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2. $\sin (4 \pi(2 k+1) / 7)=0 \rightarrow 7 m=8 k+4, m \in Z \rightarrow\left\{\begin{array}{c}k=-4+7 t \\ m=-4+8 t, t \in Z\end{array}\right.$
$\sin (4 \pi(3 k-2) / 7)=0 \rightarrow 7 n=12 k-8, m \in Z \rightarrow\left\{\begin{array}{c}k=3+7 s \\ m=4+12 s, s \in Z\end{array}\right.$
Finally, the equation $7 t-4=7 s+... | Answer: $k=3+7 s, s \in Z$ | 3+7,\inZ | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
3. Let's find all mutually prime triples $x, y, (x+y)$ of divisors of the number $a=1944=2^{3} \cdot 3^{5}$. Note that if $p>1$ is a divisor of $x$ and $y$, then $p$ divides $x+y$. Similarly, if $p>1$ is a divisor of $x$ and $x+y$, then it divides $y$, and any divisor $p>1$ of the pair $y$ and $x+y$ divides $x$. Thus, ... | The first solution $x=1, y=2, x+y=3$. This coprime triplet of divisors of the number $a=1944=2^{3} \cdot 3^{5}$ generates triplets $x=t, y=2 t, x+y=3 t$, where $t$ is any divisor of the number, where $b=324=2^{2} 3^{4}$. The largest such divisor is $t=324$ and $x \cdot y \cdot(x+y)_{\max }=6 \cdot 324^{3}=2^{7} \cdot 3... | 2^{7}\cdot3^{15} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Since $1 / 3 < 3 / 8$, the valid events are $\xi=4$ or $\xi=9$.
Case $1 . \quad \xi=4$
Transforming the equation:
$$
4 a x^{2}-16 x+a 4=0 \rightarrow a x^{2}-4 x+a=0 \rightarrow a\left(x^{2}+1\right)=4 x \rightarrow a=\frac{4 x}{x^{2}+1}
$$
The derivative of the function $a=\frac{4 x}{x^{2}+1}$ is $a^{\prime}=\f... | Answer: $a \in[8 / 5 ; 2] \cup[36 / 13 ; 3]$ | \in[8/5;2]\cup[36/13;3] | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. Express $y$ from the equation of the line $y-1=\frac{4-a x-b}{b}$ and substitute it into the equation of the circle:
$b^{2}(x-1)^{2}+(a x+b-4)^{2}-b^{2}=0 \Leftrightarrow x^{2}\left(a^{2}+b^{2}\right)+2 x\left(a(b-4)-b^{2}\right)+(b-4)^{2}=0$
The condition for the line to be tangent to the circle is $D=0 \rightarr... | Answer: 8 pairs $(a ; b)=(12 ; 5),(8 ; 6),(2 ; 0),(-4 ; 3),(5 ; 12),(6 ; 8),(0 ; 2),(3 ;-4)$ | (b)=(12;5),(8;6),(2;0),(-4;3),(5;12),(6;8),(0;2),(3;-4) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.
In the figure, a circle $K$ is tangent to the line $L_{3}\left(d_{1}<d_{2}\right)$. The radii of the circles $K_{1}, K_{2}$, and $K$ are denoted by $r_{1}=\frac{d_{1}}{2}, r_{2}=\frac{... | Answer: 1) $\left.r=\frac{\left(d_{1}+d_{2}\right) \cdot d_{2}}{2 d_{1}}=3.2\right)$ 2) $r=\frac{\left(d_{1}+d_{2}\right) \cdot d_{1}}{2 d_{2}}=\frac{3}{4}$ | r=\frac{(d_{1}+d_{2})\cdotd_{2}}{2d_{1}}=3.2\quad | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
2. Answer: $\Sigma_{a}=1680$ | Solution
Let $a=p_{1}^{r_{1}} \cdot p_{2}^{r_{2}} \cdot \ldots \cdot p_{m}^{r_{m}}, b=p_{2}^{r_{2}} \cdot \ldots \cdot p_{m}^{r_{m}} \cdot$ Here $p_{k}$ are prime divisors, $r_{k}$ are their multiplicities.
Let $q_{i}$ be any divisor of the number $b$. Then $q_{i}, p_{1} q_{i}, p_{1}^{2} q_{i}, \ldots, p_{1}^{r_{1}} ... | 1680 | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
3. Answer: 1) 167 numbers; 2) $x_{\max }=975$ | Solution
According to condition 1) $5 x=6 k+3 \rightarrow\left\{\begin{array}{l}x=6 t-3 \\ k=5 t-3\end{array} \rightarrow 1 \leq 6 t-3 \leq 1000 \rightarrow t=1,2, \ldots, 167\right.$, i.e., 167 solutions.
According to condition 2)
$x=5 m, m>0, m \in Z \rightarrow$ GCD $(6 ; 25 m)=3 \rightarrow 25 m=6 k+3 \rightarro... | 975 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
4. Answer: $400 \mathrm{~cm}^{2}$ | Solution
Fig 1
Vertices $A, B, C, D$ are located at the midpoints of the cells numbered $m$, counting from the vertices of the square $P Q M N$. The length $P B=4(m-1)$, the length $B Q=P A=... | 400\mathrm{~}^{2} | Geometry | math-word-problem | Not a problem | Yes | olympiads | false |
5. Answer: $R_{\max }=\frac{h H}{h+H}=\frac{6}{5}$
## Variant 0
A sheet of paper has the shape shown in the figure. The distance between the parallel edges of the sheet is $h$, the distance between the vertices $A$ and $B$ is $H$. Find the largest radius of a circle that can be cut out from such a sheet.
Lines $B N$ and $A Q$ are parallel, lines $B M$ and $A P$ are parallel, $B C$ and $O D$ are perpendicular to $A Q$, $O B=O D=R$. From the simila... | \frac{}{+H} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
6. On the edge $A D$ of the cube base $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ ( $A A^{\prime}, B B^{\prime}, C C^{\prime}, D D^{\prime}$ are parallel lateral edges), there is a point $M$ such that $A M: A D=2: 5$. A plane $P$ is drawn through point $M$ and vertices $A^{\prime}$ and $C^{\prime}$ of the cub... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
6. The lengths of all edges (lateral and base) of the tetrahedron $A B C D$ are 1. A point $M$ is located on the edge $A B$ such that $A M: A B=1: 2$. Find the distance between the skew lines $C M$ and $A D$.
## Solutions and answers
## Variant 1
Problem 1 Answer: $n=2$ | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false | |
6. The lateral edge of a regular triangular pyramid is inclined to the plane of its base at an angle of $75^{\circ}$. A cube is inscribed in the pyramid such that four of its vertices lie on the base of the pyramid, and the other four lie on its lateral faces. Find the ratio of the volumes of the cube and the pyramid.
... | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false | |
6. A circle can be circumscribed around quadrilateral $A B C D$. The lengths of sides $A B$ and $A D$ are equal.
On side $C D$ is point $Q$ such that $D Q=1$, and on side $B C$ is point $P$ such that $B P=5$.
Furthermore, $\measuredangle D A B=2 \measuredangle Q A P$. Find the length of segment $P Q$.
## Solutions a... | 7-\frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
2. Answer: 1) 72 divisors; 2) $d_{\max }=49000$ | Solution
Prime factorization: $a=2^{3} \cdot 3^{2} \cdot 5^{3} \cdot 7^{2}$. The divisors of the numbers $a_{1}=2^{3} \cdot 3^{2} \cdot 7^{2}$ and $a_{2}=2^{3} \cdot 5^{3} \cdot 7^{2}$ are divisors of $a$ that are not divisible by 15. The number $a_{1}$ has $4 \times 3 \times 3=36$ distinct divisors, and the number $a... | 72divisors;d_{\max}=49000 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
3. Answer: 1) 9 fractions; 2) $\left(\frac{p}{q}\right)_{\text {min }}=\frac{1}{5}$ | Solution
Transformation:
$\frac{p-1}{q}<\frac{p}{q+1} \rightarrow(p-1)(q+1)<p q \rightarrow p-q<1$
For $q=1 \rightarrow 0<p<2 \rightarrow p=1$, i.e., there is only one fraction $\frac{p}{q}=\frac{1}{1}=1$, which does not satisfy the condition of the problem.
For $q=2 \rightarrow 0<p<3 \rightarrow p=1 ; 2$, i.e., th... | 9 | Number Theory | math-word-problem | More than one problem | Yes | olympiads | false |
1. Suppose that while one worker was digging the trench, the other workers did not rest but continued to dig the trench for the next pipes. By the time the fourth worker finished digging 50 m, an additional $4 \cdot \frac{3}{4} \cdot 50=150 \mathrm{m}$
of trench had been dug. Therefore, the entire team had dug $150 \ma... | Answer: $200 \mathrm{m}$.
The translation is provided as requested, maintaining the original text's line breaks and format. | 200\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Adding the first equation to the other two, we get:
$$
x^{2}+2 x+y^{2}+4 y+z^{2}+6 z=-14
$$
Next, we complete the square for each variable. As a result, we obtain:
$$
(x+1)^{2}+(y+2)^{2}+(z+3)^{2}=0
$$
Therefore, $\mathrm{x}=-1, y=-2, z=-3$ is the only possible solution. It remains to verify this by substituting... | Answer: $x=-1, y=-2, z=-3$. | -1,-2,-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let $p_{1}, p_{2}, \ldots, p_{m}$ be the prime divisors of $a$. Among them, by condition, there are 3 and 5, for example, $\quad p_{1}=3, p_{2}=5$. Then $a=3^{s_{1}} \cdot 5^{s_{2}} \cdot p_{3}^{s_{3}} \cdot \ldots \cdot p_{m}^{s_{m}}$ and the total number of its divisors is $\left(s_{1}+1\right)\left(s_{2}+1\right)... | Answer: $a_{\text {min }}=a_{1}=2^{6} \cdot 3^{2} \cdot 5^{2}$. | 2^{6}\cdot3^{2}\cdot5^{2} | Number Theory | MCQ | Yes | Yes | olympiads | false |
4. From the first equation, we find:
$$
P=-\frac{x^{2}-x-2}{x-1}=-x+\frac{2}{x-1}, x \neq 1
$$
From the second equation, we find: $p=-\frac{x^{2}+2 x-1}{x+2}=-x+\frac{1}{x+2}, x \neq-2$.
Then, if $x$ is a common root corresponding to the parameter $p$, we have
$$
-x+\frac{2}{x-1}=-x+\frac{1}{x+2}
$$
From which we ... | Answer: $p=\frac{14}{3}, x=-5$. | \frac{14}{3},-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let's introduce the following notations: $Q$ - the center of the first circle, $F, E, B$ - the points of tangency of the circles and lines, $G$ - the point of tangency of the circles, $AD$ - the common tangent (see the figure). Draw the line $OC$ parallel to the line $FA$. Let $FA=a, OG=r$. Then $AG=a \quad(AG=FA$ a... | Answer: $r=2$.
## Variant №2 | 2 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
1. A team of five people is laying a 30 m pipe in the ground. To prepare the trench, they decided to dig in turns with the condition: each person digs for as long as the other four could dig one-third of the trench together. The 30 m trench was dug when the last member of the team had worked their allotted time. How ma... | Answer: $80 \mathrm{m}$.
Translation: Answer: $80 \mathrm{m}$. | 80\mathrm{} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
2. Solve the system $\left\{\begin{array}{c}x^{2}+y-2 z=-3 \\ 3 x+y+z^{2}=14 \\ 7 x-y^{2}+4 z=25\end{array}\right.$ | Answer: $x=2, y=-1, z=3$. | 2,-1,3 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
3. A natural number $a$ is divisible by 21 and has 105 different divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\min }=2^{6} \cdot 3^{4} \cdot 7^{2}=254016$ | 254016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. For what values of $p$ do the quadratic equations $x^{2}-(p+1) x+(p+1)=0$ and $2 x^{2}+(p-2) x-p-7=0$ have a common root? Find this root. | Answer: 1) $p=3, x=2$; 2) $p=-\frac{3}{2}, x=-1$. | p=3,2;p=-\frac{3}{2},-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A team of six people is laying a 40 m pipe in the ground. To prepare the trench, they decided to dig in turns with the condition: each person digs for as long as the other five together would dig a quarter of the trench. The 40 m trench was dug when the last member of the team had worked their allotted time. How man... | Answer: $100 \mathrm{m}$. | 100\mathrm{} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
2. Solve the system $\left\{\begin{array}{c}x^{2}-3 y+z=-4 \\ x-3 y+z^{2}=-10 \\ 3 x+y^{2}-3 z=0\end{array}\right.$ | Answer: $x=-2, y=3, z=1$. | -2,3,1 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
3. A natural number $a$ is divisible by 35 and has 75 different divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\text {min }}=2^{4} \cdot 5^{4} \cdot 7^{2}=490000$. | 490000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. For which $p$ do the quadratic equations
$$
9 x^{2}-3(p+6) x+6 p+5=0 \text{ and } 6 x^{2}-3(p+4) x+6 p+14=0
$$
have a common root? Find this root. | Answer: 1) $p=-\frac{32}{9}, x=-1$;2) $p=\frac{32}{3}, x=3$. | p=-\frac{32}{9},-1orp=\frac{32}{3},3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A team of seven people is laying a 60 m pipe in the ground. To prepare the trench, they decided to dig in turns with the condition: each person digs for as long as the other six together would dig one-sixth of the trench. The 60 m trench was dug when the last member of the team had worked their allotted time. How ma... | Answer: $130 \mathrm{M}$. | 130 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
2. Solve the system $\left\{\begin{array}{c}x^{2}-y-z=8 \\ 4 x+y^{2}+3 z=-11 \\ 2 x-3 y+z^{2}=-11\end{array}\right.$ | Answer: $x=-3, y=2, z=-1$. | -3,2,-1 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
3. A natural number $a$ is divisible by 55 and has 117 distinct divisors, including 1 and $a$. Find the smallest such $a$. | Answer: $a_{\min }=2^{12} \cdot 5^{2} \cdot 11^{2}=12390400$. | 12390400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. For what values of $p$ do the quadratic equations $x^{2}-(p+2) x+2 p+6=0$ and $2 x^{2}-(p+4) x+2 p+3=0$ have a common root? Find this root. | Answer: 1) $p=-3, x=-1$; 2) $p=9, x=3$ | p=-3,-1;p=9,3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The border of a square with a side of 9, cut out of white cardboard, is painted red. It is necessary to cut the square into 6 equal-area parts, the boundaries of which contain segments painted red with the same total length.
## Solutions
Option 1
Problem 1 | Answer: 2 km.
Solution
$S-$ the length of the path, $S_{1}$ - the length of the path on descents, $S_{2}$ - the length of the path on ascents, $S / 2=S_{1}+S_{2}$
$t=15-8-6=1-$ the time of the journey there and back. Then
$1=\frac{S_{1}}{6}+\frac{S_{2}}{3}+\frac{S}{2 \cdot 4}+\frac{S_{1}}{3}+\frac{S_{2}}{6}+\frac{S... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 2.15. Final round of the "Rosatom" Olympiad, 8th grade | # 2.15. Final round of the "Rosatom" Olympiad, 8th grade
## Answers and solutions
Problem 1 Answer: 1997 or 2011
The year of birth can be written as: $1000 x + 100 y + 10 z + u$.
By the condition: $2013 - 1000x - 100y - 10z - u = z + u$.
Case $1. x = 1 \rightarrow 2u = 1013 - 100y - 11z$
Estimate: $2u \leq 18, 11... | 47 | Other | math-word-problem | Incomplete | Yes | olympiads | false |
2.1 I.V. Savelyev Olympiad (preliminary round of the "Rosatom" Olympiad), 11th grade
## Answers and Solutions
## Problem 1
Divide the numerator by the denominator with a remainder: $x^{3}+3 x^{2}+5=(x+1)\left(x^{2}+2 x-3\right)+x+8$ $\frac{x^{3}+3 x^{2}+5}{x^{2}+2 x-3}=x+1+\frac{x+8}{x^{2}+2 x-3}$. If $\left|\frac{x... | Solution. Substitution $t=\cos 2 x$
$\cos ^{4} x-\sin ^{4} x=\left(\cos ^{2} x-\sin ^{2} x\right)\left(\cos ^{2} x+\sin ^{2} x\right)=t$
$\cos ^{6} x-\sin ^{6} x=\left(\cos ^{2} x-\sin ^{2} x\right)\left(\cos ^{4} x+\sin ^{2} x \cdot \cos ^{2} x+\sin ^{4} x\right)=$
$=t\left(\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+... | -2,2,-8 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Variant 1. $S_{1}=100, k_{1}=32, t_{1}=180, S_{2}=50, k_{2}=20, t_{2}=108, A=30$, $T=36$ | Answer: 24 cows
## Problem 4
Solution. . $x=\frac{p}{q}$, options $p= \pm 1 ; \pm 5, q=1 ; 2 ; 4$
$4 p^{2}+a p q-5 q^{2}=0 \rightarrow a=\frac{5 q}{p}-4 \frac{p}{q}$
Case 1. $q=1 \rightarrow a=\frac{5}{p}-4 p \rightarrow a= \pm 1 ; \pm 8 ; \pm 19$
Case 2. $q=2 \rightarrow a=\frac{10}{p}-2 p \rightarrow a= \pm 8$
... | 24 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
# 2.3. Qualifying round of the "Rosatom" Olympiad, 11th grade
## Answers and solutions
Problem 1 Answer 345 | Solution Variant 1. $a_{1}=\log _{1+\sqrt{5}}(1+\sqrt{5})^{2}=2$
$a_{1} a_{2}+a_{1} a_{3}+a_{2} a_{3}=2(2+d)+2(2+2 d)+(2+d)(2+2 d)=2\left(d^{2}+6 d+6\right)$
Solving the equation $x=66 \rightarrow d^{2}+6 d-27=0 \rightarrow\left[\begin{array}{c}d=3 \\ d=-9\end{array} \rightarrow\right.$
increasing $\rightarrow d=3, ... | 345 | Other | math-word-problem | Incomplete | Yes | olympiads | false |
2.4. In-person qualifying round of the "Rosatom" Olympiad, 11th grade
## Answers and solutions
Problem 1 Answer: $b_{55}=2^{31}$ | Solution.
$b_{n}: q+b_{n}+b_{n} q=\frac{7}{2} b_{n} \rightarrow \frac{1}{q}+1+q=\frac{7}{2} \rightarrow q=2$ (increasing)
$b_{55}=b_{24} q^{31}=2^{31}$
Problem 2 Answer: 7 points.
$\left(-\frac{5 \pi}{6} ; \frac{\pi}{6}\right) ;\left(-\frac{5 \pi}{6} ; \frac{5 \pi}{6}\right) ;\left(0 ; \arcsin \frac{\sqrt{5}-1}{2}\... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2.4. In-person selection round of the "Rosatom" olympiad, 11th grade
## Answers and solutions
Problem 1 Answer: $b_{1}=-1$ (when $\left.x=1 / 2\right)$ | Solution.
$f(x)=\frac{1-4 x}{x}=\frac{b_{1}(x)}{1-x} \rightarrow b_{1}(x)=\frac{(4 x-1)(x-1)}{x}=\frac{4 x^{2}-5 x+1}{x}, 0<x<1$
$b_{1}^{\prime}(x)=\frac{4 x^{2}-1}{x^{2}}=\frac{(2 x-1)(2 x+1)}{x^{2}}$
, 11th grade
## Answers and solutions
Problem 1 Answer: 1) $x=1, b=\sqrt[3]{16}$ 2) $x=3, b=16$ | Solution
The condition of arithmetic progression:
$\sqrt{4 x-3}+4=x+4 \rightarrow \sqrt{4 x-3}=x \rightarrow\left\{\begin{array}{c}x \geq 0.75 \\ 4 x-3=x^{2} \rightarrow x^{2}-4 x+3=0\end{array} \rightarrow\left[\begin{array}{l}x=1 \\ x=3\end{array}\right.\right.$
Case 1.
$x_{1}=1 \rightarrow a_{1}=1, a_{2}=\frac{5}... | 1,b=\sqrt[3]{16};3,b=16 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Selection of solutions.
$$
-\frac{7 \pi}{2} \leq \frac{2 \pi}{3}+2 \pi k \leq \pi \rightarrow-\frac{25}{12} \leq k \leq \frac{1}{6} \rightarrow k=-2 ;-1 ; 0
$$
$$
-\frac{7 \pi}{2} \leq \frac{\pi}{6}+2 \pi m \leq \pi \rightarrow-\frac{11}{6} \leq m \leq \frac{5}{12} \rightarrow m=-1 ; 0
$$
Problem 3 Answer: 1) $n_... | Solution. $x$ - the amount of money in the prize fund.
1st team. $\frac{x+1}{2}$ received, $x-\frac{x+1}{2}=\frac{x-1}{2}$ remaining.
2nd team. $\frac{x-1}{2}+\frac{1}{2}=\frac{x+1}{4}$ - received, $\frac{x-1}{2}-\frac{x+1}{4}=\frac{x-3}{4}$ - remaining
3rd team. $\frac{x-3}{8}+\frac{1}{2}=\frac{x+1}{8}$ - received,... | 39 | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
Problem 4. Answer: $x=35$ | Solution variant 1
The number $1600000000=2^{4} \cdot 2^{8} \cdot 5^{8}=2^{12} \cdot 5^{8}$
The product of numbers divisible by 5:
$5 \cdot 10 \cdot 15 \cdot 20 \cdot 25 \cdot 30 \cdot 35=2^{4} \cdot 3^{2} \cdot 7 \cdot 5^{8}$, so for numbers $x<35$
$x$ ! does not divide 1600000000.
The product of numbers divisibl... | 35 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2. $\left\{\begin{array}{c}y=x+\sqrt{2} \\ \left\{\begin{array}{l}x=3 \sqrt{2} \cos \alpha \\ y=3 \sqrt{2} \sin \alpha\end{array}\right.\end{array} \alpha \in\left[\frac{\pi}{4} ; \frac{\pi}{2}\right) \rightarrow \sin \alpha-\cos \alpha=\frac{1}{3} \rightarrow \alpha_{2}=\frac{\pi}{4}+\arcsin \frac{1}{3 \sqrt{2}}\right... | Solution. A. The plane must pass through the centers of the cubes.
B. Calculation of the area of the triangle of the centers of the bases on the plane $P$.
$S_{0,0_{2} 0_{3}}=\frac{(a+c)(a... | \frac{\sqrt{b^{2}(-)^{2}+^{2}(-b)^{2}}}{(+)(+b)}=\frac{\sqrt{2}}{4} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2.11. Final round of the "Rosatom" Olympiad, 10th grade
Answers and solutions
$\begin{array}{lll}\text { Problem } 1 \text { Answer: 1) } a=3, b=12, c=48 & \text { 2) } a=363, b=462, c=588\end{array}$
System of conditions: $\left\{\begin{array}{c}b^{2}=a c \\ a+c=2 b+27 \\ (a-2)(c-12)=(b-6)^{2}\end{array} \rightarro... | Solution.
$\measuredangle M N P=\alpha, \measuredangle M N Q=\beta, \measuredangle P N Q=\gamma-$ dihedral, $\measuredangle P N Q=\varphi-$ desired
Notation: $M N=d$
$P N=d \cdot \operator... | \cos\varphi=\frac{1}{8}(2\sqrt{2}+\sqrt{6}) | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
# 2.8. Final round of the "Rosatom" Olympiad, 7th grade
## Answers and solutions
1 Answer: 5 | Solution.
$x(x+2)+1=x^{2}+2 x+1=(x+1)^{2}=36 \rightarrow x+1=6 \rightarrow x=5$.
2 Answer: $a=0,-1,-2,-3$
Solution.
The coefficient $2 a+3 \neq 0$ for integer $a$, so the equation has a unique solution for all integer $a$. This solution is $x=\frac{4 a+9}{2 a+3}=\frac{2(2 a+3)+3}{2 a+3}=2+\frac{3}{2 a+3}$.
The int... | 2 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. Answer: into $(n-2)$ triangles. | Solution.
Let the best dissection contain $m$ triangles. We will show that this number can be reduced by at least 1 if there is a triangle among them, all vertices of which are inside the polygon (Fig. 1)
![](https://cdn.mathpix.com/cropped/2024_05_06_b6d3c8ec908d21127f56g-2.jpg?height=560&width=434&top_left_y=1262&t... | 88 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
2.13. Final round of the "Rosatom" Olympiad, 9th grade
# Answers and solutions
## Problem 1 Answer: 2
$5-x, 9-2 x, 9-x-$ modified side lengths, $05\end{array}\right.$
Problem 2 Answer: 1) $a=-8, b=9$ 2) $4 x^{2}-x+3$
$16 x^{4}+a x^{3}+25 x^{2}-6 x+b=\left(4 x^{2}+A x+B\right)^{2}=$
$=16 x^{4}+A^{2} x^{2}+B^{2}+8... | Solution. Case 1. The location of point $D$ is indicated in the figure.
Line $B K$ is parallel to $A C \rightarrow \frac{B K}{m x}=\frac{2 c}{3 c} \rightarrow B K=\frac{2 m x}{3}$
From the ... | 2 | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
1. Answer: 1) 4 people read books, went fishing, but did not collect mushrooms 2) 17 | Solution. The diagram shows the distribution of children by interests:
The letters $x, y, z, u, v, w, t$ represent the number of schoolchildren engaged in one of the activities. For example, ... | 4 | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
2. Answer: $a=\frac{3}{2}, a=-12$ | Solution. $P\left(x^{2}\right)=a x^{4}-3 x^{2}+5 \rightarrow P\left(x^{2}\right)-P(x)=a x^{2}\left(x^{2}-1\right)-3 x(x-1)=0$.
$x=0, x=1$ are solutions for any $a$. The remaining roots satisfy the equation: $a x^{2}+a x-3=0$. The original equation has three roots if $x=0$ or $x=1$ are solutions of the equation $a x^{2}... | =\frac{3}{2},=-12 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Answer: 1) $x=3, y=9$ 2) $x=11, y=1$ | Solution.
Multiply both sides of the second equation by the GCD $(x, y)$ and use the equality GCD $(x, y) \cdot \operatorname{LCM}(x, y)=x \cdot y$. Then $x \cdot(\text { GCD }(x, y))^{2}=x y \rightarrow y=(\text { GCD }(x, y))^{2}$, i.e., $y$ is a square of an integer. Out of all possible $y \in[1 ; 11]$, there are o... | 3,911,1 | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
5. Answer: $M P: P B=2: 3$ | Solution.
$\triangle A N Q \sim \triangle M N D$ with a similarity coefficient $k=3$. Then $A Q=6 y, B Q=3 y$
$\triangle Q B P \sim \Delta D M P$ with a similarity coefficient $3 / 2$. Then... | MP:PB=2:3 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
2.17. Final round of the "Rosatom" Olympiad, 7th grade
# Answers and solutions
Problem 1 Answer: 9
There exists a set of 8 buttons in which there are no three buttons of the same color: each color has two buttons. In any set of 9 buttons, there will be at least one triplet of buttons of the same color.
If we assum... | Answer: 9 buttons.
Problem 2 Answer: 1261
\[
\left\{\begin{array}{l}
a=35 n+1 \\
a=45 m+1
\end{array} \rightarrow 35 n=45 m \rightarrow 7 n=9 m \rightarrow\left\{\begin{array}{c}
n=9 t \\
m=7 t, t \in Z
\end{array} \rightarrow\right.\right.
\]
\[
a=315 t+1 \geq 1000 \rightarrow t \geq 4 \rightarrow a_{\text {min }}=... | 1261 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Each of the numbers $1,2, \ldots, 41$ appears in a $41 \times 41$ square table exactly 41 times. The sum of all numbers located above the diagonal of the table is three times the sum of the numbers located below it. Find the numbers standing at the ends of the diagonal of the table. | 1. Angle $A B M$ is equal to angle $B C A$; | notfound | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
2. Prove that the sum $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{346}$ represents a fraction, the numerator of which is divisible by 347. | Task 2. Proof
Task 3 Answer: 12 and 3
Task 4 Answer: 42 months, 12 t. p.
Task 5 Answer: $B C=\sqrt{(a+c)(2 a+c-\sqrt{a(a+c)})}=4 \sqrt{31-4 \sqrt{15}}$
Criteria for checking the final Rosatom 9th grade
1 task
There is something plausible, relating to the matter 0.5
The equation is indicated and there are calcula... | proof | Number Theory | proof | Yes | Problem not solved | olympiads | false |
3. Petya wrote four numbers on the board: $4,9,15,36$ and told Vasya that these are the sum, difference, product, and quotient of two positive numbers he came up with, written in random order. Vasya thought about it and named the numbers. Can you do it? | 3. Proportions:
$$
\frac{A M}{A B}=\frac{A B}{A C} \rightarrow A B^{2}=a(a+c) \rightarrow A B=\sqrt{a(a+c)}
$$ | notfound | Logic and Puzzles | math-word-problem | Yes | Problem not solved | olympiads | false |
4. The borrower and the lender agreed that an interest-free loan of 1365 t.r. would be repaid monthly, on the last day of the month. In the first month, a certain amount (an integer number of t. r.) is returned, and in each subsequent month, the amount returned is 1 t.r. more than in the previous month. For what maximu... | 4. The Law of Cosines:
\[
\begin{aligned}
& B C^{2}=A B^{2}+A C^{2}-A B \cdot A C=a(a+c)+(a+c)^{2}-(a+c) \sqrt{a(a+c)}= \\
& =(a+c)(2 a+c-\sqrt{a(a+c)})
\end{aligned}
\]
## Variant 2
Problem 1 Answer: 17 | 17 | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. The angle at vertex $A$ of an acute-angled triangle $A B C$ is $60^{\circ}$. Through vertices $B$ and $C$, lines perpendicular to sides $A B$ and $A C$ respectively are drawn, intersecting at point $D$. Through vertex $B$, a line perpendicular to line $A D$ is drawn and intersects side $A C$ at point $M$. The length... | 1.5
3 task
Calculated something | 1.5 | Geometry | math-word-problem | Yes | Problem not solved | olympiads | false |
3. In the decimal representation of a six-digit number $a$, there are no zeros and the sum of its digits is 18. Find the sum of all different numbers obtained from the number $a$ by cyclic permutations of its digits. In a cyclic permutation, all digits of the number, except the last one, are shifted one place to the ri... | 3. Solution. Case 1. The number $a=333333$. This number does not change under cyclic permutations, so it is the only one and the sum of the numbers is the number itself, that is, 333333.
Case 2. The number $a$ consists of three identical cycles of two digits each, for example, $a=242424$. Such numbers have two differe... | 1999998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $r_{n}$ denote the remainder of the division of a natural number $n$ by 6. For which $n$ is the equality $r_{2 n+3}=r_{5 n+6}$ true? | 4. Solution. The numbers $2 n+3$ and $5 n+6$ have the same remainder if their difference $5 n+6-(2 n+3)=3 n+3$ is divisible by 6, i.e., $3 n+3=6 k$. From this, we get $n=2 k-1, k \in \square$. | 2k-1,k\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On a sheet of paper, 14 parallel lines $L$ and 15 lines $P$ perpendicular to them are drawn. The distances between adjacent lines from $L$ from the first to the last are given: 2;4;6;2;4;6;2;4;6;2;4;6;2. The distances between adjacent lines from $P$ are also known: 3;1;2;6;3;1;2;6;3;1;2;6;3;1. Find the greatest leng... | 5. Solution. We will prove that the maximum length of the side of the square is 40. Calculate the distance from the first to the last line in $P: 3+1+2+6+3+1+2+6+3+1+2+6+3+1=40$. Therefore, the side length of the square cannot be more than 40. On the other hand, in $L$ the distance from the second line to the third lin... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Using a compass and a ruler, construct the side of a square equal in area to the given figure
## Solutions
Option 1
Problem 1 Answer: twice | Solution
$T_{1}$ - the time between consecutive meetings when moving in the same direction;
$T_{2}$ - the time between consecutive meetings when moving in opposite directions
$V_{M}$ - Masha's speed; $V_{A}$ - Sasha's speed; $R$ - the radius of the circle;
$T_{1}=\frac{2 \pi R}{V_{A}-V_{M}}, T_{2}=\frac{2 \pi R}{V_... | notfound | Geometry | math-word-problem | Incomplete | Incomplete | olympiads | false |
6. On the edge $AC$ of the triangular pyramid $ABCD$, there is a point $M$ such that $AM: MC=1: 2$. Through the midpoint of the edge $BC$ of the base of the pyramid, a plane $P$ is drawn passing through point $M$ and parallel to the lateral edge $CD$. In what ratio does the plane $P$ divide the volume of the pyramid?
... | Solution. Let there be $2 n$ chairs around the table (i.e., there are $2 n$ people sitting around the table). On the circle, points mark the chairs. The number $K$ denotes the number of chairs separating Petya and the person sitting on that chair. Then the number of chairs counted by Petya, including Vasya, is
$$
2(1+... | 7:11 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
1. Pete and Vasya invited their classmates to their birthday at Pete's house and seated everyone at a round table to drink tea. Pete noted for himself the smallest number of chairs separating him from each of the invited guests, except for Vasya. By adding up these numbers, he got 75. Find the number of chairs at the t... | Answer: 20 chairs, 6 chairs. | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\cos ^{4} 5 x+1\right)\left(\cos ^{4} 7 x+1\right)=4 \cos ^{2} 5 x \cdot \cos ^{2} 7 x$. | Answer: $x=\pi t, t \in Z$. | \pi,\inZ | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
3. Find all integer solutions to the equation $\sqrt{n+1}-\sqrt{n}=(\sqrt{2}-1)^{2023}$. | Answer: $n=\frac{1}{4}\left((\sqrt{2}+1)^{2023}-(\sqrt{2}-1)^{2023}\right)^{2}$ | \frac{1}{4}((\sqrt{2}+1)^{2023}-(\sqrt{2}-1)^{2023})^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\left(\log _{3}(2 x+1)+\sqrt{\log _{3}{ }^{2}(2 x+1)+1}\right)\left(\log _{3}(3 x+2)+\sqrt{\log _{3}{ }^{2}(3 x+2)+1}\right)=1$ | Answer: $x=-\frac{1}{6}$. | -\frac{1}{6} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. On the cells of a chessboard of size $8 \times 8$, 5 identical pieces are randomly placed. Find the probability that four of them will be located either on the same row, or on the same column, or on one of the two main diagonals. | Answer: $P(A)=\frac{18 \cdot\left(C_{8}^{4} \cdot C_{56}^{1}+C_{8}^{5}\right)}{C_{64}^{5}}=\frac{18 \cdot 56 \cdot 71}{31 \cdot 61 \cdot 63 \cdot 64}=\frac{71}{4 \cdot 31 \cdot 61}=\frac{71}{7564} \approx 0.0094$. | \frac{71}{7564} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the edge $A C$ of the triangular pyramid $A B C D$, there is a point $M$ such that $A M: M C=1: 3$. Through the midpoint of the edge $B C$ of the base of the pyramid, a plane $P$ is drawn passing through point $M$ and parallel to the lateral edge $C D$. In what ratio does the plane $P$ divide the volume of the py... | Answer: $V_{1}: V_{2}=11: 21$.
# | V_{1}:V_{2}=11:21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Pete and Vasya invited their classmates to their birthday at Pete's house and seated everyone at a round table to drink tea. Pete noted for himself the smallest number of chairs separating him from each of the invited guests, except for Vasya. By adding these numbers, he got 95. Find the number of chairs at the tabl... | Answer: 22 chairs, 5 stools. | 22 | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
2. Solve the equation $\left(\operatorname{tg}^{4} 3 x+1\right)\left(\operatorname{tg}^{4} 7 x+1\right)=4 \operatorname{tg}^{2} 3 x \cdot \operatorname{tg}^{2} 7 x$. | Answer: $x=\frac{\pi}{4}+\frac{\pi t}{2}, t \in Z$. | \frac{\pi}{4}+\frac{\pi}{2},\inZ | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
3. Find all integer solutions to the equation $\sqrt{n+1}-\sqrt{n}=(\sqrt{2}-1)^{2024}$. | Answer: $n=\frac{1}{4}\left((\sqrt{2}+1)^{2024}-(\sqrt{2}-1)^{2024}\right)^{2}$. | \frac{1}{4}((\sqrt{2}+1)^{2024}-(\sqrt{2}-1)^{2024})^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\left(\log _{4}(x+1)+\sqrt{\log _{4}^{2}(x+1)+1}\right)\left(\log _{4}(3 x+5)+\sqrt{\log _{4}^{2}(3 x+5)+1}\right)=1$. | Answer: $x=-\frac{2}{3}$. | -\frac{2}{3} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. On the cells of a chessboard of size $8 \times 8$, 6 identical pieces are randomly placed. Find the probability that five of them will be located either on the same row, or on the same column, or on one of the two main diagonals. | Answer: $P(A)=\frac{18 \cdot\left(C_{8}^{5} \cdot C_{56}^{1}+C_{8}^{6}\right)}{C_{64}^{6}}=\frac{28 \cdot 113 \cdot 18}{59 \cdot 61 \cdot 62 \cdot 21 \cdot 16}=\frac{339}{446276} \approx 0.00076$. | \frac{339}{446276} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the edge $A C$ of the triangular pyramid $A B C D$, there is a point $M$ such that $A M: M C=2: 3$. Through the midpoint of the edge $B C$ of the base of the pyramid, a plane $P$ is drawn passing through point $M$ and parallel to the lateral edge $C D$. In what ratio does the plane $P$ divide the volume of the py... | Answer: $V_{1}: V_{2}=43: 57$. | 43:57 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Pete and Vasya invited their classmates to their birthday at Pete's house and seated everyone at a round table to drink tea. Pete noted for himself the smallest number of chairs separating him from each of the invited guests, except for Vasya. By adding up these numbers, he got 114. Find the number of chairs at the ... | Answer: 24 chairs, 7 chairs. | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\operatorname{ctg}^{4} 5 x+1\right)\left(\operatorname{ctg}^{4} 9 x+1\right)=4 \operatorname{ctg}^{2} 5 x \cdot \operatorname{ctg}^{2} 9 x$. | Answer: $x=\frac{\pi}{4}+\frac{\pi t}{2}, t \in Z$. | \frac{\pi}{4}+\frac{\pi}{2},\inZ | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
3. Find all integer solutions to the equation $\sqrt{n+1}-\sqrt{n}=(\sqrt{2}-1)^{2021}$. | Answer: $n=\frac{1}{4}\left((\sqrt{2}+1)^{2021}-(\sqrt{2}-1)^{2021}\right)^{2}$. | \frac{1}{4}((\sqrt{2}+1)^{2021}-(\sqrt{2}-1)^{2021})^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On the cells of a chessboard of size $8 \times 8$, 7 identical figures are randomly placed. Find the probability that six of them will be located either on the same row, or on the same column, or on one of the two main diagonals. | Answer: $P(A)=\frac{18 \cdot\left(C_{8}^{6} \cdot C_{56}^{1}+C_{8}^{7}\right)}{C_{64}^{7}}=\frac{591}{12942004} \approx 0.000046$. | \frac{591}{12942004} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the edge $A C$ of the triangular pyramid $A B C D$, there is a point $M$ such that $A M: M C=3: 4$. Through the midpoint of the edge $B C$ of the base of the pyramid, a plane $P$ is drawn passing through point $M$ and parallel to the lateral edge $C D$. In what ratio does the plane $P$ divide the volume of the py... | Answer: $V_{1}: V_{2}=22: 27$
## National Research Nuclear University "MEPhI" Final round of the sectoral physics and mathematics olympiad for schoolchildren "Rosatom", mathematics, field trip 11th grade. | V_{1}:V_{2}=22:27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. The length of side $A D$ of the inscribed quadrilateral $A B C D$ in a circle is 5. Point $M$ divides this side in the ratio $A M: M D=1: 4$, and the lines $M C$ and $M B$ are parallel to the sides $A B$ and $C D$ respectively. Find the length of side $B C$ of the quadrilateral.
Problem 1 Answer: $P(0)_{\min }=68$. | Solution. Let $Q(x)=P(x)-2023$, then $Q(17)=Q(23)=0$, hence, by the theorem of Bezout, $Q(x)$ is divisible by $(x-17)$ and $(x-23)$. Thus, we have the representation $Q(x)=(x-17)(x-23) R(x)$, where $R(x)$ is some polynomial with integer coefficients. Then
$$
\begin{gathered}
P(x)=(x-17)(x-23) Q(x)+2023 \\
P(0)=17 \cdo... | notfound | Geometry | math-word-problem | Incomplete | Incomplete | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(29)=P(37)=2022$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\min }=949$. | 949 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$$
\log _{\sin x} \cos x+\log _{\cos x} \sin 2 x+\log _{\sin 2 x} \sin x=\log _{\cos x} \sin x+\log _{\sin 2 x} \cos x+\log _{\sin x} \sin 2 x
$$ | Answer: $x=\frac{\pi}{3}+2 \pi n, n \in Z, x=\frac{\pi}{6}+2 \pi m, m \in Z$. | \frac{\pi}{3}+2\pin,n\inZ,\frac{\pi}{6}+2\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the reduced polynomial $P(x)$ (the coefficient of the highest degree of $x$ is 1), for which the identity $x P(x-1)=(x-4) P(x)$ holds for the variable $x$. | Answer: $P(x)=x(x-1)(x-2)(x-3)$. | P(x)=x(x-1)(x-2)(x-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Petya writes on a piece of paper a string of 5 zeros and 19 ones, arranged in a completely random order. Find the mathematical expectation of the random variable - the number of zeros written before the first one appears. | Answer: $M \xi=\frac{5}{20}=\frac{1}{4}$. | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. The length of side $A D$ of the inscribed quadrilateral $A B C D$ is 3. Point $M$ divides this side in the ratio $A M: M D=1: 2$, and the lines $M C$ and $M B$ are parallel to the sides $A B$ and $C D$ respectively. Find the length of side $B C$ of the quadrilateral. | Answer: $B C=\sqrt{2}$.
# | BC=\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(11)=P(13)=2021$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\text {min }}=19$. | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$$
\log _{\sin x} \cos 2 x+\log _{\cos 2 x} \sin 3 x+\log _{\sin 3 x} \sin x=\log _{\cos 2 x} \sin x+\log _{\sin 3 x} \cos 2 x+\log _{\sin x} \sin 3 x
$$ | Answer: $x=\frac{\pi}{10}+2 \pi n, n \in Z, x=\frac{9 \pi}{10}+2 \pi m, m \in Z$. | \frac{\pi}{10}+2\pin,n\inZ,\frac{9\pi}{10}+2\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the reduced polynomial $P(x)$ (the coefficient of the highest degree of the variable is 1), for which the identity $x P(x-1)=(x-5) P(x)$ holds. | Answer: $P(x)=x(x-1)(x-2)(x-3)(x-4)$. | P(x)=x(x-1)(x-2)(x-3)(x-4) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For which triples of numbers $(x ; y ; z)$, satisfying the system $\left\{\begin{array}{l}\sin 2 x-\sin 2 y=3(x-y) \\ \cos 3 x-\cos 3 z=4(x-z)\end{array}\right.$, does the expression $\frac{\sin x+\sin y}{2-\sin z}$ take its maximum possible value? | Answer: $\left(\frac{\pi}{2}+2 \pi n ; \frac{\pi}{2}+2 \pi n ; \frac{\pi}{2}+2 \pi n\right), n \in Z$. | (\frac{\pi}{2}+2\pin;\frac{\pi}{2}+2\pin;\frac{\pi}{2}+2\pin),n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Petya writes on a piece of paper a string of 6 zeros and 29 ones, arranged in a completely random order. Find the mathematical expectation of the random variable - the number of zeros written before the first one appears. | Answer: $M \xi=\frac{6}{30}=\frac{1}{5}$. | \frac{1}{5} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. The length of side $A D$ of the inscribed quadrilateral $A B C D$ is 4. Point $M$ divides this side in the ratio $A M: M D=1: 3$, and the lines $M C$ and $M B$ are parallel to the sides $A B$ and $C D$ respectively. Find the length of side $B C$ of the quadrilateral. | Answer: $B C=\sqrt{3}$.
# | BC=\sqrt{3} | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
1. The polynomial $P(x)$ with integer coefficients satisfies the condition $P(19)=P(21)=2020$. Find the smallest possible value of $P(0)>0$ under these conditions. | Answer: $P(0)_{\min }=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$\log _{\cos x} \sin 2 x+\log _{\sin 2 x} \cos 3 x+\log _{\cos 3 x} \cos x=\log _{\sin 2 x} \cos x+\log _{\cos 3 x} \sin 2 x+\log _{\cos x} \cos 3 x$. | Answer: $x=\frac{\pi}{10}+2 \pi n, n \in Z$. | \frac{\pi}{10}+2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the reduced polynomial $P(x)$ (the coefficient of the highest degree of the variable is 1), for which the identity $x P(x-1)=(x-6) P(x)$ holds. | Answer: $P(x)=x(x-1)(x-2)(x-3)(x-4)(x-5)$. | P(x)=x(x-1)(x-2)(x-3)(x-4)(x-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For which triples of positive numbers $(x ; y ; z)$, satisfying the system $\left\{\begin{array}{l}\sin 3 x-\sin 3 y=3(x-y) \\ \cos 4 x-\cos 4 z=5(z-x)\end{array}, \quad\right.$ does the expression $x+y+z-2 \operatorname{arctg}(x+y+z) \quad$ take the smallest possible value | Answer: $\left(\frac{1}{3} ; \frac{1}{3} ; \frac{1}{3}\right)$ | (\frac{1}{3};\frac{1}{3};\frac{1}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Petya writes on a piece of paper a string of 4 zeros and 39 ones, arranged in a completely random order. Find the mathematical expectation of the random variable - the number of zeros written before the first one appears. | Answer: $M \xi=\frac{4}{40}=\frac{1}{10}$. | \frac{1}{10} | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
6. The length of side $A D$ of the inscribed quadrilateral $A B C D$ is 6. Point $M$ divides this side in the ratio $A M: M D=1: 5$, and the lines $M C$ and $M B$ are parallel to the sides $A B$ and $C D$ respectively. Find the length of side $B C$ of the quadrilateral. | Answer: $BC=\sqrt{5}$.
## Grading Criteria for the ON-SITE Final Round of the Rosatom Olympiad 2023 for 11th Grade
In all problems, the correct answer without justification is 0 | \sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Answer: 1) $x=1$ for $m=5, n=2$; 2) $x=1.5$ for $m=3, n=5$; 3) $x=\log _{2}(\sqrt{33}-1)-1$ for $m=1, n=8$ | Solution
Description of pairs (m,n): $\left\{\begin{array}{c}m=1-2 t>0 \\ n=3 t+8>0, t \in Z\end{array} \rightarrow t=-2,-1,0 \rightarrow r_{m}=2\right.$ for all permissible $t$.
Case 1. $t=-2 \rightarrow m=5, n=2 \rightarrow r_{m}=r_{n}=2 \rightarrow 2^{2 x}+2 \cdot 2^{x}-8=0 \rightarrow x=1$
Case 2. $t=-1 \rightar... | 1,1.5,\log_{2}(\sqrt{33}-1)-1 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2. Answer: 1) $\left.x=\frac{20(3 k \pm 1)}{3}, k \in Z ; 2\right) x=10 k, k \in Z$ | Solution
Transformation:
$\sin a_{n} x+\sin a_{n+1} x+\sin a_{n+2} x=2 \sin \frac{a_{n}+a_{n+2}}{2} \cos d x+\sin a_{n+1} x=\sin a_{n+1} x(2 \cos d x+1)=0$
Case 1.
$\cos \frac{\pi x}{10}=-\frac{1}{2} \rightarrow\left[\begin{array}{c}\frac{\pi x}{10}=\frac{2 \pi}{3}+2 \pi k, k \in \mathbb{Z} \\ \frac{\pi x}{10}=-\fr... | \frac{20(3k\1)}{3},k\in\mathbb{Z};10k,k\in\mathbb{Z} | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
3. Answer: $C_{12}^{3}=220$ | Solution
Let the number be represented using $k$ threes and $m$ fours, $m+k=12, m \geq 1, k \geq 1$
Then, by the divisibility rule for 9, the sum of the digits of the number must be divisible by 9:
$\left\{\begin{array}{c}3 k+4 m=9 t \\ m+k=12, m \geq 1, k \geq 1\end{array}\right.$. Notice that from the first equati... | 220 | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
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