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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 |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
![](https://cdn.mathpix.co... | Answer: 22.
Solution. Note that there are a total of 91 coins, so after all moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will be t... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
. There are exactly 10 tens: five of each parity, and thus, from this ten, five numbers will be added to each type. Therefore, the numbers of ea... | 225 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On a line, seven points A, B, C, D, E, F, G are marked in the given order. It turned out that $A G=23 \mathrm{~cm}, B F=17 \mathrm{~cm}$ and $\mathrm{CE}=9 \mathrm{~cm}$. Find the sum of the lengths of all segments with endpoints at these points. | Answer: 224 cm.
Instructions. There are five pairs of segments that sum up to AG and AG itself. There are three pairs of segments that sum up to BF and BF itself. There is one pair of segments that sum up to CE and CE itself. Therefore, the total length of all segments is $6 \cdot 23 + 4 \cdot 17 + 2 \cdot 9 = 138 + 6... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.2. Given a parallelogram $A B C D$, where $A B<A C<B C$. 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 $D$; moreover, segments $A D$ and $C E$ intersect. It turns out that $\angle A B F=\angle D C E$. Find th... | Answer: $60^{\circ}$.
Solution. Since $D$ lies outside $\omega$, angle $A B C$ is acute. Let $A^{\prime}$ be the second intersection point of $D C$ and $\omega$. Since $B C>A C$, we have $\angle D C A=\angle C A B>\angle C B A=\angle D A^{\prime} A$; thus, $A^{\prime}$ lies on the extension of segment $D C$ beyond poi... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Option 1. To a confectionery factory for cake packaging, 5 rolls of ribbon, each 60 m long, were delivered. How many cuts need to be made to get pieces of ribbon, each 1 m 50 cm long? | Answer: 195.
Solution: From one roll, 40 pieces of ribbon, each 1 m 50 cm long, can be obtained. For this, 39 cuts are needed. Therefore, a total of $5 \cdot 39=195$ cuts are required. | 195 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1
Given a rectangular grid. We will call two cells adjacent if they share a side. Let's count the number of cells that have exactly four adjacent cells. It turned out to be 23. How many cells have exactly three adjacent cells? | Answer: 48.
Solution. Let $a$ and $b$ be the sides of the rectangle. The total number of cells that have exactly 4 neighbors by side is $(a-2)(b-2)$, and on the other hand, there are 23 such cells. Since 23 is a prime number, the numbers $a-2$ and $b-2$ are equal to the numbers 1 and 23 in some order. The number of ce... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 3. Variant 1
Three rectangles A, B, and C are drawn on the sheet (see figure).
Rectangles A and B have the same width, while rectangles B and C have the same length (width - top to bottom... | Answer: 20.
Solution: Let rectangle A have a length of $a$ cm and a width of $b$ cm. If the length is increased by 3 cm, the area increases by $3 b$. Therefore, $3 b=12, b=4$. The area of rectangle B is larger than that of A by $24 \mathrm{~cm}^{2}$, so the length of rectangle B is $24: 3=8$ cm. Therefore, the length ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 4. Variant 1
The train was moving from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 15 minutes. After that, the driver had to increase the train's speed by 4 times to arrive at point B on schedule. How many minutes does the train travel from poi... | Answer: 40.
Solution. Let's represent the train's path as a segment divided into 8 parts. Denote the intervals it travels in equal time with arcs.
Since the train stopped for 15 minutes, it... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
An apple, three pears, and two bananas together weigh 920 g; two apples, four bananas, and five pears together weigh 1 kg 710 g. How many grams does a pear weigh? | Answer: 130
Solution: From the first condition, it follows that two apples, four bananas, and six pears together weigh 1 kg 840 g. Therefore, a pear weighs $1840-1710=130$ g. | 130 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7. Option 1
Diligent Masha wrote down in a row all natural numbers from 372 to 506 inclusive. Then she calculated two sums: first, the sum of all odd numbers in this row, and then the sum of all even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get? | Answer: 439.
Solution. Let's divide all numbers from 372 to 506, except 506, into pairs such that each even number corresponds to the next odd number (this is possible because the last number is 505). In each pair, the odd number will be 1 greater than the even one. There will be (505 - 371) : 2 = 67 pairs in the rang... | 439 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 11 sticks, and laid out a pattern inside the triangle such that the triangle was divided into smaller triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle wi... | Answer: 198.
Solution (1st method). All the sticks can be divided into those that go horizontally, at an angle to the right, and at an angle to the left:
In the first such row, there is 1 s... | 198 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3. Answer. $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$. | Solution.
Let $O$ be the common center of the given circles. From the condition, it follows that $BO$ and $CO$ are the bisectors of angles $ABC$ and $BCD$ ($O$ is the incenter) and, moreover... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.6. On the board, three natural numbers were written: two ten-digit numbers $a$ and $b$, as well as their sum $a+b$. What is the maximum number of odd digits that could have been written on the board?
(I. Bogdanov, P. Kozhevnikov) | Answer: 30.
Solution: Note that the number $a+b$ has no more than 11 digits, so in total, no more than 31 digits are written on the board. At the same time, all three numbers $a, b, a+b$ cannot be odd simultaneously. Therefore, one of their last three digits is even, which means that there are no more than 30 odd digi... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. From two settlements, which are 60 km apart, two cars set off in the same direction at the same time. The first car, traveling at a speed of 90 km/h, caught up with the second car after three hours. What is the speed of the second car? | Solution. The closing speed of the cars is $60: 3=20(\kappa m / h)$, so the speed of the second car is $90-20=70(km / h)$.
Answer: 70 km $/$ h.
Comment. The correct answer was obtained through correct reasoning - 7 points. The closing speed was correctly found, but the wrong answer was given - 3 points. | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. On a grid sheet of size $100 \times 100$, several non-overlapping cardboard isosceles right triangles with a leg of 1 were placed; each triangle occupies exactly half of one of the cells. It turned out that each unit segment of the grid (including boundary segments) is covered by exactly one leg of a triangle. Fi... | Answer. $49 \cdot 50=2450$ cells
Solution. Let $n=50$. We will call a triangle upper if it is located above the line containing its horizontal leg, and lower otherwise. Number the horizontal lines of the grid from bottom to top with numbers from 0 to $2n$.
Denote by $u_{k}$ (respectively $d_{k}$) the number of segmen... | 2450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. Kolya drew 10 segments and marked all their intersection points in red. Counting the red points, he noticed the following property: on each segment, there are three red points. a) Provide an example of the arrangement of 10 segments with this property. b) What is the maximum number of red points for 10 segments wi... | Answer: b) 15. Solution. a) See the example in the figure. As another example, you can take two copies of the right part of the figure from problem 7.5. b) The example given in the figure shows that 15 red points can be obtained. Let's prove that this is the maximum possible number. Number all 10 segments and write dow... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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 |
Problem 4.5. Hooligan Dima laid out a structure 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 diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire struct... | 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).
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.4. Misha made himself a homemade dartboard during the summer at the cottage. The round board is divided by circles into several sectors - darts can be thrown into it. Points are awarded for hitting a sector as indicated on the board.
Misha threw 8 darts 3 times. The second time he scored twice as many points... | Answer: 48.
Solution. The smallest possible score that can be achieved with eight darts is $3 \cdot 8=24$. Then, the second time, Misha scored no less than $24 \cdot 2=48$ points, and the third time, no less than $48 \cdot 1.5=72$.
On the other hand, $72=9 \cdot 8$ is the highest possible score that can be achieved w... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 $L M$, given that $A K=4, B L=31, M C=3$.
. 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?
 | 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 $D Z$.
 What... | Answer: 1008 numbers.
Solution. Consider an arbitrary coloring of the table. Let there be at least two successful numbers, and let $a$ be the smallest of them, and $b$ be the largest.
Divide $b$ by $a$ with a remainder: $b = qa + r$, where $0 \leq r < a$. If $r \neq 0$, then the number $qa$ is also successful, and $a... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.6. Ten-digit natural numbers $a, b, c$ are such that $a+b=c$. What is the maximum number of their 30 digits that can be odd?
(I. Bogdanov)
# | # Answer. 29.
Solution. Note that if $a+b=c$, then all three numbers $a, b$, $c$ cannot be odd simultaneously. Therefore, among them, there is at least one even number, and the last digit of this number will also be even. Thus, among the 30 digits, there is at least one even digit, and the number of odd digits is no m... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.8. Given a $1000 \times 1000$ chessboard. A figure called a cheetah, from any cell $x$, attacks all cells of a $19 \times 19$ square centered at $x$, except for the cells that are in the same row or column as $x$. What is the maximum number of non-attacking cheetahs that can be placed on the board?
(I. Bogdanov) | Answer: 100000.
Solution: Divide the board into $100^{2}$ squares of size $10 \times 10$. We will show that in each square, no more than 10 leopards can stand without attacking each other - from this it will follow that the total number of leopards cannot exceed $100^{2} \cdot 10=100000$.
Consider an arbitrary square... | 100000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. There are 30 logs, the lengths of which are 3 or 4 meters, and their total length is one hundred meters. How many cuts can be made to saw all these logs into pieces 1 meter long? (Each cut saws exactly one log). | Answer: 70.
Solution. First method The total length of the logs is 100 meters. If it were one log, 99 cuts would be needed. Since there are 30 logs, 29 cuts have already been made. Therefore, $99-29=70$ cuts remain to be made.
Second method Let's find the number of logs of each type. If all were 3-meter logs, their t... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. All natural numbers from 1 to 1000 inclusive are divided into two groups: even and odd. In which group is the sum of all digits used to write the numbers greater, and by how much? | Answer. The sum of the digits of odd numbers is 499 more.
Solution. The sum of the digits of the number 1 is equal to the sum of the digits of the number 1000; the other numbers can be divided into pairs: $2-3,4-5,6-7,8-9, \ldots, 998-999$. In each pair, the unit digit of the odd number is 1 more than that of the even... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. In the product of three natural numbers, each factor was decreased by 3. Could the product have increased by exactly 2022 in this case? | Answer. Yes, it could.
Solution. The product $1 \cdot 1 \cdot 678$ serves as an example. After the specified operation, it becomes $(-2) \cdot(-2) \cdot 675=2700=678+2022$.
Remark. The given example is the only one. Here is how to come up with it. Suppose two of the factors were 1, and the third was $a$. Their produc... | 678 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Daisies grew along the path. Between every two daisies, a cornflower grew, and then between each cornflower and daisy - a dandelion. It turned out that now 101 flowers grow along the path. How many daisies grow along the path? | Solution. If at some point there are $n$ flowers growing, then there are $n-1$ gaps between them, so $n-1$ flowers grow between them, and the total becomes $2n-1$ flowers. (The absence of this phrase does not result in a score reduction). Let's move to the solution from the end. After the dandelions appear, there are 1... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. Petya and three of his classmates started a 100-meter race at the same time, and Petya came in first. After 12 seconds from the start of the race, no one had finished yet, and the four participants had run a total of 288 meters. When Petya finished the race, the other three participants had 40 meters left to run i... | Solution: The runners were supposed to run a total of 400 meters, and in 12 seconds they ran 288 meters, meaning that in 1 second the runners run $288: 12=$ 24 meters. When Petya finished, the runners had run $400-40=360$ meters, so from the start, $360: 24=15$ seconds had passed. Therefore, Petya runs at a speed of $1... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Simplify the expression: $\frac{8}{1+a^{8}}+\frac{4}{1+a^{4}}+\frac{2}{1+a^{2}}+\frac{1}{1+a}+\frac{1}{1-a}$ and find its value at $a=2^{-\frac{1}{16}}$. | Solution. Add the last two terms, then add the obtained sum to the third term from the end, and so on. As a result, we get $\frac{16}{1-a^{16}}$. Substituting $a$ with the value $2^{-\frac{1}{16}}$, we get 32.
Answer: 32 | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the product of all roots of the equation $x^{4}+4 x^{3}-2015 x^{2}-4038 x+2018=0$. | Solution. Transform the equation to the form: $\left(x^{2}+2 x\right)^{2}-2019\left(x^{2}+2 x\right)+2018=0$. Introduce the substitution $\boldsymbol{t}=\boldsymbol{x}^{2}+2 \boldsymbol{x}$, we get the equation: $t^{2}-2019 t+2018=0$, the roots of which are $t_{1}=1, t_{2}=2018$. Then $x^{2}+2 x-1=0$ or $x^{2}+2 x-2018... | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Each participant in the school charity event brought either one encyclopedia, or three fiction books, or two reference books. In total, 150 encyclopedias were collected. After the event, two bookshelves in the library were filled, with an equal number of books on each. On the first shelf, there was one fifth of all ... | Solution. Let $x$ be the fifth part of the reference books, $y$ be the seventh part of all fiction books. Then on the first shelf, there are $-x+y+150$ books, and on the second shelf, there are $-4x+6y$ books. From the condition of the problem, the number of books on the shelves is equal, therefore, $x+y+150=4x+6y$, fr... | 267 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $L$ and $K$ are marked, respectively, and $M$ is the intersection point of segments $A K$ and $C L$. It is known that the area of triangle $A M C$ is equal to the area of quadrilateral $L B K M$. Find the angle $A M C$. | Solution. From the equality of the areas of triangle $A M C$ and quadrilateral $L B K M$, it follows that the areas of triangles $A C K$ and $C B L$ are equal (see the figure), because in this case, the same area of triangle $C M K$ is added to the equal areas. We have $S_{A C K}=\frac{1}{2} A C \cdot C K \cdot \sin \a... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 2. Clone 1
The teacher wanted to write an example for calculation on the board:
$$
1,05+1,15+1,25+1,4+1,5+1,6+1,75+1,85+1,95=?
$$
but accidentally forgot to write one comma. After this, Kolya went to the board and, correctly performing all the operations, obtained an integer result. What is it? | Answer: 27
## Solution
Let's calculate the sum
$$
1.05+1.15+1.25+1.4+1.5+1.6+1.75+1.85+1.95=13.5
$$
Therefore, in order to get an integer, the decimal point should be omitted in the number with the fractional part 0.5. As a result, instead of the number 1.5, the number 15 will appear on the board, and the result wi... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 3. Clone 1
On an island, there live knights who always tell the truth, and liars who always lie. Before a friendly match, 30 islanders gathered in T-shirts with numbers on them—arbitrary natural numbers. Each of them said: “I have a T-shirt with an odd number.” After that, they exchanged T-shirts, and each said: “I ... | # Answer: 15
## Solution
Notice that each knight exchanged a shirt with an odd number for a shirt with an even number, while each liar exchanged a shirt with an even number for a shirt with an odd number. This means that each knight exchanged with a liar, so there are no more knights than liars. On the other hand, ea... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 6. Clone 1
The figure shows a hexagon composed of identical equilateral triangles, each with an area of 10. Find the area of the shaded part.
# | # Answer: 110
## Solution
We will call the equilateral triangles that make up the original hexagon unit triangles. Consider each of the highlighted triangles inside the hexagon separately. The smallest one coincides with the unit triangle and has an area of 10. The medium triangle is inscribed in a hexagon made up of... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. When the passengers entered the empty tram, half of them took seats. How many passengers entered at the very beginning, if after the first stop their number increased by exactly $8 \%$ and it is known that the tram can accommodate no more than 70 people? | Answer: 50. From the condition, it follows that the number of passengers is divisible by 2 (since half took seats) and by 25 (8% is $2/25$ of the total number). Therefore, the initial number of passengers was divisible by 50, but it was less than 70, so it was 50.
Comment. Answer only - 0 points. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. The price of a ticket to the stadium was 25 rubles. After the ticket prices were reduced, the number of spectators at the stadium increased by $50 \%$, and the revenue from ticket sales increased by $14 \%$. What is the new price of a ticket to the stadium after the price reduction? | Answer: 19 rubles.
Let's denote: $a$ - the number of spectators coming to the stadium; $x$ - the ratio of the new ticket price to the price of 25 rubles. The new revenue is: on the one hand $(25 \cdot x) \cdot a \cdot 1.5$, on the other hand $25 \cdot 1.14 \cdot a$. From the equality $(25 \cdot x) \cdot a \cdot 1.5 = ... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. In triangle $A B C$, the bisectors $A D$ and $C E$ were drawn. It turned out that $A E + C D = A C$. Find the angle $B$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: angle $B$ is equal to $60^{\circ}$.
Let $\alpha$ and $\gamma$ be the measures of angles $A$ and $C$ of triangle $ABC$. On side $AC$, we mark segment $AF$ equal to $AE$. From the condition, we get the equality of segments $CD$ and $CF$. Let the bisectors $AD$ and $CE$ intersect segments $EF$ and $FD$
. After 30 exchanges, the number of stickers will increase by $30 * 4 = 120$. Initially, Petya had one sticker, so after 30 exchanges, he will have $1 + 120 = 121$. | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.9. On a plane, $N$ points are marked. Any three of them form a triangle, the angles of which in degrees are expressed by natural numbers. For what largest $N$ is this possible
$$
\text { (E. Bakayev) }
$$ | Answer: 180.
Fig. 1
First Solution. Example. First, we will show that the required is possible when $N=180$. Mark 180 points on the circle, dividing it into 180 equal arcs, each $2^{\circ}$... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the game "Sportlotto-Shish," the main prize is drawn according to the following rules. Each person in the studio independently writes down any number of different pairs of different integers from the set from 1 to 5. If some participants have written down the same pairs, these participants share the main prize. H... | Solution. A total of different pairs can be formed $5 * 4 / 2=10$. It is taken into account that the digits in the pair are different and the order of the digits within the pair does not matter. Each pair can be written down or not written down by the participant. A total of different sets can be formed $2^{10}=1024$. ... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In a row, 33 weights are arranged in ascending order. It is known that any four consecutive weights can be distributed on two scales so that equilibrium is achieved. The third weight weighs 9 g, and the ninth weighs 33 g. How much does the 33rd weight weigh? | Solution. Let the weights of the weights be $a_{1}<a_{2}<\cdots<a_{33}$. For all $k=1,2, \ldots, 30$ the equalities $a_{k}+a_{k+3}=a_{k+1}+a_{k+2}$, equivalent to $a_{k+3}-a_{k+2}=$ $a_{k+1}-a_{k}$, hold. Let $a_{4}-a_{3}=a_{2}-a_{1}=d$ and $a_{5}-a_{6}=a_{3}-a_{2}=\mathrm{c}$, then $a_{6}-$ $a_{5}=a_{4}-a_{3}=d, a_{7}... | 129 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The least common multiple of seven natural numbers is 2012. Find the smallest possible sum of these numbers. | Solution: Let's factorize the number 2012 into prime factors $2012=2^{2} \cdot 503$.
One of the numbers, the LCM of which is 2012, must be divisible by $2^{2}=4$, and one (possibly the same) - by 503. If the same number is divisible by both 4 and 503, then it is divisible by 2012, and the sum of seven numbers is not l... | 512 | Number Theory | 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.
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.1. A dandelion blooms in the morning, remains yellow for this and the next day, turns white on the third morning, and by evening of the third day, it has lost its petals. Yesterday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and today there are 15 yellow and 11 white. How many yellow dandel... | Solution: All yellow dandelions the day before yesterday are the white dandelions of yesterday and the white dandelions of today. Therefore, the day before yesterday there were $14+11=25$ yellow dandelions.
Answer: 25 dandelions.
Note: The number of yellow dandelions yesterday and today is not needed for solving the ... | 25 | Other | math-word-problem | Yes | Yes | olympiads | false |
# 3.1. Condition:
The number 4597 is displayed on the computer screen. In one move, it is allowed to swap any two adjacent digits, but after this, 100 is subtracted from the resulting number. What is the largest number that can be obtained by making no more than two moves? | Answer: 8357
## Solution.
The first digit cannot exceed 8, since to obtain the other digits, you need to move the nine forward by two places and subtract one from it. Note that in any other example, the first digit will be less than 8, as we can only get an eight from a nine, and all other digits are less than 9. The... | 8357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9-5. Two three-digit numbers are written on the board in a multiplication example. If the multiplication sign is replaced with 0, a seven-digit number is obtained, which is an integer multiple of the product. By what factor exactly | Answer: 73.
Solution. Let the original numbers be denoted by $a$ and $b$. Then the specified seven-digit number will have the form $10000a + b$. According to the condition, $10000a + b = nab$, from which we get $b = \frac{10000}{na - 1}$.
Note that the numbers $a$ and $a-1$ do not have common divisors, so $na - 1 = p... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. Polina has two closed boxes - a square one and a round one. She was told that the round one contains 4 white and 6 black balls, while the square one contains 10 black balls. In one move, Polina can take a ball from any box without looking and either throw it away or move it to the other box. Polina wants to make t... | Answer: 15
Solution. Suppose that in the end, both boxes still contain both white and black balls. Then the last action Polina took was to draw a ball of a certain color from one of the boxes, and there were still both black and white balls left. But she could not have done this with certainty, meaning she could have ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.2. In Olya's black box, there are 5 apples and 7 pears, and in the white box, there are 12 pears. In one move, Olya can blindly take a fruit from any box and either eat it or move it to the other box. Olya wants the contents of the boxes to be the same. What is the minimum number of moves Olya can guarantee to achiev... | # Answer: 18
Solution. Suppose that in the end both boxes still contain both apples and pears. Then the last action Olya took was to take some fruit from a box, and there were still both apples and pears left. Because of this, she could not guarantee that she would take the needed fruit, as she might have picked the w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.3. Inna has two closed boxes - a square one and a round one. She was told that the round box contains 3 white and 10 black balls, while the square box contains 8 black balls. In one move, Inna can, without looking, take a ball from any box and either throw it away or move it to the other box. Inna wants to make the c... | Answer: 17
Solution. Suppose that in both boxes, there are both white and black balls left in the end. Then the last action Inna took was to draw a ball of a certain color from one of the boxes, and there were still both black and white balls left. However, she could not have done this with certainty, i.e., she might ... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.4. In Zhenya's black box, there are 8 bananas and 10 mangoes, and in the white box - 12 mangoes. In one move, Zhenya, without looking, can take a fruit from any box and either eat it or move it to the other box. Zhenya wants the contents of the boxes to be the same. What is the minimum number of moves Zhenya can guar... | # Answer: 24
Solution. Suppose that in the end, both boxes still contain both bananas and mangoes. Then the last action was Zhenya taking some fruit from a box, and both types of fruit remained. Because of this, she could not guarantee picking the right fruit, as she might have picked the wrong one.
If removing mango... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.1. A chocolate bar weighed 250 g and cost 50 rubles. Recently, to save money, the manufacturer reduced the weight of the bar to 200 g, and increased its price to 52 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 30
Solution. We will calculate by what percentage the cost of one kilogram has increased. Before the increase, 1 kg cost 200 rubles, and after the increase, it costs $52 \cdot 5=260$ rubles. The increase is 60 rubles per kg, which originally cost 200 rubles. This is $\frac{60}{200} \cdot 100 \% = 30 \%$ | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.2. A chocolate bar weighed 400 g and cost 150 rubles. Recently, to save money, the manufacturer reduced the weight of the bar to 300 g, and increased its price to 180 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 60
Solution. We will calculate by what percentage the cost of 1 kg 200 g has increased. Before the increase, this amount cost 450 rubles, and after the increase, it costs $180 \cdot 4=720$ rubles. The increase is 270 rubles on 1.2 kg, which originally cost 450 rubles. This is $\frac{270}{450} \cdot 100 \%=$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.4. A package of milk with a volume of 1 liter cost 60 rubles. Recently, in order to save money, the manufacturer reduced the volume of the package to 0.9 liters, and increased its price to 81 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 50
Solution. We will calculate by what percentage the cost of 9 liters of milk has increased. Before the increase, 9 liters cost $9 \cdot 60=540$ rubles, and after the increase, $-81 \cdot 10=810$ rubles. The increase amounts to 270 rubles for 9 liters. This amount previously cost 540 rubles. This is $\frac{... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. Galia thought of a number, multiplied it by N, then added N to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 2021 less than the originally thought number. What is N? | Answer: 2022
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numb... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.2. Galia thought of a number, multiplied it by $\mathrm{N}$, then added $\mathrm{N}$ to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 7729 less than the originally thought number. What is N? | Answer: 7730
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numb... | 7730 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.3. Galia thought of a number, multiplied it by N, then added N to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 100 less than the originally thought number. What is N? | Answer: 101
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numbe... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 14 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 13 girls in thi... | # Answer: 15
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 7 pizzas. Then $m: 13=7: x$, from which $m x=91$. The number 91 has only one divisor greater than 13, which is 91. Therefore, $m=91, x=1$,... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.3. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 22 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 13 girls in thi... | # Answer: 23
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 11 pizzas. Then $m: 13 = 11: x$, from which $m x = 143$. The number 143 has only one divisor greater than 13, which is 143. Therefore, $m ... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. How many natural numbers $\mathrm{N}$ greater than 900 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $900, N+15,2 N$ exactly two are four-digit numbers? | Answer: 5069
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-900, \mathrm{~N}+15,2 \mathrm{~N}, 3 \mathrm{~N}$. Four-digit numbers can only be two consecutive ones. If the four-digit numbers are $\mathrm{N}-900$ and $\mathrm{N}+15$, then $2 \mathrm{... | 5069 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.2. How many natural numbers $\mathrm{N}$ greater than 300 exist such that among the numbers $4 \mathrm{~N}, \mathrm{~N}-$ $300, N+45,2 N$ exactly two are four-digit numbers? | Answer: 5410
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-300, \mathrm{~N}+45,2 \mathrm{~N}, 4 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-300$ and $\mathrm{N}+15$, then $2 \mathrm{~N}$ ... | 5410 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.3. How many natural numbers $\mathrm{N}$ greater than 700 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $700, N+35,2 N$ exactly two are four-digit numbers? | Answer: 5229
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+35$, and if we write the numbers in ascending order, we get $\mathrm{N}-700, \mathrm{~N}+35,2 \mathrm{~N}, 3 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-700$ and $\mathrm{N}+35$, then $2 \mathrm{~N}$ ... | 5229 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Misha calculated the products $1 \times 2, 2 \times 3$, $3 \times 4, \ldots, 2017 \times 2018$. For how many of them is the last digit zero? | Solution. The last digit of the product depends on the last digits of the factors. In the sequence of natural numbers, the last digits repeat every ten. In each ten, in the sequence of products, four products end in zero: ... $4 \times \ldots 5, \ldots 5 \times \ldots 6, \ldots 9 \times \ldots 0, \ldots 0 \times \ldots... | 806 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. There is a $6 \times 6$ square, all cells of which are white. In one move, it is allowed to change the color of both cells in any domino (a rectangle of two cells) to the opposite. What is the minimum number of moves required to obtain a square with a checkerboard pattern? Don't forget to explain why a smaller numbe... | Answer: 18 moves.
Solution. Note that in the chessboard coloring of a $6 \times 6$ square, there are 18 black cells. At the same time, no two of them can be turned black in one move, because they are not covered by one domino. Therefore, at least 18 moves are required (at least one move per cell). This can be done in ... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A tea set consists of six identical cups and six identical saucers. Six different sets were packed into two boxes, all saucers in one box, and all cups in the other. All items are wrapped in opaque paper and are indistinguishable by touch. Find the minimum number of items that need to be taken out of these boxes to ... | Solution. Explanation by example. 18 cups and 12 saucers may not be enough. Let's number the sets. By randomly picking 18 cups, we might get all the cups from sets №1, 2, 3.
By randomly picking 12 saucers, we might get all the saucers from sets №5, 6. It is impossible to form a cup-saucer pair. Adding one more saucer ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In quadrilateral $A B C D$, points $X, Y, Z$ are the midpoints of segments $A B, A D, B C$ respectively. It is known that $X Y$ is perpendicular to $A B$, $Y Z$ is perpendicular to $B C$, and the measure of angle $A B C$ is $100^{\circ}$. Find the measure of angle $A C D$.
. We are given that $x+y=100^{\circ}$. We need to find $z+v$.
Th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. Three runners - Anton, Seryozha, and Tolya - are participating in a 100 m race. When Anton finished, Seryozha was 10 meters behind him, and when Seryozha finished, Tolya was 10 meters behind him. What distance was Tolya from Anton when Anton finished? (It is assumed that all boys run at constant, but of course, no... | Answer: 19 m.
Solution: Since Tolya's speed is $9 / 10$ of Seryozha's speed, by the time Anton finished, Tolya had run $9 / 10$ of the distance covered by Seryozha, which is $90 \cdot 9 / 10=81$ m. Thus, when Anton finished, Tolya and Anton were 19 m apart. | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The sum of the minuend, subtrahend, and difference is 555. Can the minuend be an integer? If yes, provide an example; if no, explain why. | # Answer. Hem.
Solution. Since the sum of the subtrahend and the difference is equal to the minuend, then the sum of the minuend, subtrahend, and difference is equal to twice the minuend, i.e., the minuend is 555/2 - a non-integer. | 555 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a psychiatric hospital, there is a chief doctor and many lunatics. During the week, each lunatic bit someone (possibly even themselves) once a day. At the end of the week, it turned out that each of the patients had two bites, and the chief doctor had a hundred bites. How many lunatics are there in the hospital | Answer: 20 madmen.
Solution: Let there be n madmen in the hospital. Then, by the end of the week, on the one hand, 7n bites were made, and on the other hand, $2 n+100$. i.e.
$7 n=2 n+100$, from which $n=20$. | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked respectively such that $\angle B A M = \angle C K M = 30^{\circ}$. Find $\angle A K D$. | Answer: $75^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle B M A=\angle C M K=60^{\circ}$, and then $\angle A M K=60^{\circ}$ (see Fig. 8.5a). Further reasoning can be done in different ways.
First method. Let $A H$ be the perpendicular from vertex $A$ to $M K$. Then the right triangle... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.6. Sasha drew a square of size $6 \times 6$ cells and alternately colors one cell at a time. After coloring the next cell, he writes down the number of colored cells adjacent to it. After coloring the entire square, Sasha adds up the numbers written in all the cells. Prove that no matter in what order Sasha colors th... | Solution. Consider all unit segments that are common sides for two cells. There are exactly sixty such segments - 30 vertical and 30 horizontal. If a segment separates two shaded cells, we will say that it is "painted." Note that when Sasha writes a number in a cell, he indicates the number of segments that were not pa... | 60 | Combinatorics | proof | Yes | Yes | olympiads | false |
Task 1. In a confectionery store, three types of candies are sold: caramels for 3 rubles, toffees for 5 rubles, and chocolates for 10 rubles. Varya wanted to buy exactly 8 candies of each type and took 200 rubles with her. In the morning, she saw announcements in the store: “When paying for three chocolates, get a free... | Answer: 72 rubles.
Solution. Since Varya will buy more than six but less than nine chocolates, she will get two free caramels. Then she will need to buy another 6 caramels, for which she will get two free toffees, after which she will need to buy another 6 toffees. Then she will spend a total of $8 \cdot 10 + 6 \cdot ... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In an album, a grid rectangle $3 \times 7$ is drawn. Robot Igor was asked to trace all the lines with a marker, and it took him 26 minutes (the robot draws lines at a constant speed). How many minutes will it take him to trace all the lines of a grid square $5 \times 5 ?$ | Answer: 30 minutes.
Solution. Note that the grid rectangle $3 \times 7$ consists of four horizontal lines of length 7 and eight vertical lines of length 3. Then the total length of all lines is $4 \cdot 7 + 8 \cdot 3 = 52$. It turns out that Igor spends $26 : 52 = \frac{1}{2}$ minutes on the side of one cell.
The rec... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. A messenger was riding a horse to deliver a message to Ilya Muromets. At some point, he noticed that Ilya Muromets had passed by him (and continued walking in the opposite direction). After 10 seconds (when the horse stopped), the messenger dismounted and ran to catch up with Ilya. How many seconds will it t... | Answer: 110.
Solution. Let Ilya Muromets walk at a speed of $x$ meters per second. Then the messenger's speed is $2x$, and the horse's speed is $10x$. Therefore, 10 seconds after the meeting, the distance between the messenger and the horse will be $10 \cdot x$ (Ilya Muromets has walked) $+10 \cdot 10 x$ (the horse ha... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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