$. It is known that there exists a number $x$ such that the equations $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0$ are satisfied. What values can the sum $p + q$ take? | Solution. Let $x$ be such that both equalities from the problem statement are satisfied. Then $x^{2}+p x+q=x^{2}+q x+p$, from which $(p-q) x=p-q$. Since $p \neq q$, it follows that $x=1$. Substituting this value into any of the equations, we find that $p+q+1=0$, that is, $p+q=-1$. Answer: -1. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a football tournament, seven teams played: each team played once with each other. Teams that scored thirteen or more points advance to the next round. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round? | Solution. Evaluation. Note that the total number of games will be $\frac{7 \cdot 6}{2}=21$. Then the total number of points does not exceed $21 \cdot 3=63$. This means that no more than $\left[\frac{63}{13}\right]=4$ teams can advance to the next round.
Example. We will show that 4 teams can advance to the next round.... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. Students from 9A, 9B, and 9C gathered for a line-up. Maria Ivanovna decided to count the number of attendees from each class. She found that there were 27 students from 9A, 29 students from 9B, and 30 students from 9C. Ilya Grigoryevich decided to count the total number of attendees from all three classes,... | Answer: 29.
Solution. Let $a, b, c$ be the actual number of students from 9 "A", 9 "B", 9 "C" respectively. From Maria Ivanovna's counts, it follows that $a \leqslant 29, b \leqslant 31, c \leqslant 32$, and from Ilya Grigoryevich's counts, it follows that $a+b+c \geqslant 92$. We get that
$$
92 \leqslant a+b+c \leqs... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. An empty $3 \times 51$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 153$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 152$, the num... | Answer: (a) 152. (b) 76.
Solution. Let's color the table in black and white so that the cell where the number 1 should be is black. The third rule in the condition means that consecutive numbers in the table are placed in a "snake" pattern, and the colors of their cells alternate, that is, all odd numbers, including 1... | 152 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.5. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 34^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees doe... | Answer: (a) $107^{\circ}$. (b) $47^{\circ}$.
Fig. 5: to the solution of problem 9.5
Solution. Given $A B=B X$ and $\angle B=34^{\circ}$, therefore $\angle A X B=\angle X A B=\frac{1}{2}\lef... | 107 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. The graph of the line $y=k x+l$ intersects the $O x$ axis at point $B$, the $O y$ axis at point $C$, and the graph of the function $y=1 / x$ at points $A$ and $D$, as shown in the figure. It turns out that $A B=B C=C D$. Find $k$, given that $O C=3$.
Fig. 6: to the solution of problem 9.6
Solution. Draw a line through point $D$ parallel to the $O y$ axis; let it intersect the $O x$ axis at point $D_{1}$ (Fig. 6). It is easy ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In front of a sweet-tooth lie five boxes of candies: the first box contains 11 candies, the second - 22 candies, the third - 33 candies, the fourth - 44 candies, the fifth - 55 candies. In one move, the sweet-tooth can take four candies from one box and distribute them, one candy to each of the remaining f... | Answer: 159.
Solution. Consider the number of candies in the boxes. Initially, these are 5 natural numbers that give different remainders when divided by 5. We will prove that as long as the sweet-tooth does not take candies from any box, these 5 numbers will always give different remainders when divided by 5. For thi... | 159 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. For real numbers $x$ and $y$, define the operation $\star$ as follows: $x \star y=x y+4 y-3 x$.
Calculate the value of the expression
$$
((\ldots)(((2022 \star 2021) \star 2020) \star 2019) \star \ldots) \star 2) \star 1
$$ | Answer: 12.
Solution. Let $t=(\ldots(((2022 \star 2021) \star 2020) \star 2019) \star \ldots) \star 4$. Then the value of the expression from the problem condition is
$$
\begin{aligned}
((t \star 3) \star 2) \star 1 & =((3 t+12-3 t) \star 2) \star 1=12 \star 2 \star 1= \\
& =(24+8-36) \star 1=(-4) \star 1=-4+4+12=12
... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.2.2. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=11$, $NC=39$, $AM=12$, $MB=3$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 705. (b) 298.5. | 705 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.2.3. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=3$, $NC=39$, $AM=10$, $MB=5$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 585. (b) 202.5. | 585 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.2.4. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=9$, $NC=39$, $AM=10$, $MB=5$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 675. (b) 247.5. | 675 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.1. An empty $3 \times 51$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 153$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 152$, the n... | Answer: (a) 152. (b) 76. | 152 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.2. An empty $3 \times 53$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1,2,3, \ldots, 159$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 158$, the num... | Answer: (a) 158. (b) 79. | 158 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.3. An empty $3 \times 55$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1,2,3, \ldots, 165$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 164$, the num... | Answer: (a) 164 . (b) 82. | 82 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.4. An empty $3 \times 57$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 171$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 170$, the n... | Answer: (a) 170 . (b) 85. | 170 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.5.1. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 34^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $107^{\circ}$. (b) $47^{\circ}$. | 107 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.5.2. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 32^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees do... | Answer: (a) $106^{\circ}$. (b) $48^{\circ}$. | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.5.3. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 36^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $108^{\circ}$. (b) $46^{\circ}$. | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.5.4. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 38^{\circ}$ and $\angle X D C = 54^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $109^{\circ}$. (b) $44^{\circ}$. | 109 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.1. The numbers $a$ and $b$ are such that each of the two quadratic trinomials $x^{2} + a x + b$ and $x^{2} + b x + a$ has two distinct roots, and the product of these trinomials has exactly three distinct roots. Find all possible values of the sum of these three roots.
(S. Berlov) | Answer: 0.
Solution: From the condition, it follows that the quadratic polynomials $x^{2}+a x+b$ and $x^{2}+b x+a$ have a common root $x_{0}$, as well as roots $x_{1}$ and $x_{2}$ different from it; in particular, $a \neq b$. Then $0=\left(x_{0}^{2}+a x_{0}+b\right)-\left(x_{0}^{2}+b x_{0}+a\right)=(a-b)\left(x_{0}-1\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$. It turns out that $\angle E D A=\angle F D C$. Find the angle $A B C$.
(A. Yakub... | Answer: $60^{\circ}$.
Solution. Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, the line $\ell$ passes through the center $O$ of the circle $\omega$.
Perform a reflection about $\ell$. Since $\angle E D A = \angle F D C$, the ray $D C$ will be transformed into the ray $D A... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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