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Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$. | Answer: 4.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,371 |
Problem 8. Solve the system
$$
\left\{\begin{aligned}
x+y+z & =15 \\
x^{2}+y^{2}+z^{2} & =81 \\
x y+x z & =3 y z
\end{aligned}\right.
$$ | Answer: $(6 ; 3 ; 6),(6 ; 6 ; 3)$.
Solution. Let $y+z$ be denoted by $p$, and $yz$ by $q$. Then the system becomes
$$
\left\{\begin{aligned}
x+p & =15 \\
x^{2}+p^{2}-2 q & =81 \\
x p & =3 q .
\end{aligned}\right.
$$
From the last two equations, $(x+p)^{2}-\frac{8}{3} x p=81$, and considering the first equation, $x p... | (6;3;6),(6;6;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,372 |
Problem 9. The polyhedron shown in the figure has all dihedral angles that are right angles. Sasha claims that the shortest path along the surface of this polyhedron from vertex $X$ to vertex $Y$ has a length of 4. Is he correct?
. He scored 3 points less in Russian than in Physics, and in Physics, he scored 7 points less than in Mathematics. The golden fish that appeared in Vanya's dream promised to fulfill any number of wishes of the following types:
- add one point to each exam;
- decreas... | Answer: No.
Solution: For Vanya to score more than 100 points on more than one exam, the difference in points between these two exams must become zero. However, the difference in points between any two exams either remains unchanged or changes by 4. And initially, the difference in points does not divide by 4 for any ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,375 |
Problem 5. Three regular nonagons have a common center, their sides are respectively parallel. The sides of the nonagons are 8 cm and 56 cm. The third nonagon divides the area of the figure enclosed between the first two in the ratio $1: 7$, counting from the smaller nonagon. Find the side of the third nonagon. Answer.... | Solution. Let the side of the third nonagon be $x$. The areas of similar figures are in the ratio of the squares of corresponding sides. Therefore, the areas enclosed between the nonagons are in the ratio $\left(x^{2}-8^{2}\right):\left(56^{2}-x^{2}\right)=1: 7$, from which the answer follows. | 8\sqrt{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,376 |
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$. | Answer: 3.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,377 |
Task 1. From the relation $x_{n-1}=\frac{x_{n}+x_{n-1}+x_{n-2}}{3}$, we get: $2 x_{n-1}=x_{n-2}+x_{n}$ for all $n \geq 3$. This means that the given sequence is an arithmetic progression. Let the common difference of the progression be $d$. Then $\frac{x_{300}-x_{33}}{x_{333}-x_{3}}=\frac{(300-33) d}{330 d}=\frac{89}{1... | Answer: $\frac{89}{110}$. | \frac{89}{110} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,379 |
Problem 4. Using vectors.
$$
\begin{aligned}
& \overline{H K}=\overline{A K}-\overline{A H}=\frac{1}{2}(\overline{A Q}+\overline{A N})-\frac{1}{2}(\overline{A M}+\overline{A P})= \\
& =\frac{1}{2}\left(\frac{1}{2}(\overline{A E}+\overline{A D})+\frac{1}{2}(\overline{A B}+\overline{A C})\right)- \\
& -\frac{1}{2}\left(... | Answer: 4.8.
 | 4.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,380 |
Problem 5. Grouping the terms, we get: $\left(4 y^{2}+1\right)\left(x^{4}+2 x^{2}+2\right)=8|y|\left(x^{2}+1\right)$.
Obviously, there are no solutions when $y=0$. Dividing both sides by $2|y|\left(x^{2}+1\right)$:
$$
\left(2|y|+\frac{1}{2|y|}\right)\left(x^{2}+1+\frac{1}{x^{2}+1}\right)=4
$$
It is known ${ }^{1}$ (... | Answer: $\left(0, \frac{1}{2}\right),\left(0,-\frac{1}{2}\right)$. | (0,\frac{1}{2}),(0,-\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,381 |
Problem 6. Suppose $T$ is the period of the function $y=f(x)$. From the condition, we get:
$$
\cos x=f(x)-2 f(x-\pi)=f(x-T)-2 f(x-T-\pi)=\cos (x-T)
$$
Since this equality holds for all $x$, the number $T$ is the period of the function $y=\cos x$, i.e., $T=2 \pi n$, where $n$ is some natural number. From the given equ... | Answer: $f(x)=\frac{1}{3} \cos x$. | f(x)=\frac{1}{3}\cosx | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,382 |
Problem 8. The left side of the equation is non-negative. Therefore, $a x \geq 0$. When $a=0$, the equation has two roots -1 and 1. Hence, $a=0$ does not satisfy the condition.
Consider the case $a>0$. Then $x \geq 0$ and therefore $|x|=x$. Let's construct the graph of the function $y=|\ln x|$. The line $y=a x$ must i... | Answer. $-\frac{1}{e}<a<0, 0<a<\frac{1}{e}$. | -\frac{1}{e}<<0,0<<\frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 9,383 |
Problem 1. Prove that for each natural number $n$ the number $5^{2 n}+3^{n+2}+3^{n}$ is divisible by 11. | Solution. Note that $5^{2 n}+3^{n+2}+3^{n}=(22+3)^{n}+10 \cdot 3^{n}$. Using the binomial theorem, we can group all terms that contain $22: (22+3)^{2 n}+10 \cdot 3^{n}=22 A+3^{n}+10 \cdot 3^{n}=22 A+11 \cdot 3^{n}$. Since each term is divisible by 11, the entire number is divisible by 11.
Comment. The same solution ca... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,386 |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 13 adventurers have rubies; exactly 9 have emeralds; exactly 15 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has sapphires, then they have either emeralds or diamonds (but not both at the ... | Answer: 22.
Solution. Note that the number of adventurers who have sapphires is equal to the total number of adventurers who have emeralds or diamonds. Then, from the first condition, it follows that 9 adventurers have sapphires and emeralds, and 6 have sapphires and diamonds. That is, every adventurer who has emerald... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,387 |
Task 3. A team of workers was working on pouring the rink on the large and small fields, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 4 more workers than in the part that worked on the small field. When the pouring of the la... | Answer: 10.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+4$, and the total number of people in the team is $2n+4$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,388 |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=8$ and $AC=4$. Find the length of side $AB$ if the length of the vector $4 \overrightarrow{OA} - \overrightarrow{OB} - 3 \overrightarrow{OC}$ is 10. | Answer: 5.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,389 |
Problem 5. Solve the system of equations in real numbers:
$$
\left\{\begin{array}{l}
a+c=4 \\
a c+b+d=6 \\
a d+b c=5 \\
b d=2
\end{array}\right.
$$ | Answer: $(3,2,1,1)$ and $(1,1,3,2)$.
Solution. Let $x^{2}+a x+b$ and $x^{2}+c x+d$ be the two quadratic polynomials, the coefficients of which are the roots of the given system. Then
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=x^{4}+4 x^{3}+6 x^{2}+5 x+2$.
By t... | (3,2,1,1)(1,1,3,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,390 |
Problem 6. Solve the equation $\arcsin \frac{x \sqrt{11}}{2 \sqrt{21}}+\arcsin \frac{x \sqrt{11}}{4 \sqrt{21}}=\arcsin \frac{5 x \sqrt{11}}{8 \sqrt{21}}$. | Answer: $0, \pm \frac{21}{10}$
Solution. From the condition on the domain of the arcsine, it follows that
$$
|x| \leqslant \frac{8 \sqrt{21}}{5 \sqrt{11}}, \text { or, equivalently, } x^{2} \leqslant \frac{1344}{275} .
$$
By calculating the sine of both sides of the equation and considering that
$$
\cos \arcsin t>0... | 0,\\frac{21}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,391 |
Problem 7. For what values of the parameter $a$ does the equation
$$
\log _{2}^{2} x+(a-6) \log _{2} x+9-3 a=0
$$
have exactly two roots, one of which is four times the other? | Answer: $-2.2$
Solution. Let $t=\log _{2} x$, then the equation becomes $t^{2}+(a-6) t+(9-3 a)=0$. Notice that $3 \cdot(3-a)=9-3 a, 3+(3-a)=6-a$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -3 and $3-a$. We make the reverse substitution: $\log _{2} x=3$ or $\log _{2} x=3-a$, ... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,392 |
Problem 8. In triangle $A B C$, side $A C=42$. The bisector $C L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $2: 1$, counting from the vertex. Find the length of side $A B$, if the radius of the circle inscribed in triangle $A B C$ is 14. | Answer: 56.
Answer: Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $AI$ is the angle bisector in triangle $ALC$, by the angle bisector theorem, we have: $AC: AL = CI: IL = 2: 1$, from which $AL = AC / 2 = 21$.
Next, $AC \cdot AL \c... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,393 |
Problem 9. The function $F$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $F(n a, n b, n c)=n \cdot F(a, b, c)$, $F(a+n, b+n, c+n)=F(a, b, c)+n$, $F(a, b, c)=F(c, b, a)$ hold. Find $F(58,59,60)$. | Answer: 59.
Solution. Note that $F(-1,0,1)=F(1,0,-1)=(-1) \cdot F(-1,0,1)$, from which $F(-1,0,1)=0$. Then $F(58,59,60)=F(-1,0,1)+59=59$.
Comment. The function $F$ cannot be uniquely determined. For example, the functions $F(a, b, c)=(a+b+c) / 3$, $F(a, b, c)=b$, and $F(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,394 |
Problem 10. Let $B$ - be the set of real numbers, not containing 0 and 1. It is known that if $b \in B$, then $\frac{1}{b} \in B$ and $1-\frac{1}{b} \in B$. Can $B$ have exactly 1000 elements? | Answer: No, it cannot.
Solution. Let's look at the numbers $\left\{t, \frac{1}{t}, 1-t, \frac{1}{1-t}, \frac{t}{t-1}, \frac{t-1}{t}\right\}$. Let $\alpha: x \mapsto \frac{1}{x}, \beta: x \mapsto 1-\frac{1}{x}=\frac{x-1}{x}$. Notice that the mapping $\alpha$ transforms the numbers $t, \frac{1}{t}, 1-t, \frac{1}{1-t}, \... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,395 |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 5 adventurers have rubies; exactly 11 have emeralds; exactly 10 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has diamonds, then they have either emeralds or sapphires (but not both at the ... | Answer: 16.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or diamonds. Then, from the second condition, it follows that 5 adventurers have rubies and emeralds, and 6 have emeralds and diamonds. That is, every adventurer who has diamonds must... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,396 |
Task 3. A landscaping team worked on a large and a small football field, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 6 more workers than in the part that worked on the small field. When the landscaping of the large field wa... | Answer: 16.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+6$, and the total number of people in the team is $2n+6$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,397 |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=8$ and $AC=5$. Find the length of side $BC$ if the length of the vector $\overrightarrow{OA}+3 \overrightarrow{OB}-4 \overrightarrow{OC}$ is 10. | Answer: 4.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,398 |
Problem 5. Solve the system of equations in real numbers:
$$
\left\{\begin{array}{l}
a+c=-4 \\
a c+b+d=6 \\
a d+b c=-5 \\
b d=2
\end{array}\right.
$$ | Answer: $(-3,2,-1,1)$ and $(-1,1,-3,2)$
Solution. Let $x^{2}+a x+b$ and $x^{2}+c x+d$ be the two quadratic polynomials, the coefficients of which are the roots of the given system. Then
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=x^{4}-4 x^{3}+6 x^{2}-5 x+2$.
B... | (-3,2,-1,1)(-1,1,-3,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,399 |
Problem 6. Solve the equation $\arcsin \frac{x \sqrt{35}}{4 \sqrt{13}}+\arcsin \frac{x \sqrt{35}}{3 \sqrt{13}}=\arcsin \frac{x \sqrt{35}}{2 \sqrt{13}}$. | Answer: $0, \pm \frac{13}{12}$.
Solution. From the condition on the domain of the arcsine, it follows that
$$
|x| \leqslant \frac{2 \sqrt{13}}{\sqrt{35}}
$$
By calculating the sine of both sides of the equation and considering that $\cos \arcsin >0$, and therefore,
$$
\sin \arcsin A=A, \quad \cos \arcsin A=\sqrt{1-... | 0,\\frac{13}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,400 |
Problem 7. For what values of the parameter $a$ does the equation
$$
a \log _{3}^{2} x-(2 a+3) \log _{3} x+6=0
$$
have exactly two roots, one of which is three times the other? | Answer: 1,3.
Solution. Let $t=\log _{3} x$, then the equation takes the form $a t^{2}-(2 a+3) t+6=0$ or $t^{2}-\frac{2 a+3}{a} t+\frac{6}{a}=$ $0(a \neq 0)$. Note that $2 \cdot \frac{3}{a}=\frac{6}{a}, 2+\frac{3}{a}=\frac{2 a+3}{a}$, from which, by the theorem converse to Vieta's theorem, the roots of this equation ar... | 1,3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,401 |
Problem 8. In triangle $A B C$, side $A B=40$. The center $I$ of the inscribed circle divides the bisector $A L$ in the ratio $5: 3$, counting from the vertex. Find the radius of the circumscribed circle around triangle $A B C$, if the radius of the inscribed circle is 15. | Answer: $85 / 2=42.5$.
Solution. Noticing that $BI$ is the bisector in triangle $ALB$, by the property of the bisector of a triangle, we have: $AB: BL=AI: IL=5: 3$, from which $BL=3AB / 5=24$.
Next, $AB \cdot BL \cdot \sin \angle B=2 S_{\triangle ABL}=2 S_{\triangle AIB}+2 S_{\triangle BIL}=AB \cdot r+BL \cdot r=(AB+... | 42.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,402 |
Problem 9. The function $f$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $f(n a, n b, n c)=n \cdot f(a, b, c)$, $f(a+n, b+n, c+n)=f(a, b, c)+n$, $f(a, b, c)=f(c, b, a)$ hold. Find $f(24,25,26)$. | Answer: 25.
Solution. Note that $f(-1,0,1)=f(1,0,-1)=(-1) \cdot f(-1,0,1)$, from which $f(-1,0,1)=0$. Then $f(24,25,26)=f(-1,0,1)+25=25$.
Comment. The function $f$ cannot be uniquely determined. For example, the functions $f(a, b, c)=(a+b+c) / 3$, $f(a, b, c)=b$, and $f(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,403 |
Problem 10. Let $A$ be a set of real numbers that does not contain 0 and 1. It is known that if $a \in A$, then $\frac{1}{a} \in A$ and $\frac{1}{1-a} \in A$. Can $A$ have exactly 1000 elements? | Answer: No, it cannot.
Solution. Let's look at the numbers $\left\{t, \frac{1}{t}, 1-t, \frac{1}{1-t}, \frac{t}{t-1}, \frac{t-1}{t}\right\}$. Let $\alpha: x \mapsto \frac{1}{x}, \beta: x \mapsto \frac{1}{1-x}$. Notice that the mapping $\alpha$ transforms the numbers $t, \frac{1}{t}, 1-t, \frac{1}{1-t}, \frac{t}{t-1}, ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,404 |
Problem 1. Prove that for each natural number $n$ the number $7^{2 n}+10^{n+1}+2 \cdot 10^{n}$ is divisible by 13. | Solution. Note that $7^{2 n}+10^{n+1}+2 \cdot 10^{n}=(39+10)^{n}+12 \cdot 10^{n}$. Using the binomial theorem and grouping terms that contain $39:(39+10)^{n}+12 \cdot 10^{n}=39 A+10^{n}+12 \cdot 10^{n}=39 A+13 \cdot 10^{n}$. Since each term is divisible by 13, the entire number is divisible by 13.
Comment. The same so... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,405 |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 4 adventurers have rubies; exactly 10 have emeralds; exactly 6 have sapphires; exactly 14 have diamonds. Moreover, it is known that
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same t... | Answer: 18.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or sapphires. Then, from the second condition, it follows that 4 adventurers have rubies and emeralds, and 6 have emeralds and sapphires. That is, every adventurer who has rubies must... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,406 |
Problem 3. A team of workers was laying linoleum in a store's warehouse and in the cash hall, with the warehouse area being 3 times larger than the cash hall area. In the part of the team working in the warehouse, there were 5 more workers than in the part working in the cash hall. When the work in the warehouse was co... | Answer: 9.
Solution. Let the number of workers in the cash hall be denoted as $n$, then the number of workers in the warehouse is $n+5$, and the total number of people in the team is $2n+5$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivity o... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,407 |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=5$ and $AB=4$. Find the length of side $AC$ if the length of the vector $3 \overrightarrow{OA}-4 \overrightarrow{OB}+\overrightarrow{OC}$ is 10. | Answer: 8.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cd... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,408 |
Problem 5. Solve the system of equations in real numbers:
$$
\left\{\begin{array}{l}
a+c=-7 \\
a c+b+d=18 \\
a d+b c=-22 \\
b d=12
\end{array}\right.
$$ | Answer: $(-5,6,-2,2)$ and $(-2,2,-5,6)$
Solution. Let $x^{2}+a x+b$ and $x^{2}+c x+d$ be the two quadratic polynomials, the coefficients of which are the roots of the given system. Then
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=x^{4}-7 x^{3}+18 x^{2}-22 x+12$.... | (-5,6,-2,2)(-2,2,-5,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,409 |
Problem 6. Solve the equation $\arcsin \frac{2 x}{\sqrt{15}}+\arcsin \frac{3 x}{\sqrt{15}}=\arcsin \frac{4 x}{\sqrt{15}}$. | Answer: $0, \pm \frac{15}{16}$.
Solution. From the condition on the domain of the arcsine, it follows that
$$
|x| \leqslant \frac{\sqrt{15}}{4}
$$
By calculating the sine of both sides of the equation and considering that $\cos \arcsin >0$, and therefore,
$$
\sin \arcsin A=A, \quad \cos \arcsin A=\sqrt{1-A^{2}}
$$
... | 0,\\frac{15}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,410 |
Problem 7. For what values of the parameter $a$ does the equation
$$
\log _{4}^{2} x+(a-4) \log _{4} x+a-5=0
$$
have exactly two roots, one of which is four times the other? | Answer: 7.5.
Solution. Let $t=\log _{4} x$, then the equation takes the form $t^{2}+(a-4) t+a-5=0$. Notice that $(-1) \cdot 5-a=a-5,-1+(5-a)=4-a$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -1 and $5-a$. We make the reverse substitution: $\log _{4} x=-1$ or $\log _{4} x=5-a$... | 7.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,411 |
Problem 8. In triangle $A B C$, side $B C=35$. The center $I$ of the inscribed circle of the triangle divides the bisector $A L$ in the ratio $5: 2$, counting from the vertex. Find the length of side $A B$, if the radius of the inscribed circle of triangle $A B C$ is 10. | Answer: $A B=\frac{5}{12}(105 \pm 2 \sqrt{105})$.
Solution. Let $A H$ be the height of the triangle, $I S$ be the perpendicular from $I$ to the line $B C$. Since $I$ is the center of the inscribed circle, $I S=10$. Due to the similarity of triangles $A L H$ and $I L S, A H: I S=$ $A L: I L$, hence $A H=I S \cdot \frac... | \frac{5}{12}(105\2\sqrt{105}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,412 |
Problem 9. The function $G$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$, the equalities $G(n a, n b, n c)=n \cdot G(a, b, c)$, $G(a+n, b+n, c+n)=G(a, b, c)+n$, $G(a, b, c)=G(c, b, a)$ hold. Find $G(89,90,91)$. | Answer: 90.
Solution. Note that $G(-1,0,1)=G(1,0,-1)=(-1) \cdot G(-1,0,1)$, from which $G(-1,0,1)=0$. Then $G(89,90,91)=G(-1,0,1)+90=90$.
Comment. The function $G$ cannot be uniquely determined. For example, the functions $G(a, b, c)=(a+b+c) / 3, G(a, b, c)=b$ and $G(a, b, c)=$ median of the numbers $\{a, b, c\}$ are... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,413 |
Problem 10. Let $D$ - be the set of real numbers, not containing 0 and 1. It is known that if $d \in D$, then $1-\frac{1}{d} \in D$ and $1-d \in D$. Can $D$ have exactly 1000 elements? | Answer: No, it cannot.
Solution. Consider the numbers $\left\{t, \frac{1}{t}, 1-t, \frac{1}{1-t}, \frac{t}{t-1}, \frac{t-1}{t}\right\}$. Let $\alpha: x \mapsto 1-\frac{1}{x}=\frac{x-1}{x}, \beta: x \mapsto 1-x$. Notice that the mapping $\alpha$ transforms the numbers $t, \frac{1}{t}, 1-t, \frac{1}{1-t}, \frac{t}{t-1},... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,414 |
Problem 1. Prove that for each natural number $n$ the number $6^{2 n}+2^{n+2}+12 \cdot 2^{n}$ is divisible by 17. | Solution. Note that $6^{2 n}+2^{n+2}+12 \cdot 2^{n}=(34+2)^{n}+16 \cdot 2^{n}$. Using the binomial theorem, we can group all terms that contain 34: $(34+2)^{n}+16 \cdot 2^{n}=34 A+2^{n}+16 \cdot 2^{n}=34 A+17 \cdot 2^{n}$. Since each term is divisible by 17, the entire number is divisible by 17.
Comment. The same solu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,415 |
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 9 adventurers have rubies; exactly 8 have emeralds; exactly 2 have sapphires; exactly 11 have diamonds. Moreover, it is known that
- if an adventurer has diamonds, then they have either rubies or sapphires (but not both at the same t... | Answer: 17.
Solution. Note that the number of adventurers who have diamonds is equal to the total number of adventurers who have rubies or sapphires. Then, from the first condition, it follows that 9 adventurers have rubies and diamonds, and 2 have sapphires and diamonds. That is, every adventurer who has rubies must ... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,416 |
Problem 3. A team of lumberjacks was cutting trees on a large and a small plot, with the area of the small plot being 3 times less than that of the large plot. In the part of the team that worked on the large plot, there were 8 more lumberjacks than in the part that worked on the small plot. When the tree harvesting on... | Answer: 14.
Solution. Let the number of workers on the smaller plot be denoted as $n$, then the number of workers on the larger plot is $n+8$, and the total number of workers in the team is $2n+8$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the product... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,417 |
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=5, AC=8$, and $BC=4$. Find the length of the vector $\overrightarrow{O A}-4 \overrightarrow{O B}+3 \overrightarrow{O C}$. | Answer: 10.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \c... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,418 |
Problem 5. Solve the system of equations in real numbers:
$$
\left\{\begin{array}{l}
a+c=-1 \\
a c+b+d=-1 \\
a d+b c=-5 \\
b d=6
\end{array}\right.
$$ | Answer: $(-3,2,2,3)$ and $(2,3,-3,2)$.
Solution. Let $x^{2}+a x+b$ and $x^{2}+c x+d$ be the two quadratic polynomials, the coefficients of which are the roots of the given system. Then
$$
\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=x^{4}-x^{3}-x^{2}-5 x+6
$$
By ... | (-3,2,2,3)(2,3,-3,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,419 |
Problem 6. Solve the equation $\arcsin \frac{x \sqrt{5}}{3}+\arcsin \frac{x \sqrt{5}}{6}=\arcsin \frac{7 x \sqrt{5}}{18}$. | Answer: $0, \pm \frac{8}{7}$.
Solution. From the condition on the domain of the arcsine, it follows that
$$
|x| \leqslant \frac{18}{7 \sqrt{5}}
$$
By computing the sine of both sides of the equation and considering that $\cos \arcsin >0$, and therefore,
$$
\sin \arcsin A=A, \quad \cos \arcsin A=\sqrt{1-A^{2}}
$$
w... | 0,\\frac{8}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,420 |
Problem 7. For what values of the parameter $a$ does the equation
$$
a \log _{5}^{2} x-(2 a+5) \log _{5} x+10=0
$$
have exactly two roots, one of which is five times larger than the other? | Answer: $\frac{5}{3}, 5$.
Solution. Let $t=\log _{5} x$, then the equation takes the form $a t^{2}-(2 a+5) t+10=0$. Note that $2 \cdot \frac{5}{a}=\frac{10}{a}, 2+\frac{5}{a}=\frac{2 a+5}{a}$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -2 and $\frac{5}{a}$. We make the rever... | \frac{5}{3},5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,421 |
Problem 8. In triangle $A B C$, side $B C=28$. The bisector $B L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $4: 3$, counting from the vertex. Find the radius of the circumscribed circle around triangle $A B C$, if the radius of the inscribed circle in it is 12. | Answer: 50.
Solution. Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $CI$ is the angle bisector in triangle $BLC$, by the angle bisector theorem, we have: $BC: CL = BI: IL = 4: 3$, from which $CL = 3BC / 4 = 21$.
Next, $BC \cdot CL... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,422 |
Problem 9. The function $g$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $g(n a, n b, n c)=$ $n \cdot g(a, b, c), g(a+n, b+n, c+n)=g(a, b, c)+n, g(a, b, c)=g(c, b, a)$ hold. Find $g(14,15,16)$. | Answer: 15.
Solution. Note that $g(-1,0,1)=g(1,0,-1)=(-1) \cdot g(-1,0,1)$, from which $g(-1,0,1)=0$. Then $g(14,15,16)=g(-1,0,1)+15=15$.
Comment. The function $g$ cannot be uniquely determined. For example, the functions $g(a, b, c)=(a+b+c) / 3$, $g(a, b, c)=b$, and $g(a, b, c)=$ median of the numbers $\{a, b, c\}$ ... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,423 |
Problem 10. Let $C$ - be the set of real numbers, not containing 0 and 1. It is known that if $c \in C$, then $\frac{1}{1-c} \in C$ and $\frac{c}{1-c} \in C$. Can $C$ have exactly 1000 elements? | Answer: No, it cannot.
Solution. We will prove by induction on $n$ that if a number $t \in C$, then the number $\frac{t}{1-n t} \in C$. If $n=1$, the statement follows from the condition. Now suppose the statement is proven for $n=k$, we will prove it for $n=k+1$. So, let $t \in C$. By the induction hypothesis, $\frac... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,424 |
Task 1. Represent in the form of an irreducible fraction
$$
7 \frac{19}{2015} \times 6 \frac{19}{2016}-13 \frac{1996}{2015} \times 2 \frac{1997}{2016}-9 \times \frac{19}{2015}
$$ | Answer: $19 / 96$.
Solution. First solution. Let $a=19 / 2015, b=19 / 2016$. Then
$$
\begin{aligned}
& 7 \frac{19}{2015} \times 6 \frac{19}{2016}-13 \frac{1996}{2015} \times 2 \frac{1997}{2016}-9 \times \frac{19}{2015}=(7+a)(6+b)-(14-a)(3-b)-9 a= \\
& \quad=42+6 a+7 b+a b-42+3 a+14 b-a b-9 a=21 b=21 \cdot \frac{19}{2... | \frac{19}{96} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,425 |
Task 2. In the $1^{\text{st}}$ grade class, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the ans... | Answer: 15 boys and 12 girls.
Solution. First solution. Let's denote the children who gave the answers (13,11), (17,11), (14,14) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of the girls has the same parity as $m$, and in the answers of the boys - the... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,426 |
Problem 3. For which natural numbers $n>1$ will there be $n$ consecutive natural numbers whose sum is equal to $2016?$ | Answer: For $n$ equal to $3, 7, 9, 21$ or 63.
Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a + (n-1) n / 2 = 2016$, or, after algebraic transformations, $n(2a + n - 1) = 4032 = 2^6 \cdot 3^2 \cdot 7$.
Note that $n$ and $... | 3,7,9,21,63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,427 |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 4$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $4 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016. | Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 4$, from which $\overrightarrow{B L}=5 / 9 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{5}{9} \cdot \overrightarrow{B C}=\overrigh... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,428 |
Problem 5. Solve the system of equations
$$
\left\{\begin{aligned}
x^{2}-x y+y^{2} & =19 \\
x^{4}+x^{2} y^{2}+y^{4} & =931
\end{aligned}\right.
$$ | Answer: $(-5, -3), (-3, -5), (3, 5), (5, 3)$.
Solution. Note that $x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-x^{2} y^{2}=\left(x^{2}+x y+y^{2}\right)\left(x^{2}-x y+y^{2}\right)$. Thus, the original system is equivalent to $\left\{\begin{array}{l}x^{2}-x y+y^{2}=19 \\ x^{2}+x y+y^{2}=49\end{array}\right.$.
... | (-5,-3),(-3,-5),(3,5),(5,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,429 |
Problem 6. Calculate $2 \operatorname{arctg} 2+\arcsin \frac{4}{5}$. | Answer: $\pi$.
Solution. First solution. Let $\operatorname{arctg} 2$ be denoted by $\alpha$, and $\arcsin 4 / 5$ by $\beta$. Note that $\beta \in(0, \pi / 2)$, and $\operatorname{tg}^{2} \beta=\sin ^{2} \beta /\left(1-\sin ^{2} \beta\right)=4^{2} / 3^{2}$, from which $\operatorname{tg} \beta=4 / 3$; also $\operatorna... | \pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,430 |
Problem 7. Let $OP$ be the diameter of circle $\Omega$, and $\omega$ be a circle centered at point $P$ with a radius smaller than that of $\Omega$. Circles $\Omega$ and $\omega$ intersect at points $C$ and $D$. Chord $OB$ of circle $\Omega$ intersects the second circle at point $A$. Find the length of segment $AB$, if ... | Answer: $\sqrt{5}$.
Solution. Let $N-$ be the second intersection point of line $O A$ with $\omega ; M$ - the second intersection point of line $D B$ with $\omega$.
Note that due to symmetry with respect to line $O P$, arcs $O C$ and $O D$ are equal. Therefore, $\angle A B C=$ $\angle D B A$. Let this angle be $\alph... | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,431 |
Problem 8. For what values of the parameter $a$ does the equation $x^{3}+a x^{2}+13 x-6=0$ have a unique solution | Answer: $(-\infty,-8) \cup(-20 / 3,61 / 8)$.
Solution. Note that $x=0$ is not a solution to the original equation. Therefore, it is equivalent to the equation $a=\frac{-x^{3}-13 x+6}{x^{2}}(*)$. Let's denote the right side by $f(x)$.
Note that $f(x) \xrightarrow{\rightarrow}+\infty$ as $x \rightarrow-\infty$ and $f(x... | (-\infty,-8)\cup(-20/3,61/8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,432 |
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 256 wrestlers is held on an Olympic system: at ... | Answer: 16
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 256 participants, there are 8 rounds, so the number of the tournament winner does not exceed $1+2 \cdot... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,433 |
Task 1. Represent in the form of an irreducible fraction
$$
6 \frac{16}{2015} \times 9 \frac{17}{2016}-2 \frac{1999}{2015} \times 17 \frac{1999}{2016}-27 \times \frac{16}{2015}
$$ | Answer: $17 / 224$.
Solution. First solution. Let $a=16 / 2015, b=17 / 2016$. Then
$$
\begin{aligned}
& 6 \frac{16}{2015} \times 9 \frac{17}{2016}-2 \frac{1999}{2015} \times 17 \frac{1999}{2016}-27 \times \frac{16}{2015}=(6+a)(9+b)-(3-a)(18-b)-27 a= \\
& \quad=54+9 a+6 b+a b-54+18 a+3 b-a b-27 a=9 b=9 \cdot \frac{17}... | \frac{17}{224} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,435 |
Task 2. In $1^{st}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers received wer... | Answer: 16 boys and 14 girls.
Solution. First solution. Let's denote the children who gave the answers (15,18), (15,10), (12,13) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of girls has the same parity as $m$, while in the answers of boys, it has the... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,436 |
Problem 3. Several (more than one) consecutive natural numbers are written on the board, the sum of which is 2016. What can the smallest of these numbers be? | Answer: $1,86,220,285$ or 671.
Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a+n-1$ have differen... | 1,86,220,285,671 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,437 |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=4: 3$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $3 \cdot \overrightarrow{A B}+4 \cdot \overrightarrow{A C}$ is 2016. | Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=4: 3$, from which $\overrightarrow{B L}=4 / 7 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrigh... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,438 |
Problem 5. Solve the system of equations
$$
\left\{\begin{aligned}
x^{2}-x y+y^{2} & =7 \\
x^{4}+x^{2} y^{2}+y^{4} & =91
\end{aligned}\right.
$$ | Answer: $(-3 ;-1),(-1 ;-3),(1 ; 3),(3 ; 1)$.
Solution. Note that $x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-x^{2} y^{2}=\left(x^{2}+x y+y^{2}\right)\left(x^{2}-x y+y^{2}\right)$. Thus, the original system is equivalent to $\left\{\begin{array}{l}x^{2}-x y+y^{2}=7 \\ x^{2}+x y+y^{2}=13\end{array}\right.$.
L... | (-3,-1),(-1,-3),(1,3),(3,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,439 |
Problem 6. Calculate $2 \operatorname{arctg} 3+\arcsin \frac{3}{5}$. | Answer: $\pi$.
Solution. First solution. Let $\operatorname{arctg} 3$ be denoted by $\alpha$, and $\arcsin 3 / 5$ by $\beta$. Note that $\beta \in(0, \pi / 2)$, $\mathrm{a} \operatorname{tg}^{2} \beta=\sin ^{2} \beta /\left(1-\sin ^{2} \beta\right)=3^{2} / 4^{2}$, from which $\operatorname{tg} \beta=3 / 4$; also $\ope... | \pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,440 |
Problem 8. For what values of the parameter $a$ does the equation $x^{3}+6 x^{2}+a x+8=0$ have no more than two solutions? | Answer: $[-15 ;+\infty)$.
Solution. Note that $x=0$ is not a solution to the original equation. Therefore, it is equivalent to the equation $a=\frac{-x^{3}-6 x^{2}-8}{x}(*)$. Let's denote the right side by $f(x)$.
Notice that $f(x) \xrightarrow{x}-\infty$ as $x \rightarrow-\infty$ and $f(x) \rightarrow-\infty$ as $x ... | [-15;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,441 |
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 512 wrestlers is held on an Olympic system: at ... | Answer: 18.
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 512 participants, there are 9 rounds, so the number of the tournament winner does not exceed $1+2 \cdo... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,442 |
Task 1. Represent in the form of an irreducible fraction
$$
6 \frac{3}{2015} \times 8 \frac{11}{2016}-11 \frac{2012}{2015} \times 3 \frac{2005}{2016}-12 \times \frac{3}{2015}
$$ | Answer: $11 / 112$.
Solution. First solution. Let $a=3 / 2015, b=11 / 2016$. Then
$$
\begin{aligned}
& 6 \frac{3}{2015} \times 8 \frac{11}{2016}-11 \frac{2012}{2015} \times 3 \frac{2005}{2016}-12 \times \frac{3}{2015}=(6+a)(8+b)-(12-a)(4-b)-12 a= \\
& =48+8 a+6 b+a b-48+4 a+12 b-a b-12 a=18 b=18 \cdot \frac{11}{2016}... | \frac{11}{112} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,443 |
Task 2. In $1^{\text {st }}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the answers rec... | Answer: 13 boys and 16 girls.
Solution. First solution. Let's denote the children who gave the answers (12,18), (15,15), (11,15) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. T... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,444 |
Problem 3. Several (more than one) consecutive natural numbers are written on the board, the sum of which is 2016. What can the largest of these numbers be? | Answer: 63, 106, 228, 291, or 673.
Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a + (n-1) n / 2 = 2016$, or, after algebraic transformations, $n(2 a + n - 1) = 4032 = 2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a + ... | 63,106,228,291,673 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,445 |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016. | Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 2$, from which $\overrightarrow{B L}=5 / 7 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrig... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,446 |
Problem 5. Solve the system of equations
$$
\left\{\begin{aligned}
x^{2}+x y+y^{2} & =37 \\
x^{4}+x^{2} y^{2}+y^{4} & =481
\end{aligned}\right.
$$ | Answer: $(-4, -3), (-3, -4), (3, 4), (4, 3)$.
Solution. Note that $x^{4} + x^{2} y^{2} + y^{4} = (x^{2} + y^{2})^{2} - x^{2} y^{2} = (x^{2} + xy + y^{2})(x^{2} - xy + y^{2})$. Thus, the original system is equivalent to $\left\{\begin{array}{l}x^{2} - xy + y^{2} = 13 \\ x^{2} + xy + y^{2} = 37\end{array}\right.$.
Let'... | (-4,-3),(-3,-4),(3,4),(4,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,447 |
Problem 6. Calculate $2 \operatorname{arctg} 4+\arcsin \frac{8}{17}$. | Answer: $\pi$.
Solution. First solution. Let $\operatorname{arctg} 4$ be denoted by $\alpha$, and $\arcsin 8 / 17$ by $\beta$. Note that $\beta \in(0, \pi / 2)$, and $\operatorname{tg}^{2} \beta=\sin ^{2} \beta /\left(1-\sin ^{2} \beta\right)=8^{2} / 15^{2}$, from which $\operatorname{tg} \beta=8 / 15$; also $\operato... | \pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,448 |
Problem 8. For what values of the parameter a does the equation $x^{3}+6 x^{2}+a x+8=0$ have exactly three solutions? | Answer: $(-\infty ;-15)$.
Solution. Note that $x=0$ is not a solution to the original equation. Therefore, it is equivalent to the equation $a=\frac{-x^{3}-6 x^{2}-8}{x}(*)$. Let's denote the right side by $f(x)$.
Note that $f(x) \xrightarrow{x}-\infty$ as $x \rightarrow-\infty$ and $f(x) \rightarrow-\infty$ as $x \r... | (-\infty;-15) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,449 |
Task 1. Represent in the form of an irreducible fraction
$$
6 \frac{7}{2015} \times 4 \frac{5}{2016}-7 \frac{2008}{2015} \times 2 \frac{2011}{2016}-7 \times \frac{7}{2015}
$$ | Answer: $5 / 144$.
Solution. First solution. Let $a=7 / 2015, b=5 / 2016$. Then
$$
\begin{aligned}
& 6 \frac{7}{2015} \times 4 \frac{5}{2016}-7 \frac{2008}{2015} \times 2 \frac{2011}{2016}-7 \times \frac{7}{2015}=(6+a)(4+b)-(8-a)(3-b)-7 a= \\
& =24+4 a+6 b+a b-24+3 a+8 b-a b-7 a=14 b=14 \cdot \frac{5}{2016}=\frac{14 ... | \frac{5}{144} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,451 |
Task 2. In the $1^{\text{st}}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers r... | Answer: 14 boys and 15 girls.
Solution. First solution. Let's denote the children who gave the answers $(10,14),(13,11),(13,19)$ as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. T... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,452 |
Problem 3. How many natural numbers $n>1$ exist, for which there are $n$ consecutive natural numbers, the sum of which is equal to 2016? | Answer: 5.
Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a+n-1$ have different parity. Therefore,... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,453 |
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=7: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+7 \cdot \overrightarrow{A C}$ is 2016. | Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=7: 2$, from which $\overrightarrow{B L}=7 / 9 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{7}{9} \cdot \overrightarrow{B C}=\overrig... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,454 |
Problem 5. Solve the system of equations
$$
\left\{\begin{array}{rl}
x^{2}+x y+y^{2} & =23 \\
x^{4}+x^{2} y^{2}+y^{4} & =253
\end{array} .\right.
$$ | Answer: $\left(-\frac{\sqrt{29}+\sqrt{5}}{2} ;-\frac{\sqrt{29}-\sqrt{5}}{2}\right),\left(-\frac{\sqrt{29}-\sqrt{5}}{2} ;-\frac{\sqrt{29}+\sqrt{5}}{2}\right),\left(\frac{\sqrt{29}-\sqrt{5}}{2} ; \frac{\sqrt{29}+\sqrt{5}}{2}\right),\left(\frac{\sqrt{29}+\sqrt{5}}{2} ; \frac{\sqrt{29}-\sqrt{5}}{2}\right)$
Solution. Note ... | (-\frac{\sqrt{29}+\sqrt{5}}{2};-\frac{\sqrt{29}-\sqrt{5}}{2}),(-\frac{\sqrt{29}-\sqrt{5}}{2};-\frac{\sqrt{29}+\sqrt{5}}{2}),(\frac{\sqrt{29}-\sqrt{5}}{2};\frac{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,455 |
Problem 6. Calculate $2 \operatorname{arctg} 5+\arcsin \frac{5}{13}$. | Answer: $\pi$.
Solution. First solution. Let $\operatorname{arctg} 5$ be denoted by $\alpha$, and $\arcsin 5 / 13$ by $\beta$. Note that $\beta \in(0, \pi / 2)$, and $\operatorname{tg}^{2} \beta=\sin ^{2} \beta /\left(1-\sin ^{2} \beta\right)=5^{2} / 12^{2}$, from which $\operatorname{tg} \beta=5 / 12$; also $\operato... | \pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,456 |
Problem 8. For what values of the parameter $a$ does the equation $x^{3}+a x^{2}+13 x-6=0$ have more than one solution? | Answer: $[-8 ;-20 / 3] \cup[61 / 8 ;+\infty)$
Solution. Note that $x=0$ is not a solution to the original equation. Therefore, it is equivalent to the equation $a=\frac{-x^{3}-13 x+6}{x^{2}}(*)$. Let's denote the right side by $f(x)$.
Note that $f(x) \xrightarrow{\rightarrow}+\infty$ as $x \rightarrow-\infty$ and $f(... | [-8;-20/3]\cup[61/8;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,457 |
Problem 6. Suppose $T$ is the period of the function $y=f(x)$. From the condition, we get:
$$
\cos x=f(x)-3 f(x-\pi)=f(x-T)-3 f(x-T-\pi)=\cos (x-T)
$$
Since this equality holds for all $x$, the number $T$ is the period of the function $y=\cos x$, i.e., $T=2 \pi n$, where $n$ is some natural number. From the given equ... | Answer: $f(x)=\frac{1}{4} \cos x$. | f(x)=\frac{1}{4}\cosx | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,458 |
Problem 7. Let's introduce the notation as shown in the figure. By extending the lateral sides, denote their point of intersection as $E$. The point of intersection of the diagonals is called $F$. Right triangles $B F C$ and $D F A$ are similar, and if the legs of the first are denoted as $x$ and $y$, then the correspo... | Answer: $\frac{4}{\sqrt{3}}$. | \frac{4}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,459 |
Problem 9. Each knight gives one affirmative answer to four questions, while a liar gives three. In total, there were $95+115+157+133=500$ affirmative answers. If all the residents of the city were knights, the total number of affirmative answers would be 200. The 300 extra "yes" answers come from the lies of the liars... | Answer: in quarter A, by 55 people, in quarter B by 35 people and in quarter $\Gamma$ by 17 people. | inquarterA,55people,inquarterB35peopleinquarter\Gamma17people | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,460 |
7.2. Arseny has 10 ten-liter buckets, each containing $1, 2, 3, \ldots 9, 10$ liters of water respectively. Arseny is allowed to take any two buckets and pour from the first into the second exactly as much water as is already in the second bucket. Can Arseny collect all the water in one bucket? | Answer: No
Solution: in total, there are $1+2+\ldots+10=55$ liters of water in the buckets - an odd number. And with any pouring, we double the amount of water in the bucket, i.e., in particular, make it even. Therefore, it is impossible to end up with an odd amount of water in the last bucket.
Criteria: noticed that... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,463 |
7.3. Egor borrowed 28 rubles from Nikita, and then returned them in four payments. It turned out that Egor always returned a whole number of rubles, and the amount of each payment increased and was divisible by the previous one. What was the last amount Egor paid? | Answer: 18 rubles
Solution:
1) If Egor paid $a$ rubles the first time, then the second time - not less than $2a$, the third - not less than $4a$, the fourth - not less than $8a$, and in total - not less than $15a$. Since $15a \leq 28$, we get that $a=1$.
2) The second time he paid 2 or 3 rubles (because if 4, then he... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,464 |
7.4. On the surface of a pentagonal pyramid (see fig.), several gnomes live in pairwise distinct points, and they can live both inside the faces and on the edges or at the vertices. It turned out that on each face (including the vertices and edges that bound it) a different number of gnomes live. What is the minimum nu... | Answer: 6
Solution: There are 6 faces in total, so at least 5 gnomes live on the most "populated" one. If there are exactly 5 gnomes in total, then all of them live on one face (let's call it face $A$), so the faces where 4 and 3 gnomes live (let's call them $B$ and $C$ respectively) are adjacent to it. Then, on the e... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,465 |
7.5. At a round table, 47 deputies from 12 different regions are seated, and they try to ensure that among any 15 consecutive people, there are representatives from all regions. Will the deputies be able to achieve their goal?
# | # Answer: No
Solution: Suppose they can. Note that there will be a region from which no more than three deputies have arrived. Otherwise, there would be at least $12 * 4=48>47$ deputies in total.
Consider these three (or fewer) deputies. They sit in a circle, and between any two adjacent deputies, there are no more t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,466 |
10.4. Ten chess players over nine days played a full round-robin tournament, during which each of them played exactly one game with each other. Each day, exactly five games were played, with each chess player involved in exactly one of them. For what maximum $n \leq 9$ can it be claimed that, regardless of the schedule... | Answer. $n=5$.
Solution. By the end of the eighth day, each chess player has played 8 games, meaning they have not played one. The unplayed games divide the chess players into 5 non-intersecting pairs. By the Pigeonhole Principle, among any six chess players, there will always be two belonging to the same pair, that i... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,467 |
8.2. Arseny has 2018 buckets, into which 1, 2, 3, ... 2017, 2018 liters of water are poured respectively. Arseny is allowed to take any two buckets and pour from the first into the second exactly as much water as is already in the second bucket. Can Arseny collect all the water in one bucket? All buckets are large enou... | # Answer: No
Solution: in total, there are $1+2+\ldots+2018=2018 * 2019 / 2=1009 * 2019$ liters of water in the buckets - an odd number. And with any pouring, we double the amount of water in the bucket, i.e., in particular, we make it even. Therefore, it is impossible to end up with an odd amount of water in the last... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,469 |
8.3. In an equilateral triangle $ABC$, through a random point inside it, three lines are drawn: parallel to $AB$ until intersecting with $BC$ and $CA$; parallel to $BC$ until intersecting with $AB$ and $CA$; parallel to $CA$ until intersecting with $BC$ and $AB$. Prove that the sum of the three obtained segments is equ... | Solution: Let an arbitrary point $P$ be chosen inside the triangle. Draw the segments and label them as shown in the figure. It is obvious that triangles $D E P, P F G, P I H$ are equilateral, as all angles in them are 60 degrees.
Next, notice that $BFPE$ is a parallelogram, since the opposite sides in it are pairwise... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,470 |
8.4. Nikita and Egor are running on a circular road, having started from the same place in opposite directions. It is known that Nikita runs a lap 12 seconds faster than Egor, but still takes more than 30 seconds to do so. It turned out that for the seventh time after the start, they met at the same place where they be... | Answer: Nikita in 36 seconds, Egor in 48 seconds.
Solution: Note that between any two meetings, the boys run a total of one lap. Therefore, by the 7th meeting, they have run 7 laps together. Also, note that each of them ran an integer number of laps, as they met at the starting line. Since Nikita ran faster, he ran mo... | Nikita:36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,471 |
8.5. At a round table, 410 deputies sat down, each of whom was either a knight, who always tells the truth, or a liar, who always lies. Each of the deputies said: "Among my twenty neighbors to the left and twenty neighbors to the right, there are exactly 20 liars in total." It is known that at least half of the people ... | Answer: None.
Solution: Let's divide all the people sitting at the table into ten groups of 41 people each. Then, at least one group will have at least 21 liars. Otherwise, in each group, there would be a maximum of 20, i.e., no more than \(20 \times 10 = 200\) in total, which is less than half of the total number. Co... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,472 |
7.1. On a circular route, two buses operate with the same speed and a movement interval of 21 minutes. What will be the movement interval if 3 buses operate on this route at the same constant speed? | Answer: 14.
Solution: Since the interval of movement with two buses on the route is 21 minutes, the length of the route in "minutes" is 42 minutes. Therefore, the interval of movement with three buses on the route is $42: 3=14$ minutes.
Criteria: only answer, answer with verification - 3 points. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,473 |
7.2. Kay has an ice plate in the shape of a "corner" (see figure). The Snow Queen demanded that Kay cut it into four equal parts. How can he do this?
 | Answer: for example, the corner can be cut as follows:
Criteria: any correct cutting is scored 7 points. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,474 |
7.4. Natasha and Inna bought the same box of tea bags. It is known that one tea bag is enough for two or three cups of tea. Natasha's box was enough for only 41 cups of tea, while Inna's was enough for 58 cups. How many tea bags were in the box? | Answer: 20.
Solution: Let there be $n$ tea bags in the box. Then the number of brewings can vary from $2n$ to $3n$. Therefore, 58 is not greater than $3n$, which means $19<n$. Additionally, 41 is not less than $2n$, so $n<21$. Since the number of tea bags must be a natural number that is less than 21 but greater than ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,476 |
7.5. On a line, one hundred points are marked: green, blue, and red. It is known that between any two red points there is a blue one, and between any two blue points there is a green one. In addition, there are no fewer red points than blue ones, and no fewer blue points than green ones. How many points are painted blu... | Answer: 33.
Solution: Let the number of red points be $n$, then the number of blue points is not less than $n-1$ (the number of intervals between "adjacent" red points), and since by condition the number of red points is not less than the number of blue points, the number of blue points is either $n$ or $n-1$. Similar... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,477 |
8.2. On a circular route, 12 trams run in one direction at the same speed and at equal intervals. How many trams need to be added so that at the same speed, the intervals between trams decrease by one fifth?
# | # Answer: 3.
Solution: Let's take the entire distance as 60 arbitrary units, which means the trams are currently 5 arbitrary units apart. We want this distance to be reduced by $1 / 5$, making it equal to 4 arbitrary units. For this, we need $60: 4=15$ trams, which is 3 more than the current number.
Criteria: only th... | 3 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,479 |
8.3. The square of the sum of the digits of the number $A$ is equal to the sum of the digits of the number $A^{2}$. Find all such two-digit numbers $A$ and explain why there are no others. | Answer: $10,20,11,30,21,12,31,22,13$.
Solution: Let's consider what the sums of the digits of the original number and the number squared can be. The square of a two-digit number can result in no more than a four-digit number. The sum of the digits of a four-digit number is no more than $9 \cdot 4=36$, and 36 is only a... | 10,20,11,30,21,12,31,22,13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,480 |
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