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Problem 9.7. Two parallel lines are drawn through points $A(0 ; 14)$ and $B(0 ; 4)$. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.
What is $\frac{A L-... | Answer: 3.5.
Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.
The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,024 |
Problem 9.8. A group of kids decided to play a computer game. Any two people either play together or against each other; moreover, if player $A$ plays together with $B$, and $B$ plays against $C$, then $A$ also plays against $C$. How many kids are in the group if each player had exactly 15 opponents? List all possible ... | Answer: $16,18,20,30$.
Solution. From the condition of the problem, it follows that all players are divided into several groups such that any two within a group play together, while any two people from different groups play against each other. For each player, the number of opponents is the difference between the tota... | 16,18,20,30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,025 |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,026 |
Problem 10.2. How many real numbers $x$ exist such that the value of the expression $\sqrt{123-\sqrt{x}}$ is an integer? | Answer: 12.
Solution. From the condition, it follows that the value $s=123-\sqrt{x} \leqslant 123$ is a square of an integer. Since $11^{2}<123<12^{2}$, this value can take one of 12 values $0^{2}$, $1^{2}, 2^{2}, \ldots, 11^{2}$. And for each of these 12 values of $s$, there is a unique value of $x=(123-s)^{2}$ (obvi... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,027 |
Problem 10.3. At exactly noon, a truck left the village and headed for the city, at the same time, a car left the city and headed for the village. If the truck had left 45 minutes earlier, they would have met 18 kilometers closer to the city. And if the car had left 20 minutes earlier, they would have met $k$ kilometer... | Answer: 8.
Solution. We will express all distances in kilometers, time in hours, and speed in kilometers per hour. Let the distance between the village and the city be $S$, the speed of the truck be $x$, and the speed of the car be $y$. The time until the first meeting is $\frac{S}{x+y}$, so the distance from the vill... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,028 |
Problem 10.4. Consider the sequence
$$
a_{n}=\cos (\underbrace{100 \ldots 0^{\circ}}_{n-1})
$$
For example, $a_{1}=\cos 1^{\circ}, a_{6}=\cos 100000^{\circ}$.
How many of the numbers $a_{1}, a_{2}, \ldots, a_{100}$ are positive? | Answer: 99.
Solution. Note that for an integer $x$ divisible by 40, the cosines of the angles $x^{\circ}$ and $10 x^{\circ}$ coincide, since the difference between these angles $9 x^{\circ}$ is divisible by $360^{\circ}$. Therefore, $\cos \left(\left(10^{k}\right)^{\circ}\right)=$ $\cos \left(\left(10^{k+1}\right)^{\c... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,029 |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
 than X's (crosses), then in the entire table there are no fewer O's than X's. Similarly, considering the rows, it follows that in the entire table there are no fewer X's than O's. Therefore, there are an equal number of X's and O's in the entir... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,031 |
Problem 10.7. Olya drew $N$ different lines on the plane, any two of which intersect. It turned out that among any 15 lines, there are always two lines such that the angle between them is $60^{\circ}$. What is the largest $N$ for which this is possible? | Answer: 42.
Solution. First, we will show that 42 such lines exist. Draw three lines so that the angle between any two of them is $60^{\circ}$, and take 14 such triplets at different angles:
... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,032 |
Problem 10.8. Real numbers $x$ and $y$ are such that $x^{3}+21 x y+y^{3}=343$. What can $x+y$ be? List all possible options. | Answer: $7, -14$.
Solution. By directly expanding the brackets, it is easy to verify the identity
$$
a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)
$$
This means that the expression on the left side can be zero either when $a+b+c=0$,
or when $a^{2}+b^{2}+c^{2}-a b-b c-c a=0$, which is on... | 7,-14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,033 |
1. Let's divide the sequence of natural numbers into groups:
$(1),(2,3),(4,5,6)(7,8,9,10), \ldots$
Denote $S_{n}$ as the sum of the $n$-th group of numbers. Find $S_{16}-S_{4}-S_{1}$. | Solution. Note that in the $n$-th group there are $n$ terms and the first one is $\frac{n(n-1)}{2}+$ 1. The last term of the $n$-th group is $\frac{n(n-1)}{2}+1+(n-1)=\frac{n^{2}+n}{2}$. Therefore, $S_{n}=\left(\frac{n(n-1)}{2}+1+\frac{n(n+1)}{2}\right) \cdot \frac{n}{2}=\frac{n\left(n^{2}+1\right)}{2}$. Hence, $S_{16}... | 2021 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,034 |
2. Prove that if for any value of $x$ and constant $c$ the equality $f(x+c)=\frac{2}{1+f(x)}-1$ holds, then $f(x)$ is a periodic function. | Solution. From the equality $f(x+c)=\frac{2}{1+f(x)}-1$ it easily follows that $f(x+c)=\frac{1-f(x)}{1+f(x)}$. Therefore, $f(x+2c)=\frac{1-f(x+c)}{1+f(x+c)}=$ $\frac{1-\frac{1-f(x)}{1+f(x)}}{1+\frac{1-f(x)}{1+f(x)}}=\frac{1+f(x)-1+f(x)}{1+f(x)+1-f(x)}=f(x)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,035 |
3. Chord $AB$ of a circle with radius $R$ is extended to segment $BC = AB$, point $C$ is connected to the center of the circle $O$, and $CO$ intersects the circle at point $D$. Prove that $CD = 4R \sin 18^{\circ}$, given that a square can be inscribed in the circle on $AB$. | Solution. Let the extension of $C O$ intersect the circle at point $E$. Then, by the theorem of two secants, $C A \cdot C B=C E \cdot C D$. Since $A B=$ $B C=R \sqrt{2}$, and $C E=2 R+C D$, we have
$$
\begin{gathered}
2 R \sqrt{2} \cdot R \sqrt{2}=(2 R+C D) \cdot C D \\
C D^{2}+2 R \cdot C D-4 R^{2}=0
\end{gathered}
$... | CD=R(\sqrt{5}-1) | Geometry | proof | Yes | Yes | olympiads | false | 14,036 |
4. Find all solutions of the equation $x^{2}-12 \cdot[x]+20=0$, where $[x]-$ is the greatest integer not exceeding $x$. | $$
\begin{aligned}
& \text { Solution. Let } [x]=a. \text { Then } a \leq x < a+1 \text { and } x^{2}+20=12 a . \text { We have: } \\
& \left\{\begin{array}{l}
x^{2}+20=12 a \\
a \leq x < a+1
\end{array} \Rightarrow\left\{\begin{array}{l}
x^{2}+20=12 a \\
a \leq x < a+1
\end{array} \Rightarrow\right.\right. \\
& \Right... | 2,2\sqrt{19},2\sqrt{22},10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,037 |
5. In each cell of a $6 \times 6$ table, numbers +1 or -1 were placed such that the product of all numbers in any row and any column is positive. In how many ways can this be done? | Solution. Fill a $5 \times 5$ square with numbers +1 and -1 arbitrarily. Choose the numbers in the last column so that the product in the first 5 rows is positive, and similarly choose the numbers in the last row for the first 5 columns so that the product is positive. It remains to choose the last number.
Let the num... | 2^{25} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,038 |
# 1.1. Condition:
Twelve figures are made of matches - 3 triangles, 4 squares, and 5 pentagons. The figures have no common sides. Petya and Vasya take turns removing one match at a time. Vasya wants to leave as few untouched figures as possible, while Petya wants to leave as many untouched figures as possible. How man... | Answer: 6
Solution. Vasya will be able to take matches from a maximum of 5 different figures. This means that if Petya takes matches only from those figures that have already lost at least one match, then $12-5=7$ figures will remain untouched. Petya can fulfill the condition, as otherwise, Vasya would have taken a ma... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,039 |
# 2.1. Condition:
Andrey, Boris, Svetlana, and Larisa are four people of different ages, forming two family couples. It is known that the oldest is the husband of Larisa, and Andrey is younger than Svetlana but older than Larisa. Choose all correct statements:
## Answer options:
$\square$ Andrey is older than Boris ... | # Solution.
Since Andrey is younger than Svetlana, he cannot be the oldest. Since the oldest is Larisa's husband, it can only be Boris. Therefore, Andrey is married to Svetlana. In order from youngest to oldest, they will be arranged as follows: Larisa, Andrey, Svetlana, Boris. | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 14,040 | |
# 2.2. Condition:
Roman, Oleg, Ekaterina, and Zhanna are four people of different ages, forming two family couples. It is known that each husband is older than his wife, and Zhanna is older than Oleg. Select all correct statements:
## Answer options:
Roman is older than Oleg and married to Ekaterina
$\square$ Roman... | # Solution.
Since Jeanne is older than Oleg, and each husband is older than his wife, Jeanne is married to Roman, and Oleg is married to Ekaterina. In order from youngest to oldest, they will line up as follows: Ekaterina, Oleg, Jeanne, Roman. | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 14,041 | |
# 3.1. Condition:
Danil took a white cube and numbered its faces with numbers from 1 to 6, writing each one exactly once. It turned out that the sum of the numbers on one pair of opposite faces is 11. What can the sum of the numbers on none of the remaining pairs of opposite faces NOT be?
## Answer options:
$\square... | # Solution.
If the sum of the numbers on opposite faces is 11, then the numbers on these faces are 5 and 6. The remaining numbers can be paired in three ways: $(1,2)$ and $(3,4); (1,3)$ and $(2,4); (1,4)$ and $(2,3)$, i.e., the sums that CAN be obtained are: $3,7,4,6$ and 5. Therefore, the sums that CANNOT be obtained... | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 14,042 |
# 5.1. Condition:
Polina came to the cafeteria and saw that 2 puff pastries cost 3 times more than 2 pies. Polina didn't have enough money for 2 puff pastries, but she did have enough for 1 pie and 1 puff pastry. After the purchase, she wondered how many times more money she spent buying 1 puff pastry and 1 pie instea... | Answer: 2
## Solution.
2 puff pastries are 3 times more expensive than 2 pies, so one puff pastry is 3 times more expensive than one pie, which means one puff pastry costs as much as 3 pies. Therefore, 1 puff pastry and 1 pie cost as much as 4 pies. Then they are 2 times more expensive than 2 pies. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,047 |
# 7.1. Condition:
On the Christmas tree, there is a garland of 100 bulbs. It is known that the first and third bulbs are yellow. Moreover, among any five consecutive bulbs, exactly two are yellow and exactly three are blue. Father Frost can see only part of the garland from behind the tree. Help him find out the color... | # Answer:
$97 ; 99 ; 100$ - Blue color
98 - Yellow color
## Solution.
Consider the first five light bulbs. From the condition, it follows that blue light bulbs are in the 2nd, 4th, and 5th positions. Note that the color of the sixth light bulb must be the same as the first, since for the first five light bulbs and ... | 97;99;100-Bluecolor,98-Yellowcolor | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,049 |
# 7.3. Condition:
On the Christmas tree, there is a garland of 100 bulbs. It is known that the first and fifth bulbs are yellow. Moreover, among any five consecutive bulbs, exactly two are yellow and exactly three are blue. Father Frost can see only part of the garland from behind the tree. Help him determine the colo... | # Translation:
97; 98; 99; - Blue color
100 - Yellow color
# | 97;98;99-Bluecolor,100-Yellowcolor | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,051 |
# 7.4. Condition:
On the Christmas tree, there is a garland of 100 bulbs. It is known that the third and fourth bulbs are yellow. Additionally, among any five consecutive bulbs, exactly two are yellow and exactly three are blue. Father Frost can see only part of the garland from behind the tree. Help him determine the... | # Response:
97; 100 - Blue color
98; 99 - Yellow color
# | 97;100-Bluecolor,98;99-Yellowcolor | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,052 |
# 8.1. Condition:
A five-digit number is called a hill if the first three digits are in ascending order and the last three digits are in descending order. For example, 13760 and 28932 are hills, while 78821 and 86521 are not hills. How many hills exist that are greater than the number $77777?$
# | # Answer: 36
## Solution.
Suitable slides must start with 7, so the next digit is either 8 or 9. If the second position is occupied by the digit 9, the third digit will not be greater than the second, so the condition is only satisfied by 8. Therefore, the third position is occupied by 9. The last two positions can b... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,053 |
# 8.2. Condition:
A five-digit number is called a pit if the first three digits are in descending order, and the last three digits are in ascending order. For example, 73016 and 98346 are pits, while 88012 and 56821 are not pits. How many pits are there that are less than the number $22222?$ | Answer: 36
## Solution.
Suitable pits must start with 2, so the next is either 1 or 0. If the second position is occupied by the digit 1, then the third digit will not be greater than the second, so the condition is only satisfied by 0. Therefore, the third position is 0. The last two places can be occupied by any tw... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,054 |
9.1. If the discriminant of the quadratic polynomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic polynomial $g(x)=(a+1) x^{2}+2(b+2) x+c+4$, the result is 24. Find $f(-2)$. | Answer: 6.
Solution: We have: $D_{1}-D_{2}=4\left(b^{2}-a c-(b+2)^{2}+(a+1)(c+4)\right)=4(-4 b+4 a+c)=$ $4 f(-2)$.
Comment: Correct answer without justification - 0 points. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,055 |
9.2. Six people - liars and knights - sat around a table. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, 3 people said: "I have one coin," while the other 3 said: "I have no coins." What is the maximum ... | Answer: 4.
Solution: After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 6. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs from t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,056 |
9.3. On the board, there are $N$ prime numbers (not necessarily distinct). It turns out that the sum of any three numbers on the board is also a prime number. For what largest $N$ is this possible | Answer: $N=4$.
Solution. Consider the remainders when the $N$ written numbers are divided by 3. All three remainders cannot occur, because in this case, the sum of three numbers with different remainders will be divisible by 3 (and will be greater than 3), so it will not be a prime number. Therefore, there can be no m... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,057 |
9.4. Vasya cut out a triangle from cardboard and numbered its vertices with the digits 1, 2, and 3. It turned out that if Vasya's triangle is rotated 15 times clockwise around its vertex numbered 1 by the angle at this vertex, the triangle will return to its original position. If Vasya's triangle is rotated 6 times clo... | Answer: 5.
Solution. Let the angle at the first vertex be $\alpha$ degrees, at the second $\beta$ degrees, and at the third $\gamma$ degrees. When rotating around the vertices, Vasya's triangle could make several full turns before returning to its starting position. Suppose that during the rotation around the first ve... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,058 |
9.5. In a rectangular non-isosceles triangle \(ABC\) with a right angle at \(C\), the bisector \(CL\) is drawn. Point \(K\) is chosen on the hypotenuse of this triangle such that \(AL = BK\). The perpendicular to \(AB\) passing through point \(K\) intersects ray \(CL\) at point \(N\). Prove that \(KN = AB\). | Solution. Let's mark the midpoint of $AB$, point $M$, which is also the midpoint of $LK$, since $AL = BK$. The perpendicular bisector of side $AB$ intersects the bisector $CL$ at point $D$ on the circumcircle $\omega$ of triangle $ABC$. Indeed, the bisector $CL$ passes through the midpoint of the arc $BA$, and the perp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,059 |
11.1. The numbers $2^{2019}$ and $5^{2019}$ are written consecutively. How many digits are written in total? | Answer: 2020.
Solution. Let the number $2^{2019}$ contain $\mathrm{m}$ digits, and the number $5^{2019}$ contain $\mathrm{n}$ digits. Then the following inequalities hold: $10^{\mathrm{m}-1}<2^{2019}<10^{\mathrm{m}}, 10^{\mathrm{n}-1}<5^{2019}<10^{\mathrm{n}}$ (the inequalities are strict because the power of two or f... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,060 |
11.2. In a quadrilateral, the diagonals are perpendicular. A circle can be inscribed in it and a circle can be circumscribed around it. Can we assert that it is a square? | Answer: No.
Solution. Consider in the circle the diameter $A C$ and a chord $B D$ perpendicular to it, not passing through the center (see figure).
We will show that the quadrilateral $A B ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,061 |
11.3. Prove that if the cotangents of the angles of a triangle form an arithmetic progression, then the squares of the sides of this triangle also form an arithmetic progression. | Solution. From the cosine theorem: $a^{2}-b^{2}=a c \times \cos B-b c \times \cos A$.
From the triangle area theorem: $b c=2 S / \sin A, a c=2 S / \sin B$.
From these relations (and similar ones):
$b^{2}-a^{2}=2 S \times(\operatorname{ctg} A-\operatorname{ctg} B)$,
$c^{2}-b^{2}=2 S \times(\operatorname{ctg} B-\oper... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,062 |
11.4. The equation $(x+a)(x+b)=9$ has a root $a+b$. Prove that $a b \leq 1$. | Solution. Substituting the root $x=a+b$ into the equation, we get the equality $(a+b+a)(a+b+b)=(2a+b)(2b+a)=9$.
Then $9=5ab+2(a^2+b^2)>5ab+2 \cdot 2ab=9ab$, from which $ab \leq 1$. (We used the inequality $a^2+b^2>2ab$, which is equivalent to $(a-b)^2>0$.)
Comment. The solution applies the inequality of means for num... | \leq1 | Algebra | proof | Yes | Yes | olympiads | false | 14,063 |
11.5. Each cell of a $7 \mathrm{x} 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells and not less than the number of green cells, and in each column, the number of yellow cells is not less than the n... | Answer: 8.
Solution. 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones. In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones. Thus... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,064 |
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 | 14,066 |
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
. When Petya was at the 22nd... | Answer. At the 64th lamppost.
Solution. There are a total of 99 intervals between the lampposts. From the condition, it follows that while Petya walks 21 intervals, Vasya walks 12 intervals. This is exactly three times less than the length of the alley. Therefore, Petya should walk three times more to the meeting poin... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,070 |
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 | 14,071 |
5-4. Andrei, Boris, Vladimir, and Dmitry each made two statements. For each boy, one of his statements turned out to be true, and the other - false.
Andrei: "Boris is not the tallest among us four." "Vladimir is the shortest among us four."
Boris: "Andrei is the oldest in the room." "Andrei is the shortest in the roo... | Answer: Vladimir.
Solution. The first statement of Dmitry is incorrect, as it contradicts the condition of the problem. Therefore, Dmitry is the oldest. Consequently, the first statement of Boris is incorrect, so Andrey is the shortest. From this, it follows that the second statement of Andrey is incorrect, so Boris i... | Vladimir | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,072 |
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 | 14,073 |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. 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.
We will call suc... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,074 |
5-7. On the faces of a die, the numbers $6,7,8,9,10,11$ are written. The die was rolled twice. The first time, the sum of the numbers on the four "vertical" (that is, excluding the bottom and top) faces was 33, and the second time - 35. What number can be written on the face opposite the face with the number 7? Find al... | Answer: 9 or 11.
Solution. The total sum of the numbers on the faces is $6+7+8+9+10+11=51$. Since the sum of the numbers on four faces the first time is 33, the sum of the numbers on the two remaining faces is $51-33=18$. Similarly, the sum of the numbers on two other opposite faces is $51-35=16$. Then, the sum on the... | 9or11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,075 |
6-3. The red segments in the figure have equal length. They overlap by equal segments of length $x$ cm. What is $x$ in centimeters?
 | Answer: 2.5.
Solution. Adding up the lengths of all the red segments, we get 98 cm. Why is this more than 83 cm - the distance from edge to edge? Because all overlapping parts of the red segments have been counted twice. There are 6 overlapping parts, each with a length of $x$. Therefore, the difference $98-83=15$ equ... | 2.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,077 |
7-3. A secret object is a rectangle measuring $200 \times 300$ meters. Outside the object, there is a guard at each of the four corners. An intruder approached the perimeter of the secret object from the outside, and all the guards ran to him along the shortest paths along the external perimeter (the intruder remained ... | Answer: 150.
Solution. Note that no matter where the violator is, two guards in opposite corners will run a distance equal to half the perimeter in total.
Therefore, all four guards will ru... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,079 |
7-4. In a giraffe beauty contest, two giraffes, Tall and Spotted, made it to the final. 135 voters are divided into 5 districts, each district is divided into 9 precincts, and each precinct has 3 voters. The voters choose the winner by majority vote in their precinct; in the district, the giraffe that wins the majority... | Answer: 30.
Solution. For High to win the final, he must win in 3 districts. To win in a district, High must win in 5 precincts of that district. In total, he needs to win in at least $3 \cdot 5=15$ precincts. To win in a precinct, at least 2 voters must vote for him. Therefore, at least 30 voters are needed.
Comment... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,080 |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,082 |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,085 |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | 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 | 14,086 |
11.1. Can a rectangle, each side of which is greater than 1, be formed from all rectangles of sizes $1 \times 1, 1 \times 3, 1 \times 5, \ldots, 1 \times 2019$, taken exactly once? | Answer: Yes.
Solution. Note that there are 1010 rectangles, which means their number is even. We can pair the rectangles so that for their longer sides, the following equalities hold: $1+2019=3+2017=5+2015=\ldots=1009+1011$. By placing the rectangles in each pair with their shorter sides together, we get 505 rectangle... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,087 |
11.2. In a triangular pyramid $S A B C$, the lateral edge $S A$ is perpendicular to the base $A B C$. It is known that the bisectors of the plane angles $B A C$ and $B S C$ intersect. Prove that the angles $A B C$ and $A C B$ are equal. | Solution. The bisectors of the plane angles $BAC$ and $BSC$ intersect at point $D$, which lies on the edge $BC$ (see Fig. 11.2).
By the property of the bisector of a triangle: $BD: DC = AB: AC$ and $BD: DC = SB: SC$. Therefore, $AB: AC = SB: SC$. Rewrite this proportion as $AB: SB = AC: SC$. Then, in the right triangl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,088 |
11.3. Solve the equation: $|\sin x-\sin y|+\sin x \cdot \sin y=0$. | Answer: $x=\pi k, k \in Z ; y=\pi n, n \in Z$.
Solution. If $\sin x=0$, then $\sin y=0$, and vice versa, if $\sin y=0$, then $\sin x=0$. Therefore, $x=\pi k, k \in Z ; y=\pi n, n \in Z$. We will prove that there are no other solutions.
First method. Suppose $\sin x \neq 0$ and $\sin y \neq 0$. Then the given equation... | \pik,k\inZ;\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,089 |
11.4. What is the minimum number of cells that need to be marked on an $8 \times 9$ board so that among any five consecutive cells in a row, column, or diagonal, there is a marked cell
Fig. 11.4a | Answer: 14 cells.
Solution. Example.
See Fig. 11.4a.
Estimate. Let's highlight 14 rectangles of size $\quad 1 \times 5 \quad$ on the board, touching only two central cells (see Fig. 11.4b). In each of them, there must be at least one marked cell, meaning there are at least 14 marked cells.
. Then $A D P I$ is a parallelogram and $A D = P I$.
Then the required equality can be written as: $A B \cdot A C = 4 R^{2} - A D^{2} = A P^{2} - P I^{2}$ (1). Moreover, since $A P$ is the diameter... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,092 |
3. On the segment $AB$, a point $M$ is marked. Points $P$ and $Q$ are the midpoints of segments $AM$ and $BM$ respectively, and point $O$ is the midpoint of segment $PQ$. Choose a point $C$ such that angle $ACB$ is a right angle. Let $MD$ and $ME$ be the perpendiculars dropped from point $M$ to lines $CA$ and $CB$, and... | Solution. Note that CDME is a rectangle. Its diagonals are bisected by the point of intersection, so point $F$ is the midpoint of segment $C M$. Next, segments $P F$ and $F Q$ are the midlines of triangles $A C M$ and $B C M$ respectively. Therefore, they are parallel to the mutually perpendicular segments $A C$ and $C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,093 |
4. Do there exist six distinct natural numbers a, b, c, d, e, f such that the equality ( $a+b+c+d+e+f):(1 / a+1 / b+1 / c+1 / d+1 / e+1 / f)=2012$ holds? (B. Trushin) | Answer. Yes, they do exist. Solution. For example, let $a=1, b=2012, c=2, d=1006$, $e=4, f=503$; then $a b=c d=e f=2012$. Therefore,
$1 / a+1 / b+1 / c+1 / d+1 / e+1 / f=(a+b) / a b+(c+d) / c d+(e+f) / e f=(a+b+c+d+e+f) / 2012$, from which the required equality follows.
Criteria. Answer "they exist" without an exampl... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,094 |
10.1. Find at least one four-digit number that has the following property: if the sum of all its digits is multiplied by the product of all its digits, the result is 3990. (I. Rubanov) | Solution. Note that $3990=2 \cdot 3 \cdot 5 \cdot 7 \cdot 19=1 \cdot 6 \cdot 5 \cdot 7 \cdot 19$ and $6+5+7+1=19$. Therefore, any four-digit number that contains one 1, one 6, one 5, and one 7 will work, for example, 1567.
Comment. To receive full credit, it is sufficient to provide any correct example. | 1567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,095 |
10.2. Set $A$ consists of $n$ distinct natural numbers, the sum of which is equal to $n^{2}$. Set $B$ also consists of $n$ distinct natural numbers, the sum of which is equal to $n^{2}$. Prove that there will be a number that belongs to both set $A$ and set $B$. | Solution. Suppose the opposite: sets $A$ and $B$ do not intersect. Then their union contains $2n$ distinct natural numbers. Consequently, the sum $S$ of all elements in the union of sets $A$ and $B$ will be no less than the sum $1+2+\ldots+2n=n(2n+1)$. On the other hand, by the condition $S=2n^2$, which is less than $n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,096 |
10.3. Kolya and Dima are playing a game on an $8 \times 8$ board, taking turns, with Kolya starting. Kolya draws crosses in the cells, while Dima covers pairs of adjacent cells (dominoes) with $1 \times 2$ rectangles. On his turn, Kolya must place one cross in any empty cell (i.e., a cell that does not yet have a cross... | Answer. Dima.
First solution. We will outline a strategy for Dima. Before the game starts, we divide the board into 32 rectangles of $1 \times 2$. In the first 16 moves, the opponent can place their crosses in no more than 16 of these rectangles, so we can cover the remaining 16 rectangles with our dominoes, and all t... | Dima | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,097 |
10.4. Let $p$ be a prime number greater than 3. Prove that there exists a natural number $y$, less than $p / 2$, such that the number $p y + 1$ cannot be represented as the product of two integers, each greater than $y$.
(M. Antipov) | Solution. Let $p=2k+1$. Assume the opposite: for each of the numbers $y=1,2, \ldots, k$ there exists a factorization $p y+1=a_{y} b_{y}$, where $a_{y}>y, b_{y}>y$. Note that each of the numbers $a_{y}$ and $b_{y}$ is strictly greater than 1, and that $a_{y} > p y + 1$. Therefore, each of the $p-1$ numbers in the set $a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,098 |
10.5. Quadrilateral $ABCD$ is circumscribed around a circle $\omega$. Prove that the diameter of the circle $\omega$ does not exceed the length of the segment connecting the midpoints of sides $BC$ and $AD$.
(O. Yuzhakov) | # Solution.
We have $S_{A B C D}=p r$, where $p$ is the semiperimeter of the quadrilateral, and $r$ is the radius of $\omega$. From the tangential property, it follows that $A B+C D=B C+D A$, hence
$$
S_{A B C D}=(B C+A D) \cdot r
$$
On the other hand, if $M$ and $N$ are the midpoints of sides $B C$ and $A D$ respect... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,099 |
1. Kolya says that two chocolate bars are more expensive than five gums, Sasha - that three chocolate bars are more expensive than eight gums. When this was checked, only one of them was right. Is it true that 7 chocolate bars are more expensive than 19 gums? Don't forget to justify your answer. | Solution. Let the price of a chocolate bar be $c$, and the price of a piece of gum be $g$. Kolya says that $2 c>5 g$ or $6 c>15 g$, while Sasha says that $3 c>8 g$ or $6 c>16 g$. If Sasha were correct, Kolya would also be correct, which contradicts the condition. Therefore, Kolya is right, but not Sasha, and in fact $3... | false | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,100 |
3. Pete and Vasya are playing a game on a $9 \times 9$ grid. At the beginning of the game, a token is placed in the central cell. Pete and Vasya take turns moving it to an adjacent cell. Pete, on his turn, can either continue in the direction of Vasya's previous move or turn right. Vasya, on his turn, can either contin... | Answer: No, he cannot. It is clear that only the player who places a chip in a corner on their turn can win. Let's color the cells of the board in a checkerboard pattern. With each move, the color of the cell where the chip is placed changes. Since all the corner cells of the $9 \times 9$ board are the same color as th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,102 |
5. Points $X$ and $Y$ are the midpoints of the diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$. The lines $BC$ and $AD$ intersect at point $P$. Prove that the area of triangle $PXY$ is four times smaller than the area of quadrilateral $ABCD$.
$ does the inequality $|x|+|y| \leq n$ have? | # Solution
Graphical interpretation of the inequality - a square with sides on the lines $x+y=n$, $-x+y=n, x-y=n,-x-y=n$. The side lying on the line $x+y=n$ corresponds to $\mathrm{n}+1$ points with integer coordinates, the side lying on the line $-x+y=n$ corresponds to $\mathrm{n}+1$ points with integer coordinates, ... | 1+2(n+1)n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,108 |
5. Represent as an algebraic expression (simplify) the sum $8+88+888+8888+\ldots+8 \ldots 8$, if the last term in its notation contains p times the digit eight.
# | # Solution
$8+88+888+8888+\ldots+8 \ldots 8=8(1+11+111+\ldots+1 \ldots 1)$
Notice that
$$
\begin{gathered}
10-1=9 * 1 \\
100-1=9 * 11 \\
\ldots \ldots \ldots . \\
10^{n}-1=9 * 1 \ldots 1
\end{gathered}
$$
Then $\frac{8}{9}\left(\frac{10 \times 10^{n}-10}{9}-n\right)=\frac{8}{81}\left(10^{n+1}-10-9 n\right)$.
## Gr... | \frac{8}{81}(10^{p+1}-10-9p) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,109 |
# 4. Having covered half of the distance, the train reduced its speed by $25 \%$ and therefore arrived at the destination 0.5 hours late. How long did it take the train to cover the entire distance? | Solution Let the total distance be S. Let X km/h be the speed of the train when it travels the first half of the distance. Then $0.75 \mathrm{x}$ km/h is the speed of the train when it travels the second half of the distance. Therefore, $\frac{0.5 S}{\mathrm{x}}$ hours were spent by the train on the first half of the j... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,111 |
8.4 In parallelogram $\mathrm{ABCD}$, points $\mathrm{M}$ and $\mathrm{N}$ are the midpoints of sides BC and CD, respectively. Can rays AM and AN divide angle BAD into three equal parts | Solution: See fig. Suppose they can.
In triangle $\mathrm{ABC}$, segments $\mathrm{AM}$ and $\mathrm{BO}$ are medians. Therefore, $B E=\frac{2}{3} B O, \quad E O=\frac{1}{3} B O$. Similarly,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,112 |
9.1 Solve the equation $x(x-1)(x-2)+(100-x)(99-x)(98-x)=0$. | Answer: no solutions. Note that the given equation is quadratic (the coefficients of $x^{3}$ cancel out). The vertex of this quadratic trinomial is at the point with abscissa $x_{0}=\frac{100}{2}=50$; this can be verified either by direct calculation or in the following way: if we introduce the substitution $t=x-$ 50, ... | nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,113 |
8.1. The hikers had several identical packs of cookies. At the midday break, they opened two packs and divided the cookies equally among all the participants of the hike. One cookie was left over, and the hikers fed it to a squirrel. At the evening break, they opened another three packs and also divided the cookies equ... | Solution: Let there be $n$ tourists in the group, and $m$ cookies in each pack (where $m$ and $n$ are natural numbers, and $n > 13$). According to the problem, the numbers $x = 2m - 1$ and $y = 3m - 13$ are divisible by $n$. Therefore, the number $3x - 2y = 23$ is also divisible by $n$. Since 23 is a prime number and h... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,114 |
8.2. The work was divided equally among the workers in the team. After the first day, they counted how many people had completed no less than 30 percent of their share; it turned out that 70 percent of all workers had done so. When they counted only those who had completed no less than 70 percent of their share, it tur... | Solution: Let's provide an example where the condition of the problem is met, but a third of the work is not done. Suppose there were 10 workers, and the work involves creating 300 identical parts. Then each worker needs to make 30 parts, and a third of the work consists of 100 parts. Suppose three workers made 21 part... | No | Other | math-word-problem | Yes | Yes | olympiads | false | 14,115 |
8.3. In a convex 6-sided polygon $A B C D E F$, all angles are equal to $120^{\circ}$. The lengths of four of its sides are known: $B C=3, C D=5, D E=4$ and $A F=6$. Find the lengths of the two remaining sides of the hexagon. Justify your answer.
\left(b^{2}-1\right)\left(c^{2}-1\right)$? Find all possible values and prove that there are no others. | Solution: Subtract 1 from both sides of each equation, we get:
$$
\begin{gathered}
a^{2}+1=b^{4}-1=\left(b^{2}-1\right)\left(b^{2}+1\right) \\
b^{2}+1=\left(c^{2}-1\right)\left(c^{2}+1\right), \quad c^{2}+1=\left(a^{2}-1\right)\left(a^{2}+1\right)
\end{gathered}
$$
Multiply the left and right parts of all three equat... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,117 |
8.5. When fully fueled, a motorboat can travel exactly 40 km upstream or exactly 60 km downstream. What is the greatest distance the motorboat can travel along the river if the fuel must be enough for the round trip, back to the starting point? Justify your answer. | # Solution:
Method 1. Let's launch a raft along with the boat, which will move with the current, and observe the boat from this raft. Then, no matter which direction the boat goes, upstream or downstream, it will be at the same distance from the raft by the time the fuel runs out. This means that the raft will be at t... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,118 |
8.6. Several eighth-graders were solving problems. The teacher did not record in the journal how many students there were in total and how many problems each of them solved. However, he remembers that, on the one hand, each student solved more problems than one-fifth of what the others solved. On the other hand, he kno... | Solution: Let the number of eighth-graders be $n$. Let the $i$-th eighth-grader $(i=1, \ldots, n)$ have solved $a_{i}$ problems. According to the condition, for any $i$, the inequalities $a_{i}>\frac{1}{5} \cdot\left(S-a_{i}\right)$ and $a_{i}$ solved by each student | 7 points |
| Correct and justified answer without ... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 14,119 |
1. On the board, there is a table containing 100 columns and 2 rows. Sasha and Zhenya take turns filling in one cell of the table, writing either -1 or 1. Sasha makes the first move. After the table is completely filled, the products of all 100 numbers in each row are calculated. Similarly, for each column, the product... | Solution. Consider pairs of cells in each row of the table, where each pair will consist of a cell with an odd and an even number. The winning strategy for Zhenya is to give Sasha the opportunity to make the first move in each such pair. Let Sasha fill one (any) of the two cells. In response, Zhenya needs to fill the s... | Can | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,120 |
2. It is known that $a<b<c$. Prove that the equation
$$
3 x^{2}-2(a+b+c) x+a b+b c+a c=0
$$
has two distinct roots $x_{1}$ and $x_{2}$, such that $a<x_{1}<b, b<x_{2}<c$. | Solution. Consider the function $f(x)=3 x^{2}-2(a+b+c) x+a b+b c+a c$ and substitute into it the values $x=a, x=b, x=c$ in turn.
$$
f(a)=3 a^{2}-2(a+b+c) a+a b+b c+a c=a^{2}-a b-a c+b c=(a-b)(a-c)
$$
Given the inequality in the condition, $f(a)>0$.
Similarly,
$$
f(b)=3 b^{2}-2(a+b+c) b+a b+b c+a c=b^{2}-a b-b c+a c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,121 |
3. Two lines $l_{1}$ and $l_{2}$, the angle between which is $\alpha$, are common internal tangents to circles of radii $R_{1}$ and $R_{2}$. A common external tangent to the circles intersects the lines $l_{1}$ and $l_{2}$ at points $A$ and $B$ respectively. Find the length of the segment $A B$. | Solution.
The notations are shown in the figure. The external tangent is the line $A B$. Let the length of the segment $A B$ be denoted by $a$. Suppose the segments of the tangents drawn fr... | (R_{1}+R_{2})\operatorname{ctg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,122 |
4. Prove the equality
\[
\begin{gathered}
2018 \cdot 1 + 2017 \cdot 2 + 2016 \cdot 3 + \ldots + (2019 - k) \cdot k + \ldots + 1 \cdot 2018 = \\
= \frac{2019 \cdot 2018}{2} + \frac{2018 \cdot 2017}{2} + \frac{2017 \cdot 2016}{2} + \ldots + \frac{n \cdot (n-1)}{2} + \ldots + \frac{2 \cdot 1}{2}
\end{gathered}
\] | Solution.
$2018 \cdot 1+2017 \cdot 2+2016 \cdot 3 \ldots+(2019-k) \cdot k+\ldots+1 \cdot 2018=$ $=2018+2017+2017+2016+2016+2016+\ldots+$
$+\underbrace{(2019-k)+(2019-k)+\ldots+(2019-k)}_{k \text { times }}+\ldots+\underbrace{1+1+\ldots+1}_{2018 \text { times }}=$
$=(2018+2017+2016+\ldots+1)+(2017+2016+\ldots+1)+(201... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,123 |
5. We took ten consecutive natural numbers greater than 1, multiplied them, found all the prime divisors of the resulting number, and multiplied these prime divisors (taking each exactly once). What is the smallest number that could have resulted? Fully justify your answer.
Solution. We will prove that among ten conse... | Answer: 2310.
## Municipal Stage of the All-Russian Mathematics Olympiad 2018-2019 Academic Year
10th Grade
Grading Criteria
| Score | Points for |
| :---: | :---: | :--- |
| 7 | Complete solution, a correct algorithm of actions for Zhenya is provided, and all conditions are verified. |
| 6 | Complete solution, a c... | 2310 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,124 |
1. How many 4-digit numbers exist where the digit in the thousands place is greater than the digit in the hundreds place? | Solution. The digit in the thousands place can take one of 9 possible values: $1,2,3, \ldots, 9$ (we cannot take 0, since the number is four-digit). For each of these options, we can specify the corresponding number of options for the hundreds digit: 1, $2,3, \ldots, 9$. That is, there are a total of 45 options. The ot... | 4500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,125 |
3. Two excavators were hired to dig a tunnel. One of them digs one and a half times faster than the other, and they are paid the same for each hour of work. There are two options for the work: digging from both ends until they meet in the middle, or each digging half of the tunnel. Which option will be cheaper? | Solution. Since the cost of an hour of work is the same, but the fast digger will dig more, the meter of the tunnel dug by the fast digger will be cheaper. In the "until they meet" option, the fast digger will dig more than half of the tunnel, so this option will be cheaper. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,127 |
4. Vasya wrote several integers on the board. Petya wrote the square of each of Vasya's numbers under each of them. Then Masha added up all the numbers written on the board and got 2021. Prove that someone made a mistake. | Solution. The sum of any integer $n$ and its square is $n+n^{2}=n(n+1)$ and is even, as the product of two consecutive integers. Thus, Masha should have added only even numbers, so her answer should also have been an even number. Therefore, someone among the kids must have made a mistake. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 14,128 |
5. Rectangles $A B N F$ and $C M K D$ are overlapped such that their common part is rectangle $M N F K$. It is given that $B N=8$ cm and $K D=9$ cm. Find the area of rectangle $M N F K$, if the area of $A B M K$ is 25 cm $^{2}$, and the area of $C D F N$ is $32 \mathrm{~cm}^{2}$.
, then increased by 1, for the second - the number is multiplied by 5, then increased by 1. Can we obtain the number 2022 from the number 2023 by performing on... | # Solution:
For the second operation, you can get a number where the units digit is 1 or 6. Therefore, 2022 can only be obtained by the first operation.
$10 x+2=\frac{y}{2}+1$, where $x$ is some integer and $y$ is the number obtained before applying the first operation. We get $y=20 x+2$. Thus, it can be stated that ... | No,itcannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,130 |
10.2. It is known that for numbers $x, y, z$ the equality $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$ holds. Prove that some two of these numbers are opposites of each other. | # Solution:
Assume that there are no two numbers that are opposites of each other, then $x+y \neq 0, x+z \neq 0$ and $y+z \neq 0$.
$\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}=>\frac{x+y}{x y}=\frac{z-x-y-z}{z(x+y+z)} \Rightarrow \frac{x+y}{x y}=\frac{-(x+y)}{z(x+y+z)}$
Using the fact that $x+y \neq 0$, divi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,131 |
10.3. For which prime numbers $p$ is the value of the expression $22 p^{2}+23$ also a prime number? Consider all remainders when dividing by 3.
1) If $p=3 k$ (divides evenly), then $p=3$, as only one prime number divides evenly by 3. Then $22 p^{2}+23=221.221=17 \cdot 13$, which is composite.
2) If $p=3 k \pm 1$ (give... | Answer: under no circumstances.
| 1. Full solution of the problem | 7 |
| :--- | :---: |
| 2. It is proven that if p does not divide 3, then the value of the expression is composite, but it is claimed that 221 is prime. An incorrect answer is obtained. | 3 |
| 3. Only the case where $p=3$ is considered. | 1 |
| 4. Inc... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,132 |
10.4. Let point $P$ be the intersection of the diagonals of a convex quadrilateral $K L M N$. The areas of triangles $K L M, L M N$, and $N K P$ are $8 \mathrm{~m}^{2}, 9 \mathrm{~m}^{2}$, and $10 \mathrm{~m}^{2}$, respectively. Find the area of quadrilateral $K L M N$. | # Solution:
Let $S_{P L M}=x$, and $\angle N P M=\angle K P L=\alpha$, then $S_{K L P}=8-x, S_{N P M}=9-x, \quad \angle K P N=\angle L P M=180^{\circ}-$
$$
\left\{\begin{array}{l}
x=\frac{1}{2} \sin \left(180^{\circ}-\alpha\right) P L \cdot P M \\
9-x=\frac{1}{2} \sin \alpha \cdot P M \cdot P N \\
10=\frac{1}{2} \sin... | 24\mathrm{}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,133 |
10.5. What is the smallest value that the GCD $(x, y)$ can take, where $x$ and $y$ are natural numbers, if the LCM $(x, y)=(x-y)^{2}$? | Solution:
When $x=4$ and $y=2$, LCM $(x, y)=(x-y)^{2}=4$, and GCD $(x, y)=2$. Thus, 2 is achievable.
Assume that GCD $(x, y)=1$ (the numbers are coprime). Then LCM $(x, y)=x y$. From the condition, we get that $x y=x^{2}-2 x y+y^{2}$. We obtain $x^{2}-3 x y+y^{2}=0$. Solving the obtained quadratic equation with respe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,134 |
1.1. Around a circle, Petya and Vasya are riding bicycles at constant speeds. Petya's speed is 8 km/h, and Vasya's speed is 10 km/h. Initially, they were riding in opposite directions (Petya - clockwise, and Vasya - counterclockwise), and then Petya changed his direction (started moving counterclockwise) and simultaneo... | # Answer: 3
Solution. Let $d$ km be the length of the circumference. Then, initially, the time between two consecutive meetings of the cyclists is $t_{1}=d /(8+10)=d / 18$ hours. Then, Petya's speed increased to 16 km/h, and the cyclists started moving in the same direction, so the time between two consecutive meeting... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,137 |
2.1. Sasha solved the quadratic equation $3 x^{2}+b x+c=0$ (where $b$ and $c$ are some real numbers). In his answer, he got exactly one root: $x=-4$. Find $b$. | Answer: 24
Solution. By Vieta's theorem $x_{1}+x_{2}=-b / 3$. In our case $x_{1}=x_{2}=-4$, hence $-b / 3=-8$ and $b=24$.
Remark. Another solution can be obtained by noticing that the equation should reduce to a perfect square $3(x+4)^{2}=0$. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,138 |
3.1. Find the sum
$$
\sqrt[7]{(-7)^{7}}+\sqrt[8]{(-8)^{8}}+\sqrt[9]{(-9)^{9}}+\ldots+\sqrt[100]{(-100)^{100}}
$$
(Each term is of the form $\left.\sqrt[k]{(-k)^{k}}\right)$ | Answer: 47
Solution. Note that $\sqrt[k]{(-k)^{k}}$ equals $k$ when $k$ is even, and equals $-k$ when $k$ is odd. Therefore, our sum is $-7+8-9+10-\ldots-99+100=(-7+8)+(-9+10)+\ldots+(-99+100)$. In the last expression, there are 47 parentheses, each of which equals 1. | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,139 |
4.1. For triangle $A B C$, the following is known: $A B=12, B C=10, \angle A B C=120^{\circ}$. Find $R^{2}$, where $R-$ is the radius of the smallest circle in which this triangle can be placed.
Solution. Since segment $AC$ fits inside the circle, $2R \geqslant AC$. On the other hand, the circle constructed with $AC$ as its diameter covers the triangle $ABC$, since $\angl... | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,140 |
5.1. Let's consider the number 616. The sum of its digits is 13, and the product of its digits is 36. What is the largest natural number whose sum of digits is 13 and the product of its digits is 36? | Answer: 3322111
Solution. In the decimal representation of this number, there cannot be any zeros. Let's list all the digits of this number that are greater than 1; their product is still 36. Since 36 can be factored into prime factors as $36=2 \cdot 2 \cdot 3 \cdot 3$, the possible sets of digits greater than 1 are: ... | 3322111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,141 |
6.1. On a plane, 55 points are marked - the vertices of a certain regular 54-gon and its center. Petya wants to paint a triplet of the marked points in red so that the painted points are the vertices of some equilateral triangle. In how many ways can Petya do this? | Answer: 72
Solution. Assume that our 54-gon is inscribed in a circle, and its vertices divide this circle into 54 arcs of length 1.
If one of the painted points is the center, then the other two painted points must be the ends of an arc of size $54 / 6=9$. There are 54 such possibilities.
Otherwise, all three painte... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,142 |
7.1. In a row, the numbers are written: $100^{100}, 101^{101}, 102^{102}, \ldots, 234^{234}$ (i.e., the numbers of the form $n^{n}$ for natural numbers n from 100 to 234). How many of the listed numbers are perfect squares? (A perfect square is the square of an integer.) | # Answer: 71
Solution. Consider a number of the form $m^{k}$, where $m$ and $k$ are natural numbers. If $k$ is even, then $m^{k}$ is a perfect square. If $k$ is odd, then $m^{k}$ is a perfect square if and only if $m$ is a perfect square. Thus, the answer to our problem is the total number of even numbers and odd perf... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,143 |
8.1. On the coordinate plane, a parallelogram $O A B C$ is drawn, with its center located at the point $\left(\frac{19}{2}, \frac{15}{2}\right)$, and points $A, B$, and $C$ have natural coordinates. Find the number of such parallelograms. (Here, $O$ denotes the origin - the point $(0,0)$; two parallelograms with the sa... | Answer: 126.
Solution. Note that $O A B C$ is a parallelogram if and only if the point $E\left(\frac{19}{2}, \frac{15}{2}\right)$ is the midpoint of segments $O B$ and $A C$, and moreover, points $O, A, B, C$ do not lie on the same line.
![](https://cdn.mathpix.com/cropped/2024_05_06_6bdc1b7f380f62058621g-09.jpg?heig... | 126 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,144 |
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