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7. The distance from the point of intersection of the diameter of a circle with a chord of length 18 cm to the center of the circle is $7 \mathrm{~cm}$. This point divides the chord in the ratio $2: 1$. Find the radius.
$$
A B=18, E O=7, A E=2 B E, R=?
$$ | Solution: $2 B E \cdot B E=(R-7)(7+R)$
$$
A E \cdot B E=D E \cdot E C, \quad A E+B E=18, \quad B E=6
$$
$$
2 \cdot 6 \cdot 6=\left(R^{2}-7^{2}\right), R^{2}=72+49=121=11^{2}
$$
Answer: $R... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $7^{-x}-3 \cdot 7^{1+x}=4$. | Solution: $7^{-x}-3 \cdot 7^{1+x}=4, \frac{1}{7^{x}}-21 \cdot 7^{x}=4,-21\left(7^{x}\right)^{2}-4 \cdot 7^{x}+1=0$,
$\left(7^{x}\right)=\frac{4 \pm \sqrt{16+84}}{-42}=\frac{4 \pm 10}{-42}=\left[\begin{array}{l}-\frac{14}{42}=-\frac{1}{3} \text {, not valid } \\ \frac{1}{7}\end{array}, x=-1\right.$. Answer: $x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}$. | Solution: $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}=A, A^{2}=12-2 \sqrt{36-20}=4, A= \pm \sqrt{4}=2$. Answer: 2 . | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $9 \cdot 3^{2 x-1}+3^{x}-30=0$. | Solution: $9 \cdot 3^{2 x-1}+3^{x}-30=0,3 \cdot 3^{2 x}+3^{x}-30=0$, $3^{x}=\frac{-1 \pm \sqrt{1+360}}{6}=\frac{-1 \pm 19}{6}=\left[\begin{array}{l}3 \\ -\frac{10}{3} \text {, not valid }\end{array}, x=1\right.$.
Answer: $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$. | Solution: $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$, Domain of Definition $\frac{x^{2}-16}{x-3} \geq 0, x>3 \Rightarrow x \geq 4$,
$\sqrt{x^{2}-16}+\sqrt{x^{2}-9}=7, x^{2}-16=t, \sqrt{t}+\sqrt{t+7}=7, t+2 \sqrt{t(t+7)}+t+7=49$,
$2 \sqrt{t(t+7)}=42-2 t, \sqrt{t^{2}+7 t}=21-t, t \leq 21, t^{2}+7 t=21... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Two adjacent faces of a tetrahedron, which are isosceles right triangles with a hypotenuse of 2, form a dihedral angle of 60 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron onto the plane containing
=\frac{x^{2}}{(1+\sqrt{x+1})^{2}}
$$ | Solution 1. Rewrite the equation as $2(x-6)(1+\sqrt{x+1})^{2}=x^{2}$ and make the substitution $\sqrt{x+1}=y$. Then $x=y^{2}-1$ and $y \geq 0$. After the substitution, we get $2\left(y^{2}-7\right)(y+1)^{2}=$ $\left(y^{2}-1\right)^{2}$. Factor the right side as $(y+1)^{2}(y-1)^{2}$, move it to the left, and factor out ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. How many positive numbers are there among the 2014 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}$, $\sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$ | Solution. For $n>3$ we have
$10^{n}-1000=10^{3}\left(10^{n-3}-1\right)=25 \cdot 40 \cdot\left(10^{n-3}-1\right)$.
Since $10^{n-3}-1$ is divisible by 9, it follows that $10^{n}-1000$ is divisible by 360. Therefore, all terms of the sequence, starting from the fourth, coincide with $\sin 1000^{\circ}=\sin \left(3 \cdot... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Mom baked a cake for Anya's birthday, which weighs an integer number of grams. Before decorating it, Mom weighed the cake on digital scales that round the weight to the nearest ten grams (if the weight ends in 5, the scales round it down). The result was 1440 g. When Mom decorated the cake with identical candle... | Answer: 4 years.
Solution: From the condition, it follows that the cake weighed from 1436 to 1445 grams before decoration, and from 1606 to 1615 grams after decoration. Therefore, the total weight of the candles is between 161 and 179 grams. Since the weight of each candle is given as 40 grams, the weight of one candl... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Timofey placed 10 grid rectangles on a grid field, with areas of $1, 2, 3, \ldots, 10$ respectively. Some of the rectangles overlapped each other (possibly completely, or only partially). After this, he noticed that there is exactly one cell covered exactly once; there are exactly two cells covered exactly t... | Answer: 5 cells.
Solution. Estimation. There are a total of $1+2+\ldots+10=55$ cell coverings. The ten cells described in the condition are covered in total exactly $1 \cdot 1+2 \cdot 2+3 \cdot 3+4 \cdot 4=30$ times. There remain $55-30=25$ cell coverings. Therefore, the number of cells covered at least five times is ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) How many zeros does the number $4^{5^{6}}+6^{5^{4}}$ end with in its decimal representation?
# | # Answer: 5
Solution. Consider the number $4^{25}+6$. Let's check that it is divisible by 5 and not divisible by 25. We have
$$
\begin{gathered}
4^{25}+6 \equiv(-1)^{25}+6 \equiv 0 \quad(\bmod 5) \\
4^{25} \equiv 1024^{5}+6 \equiv(-1)^{5}+6 \equiv 5 \quad(\bmod 25)
\end{gathered}
$$
We will write $5^{t} \| c$, if $5... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\cos \alpha = \frac{2}{5}$? | Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $8$ and $\cos \alpha = \frac{3}{4}$? | Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $6$ and $\cos \alpha = \frac{2}{3}$? | Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\cos \alpha = \frac{1}{4}$? | Answer: 6.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $9$ and $\cos \alpha = \frac{1}{3}$? | Answer: 6.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Find all pairs of natural numbers $x$ and $y$ such that
$$
\log _{2} a x+\log _{2} b y=\log _{2}\left(b x+a y+p_{1} p_{2}-1\right), \text { where } p_{1}, p_{2} \in \mathbb{P} \quad a, b>2
$$
In your answer, write the smallest possible value of $x+y$. | Solution: Let's get rid of the logarithms in the equation
$$
a x b y=b x+a y+p_{1} p_{2}-1
$$
Transform and factorize
$$
(a y-1)(b x-1)=p_{1} p_{2}
$$
Since $p_{1}, p_{2} \in \mathbb{P}$, we get four possible solutions
$$
\left\{\begin{array} { l }
{ x = \frac { p _ { 1 } p _ { 2 } + 1 } { b } } \\
{ y = \frac { ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Find the sum of the real roots of the equation
$$
2 \cdot 3^{3 x}-a \cdot 3^{2 x}-3(a+4) \cdot 3^{x}+18=0
$$ | Solution: Let's make the substitution $t=3^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation
$$
2 t^{3}-a t^{2}-3(a+4) t+18=0
$$
It is obvious that the number $t_{1}=-3$ is a root of the equation and the corresponding $x_{1}$ is not real. We get
$$
(t+3)\left(2 t^{2}-(a+6) t+6\right)=0... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $15$, and $\sin \alpha = \frac{\sqrt{21}}{5}$? | Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\sin \alpha = \frac{\sqrt{24}}{5}$? | Answer: 4.
## Solution:
Consider the point $B_{1}$ symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\sin \alpha = \frac{\sqrt{21}}{5}$? | Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\sin \alpha = \frac{\sqrt{35}}{6}$? | Answer: 4.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $16$, and $\sin \alpha = \frac{\sqrt{55}}{8}$? | Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Find all values of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2|x+7|+|y-4|=c \\
|x+4|+2|y-7|=c
\end{array}\right.
$$ | Answer: $c=3$.
Solution. Let $\left(x_{0} ; y_{0}\right)$ be the unique solution of the system. Then $\left(-y_{0} ;-x_{0}\right)$ also satisfies the conditions of the system. This solution coincides with the first, so $y_{0}=-x_{0}$. The equation $2|x+7|+|x+4|=$ $c$ has a unique solution $x_{0}=-7$ when $c=|7-4|$, si... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Find the remainder when $20^{16}+201^{6}$ is divided by 9. | Answer: 7
Solution. $201^{16}$ is divisible by 9. 20 gives a remainder of $2.2^{6}$ gives a remainder of $1,2^{16}=2^{6} \cdot 2^{6} \cdot 2^{4}$ gives the same remainder as 16. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 70 negative numbers were recorded. What is the ... | Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=70+62$, which means $x=12$.
2) Let there be $y$ people with "positive tempe... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 68 positive and 64 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=64+68$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 78 positive and 54 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=54+78$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=21
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 12
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=12$, $B C^{2}=9, A C^{2}=21$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 92 positive and 40 negative numbers were recorded. What is the ... | Answer: 2
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=40+92$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 50 positive and 60 negative numbers were recorded. What is the ... | Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=60+50$, which means $x=11$.
2) Let there be $y$ people with "positive tempe... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 54 positive and 56 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=56+54$, which means $x=11$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 48 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+62$, which means $x=11$.
2) Let there be $y$ people with "positive t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 42 positive and 48 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+42$, which means $x=10$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 48 positive and 42 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each of them gave $x-1$ answers, so $x(x-1)=42+48$, which means $x=10$.
2) Let there be $y$ people with "posi... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 36 positive and 36 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=36+36$, which means $x=9$.
2) Let there be $y$ people with "positive te... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Given various natural numbers $a, b, c, d$, for which the following conditions are satisfied: $a>d, a b=c d$ and $a+b+c+d=a c$. Find the sum of all four numbers. | Answer: 12
Solution. From the relations $a>d$ and $ab=cd$, we have the inequality $ba+d$ and $\frac{ac}{2} \geqslant 2c > b+c$. Adding them, we get a contradiction with the condition.
Therefore, without loss of generality, assume that $a \geqslant 3$. Since $a$ is a natural number, then $a \in\{1,2,3\}$. Moreover, $d... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) The area of a sector of a circle is 100. For what value of the radius of the circle will the perimeter of this sector be minimal? If the answer is not an integer, round to the tenths.
Answer: 10 | Solution. The area of the sector corresponding to $\alpha$ radians $S_{\alpha}=\frac{1}{2} R^{2} \alpha=100$. The length of the perimeter is $L(R)=2 R+R \alpha=2 R+\frac{200}{R}$, which reaches a minimum at the point $R=10$. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Find the smallest value of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2(x+7)^{2}+(y-4)^{2}=c \\
(x+4)^{2}+2(y-7)^{2}=c
\end{array}\right.
$$ | Answer: $c=6.0$.
Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$
\begin{aligned}
& \left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right) \geqslant(|x+\alpha|+|y-\beta|)^{2} \\
& \left(1+\frac{1}{2}\right)\left((x+\beta)^{2}+2(y-\alpha)^{2}\right) \geqslant(|x+\beta|+|y-\alpha|)^{2}
\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Several people played a round-robin table tennis tournament. At the end of the tournament, it turned out that for any four participants, there would be two who scored the same number of points in the games between these four participants. What is the maximum number of tennis players that could have participated in t... | Answer: 7 tennis players.
Solution. Let $n \geqslant 8$ be the number of tennis players in the tournament. Then the total number of matches played is $\frac{1}{2} n(n-1)$, and thus the total number of victories is also $\frac{1}{2} n(n-1)$. Therefore, there must be a participant who has won at least $\frac{n-1}{2}$ ma... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. On the side $AB$ of triangle $ABC$, a point $M$ is taken. It starts moving parallel to $BC$ until it intersects with $AC$, then it moves parallel to $AB$ until it intersects with $BC$, and so on. Is it true that after a certain number of such steps, point $M$ will return to its original position? If this is ... | Solution. Let the length of side $AB$ be 1, and let point $M$ be at a distance $a$ from point $B$. From the properties of the parallelogram, the small triangles are equal, so after 3 steps, point $M$ will be at a distance $a$ from point $A$, which is at a distance $1-a$ from point $B$. After another 3 steps, the point ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$. | Solution. Let
$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$
Replace the element $x_{1}$ with the sum of the others. Then
$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$
Reasoning similarly for the other elements, we get that
$$
2 x_{k}=S, \q... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x+3}{2}\right]^{2}-x=1$. | # Solution.
From the equation, it follows that $x=\left[\frac{x+3}{2}\right]^{2}-1$ is an integer. Therefore, $x=n \in \mathbb{Z}$, but we need to consider the cases of even and odd $n$ separately.
First, let's move all terms to the left side of the equation.
1) If $x=2 k$.
$$
\left[\frac{2 k+3}{2}\right]^{2}-2 k-1... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 2.
What is the last digit of the value of the sum $2019^{2020}+2020^{2019} ?$ | # Solution.
The number $2019^{n}$ for $n \in \mathbb{N}$ ends in 9 if $n$ is odd, and in 1 if $n$ is even. Therefore, $2019^{2020}$ ends in 1. The number $2020^{n}$ ends in 0 for any $n \in \mathbb{N}$, so $2020^{2019}$ ends in 0. Thus, the sum $2019^{2020} +$ $2020^{2019}$ ends in the digit 1.
Answer. The digit 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
On the coordinate plane, a square $K$ is marked with vertices at points $(0,0)$ and $(10,10)$. Inside this square, draw the set $M$ of points $(x, y)$ whose coordinates satisfy the equation
$$
[x]=[y],
$$
where $[a]$ denotes the integer part of the number $a$ (i.e., the greatest integer not exceeding $a... | Solution. Let $n \leq x < n+1$, where $n$ is an integer from 0 to 9. Then $[x]=n$ and $[y]=n$. The solution to the latter equation is all $y \in [n, n+1)$. Thus, the solution will be the union of unit squares
$$
\{x \in [n, n+1), y \in [n, n+1), n \in \mathbb{Z}\}
$$
Inside the square $K$ specified in the problem, te... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the code of the letter. Then, the weight of a word can be defined as the sum of the codes of all the letters in that ... | # Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have
$k(C)+k(T)+k(O) \geq k(\amalg)+k(\mathrm{E})+k(C)+k(T)+k(\mathrm{~b})+k(C)+k(O)+k(T)$, which is equivalent to
$$
k(\amalg)+k(\mathrm{E})+k(C)+k(T)+k(\mathrm{~b})=0 .
$$
Thus,
$$
k(\amalg)=k(\mathrm{E})=k(C)=k(T)=k(\mathrm{~b})=0 .
$$
The... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. A triangle was cut into two triangles. Find the greatest value of $N$ such that among the 6 angles of these two triangles, exactly $N$ are the same. | Solution. For $N=4$, an example is an isosceles right triangle divided into two isosceles right triangles: four angles of $45^{\circ}$. Suppose there are five equal angles. Then in one of the triangles, all three angles are equal, meaning all of them, and two angles of the other triangle are $60^{\circ}$. But then both... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements
from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$. | # Solution. Let
$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$
Replace the element $x_{1}$ with the sum of the others. Then
$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$
Reasoning similarly for the other elements, we get that
$$
2 x_{k}=S, ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A table of numbers with 20 rows and 15 columns, $A_{1}, \ldots, A_{20}$ are the sums of the numbers in the rows, $B_{1}, \ldots, B_{15}$ are the sums of the numbers in the columns.
a) Is it possible that $A_{1}=\cdots=A_{20}=B_{1}=\cdots=B_{15}$?
b) If the equalities in part a) are satisfied, what is the sum $A_{1... | Let $A_{i}=B_{j}=X$ for $i=1, \ldots 20$ and $j=1, \ldots, 15$. Consider the sum $S$ of all elements in the table. We have $S=20 X=15 X, X=0$ and $A_{1}+\cdots+A_{20}+B_{1}+\cdots+B_{15}=0$. An example of such a table is, for instance, a table consisting entirely of zeros. There is no need to consider other cases.
Ans... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Early in the morning, the pump was turned on and they started filling the pool. At 10 am, another pump was connected and by 12 pm the pool was half full. By 5 pm, the pool was full. What could be the latest time the first pump was turned on? | Solution. Let the volume of the pool be $V$. Denote by $x$ and $y$ the capacities of the pumps, and by $t$ the time the first pump operates before the second pump is turned on.
Then $t x + 2 x + 2 y = V / 2.5 x + 5 y = V / 2$.
From this, $t x + 2 x + 2 y = 5 x + 5 y$ or $t x = 3 x + 3 y$.
In the end, $t = 3 + 3 y / ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x-1}{2}\right]^{2}+2 x+2=0$. | Solution. From the equation, it follows that $2 x=2-\left[\frac{x-1}{2}\right]^{2}-$ is an integer. Therefore, either $2 x=n \in \mathbb{Z}$, or $x=n+\frac{1}{2} \quad(n \in \mathbb{Z})$. In this case, we need to separately consider the cases of even and odd $n$.
1) If $x=2 k$, then
$$
\begin{aligned}
& {\left[\frac{... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The Atelier "Heavy Burden" purchased a large batch of cast iron buttons. If they sew two buttons on each coat or if they sew three buttons on each coat, in each case 1 piece will remain from the entire batch. If, however, they sew four buttons on each coat or if they sew five buttons on each coat, in each case 3 pie... | # Solution
Let $a$ be the desired number. From the condition, it follows that the number $a-1$ is divisible by 2 and 3. Therefore, $a=6k+1$. Also, the number $a-3$ is divisible by 4 and 5. Therefore, $a=20n+3$. We solve the equation
$$
6k+1=20n+3
$$
Or, equivalently,
$$
3k=10n+1
$$
Its general solution is
$$
k=7+... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
Four brigades were developing an open coal deposit for three years, working with a constant productivity for each brigade. In the second year, due to weather conditions, work was not carried out for four months, and for the rest of the time, the brigades worked in rotation (one at a time). The ratio of the w... | Solution. Let the $i$-th brigade mine $x_{i}$ coal per month. Then we have the system
$$
\left\{\begin{array}{l}
4 x_{1}+x_{2}+2 x_{3}+5 x_{4}=10 \\
2 x_{1}+3 x_{2}+2 x_{3}+x_{4}=7 \\
5 x_{1}+2 x_{2}+x_{3}+4 x_{4}=14
\end{array}\right.
$$
By adding twice the first equation to thrice the second and subtracting the thi... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 2.
What is the last digit of the value of the sum $5^{2020}+6^{2019} ?$ | # Solution.
The number 5 to any power ends in 5.
The number 6 to any power ends in 6.
Therefore, the sum $5^{2020}+6^{2019}$ ends in the digit 1.
Answer. The digit 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5.
The carriages of the express train "Moscow - Yalta" must be numbered consecutively, starting from one. But in a hurry, two adjacent carriages received the same number. As a result, it turned out that the sum of the numbers of all carriages is 111. How many carriages are in the train and which number was u... | # Solution.
Let's start calculating the sums sequentially
$$
\begin{aligned}
& 1+2=3 \\
& 1+2+3=6 \\
& \cdots \\
& 1+2+\ldots+14=105 \\
& 1+2+\ldots+14+15=120
\end{aligned}
$$
From this, it is clear that the first 14 wagons had the first 14 sequential numbers, which sum up to 105, and one more wagon had a number equ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1.
Three electric generators have powers $x_{1}, x_{2}, x_{3}$, the total power of all three does not exceed 2 MW. In the power system with such generators, a certain process is described by the function
$$
f\left(x_{1}, x_{2}, x_{3}\right)=\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{... | # Solution.
It is clear that the minimum value of the function is zero (achieved when $\left.x_{1}=x_{2}=x_{3}=0\right)$.
Let's find the maximum. We can assume that $x_{1} \geq x_{2} \geq x_{3} \geq 0$. We will prove two inequalities:
$$
\begin{gathered}
\sqrt{x_{1}^{2}+x_{2} x_{3}} \leq x_{1}+\frac{x_{3}}{2} \\
\sq... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
Two swimmers are training in a rectangular quarry. The first swimmer finds it more convenient to exit at a corner of the quarry, so he swims along the diagonal to the opposite corner and back. The second swimmer finds it more convenient to start from a point that divides one of the quarry's shores in the ... | # Solution.
Let's draw a rectangular quarry $A D C D$ and the inscribed quadrilateral route of the second swimmer $N L M K$.
Reflect the drawing symmetrically first with respect to side $C D$, then with respect to side $C B^{\prime}$, and finally with respect to side $A^{\prime} B^{\prime}$.
 is even. We will swap the positions of two adjacent children so that all boys gat... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. There are 4 numbers, not all of which are the same. If you take any two of them, the ratio of the sum of these two numbers to the sum of the other two numbers will be the same value $\mathrm{k}$. Find the value of $\mathrm{k}$. Provide at least one set of four numbers that satisfy the condition. Describe all... | Solution. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be such numbers. Write the relations for the sums of pairs of numbers:
$$
\begin{aligned}
& \frac{x_{1}+x_{2}}{x_{3}+x_{4}}=\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=k \\
& \frac{x_{1}+x_{3}}{x_{2}+x_{4}}=\frac{x_{2}+x_{4}}{x_{1}+x_{3}}=k \\
& \frac{x_{1}+x_{4}}{x_{2}+x_{3}}=\frac{x_{2... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1.
The Triassic Discoglossus tadpoles have five legs, while the saber-toothed frog tadpoles grow several tails (all have the same number of tails). An employee of the Jurassic Park scooped up several tadpoles with water. It turned out that the captured tadpoles had a total of 100 legs and 64 tails. How many tai... | # Solution.
Let $x$ be the number of tails of the saber-toothed frog's pollywog. Suppose $n$ five-legged and $k$ many-tailed pollywogs were caught. Counting the total number of legs and tails gives the equations
$$
\left\{\begin{aligned}
5 n+4 k & =100 \\
n+x k & =64
\end{aligned}\right.
$$
From the first equation, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 5.
Once upon a time, Baba Yaga and Koschei the Deathless tried to divide a magical powder that turns everything into gold equally. Baba Yaga took out a scale and weighed all the powder. The scales showed 6 zolotniks. Then she started removing the powder until the scales showed 3 zolotniks. However, Koschei susp... | # Solution.
Let $A$ be the weight of the first part (the one that remained on the scales), $B$ be the weight of the second part (the one that was poured off), and $d$ be the error of the scales.
Then the result of the first weighing (of the entire powder) gives
$$
A+B+d=6,
$$
the result of the second weighing (afte... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4.
At 9:00 AM, ships "Anin" and "Vinin" departed from port O to port E. At the same moment, ship "Sanin" departed from port E to port O. All three vessels are traveling on the same course ("Sanin" heading towards "Anin" and "Vinin") at constant but different speeds. At 10:00 AM, "Vinin" was at the same dista... | # Solution.
For simplicity, let's assume the distance between the ports is 1. Let $x$ be the speed of "Anina", $y$ be the speed of "Vanina", and $z$ be the speed of "Sanina". Then at 10 o'clock (i.e., one hour after departure),
$$
2 y = x + 1 - z,
$$
and one and a half hours after departure,
$$
2(1 - 1.5 z) = 1.5 x... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. (15 points) In one move, you can choose a natural number $x$ and cross out all natural numbers $y$ such that $|x-y|$ is a natural composite number. At the same time, $x$ can be chosen from already crossed out numbers.
What is the minimum number of moves needed to cross out all numbers from the natural num... | Answer: 2.
Solution. First, note that one move is not enough, because if we choose some natural number $x$, then the number $y=x$ will remain unchecked. Let's prove that two moves are enough.
We will look for two suitable numbers $x_{1}$ and $x_{2}$ of different parity. Then one of the differences $\left|y-x_{1}\righ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 7. Problem 7.1*
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha, in any order, tells the digits that stand in these places. The order in which to tell the digits is chosen by Misha. For examp... | # 7. Problem $7.1^{*}$
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha reports the digits in these positions in any order. The order in which to report the digits is chosen by Misha. For exampl... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. (15 points) Do there exist polynomials $P(x), Q(x)$, and $R(x)$ with real coefficients such that the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have the same degree, while the polynomials $P(x)+Q(x), P(x)+R(x)$, and $Q(x)+R(x)$ have pairwise distinct degrees? (We consider that t... | Answer: $: 2$.
Solution. It is sufficient to take, for example, $P(x)=x^{2}, Q(x)=-x^{2}+1, R(x)=x^{2}+x$. Then the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have degree 4 (these are products of two polynomials of degree 2); the polynomial $P(x)+Q(x)$ equals 1 and has degree 0; the polynomi... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 10.4. (15 points) Once, 45 friends living in different parts of the world decided to exchange news with each other. To do this, they plan to arrange $k$ video meetings, at each of which every person will share their news, as well as all the news from other people they have learned previously.
For the video mee... | Answer: 5 days.
Solution. Let's provide an example of a situation where 4 days would not be enough. Suppose each of the 45 people has their own unique pair of days when they cannot participate in the meeting. Since the number of ways to choose a pair of days from the 10 offered is $\mathrm{C}_{10}^{2}=45$, for any pai... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 10.5. (20 points) Find all composite natural numbers $n$ that have the following property: each natural divisor of the number $n$ (including $n$ itself), decreased by 1, is a square of an integer. | Answer: 10.
Solution. Suppose that $n$ is divisible by the square of some prime number $p$. Then it has a divisor $p^{2}=b^{2}+1$; but two squares of integers can only differ by 1 if they are 0 and 1.
Let $n$ be divisible by some two prime numbers $p$ and $q$. Without loss of generality, we can assume that $p>q$. Fro... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In an acute-angled triangle $A B C$, the altitude $A A_{1}$ is drawn. $H$ is the orthocenter of triangle $A B C$. It is known that $A H=3, A_{1} H=2$, and the radius of the circumcircle of triangle $A B C$ is 4. Find the distance from the center of this circle to $H$. | Solution. Draw the altitudes $B B_{1}$ and $C C_{1}$ in the triangle. Then the quadrilateral $A C_{1} H B_{1}$ is cyclic, since its opposite angles $C_{1}$ and $B_{1}$ are right angles. Therefore: $\angle B H C=$ $\angle C_{1} H B_{1}=180^{\circ}-\angle C_{1} A B_{1}$. Reflect point $H$ symmetrically with respect to th... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9-4. From $n$ regular hexagons with side 1, a polygon on the plane was made by gluing hexagons along their sides. Any two hexagons either have exactly one common side or have no common points at all. There are no holes inside the polygon. Moreover, each hexagon has at least one side lying on the boundary of the... | Answer: $2 n+6$ for $n \geq 2$, $6$ for $n=1$.
Solution. It is easy to understand that if a point is an endpoint for at least one side of a hexagon, it is an endpoint for two or three sides: more than three is impossible because all angles are $120^{\circ}$.
Let's draw a bit. Imagine that in each point that is an end... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9-5. A divisor of a natural number is called proper if it is different from 1 and the number itself. Find all natural numbers for which the difference between the sum of the two largest proper divisors and the sum of the two smallest proper divisors is a prime number. | Answer: $12(2,3,4,6)$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.4. Along the shore of a circular lake with a perimeter of 1 km, two salmons are swimming - one at a constant speed of $500 \mathrm{m} /$ min clockwise, the other at a constant speed of 750 m/min counterclockwise. Along the edge of the shore, a bear is running, always moving along the shore at a speed of 200 m/min in ... | Solution. Note that the bear runs to the nearest salmon if and only if it runs from a point on the shore that is equidistant from the salmons and not separated from the bear by the salmons. Since both points equidistant from the salmons move counterclockwise at a speed of $|750-500| / 2=125<200$ m/min, they will never ... | 7 | Other | math-word-problem | Yes | Yes | olympiads | false |
# 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively, such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30. | # 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30.
## Answer: 7
# | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)
Answer: 1. | Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not in... | Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7-5. In the garden of the oracle, there live four turtles. A visitor can choose any subset of turtles in a move and ask the oracle how many of these turtles are males (the oracle's answers are always truthful). What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3. | Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question fo... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. (20 points) Quadratic trinomials $P(x)$ and $Q(x)$ with real coefficients are such that together they have 4 distinct real roots, and each of the polynomials $P(Q(x))$ and $Q(P(x))$ has 4 distinct real roots. What is the smallest number of distinct real numbers that can be among the roots of the polynomia... | # Answer: 6.
Solution. Note that if among the roots of the polynomial $P(Q(x))$ there is a root of $Q(x)$, say, the number $x_{0}$, then $P\left(Q\left(x_{0}\right)\right)=P(0)=0$, from which 0 is a root of $P(x)$. Similarly, if among the roots of $Q(P(x))$ there is a root of $P(x)$, then 0 is a root of $Q(x)$. Howeve... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8-4. In the garden of the oracle, there live four turtles. A visitor can choose any subset of turtles in a move and ask the oracle how many of these turtles are males (the oracle's answers are always truthful). What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3. | Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question fo... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8-5. 8 8-5. A divisor of a natural number is called proper if it is different from 1 and the number itself. Find all natural numbers for which the difference between the sum of the two largest proper divisors and the sum of the two smallest proper divisors is a prime number. | Answer: $12(2,3,4,6)$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Natural numbers $x$ and $y$ are such that the following equality holds:
$$
x^{2}-3 x=25 y^{2}-15 y
$$
How many times greater is the number $x$ than the number $y$? | Answer. 5
Solution. The equality $x^{2}-3 x=25 y^{2}-15 y$ is equivalent to the equality $(5 y-x)(5 y+x-3)=$ 0. Since $x$ and $y$ are natural numbers, $x, y \geq 1$. Therefore, the second bracket $5 y+x-3 \geq 3$, in particular, is non-zero. Therefore, the first bracket $5 y-x$ is zero, meaning $x$ is 5 times $y$.
Cr... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha played 7 ma... | Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and M... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. How many different (i.e., not equal to each other) acute triangles with integer side lengths and a perimeter of 24 exist? List the lengths of the three sides of all these triangles and prove that there are no others.
ANSWER: 6.
CRITERIA
b - correct criterion for acuteness and either a correct list of acute tria... | Solution. Note that a triangle with sides $a \geqslant b \geqslant c$ is obtuse if and only if $a^{2}>b^{2}+c^{2}$. Indeed, if $A B C$ is a triangle with sides $a \geqslant b \geqslant c$ opposite vertices $A, B$, and $C$ respectively, and $A B^{\prime} C$ is a triangle with a right angle $\angle A$ and $A B^{\prime}=A... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. (20 points) There were $n$ identical-looking coins weighing $x_{1}, x_{2}, \ldots, x_{n}$ grams (the weights of the coins are pairwise distinct positive real numbers), and also weightless stickers with numbers $x_{1}, x_{2}, \ldots, x_{n}$. At night, a lab assistant weighed the coins and labeled them with ... | Answer: 2 yes
Solution. Consider the weights 10, 20, 25, 30, 35, 201, 203, 207 (weights will be measured in grams hereafter). We will perform two checks:
$$
\begin{aligned}
10+20+30 & =25+35 \\
10+25+201 & <30+207
\end{aligned}
$$
First, consider the first weighing. We will prove that if some three coins balance som... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha - 7 matches... | Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and M... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 11-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)
Answer: 1. | Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not in... | Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On side $AB$ of triangle $ABC$, a point $D$ is taken. In triangle $ADC$, the angle bisectors $AP$ and $CQ$ are drawn. On side $AC$ of triangle $ADC$, a point $R$ is taken such that $PR \perp CQ$. It is known that the angle bisector of angle $D$ of triangle $BCD$ is perpendicular to segment $PB$, $AB=18$, $AP=12$. Fi... | Solution. Let $A B=n, A P=m, A D=c, A C=b, D P=p, C P=q, B D=D P=p, C P=C R=q$.
Using the property of the angle bisector and the formula for the length of the angle bisector, we get $\left\{\begin{array}{l}\frac{c}{b}=\frac{p}{q}, \\ c+p=A B=n, \\ m^{2}=b c-p q,\end{array}\right.$
$\left\{\begin{array}{l}c=\frac{b p}... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given three points $A, B, C$, forming a triangle with angles $30^{\circ}, 45^{\circ}, 105^{\circ}$. Two of these points are chosen, and the perpendicular bisector of the segment connecting them is drawn, after which the third point is reflected across this perpendicular bisector. This results in a fourth point $D$. ... | Answer: 12 points.
## Solution. FIRST WAY. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
# | # 1. Problem 1*
Let $S$ be the sum of the digits of the number $11^{2017}$. Find the remainder when $S$ is divided by 9. Points for the problem: 8.
## Answer: 2
# | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Each move of a chess knight is a move of one square horizontally and two squares vertically, or vice versa - one square vertically and two squares horizontally. (In the diagram on the right, the knight marked with the letter $\mathrm{K}$ can move to any of the shaded squares in one move.)
A chess knight is placed i... | Answer: 2
Solution. Let's number the cells as shown in Figure 1. The knight can move from any cell to any cell with the same number, and cannot move to other cells.
| $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{4}$ | $\mathbf{2}$ | $\mathbf{3}$ | $... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
27. On a perfectly competitive market for a good, the market demand and market supply functions are given respectively as: $Q^{d}=60-14 P$ and $Q^{s}=20+6 P$. It is known that if the amount of labor used by a typical firm is $\mathbf{L}$ units, then the marginal product of labor is $\frac{160}{L^{2}}$. Determine how ma... | Solution: find the equilibrium price of the good $60-14 P=20+6 P, P=2$. Optimal choice of the firm: MPL $x P=w ; \frac{160}{L^{2}} \times 2=5 ; \boldsymbol{L}=\boldsymbol{8}$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given three points $A, B, C$, forming a triangle with angles $30^{\circ}, 45^{\circ}, 105^{\circ}$. Two of these points are chosen, and the perpendicular bisector of the segment connecting them is drawn, after which the third point is reflected across this perpendicular bisector. This results in a fourth point $D$. ... | Answer: 12 points.
## Solution. FIRST WAY. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (3 points) In triangle $A B C$, a square $K L M N$ with side length 1 is inscribed: points $K$ and $L$ lie on side $A C$, points $M$ and $N$ lie on sides $A B$ and $B C$ respectively. The area of the square is half the area of the triangle. Find the length of the height $B H$ of triangle $A B C$. | Answer: 2.
Solution 1:
$A C \cdot B H=2 S_{A B C}=4 S_{K L M N}=4$.
Points $A, L, K, C$ lie on a straight line in that exact order. Moreover, triangle $A B C$ is clearly acute. From the condition, it follows that
$$
\begin{gathered}
S_{A M L}+S_{C K N}+S_{B M N}=S_{K L M N} \\
\frac{A L \cdot 1}{2}+\frac{C K \cdot ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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 $\mathrm{AD}$. Point $\mathrm{L}$ on side $\mathrm{CD}$ is such that $\mathrm{CL}=\mathrm{AK}$. Point $\mathrm{M}$ is the intersection of ... | Answer: 8
## Examples of answer recording: 14 0.5
# | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In a watch repair shop, there is a certain number of electronic watches (more than one), displaying time in a 12-hour format (the number of hours on the watch face changes from 1 to 12). All of them run at the same speed, but show completely different times: the number of hours on the face of any two different watch... | Answer: 11
## Examples of answer notation:
14
# | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
In triangle $A B C$, the median $B M$ is drawn, in triangle $M C B$ - the median $B N$, in triangle $B N A$ - the median $N K$. It turned out that $N K \perp B M$. Find $A C: B C$.
Answer: 2
# | # Solution:
Let $\vec{b}=\overrightarrow{C B}$ and $\vec{a}=\overrightarrow{C A}$.
$\overrightarrow{N K}=\overrightarrow{C K}-\overrightarrow{C N}=\frac{\vec{a}+\vec{b}}{2}-\frac{\vec{a}}{4}=\frac{\vec{a}+2 \vec{b}}{4}$.
$\overrightarrow{B M}=\overrightarrow{C M}-\overrightarrow{C B}=\frac{\vec{a}}{2}-\vec{b}$
$0=\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (4 points)
Natural numbers $a, b, c$ are such that $\operatorname{GCD}(\operatorname{LCM}(a, b), c) \cdot \operatorname{LCM}(\operatorname{GCD}(a, b), c)=200$.
What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(a, b), c)$ can take? | Answer: 10
## Solution:
Note that $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $z$, and $z$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$, so $\operatorname{LCM}(\operatorname{GCD}(x, y)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.
$200=2^{3} \cdot 5^{2}... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The three heights of a tetrahedron are three, four, and four times greater than the radius of its inscribed sphere, respectively. How many times greater is the fourth height than the radius of the inscribed sphere? | Answer: 6
## Examples of answer notations:
14
$1 / 4$
# | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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