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2. $A B C D E$ - a pentagon inscribed in circle $S$. Lines $D E$ and $A B$ are parallel, $B C=C D$, $A D=D E=20, \cos C B D=5 / 8$. Find the radius of circle $S$. | Answer: 16.
## Examples of how to write answers:
## $1 / 4$
0.25
1
# | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,875 |
3. On the board, there was a formula for a fractional-linear function of the form $f(x)=\frac{a x+b}{c x+2}$, where $a, b$, and $c$ are some real numbers (not necessarily distinct). Vasya swapped the numbers $b$ and $c$, and it turned out that the formula now written on the board is the inverse of the function $f(x)$. ... | Answer: $2 ;-2|2,-2|-2 ; 2$
## Examples of writing answers:
$1 / 4$
$-0.25 ; 10$
## Grade 9
# | 2;-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,876 |
1. (2 points) Can a square grid $8 \times 8$ be cut into 7 figures of equal perimeter? | Answer: Yes.
Solution: Here is one of the possible examples, of course, there can be others.
 | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,877 |
2. (3 points) Boy Vasya looked at the clock and saw some time with the number of hours greater than zero. The digits he saw he read as a three- or four-digit number (for example, 5:07 would have turned into 507 for him). Then he calculated the number of minutes that had passed since midnight. Could it have turned out t... | # Answer: No.
## Solution:
Let the time on the clock be $a$ hours and $b$ minutes. Then the number that Vasya saw is $100a + b$, and the number of minutes he calculated is $60a + b$. If $100a + b$ is divisible by $60a + b$, then their difference, $40a$, is also divisible by $60a + b$, but this cannot be, since for $a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,878 |
3. (3 points) Ani had two triangles, each of which was made up of exactly three sticks of natural length. Then, all these sticks were used to form a square with a side length of 5. What pairs of triangles could Ani have had? List all possible options and prove that there could be no others.
# | # Answer:
1) $5,5,5$ and $1,2,2$
2) $5,5,3$ and $3,2,2$;
3) $5,5,2$ and $3,3,2$;
4) $5,5,1$ and $4,3,2$;
5) $5,5,1$ and $1,4,4$;
## Solution:
Let's consider the reverse process: suppose we have four sticks of length 5, forming a square, and we need to turn them into six sticks that form triangles. To do this, we nee... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,879 |
4. (3 points) On the Island of Misfortune, there live knights, who always tell the truth, and liars, who always lie. One day, $n$ islanders gathered in a room. $\quad 30$
The first one said: "Exactly every first person present in this room is a liar."
The second one said: "Exactly every second person present in this ... | # Answer: 2
## Solution:
Let's note that among these people, there is definitely one knight, otherwise it would mean that the first islander is a liar telling the truth, which is impossible. Moreover, the first person is definitely a liar.
Furthermore, there is exactly one knight, since all the speakers contradict e... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,880 |
5. (3 points) Anya, Vanya, Danya, Sanya, and Tanya were collecting apples. It turned out that each of them collected a whole percentage of the total number of apples collected, and all these numbers are different and greater than zero. What is the minimum number of apples that could have been collected?
Answer: 20 | Solution:
Example: there are many different examples, for instance $1+2+3+4+10$ or $2+3+4+5+6=20$
Estimate (proof that there are no fewer than 20 apples):
Let Anna have collected the fewest apples, and this number is at least 2. Then we get that the total collected is no less than $2+3+4+5+6=20$ apples.
If Anna has... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,881 |
6. (3 points) A number written on the board is allowed to be multiplied by 5 or to have its digits rearranged (it is not allowed to place zero in the first position). Is it possible to obtain the hundred-digit number $222 \ldots 2225$ from the number 1 in this way?
# | # Answer: No.
## Solution:
Consider the reverse process: we will try to obtain the number $222 \ldots 2225$ from 1 using the specified operations.
Notice that the only way to get a number divisible by 5 by rearranging the digits is the number $222 \ldots 2225$ itself. Dividing it by 5, we get 44...445. In this numbe... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,882 |
7. (3 points) Given an equilateral triangle $A B C$. Points $D$ and $E$ are marked on sides $A B$ and $B C$ respectively. Prove that the length of the broken line $A E D C$ is not less than $2 A B$.
# | # Solution:
Construct equilateral triangles $A^{\prime} B C$ and $A^{\prime} B C^{\prime}$ as shown in the figure. On the segment $A^{\prime} B$, take a point $D^{\prime}$ such that $B D = B D^{\prime}$. By considering several pairs of congruent triangles, it is easy to prove that the length of the broken line $A E D ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,883 |
8. (4 points) In an $8 \times 8$ table, integers from 0 to 10 are arranged (naturally, numbers can repeat, not all specified numbers necessarily appear). It is known that in each rectangle $3 \times 2$ or $2 \times 3$, the sum of the numbers is 10. Find the smallest possible value of the sum of the numbers in the entir... | # Answer: 105
## Solution:
Let's look at the left part of the figure below. Consider the rectangles $3 \times 2$ and $2 \times 3$, which contain the left upper $2 \times 2$ square. The remaining parts of these two rectangles are the leftmost upper rectangles $1 \times 2$ and $2 \times 1$, shaded in gray. Since the su... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,884 |
1. (2 points) Can a $10 \times 10$ grid square be cut into 7 figures of equal perimeter? Answer: Yes. | Solution: Here is one of the possible examples, of course, there can be others.
 | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,885 |
2. (3 points) Boy Vasya looked at the clock and saw some time with the number of hours greater than zero. The digits he saw he read as a three- or four-digit number (for example, 5:07 would turn into 507 for him). Then he calculated the number of minutes that had passed since midnight. Could it be that twice the second... | # Answer: No.
## Solution:
Let the time on the clock be $a$ hours and $b$ minutes. Then the number Vasya saw is $100a + b$, and the number of minutes he calculated is $60a + b$. If $120a + 2b$ is divisible by $100a + b$, then their difference, $20a + b$, is also divisible by $100a + b$. However, this is not possible,... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,886 |
3. (3 points) Ani had two triangles, each of which was made up of exactly three sticks of natural length. Then, all these sticks were used to form a square with a side length of 4. What pairs of triangles could Ani have had? List all possible options and prove that there could be no others.
---
Translate the above te... | Answer:
1) $4,4,3$ and $1,2,2$;
2) $4,4,2$ and $2,2,2$;
3) $4,4,1$ and $3,2,2$;
4) $4,4,1$ and $3,3,1$;
## Solution:
Let's consider the reverse process: suppose we have four sticks of length 4, forming a square, and we need to turn them into six sticks that form triangles. For this, we need to divide one of the stic... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,887 |
4. (3 points) On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, $n$ islanders gathered in a room.
The first one said: "Exactly every second person in this room is a liar."
The second one said: "Exactly every third person in this room is a liar."
and so on
... | # Answer: 2
Solution:
According to the condition, there is at least one knight among these people. Therefore, there must be at least two people, otherwise the first person would be a knight who is lying.
Moreover, there is exactly one knight, as all the speakers contradict each other. Thus, there cannot be more than... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,888 |
5. (3 points) Anya, Vanya, Danya, Manya, Sanya, and Tanya were collecting apples. It turned out that each of them collected a whole percentage of the total number of apples collected, and all these numbers are different and greater than zero. What is the minimum number of apples that could have been collected?
Answer:... | Solution:
Example: there are many different examples, for instance $1+2+3+4+5+10$.
Evaluation (proof that there are no fewer than 25 apples):
Let Anya have collected the fewest apples, and this number is at least 2. Then we get that the total collected is no less than $2+3+4+5+6+7=27$ apples.
If Anya has 1 apple an... | 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,889 |
7. (3 points) Given a square $A B C D$. Points $F$ and $E$ are marked on sides $A B$ and $B C$ respectively. Prove that the perimeter of triangle $D E F$ is not less than $2 A C$.
# | # Solution:
Construct squares $A^{\prime} B C D^{\prime}$ and $A^{\prime} B C^{\prime} D^{\prime}$ as shown in the figure. On segment $A^{\prime} B$, take a point $F^{\prime}$ such that $B F = B F^{\prime}$. By considering several pairs of congruent triangles, it is easy to prove that the perimeter of triangle $D E F$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,890 |
8. (4 points) In an $11 \times 11$ table, integers from 0 to 10 are arranged (naturally, numbers can repeat, not all specified numbers necessarily appear). It is known that in each $3 \times 2$ or $2 \times 3$ rectangle, the sum of the numbers is 10. Find the smallest possible value of the sum of the numbers in the ent... | # Solution:
The right part of the figure below shows the table divided into 20 rectangles and one cell. This means that the sum of the numbers in the entire table is no less than the sum of the numbers in these 20 rectangles, which is 200.
An example is provided in the left part of the figure, cells filled with zeros... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,891 |
# Problem 8. (4 points)
Solve in integers: $6 x^{2}+5 x y+y^{2}=6 x+2 y+7$
Indicate the answer for which the value of $|x|+|y|$ is the greatest. Write the answer in the form ( $x ; y$ ). | # Problem 8. (4 points)
Solve in integers: $6 x^{2}+5 x y+y^{2}=6 x+2 y+7$
Indicate the answer for which the value of $|x|+|y|$ is the greatest. Write the answer in the form ( $x ; y$ ).
Answer: (-8; 25) | (-8;25) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,899 |
2. (2 points) In the quarry, there was a pile of 20160000 sand grains. A truck carried away from the quarry a quantity of sand grains per trip, which was some power of 8 (including possibly $8^{0}=1$). Could it have carried away the entire pile of sand in exactly 1000 trips? | Answer: No.
Solution: Note that 8 to any power always gives a remainder of 1 when divided by 7. Since 20160000 is divisible by 7, this means that the number of trips needed to transport the entire pile must be a multiple of 7, and 1000 is not divisible by 7. | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,902 |
3. (3 points) The number 2017 is written on the board. Petya and Vasya are playing the following game: in one move, they can subtract any natural divisor of the number written on the board, except the number itself, and write the result of this subtraction on the board instead of the original number. Petya starts. The ... | # Answer: Vasya
Solution: It is impossible to make a move only if the number on the board is one. At the same time, an odd number always turns into an even number. Vasya, receiving an even number, can always subtract 1 and get an odd number again. Thus, Vasya can always receive even numbers from Petya and give him odd... | Vasya | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,903 |
7. (4 points) Given a triangle $A B C$, where $A B=2, B C=8, A C=8$. From point $B$, a bisector is drawn which intersects the circumcircle of this triangle at point $D$. Find the value of $D I$, where $I$ is the center of the inscribed circle of triangle $A B C$. | Answer: $\frac{16}{3}$.
Solution: According to the trident lemma, $D I=A D$, and by the Law of Sines in triangle $A D B$, $A D=\frac{A B \sin \angle A B D}{\sin \angle A D B}=\frac{A B}{\sin \angle A C B} \sin \angle \frac{A B C}{2}$. Expressing $\frac{A B}{\sin \angle A C B}$ using the Law of Sines in triangle $A B C... | \frac{16}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,904 |
8. (4 points) At a knight's tournament, each knight gave each of his acquainted ladies as many flowers as she had acquainted knights, except for him. After this, every two knights arranged as many duels as they had common acquainted ladies. What was more: the given flowers or the arranged duels, and by how many times?
... | Solution: For each trio consisting of a lady and two of her acquainted knights, one duel will take place. Regarding the flowers, the first knight will give the lady one flower for introducing him to the second, and the second will give one for being introduced to the first, making a total of two flowers.
## Grade 9 2... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,905 |
1. (2 points) According to the standards of the International Crabbing Federation, the crabbing field consists of two areas, one of which is square, and the second has the same width, with a length of 25 to 30 meters inclusive. All dimensions must be whole numbers of meters, and the total area of the field must be betw... | Solution:
Let the side of the square be $x$. When the length of the rectangle is minimal and the area of the field is maximal, we find the maximal $x$ and vice versa, when the length of the rectangle is maximal and the area of the field is minimal, we find the minimal $x$.
We solve the equation $x^{2}-30 x=250$, obta... | 7or8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,906 |
2. (2 points) In the quarry, there was a pile of 20160000 sand grains. A truck carried away from the quarry a quantity of sand grains per trip, which was some power of 9 (including possibly $9^{0}=1$). Could it have carried away the entire pile of sand in exactly 2000 trips? | Answer: Yes.
Solution: $20160000=2240000 \cdot 9=248888 \cdot 9^{2}+8 \cdot 9=27654 \cdot 9^{3}+2 \cdot 9^{2}+8 \cdot 9==3072 \cdot 9^{4}+6 \cdot 9^{3}+2 \cdot 9^{2}+8 \cdot 9=$ $341 \cdot 9^{5}+3 \cdot 9^{4}+6 \cdot 9^{3}+2 \cdot 9^{2}+8 \cdot 9==37 \cdot 9^{6}+8 \cdot 9^{5}+3 \cdot 9^{4}+6 \cdot 9^{3}+2 \cdot 9^{2}+... | Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,907 |
3. (3 points) The number 2016 is written on the board. Petya and Vasya are playing the following game: in one move, they can subtract any natural divisor of the number written on the board, except the number itself, and write the result of this subtraction on the board instead of the original number. Petya starts. The ... | Answer: Petya
Solution: It is impossible to make a move only if the number on the board is one. At the same time, an odd number always turns into an even number. Petya, receiving an even number, can always subtract 1 and get an odd number again. Thus, Petya can always receive even numbers from Vasya and give him odd n... | Petya | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,908 |
7. (4 points) Given a triangle $ABC$, where $AB=4, \quad BC=4, \quad AC=1$. A bisector is drawn from point $A$, intersecting the circumcircle of this triangle at point $D$. Find the value of $DI$, where $I$ is the incenter of triangle $ABC$.
Answer: $\frac{8}{3}$. | Solution: According to the trident lemma, $D I=C D$, and by the law of sines in triangle $A D C$, $C D=\frac{A C \sin \angle C A D}{\sin \angle A D C}=\frac{A C}{\sin \angle A B C} \sin \angle \frac{C A B}{2}$. Expressing $\frac{A C}{\sin \angle A B C}$ using the law of sines in triangle $A B C$, we get $C D=\frac{B C}... | \frac{8}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,909 |
8. (4 points) At an art school, there were painters and photographers. Some of them knew each other, while others did not. Each painter drew a pair portrait for every two of their acquaintances who were photographers. Each photographer took a photo of each of their acquaintances who were painters, in turn, together wit... | Solution: For each triplet of artist $A$ and two photographers $B$ and $C$, one can find a photo of $A$ and $C$ taken by $B$, a photo of $B$ and $A$ taken by $C$, and a portrait of $B$ and $C$ painted by $A$. This means that there are twice as many photos. | Photographs\\twice\\many | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,910 |
1. (1 point) Do there exist four such distinct multiples of 3 that the sum of any three of them is divisible by the fourth? | Answer: Yes, they do exist.
Solution:
For example, the numbers $3, 6, 9$ and 18.
To verify, we only need to ensure that the sum of these numbers, which is 36, is divisible by each of them. | Yes,theydoexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,911 |
2. (2 points) Anya was making pancakes, planning for each of the three members of her family to get an equal number of pancakes. But something went wrong: every third pancake Anya couldn't flip; $40 \%$ of the pancakes that Anya managed to flip burned; and $\frac{1}{5}$ of the edible pancakes Anya dropped on the floor.... | Solution:
In total, Anya managed to save $\frac{2}{3} \cdot 0.6 \cdot 0.8=0.32$ of the number of pancakes. | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,912 |
3. (3 points) There are 20 plates arranged in a circle, and on them are 40 buns. In one move, it is allowed to take 2 buns from one plate and place them on 2 adjacent plates. Can it be achieved that all plates have an equal number of buns, regardless of the initial arrangement of the buns? | Answer: No, not in any case.
## Solution:
Let's color the plates black and white so that each black plate is adjacent to two white plates and vice versa.
When performing the operation described in the condition, the parity of the number of pies on the black plates does not change. In the required final arrangement, ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,913 |
4. (3 points) Given triangle $A B C$. Points $K, L$, and $M$ are placed on the plane such that triangles $A B K, L B C$, and $A M C$ are equal to $A B C$. What inequality sign should be placed between the semiperimeter of triangle $K L M$ and the perimeter of triangle $A B C$?
The vertices of the triangles are given i... | Answer: The perimeter of $A B C$ is not less than the semiperimeter of $K L M$.
Solution:
$A K+A M \geqslant K M$ by the triangle inequality in triangle $A K M$ (equality is achieved when this triangle is degenerate, i.e., point $A$ lies on segment $K M$). Similarly, $B L+B K \geqslant K L$ and $C L+C M \geqslant L M... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,914 |
5. (3 points) In a room, knights who always tell the truth and liars who always lie have gathered (both are definitely present). They were asked the question: "How many liars are in the room?". To this question, all possible answers from 1 to 100 were received (some possibly multiple times). How many liars could there ... | Answer: 99 or 100.
## Solution:
Since a total of 100 different answers were received and one of them is definitely correct, the remaining 99 answers are incorrect. This means there are at least 99 liars in the room. There cannot be more than 100 liars, as the correct answer does not exceed 100.
Let's now confirm tha... | 99or100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,915 |
6. (3 points) Vasya has a drawing tool - a triangle with angles $40^{\circ}$, $70^{\circ}$ and $70^{\circ}$. How can he use it to construct an equilateral triangle? | Solution:
Construct two triangles $A B C$ and $A B D$ with a common side $A B$ and angles at point $A$ of $70^{\circ}$. Then construct triangles $A C K$ and $A D L$ with angles at point $A$ of $40^{\circ}$, placed inside the previously constructed angles of $80^{\circ}$, such that $A K=A L$ and $C K=D L$.
Then triang... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,916 |
7. (4 points) With the number written on the board, one of the following operations is allowed:
1) If there is a digit in the original number that is not 9 and has two neighboring digits greater than 0, you can increase this digit by 1, and decrease the neighboring digits by 1.
2) Subtract 1 from any non-zero digit exc... | Answer: 3
Solution:
The first operation corresponds to subtracting a number of the form $910 \cdot 0$, the second and third - subtracting $70 \cdot 0$. Both these operations do not change the remainder of the original number when divided by 7, since 91 is divisible by 7.
$1001=c d o t 91$, and $888888=888 \cdot 1001... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,917 |
8. (5 points) From a $12 \times 12$ grid, a $4 \times 4$ square was cut out, lying at the intersection of the third to sixth horizontal lines and the same vertical lines. What is the maximum number of non-attacking rooks that can be placed on this field, if the rooks do not attack through the cut-out cells?
## Answer:... | # Solution:
The available part of the board is divided into the following four areas, in each of which no more than the specified number of rooks can be placed (see the left figure), as in each rectangle, no more rooks can be placed than the length of its shorter side. Therefore, the total number of rooks does not exc... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,918 |
1. (1 point) Do there exist four such distinct even numbers that the sum of any three of them is divisible by the fourth? | Answer: Yes, they do exist.
Solution:
For example, the numbers $2, 4, 6$ and 12.
To verify, it is sufficient to ensure that the sum of these numbers, which is 24, is divisible by each of them. | 2,4,6,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,919 |
2. (2 points) Anya was making pancakes, planning for each of the five members of her family to get an equal number of pancakes. But something went wrong: every fifth pancake Anya couldn't flip; $49\%$ of the pancakes that Anya managed to flip burned; and $\frac{1}{6}$ of the edible pancakes Anya dropped on the floor. W... | # Solution:
In total, Anya managed to save $0.8 \cdot 0.51 \cdot \frac{5}{6}=0.34$ of the pancakes. | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,920 |
3. (3 points) There are 30 plates arranged in a circle, and 60 buns are distributed among them. In one move, it is allowed to take 2 buns from one plate and place them on 2 adjacent plates. Can you achieve an equal number of buns on all plates from any initial arrangement?
Answer: No, not from any initial arrangement.... | # Solution:
Let's color the plates black and white so that each black plate is adjacent to two white plates and vice versa.
When performing the operation described in the condition, the parity of the number of buns on the black plates does not change. In the required final arrangement, this number is even, so if we c... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,921 |
4. (3 points) Given triangle $A B C$. Points $K, L$ and $M$ are placed on the plane such that triangles $K A M, C L M$ and $K L B$ are equal to triangle $K L M$. What inequality sign should be placed between the perimeter of triangle $K L M$ and the semiperimeter of triangle $A B C?$
The vertices of the triangles are ... | Answer: The perimeter of $KLM$ is not less than the semiperimeter of $ABC$.
Solution:
$AK + KB \geqslant AB$ by the triangle inequality in triangle $ABK$ (equality is achieved when this triangle is degenerate, i.e., point $K$ lies on segment $AB$). Similarly, $BL + CL \geqslant BC$ and $AM + CM \geqslant AC$.
Thus, ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,922 |
5. (3 points) In a room, knights who always tell the truth and liars who always lie have gathered (both are definitely present). They were asked the question: "How many liars are in the room?". To this question, all possible answers from 1 to 200 were received (some possibly multiple times). How many liars could there ... | Answer: 199 or 200.
Solution:
Since a total of 200 different answers have been received and one of them is definitely correct, the remaining 199 answers are incorrect. Therefore, there are at least 199 liars in the room. There cannot be more than 200, as the correct answer does not exceed 200.
Let's now confirm that... | 199or200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,923 |
6. (3 points) Vasya has a drawing tool - a triangle with angles $80^{\circ}$, $50^{\circ}$ and $50^{\circ}$. How can he use it to construct an equilateral triangle?
# | # Solution:
Construct two triangles $A B C$ and $A B D$ with a common side $A B$ and angles at point $A$ of $80^{\circ}$. Then construct triangles $A C K$ and $A D L$ with angles at point $A$ of $50^{\circ}$, placed inside the previously constructed angles of $80^{\circ}$, such that $A K=A L$ and $C K=D L$.
Then tria... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,924 |
7. (4 points) With the number written on the board, one of the following operations is allowed:
1) If there is a digit in the original number that is not equal to 9 and has two neighboring digits greater than 0, you can increase this digit by 1, and decrease the neighboring digits by 1.
2) Subtract 1 from any non-zero ... | # Answer: 3
Solution:
The first operation corresponds to subtracting a number of the form $910 \cdot 0$, the second and third - subtracting $70 \cdot 0$. Both these operations do not change the remainder of the original number when divided by 7, since 91 is divisible by 7.
$1001 = 91 \cdot 11$, and $999999 = 999 \cd... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,925 |
8. (5 points) From a $12 \times 12$ grid, a $4 \times 4$ square was cut out, lying at the intersection of the fourth to seventh horizontal and vertical lines. What is the maximum number of non-attacking rooks that can be placed on this field, if the rooks do not attack through the cut-out cells?
## Answer: 15
# | # Solution:
The available part of the board is divided into the following four areas, in each of which no more than the specified number of rooks can stand (see the left figure), as in each rectangle, no more rooks can stand than the length of its shorter side. Therefore, the total number of rooks does not exceed 15.
... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,926 |
1. (2 points) A numerical sequence is defined by the condition $x_{n+1}=3 x_{n}+4 x_{n-1}$. Can it be periodic but not constant? | Answer: Yes, it can.
Solution: For example, the sequence $(-1)^{n}$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,927 |
2. (2 points) Point $M$ lies on the side of a regular hexagon with side length 10. Find the sum of the distances from point $M$ to the lines containing the other sides of the hexagon. | Answer: $30 \sqrt{3}$.
Solution: The distance from a point on one side to the opposite side is equal to the distance between these sides and, consequently, the distance between two vertices of the hexagon located one apart. This distance is obviously equal to $2 \cdot \frac{\sqrt{3}}{2} \cdot 12=10 \sqrt{3}$.
Also, i... | 30\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,928 |
3. (2 points) For any quadratic trinomial $f(x)$, do there exist distinct numbers $a, b$, $c$ and $d$ such that $f(a)=b, f(b)=c, f(c)=d$ and $f(d)=a$? | Answer: No, not for any.
Solution: A counterexample is, in particular, $y=x^{2}$.
Firstly, none of the numbers $a, b, c$ and $d$ can be negative, since negative numbers cannot be values of this quadratic polynomial.
Secondly, the numbers 0 and 1 also do not work, as $f(x)$ does not map any of our numbers to themselv... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,929 |
4. (3 points) On the board, 15 different non-integer numbers were written. For each number $x$ of these fifteen, Vasya wrote down separately $[x]$ and $\frac{1}{\{x\}}$ in his notebook. What is the smallest number of different numbers Vasya could have obtained?
$[x]$ and $\{x\}$ denote the integer and fractional parts... | Answer: 4.
Solution:
Let's assume there are no more than 3 resulting numbers. Then, the original numbers can have a maximum of 3 different integer parts and 3 different fractional parts, which means there are no more than $3 \cdot 3=9$ variants of the original numbers.
An example for 4 numbers can be constructed in ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,930 |
5. (3 points) Let $p, q$ and $r$ be odd prime numbers. Prove that $p^{3}+q^{3}+3 p q r \neq r^{3}$. | Solution:
Let it be $p^{3}+q^{3}+3 p q r-r^{3}=0$. We factorize it. It becomes $(p+q-r)\left(p^{2}+\right.$ $\left.q^{2}+r^{2}+p r+q r-p q\right)=(p+q-r) \cdot \frac{1}{2} \cdot\left((p-q)^{2}+(p+r)^{2}+(q+r)^{2}\right)$.
The second bracket is always positive, so the first one must be 0, from which $r=p+q$, which is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,931 |
6. (4 points) Given three circles with radii 2, 3, and 5, each pair of circles touching each other externally at points $A, B$, and $C$. Find the radius of the inscribed circle of triangle $A B C$. | Answer: $\frac{9-\sqrt{21}}{2 \sqrt{7}}=\frac{9 \sqrt{7}-7 \sqrt{3}}{14}$
Solution:
Let's introduce the following notations: for the first circle, the center is point $O_{1}$, the radius is $a$; for the second circle, the center is point $O_{2}$, the radius is $b$; for the third circle, the center is point $O_{3}$, t... | \frac{9-\sqrt{21}}{2\sqrt{7}}=\frac{9\sqrt{7}-7\sqrt{3}}{14} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,932 |
7. (4 points) Find what $x+y$ can be equal to, given that $x^{3}-6 x^{2}+15 x=12$ and $y^{3}-6 y^{2}+15 y=16$. | Answer: 4.
Solution:
Let $u=x-2$ and $v=y-2$. Then the original equations transform into $u^{3}+3 u=-2$ and $v^{3}+3 v=2$. Adding these equations, we get $(u+v)\left(u^{2}-u v+v^{2}+3\right)=0$. The second bracket is always positive, so the first one must be zero, from which we have $x+y=u+v+4=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,933 |
8. (5 points) On a $9 \times 9$ checkerboard, there are 324 chips. We will call two cells neighbors if they share a side, or if they are the two end cells of the same row or column. Thus, each cell will have exactly 4 neighbors.
In one move, it is allowed to take 4 chips from one cell and place them on the 4 neighbori... | Answer: Not for any.
Solution 1:
Let's number the rows and columns from 0 to 10. The numbers of the row and column where a chip is located will be called its coordinates.
Notice that with the given operation, the remainder of the sum of all coordinates of all chips divided by 11 does not change. Initially, this rema... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,934 |
1. (2 points) A numerical sequence is defined by the condition $x_{n+1}=2 x_{n}+3 x_{n-1}$. Can it be periodic but not constant? | Answer: Yes, it can.
Solution: For example, the sequence $(-1)^{n}$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,935 |
2. (2 points) Point $M$ lies on the side of a regular hexagon with side length 12. Find the sum of the distances from point $M$ to the lines containing the other sides of the hexagon. | Answer: $36 \sqrt{3}$.
Solution: The distance from a point on one side to the opposite side is equal to the distance between these sides and, consequently, the distance between two vertices of the hexagon located one apart. This distance is obviously equal to $2 \cdot \frac{\sqrt{3}}{2} \cdot 12=12 \sqrt{3}$.
Also, i... | 36\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,936 |
3. (2 points) For any quadratic trinomial $f(x)$, do there exist distinct numbers $a, b$, and $c$ such that $f(a)=b, f(b)=c, f(c)=a$? | Answer: No, not for any.
Solution: A counterexample is, in particular, $y=x^{2}$.
Firstly, none of the numbers $a, b$, and $c$ can be negative, since negative numbers cannot be values of this quadratic polynomial.
Secondly, the numbers 0 and 1 also do not work, because $f(x)$ does not map any of our numbers to thems... | proof | Algebra | proof | Yes | Yes | olympiads | false | 20,937 |
4. (3 points) On the board, 20 different non-integer numbers were written. For each number $x$ from these twenty, Vasya wrote down separately $[x]$ and $\frac{1}{\{x\}}$ in his notebook. What is the smallest number of different numbers Vasya could have obtained?
$[x]$ and $\{x\}$ denote the integer and fractional part... | Answer: 5.
Solution:
Let's assume there are no more than 4 numbers in total. Then the original numbers can have a maximum of 4 different integer parts and 4 different fractional parts, which means there are no more than $4 \cdot 4=16$ variants of the original numbers.
An example for 5 numbers can be constructed in a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,938 |
5. (3 points) Let $p, q$ and $r$ be distinct prime numbers and $p^{3}+q^{3}+3 p q r=r^{3}$. Prove that the smallest of these three numbers is 2. | Solution:
Factorize the expression $p^{3}+q^{3}+3 p q r-r^{3}$. We get $(p+q-r)\left(p^{2}+q^{2}+r^{2}+\right.$ $p r+q r-p q)=(p+q-r) \cdot \frac{1}{2} \cdot\left((p-q)^{2}+(p+r)^{2}+(q+r)^{2}\right)$.
The second bracket is always positive, so the first one must be 0, from which we get $r=p+q$. Therefore, either $p$ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 20,939 |
6. (4 points) Given three circles with radii 1, 2, and 3, each pair of which touch each other externally at points $A, B$, and $C$. Find the radius of the inscribed circle of triangle $A B C$. | Answer: $\frac{(-30+15 \sqrt{2}+6 \sqrt{5}+3 \sqrt{10}))}{30}$.
Solution:
Let's introduce the following notations: for the first circle, the center is point $O_{1}$, the radius is $a$; for the second circle, the center is point $O_{2}$, the radius is $b$; for the third circle, the center is point $O_{3}$, the radius ... | \frac{(-30+15\sqrt{2}+6\sqrt{5}+3\sqrt{10}))}{30} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,940 |
7. (4 points) Find what $x+y$ can be equal to, given that $x^{3}+6 x^{2}+16 x=-15$ and $y^{3}+6 y^{2}+16 y=-17$. | Answer: -4
Solution: Let $u=x+2$ and $v=y+2$. Then the original equations transform into $u^{3}+4 u=1$ and $v^{3}+4 v=-1$. Adding these equations, we get $(u+v)\left(u^{2}-u v+v^{2}+4\right)=0$. The second bracket is always positive, so the first one must be zero, from which we have $x+y=u+v-4=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,941 |
8. (5 points) On a $11 \times 11$ checkerboard, there are 484 chips. We will call two cells neighbors if they share a side, or if they are the two end cells of the same row or column. Thus, each cell will have exactly 4 neighbors.
In one move, it is allowed to take 4 chips from one cell and place them on the 4 neighbo... | Answer: Not for any.
Solution 1:
Let's number the rows and columns with numbers from 0 to 10. The numbers of the row and column where a chip is located will be called its coordinates.
Notice that with the given operation, the remainder of the sum of all coordinates of all chips divided by 11 does not change. Initial... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,942 |
3. Point $O$ is the center of a certain circle, $A$ is a point outside the circle, $B$ is a point on the circle such that $AB$ is a tangent. $AO=6$. Find the greatest possible value of the area of triangle $AOB$. | Answer: 9.
## Examples of answer notations:
17
$1 / 7$
1.7 | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,943 |
1. On the plane, points $\mathrm{A}(-4,2), \mathrm{B}(-1,3), \mathrm{C}(2,1), \mathrm{D}(0,-3), \mathrm{E}(-4,-2)$ are given. What is the area of figure ABCDE? | Answer: $25.5 \| 51 / 2$ | 25.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,944 |
3. On the plane, points $\mathrm{A}(-4,0), \mathrm{B}(-1,3), \mathrm{C}(2,2), \mathrm{D}(1,0), \mathrm{E}(-3,-3)$ are given. What is the area of figure $\mathrm{ABCDE}$? | Answer: $18.5 \| 37 / 2$
## Examples of answer notation:
17
$1 / 7$
1.7 | 18.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,946 |
3. A five-digit number ABCDE, consisting of different digits, is divisible by both the three-digit number CDE and the two-digit number AB. Find the smallest possible value of ABCDE. | Answer: 12480
## Examples of answer recording:
12345
# | 12480 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,947 |
1. On Misfortune Island, a table tennis tournament was held in a round-robin format (i.e., everyone played against everyone else once). After each match, both participants approached the Chief Referee separately and reported the result. Among the participants, there were only three types of people: knights, who always ... | Answer: 16
## Examples of answer recording:
17
# | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,948 |
3. Solve the equation $\mathrm{p}^{3}-\mathrm{q}^{3}=2716$ in prime numbers. In your answer, specify the pair $(p ; q)$.
If there are several such pairs, specify the one for which $p-q$ is maximal. | Answer: $17 ; 13 \mid(17 ; 13)$
## Examples of answer notation:
$(37 ; 11)$
# | (17;13) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,952 |
1. Given a rectangular parallelepiped with sides 6.6 and $x$. For what values of the number $x$ is the product of the surface area and the perimeter (the sum of the lengths of all edges) of this parallelepiped not greater than $224 / 3$ of its volume? | Write the answer in the form of an interval.
Answer: $[4 ; 9] \|[4,9]$ | [4,9] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,954 |
3. Given a rectangular parallelepiped with sides 2.6 and $x$. For what values of the number $x$ is the product of the surface area and the perimeter (the sum of the lengths of all edges) of this parallelepiped no more than 88 times its volume? | Write the answer in the form of an interval.
Answer: $[3 ; 4] \|[3,4]$
## Examples of writing answers:
$[1 ; 5]$
# | [3;4] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,956 |
1. A circle with radius 12 and center at point $O$ and a circle with radius 3 touch internally at point $H$. The line $X H$ is their common tangent, and the line $O X$ is tangent to the smaller circle. Find the square of the length of the segment $O X$. | Answer: 162
2.. A circle with radius 12 and center at point $O$ and a circle with radius 4 touch internally at point $H$. Line $X H$ is their common tangent, and line $O X$ is tangent to the smaller circle. Find the square of the length of segment $O X$.
Answer: 192 | 162 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,957 |
3. A circle of radius 20 with center at point $O$ and a circle of radius 8 touch internally at point $H$. The line $X H$ is their common tangent, and the line $O X$ is tangent to the smaller circle. Find the square of the length of the segment $O X$. | Answer: 720
## Examples of how to write answers:
17
$1 / 7$
1.7 | 720 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,958 |
3. It is known that $\log _{b} c=2\left(\log _{a} c+\log _{b} a\right)$. Find $\log _{c} a+\log _{a} b$. | Answer: $0.5 \| 1 / 2$
Allowed for input are digits, a dot or comma, and the division sign
# | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,959 |
3. There are 12 students in the class. Each of them was asked how many friends they have in the class. Each number \( n \) was either mentioned exactly \( n \) times or not at all. What is the maximum value that the sum of all the numbers mentioned can take? | Answer: 90
Only digits are allowed as input.
## Problem 5. (3 points) | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,962 |
3. The numbers $x, y, z$ satisfy the relations $\sin (x+y+z)=\frac{9}{41}$ and $\cos x \cos y \cos z=\frac{5}{41}$. Find $\operatorname{tg} x \operatorname{tg} y+\operatorname{tg} y \operatorname{tg} z+\operatorname{tg} z \operatorname{tg} x$.
If there are multiple possible answers, write them in any order separated b... | Answer: $-7 ; 9 | 9 ;-7$
Allowed for input are digits, a dot or comma, division sign, semicolon
# | -7;9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,966 |
3. Solve the equation $p^{2}+q^{2}=r^{2}+6 p+4 q$ in prime natural numbers. In your answer, specify $p+q+r$. If there are multiple possible answers, list them in any order separated by a semicolon.
| Answer: $12 ; 10 \| 10 ; 12$
Allowed input: digits, semicolon
# | 12;10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,969 |
3. Each vertex of a tetrahedron with a volume of 216 was symmetrically reflected relative to the point of intersection of the medians of the opposite face. The four resulting points formed a new tetrahedron. Find its volume. | Answer: 1000
Allowed for input are digits, a dot or comma, and a division sign
# | 1000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,970 |
3. Circle $S_{1}$ intersects circle $S_{2}$ at points $A$ and $B$ and is tangent to circle $S_{3}$ at point $Z$. The common tangent of circles $S_{1}$ and $S_{3}$ intersects line $A B$ at point $C$. Also, through $C$ passes line $X Y$, which is tangent to circle $S_{2}$ at point $X$ and to circle $S_{3}$ at point $Y$.
... | Answer: 12
Allowed for input are digits, a period or comma, a division sign, a semicolon
# | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,971 |
3. Given a rectangle $\mathrm{ABCD}$. The length of side $\mathrm{BC}$ is one and a half times less than the length of side $\mathrm{AB}$. Point $\mathrm{K}$ is the midpoint of side AD. Point $\mathrm{L}$ on side $\mathrm{CD}$ is such that $\mathrm{CL}=\mathrm{AK}$. Point $\mathrm{M}$ is the intersection of line BL and... | Answer: 8
## Examples of answer recording:
# | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,975 |
3. A $10 x 10$ grid is filled with non-negative numbers. It is known that in each (vertical or horizontal) strip of 1 x 3, the sum of the numbers is 9. What is the maximum value that the sum of all numbers in the grid can take? | Answer: 306
Examples of writing answers:
14 | 306 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,976 |
1. Given a fractional-linear function of the form $f(x)=\frac{a x+b}{c x+d}$, where $a, b, c$ and $d$ are different natural numbers from 1 to 10. What is the smallest value that $f(4)$ can take? | Write the answer in the form of a proper fraction, do not round.
Example of answer recording: $27 / 34$
Answer: $6 / 49$ | \frac{6}{49} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,977 |
2. Given a fractional-linear function of the form $f(x)=\frac{a x+b}{c x+d}$, where $a, b, c$ and $d$ are distinct natural numbers from 1 to 15. What is the smallest value that $f(3)$ can take? | Write the answer in the form of a proper fraction, do not round.
Example of answer recording: $27 / 34$
Answer: $5 / 59$ | \frac{5}{59} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,978 |
3. Given a fractional-linear function of the form $f(x)=\frac{a x+b}{c x+d}$, where $a, b, c$ and $d$ are distinct natural numbers from 1 to 20. What is the smallest value that $f(2)$ can take? | Write the answer in the form of a proper fraction, do not round.
Example of answer recording: $27 / 34$
Answer: $4 / 59$ | \frac{4}{59} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,979 |
3. The function $f(x)$ is such that $f(x+1)+f(x-1)=5.2 f(x)$. It is known that $f(0)=-98$, and $\mathrm{f}(3)=249.2$. Find $f(1)$. | Answer: -10
## Examples of how to write answers:
17
$-1.7$
$1 / 7$
# | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,980 |
1. The graphs of a quadratic trinomial with a positive leading coefficient 2 and its derivative intersect at the vertex of the parabola with abscissa $x_{0}$ and at another point with abscissa $x_{1}$. Find the total area of both regions bounded by the graphs of the trinomial, its derivative, and the line symmetric to ... | Answer: 16
## Examples of answer notations:
17
$-1.7$
$1 / 7$
# | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,982 |
3. Find the total length of all intervals of positive solutions to the inequality $5^{[x]}\{x\}<$ 15. Square brackets denote the integer part, and curly braces denote the fractional part. | Answer: $2.75 \| 11 / 4$
## Examples of answer notations:
17
1.7
$1 / 7$
# | \frac{11}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 20,986 |
3. Around a circle, 129 (not necessarily integer) numbers from 5 to 25 inclusive were written. From each number, the logarithm to the base of the next number in the clockwise direction was taken, after which all the obtained logarithms were added. What is the maximum value that the sum of these logarithms can take? | Answer: 161
## Examples of answer notations:
17
$-1.7$
$1 / 7$
## 10th grade
# | 161 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,987 |
1. (2 points) Boy Vasya wrote down the non-zero coefficients of the polynomial $P(x)$ of the tenth degree in his notebook. Then he calculated the derivative of the resulting polynomial and wrote down its non-zero coefficients, and so on, until he obtained a constant, which he also wrote down.
What is the smallest numb... | Answer: 10
Solution: Evaluation: since the polynomial has a degree of 10, it definitely has a non-zero coefficient for $x^{10}$, let's call it $a$. Then the leading coefficient of the derivative of this polynomial is $10a$, the leading coefficient of the second derivative is $10 \cdot 9a$, and so on, the leading coeff... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,988 |
3. (3 points) In a certain country, there are 100 cities. The Ministry of Aviation requires that every two cities be connected by a two-way flight operated by exactly one airline, and that it should be possible to travel from any city to any other city using the flights of each airline (possibly with layovers). What is... | Solution: Evaluation: in order to connect all cities, the airline must have at least 99 flights. In total, there are $100 \cdot 99 / 2$ pairs of cities, which means there are no more than 50 airlines.
Example: let's denote the cities as $a_{1}, a_{2}, \ldots a_{50}, b_{1}, b_{2}, \ldots, b_{50}$ in some way. Then the ... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,990 |
4. (3 points) Prove that for $n=6002$ the sum of binomial coefficients with a step of 6, i.e., $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$, gives a remainder of 1 when divided by 3.
$C_{n}^{k}-$ the number of ways to choose $k$ items from $n$, which is $\frac{n!}{k!(n-k)!}$ if $0 \leqslant k \leqslant n$ and 0 in all oth... | # Solution:
It is easy to verify that $C_{n}^{k}=C_{n-3}^{k-3}+3 C_{n-3}^{k-2}+3 C_{n-3}^{k-1}+C_{n-3}^{k}$, therefore, $C_{n}^{k}$ gives the same remainder when divided by 3 as $C_{n-3}^{k-3}+C_{n-3}^{k}$.
Thus, $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$ gives the same remainder when divided by 3 as $C_{n-3}^{1}+C_{n-... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false | 20,991 |
5. (3 points) Three circles with radii 1, 1, and \(2 \sqrt{\frac{13-6 \sqrt{3}}{13}}\) are arranged such that the triangle formed by their centers is equilateral with a side length of \(\sqrt{3}\). Find the radius of the circumcircle of the triangle, each vertex of which is the point of intersection of two of these cir... | # Solution:
Let $A, B, C$ be the points of intersection of the first and second, second and third, and first and third circles, respectively, and $O_{1}, O_{2}, O_{3}$ be the centers of these circles.
The radii of the first and second circles are equal, so point $A$ will lie on the perpendicular bisector of $O_{1} O_... | 4\sqrt{3}-6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,992 |
6. (3 points) Find the total length of intervals on the number line where the inequalities $x<1$ and $\sin \log _{2} x<0$ are satisfied. | Answer:
$$
\frac{2^{\pi}}{1+2^{\pi}}
$$
## Solution:
Note that we are only interested in positive $x$.
If $\sin \log _{2} x<0$, then $2(k-1) \pi<\log _{2} x<2 k \pi$ for some integer $k$, which means $2^{2(k-1) \pi}<x<2^{2 k \pi}$. Since $x<1$, we are only interested in non-positive $k$. Let $n=-k$. The number $n$ ... | \frac{2^{\pi}}{1+2^{\pi}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 20,993 |
7. (4 points) Four out of six midpoints of the edges of a certain tetrahedron form a regular tetrahedron with an edge length of 1. Find the edges of the original tetrahedron. | Answer: Five edges of length 2, one edge of length $2 \sqrt{2}$.
Solution:
Let the vertices of the tetrahedron be $A, B, C, D$. Denote the midpoint of edge $A B$ as $M_{A B}$, and similarly for the other edges.
Suppose the midpoints of the skew edges of the original tetrahedron do not fall into the regular tetrahedr... | Five\edges\of\length\2,\one\edge\of\length\2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,994 |
8. (5 points) Prove that for positive $x, y, z$ the inequality
$$
(x+y+z)(4 x+y+2 z)(2 x+y+8 z) \geqslant \frac{375}{2} x y z
$$
holds. | Solution:
Consider the first two parentheses and note that
$$
(x+y+z)(4 x+y+2 z)=(2 x+y)^{2}+3 z(2 x+y)+2 z^{2}+x y
$$
Then we can rewrite the required inequality as
$$
\left(\frac{(2 x+y)^{2}+3 z(2 x+y)+2 z^{2}}{x y}+1\right) \cdot \frac{2 x+y+8 z}{z} \geqslant \frac{375}{2}
$$
Now fix $z$ and $2 x+y$ and shift $... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 20,995 |
1. (2 points) Boy Vasya wrote down the non-zero coefficients of the polynomial $P(x)$ of the ninth degree in his notebook. Then he calculated the derivative of the resulting polynomial and wrote down its non-zero coefficients, and so on, until he obtained a constant, which he also wrote down.
What is the smallest numb... | Solution: Evaluation: since the polynomial has degree 9, it definitely has a non-zero coefficient for $x^{9}$, let's call it $a$. Then the leading coefficient of the derivative of this polynomial is $9a$, the leading coefficient of the second derivative is $9 \cdot 8a$, and so on, the leading coefficients of the eighth... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,996 |
3. (3 points) In a certain country, there are 50 cities. The Ministry of Aviation requires that every two cities be connected by a two-way flight operated by exactly one airline, and that it should be possible to travel from any city to any other using the flights of each airline (possibly with layovers). What is the m... | Solution: Evaluation: in order to connect all cities, the airline must have at least 49 flights. In total, there are $50 \cdot 49 / 2$ pairs of cities, so there are no more than 25 airlines.
Example: let the cities be denoted as $a_{1}, a_{2}, \ldots a_{25}, b_{1}, b_{2}, \ldots, b_{25}$ in some way. Then the airline ... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,998 |
5. (3 points) Three circles with radii $\sqrt{3}, \sqrt{3}$, and $2 \sqrt{\frac{39-18 \sqrt{3}}{13}}$ are arranged such that the triangle formed by their centers is an equilateral triangle with side length 3. Find the radius of the circumscribed circle around the triangle, each vertex of which is the point of intersect... | Answer: $12-6 \sqrt{3}$
Solution:
Let $A, B, C$ be the points of intersection of the first and second, second and third, and first and third circles, respectively, and $O_{1}, O_{2}, O_{3}$ be the centers of these circles.
The radii of the first and second circles are equal, so point $A$ will lie on the perpendicula... | 12-6\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,999 |
6. (3 points) Find the total length of intervals on the number line where the inequalities $|x|<1$ and $|x|<3^{-\pi}$ are satisfied.
Answer:
$$
\frac{2}{1+3^{\pi}}
$$ | Solution:
It is sufficient to solve the problem for positive $x$, and then multiply the result by 2.
If $\sin \log _{3} x>0$, then $2 k \pi<\log _{3} x<(2 k+1) \pi$ for some integer $k$, which means $3^{2 k \pi}<x<3^{(2 k+1) \pi}$. Since $x<1$, we are only interested in negative $k$. Let $n=-k$. The number $n$ takes ... | \frac{2}{1+3^{\pi}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 21,000 |
7. (4 points) Four out of six midpoints of the edges of a certain tetrahedron form a regular tetrahedron with edge $\frac{1}{2}$. Find the edges of the original tetrahedron. | Answer: Five edges of length 1, one edge of length $\sqrt{2}$.
Solution:
Let the vertices of the tetrahedron be $A, B, C, D$. The midpoint of edge $A B$ is denoted as $M_{A B}$, and similar notations are introduced for the other edges.
Suppose the midpoints of the skew edges of the original tetrahedron do not fall i... | Five\edges\of\length\1,\one\edge\of\length\\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,001 |
1. (2 points) Boy Vasya wrote down the non-zero coefficients of the polynomial $P(x)$ of the eighth degree in his notebook. Then he calculated the derivative of the resulting polynomial and wrote down its non-zero coefficients, and so on, until he obtained a constant, which he also wrote down.
What is the smallest num... | Answer: 8
Solution: Evaluation: since the polynomial has degree 8, it definitely has a non-zero coefficient for $x^{8}$, let's call it $a$. Then the leading coefficient of the derivative of this polynomial is $8a$, the leading coefficient of the second derivative is $8 \cdot 7a$, and so on, the leading coefficients of... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,002 |
3. (3 points) In a certain country, there are 200 cities. The Ministry of Aviation requires that every two cities be connected by a two-way flight operated by exactly one airline, and that it should be possible to travel from any city to any other using the flights of each airline (possibly with layovers). What is the ... | # Answer: 100
Solution: Evaluation: for the airline to connect all cities, it must have at least 199 flights. The total number of city pairs is $200 \cdot 199 / 2$, so there are no more than 100 airlines.
Example: let's denote the cities as $a_{1}, a_{2}, \ldots a_{100}, b_{1}, b_{2}, \ldots, b_{100}$ in some way. Th... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,004 |
4. (3 points) Prove that for $n=9002$ the sum of binomial coefficients with a step of 6, i.e., $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$, gives a remainder of 1 when divided by 3.
$C_{n}^{k}-$ the number of ways to choose $k$ items from $n$, which is $\frac{n!}{k!(n-k)!}$ if $0 \leqslant k \leqslant n$ and 0 in all oth... | Solution:
It is easy to verify that $C_{n}^{k}=C_{n-3}^{k-3}+3 C_{n-3}^{k-2}+3 C_{n-3}^{k-1}+C_{n-3}^{k}$, therefore, $C_{n}^{k}$ gives the same remainder when divided by 3 as $C_{n-3}^{k-3}+C_{n-3}^{k}$.
Thus, $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$ gives the same remainder when divided by 3 as $C_{n-3}^{1}+C_{n-3}... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false | 21,005 |
5. (3 points) Three circles with radii $\sqrt{13}, \sqrt{13}$, and $2 \sqrt{13-6 \sqrt{3}}$ are arranged such that the triangle formed by their centers is an equilateral triangle with side length $\sqrt{39}$. Find the radius of the circumscribed circle around the triangle, each vertex of which is the point of intersect... | Solution:
Let $A, B, C$ be the points of intersection of the first and second, second and third, and first and third circles, respectively, and $O_{1}, O_{2}, O_{3}$ be the centers of these circles.
The radii of the first and second circles are equal, so point $A$ will lie on the perpendicular bisector of $O_{1} O_{2... | 4\sqrt{39}-6\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,006 |
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