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5. Point $M$ divides the diagonal $A C$ of square $A B C D$ in the ratio $M C: A M=1: 4$. A line passing through point $M$ divides the square into two parts, the areas of which are in the ratio $1: 11$. In what ratio does this line divide the perimeter of the square? | Answer: $\frac{5}{19}$.
# | \frac{5}{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For Kostya's birthday, a cake with a base in the shape of a $3 \times 5$ dm rectangle was bought, divided into 15 equal pieces with square bases of $1 \times 1$ dm. Kostya decided to cut the cake with a knife along a straight line to create the maximum number of separate pieces, not necessarily equal. What is the ma... | Answer: 22 pieces and 3 cuts. | 22 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
2. Find the integer values of $a$ for which the equation $(x-a-4)^{2}+2 x-2 a-16=0$ has only one root in the interval $[1 ; 8]$. | Answer: $[-5 ; 0] \cup[3 ; 8]$. | [-5;0]\cup[3;8] | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. Point $M$ divides the diagonal $A C$ of square $A B C D$ in the ratio $A M: M C=3: 2$. A line passing through point $M$ divides the square into two parts, the areas of which are in the ratio $9: 11$. In what ratio does this line divide the perimeter of the square? | Answer: $\frac{19}{21}$.
# | \frac{19}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For Masha's birthday, a cake with a base in the shape of a $2 \times 5$ dm rectangle was bought, divided into 10 equal pieces with square bases of $1 \times 1$ dm. Masha decided to cut the cake with a knife along a straight line to create the maximum number of separate pieces, not necessarily equal. What is the maxi... | Answer: 16 pieces and 2 cuts. | 16 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
5. Point $M$ divides the diagonal $A C$ of square $A B C D$ in the ratio $A M: M C=2: 1$. A line passing through point $M$ divides the square into two parts, the areas of which are in the ratio $9: 31$. In what ratio does this line divide the perimeter of the square? | Answer: $\frac{27}{53}$. | \frac{27}{53} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1.
$\left\{\begin{array}{c}2 p^{2}-2 q=0 \\ q^{2}+p q-2 q=0 \rightarrow q(q+p-2)=0\end{array}\right.$
Case $1: q=0 \rightarrow p=0$
Case $2: q \neq 0 \rightarrow q=2-p \rightarrow 2 p^{2}+2 p-4=0 \rightarrow p^{2}+p-2=0 \rightarrow\left\{\begin{array}{c}p=1 \rightarrow q=1 \\ p=-2 \rightarrow q=4\end{array... | Solution: $a$ - divisor of $23 \rightarrow a=\{ \pm 1 ; \pm 23\}$
Case $a=1 . \rightarrow x^{2}+b x+23=0 \rightarrow x_{1,2}=\left\{\begin{array}{c}1 \\ 23 \\ b=-24\end{array} \quad x_{1,2}=\left\{\begin{array}{c}-1 \\ -23 \\ b=24\end{array}\right.\right.$
$(2 a+3 b)_{\max }=74$
Case $a=-1 . \rightarrow x^{2}-b x-23... | notfound | Algebra | math-word-problem | More than one problem | Yes | olympiads | false |
1. Find the number of divisors of the number $a=2^{3} \cdot 3^{2} \cdot 5^{2}$, which are divisible by 3. Find the sum of such divisors. | 1. Solution. Among the divisors of the number $a=2^{3} \cdot 3^{2} \cdot 5^{2}$, which are divisible by 3, the number 2 can be chosen with exponents $0,1,2,3$ (4 options), the number 3 - with exponents 1,2 (2 options), and the number 5 - with exponents $0,1,2$ (3 options). Therefore, the total number of divisors is $4 ... | 5580 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. How many integers $b$ exist such that the equation $x^{2}+b x-9600=0$ has an integer solution that is a multiple of both 10 and 12? Specify the largest possible $b$. | 2. Solution. Since the desired integer solution $x$ is divisible by 10 and 12, it is divisible by 60, hence it can be written in the form $x=60 t, t \in Z$. Substitute $x=60 t$ into the original equation:
$3600 t^{2}+60 b t-9600=0$. Express $b: b=\frac{160}{t}-60 t$. For $b$ to be an integer, $t$ must be a divisor of ... | 9599 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How many three-digit positive numbers $x$ exist that are divisible by 3 and satisfy the equation $GCD(15, GCD(x, 20))=5$? Find the largest one. | 3. Solution. Since 20 is not divisible by 3, $NOD(x, 20)$ is also not divisible by 3. Therefore, the equation $NOD(15, NOD(x, 20))=5$ is equivalent to the condition that $NOD(x, 20)$ is divisible by 5, which is possible if and only if $x$ is divisible by 5. Thus, the condition of the problem is equivalent to $x$ being ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The quadratic trinomial $p(x)=a x^{2}+b x+c, a>0$ when divided by ( $x-1$ ) gives a remainder of 4, and when divided by ( $x-2$ ) - a remainder of 15. Find the maximum possible value of the ordinate of the vertex of the parabola $y=p(x)$ under these conditions. For what value of $x$ is it achieved? | 4. Solution. By the Remainder Theorem, the remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$.
Therefore, the conditions of the problem are equivalent to the system:
$\left\{\begin{array}{l}4 a+2 b+c=15 \\ a+b+c=4\end{array} \Rightarrow\left\{\begin{array}{l}b=11-3 a \\ c=2 a-7\end{array}\right.\righ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In a right triangle $ABC$, a height $CH$ is drawn from the vertex of the right angle. From point $N$ on the leg $BC$, a perpendicular $NM$ is dropped to the hypotenuse, such that line $NA$ is perpendicular to $CM$ and $MH: CH=1: \sqrt{3}$. Find the acute angles of triangle $ABC$. | 5. Solution. Since $\square A C B=90^{\circ}$ and $\square N M A=90^{\circ}$, a circle can be circumscribed around quadrilateral $A C N M$.
Consider triangle $M C H$. It is a right triangle, ... | \BAC=60,\ABC=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 2.10. Qualifying round of the "Rosatom" Olympiad, 10th grade | 2.10. Qualifying round of the "Rosatom" Olympiad, 10th grade
# Answers and solutions
## Problem 1
Case 1. $\left(x^{2}+m x+3\right)^{2}=x^{4}+2 m x^{3}+\left(m^{2}+6\right) x^{2}+6 m x+9$
$P(1)=9 \rightarrow 1+a+b+c+9=9 \rightarrow a+b+c=-1$
Comparison of coefficients $\left\{\begin{array}{c}a=2 m \\ b=m^{2}+6 \\... | notfound | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false |
1. The robot can perform maneuvers of two types: stepping forward in a straight line for 644 steps and stopping, or moving backward along the same line for 308 steps and stopping. The battery charge allows no more than 13 such maneuvers. The robot's task is to stop as close as possible to an object located 724 steps ah... | 1. construct parallelograms $N C Q E$ and $M A Q F$; | notfound | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
2. For what value of $a$ does the equation $|\cos (2 x+3 y)|+|\cos (3 x-2 y)|-|a+4|=2$ have a solution? Find these solutions. Find the minimum $R$, such that any circle of radius $R$ on the plane contains at least one point with coordinates $(x ; y)$ - solutions of the equation. | 2. construct triangle $E F Q$ | notfound | Algebra | math-word-problem | Incomplete | Problem not solved | olympiads | false |
3. Students were allowed to take any number of pens from a box upon entering the school. It turned out that not less than $65 \%$ of the pens received by any group of seven people ended up with one student from this group. Prove that there is a student in the school who took more than $62 \%$ of the pens taken by all t... | Task 3. Proof.
Task 4 Answer: $P(A)=1-3 \ln \frac{4}{3} \approx 0.14$
Task 5 Answer: $b \in(-\infty ;-7) \cup(5 ;+\infty)$
Task 6 Answer: $P Q=3$ | proof | Combinatorics | proof | Yes | Problem not solved | olympiads | false |
2.1
$\cos 4 x=0 \rightarrow 4 x=\frac{\pi}{2}+\pi k \rightarrow 2 x=\frac{\pi}{4}+\frac{\pi k}{2}, 6 x=\frac{3 \pi}{4}+\frac{3 \pi k}{2}, 10 x=\frac{5 \pi}{4}+\frac{5 \pi k}{2} k=0,1,2 \ldots$
For all $k$, the terms of the progression are the same and equal to $\lg \frac{\sqrt{2}}{2}$, therefore such $x$ are not solu... | Solution
Substitution $t=\sin ^{2} x \geq 0$. Cubic equation $32 t^{3}-48 t^{2}+22 t-3=0$
has solutions $\left[\begin{array}{l}t=1 / 2 \\ t=3 / 4 \\ t=1 / 4\end{array} \rightarrow \sin x= \pm \frac{\sqrt{2}}{2} ; \pm \frac{\sqrt{3}}{2} ; \pm \frac{1}{2}\right.$
![](https://cdn.mathpix.com/cropped/2024_05_06_5fa80a85... | \sin2x_{1}\cdot\sin10x_{1}=-\ | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2.7. Final round of the "Rosatom" olympiad, 11th grade (Sarov, Kirov, Tambov, Voronezh, Saint Petersburg)
Answers and solutions
Problem 1 Answer: 1) $x=2, a=3 \quad$ 2) $x=-\frac{2}{9}$; $a=\frac{1}{2}$ | Solution. Condition of geometric progression: $3|x| \sqrt{x+2}=5 x+2 \rightarrow$
$9 x^{3}-7 x^{2}-20 x-4=0 \rightarrow x_{1}=-1$ (not valid), $x_{2}=2, x_{3}=-\frac{2}{9}$
Case $x_{2}=2 \rightarrow q_{2}=\left.\frac{\sqrt{5 x+2}}{3|x|}\right|_{x=2}=\frac{1}{\sqrt{3}} \rightarrow d=\log _{\alpha} q=\log _{a} \frac{1}... | Algebra | math-word-problem | Incomplete | Yes | olympiads | false | |
2.8. Final round of the "Rosatom" Olympiad, 11th grade (Sergiev Posad, Obninsk)
Answers and solutions
Problem 1 Answer: $b_{3}= \pm 45$ | Solution Variant 1. $b_{1}^{2}\left(1+q^{2}+q^{4}\right)=2275=25 \cdot 13 \cdot 7$
Among the divisors of $b_{1}$, there are no divisors not equal to $5 \rightarrow b_{1}= \pm 5$
$q^{4}+q^{2}+1=91 \rightarrow q_{1}{ }^{2}=9, q_{2}{ }^{2}-10 \rightarrow b_{3}=b_{1} \cdot q^{2}= \pm 45$
Problem 2 Answer: $4096 \pi$
So... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2.9. Final round of the "Rosatom" Olympiad, 11th grade (Lesnoy, Snezhinsk, Kurchatov, Kovrov, Ozersk, Kaliningrad, Smolensk, Ryazan, Novovoronezh, Seversk, Kaluga)
Answers and solutions
Problem 1 Answer: 1) $x_{1}=\frac{4 \pi}{5}, y_{1}=-\frac{\pi}{5}$
2) $x_{2}=\frac{12 \pi}{5}, y_{2}=-\frac{3 \pi}{5} 3$ ) $x_{3}=\fr... | Solution.
A.
$4 \log _{\pi-x-3 y}(2 x+y) \cdot \log _{3 x+5 y}(\pi-x-3 y)=4 \log _{3 x+5 y}^{2}(2 x+y) \rightarrow$
$\left\{\begin{array}{c}\log _{3 x+5 y}(2 x+y)=\log _{3 x+5 y}^{2}(2 x+y) \\ x+3 y<0,2 x+y \neq 1 \\ x+3 y<0 \\ x+3 y<0\end{array} \rightarrow\left\{\begin{array}{c}x=\frac{4 \pi(2 k+1)}{5} \\ y=-\frac... | Algebra | math-word-problem | Incomplete | Yes | olympiads | false | |
9. $k=9 t+8$, the number $5(9 t+8)(9 \cdot 25 t+222)$ is not divisible by 9;
Problem 4 Answer: $R_{\text {min }}=10$, solutions $(-12 ; 3)$ and $(8 ; 3)$ | Solution.
\((x+8)(x-4)=4^{y} \rightarrow 1) x+8=2^{n}, x-4=2^{m}, m, n \in Z, m+n=2 y\)
\(x>4 \rightarrow 2^{n}-8>4 \rightarrow n \geq 4, 2^{m}+4>4 \rightarrow m \geq 0\)
\(2^{n}-2^{m}=12 \rightarrow 2^{m}(2^{n-m}-1)=2^{2} \cdot 3 \rightarrow m=2, n-m=2 \rightarrow n=4\)
\(y_{1}=3, x_{1}=8\) - first solution.
2) \... | 10 | Number Theory | math-word-problem | More than one problem | Yes | olympiads | false |
5. On the sides of triangle $A B C$ with side lengths 2, 3, and 4, external squares $A B B_{1} A_{1}, B C C_{2} B_{2}, C A A_{3} C_{3}$ are constructed. Find the sum of the squares of the side lengths of the hexagon $A_{1} B_{1} B_{2} C_{2} C_{3} A_{3}$.
## Answers and Solutions
Problem 1 Answer: 165 m. | Solution
$m$ - the number of steps Petya takes, $n$ - the number of steps Vova takes
$0.75 m=0.55 n=L-$ path length
$k$ - the step number of Petya leading to the coincidence of the footprint,
$i$ - the step number of Vova leading to the coincidence of the footprint,
$0.75 k=0.55 i \rightarrow 15 k=11 i \rightarrow... | 116 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The sum of the squares of the lengths of the sides of triangle $ABC$ is 36, and its area is $5 \sqrt{3}$.
Points $P, Q$, and $R$ are the vertices of isosceles triangles with an angle of $120^{\circ}$, constructed on the sides of triangle $ABC$ as bases (externally). Find the lengths of the sides of triangle $PQR$.
... | Solution
Given $x \neq 0, y \neq 0$. Divide the left and right sides of the equations: $\frac{y^{3}}{x^{3}}=\frac{2 x+5 y}{5 x+2 y}=\frac{2+5 y / x}{5+2 y / x}$.
Substitution: $t=\frac{y}{x} \rightarrow t^{3}=\frac{5 t+2}{2 t+5} \rightarrow 2 t^{4}+5 t^{3}-5 t-2=0, t \neq-\frac{5}{2}$.
Factorize the polynomial: $(t-... | PQ=PR=QR=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. $x=p_{3}^{t}, y=p_{2}^{r} \cdot p_{1}^{q}, q=1,2, \ldots, k, r=1,2, \ldots, m, t=1,2, \ldots, n$. There are $k \cdot m \cdot n$ such pairs.
Then the total number $N$ of different pairs of coprime divisors of the number $a$ is $N=(k+1)(m+1)(n+1)+k m+k n+m n+3 m n k=4 m n k+2(k m+k n+m n)+(m+n+k)+1$
In the variant $... | Solution. Let $a$ and $b$ (with $a > b$) be the lengths of the bases of the trapezoid (integers), and $p$ be a common divisor of $a$ and $b$.
Then $a = p \cdot a_1$, $b = p \cdot b_1$. The trapezoid can be divided into equal triangles only as shown in the figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_82168e1... | 5 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
# 2.14. Qualifying round of the "Rosatom" Olympiad, 8th grade | # 2.14. Qualifying round of the "Rosatom" Olympiad, 8th grade
## Answers and solutions
Problem 1 Answer: father - 10 million, Petya - 20 million, Kolya - 5 million.
## Problem 2
$7 \cdot 49^{n} + 40 \cdot 2^{n} = 7 \cdot (47 + 2)^{n} + 40 \cdot 2^{n}$. There exists a natural number $k$,
such that $(47 + 2)^{n} = 4... | 10,20,5 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
1. A parent decided to divide their business between their sons Petya and Kolya according to the following rules: if Petya wants to participate in the business, he will receive a share that is twice the father's share. If Kolya participates in the business, his share will be half of the father's share. It so happened t... | 1. Construct point $P$ - the midpoint of $M N$. | notfound | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
2. 2013-2014 academic year
## | 2. On the line $A P$, we lay off the segment $P C$ equal in length to $\frac{q}{q+2 p} \cdot A P = \frac{3}{7} A P$.
 | \frac{3}{7}AP | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. Find the positive integers $x$ and $y$ for which
$$
\left\{\begin{array}{c}
\sqrt{x}+\sqrt{y}=53 \\
\text{LCM}(x, y)=22500
\end{array}\right.
$$ | 3. Draw the lines $C M$ and $C N$ and point $O$ - the midpoint of segment $A C$. | notfound | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
4. Dima drew a parallelogram $A B C D$ and points $M$ and $N$ on sides $B C$ and $C D$ respectively such that $B M: M C=D N: N C=2: 3$. After that, he erased everything except points $A, M$ and $N$ using a cloth. Vova restored the drawing using a ruler and a compass. How did he do it? | 4. Points $B$ and $D$ are the intersection points of lines $C M$ and $C N$ with the line passing through point $O$ and parallel to line $M N$.
## Problem 5
Numbers $A$, when divided by $n$, have a remainder of $r$, when divided by $n+1$ - a remainder of $r+1$, and when divided by $n+2$ - a remainder of $r+2$, have th... | 1710 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha, Masha, Tolya, and Olya ate 88 candies, and each of them ate at least one candy. Masha and Tolya ate 57 candies, but Pasha ate the most candies. How many candies did Olya eat? | Solution. Either Masha or Tolya ate no less than 29 candies, then Pasha ate no less than 30 candies. Then the number of candies eaten by Pasha, Masha, and Tolya is no less than $57+30=87$. Since there are a total of 88 candies, and Olya did not refuse any candies, she gets one candy.
Answer: 1 candy. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Find the fraction of the form $\frac{n}{23}$ that is closest to the fraction $\frac{37}{57}$ ( $n$ - integer). | Solution. After bringing to a common denominator, we need to find the smallest deviation of the fraction $\frac{57 n}{1311}$ from $\frac{851}{1311}$. Dividing 851 by 57 with a remainder gives $851=14 \cdot 57+53$. If $n=14$, then the deviation of the number $57 n$ from 851 is 53, and for $n=15$ this deviation is 4. For... | \frac{15}{23} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The difference of two natural numbers is 5 times less than their sum and 24 times less than their product. Find these numbers. | Solution. Let $x$ be the larger of the two numbers, and $y$ be the smaller. The condition of the problem is $\left\{\begin{array}{c}x+y=5(x-y) \\ x y=24(x-y)\end{array} \rightarrow\left\{\begin{array}{c}2 x=3 y \\ \frac{3}{2} y^{2}=12 y\end{array}\right.\right.$.
$y=0$ is not a natural number, the second solution is $... | 12,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Pete is trying to lay out a square on a table using identical cardboard rectangles measuring $14 \times 10$. Will he be able to do this? Propose your version of constructing such a square. What is the minimum number of rectangles he will need? | Solution. Let $n$ be the number of rectangles that form the square. Then the area of the square is $S=14 \cdot 10 \cdot n$. If the side of the square is formed by $m$ sides of length 14 and $k$ sides of length 10, then the length of the side of the square is $14 m + 10 k$, and its area is $S=(14 m + 10 k)^{2}$. Equatin... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a rectangular table, the letters of the word "олимпиада" (olympiada) are arranged in a certain order.
| $\mathrm{O}$ | Л | И | M | П | И | A | Д | A |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Л | И | M | П | И | A | Д | A | O |
| И | M | П | И | A | Д | A | O | Л |
| M | П | И | A | Д |... | Solution. In each cell of the table shown below
| $\mathrm{O}$ | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| 1 | 3 | 6 | 10 | 15 | 21 | 28 | | |
| 1 | 4 | 10 | 20 | 35 | 56 | | | |
the number of different paths leading to... | 93 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. $[x]$ - the integer part of the number $x$ (the greatest integer not exceeding $x$), $\{x\}=x-[x]-$ the fractional part of the number $x$.
For which $a$ does the system $\left\{\begin{array}{c}2 x-[x]=4 a+1 \\ 4[x]-3\{x\}=5 a+15\end{array}\right.$ have a solution? Find such $x$. | Solution. Let $\left\{\begin{array}{l}u=[x] \\ v=\{x\}\end{array} \rightarrow u+v=x, 0 \leq v<1\right.$.
Then the system will take the form: $\left\{\begin{array}{c}u+2 v=4 a+1 \\ 4 u-3 v=5 a+15\end{array} \rightarrow\left\{\begin{array}{c}u=2 a+3 \\ v=a-1\end{array}\right.\right.$.
Since $0 \leq v<1$, then $0 \leq a... | 1) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. One of the parrots always tells the truth, another always lies, and the third one - the trickster - sometimes tells the truth and sometimes lies. When asked: "Who is Kesha?" - they answered: Gоsha: The liar. Keshа: - I am the trickster! Roma: - The absolutely honest parrot. Who among the parrots is the liar, and who... | Solution. Since Kesh said that he is a trickster, he cannot be honest. Since Roma said that Kesh is an absolutely honest parrot, Roma cannot be honest either. Therefore, Gosha is the one telling the truth. This means Kesh is a liar. Then Roma is a trickster.
## Answer: Kesh is a liar, Roma is a trickster. | Kesh | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. For which numbers $n$ are the numbers $a_{n}=7 n-3$ divisible by 5 but not divisible by 3? | Solution. $a_{n}=7 n-3: 5 \rightarrow 7 n-3=5 k, k \in Z$. We obtain the linear Diophantine equation $7 n-5 k=3, k \in Z$. We have
$$
\begin{aligned}
& k=\frac{7 n-3}{5}=n+\frac{2 n-3}{5} \rightarrow \frac{2 n-3}{5}=l \in Z \rightarrow n=\frac{5 l+3}{2}=2 l+1+\frac{l+1}{2} \rightarrow \\
& \rightarrow \frac{l+1}{2}=t ... | 5-1,\neq3-1,,\inN | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Represent the number 43 as the sum of three prime numbers. In how many ways can this be done? Note that the number 1 is not considered a prime number. | Solution.
$$
43=a+b+c, a, b, c \text { - prime numbers. }
$$
Prime numbers from 2 to 43
| 2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | 37 | 41 | 43 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
$a=2 \rightarrow 43-4=41 \rightarrow b=2 \righ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The sum of the bases of the trapezoid is 4. Find the greatest possible length of the segment passing through the intersection of the diagonals of the trapezoid, parallel to its bases.
 | Solution. Let $A D=a, B C=b, \frac{b}{a}=k$.
$\square B O C \square A O D: \frac{O C}{A O}=\frac{O B}{D O}=\frac{b}{a}=k \rightarrow O C=k \cdot A O, A C=A O+O C=(1+k) A O$.
$\square A O M \square A C B: \quad \frac{M O}{b}=\frac{A O}{A C}=\frac{1}{1+k} \rightarrow M O=\frac{b}{1+k}$.
$\square O C N \square \triangl... | 2 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
6. Along the diagonals of the bases $AC$ and $B_{1}D_{1}$ of the cube $ABCD A_{1}B_{1}C_{1}D_{1}$ with edge $a$, two ants, Gosha and Lesha, are crawling. They started moving simultaneously from points $A$ and $B_{1}$ respectively at a constant speed, with Lesha's speed being twice that of Gosha's, and they finished whe... | # Solution:
$A$ - the position of the bus on the road at the moment Petya saw it;
$P$ - the position of Petya on the road at the moment he saw the bus;
$B$ - the position of the last stop ... | \frac{8}{15} | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
1. Petya came to the bus stop of the bus going to school with stops equally spaced from each other, and, not seeing the bus on the road, decided to run and get on the bus at the next stops along the way to school. Petya ran in such a way that at any moment he could notice the appearance of the bus on the road behind hi... | Answer: $a_{\max }=0.6$ km | 0.6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The coordinates $(x ; y)$ of points in the square $\{(x ; y):-\pi \leq x \leq \pi, 0 \leq y \leq 2 \pi\}$ satisfy the system of equations $\left\{\begin{array}{c}\sin x+\sin y=\sin 2 \\ \cos x+\cos y=\cos 2\end{array}\right.$. How many such points are there in the square? Find the coordinates $(x ; y)$ of the point ... | Answer: 1) two points
$$
\text { 2) } x=2+\frac{\pi}{3}, y=2-\frac{\pi}{3}
$$ | 2)2+\frac{\pi}{3},2-\frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Kuzya the flea can make jumps along a straight line $L$. The starting point for the jumps is at point $A$ on the line $L$, the length of each jump is $h$, and the direction of each jump is chosen randomly and equally likely. Find the probability that, after making from three to nine random jumps, Kuzya will be at a ... | Answer: $P(A)=\frac{2}{7} \cdot \frac{47}{64}=\frac{47}{224}$ | \frac{47}{224} | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
6. Along the diagonals of the bases $AC$ and $B_{1}D_{1}$ of the cube $ABCD A_{1}B_{1}C_{1}D_{1}$ with edge $a$, two ants, Gosha and Lesha, are crawling. They started moving simultaneously from points $A$ and $B_{1}$ respectively at a constant speed, with Lesha's speed being three times the speed of Gosha's movement, a... | Answer: $d_{\min }=a \sqrt{\frac{6}{5}}$ | \sqrt{\frac{6}{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Petya came to the bus stop of the bus going to school with stops equally spaced from each other, and, not seeing the bus on the road, decided to run and get on the bus at the next stops along the way to school. Petya ran in such a way that at any moment he could notice the appearance of the bus on the road behind hi... | Answer: $a_{\max }=0.25$ km | 0.25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The coordinates $(x ; y)$ of points in the square $\{(x ; y):-\pi \leq x \leq 2 \pi, 0 \leq y \leq 3 \pi\}$ satisfy the system of equations $\left\{\begin{array}{l}\sin x+\sin y=\sin 3 \\ \cos x+\cos y=\cos 3\end{array}\right.$. Find the coordinates $(x ; y)$ of all such points. | Answer: 1) five points;
$$
\text { 2) }\left\{\begin{array}{l}
x_{1}=3-\frac{\pi}{3} \\
y_{1}=3+\frac{\pi}{3}
\end{array} ;\left\{\begin{array}{l}
x=3-\frac{5 \pi}{3} \\
y=3+\frac{5 \pi}{3}
\end{array} ;\left\{\begin{array}{l}
x_{1}=3-\frac{5 \pi}{3} \\
y_{1}=3-\frac{\pi}{3}
\end{array} ;\left\{\begin{array}{l}
x_{1}=... | {\begin{pmatrix}x_{1}=3-\frac{\pi}{3}\\y_{1}=3+\frac{\pi}{3}\end{pmatrix};{\begin{pmatrix}3-\frac{5\pi}{3}\\3+\frac{5\pi}{3}\end{pmatrix};{\begin{pmatrix}x_{1}=3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Kuzya the flea can make jumps along a straight line $L$. The starting point for the jumps is at point $A$ on the line $L$, the length of each jump is $h$, and the direction of each jump is chosen randomly and equally likely. Find the probability that, after making from two to five random jumps, Kuzya will be at a di... | Answer: $\quad P(A)=\frac{5}{8}$ | \frac{5}{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Along the diagonals of the bases $AC$ and $B_{1}D_{1}$ of the cube $ABCD A_{1}B_{1}C_{1}D_{1}$ with edge $a$, two ants, Gosha and Lesha, are crawling. They started moving simultaneously from points $A$ and $B_{1}$ respectively at a constant speed, with Lesha's speed being four times the speed of Gosha's movement, an... | Answer: $d_{\text {min }}=a \sqrt{\frac{43}{34}}$ | \sqrt{\frac{43}{34}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Petya came to the bus stop of the bus going to school with stops equally spaced from each other, and, not seeing the bus on the road, decided to run and get on the bus at the next stops along the way to school. Petya ran in such a way that at any moment he could notice the appearance of the bus on the road behind hi... | Answer: $a_{\max }=0.8$ km | 0.8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The coordinates $(x ; y)$ of points in the square $\{(x ; y):-2 \pi \leq x \leq 3 \pi,-\pi \leq y \leq 4 \pi\}$ satisfy the system of equations $\left\{\begin{array}{l}\sin x+\sin y=\sin 4 \\ \cos x+\cos y=\cos 4\end{array}\right.$. How many such points are there? Find the coordinates $(x ; y)$ of the points with th... | Answer: 1) 13 points; 2) $\left\{\begin{array}{l}x=4+\frac{5 \pi}{3} \\ y=4+\frac{\pi}{3}\end{array} ;\left\{\begin{array}{l}x=4+\frac{5 \pi}{3} \\ y=4+\frac{7 \pi}{3}\end{array} ;\left\{\begin{array}{l}x=4+\frac{5 \pi}{3} \\ y=4-\frac{5 \pi}{3}\end{array}\right.\right.\right.$ | {\begin{pmatrix}4+\frac{5\pi}{3}\\4+\frac{\pi}{3}\end{pmatrix};{\begin{pmatrix}4+\frac{5\pi}{3}\\4+\frac{7\pi}{3}\end{pmatrix};{\begin{pmatrix}4+\frac{5\pi}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Kuzya the flea can make jumps along a straight line $L$. The starting point for the jumps is at point $A$ on the line $L$, the length of each jump is $h$, and the direction of each jump is chosen randomly and equally likely. Find the probability that, after making from five to ten random jumps, Kuzya will be at a di... | Answer: $P(A)=\frac{2}{6} \cdot \frac{108}{256}=\frac{9}{64}$ | \frac{9}{64} | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
6. Along the diagonals of the bases $AC$ and $B_{1}D_{1}$ of the cube $ABCD A_{1}B_{1}C_{1}D_{1}$ with edge $a$, two ants, Gosha and Lesha, are crawling. They started moving simultaneously from points $A$ and $B_{1}$ respectively at a constant speed, with Lesha's speed being five times the speed of Gosha's movement, an... | Answer: $d_{\text {min }}=a \sqrt{\frac{17}{13}}$
## Final Rosatom 11th grade criteria (set №1) | \sqrt{\frac{17}{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Point $M$ lies on the edge $AB$ of the cube $ABCD A_{1} B_{1} C_{1} D_{1}$. A rectangle $MN LK$ is inscribed in the square $ABCD$ such that one of its vertices is point $M$, and the other three are located on different sides of the square base. The rectangle $M_{1} N_{1} L_{1} K_{1}$ is the orthogonal projection of ... | # Solution:
Let $O$ be an arbitrary point of delivery of goods $d_{1}, d_{2}, \ldots, d_{7}$ - distances from the delivery point to the houses;
The total distance:
$\Sigma=1 \cdot d_{1}+2 d_{2}+3 d_{3}+4 d_{4}+5 d_{5}+6 d_{6}+21 d_{7}=\left(d_{1}+d_{7}\right)+2\left(d_{2}+d_{7}\right.$
![](https://cdn.mathpix.com/c... | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
Problem 3. Answer: $n=78$
# | # Solution:
Given $a_{1}, a_{2} \neq 0$. The elements of the sequence are defined as long as $a_{n+1} \neq 0$. For other indices, the terms of the sequence are not defined. Let $a_{n+1} \neq 0, a_{n+2}=0$. Then for all $k \leq n \Rightarrow a_{k+2} a_{k+1}=a_{k+1} a_{k}-2$. The sequence $b_{k}=a_{k+1} a_{k}$ satisfies... | 78 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
1. Customers of the online store "Ali-Express" live in six houses located at the vertices of a convex hexagon. One order was placed by residents of the first house, two orders by residents of the second house, and so on, with five orders from the residents of the fifth house. However, the residents of the last sixth ho... | Answer: the goods should be delivered to the sixth house | the\sixth\house | Geometry | math-word-problem | Yes | Problem not solved | olympiads | false |
2. Find all numbers $C$ for which the inequality $|\alpha \sin x+\beta \cos 4 x| \leq C$ holds for all $x$ and any $(\alpha ; \beta)$ such that $|\alpha| \leq 1,|\beta| \leq 1$. | Answer: $\quad C \geq 2$ | 2 | Inequalities | math-word-problem | Yes | Incomplete | olympiads | false |
4. Pete throws a die several times onto the table and counts the sum of the points that come up. For any natural number $n$, the event $A_{n}$ occurs if this sum equals $n$. Find the probability of the event $A_{10}$. | Answer: $P(A)=\frac{1}{9} \cdot\left(\frac{3}{6^{2}}+\frac{27}{6^{3}}+\frac{80}{6^{4}}+\frac{126}{6^{5}}+\frac{126}{6^{6}}+\frac{84}{6^{7}}+\frac{36}{6^{8}}+\frac{9}{6^{9}}+\frac{1}{6^{10}}\right)$ | \frac{1}{9}\cdot(\frac{3}{6^{2}}+\frac{27}{6^{3}}+\frac{80}{6^{4}}+\frac{126}{6^{5}}+\frac{126}{6^{6}}+\frac{84}{6^{7}}+\frac{36}{6^{8}}+\frac{9}{6^{9}}+\frac{1}{6^{} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the plane, there is a set of points $M$, the coordinates $X$ and $y$ of which are related by the equation
$$
\sin (x+2 y)=\sin x+\sin 2 y
$$
A circle of radius $R$, located on the same plane, does not intersect with the set $M$.
What values can the radius of such a circle take | Answer: $R \in\left(0 ; \frac{3-\sqrt{5}}{2} \pi\right)$ | R\in(0;\frac{3-\sqrt{5}}{2}\pi) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Customers of the online store "Ali-Express" live in five houses located at the vertices of a convex pentagon. One order was placed by residents of the first house, two orders by residents of the second house, and so on, with four orders from the residents of the fourth house. However, the residents of the last fifth... | Answer: the goods should be delivered to the fifth house. | 5 | Geometry | math-word-problem | Yes | Problem not solved | olympiads | false |
4. Pete throws a die several times onto the table and counts the sum of the points that come up. For any natural number $n$, the event $A_{n}$ occurs if this sum equals $n$. Find the probability of the event $A_{9}$. | Answer: $\quad P(A)=\frac{1}{8} \cdot\left(\frac{4}{6^{2}}+\frac{25}{6^{3}}+\frac{56}{6^{4}}+\frac{70}{6^{5}}+\frac{56}{6^{6}}+\frac{28}{6^{7}}+\frac{8}{6^{8}}+\frac{1}{6^{9}}\right)$ | \frac{1}{8}\cdot(\frac{4}{6^{2}}+\frac{25}{6^{3}}+\frac{56}{6^{4}}+\frac{70}{6^{5}}+\frac{56}{6^{6}}+\frac{28}{6^{7}}+\frac{8}{6^{8}}+\frac{1}{6^{9}}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the plane, there is a set of points $M$, the coordinates $x$ and $y$ of which are related by the equation
$$
\sin (3 x+4 y)=\sin 3 x+\sin 4 y
$$
A circle of radius $R$, located on the same plane, does not intersect the set $M$.
What values can the radius of such a circle take? | Answer: $R \in\left(0 ; \frac{\pi}{6}\right)$ | R\in(0;\frac{\pi}{6}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Point $M$ lies on the edge $A B$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. A rectangle $M N L K$ is inscribed in the square $A B C D$ such that one of its vertices is point $M$, and the other three are located on different sides of the square base. The rectangle $M_{1} N_{1} L_{1} K_{1}$ is the orthogonal proje... | Answer: $A M: M B=1: 4$ | AM:MB=1:4 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
1. Customers of the online store "Ali-Express" live in four houses located at the vertices of a convex quadrilateral. One order was placed by residents of the first house, two orders by residents of the second house, three orders by residents of the third house, and six orders by residents of the last fourth house. The... | Answer: the goods should be delivered to the fourth house. | 4 | Geometry | math-word-problem | Yes | Problem not solved | olympiads | false |
2. Find all numbers $C$ for which the inequality $|\alpha \sin 2 x+\beta \cos 8 x| \leq C$ holds for all $x$ and any $(\alpha ; \beta)$ such that $\alpha^{2}+\beta^{2} \leq 16$. | Answer: $C \geq 4 \sqrt{2}$ | C\geq4\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Pete throws a die several times onto the table and counts the sum of the points that come up. For any natural number $n$, the event $A_{n}$ occurs if this sum equals $n$. Find the probability of the event $A_{8}$. | Answer: $\quad P(A)=\frac{1}{7} \cdot\left(\frac{5}{6^{2}}+\frac{21}{6^{3}}+\frac{35}{6^{4}}+\frac{35}{6^{5}}+\frac{21}{6^{6}}+\frac{7}{6^{7}}+\frac{1}{6^{8}}\right)$ | \frac{1}{7}\cdot(\frac{5}{6^{2}}+\frac{21}{6^{3}}+\frac{35}{6^{4}}+\frac{35}{6^{5}}+\frac{21}{6^{6}}+\frac{7}{6^{7}}+\frac{1}{6^{8}}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the plane, there is a set of points $M$, the coordinates $x$ and $y$ of which are related by the equation
$$
\sin (12 x+5 y)=\sin 12 x+\sin 5 y .
$$
A circle of radius $R$, located on the same plane, does not intersect the set $M$.
What values can the radius of such a circle take? | Answer: $R \in\left(0 ; \frac{\pi}{15}\right)$ | R\in(0;\frac{\pi}{15}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Point $M$ lies on the edge $A B$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. A rectangle $M N L K$ is inscribed in the square $A B C D$ such that one of its vertices is point $M$, and the other three are located on different sides of the square base. The rectangle $M_{1} N_{1} L_{1} K_{1}$ is the orthogonal proje... | Answer: $A M: M B=1: 2$
## Evaluation Criteria Final Rosatom 11th grade | AM:MB=1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Rewrite the original equation
$$
\log _{2}\left(\frac{x}{17}+\frac{y}{5}\right)=\log _{2} \frac{x}{17}+\log _{2} \frac{y}{5}
$$
as
$$
\log _{2}\left(\frac{x}{17}+\frac{y}{5}\right)=\log _{2} \frac{x y}{85}
$$
From this, we get $\frac{x}{17}+\frac{y}{5}=\frac{x y}{85}$ or $5 x+17 y=x y$. We can rewrite the last e... | Answer: $\left\{\begin{array}{l}x=18, \\ y=90 ;\end{array}\left\{\begin{array}{l}x=22, \\ y=22 ;\end{array}\left\{\begin{array}{l}x=34, \\ y=10 ;\end{array}\left\{\begin{array}{c}x=102, \\ y=6 .\end{array}\right.\right.\right.\right.$ | {\begin{pmatrix}18,\\90;\end{pmatrix}{\begin{pmatrix}22,\\22;\end{pmatrix}{\begin{pmatrix}34,\\10;\end{pmatrix}{\begin{pmatrix}102,\\60\end} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. For all x, the equality must hold
$$
\begin{gathered}
f(x+5 \pi)=f(x) \text {, i.e., } \\
\sin (n x+5 \pi n) \cos \frac{6(x+5 \pi)}{n+1}=\sin n x \cos \frac{6 x}{n+1} .
\end{gathered}
$$
For \( n=0 \), this equality is satisfied, and therefore we can ignore this value in further analysis.
Case 1: \( n \) is even.... | Answer:
$$
n \in\{-31 ; -16 ; -11 ; -7 ; -6 ; -4 ; -3 ; -2 ; 0 ; 1 ; 2 ; 4 ; 5 ; 9 ; 14 ; 29\} .
$$ | n\in{-31;-16;-11;-7;-6;-4;-3;-2;0;1;2;4;5;9;14;29} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The number of correct answers by Sasha is distributed according to the binomial law with a success probability of $1 / 2$
$$
P_{1}(k)=C_{5}^{k}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{5-k}=\frac{C_{5}^{k}}{32}
$$
| $k$ | 0 | 1 | 2 | 3 | 4 | 5 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $P_... | Answer: $\frac{65}{324}$. | \frac{65}{324} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. From the equation $2 \cos ^{2} a_{n}=\cos a_{n+1}$, it follows that for all $n$, the value of $\cos a_{n+1} \geq 0$ and $\left|\cos a_{n}\right|=\sqrt{\frac{\cos a_{n+1}}{2}} \leq \frac{1}{\sqrt{2}}$, i.e., on the unit circle, all values of $a_{n}$ fall within the interval $[\pi / 4 ; \pi / 2]$.
 a_{1}=\frac{\pi}{3}+2 \pi m, m \in Z ; d=2 \pi k, k \in Z, k \neq 0$;
$$
\text { 3) } a_{1}=-\frac{\pi}{3}+2 \pi m, m \in Z ; d=2 \pi k, k \in Z, k \neq 0 \text {. }
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false |
2. For which integers $n$ does the function $f(x)=\cos ((n+1) x) \cdot \sin \frac{8 x}{n-2}$ have a period $T=3 \pi$? | Answer: $n \in\{3 ; 1 ; 5,-1 ; 10 ;-6 ; 26 ;-22\}$. | n\in{3;1;5,-1;10;-6;26;-22} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. The arithmetic progression $\left\{a_{n}\right\}$ with a non-zero difference is such that the sequence $b_{n}=a_{n} \cdot \cos a_{n}$ is also an arithmetic progression with a non-zero difference. Find the possible values of the first term and the difference of the progression $\left\{a_{n}\right\}$, if for all $n$ t... | Answer:
$$
\text { 1) } a_{1}=-\frac{\pi}{6}+2 \pi m, m \in Z ; d=2 \pi k, k \in Z, k \neq 0
$$
$$
\text { 2) } a_{1}=-\frac{5 \pi}{6}+2 \pi m, m \in Z ; d=2 \pi k, k \in Z, k \neq 0 \text {. }
$$ | 1)a_{1}=-\frac{\pi}{6}+2\pi,\inZ;=2\pik,k\inZ,k\neq0\\2)a_{1}=-\frac{5\pi}{6}+2\pi,\inZ;=2\pik,k\inZ,k\neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the sides $A B$ and $A C$ of an acute-angled triangle $A B C$, two equal rectangles $A M N B$ and $A P Q C$ are constructed outwardly. Find the distance between the vertices $N$ and $Q$ of the rectangles, if the lengths of the sides $A B$ and $A C$ are $\sqrt{3}$ and 1 respectively, and the angle at vertex $A$ of... | Answer: $2 \sqrt{2+\sqrt{3}}$.
# | 2\sqrt{2+\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the integers $y$ and even $x$, for which
$$
\log _{2}\left(\frac{x}{7}+\frac{y}{6}\right)=\log _{2} \frac{x}{7}+\log _{2} \frac{y}{6}
$$ | Answer: $\left\{\begin{array}{l}x=8, \\ y=48 ;\end{array}\left\{\begin{array}{l}x=10, \\ y=20 ;\end{array}\left\{\begin{array}{l}x=14, \\ y=12 ;\end{array}\left\{\begin{array}{c}x=28, \\ y=8 .\end{array}\right.\right.\right.\right.$ | Algebra | math-word-problem | Yes | Incomplete | olympiads | false | |
2. For which integers $n$ does the function $f(x)=\sin ((2 n+1) x) \cdot \sin \frac{5 x}{n-1}$ have a period $T=7 \pi$? | Answer: $n \in\{2 ; 0 ; 6,-4 ; 8 ;-6 ; 36 ;-34\}$. | n\in{2;0;6,-4;8;-6;36;-34} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
4. Katya and Pasha ask each other four tricky questions and answer them without thinking, randomly. The probability that Katya will lie to a question asked by Pasha is $\frac{1}{3}$, regardless of the question number. Pasha gives a truthful answer to Katya's question with a probability of $\frac{3}{5}$, independently o... | Answer: $\frac{48}{625}$. | \frac{48}{625} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On the sides $A B$ and $A C$ of an acute-angled triangle $A B C$, two equal rectangles $A M N B$ and $A P Q C$ are constructed outwardly. Find the distance between the vertices $N$ and $Q$ of the rectangles, if the lengths of the sides $A B$ and $A C$ are $2 \sqrt{2}$ and 1, respectively, and the angle at vertex $A$... | Answer: $3 \sqrt{2+\sqrt{2}}$.
# | 3\sqrt{2+\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find integers $x$ and $y$ such that $x+y$ is even and
$$
\log _{2}\left(\frac{x}{5}+\frac{y}{8}\right)=\log _{2} \frac{x}{5}+\log _{2} \frac{y}{8}
$$ | Answer: $\left\{\begin{array}{c}x=6, \\ y=48 ;\end{array}\left\{\begin{array}{l}x=10, \\ y=16 ;\end{array}\left\{\begin{array}{l}x=13, \\ y=13 ;\end{array}\left\{\begin{array}{c}x=45 \\ y=9\end{array}\right.\right.\right.\right.$ | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false | |
2. For which integers $n$ does the function $f(x)=\cos ((n-1) x) \cdot \cos \frac{15 x}{2 n+1}$ have a period $T=\pi$? | Answer: $n \in\{0 ;-2 ; 2 ;-8\}$. | n\in{0,-2,2,-8} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
6. On the sides $A B$ and $A C$ of an acute-angled triangle $A B C$, two equal rectangles $A M N B$ and $A P Q C$ are constructed outside. Find the distance between the vertices $N$ and $Q$ of the rectangles, if the lengths of the sides $A B$ and $A C$ are 5 and $\sqrt{11}$, respectively, and the angle at vertex $A$ of... | Answer: $\sqrt{72+48 \sqrt{2}}$.
## Final round of the "Rosatom" Olympiad, 11th grade, CIS, February 2020
# | \sqrt{72+48\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The expression $\eta(x)-\eta(x-7 \pi)=\left\{\begin{array}{l}1, x \in[0,7 \pi), \\ 0, x \in(-\infty ; 0) \cup[7 ;+\infty) .\end{array}\right.$
Case 1. $x \in[0,7 \pi)$. The equation takes the form
$$
\sin x=1+\cos x
$$
Using the method of introducing an auxiliary argument, we get
$$
\sin \left(x-\frac{\pi}{4}\ri... | Answer: 1) $x=\frac{\pi}{2}+2 \pi k, k=0,1,2,3$; 2) $x=\pi(2 m+1), m \in Z$. | \frac{\pi}{2}+2\pik,k=0,1,2,3;\pi(2+1),\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let's calculate the number of ways to write 3 digits:
$$
10 \cdot 10 \cdot 10=1000
$$
We will derive the formula for the number $A$, which is divisible by 5 and has a remainder of 3 when divided by 7. This number has the form
$$
A=5 k=7 n+3
$$
Solving the equation $5 k-7 n=3$ in integers, we find
$$
\left\{\beg... | Answer: $\quad P=0.026$. | 0.026 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Transform the left side of the equation $(\sin x+\cos y)(\cos x-\sin y)=$ $=\sin x \cos x+\cos y \cos x-\sin x \sin y-\cos y \sin y=$ $=\cos (x+y)+\frac{1}{2}(\sin 2 x-\sin 2 y)=\cos (x+y)+\sin (x-y) \cos (x+y)$.
Substitute the obtained expression into the original equation
$$
\cos (x+y)+\sin (x-y) \cos (x+y)=1+\s... | Answer: $(x+y)_{\min }=2 \pi$. | (x+y)_{\}=2\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Let the length of side $AB$ of parallelogram $ABCD$ be denoted by $a$. Drop a perpendicular $BH$ from vertex $B$ to line $CM$. Consider the right triangle $BHC$. According to the problem, $BH = AB = a$. Since $BC \parallel AB$, we have $\angle BHC = \angle CMD = 30^{\circ}$. Therefore, $BC = 2BH = 2a$. Now consider ... | Answer:
1) $60^{\circ}, 120^{\circ}$; 2) $S_{A B C D}=\sqrt{3}$
# | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$$
\cos (2 x(\eta(x+3 \pi)-\eta(x-8 \pi))=\sin x+\cos x
$$
where $\eta(x)=\left\{\begin{array}{l}1, x \geq 0 \\ 0, x<0\end{array}\right.$ is the Heaviside function. | Answer: 1) $x=2 \pi k, k \in Z$; 2) $x=-\frac{\pi}{2}+2 \pi k, k=-1,0,1,2,3,4$.
$$
\begin{aligned}
& \text { 3) } x=\frac{\pi}{2}+2 \pi k, k \in Z, k \neq-1,0,1,2,3 \\
& \text { 4) } x=-\frac{\pi}{4}+\pi m, m=-2,-1, \ldots, 8
\end{aligned}
$$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Pete, Vasya, and Ivan each wrote a random digit on their own card and passed the cards to Masha so that she couldn't see the written digits. Masha randomly shuffled the cards and laid them out in a row on the table. Find the probability that a three-digit number on the table is divisible by 3 and leaves a remainder ... | Answer: $\quad P=0.037$. | 0.037 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Find the greatest negative value of the expression $x-y$ for all pairs of numbers $(x ; y)$ satisfying the equation
$$
(\sin x+\sin y)(\cos x-\cos y)=\frac{1}{2}+\sin (x-y) \cos (x+y)
$$ | Answer: $(x+y)_{\max }=-\frac{\pi}{6}$. | -\frac{\pi}{6} | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
6. Point $M$ is the midpoint of side $A D$ of parallelogram $A B C D$. Line $C M$ is inclined to the base $A D$ at an angle of $45^{\circ}$. Vertex $B$ is equidistant from line $C M$ and vertex $A$. Find the angles of the parallelogram. Find the area of the parallelogram if the length of the base $A D$ is 1. | Answer: 1) $75^{0}, 105^{0}$; 2) $S_{\text {ABCD }}=\frac{\sqrt{3}+1}{4}$. | 75^{0},105^{0};S_{\text{ABCD}}=\frac{\sqrt{3}+1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\operatorname{tg}\left(x(\eta(x-2 \pi)-\eta(x-5 \pi))=\frac{1}{\cos ^{2} x}-1\right.$, where $\eta(x)=\left\{\begin{array}{l}1, x \geq 0 \\ 0, x<0\end{array}\right.$ is the Heaviside function. | Answer: 1) $x=\pi k, k \in Z$; 2) $x=\frac{\pi}{4}+\pi m, m=2,3,4$. | \pik,k\inZ;\frac{\pi}{4}+\pi,=2,3,4 | Algebra | math-word-problem | Incomplete | Problem not solved | olympiads | false |
5. Find the smallest positive value of the expression $x+y$ for all pairs of numbers ( $x ; y$ ) that satisfy the equation $(\tan x-2)(\tan y-2)=5$.
翻译完成，保留了原文的格式和换行。 | Answer: $(x+y)_{\min }=\pi-\operatorname{arctg} \frac{1}{2}$. | \pi-\operatorname{arctg}\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Point $M$ is the midpoint of side $A D$ of parallelogram $A B C D$. Line $C M$ is inclined to the base $A D$ at an angle of $15^{0}$. Vertex $B$ is equidistant from line $C M$ and vertex $A$. Find the angles of the parallelogram. Find the area of the parallelogram if the length of the base $A D$ is 6. | Answer: 1) $45^{\circ}, 135^{\circ} ; 2) S_{\text {ABCD }}=9(\sqrt{3}-1)$. | 45,135;S_{\text{ABCD}}=9(\sqrt{3}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$$
\operatorname{ctg}\left(2 x(\eta(x+3 \pi)-\eta(x-6 \pi))=\frac{1}{\sin x}-\frac{1}{\cos x}\right.
$$
where $\eta(x)=\left\{\begin{array}{l}1, x \geq 0 \\ 0, x<0\end{array}\right.$ is the Heaviside function. | Answer: $x=\frac{\pi}{4}+\pi k, k=-3,-2, \ldots, 5$. | \frac{\pi}{4}+\pik,k=-3,-2,\ldots,5 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. Find the greatest negative value of the expression $x-y$ for all pairs of numbers $(x ; y)$ satisfying the equation $(1-\operatorname{ctg} x)(1+\operatorname{ctg} y)=2$. | Answer: $(x+y)_{\max }=-\frac{3 \pi}{4}$. | -\frac{3\pi}{4} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
6. Point $M$ is the midpoint of side $A D$ of parallelogram $A B C D$. Line $C M$ is inclined to the base $A D$ at an angle of $75^{\circ}$. Vertex $B$ is equidistant from line $C M$ and vertex $A$. Find the angles of the parallelogram. Find the area of the parallelogram if the length of the base $A D$ is 4. | Answer: 1) $75^{\circ}, 105^{\circ} ; 2) S_{A B C D}=4(\sqrt{3}+2)$.
## Final round of the "Rosatom" olympiad, 11th grade, Moscow, March 2020
# | 75,105;S_{ABCD}=4(\sqrt{3}+2) | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Two points $A(x ; y)$ and $B\left(x^{\prime} ; y^{\prime}\right)$ are symmetric with respect to the line $x+y=0$ if $x^{\prime}=-y, y^{\prime}=-x$. This leads to the system:
$$
\left\{\begin{array}{l}
\sin a=-\cos 3 a \\
\sin 3 a=-\cos a
\end{array}\right.
$$
Let's solve the first equation of the system:
$$
\begi... | Answer: $a=-\frac{\pi}{8}+\frac{\pi n}{2}, n \in Z$. | -\frac{\pi}{8}+\frac{\pin}{2},n\inZ | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
4. According to the problem, the random variable $\xi=\frac{C M}{C B}$ is uniformly distributed on the interval $[0 ; 1]$. If $x$ is the value of the random variable $\xi$, then
$$
\mathrm{s}(x)=\mathrm{S}_{P Q M}=S_{A B C} \cdot\left(1-\frac{1}{6}-\frac{3 x}{4}-\frac{(1-x)}{3}\right)=\frac{S_{A B C}}{12}(6-5 x) \righ... | Answer: 1) $\left.P(A)=\frac{2}{5} ; 2\right) M_{X}=\frac{7}{24}$. | P(A)=\frac{2}{5};M_{X}=\frac{7}{24} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Note that for any $n \in Z$
$$
(n+1)^{3}+(n-1)^{3}+(-n)^{3}+(-n)^{3}=6 n
$$
i.e., any integer of the form $a=6 n$ can be represented as the sum of the cubes of four, and thus, with zero, five integers. Numbers of the form $a=6 n \pm 1$ can be represented in the form
$$
a=(n+1)^{3}+(n-1)^{3}+(-n)^{3}+(-n)^{3}+( \p... | Answer: $2020=(339)^{3}+337^{3}+(-338)^{3}+(-338)^{3}+(-2)^{3}$. | 2020=(339)^{3}+337^{3}+(-338)^{3}+(-338)^{3}+(-2)^{3} | Number Theory | proof | Yes | Yes | olympiads | false |
2. For what values of $a$ is the point with coordinates $(\sin a ; \sin 3 a)$ symmetric to the point with coordinates $(\cos a ; \cos 3 a)$ with respect to the line with the equation $x+y=0$? | Answer: $a= \pm \frac{\pi}{3}+2 \pi n, a=(2 n-1) \pi, n \in Z$. | =\\frac{\pi}{3}+2\pin,=(2n-1)\pi,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Points $P, Q$ are located on the sides $A B$ and $A C$ of triangle $A B C$ such that $A P: P B=2: 1, A Q: Q C=1: 3$. Point $M$ is chosen on side $B C$ completely at random. Find the probability that the area of triangle $A B C$ does not exceed the area of triangle $P Q M$ by more than three times. Find the expected ... | Answer: 1) $P(A)=\frac{9}{16}$; 2) $M_{X}=\frac{4}{15}$. | P(A)=\frac{9}{16};M_{X}=\frac{4}{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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